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Haus Melcher Electromagnetic-Fields-and-Energy

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A university-level electromagnetism textbook, Electromagnetic Fields and Energy by Haus and Melcher, kept in a folder of downloaded physics books. The preface describes a circuit-analogy organization: EQS (RC) and MQS (L-R) systems, then electrodynamics, waves, waveguides and lossy media. It also covers classroom demonstrations, numerical methods, and magnetization via magnetic dipoles (Chu formulation).

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Sec. 0.1 Preface 1 0.1 PREF ACE The text is aimed at an audience that has seen Maxw ell’s equations in integral or diﬀeren tial form (second-term Freshman Physics) and had some exposure to integral theorems and diﬀeren tial operators (second term Freshman Calculus). The ﬁrst two chapters and supporting problems and appendices are a review of this material. In Chap. 3, a simple and physically appealing argumen t is presen ted to show that Maxw ell’s equations predict the time evolution ofa ﬁeld, produced by free charges, given the initial charge densities and velocities, and electric and magnetic ﬁelds. This is a form of the uniqueness theorem that is established more rigorously later. Aspartofthisdevelopmen t,itisshownthata ﬁeldis completely speciﬁed by its divergence and its curl throughout all of space, a proof that explains the general form of Maxw ell’s equations. With this background, Maxw ell’s equations are simpliﬁed into their electro quasistatic (EQS) and magneto quasistatic (MQS) forms. The stage is set for taking a structured approac h that gives a physical overview while developing the mathe matical skills needed for the solution of engineering problems. The text builds on and reinforces an understanding of analog circuits. The ﬁelds are never static. Their dynamics are often illustrated with step and sinusoidal steady state responses in systems where the spatial dependence hasbeen encapsu lated in time-dep enden t coeﬃcien ts (of solutions to partial diﬀeren tial equations) satisfying ordinary diﬀeren tial equations. However, the connection with analog cir cuits goeswellbeyondthe same approac hto solving diﬀeren tial equations as used in circuit theory . The appro ximations inheren t in the developmen t of circuit theory from Maxw ell’s equations are brough t out very explicitly , so that the studen t ap preciates under what conditions the assumptions implicit in circuit theory cease to be applicable. To appreciate the organization of material in this text, it may be helpful to makea more subtle connection with electrical analog circuits. We think of circuit theory asbeing analogous to ﬁeld theory . In this analogy , our developmen tbegins with capacitors– charges and their associated ﬁelds, equip otentials used to repre sentperfect conductors. It continues with resistors– steady conduction to represen t losses. Then these elemen ts are combined to represen t charge relaxation, i.e. “RC” systems dynamics (Chaps. 4-7). Because EQS ﬁelds are not necessarily static, the studen t can appreciate R-C type dynamics, where the distribution of free charge is determined by the continuum analog of R-C systems. Using the same approac h, we then take up the continuum generalization of L-R systems (Chaps. 8–10). As before, we ﬁrst are given the source (the curren t densit y) and ﬁnd the magnetic ﬁeld. Then we consider perfectly conducting systems and once again take theboundary valuepointof view. With the addition of ﬁnite conductivit y to this continuum analog of systems of inductors, we arrive at the dynamics of systems that are L-R-lik e in the circuit analogy . Based onan appreciation ofthe connection between sources and ﬁelds aﬀorded by these quasistatic developmen ts, it is natural to use the study of electric and magnetic energy storage and dissipation as an entree into electro dynamics (Chap. 11). Centralto electro dynamics are electromagnetic wavesin loss-free media (Chaps. 12–14). In this limit, the circuit analog is a system of distributed diﬀeren tial induc- 2 Chapter 0 tors and capacitors, an L-C system. Following the same pattern used for EQS and MQS systems, ﬁelds are ﬁrst found for given sources– antennae and arrays. The boundary valuepointof view then brings in micro waveand optical waveguides and transmission lines. We conclude with the electro dynamics of lossy material, the generalization of L-R-C systems (Chaps. 14–15). Drawing on what has been learned for EQS, MQS, and electro dynamic systems, for example, on the physical signiﬁcance of the dominan t characteristic times, we form a perspective as to how electromagnetic ﬁelds are exploited in practica l systems. In the circuit analogy , these characteristic√ times are RC, L/R, and 1/ LC. One beneﬁt of the ﬁeld theory point of view is that it shows the inﬂuence of physical scale and conﬁguration on the dynamics represen tedby these times. The circuit analogy givesahintas√to why it is so often possible toview theworldas either EQS orMQS. Thetime 1/ √LC is the geometric mean of RC and L/R. Either RC or L/R is smaller than 1/ LC, but not both. For large R,RC dynamics comes ﬁrstasthe frequency is raised (EQS), followedby electro dynamics. For small R,L/R dynamics comes ﬁrst (MQS), again followedby electro dynamics. Implicit is the enormous diﬀerence between what is mean tbya “perfect conductor” in systems appropriately modeled as EQS and MQS. This organization of the material is intended to bring the studen t to the realization that electric, magnetic, and electromagnetic devices and systems canbe broken into parts, often describ ed by one or another limiting form of Maxw ell’s equations. Recognition of these limits is part of the art and science of modeling, of making the simpliﬁcations necessary to make the device or system amenable to analytic treatmen t or computer analysis and of eﬀectiv ely using appropriate simpliﬁcations of the laws to guide in the process of invention. With the EQS appro ximation comes the opportunit y to treat such devices as transistors, electrostatic precipitators, and electrostatic sensors and actuators, while relays, motors, and magnetic recording media are examples of MQS systems. Transmission lines, antenna arrays, and dielectric waveguides (i.e., optical ﬁbers) are examples where the full, dynamic Maxw ell’s equations mustbe used. In connection with examples, about 40 demonstrations are describ ed in this text. These are designed to make the mathematical results take on physical mean ing. Based upon relativ ely simple conﬁgurations and arrangemen ts of equipmen t, they incorp orate no more complexit y then required to make a direct connection between what has been deriv ed and what is observ ed. Their purpose is to help the studen t observ ephysically what hasbeen describ ed symbolically . Often coming with aplotofthe theoretical predictions thatcanbe compared todata takeninthe classro om, they give the opportunit y to test the range of validit y of the theory and to prom ulgate a quantitativ e approac h to dealing with the physical world. More detailed consideration of the demonstrations canbe the basis for special projects, often bringing in computer modeling. For the studen thaving only the text asa resource, the descriptions of the experimen ts stand on their own as a connection between the abstractions and the physical realit y. For those fortunate enough to have some of the demonstrations used in the classro om, they serve as documen ta tion of what was done. All too often, studen ts fail to proﬁt from demonstrations because conventional note taking fails to do justice to the presen tation. The demonstrations included in the text are of physical phenomena more than of practical applications. To ﬁll out the classro om experience, to provide the Sec. 0.1 Preface 3 engineering motiv ation, applications should also be exempliﬁed. In the subject as we teach it, and as a practical matter, these are more of the nature of “show and tell” than of working demonstrations, often reﬂecting the curren t experience and interests of the instructor and usually involving more complexit y than appropriate for more than a qualitativ e treatmen t. The text provides a natural frame of reference for developing numerical ap proac hes to the details of geometry and nonlinearit y,beginning with the metho d of momen ts as the superposition integral approac h to boundary value problems and culminating in energy metho ds as a basis for the ﬁnite elemen t approac h. Profes sor J. L. Kirtley and Dr. S. D. Umans are curren tly spearheading our eﬀorts to expose the studen t to the “muscle” provided by the computer for making practical use of ﬁeld theory while helping the studen t gain physical insigh t. Work stations, ﬁnite elemen t packages, and the like make itpossible to take detailed accoun t of geometric eﬀects in routine engineering design. However, no matter how advanced the computer packages available to the studen t maybecome in the future, it will remain essen tial that a studen t comprehend the physical phenomena at work with the aid of special cases. This is the reason for the emphasis of the text on simple ge ometries toprovidephysical insigh tintotheprocesses atworkwhen ﬁelds interact with media. The mathematics of Maxw ell’s equations leads the studen t to a good under- standing of the gradien t, divergence, and curl operators. This mathematical con versance will help the studen t enter other areas– such as ﬂuid and solid mechanics, heat and mass transfer, and quantum mechanics– that also use the language of clas sical ﬁelds. So that the material serves this larger purpose, there is an emphasis on source-ﬁeld relations, on scalar and vector potentials to represen t the irrotational and solenoidal parts of ﬁelds, and on that understanding of boundary conditions that accoun ts for ﬁnite system size and ﬁnite time rates of change. Maxw ell’s equations form an intellectual ediﬁce that is unsurpassed by any other discipline ofphysics. Veryfew equations encompass suchagamutofphysical phenomena. Conceiv ed before the introduction of relativit y Maxw ell’s equations not only surviv ed the formulation of relativit y, but were instrumen tal in shaping it. Because they are linear in the ﬁelds, the replacemen t of the ﬁeld vectors by operators is all that is required to make them quantum theoretically correct; thus, they also surviv ed the introduction of quantum theory . The introduction of magnetizable materials deviates from the usual treatmen t in that we use paired magnetic charges, magnetic dipoles, as the source of magneti zation. The often-used alternativ e is circulating Amp`erian curren ts. The magnetic charge approac h is based on the Chu formulation of electro dynamics. Chu exploited the symmetry of the equations obtained in this way to facilitate the study of mag netism by analogy with polarization. Astheyearswentby,itwas unavoidable that this approac hwould be criticized, because thedipole momen tofthe electron, the main source of ferromagnetism, is associated with the spin of the electron, i.e., seems tobe more appropriately pictured by circulating curren ts. Tellegen in particular, of Tellegen-theorem fame, took issue with this ap proac h. Whereas he conceded thatachoice betweentwo approac hes thatgive iden tical answ ers is a matter of taste, he gave a deriv ation of the force on a curren t loop (the Amp`erian model of a magnetic dipole) and showed that it gave a diﬀeren t answ er from that on a magnetic dipole. The diﬀerence was small, the correction term was relativistic in nature; thus, it would have been diﬃcult to detect the 4 Chapter 0 eﬀect in macroscopic measuremen ts. It occurred only in the presence of a time- varying electric ﬁeld. Yet this criticism, ifvalid, would have made the treatmen tof magnetization in terms of magnetic dipoles highly suspect. The resolution of this issue followed a careful investigation of the force exerted on a curren t loop on one hand, and a magnetic dipole on the other. It turned out that Tellegen’s analysis, inpostulating a constan t circulating curren t around the loop, was in error. A time-v arying electric ﬁeld causes changes in the circulating curren t that, when taken into accoun t, causes an additional force that cancels the critical term. Both models of a magnetic dipole yield the same force expression. The diﬃcult y in the analysis arose because the curren t loop contains “moving parts,” i.e., a circulating curren t, and therefore requires the use of relativistic corrections in the rest-frame of the loop. Hence, the curren t loop model is inheren tly much harder to analyze than the magnetic charge–dip ole model. The resolution of the force parado x also helped clear up the question of the symmetry ofthe energy momen tum tensor. Ataboutthe same time as thisworkwas in progress, Shockley and James at Stanford indep enden tly raised related questions thatledtoalivelyexchange betweenthem and Coleman andVanVleckatHarvard. Shockley used the term “hidden momen tum” for contributions to the momen tum of the electromagnetic ﬁeld in the presence of magnetizable materials. Coleman and Van Vleck showed that a proper formulation based on the Dirac equation (i.e., a relativistic description) automatically includes such terms. With all this theoretical workbehind us,we are comfortable with the useof the magnetic charge– dipole model for the source of magnetization. The studen t is not introduced to the intricacies of the issue, although brief mention is made of them in the text. Aspartof curriculum developmen toveraperiodaboutequal intime totheage ofatypical studen t studying this material (the authors began their collab oration in 1968) this text ﬁts into an evolution of ﬁeld theory with its origins in the “Radiation Lab” days during and following WorldWarII. Quasistatics, prom ulgated in texts by Professors Richard B. Adler, L.J.Chu,andRobertM.Fano,isamajor theme inthis textaswell.However,the notion hasbeen broadened and made more rigorous and useful by recognizing that electromagnetic phenomena that are “quasistatic,” in the sense that electromagnetic wave phenomena canbe ignored, can nevertheless be rate dependen t.As used in this text, a quasistatic regime includes dynamical phenomena with characteristic times longer than those associated with electromagnetic waves. (A model in which no time-rate processes are included is termed “quasistationary” for distinction.) In recognition of the lineage of our text, it is dedicated to Professors R. B. Adler, L.J.Chu andR.M.Fano. Professor Adler, aswell as Professors J. Moses, G. L. Wilson, and L. D. Smullin, who headed the departmen t during the period of developmen t, have been a source of intellectual, moral, and ﬁnancial support. Our inspiration has also come from colleagues in teaching– facult y and teaching assistan ts, and those studen ts who provided insigh t concerning the many evolutions of the “notes.” The teaching of Professor Alan J. Grodzinsky , whose latterda y lectures havebeena mainsta yforthe course, is reﬂected inthe text itself. A partial list of others who contributed to the curriculum developmen t includes Professors J. A. Kong, J. H. Lang, T.P. Orlando, R. E.Parker, D. H. Staelin, and M. Zahn (who helped with a ﬁnal reading of the text). With “macros” written by Ms.Amy Hendric kson, the textwas“Tex’t” byMs. Cindy Kopf, who managed tomakethe ﬁnal publication process a pleasure for the authors. 1 MAXWELL’S INTEGRAL LAWS IN FREE SPACE 1.0 INTRODUCTION Practical, intellectual, and cultural reasons motivate the study of electricity and magnetism. The operation of electrical systems designed to perform certain engi- neering tasks depends, at least in part, on electrical, electromechanical, or electro- chemical phenomena. The electrical aspects of these applications are described by Maxwell’s equations. As a description of the temporal evolution of electromagnetic ﬁelds in three-dimensional space, these same equations form a concise summary of a wider range of phenomena than can be found in any other discipline. Maxwell’s equations are an intellectual achievement that should be familiar to every student of physical phenomena. As part of the theory of ﬁelds that includes continuum me- chanics, quantum mechanics, heat and mass transfer, and many other disciplines, our subject develops the mathematical language and methods that are the basis for these other areas. For those who have an interest in electromechanical energy conversion, trans- mission systems at power or radio frequencies, waveguides at microwave or optical frequencies, antennas, or plasmas, there is little need to argue the necessity for becoming expert in dealing with electromagnetic ﬁelds. There are others who may require encouragement. For example, circuit designers may be satisﬁed with circuit theory, the laws of which are stated in terms of voltages and currents and in terms of the relations imposed upon the voltages and currents by the circuit elements. However, these laws break down at high frequencies, and this cannot be understood without electromagnetic ﬁeld theory. The limitations of circuit models come into play as the frequency is raised so high that the propagation time of electromagnetic ﬁelds becomes comparable to a period, with the result that “inductors” behave as “capacitors” and vice versa. Other limitations are associated with loss phenom- ena. As the frequency is raised, resistors and transistors are limited by “capacitive” eﬀects, and transducers and transformers by “eddy” currents. 1 2 Maxwell’s Integral Laws in Free Space Chapter 1 Anyone concerned with developing circuit models for physical systems requires a ﬁeld theory background to justify approximations and to derive the values of the circuit parameters. Thus, the bioengineer concerned with electrocardiography or neurophysiology must resort to ﬁeld theory in establishing a meaningful connection between the physical reality and models, when these are stated in terms of circuit elements. Similarly, even if a control theorist makes use of a lumped parameter model, its justiﬁcation hinges on a continuum theory, whether electromagnetic, mechanical, or thermal in nature. Computer hardware may seem to be another application not dependent on electromagnetic ﬁeld theory. The software interface through which the computer is often seen makes it seem unrelated to our subject. Although the hardware is generally represented in terms of circuits, the practical realization of a computer designed to carry out logic operations is limited by electromagnetic laws. For exam- ple, the signal originating at one point in a computer cannot reach another point within a time less than that required for a signal, propagating at the speed of light, to traverse the interconnecting wires. That circuit models have remained useful as computation speeds have increased is a tribute to the solid state technology that has made it possible to decrease the size of the fundamental circuit elements. Sooner or later, the fundamental limitations imposed by the electromagnetic ﬁelds deﬁne the computation speed frontier of computer technology, whether it be caused by electromagnetic wave delays or electrical power dissipation. Overview of Subject. As illustrated diagrammatically in Fig. 1.0.1, we start with Maxwell’s equations written in integral form. This chapter begins with a deﬁnition of the ﬁelds in terms of forces and sources followed by a review of each of the integral laws. Interwoven with the development are examples intended to develop the methods for surface and volume integrals used in stating the laws. The examples are also intended to attach at least one physical situation to each of the laws. Our objective in the chapters that follow is to make these laws useful, not only in modeling engineering systems but in dealing with practical systems in a qualitative fashion (as an inventor often does). The integral laws are directly useful for (a) dealing with ﬁelds in this qualitative way, (b) ﬁnding ﬁelds in simple conﬁgurations having a great deal of symmetry, and (c) relating ﬁelds to their sources. Chapter 2 develops a diﬀerential description from the integral laws. By follow- ing the examples and some of the homework associated with each of the sections, a minimum background in the mathematical theorems and operators is developed. The diﬀerential operators and associated integral theorems are brought in as needed. Thus, the divergence and curl operators, along with the theorems of Gauss and Stokes, are developed in Chap. 2, while the gradient operator and integral theorem are naturally derived in Chap. 4. Static ﬁelds are often the ﬁrst topic in developing an understanding of phe- nomena predicted by Maxwell’s equations. Fields are not measurable, let alone of practical interest, unless they are dynamic. As developed here, ﬁelds are never truly static. The subject of quasistatics, begun in Chap. 3, is central to the approach we will use to understand the implications of Maxwell’s equations. A mature un- derstanding of these equations is achieved when one has learned how to neglect complications that are inconsequential. The electroquasistatic (EQS) and magne- Sec. 1.0 Introduction 3 4 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.0.1 Outline of Subject. The three columns, respectively for electro- quasistatics, magnetoquasistatics and electrodynamics, show parallels in de- velopment. toquasistatic (MQS) approximations are justiﬁed if time rates of change are slow enough (frequencies are low enough) so that time delays due to the propagation of electromagnetic waves are unimportant. The examples considered in Chap. 3 give some notion as to which of the two approximations is appropriate in a given situa- tion. A full appreciation for the quasistatic approximations will come into view as the EQS and MQS developments are drawn together in Chaps. 11 through 15. Although capacitors and inductors are examples in the electroquasistatic and magnetoquasistatic categories, respectively, it is not true that quasistatic sys- tems can be generally modeled by frequency-independent circuit elements. High- frequency models for transistors are correctly based on the EQS approximation. Electromagnetic wave delays in the transistors are not consequential. Nevertheless, dynamic eﬀects are important and the EQS approximation can contain the ﬁnite time for charge migration. Models for eddy current shields or heaters are correctly based on the MQS approximation. Again, the delay time of an electromagnetic wave is unimportant while the all-important diﬀusion time of the magnetic ﬁeld Sec. 1.0 Introduction 5 is represented by the MQS laws. Space charge waves on an electron beam or spin waves in a saturated magnetizable material are often described by EQS and MQS laws, respectively, even though frequencies of interest are in the GHz range. The parallel developments of EQS (Chaps. 4–7) and MQS systems (Chaps. 8– 10) is emphasized by the ﬁrst page of Fig. 1.0.1. For each topic in the EQS column to the left there is an analogous one at the same level in the MQS column. Although the ﬁeld concepts and mathematical techniques used in dealing with EQS and MQS systems are often similar, a comparative study reveals as many contrasts as direct analogies. There is a two-way interplay between the electric and magnetic studies. Not only are results from the EQS developments applied in the description of MQS systems, but the examination of MQS situations leads to a greater appreciation for the EQS laws. At the tops of the EQS and the MQS columns, the ﬁrst page of Fig. 1.0.1, general (contrasting) attributes of the electric and magnetic ﬁelds are identiﬁed. The developments then lead from situations where the ﬁeld sources are prescribed to where they are to be determined. Thus, EQS electric ﬁelds are ﬁrst found from prescribed distributions of charge, while MQS magnetic ﬁelds are determined given the currents. The development of the EQS ﬁeld solution is a direct investment in the subsequent MQS derivation. It is then recognized that in many practical situations, these sources are induced in materials and must therefore be found as part of the ﬁeld solution. In the ﬁrst of these situations, induced sources are on the boundaries of conductors having a suﬃciently high electrical conductivity to be modeled as “perfectly” conducting. For the EQS systems, these sources are surface charges, while for the MQS, they are surface currents. In either case, ﬁelds must satisfy boundary conditions, and the EQS study provides not only mathematical techniques but even partial diﬀerential equations directly applicable to MQS problems. Polarization and magnetization account for ﬁeld sources that can be pre- scribed (electrets and permanent magnets) or induced by the ﬁelds themselves. In the Chu formulation used here, there is a complete analogy between the way in which polarization and magnetization are represented. Thus, there is a direct transfer of ideas from Chap. 6 to Chap. 9. The parallel quasistatic studies culminate in Chaps. 7 and 10 in an examina- tion of loss phenomena. Here we learn that very diﬀerent answers must be given to the question “When is a conductor perfect?” for EQS on one hand, and MQS on the other. In Chap. 11, many of the concepts developed previously are put to work through the consideration of the ﬂow of power, storage of energy, and production of electromagnetic forces. From this chapter on, Maxwell’s equations are used with- out approximation. Thus, the EQS and MQS approximations are seen to represent systems in which either the electric or the magnetic energy storage dominates re- spectively. In Chaps. 12 through 14, the focus is on electromagnetic waves. The develop- ment is a natural extension of the approach taken in the EQS and MQS columns. This is emphasized by the outline represented on the right page of Fig. 1.0.1. The topics of Chaps. 12 and 13 parallel those of the EQS and MQS columns on the previous page. Potentials used to represent electrodynamic ﬁelds are a natural gen- eralization of those used for the EQS and MQS systems. As for the quasistatic ﬁelds, the ﬁelds of given sources are considered ﬁrst. An immediate practical application is therefore the description of radiation ﬁelds of antennas. 6 Maxwell’s Integral Laws in Free Space Chapter 1 The boundary value point of view, introduced for EQS systems in Chap. 5 and for MQS systems in Chap. 8, is the basic theme of Chap. 13. Practical examples include simple transmission lines and waveguides. An understanding of transmission line dynamics, the subject of Chap. 14, is necessary in dealing with the “conventional” ideal lines that model most high-frequency systems. They are also shown to provide useful models for representing quasistatic dynamical processes. To make practical use of Maxwell’s equations, it is necessary to master the art of making approximations. Based on the electromagnetic properties and dimen- sions of a system and on the time scales (frequencies) of importance, how can a physical system be broken into electromagnetic subsystems, each described by its dominant physical processes? It is with this goal in mind that the EQS and MQS approximations are introduced in Chap. 3, and to this end that Chap. 15 gives an overview of electromagnetic ﬁelds. 1.1 THE LORENTZ LAW IN FREE SPACE There are two points of view for formulating a theory of electrodynamics. The older one views the forces of attraction or repulsion between two charges or currents as the result of action at a distance. Coulomb’s law of electrostatics and the corresponding law of magnetostatics were ﬁrst stated in this fashion. Faraday[1]introduced a new approach in which he envisioned the space between interacting charges to be ﬁlled with ﬁelds, by which the space is activated in a certain sense; forces between two interacting charges are then transferred, in Faraday’s view, from volume element to volume element in the space between the interacting bodies until ﬁnally they are transferred from one charge to the other. The advantage of Faraday’s approach was that it brought to bear on the electromagnetic problem the then well-developed theory of continuum mechanics. The culmination of this point of view was Maxwell’s formulation[2]of the equations named after him. From Faraday’s point of view, electric and magnetic ﬁelds are deﬁned at a point reven when there is no charge present there. The ﬁelds are deﬁned in terms of the force that would be exerted on a test charge qif it were introduced at r moving at a velocity vat the time of interest. It is found experimentally that such a force would be composed of two parts, one that is independent of v, and the other proportional to vand orthogonal to it. The force is summarized in terms of the electric ﬁeld intensity Eandmagnetic ﬂux density µoHby the Lorentz force law . (For a review of vector operations, see Appendix 1.) f=q(E+v×µoH) (1) The superposition of electric and magnetic force contributions to (1) is illus- trated in Fig. 1.1.1. Included in the ﬁgure is a reminder of the right-hand rule used to determine the direction of the cross-product of vandµoH. In general, EandH are not uniform, but rather are functions of position rand time t:E=E(r, t) and µoH=µoH(r, t). In addition to the units of length, mass, and time associated with mechanics, a unit of charge is required by the theory of electrodynamics. This unit is the Sec. 1.1 The Lorentz Law in Free Space 7 Fig. 1.1.1 Lorentz force fin geometric relation to the electric and magnetic ﬁeld intensities, EandH, and the charge velocity v: (a) electric force, (b) magnetic force, and (c) total force. coulomb. The Lorentz force law, (1), then serves to deﬁne the units of Eand of µoH. units of E=newton coulomb=kilogram meter /(second)2 coulomb(2) units of µoH=newton coulomb meter /second=kilogram coulomb second(3) We can only establish the units of the magnetic ﬂux density µoHfrom the force law and cannot argue until Sec. 1.4 that the derived units of Hare ampere/meter and hence of µoare henry/meter. In much of electrodynamics, the predominant concern is not with mechanics but with electric and magnetic ﬁelds in their own right. Therefore, it is inconvenient to use the unit of mass when checking the units of quantities. It proves useful to introduce a new name for the unit of electric ﬁeld intensity– the unit of volt/meter. In the summary of variables given in Table 1.8.2 at the end of the chapter, the fundamental units are SI, while the derived units exploit the fact that the unit of mass, kilogram = volt-coulomb-second2/meter2and also that a coulomb/second = ampere. Dimensional checking of equations is guaranteed if the basic units are used, but may often be accomplished using the derived units. The latter communicate the physical nature of the variable and the natural symmetry of the electric and magnetic variables. Example 1.1.1. Electron Motion in Vacuum in a Uniform Static Electric Field In vacuum, the motion of a charged particle is limited only by its own inertia. In the uniform electric ﬁeld illustrated in Fig. 1.1.2, there is no magnetic ﬁeld, and an electron starts out from the plane x= 0 with an initial velocity vi. The “imposed” electric ﬁeld is E=ixEx, where ixis the unit vector in the x direction and Exis a given constant. The trajectory is to be determined here and used to exemplify the charge and current density in Example 1.2.1. 8 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.1.2 An electron, subject to the uniform electric ﬁeld intensity Ex, has the position ξx, shown as a function of time for positive and negative ﬁelds. With mdeﬁned as the electron mass, Newton’s law combines with the Lorentz law to describe the motion. md2ξx dt2=f=−eEx (4) The electron position ξxis shown in Fig. 1.1.2. The charge of the electron is custom- arily denoted by e(e= 1.6×10−19coulomb) where eis positive, thus necessitating an explicit minus sign in (4). By integrating twice, we get ξx=−1 2e mExt2+c1t+c2 (5) where c1andc2are integration constants. If we assume that the electron is at ξx= 0 and has velocity viwhen t=ti, it follows that these constants are c1=vi+e mExti; c2=−viti−1 2e mExt2 i (6) Thus, the electron position and velocity are given as a function of time by ξx=−1 2e mEx(t−ti)2+vi(t−ti) (7) dξx dt=−e mEx(t−ti) +vi (8) With xdeﬁned as upward and Ex>0, the motion of an electron in an electric ﬁeld is analogous to the free fall of a mass in a gravitational ﬁeld, as illustrated by Fig. 1.1.2. With Ex<0, and the initial velocity also positive, the velocity is a monotonically increasing function of time, as also illustrated by Fig. 1.1.2. Example 1.1.2. Electron Motion in Vacuum in a Uniform Static Magnetic Field The magnetic contribution to the Lorentz force is perpendicular to both the particle velocity and the imposed ﬁeld. We illustrate this fact by considering the trajectory Sec. 1.1 The Lorentz Law in Free Space 9 Fig. 1.1.3 (a) In a uniform magnetic ﬂux density µoHoand with no initial velocity in the ydirection, an electron has a circular orbit. (b) With an initial velocity in the ydirection, the orbit is helical. resulting from an initial velocity vizalong the zaxis. With a uniform constant magnetic ﬂux density µoHexisting along the yaxis, the force is f=−e(v×µoH) (9) The cross-product of two vectors is perpendicular to the two vector factors, so the acceleration of the electron, caused by the magnetic ﬁeld, is always perpendicular to its velocity. Therefore, a magnetic ﬁeld alone cannot change the magnitude of the electron velocity (and hence the kinetic energy of the electron) but can change only the direction of the velocity. Because the magnetic ﬁeld is uniform, because the velocity and the rate of change of the velocity lie in a plane perpendicular to the magnetic ﬁeld, and, ﬁnally, because the magnitude of vdoes not change, we ﬁnd that the acceleration has a constant magnitude and is orthogonal to both the velocity and the magnetic ﬁeld. The electron moves in a circle so that the centrifugal force counterbalances the magnetic force. Figure 1.1.3a illustrates the motion. The radius of the circle is determined by equating the centrifugal force and radial Lorentz force eµo|v|Ho=mv2 r(10) which leads to r=m e|v| µoHo(11) The foregoing problem can be modiﬁed to account for any arbitrary initial angle between the velocity and the magnetic ﬁeld. The vector equation of motion (really three equations in the three unknowns ξx, ξy, ξz) md2¯ξ dt2=−e¡d¯ξ dt×µoH¢ (12) is linear in ¯ξ, and so solutions can be superimposed to satisfy initial conditions that include not only a velocity vizbut one in the ydirection as well, viy. Motion in the same direction as the magnetic ﬁeld does not give rise to an additional force. Thus, 10 Maxwell’s Integral Laws in Free Space Chapter 1 theycomponent of (12) is zero on the right. An integration then shows that the y directed velocity remains constant at its initial value, viy. This uniform motion can be added to that already obtained to see that the electron follows a helical path, as shown in Fig. 1.1.3b. It is interesting to note that the angular frequency of rotation of the electron around the ﬁeld is independent of the speed of the electron and depends only upon the magnetic ﬂux density, µoHo. Indeed, from (11) we ﬁnd v r≡ωc=e mµoHo (13) For a ﬂux density of 1 volt-second/meter (or 1 tesla), the cyclotron frequency isfc= ωc/2π= 28 GHz. (For an electron, e= 1.602×10−19coulomb and m= 9.106×10−31 kg.) With an initial velocity in the zdirection of 3 ×107m/s, the radius of gyration in the ﬂux density µoH= 1 tesla is r=viz/ωc= 1.7×10−4m. 1.2 CHARGE AND CURRENT DENSITIES In Maxwell’s day, it was not known that charges are not inﬁnitely divisible but occur in elementary units of 1 .6×10−19coulomb, the charge of an electron. Hence, Maxwell’s macroscopic theory deals with continuous charge distributions. This is an adequate description for ﬁelds of engineering interest that are produced by ag- gregates of large numbers of elementary charges. These aggregates produce charge distributions that are described conveniently in terms of a charge per unit volume, a charge density ρ. Pick an incremental volume and determine the net charge within. Then ρ(r, t)≡net charge in ∆V ∆V(1) is the charge density at the position rwhen the time is t. The units of ρare coulomb/meter3. The volume ∆ Vis chosen small as compared to the dimensions of the system of interest, but large enough so as to contain many elementary charges. The charge density ρis treated as a continuous function of position. The “graini- ness” of the charge distribution is ignored in such a “macroscopic” treatment. Fundamentally, current is charge transport and connotes the time rate of change of charge. Current density is a directed current per unit area and hence measured in (coulomb/second)/meter2. A charge density ρmoving at a velocity v implies a rate of charge transport per unit area, a current density J, given by J=ρv (2) One way to envision this relation is shown in Fig. 1.2.1, where a charge density ρhaving velocity vtraverses a diﬀerential area δa. The area element has a unit normal n, so that a diﬀerential area vector can be deﬁned as δa=nδa. The charge that passes during a diﬀerential time δtis equal to the total charge contained in the volume v·δadt. Therefore, d(δq) =ρv·δadt (3) Sec. 1.2 Charge and Current Densities 11 Fig. 1.2.1 Current density Jpassing through surface having a normal n. Fig. 1.2.2 Charge injected at the lower boundary is accelerated up- ward by an electric ﬁeld. Vertical distributions of (a) ﬁeld intensity, (b) velocity and (c) charge density. Divided by dt, we expect (3) to take the form J·δa, so it follows that the current density is related to the charge density by (2). The velocity vis the velocity of the charge. Just how the charge is set into motion depends on the physical situation. The charge might be suspended in or on an insulating material which is itself in motion. In that case, the velocity would also be that of the material. More likely, it is the result of applying an electric ﬁeld to a conductor, as considered in Chap. 7. For charged particles moving in vacuum, it might result from motions represented by the laws of Newton and Lorentz, as illustrated in the examples in Sec.1.1. This is the case in the following example. Example 1.2.1. Charge and Current Densities in a Vacuum Diode Consider the charge and current densities for electrons being emitted with initial velocity vfrom a “cathode” in the plane x= 0, as shown in Fig. 1.2.2a.1 Electrons are continuously injected. As in Example 1.1.1, where the motions of the individual electrons are considered, the electric ﬁeld is assumed to be uniform. In the next section, it is recognized that charge is the source of the electric ﬁeld. Here it is assumed that the charge used to impose the uniform ﬁeld is much greater than the “space charge” associated with the electrons. This is justiﬁed in the limit of a low electron current. Any one of the electrons has a position and velocity given by (1.1.7) and (1.1.8). If each is injected with the same initial velocity, the charge and current densities in any given plane x= constant would be expected to be independent of time. Moreover, the current passing any x-plane should be the same as that passing any other such plane. That is, in the steady state, the current density is independent 1Here we picture the ﬁeld variables Ex, vx, and ρas though they were positive. For electrons, ρ <0, and to make vx>0, we must have Ex<0. 12 Maxwell’s Integral Laws in Free Space Chapter 1 of not only time but xas well. Thus, it is possible to write ρ(x)vx(x) =Jo (4) where Jois a given current density. The following steps illustrate how this condition of current continuity makes it possible to shift from a description of the particle motions described with time as the independent variable to one in which coordinates ( x, y, z ) (or for short r) are the independent coordinates. The relation between time and position for the electron described by (1.1.7) takes the form of a quadratic in ( t−ti) 1 2e mEx(t−ti)2−vi(t−ti) +ξx= 0 (5) This can be solved to give the elapsed time for a particle to reach the position ξx. Note that of the two possible solutions to (5), the one selected satisﬁes the condition that when t=ti, ξx= 0. t−ti=vi−p v2 i−2e mExξx e mEx(6) With the beneﬁt of this expression, the velocity given by (1.1.8) is written as dξx dt=r v2 i−2e mExξx (7) Now we make a shift in viewpoint. On the left in (7) is the velocity vxof the particle that is at the location ξx=x. Substitution of variables then gives vx=q v2 i−2e mExx (8) so that xbecomes the independent variable used to express the dependent variable vx. It follows from this expression and (4) that the charge density ρ=Jo vx=Jop v2 i−2e mExx(9) is also expressed as a function of x. In the plots shown in Fig. 1.2.2, it is assumed thatEx<0, so that the electrons have velocities that increase monotonically with x. As should be expected, the charge density decreases with xbecause as they speed up, the electrons thin out to keep the current density constant. 1.3 GAUSS’ INTEGRAL LAW OF ELECTRIC FIELD INTENSITY The Lorentz force law of Sec. 1.1 expresses the eﬀect of electromagnetic ﬁelds on a moving charge. The remaining sections in this chapter are concerned with the reaction of the moving charges upon the electromagnetic ﬁelds. The ﬁrst of Sec. 1.3 Gauss’ Integral Law 13 Fig. 1.3.1 General surface Senclosing volume V. Maxwell’s equations to be considered, Gauss’ law , describes how the electric ﬁeld intensity is related to its source. The net charge within an arbitrary volume Vthat is enclosed by a surface Sis related to the net electric ﬂux through that surface by I S/epsilon1oE·da=Z Vρdv (1) With the surface normal deﬁned as directed outward, the volume is shown in Fig. 1.3.1. Here the permittivity of free space ,/epsilon1o= 8.854×10−12farad/meter, is an empirical constant needed to express Maxwell’s equations in SI units. On the right in (1) is the net charge enclosed by the surface S. On the left is the summation over this same closed surface of the diﬀerential contributions of ﬂux /epsilon1oE·da. The quantity /epsilon1oEis called the electric displacement ﬂux density and, [from (1)], has the units of coulomb/meter2. Out of any region containing net charge, there must be a net displacement ﬂux. The following example illustrates the mechanics of carrying out the volume and surface integrations. Example 1.3.1. Electric Field Due to Spherically Symmetric Charge Distribution Given the charge and current distributions, the integral laws fully determine the electric and magnetic ﬁelds. However, they are not directly useful unless there is a great deal of symmetry. An example is the distribution of charge density ρ(r) =nρor R;r < R 0; r > R(2) in the spherical coordinate system of Fig. 1.3.2. Here ρoandRare given constants. An argument based on the spherical symmetry shows that the only possible com- ponent of Eis radial. E=irEr(r) (3) Indeed, suppose that in addition to this rcomponent the ﬁeld possesses a φcom- ponent. At a given point, the components of Ethen appear as shown in Fig. 1.3.2b. Rotation of the system about the axis shown results in a component of Ein some new direction perpendicular to r. However, the rotation leaves the source of that ﬁeld, the charge distribution, unaltered. It follows that Eφmust be zero. A similar argument shows that Eθalso is zero. 14 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.3.2 (a) Spherically symmetric charge distribution, showing ra- dial dependence of charge density and associated radial electric ﬁeld intensity. (b) Axis of rotation for demonstration that the components ofEtransverse to the radial coordinate are zero. The incremental volume element is dv= (dr)(rdθ)(rsinθdφ) (4) and it follows that for a spherical volume having arbitrary radius r, Z Vρdv=(Rr 0Rπ 0R2π 0£ ρor/prime R¤ (r/primesinθdφ)(r/primedθ)dr/prime=πρo Rr4;r < RRR 0Rπ 0R2π 0£ ρor/prime R¤ (r/primesinθdφ)(r/primedθ)dr/prime=πρoR3;R < r(5) To evaluate the left-hand side of (1), note that n=ir; da=ir(rdθ)(rsinθdφ) (6) Thus, for the spherical surface at the arbitrary radius r, I S/epsilon1oE·da=Zπ 0Z2π 0/epsilon1oEr(rsinθdφ)(rdθ) =/epsilon1oEr4πr2(7) With the volume and surface integrals evaluated in (5) and (7), Gauss’ law, (l), shows that /epsilon1oEr4πr2=πρo Rr4⇒Er=ρor2 4/epsilon1oR; r < R (8a) /epsilon1oEr4πr2=πρoR3⇒Er=ρoR3 4/epsilon1or2; R < r (8b) Inside the spherical charged region, the radial electric ﬁeld increases with the square of the radius because even though the associated surface increases like the square Sec. 1.3 Gauss’ Integral Law 15 Fig. 1.3.3 Singular charge distributions: (a) point charge, (b) line charge, (c) surface charge. Fig. 1.3.4 Filamentary volume element having cross-section daused to de- ﬁne line charge density. of the radius, the enclosed charge increases even more rapidly. Figure 1.3.2 illus- trates this dependence, as well as the exterior ﬁeld decay. Outside, the surface area continues to increase in proportion to r2, but the enclosed charge remains constant. Singular Charge Distributions. Examples of singular functions from circuit theory are impulse and step functions. Because there is only the one independent variable, namely time, circuit theory is concerned with only one “dimension.” In three-dimensional ﬁeld theory, there are three spatial analogues of the temporal impulse function. These are point, line, and surface distributions of ρ, as illustrated in Fig. 1.3.3. Like the temporal impulse function of circuit theory, these singular distributions are deﬁned in terms of integrals. Apoint charge is the limit of an inﬁnite charge density occupying zero volume. With qdeﬁned as the net charge, q= limρ→∞ V→0Z Vρdv (9) the point charge can be pictured as a small charge-ﬁlled region, the outside of which is charge free. An example is given in Fig. 1.3.2 in the limit where the volume 4 πR3/3 goes to zero, while q=πρoR3remains ﬁnite. Aline charge density represents a two-dimensional singularity in charge den- sity. It is the mathematical abstraction representing a thin charge ﬁlament. In terms of the ﬁlamentary volume shown in Fig. 1.3.4, the line charge per unit length λl (the line charge density) is deﬁned as the limit where the cross-sectional area of the volume goes to zero, ρgoes to inﬁnity, but the integral 16 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.3.5 Volume element having thickness hused to deﬁne surface charge density. Fig. 1.3.6 Point charge qat origin of spherical coordinate system. λl= limρ→∞ A→0Z Aρda (10) remains ﬁnite. In general, λlis a function of position along the curve. The one-dimensional singularity in charge density is represented by the surface charge density . The charge density is very large in the vicinity of a surface. Thus, as a function of a coordinate perpendicular to that surface, the charge density is a one-dimensional impulse function. To deﬁne the surface charge density, mount a pillbox as shown in Fig. 1.3.5 so that its top and bottom surfaces are on the two sides of the surface. The surface charge density is then deﬁned as the limit σs= limρ→∞ h→0Zξ+h 2 ξ−h 2ρdξ (11) where the ξcoordinate is picked parallel to the direction of the normal to the surface, n. In general, the surface charge density σsis a function of position in the surface. Illustration. Field of a Point Charge A point charge qis located at the origin in Fig. 1.3.6. There are no other charges. By the same arguments as used in Example 1.3.1, the spherical symmetry of the charge distribution requires that the electric ﬁeld be radial and be independent of θandφ. Evaluation of the surface integral in Gauss’ integral law, (1), amounts to multiplying /epsilon1oErby the surface area. Because all of the charge is concentrated at the origin, the volume integral gives q, regardless of radial position of the surface S. Thus, 4πr2/epsilon1oEr=q⇒E=q 4π/epsilon1or2ir (12) Sec. 1.3 Gauss’ Integral Law 17 Fig. 1.3.7 Uniform line charge distributed from −inﬁnity to + in- ﬁnity along zaxis. Rotation by 180 degrees about axis shown leads to conclusion that electric ﬁeld is radial. is the electric ﬁeld associated with a point charge q. Illustration. The Field Associated with Straight Uniform Line Charge A uniform line charge is distributed along the zaxis from z=−∞toz= +∞, as shown in Fig. 1.3.7. For an observer at the radius r, translation of the line source in the zdirection and rotation of the source about the zaxis (in the φdirection) results in the same charge distribution, so the electric ﬁeld must only depend on r. Moreover, Ecan only have a radial component. To see this, suppose that there were a zcomponent of E. Then a 180 degree rotation of the system about an axis perpendicular to and passing through the zaxis must reverse this ﬁeld. However, the rotation leaves the charge distribution unchanged. The contradiction is resolved only if Ez= 0. The same rotation makes it clear that Eφmust be zero. This time, Gauss’ integral law is applied using for Sthe surface of a right circular cylinder coaxial with the zaxis and of arbitrary radius r. Contributions from the ends are zero because there the surface normal is perpendicular to E. With the cylinder taken as having length l, the surface integration amounts to a multiplication of /epsilon1oErby the surface area 2 πrlwhile, the volume integral gives lλl regardless of the radius r. Thus, (1) becomes 2πrl/epsilon1 oEr=λll⇒E=λl 2π/epsilon1orir (13) for the ﬁeld of an inﬁnitely long uniform line charge having density λl. Example 1.3.2. The Field of a Pair of Equal and Opposite Inﬁnite Planar Charge Densities Consider the ﬁeld produced by a surface charge density + σooccupying all the x−y plane at z=s/2 and an opposite surface charge density −σoatz=−s/2. First, the ﬁeld must be zdirected. Indeed there cannot be a component of Etransverse to the zaxis, because rotation of the system around the zaxis leaves the same source distribution while rotating that component of E. Hence, no such component exists. 18 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.3.8 Sheets of surface charge and volume of integration with upper surface at arbitrary position x. With ﬁeld Eodue to external charges equal to zero, the distribution of electric ﬁeld is the discontinu- ous function shown at right. Because the source distribution is independent of xandy,Ezis independent of these coordinates. The zdependence is now established by means of Gauss’ integral law, (1). The volume of integration, shown in Fig. 1.3.8, has cross-sectional area A in the x−yplane. Its lower surface is located at an arbitrary ﬁxed location below the lower surface charge distribution, while its upper surface is in the plane denoted byz. For now, we take Ezas being Eoon the lower surface. There is no contribution to the surface integral from the side walls because these have normals perpendicular toE. It follows that Gauss’ law, (1), becomes A(/epsilon1oEz−/epsilon1oEo) = 0; − ∞ < z < −s 2⇒Ez=Eo A(/epsilon1oEz−/epsilon1oEo) =−Aσo;−s 2< z <s 2⇒Ez=−σo /epsilon1o+Eo A(/epsilon1oEz−/epsilon1oEo) = 0;s 2< z < ∞ ⇒ Ez=Eo(14) That is, with the upper surface below the lower charge sheet, no charge is enclosed by the surface of integration, and Ezis the constant Eo. With the upper surface of integration between the charge sheets, EzisEominus σo//epsilon1o. Finally, with the upper integration surface above the upper charge sheet, Ezreturns to its value of Eo. The external electric ﬁeld Eomust be created by charges at z= +∞, much as the ﬁeld between the charge sheets is created by the given surface charges. Thus, if these charges at “inﬁnity” are absent, Eo= 0, and the distribution of Ezis as shown to the right in Fig. 1.3.8. Illustration. Coulomb’s Force Law for Point Charges It is worthwhile to see that for charges at rest, Gauss’ integral law and the Lorentz force law give the familiar action at a distance force law. The force on a charge q is given by the Lorentz law, (1.1.1), and if the electric ﬁeld is caused by a second charge at the origin in Fig. 1.3.9, then f=qE=q1q2 4π/epsilon1or2ir (15) Coulomb’s famous statement that the force exerted by one charge on another is proportional to the product of their charges, acts along a line passing through each Sec. 1.3 Gauss’ Integral Law 19 Fig. 1.3.9 Coulomb force induced on charge q2due to ﬁeld from q1. Fig. 1.3.10 Like-charged particles on ends of thread are pushed apart by the Coulomb force. charge, and is inversely proportional to the square of the distance between them, is now demonstrated. Demonstration 1.3.1. Coulomb’s Force Law The charge resulting on the surface of adhesive tape as it is pulled from a dispenser is a common nuisance. As the tape is brought toward a piece of paper, the force of attraction that makes the paper jump is an aggravating reminder that there are charges on the tape. Just how much charge there is on the tape can be approximately determined by means of the simple experiment shown in Fig. 1.3.10. Two pieces of freshly pulled tape about 7 cm long are folded up into balls and stuck on the ends of a thread having a total length of about 20 cm. The middle of the thread is then tied up so that the charged balls of tape are suspended free to swing. (By electrostatic standards, our ﬁngers are conductors, so the tape should be manipulated chopstick fashion by means of plastic rods or the like.) It is then easy to measure approximately landr, as deﬁned in the ﬁgure. The force of repulsion that separates the “balls” of tape is presumably predicted by (15). In Fig. 1.3.10, the vertical component of the tension in the thread must balance the gravitational force Mg (where gis the gravitational acceleration and Mis the mass). It follows that the horizontal component of the thread tension balances the Coulomb force of repulsion. q2 4π/epsilon1or2=Mg(r/2) l⇒q=r Mgr32π/epsilon1o l(16) As an example, tape balls having an area of A= 14 cm2, (7 cm length of 2 cm wide tape) weighing 0.1 mg and dangling at a length l= 20 cm result in a distance of separation r= 3 cm. It follows from (16) (with all quantities expressed in SI units) that q= 2.7×10−9coulomb. Thus, the average surface charge density is q/A= 1.9×10−6coulomb/meter or 1 .2×1013electronic charges per square meter. If 20 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.3.11 Pillbox-shaped incremental volume used to deduce the jump condition implied by Gauss’ integral law. these charges were in a square array with spacing sbetween charges, then σs=e/s2, and it follows that the approximate distance between the individual charge in the tape surface is 0 .3µm. This length is at the limit of an optical microscope and may seem small. However, it is about 1000 times larger than a typical atomic dimension.2 Gauss’ Continuity Condition. Each of the integral laws summarized in this chapter implies a relationship between ﬁeld variables evaluated on either side of a surface. These conditions are necessary for dealing with surface singularities in the ﬁeld sources. Example 1.3.2 illustrates the jump in the normal component of Ethat accompanies a surface charge. A surface that supports surface charge is pictured in Fig. 1.3.11, as having a unit normal vector directed from region (b) to region (a). The volume to which Gauss’ integral law is applied has the pillbox shape shown, with endfaces of area Aon opposite sides of the surface. These are assumed to be small enough so that over the area of interest the surface can be treated as plane. The height hof the pillbox is very small so that the cylindrical sideface of the pillbox has an area much smaller than A. Now, let happroach zero in such a way that the two sides of the pillbox remain on opposite sides of the surface. The volume integral of the charge density, on the right in (1), gives Aσs. This follows from the deﬁnition of the surface charge density, (11). The electric ﬁeld is assumed to be ﬁnite throughout the region of the surface. Hence, as the area of the sideface shrinks to zero, so also does the contribution of the sideface to the surface integral. Thus, the displacement ﬂux through the closed surface consists only of the contributions from the top and bottom surfaces. Applied to the pillbox, Gauss’ integral law requires that n·(/epsilon1oEa−/epsilon1oEb) =σs (17) where the area Ahas been canceled from both sides of the equation. The contribution from the endface on side (b) comes with a minus sign because on that surface, nis opposite in direction to the surface element da. Note that the ﬁeld found in Example 1.3.2 satisﬁes this continuity condition atz=s/2 and z=−s/2. 2An alternative way to charge a particle, perhaps of low density plastic, is to place it in the corona discharge around the tip of a pin placed at high voltage. The charging mechanism at work in this case is discussed in Chapter 7 (Example 7.7.2). Sec. 1.4 Amp` ere’s Integral Law 21 Fig. 1.4.1 Surface Sis enclosed by contour Chaving positive direction de- termined by the right-hand rule. With the ﬁngers in the direction of ds, the thumb passes through the surface in the direction of positive da. 1.4 AMP `ERE’S INTEGRAL LAW The law relating the magnetic ﬁeld intensity Hto its source, the current density J, is I CH·ds=Z SJ·da+d dtZ S/epsilon1oE·da (1) Note that by contrast with the integral statement of Gauss’ law, (1.3.1), the surface integral symbols on the right do not have circles. This means that the integrations are over open surfaces, having edges denoted by the contour C. Such a surface Senclosed by a contour Cis shown in Fig. 1.4.1. In words, Amp` ere’s integral law as given by (1) requires that the line integral (circulation) of the magnetic ﬁeld intensity Haround a closed contour is equal to the net current passing through the surface spanning the contour plus the time rate of change of the net displacement ﬂux density /epsilon1oEthrough the surface (the displacement current ). The direction of positive dais determined by the right-hand rule, as also illustrated in Fig. 1.4.1. With the ﬁngers of the right-hand in the direction of ds, the thumb has the direction of da. Alternatively, with the right hand thumb in the direction of ds, the ﬁngers will be in the positive direction of da. In Amp` ere’s law, Happears without µo. This law therefore establishes the basic units of Has coulomb/(meter-second). In Sec. 1.1, the units of the ﬂux den- sityµoHare deﬁned by the Lorentz force, so the second empirical constant, the permeability of free space , isµo= 4π×10−7henry/m (henry = volt sec/amp). Example 1.4.1. Magnetic Field Due to Axisymmetric Current A constant current in the zdirection within the circular cylindrical region of radius R, shown in Fig. 1.4.2, extends from −inﬁnity to + inﬁnity along the zaxis and is represented by the density J=½ Jo¡r R¢ ;r < R 0; r > R(2) 22 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.4.2 Axially symmetric current distribution and associated ra- dial distribution of azimuthal magnetic ﬁeld intensity. Contour Cis used to determine azimuthal H, while C/primeis used to show that the z-directed ﬁeld must be uniform. where JoandRare given constants. The associated magnetic ﬁeld intensity has only an azimuthal component. H=Hφiφ (3) To see that there can be no rcomponent of this ﬁeld, observe that rotation of the source around the radial axis, as shown in Fig. 1.4.2, reverses the source (the current is then in the −zdirection) and hence must reverse the ﬁeld. But an rcomponent of the ﬁeld does not reverse under such a rotation and hence must be zero. The HφandHzcomponents are not ruled out by this argument. However, if they exist, they must not depend upon the φandzcoordinates, because rotation of the source around the zaxis and translation of the source along the zaxis does not change the source and hence does not change the ﬁeld. The current is independent of time and so we assume that the ﬁelds are as well. Hence, the last term in (1), the displacement current, is zero. The law is then used with S, a surface having its enclosing contour Cat the arbitrary radius r, as shown in Fig. 1.4.2. Then the area and line elements are da=rdφdr iz; ds=iφrdφ (4) and the right-hand side of (1) becomes Z SJ·da=(R2π 0Rr 0Jor Rrdφdr =Jor32π 3R; r < RR2π 0RR 0Jor Rrdφdr =JoR22π 3; R < r(5) Integration on the left-hand side amounts to a multiplication of the φindependent Hφby the length of C. I CH·ds=Z2π 0Hφrdφ=Hφ2πr (6) Sec. 1.4 Amp` ere’s Integral Law 23 Fig. 1.4.3 (a) Line current enclosed by volume having cross-sectional area A. (b) Surface current density enclosed by contour having thickness h. These last two expressions are used to evaluate (1) and obtain 2πrH φ=Jor32π 3R⇒Hφ=Jor2 3R; r < R 2πrH φ=JoR22π 3⇒Hφ=JoR2 3r; r < R (7) Thus, the azimuthal magnetic ﬁeld intensity has the radial distribution shown in Fig. 1.4.2. Thezcomponent of His, at most, uniform. This can be seen by applying the integral law to the contour C/prime, also shown in Fig. 1.4.2. Integration on the top and bottom legs gives zero because Hr= 0. Thus, to make the contributions due to Hz on the vertical legs cancel, it is necessary that Hzbe independent of radius. Such a uniform ﬁeld must be caused by sources at inﬁnity and is therefore set equal to zero if such sources are not postulated in the statement of the problem. Singular Current Distributions. The ﬁrst of two singular forms of the current density shown in Fig. 1.4.3a is the line current . Formally, it is the limit of an inﬁnite current density distributed over an inﬁnitesimal area. i= lim |J|→∞ A→0Z AJ·da (8) With ia constant over the length of the line, a thin wire carrying a current i conjures up the correct notion of the line current. However, in general, the current imay depend on the position along the line if it varies with time as in an antenna. The second singularity, the surface current density , is the limit of a very large current density Jdistributed over a very thin layer adjacent to a surface. In Fig. 1.4.3b, the current is in a direction parallel to the surface. If the layer extends between ξ=−h/2 and ξ= +h/2, the surface current density Kis deﬁned as K= lim |J|→∞ h→0Zh 2 −h 2Jdξ (9) 24 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.4.4 Uniform line current with contours for determining H. Axis of rotation is used to deduce that radial component of ﬁeld must be zero. By deﬁnition, Kis a vector tangential to the surface that has units of am- pere/meter. Illustration. Hﬁeld Produced by a Uniform Line Current A uniform line current of magnitude iextends from −inﬁnity to + inﬁnity along thezaxis, as shown in Fig. 1.4.4. The symmetry arguments of Example 1.4.1 show that the only component of His azimuthal. Application of Amp` ere’s integral law, (1), to the contour of Fig. 1.4.4 having arbitrary radius rgives a line integral that is simply the product of Hφand the circumference 2 πrand a surface integral that is simply i, regardless of the radius. 2πrH φ=i⇒Hφ=i 2πr(10) This expression makes it especially clear that the units of Hare ampere/meter. Demonstration 1.4.1. Magnetic Field of a Line Current At 60 Hz, the displacement current contribution to the magnetic ﬁeld of the exper- iment shown in Fig. 1.4.5 is negligible. So long as the ﬁeld probe is within a distance rfrom the wire that is small compared to the distance to the ends of the wire or to the return wires below, the magnetic ﬁeld intensity is predicted quantitatively by (10). The curve shown is typical of demonstration measurements illustrating the radial dependence. Because the Hall-eﬀect probe fundamentally exploits the Lorentz force law, it measures the ﬂux density µoH. A common unit for ﬂux density is the Gauss. For conversion of units, 10,000 gauss = 1 tesla, where the tesla is the SI unit. Illustration. Uniform Axial Surface Current At the radius Rfrom the zaxis, there is a uniform zdirected surface current density Kothat extends from - inﬁnity to + inﬁnity in the zdirection. The sym- metry arguments of Example 1.4.1 show that the resulting magnetic ﬁeld intensity Sec. 1.4 Amp` ere’s Integral Law 25 Fig. 1.4.5 Demonstration of peak magnetic ﬂux density induced by line current of 6 ampere (peak). Fig. 1.4.6 Uniform current density Koiszdirected in circular cylin- drical shell at r=R. Radially discontinuous azimuthal ﬁeld shown is determined using the contour at arbitrary radius r. is azimuthal. To determine that ﬁeld, Amp` ere’s integral law is applied to a contour having the arbitrary radius r, shown in Fig. 1.4.6. As in the previous illustration, the line integral is the product of the circumference and Hφ. The surface integral gives nothing if r < R , but gives 2 πRtimes the surface current density if r > R . Thus, 2πrH φ=n0; r < R 2πRK o;r > R⇒Hφ=n0; r < R KoR r;r > R(11) Thus, the distribution of Hφis the discontinuous function shown in Fig. 1.4.6. The ﬁeld tangential to the surface current undergoes a jump that is equal in magnitude 26 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.4.7 Amp` ere’s integral law is applied to surface S/primeenclosed by a rect- angular contour that intersects a surface Scarrying the current density K. In terms of the unit normal to S,n, the resulting continuity condition is given by (16). to the surface current density. Amp` ere’s Continuity Condition. A surface current density in a surface S causes a discontinuity of the magnetic ﬁeld intensity. This is illustrated in Fig. 1.4.6. To obtain a general relation between ﬁelds evaluated to either side of S, a rectan- gular surface of integration is mounted so that it intersects Sas shown in Fig. 1.4.7. The normal to Sis in the plane of the surface of integration. The length lof the rectangle is assumed small enough so that the surface of integration can be consid- ered plane over this length. The width wof the rectangle is assumed to be much smaller than l. It is further convenient to introduce, in addition to the normal n toS, the mutually orthogonal unit vectors isandinas shown. Now apply the integral form of Amp` ere’s law, (1), to the rectangular surface of area lw. For the right-hand side we obtain Z S/primeJ·da+Z S/prime∂ ∂t/epsilon1oE·da/similarequalK·inl (12) Only Jgives a contribution, and then only if there is an inﬁnite current density over the zero thickness of S, as required by the deﬁnition of the surface current density, (9). The time rate of change of a ﬁnite displacement ﬂux density integrated over zero area gives zero, and hence there is no contribution from the second term. The left-hand side of Amp` ere’s law, (1), is a contour integral following the rectangle. Because whas been assumed to be very small compared with l, and H is assumed ﬁnite, no contribution is made by the two short sides of the rectangle. Hence, lis·(Ha−Hb) =K·inl (13) From Fig. 1.4.7, note that is=in×n (14) Sec. 1.5 Charge Conservation in Integral 27 The cross and dot can be interchanged in this scalar triple product without aﬀecting the result (Appendix 1), so introduction of (14) into (13) gives in·n×(Ha−Hb) =in·K (15) Finally, note that the vector inis arbitrary so long as it lies in the surface S. Since it multiplies vectors tangential to the surface, it can be omitted. n×(Ha−Hb) =K (16) There is a jump in the tangential magnetic ﬁeld intensity as one passes through a surface current. Note that (16) gives a prediction consistent with what was found for the illustration in Fig. 1.4.6. 1.5 CHARGE CONSERVATION IN INTEGRAL FORM Embedded in the laws of Gauss and Amp` ere is a relationship that must exist between the charge and current densities. To see this, ﬁrst apply Amp` ere’s law to a closed surface, such as sketched in Fig. 1.5.1. If the contour Cis regarded as the“drawstring” and Sas the “bag,” then this limit is one in which the “string” is drawn tight so that the contour shrinks to zero. Thus, the open surface integrals of (1.4.1) become closed, while the contour integral vanishes. I SJ·da+d dtI S/epsilon1oE·da= 0 (1) But now, in view of Gauss’ law, the surface integral of the electric displacement can be replaced by the total charge enclosed. That is, (1.3.1) is used to write (1) as I SJ·da+d dtZ Vρdv= 0 (2) This is the law of conservation of charge. If there is a net current out of the volume shown in Fig. 1.5.2, (2) requires that the net charge enclosed be decreasing with time. Charge conservation, as expressed by (2), was a compelling reason for Maxwell to add the electric displacement term to Amp` ere’s law. Without the displacement current density, Amp` ere’s law would be inconsistent with charge conservation. That is, if the second term in (1) would be absent, then so would the second term in (2). If the displacement current term is dropped in Amp` ere’s law, then net current cannot enter, or leave, a volume. The conservation of charge is consistent with the intuitive picture of the rela- tionship between charge and current developed in Example 1.2.1. Example 1.5.1. Continuity of Convection Current 28 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.5.1 Contour Cenclosing an open surface can be thought of as the drawstring of a bag that can be closed to create a closed surface. Fig. 1.5.2 Current density leaves a volume Vand hence the net charge must decrease. Fig. 1.5.3 In steady state, charge conservation requires that the cur- rent density entering through the x= 0 plane be the same as that leaving through the plane at x=x. The steady state current of electrons accelerated through vacuum by a uniform electric ﬁeld is described in Example 1.2.1 by assuming that in any plane x= con- stant the current density is the same. That this must be true is now seen formally by applying the charge conservation integral theorem to the volume shown in Fig. 1.5.3. Here the lower surface is in the injection plane x= 0, where the current density is known to be Jo. The upper surface is at the arbitrary level denoted by x. Because the steady state prevails, the time derivative in (2) is zero. The remaining surface integral has contributions only from the top and bottom surfaces. Evaluation of these, with the recognition that the area element on the top surface is ( ixdydz) while it is ( −ixdydz) on the bottom surface, makes it clear that AJx−AJo= 0⇒ρvx=Jo (3) This same relation was used in Example 1.2.1, (1.2.4), as the basis for converting from a particle point of view to the one used here, where ( x, y, z ) are independent oft. Example 1.5.2. Current Density and Time-Varying Charge Sec. 1.5 Charge Conservation in Integral 29 Fig. 1.5.4 With the given axially symmetric charge distribution pos- itive and decreasing with time ( ∂ρ/∂t < 0), the radial current density is positive, as shown. With the charge density a given function of time with an axially symmetric spatial distribution, (2) can be used to deduce the current density. In this example, the charge density is ρ=ρo(t)e−r/a(4) and can be pictured as shown in Fig. 1.5.4. The function of time ρois given, as is the dimension a. As the ﬁrst step in ﬁnding J, we evaluate the volume integral in (2) for a circular cylinder of radius rhaving zas its axis and length lin the zdirection. Z Vρdv=Zl 0Z2π 0Zr 0ρoe−r adr(rdφ)dz = 2πla2£ 1−e−r a¡ 1 +r a¢¤ ρo(5) The axial symmetry demands that Jis in the radial direction and indepen- dent of φandz. Thus, the evaluation of the surface integral in (2) amounts to a multiplication of Jrby the area 2 πrl, and that equation becomes 2πrlJ r+ 2πla2£ 1−e−r a¡ 1 +r a¢¤dρo dt= 0 (6) Finally, this expression can be solved for Jr. Jr=a2 r£ e−r a¡ 1 +r a¢ −1¤dρo dt(7) Under the assumption that the charge density is positive and decreasing, so that dρo/dt < 0, the radial distribution of Jris shown at an instant in time in 30 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.5.5 When a charge qis introduced into an essentially grounded metal sphere, a charge −qis induced on its inner surface. The inte- gral form of charge conservation, applied to the surface S, shows that i=dq/dt . The net excursion of the integrated signal is then a direct measurement of q. Fig. 1.5.4. In this case, the radial current density is positive at any radius rbecause the net charge within that radius, given by (5), is decreasing with time. The integral form of charge conservation provides the link between the current carried by a wire and the charge. Thus, if we can measure a current, this law provides the basis for measuring the net charge. The following demonstration illustrates its use. Demonstration 1.5.1. Measurement of Charge In Demonstration 1.3.1, the net charge is deduced from mechanical measurements and Coulomb’s force law. Here that same charge is deduced electrically. The “ball” carrying the charge is stuck to the end of a thin plastic rod, as in Fig. 1.5.5. The objective is to measure this charge, q, without removing it from the ball. We know from the discussion of Gauss’ law in Sec. 1.3 that this charge is the source of an electric ﬁeld. In general, this ﬁeld terminates on charges of opposite sign. Thus, the net charge that terminates the ﬁeld originating from qis equal in magnitude and opposite in sign to q. Measurement of this “image” charge is tantamount to measuring q. How can we design a metal electrode so that we are guaranteed that all of the lines of Eoriginating from qwill be terminated on its surface? It would seem that the electrode should essentially surround q. Thus, in the experiment shown in Fig. 1.5.5, the charge is transported to the interior of a metal sphere through a hole in its top. This sphere is grounded through a resistance Rand also surrounded by a grounded shield. This resistance is made low enough so that there is essentially no electric ﬁeld in the region between the spherical electrode, and the surrounding shield. As a result, there is negligible charge on the outside of the electrode and the net charge on the spherical electrode is just that inside, namely −q. Now consider the application of (2) to the surface Sshown in Fig. 1.5.5. The surface completely encloses the spherical electrode while excluding the charge qat its center. On the outside, it cuts through the wire connecting the electrode to the resistance R. Thus, the volume integral in (2) gives the net charge −q, while Sec. 1.6 Faraday’s Integral Law 31 contributions to the surface integral only come from where Scuts through the wire. By deﬁnition, the integral of J·daover the cross-section of the wire gives the current i(amps). Thus, (2) becomes simply i+d(−q) dt= 0⇒i=dq dt(8) This current is the result of having pushed the charge through the hole to a position where all the ﬁeld lines terminated on the spherical electrode.3 Although small, the current through the resistor results in a voltage. v/similarequaliR=Rdq dt(9) The integrating circuit is introduced into the experiment in Fig. 1.5.5 so that the oscilloscope directly displays the charge. With this circuit goes a gain Asuch that vo=AZ vdt=ARq (10) Then, the voltage voto which the trace on the scope rises as the charge is inserted through the hole reﬂects the charge q. This measurement of qcorroborates that of Demonstration 1.3.1. In retrospect, because SandVare arbitrary in the integral laws, the experi- ment need not be carried out using an electrode and shield that are spherical. These could just as well have the shape of boxes. Charge Conservation Continuity Condition. The continuity condition asso- ciated with charge conservation can be derived by applying the integral law to the same pillbox-shaped volume used to derive Gauss’ continuity condition, (1.3.17). It can also be found by simply recognizing the similarity between the integral laws of Gauss and charge conservation. To make this similarity clear, rewrite (2) putting the time derivative under the integral. In doing so, d/dt must again be replaced by ∂/∂t, because the time derivative now operates on ρ, a function of tandr. I SJ·da+Z V∂ρ ∂tdV= 0 (11) Comparison of (11) with Gauss’ integral law, (1.3.1), shows the similarity. The role of/epsilon1oEin Gauss’ law is played by J, while that of ρis taken by −∂ρ/∂t . Hence, by analogy with the continuity condition for Gauss’ law, (1.3.17), the continuity condition for charge conservation is 3Note that if we were to introduce the charged ball without having the spherical electrode essentially grounded through the resistance R, charge conservation (again applied to the surface S) would require that the electrode retain charge neutrality. This would mean that there would be a charge qon the outside of the electrode and hence a ﬁeld between the electrode and the surrounding shield. With the charge at the center and the shield concentric with the electrode, this outside ﬁeld would be the same as in the absence of the electrode, namely the ﬁeld of a point charge, (1.3.12). 32 Maxwell’s Integral Laws in Free Space Chapter 1 Fig. 1.6.1 Integration line for deﬁnition of electromotive force. n·(Ja−Jb) +∂σs ∂t= 0(12) Implicit in this condition is the assumption that Jis ﬁnite. Thus, the condition does not include the possibility of a surface current. 1.6 FARADAY’S INTEGRAL LAW The laws of Gauss and Amp` ere relate ﬁelds to sources. The statement of charge conservation implied by these two laws relates these sources. Thus, the previous three sections either relate ﬁelds to their sources or interrelate the sources. In this and the next section, integral laws are introduced that do not involve the charge and current densities. Faraday’s integral law states that the circulation of Earound a contour C is determined by the time rate of change of the magnetic ﬂux linking the surface enclosed by that contour (the magnetic induction). I CE·ds=−d dtZ SµoH·da (1) As in Amp` ere’s integral law and Fig. 1.4.1, the right-hand rule relates dsand da. Theelectromotive force , or EMF, between points (a) and (b) along the path Pshown in Fig. 1.6.1 is deﬁned as Eab=Z(b) (a)E·ds (2) We will accept this deﬁnition for now and look forward to a careful development of the circumstances under which the EMF is measured as a voltage in Chaps. 4 and 10. Electric Field Intensity with No Circulation. First, suppose that the time rate of change of the magnetic ﬂux is negligible, so that the electric ﬁeld is essentially Sec. 1.6 Faraday’s Integral Law 33 Fig. 1.6.2 Uniform electric ﬁeld intensity Eo, between plane parallel uniform distributions of surface charge density, has no circulation about contours C1andC2. free of circulation. This means that no matter what closed contour Cis chosen, the line integral of Emust vanish. I CE·ds= 0 (3) We will ﬁnd that this condition prevails in electroquasistatic systems and that all of the ﬁelds in Sec. 1.3 satisfy this requirement. Illustration. A Field Having No Circulation A static ﬁeld between plane parallel sheets of uniform charge density has no circu- lation. Such a ﬁeld, E=Eoix, exists in the region 0 < y < s between the sheets of surface charge density shown in Fig. 1.6.2. The most convenient contour for testing this claim is denoted C1in Fig. 1.6.2. Along path 1, E·ds=Eody, and integration from y= 0 to y=sgives sEofor the EMF of point (a) relative to point (b). Note that the EMF between the plane parallel surfaces in Fig. 1.6.2 is the same regardless of where the points (a) and (b) are located in the respective surfaces. On segments 2 and 4, Eis orthogonal to ds, so there is no contribution to the line integral on these two sections. Because dshas a direction opposite to Eon segment 3, the line integral is the integral from y= 0 to y=sofE·ds=−Eody. The result of this integration is −sEo, so the contributions from segments 1 and 3 cancel, and the circulation around the closed contour is indeed zero.4 In this planar geometry, a ﬁeld that has only a ycomponent cannot be a function of xwithout incurring a circulation. This is evident from carrying out this integration for such a ﬁeld on the rectangular contour C1. Contributions to paths 1 and 3 cancel only if Eis independent of x. Example 1.6.1. Contour Integration To gain some appreciation for what it means to require of Ethat it have no circu- lation, no matter what contour is chosen, consider the somewhat more complicated contour C2in the uniform ﬁeld region of Fig. 1.6.2. Here, C2is composed of the 4In setting up the line integral on a contour such as 3, which has a direction opposite to that in which the coordinate increases, it is tempting to double-account for the direction of dsnot only be recognizing that ds=−iydy, but by integrating from y=stoy= 0 as well. 34 Maxwell’s Integral Laws in Free Space Chapter 1 semicircle (5) and the straight segment (6). On the latter, Eis perpendicular to ds and so there is no contribution there to the circulation. I CE·ds=Z 5E·ds+Z 6E·ds=I 5E·ds (4) On segment 5, the vector diﬀerential dsis ﬁrst written in terms of the unit vector iφ, and that vector is in turn written (with the help of the vector decomposition shown in the ﬁgure) in terms of the Cartesian unit vectors. ds=iφRdφ;iφ=iycosφ−ixsinφ (5) It follows that on the segment 5 of contour C2 E·ds=EocosφRdφ (6) and integration gives I CE·ds=Zπ 0EocosφRdφ = [EoRsinφ]π 0= 0 (7) So for contour C2, the circulation of Eis also zero. When the electromotive force between two points is path independent, we call it the voltage between the two points. For a ﬁeld having no circulation, the EMF must be independent of path. This we will recognize formally in Chap. 4. Electric Field Intensity with Circulation. The second limiting situation, typical of the magnetoquasistatic systems to be considered, is primarily concerned with the circulation of E, and hence with the part of the electric ﬁeld generated by the time-varying magnetic ﬂux density. The remarkable fact is that Faraday’s law holds for any contour, whether in free space or in a material. Often, however, the contour of interest coincides with a conducting wire, which comprises a coil that links a magnetic ﬂux density. Illustration. Terminal EMF of a Coil A coil with one turn is shown in Fig. 1.6.3. Contour (1) is inside the wire, while (2) joins the terminals along a deﬁned path. With these contours constituting C, Faraday’s integral law as given by (1) determines the terminal electromotive force. If the electrical resistance of the wire can be regarded as zero, in the sense that the electric ﬁeld intensity inside the wire is negligible, the contour integral reduces to an integration from (b) to (a).5In view of the deﬁnition of the EMF, (2), this integration gives the negative of the EMF. Thus, Faraday’s law gives the terminal EMF as Eab=d dtλf; λf≡Z SµoH·da (8) 5With the objectives here limited to attaching an intuitive meaning to Faraday’s law, we will give careful attention to the conditions required for this terminal relation to hold in Chaps. 8, 9, and 10. Sec. 1.6 Faraday’s Integral Law 35 Fig. 1.6.3 Line segment (1) through a perfectly conducting wire and (2) joining the terminals (a) and (b) form closed contour. Fig. 1.6.4 Demonstration of voltmeter reading induced at terminals of a coil in accordance with Faraday’s law. To plot data on graph, normalize voltage to Voas deﬁned with (11). Because Iis the peak current, vis the peak voltage. where λf, the total ﬂux of magnetic ﬁeld linking the coil, is deﬁned as the ﬂux linkage. Note that Faraday’s law makes it possible to measure µoHelectrically (as now demonstrated). Demonstration 1.6.1. Voltmeter Reading Induced by Magnetic Induction The rectangular coil shown in Fig. 1.6.4 is used to measure the magnetic ﬁeld intensity associated with current in a wire. Thus, the arrangement and ﬁeld are the same as in Demonstration 1.4.1. The height and length of the coil are handlas shown, and because the coil has Nturns, it links the ﬂux enclosed by one turn N times. With the upper conductors of the coil at a distance Rfrom the wire, and the magnetic ﬁeld intensity taken as that of a line current, given by (1.4.10), evaluation of (8) gives λf=µoNZz+l zZR+h Ri 2πrdrdz =" µoNl 2πlnµ 1 +h R¶# i (9) In the experiment, the current takes the form i=Isinωt (10) 36 Maxwell’s Integral Laws in Free Space Chapter 1 where ω= 2π(60). The EMF between the terminals then follows from (8) and (9) as v=Voln¡ 1 +h R¢ cosωt; Vo≡µoNlωI 2π(11) A voltmeter reads the electromotive force between the two points to which it is connected, provided certain conditions are satisﬁed. We will discuss these in Chap. 8. In a typical experiment using a 20-turn coil with dimensions of h= 8 cm, l= 20 cm, I= 6 amp peak, the peak voltage measured at the terminals with a spacing R= 8 cm is v= 1.35 mV. To put this data point on the normalized plot of Fig. 1.6.4, note that R/h= 1 and the measured v/Vo= 0.7. Faraday’s Continuity Condition. It follows from Faraday’s integral law that the tangential electric ﬁeld is continuous across a surface of discontinuity, provided that the magnetic ﬁeld intensity is ﬁnite in the neighborhood of the surface of discontinuity. This can be shown by applying the integral law to the incremental surface shown in Fig. 1.4.7, much as was done in Sec. 1.4 for Amp` ere’s law. With J set equal to zero, there is a formal analogy between Amp` ere’s integral law, (1.4.1), and Faraday’s integral law, (1). The former becomes the latter if H→E,J→ 0, and /epsilon1oE→ −µoH. Thus, Amp` ere’s continuity condition (1.4.16) becomes the continuity condition associated with Faraday’s law. n×(Ea−Eb) = 0 (12) At a surface having the unit normal n, the tangential electric ﬁeld intensity is continuous. 1.7 GAUSS’ INTEGRAL LAW OF MAGNETIC FLUX The net magnetic ﬂux out of any region enclosed by a surface Smust be zero. I SµoH·da= 0 (1) This property of ﬂux density is almost implicit in Faraday’s law. To see this, consider that law, (1.6.1), applied to a closed surface S. Such a surface is obtained from an open one by letting the contour shrink to zero, as in Fig. 1.5.1. Then Faraday’s integral law reduces to d dtI SµoH·da=0 (2) Gauss’ law (1) adds to Faraday’s law the empirical fact that in the beginning, there was no closed surface sustaining a net outward magnetic ﬂux. Illustration. Uniqueness of Flux Linking Coil Sec. 1.7 Magnetic Gauss’ Law 37 Fig. 1.7.1 Contour Cfollows loop of wire having terminals a−b. Because each has the same enclosing contour, the net magnetic ﬂux through surfaces S1andS2must be the same. Fig. 1.7.2 (a) The ﬁeld of a line current induces a ﬂux in a horizon- tal rectangular coil. (b) The open surface has the coil as an enclosing contour. Rather than being in the plane of the contour, this surface is composed of the ﬁve segments shown. An example is shown in Fig. 1.7.1. Here a wire with terminals a−bfollows the contour C. According to (1.6.8), the terminal EMF is found by integrating the normal magnetic ﬂux density over a surface having Cas its edge. But which surface? Figure 1.7.1 shows two of an inﬁnite number of possibilities. The terminal EMF can be unique only if the integrals over S1andS2result in the same answer. Taken together, S1andS2form a closed surface. The magnetic ﬂux continuity integral law, (1), requires that the net ﬂux out of this closed surface be zero. This is equivalent to the statement that the ﬂux passing through S1in the direction of da1must be equal to that passing through S2in the direction of da2. We will formalize this statement in Chap. 8. Example 1.7.1. Magnetic Flux Linked by Coil and Flux Continuity In the conﬁguration of Fig. 1.7.2, a line current produces a magnetic ﬁeld intensity that links a one-turn coil. The left conductor in this coil is directly below the wire at a distance d. The plane of the coil is horizontal. Nevertheless, it is convenient to specify the position of the right conductor in terms of a distance Rfrom the line current. What is the net ﬂux linked by the coil? The most obvious surface to use is one in the same plane as the coil. However, 38 Maxwell’s Integral Laws in Free Space Chapter 1 in doing so, account must be taken of the way in which the unit normal to the surface varies in direction relative to the magnetic ﬁeld intensity. Selection of another surface, to which the magnetic ﬁeld intensity is either normal or tangential, simpliﬁes the calculation. On surfaces S2andS3, the normal direction is the direction of the magnetic ﬁeld. Note also that because the ﬁeld is tangential to the end surfaces, S4 andS5, these make no contribution. For the same reason, there is no contribution from S6, which is at the radius rofrom the wire. Thus, λf≡Z SµoH·da=Z S2µoH·da+Z S3µoH·da (3) OnS2the unit normal is iφ, while on S3it is−iφ. Therefore, (3) becomes λf=Zl 0ZR roµoHφdrdz−Zl 0Zd roµoHφdrdz (4) With the ﬁeld intensity for a line current given by (1.4.10), it follows that λf=µoli 2π¡ lnR ro−lnd ro¢ =µoli 2πln¡R d¢ (5) That rodoes not appear in the answer is no surprise, because if the surface S1had been used, rowould not have been brought into the calculation. Magnetic Flux Continuity Condition. With the charge density set equal to zero, the magnetic continuity integral law (1) takes the same form as Gauss’ integral law (1.3.1). Thus, Gauss’ continuity condition (1.3.17) becomes one representing the magnetic ﬂux continuity law by making the substitution /epsilon1oE→µoH. n·(µoHa−µoHb) = 0 (6) The magnetic ﬂux density normal to a surface is continuous. 1.8 SUMMARY Electromagnetic ﬁelds, whether they be inside a transistor, on the surfaces of an antenna or in the human nervous system, are deﬁned in terms of the forces they produce. In every example involving electromagnetic ﬁelds, charges are moving somewhere in response to electromagnetic ﬁelds. Hence, our starting point in this introductory chapter is the Lorentz force on an elementary charge, (1.1.1). Repre- sented by this law is the eﬀect of the ﬁeld on the charge and current (charge in motion). The subsequent sections are concerned with the laws that predict how the ﬁeld sources, the charge, and current densities introduced in Sec. 1.2, in turn give rise to the electric and magnetic ﬁelds. Our presentation is aimed at putting these Sec. 1.8 Summary 39 laws to work. Hence, the empirical origins of these laws that would be evident from a historical presentation might not be fully appreciated. Elegant as they appear, Maxwell’s equations are no more than a summary of experimental results. Each of our case studies is a potential test of the basic laws. In the interest of being able to communicate our subject, each of the basic laws is given a name. In the interest of learning our subject, each of these laws should now be memorized. A summary is given in Table 1.8.1. By means of the examples and demonstrations, each of these laws should be associated with one or more physical consequences. From the Lorentz force law and Maxwell’s integral laws, the units of variables and constants are established. For the SI units used here, these are summarized in Table 1.8.2. Almost every practical result involves the free space permittivity /epsilon1oand/or the free space permeability µo. Although these are summarized in Table 1.8.2, conﬁdence also comes from having these natural constants memorized. A common unit for measuring the magnetic ﬂux density is the Gauss, so the conversion to the SI unit of Tesla is also given with the abbreviations. A goal in this chapter has also been the use of examples to establish the mathematical signiﬁcance of volume, surface, and contour integrations. At the same time, important singular source distributions have been deﬁned and their associated ﬁelds derived. We will make extensive use of point, line, and surface sources and the associated ﬁelds. In dealing with surface sources, a continuity condition should be associated with each of the integral laws. These are summarized in Table 1.8.3. The continuity conditions should always be associated with the integral laws from which they originate. As terms are added to the integral laws to account for macroscopic media, there will be corresponding changes in the continuity condi- tions. R E F E R E N C E S [1] M. Faraday, Experimental Researches in Electricity , R. Taylor Publisher (1st-9th series), 1832-1835, 1 volume, various pagings; “From the Philosophical Transactions 1832-1835,” London, England. [2] J.C. Maxwell, A Treatise on Electricity and Magnetism , 3rd ed., 1891, reissued by Dover, N.Y. (1954). 40 Maxwell’s Integral Laws in Free Space Chapter 1 TABLE 1.8.1 SUMMARY OF MAXWELL’S INTEGRAL LAWS IN FREE SPACE NAME INTEGRAL LAW EQ. NUMBER Gauss’ LawH S/epsilon1oE·da=R Vρdv 1.3.1 Ampere’s LawH CH·ds=R SJ·da+d dtR S/epsilon1oE·da 1.4.1 Faraday’s LawH CE·ds=−d dtR SµoH·da 1.6.1 Magnetic Flux ContinuityH SµoH·da= 0 1.7.1 Charge ConservationH SJ·da+d dtR Vρdv= 0 1.5.2 Sec. 1.8 Summary 41 TABLE 1.8.2 DEFINITIONS AND UNITS OF FIELD VARIABLES AND CONSTANTS (basic unit of mass, kg, is replaced by V-C-s2/m2) VARIABLE OR PARAMETERNOMENCLATURE BASIC UNITSDERIVED UNITS Electric Field Intensity E V/m V/m Electric Displacement Flux Density/epsilon1oE C/m2C/m2 Charge Density ρ C/m3C/m3 Surface Charge Density σs C/m2C/m2 Magnetic Field Intensity H C/(ms) A/m Magnetic Flux Density µoH Vs/m2T Current Density J C/(m2s) A/m2 Surface Current Density K C/(ms) A/m Free Space Permittivity /epsilon1o= 8.854×10−12C/(Vm) F/m Free Space Permeability µo= 4π×10−7Vs2/(Cm) H/m UNIT ABBREVIATIONS Amp` ere A Kilogram kg Volt V Coulomb C Meter m Farad F Second s Henry H Tesla T (104Gauss) 42 Maxwell’s Integral Laws in Free Space Chapter 1 TABLE 1.8.3 SUMMARY OF CONTINUITY CONDITIONS IN FREE SPACE NAME CONTINUITY CONDITION EQ. NUMBER Gauss’ Law n·(/epsilon1oEa−/epsilon1oEb) =σs 1.3.17 Amp` ere’s Law n×(Ha−Hb) =K 1.4.16 Faraday’s Law n×(Ea−Eb) = 0 1.6.14 Magnetic Flux Continuityn·(µoHa−µoHb) = 0 1.7.6 Charge Conservationn·(Ja−Jb) +∂σs ∂t= 0 1.5.12 Sec. 1.2 Problems 43 P R O B L E M S 1.1 The Lorentz Law in Free Space∗ 1.1.1∗Assuming in Example 1.1.1 that vi= 0 and that Ex<0, show that by the time the electron has reached the position x=h, its velocity isp −2eExh/m. In an electric ﬁeld of only Ex= 1v/cm = 10−2v/m, show that by the time it reaches h= 10−2m, the electron has reached a velocity of 5.9×103m/s. 1.1.2 An electron moves in vacuum under the same conditions as in Example 1.1.1 except that the electric ﬁeld takes the form E=Exix+Eyiywhere ExandEyare given constants. When t= 0, the electron is at ξx= 0 and ξy= 0 and the velocity dξx/dt=vianddξy/dt= 0. (a) Determine ξx(t) and ξy(t). (b) For Ex>0, when and where does the electron return to the plane x= 0? 1.1.3∗An electron, having velocity v=viiz, experiences the ﬁeld H=Hoiyand E=Eoix, where HoandEoare constants. Show that the electron retains this velocity if Eo=viµoHo. 1.1.4 An electron has the initial position x= 0, y= 0, z=zo. It has an initial velocity v=voixand moves in the uniform and constant ﬁelds E=Eoiy,H=Hoiy. (a) Determine the position of the electron in the ydirection, ξy(t). (b) Describe the trajectory of the electron. 1.2 Charge and Current Densities 1.2.1∗The charge density is ρor/Rcoulomb/m3throughout the volume of a spher- ical region having radius R, with ρoa constant and rthe distance from the center of the region (the radial coordinate in spherical coordinates). Show that the total charge associated with this charge density is q=πρoR3 coulomb. 1.2.2 In terms of given constants ρoanda, the net charge density is ρ= (ρo/a2) (x2+y2+z2) coulomb/m3. What is the total charge q(coulomb) in the cubical region −a < x < a, −a < y < a, −a < z < a ? ∗An asterisk on a problem number designates a “show that” problem. These problems are especially designed for self study. 44 Maxwell’s Integral Laws in Free Space Chapter 1 1.2.3∗With Joandagiven constants, the current density is J= (Jo/a2)(y2+ z2)[ix+iy+iz]. Show that the total current ipassing through the surface x= 0,−a < y < a, −a < z < a isi= 8Joa2/3 amp. 1.2.4 In cylindrical coordinates ( r, φ, z ) the current density is given in terms of constants JoandabyJ=Jo(r/a)2iz(amp/m2). What is the net current i(amp) through the surface z= 0, r < a ? 1.2.5∗In cylindrical coordinates, the electric ﬁeld in the annular region b < r < a isE=irEo(b/r), where Eois a given negative constant. When t= 0, an electron having mass mand charge q=−ehas no velocity and is positioned atr=ξr=b. (a) Show that, in vacuum, the radial motion of the electron is governed by the diﬀerential equation mdv r/dt=−eEob/ξr, where vr=dξr/dt. Note that these expressions combine to provide one second-order dif- ferential equation governing ξr. (b) By way of providing one integration of this equation, multiply the ﬁrst of the ﬁrst-order expressions by vrand (with the help of the second ﬁrst-order expression) show that the resulting equation can be written asd[1 2mv2 r+eEob lnξ r]/dt= 0. That is, the sum of the kinetic and potential energies (the quantity in brackets) remains constant. (c) Use the result of (b) to ﬁnd the electron velocity vr(r). (d) Assume that this is one of many electrons that ﬂow radially outward from the cathode at r=btor=aand that the number of electrons passing radially outward at any location ris independent of time. The system is in the steady state so that the net current ﬂowing outward through a surface of radius rand length l, i= 2πrlJ r, is the same at one radius ras at another. Use this fact to determine the charge density ρ(r). 1.3 Gauss’ Integral Law 1.3.1∗Consider how Gauss’ integral law, (1), is evaluated for a surface that is not naturally symmetric. The charge distribution is the uniform line charge of Fig. 1.3.7 and hence Eis given by (13). However, the surface integral on the left in (1) is to be evaluated using a surface that has unit length in the zdirection and a square cross-section centered on the zaxis. That is, the surface is composed of the planes z= 0, z= 1, x=±a, and y=±a. Thus, we know from evaluation of the right-hand side of (1) that evaluation of the surface integral on the left should give the line charge density λl. (a) Show that the area elements daon these respective surfaces are ±izdxdy,±ixdydz, and±iydxdz. Sec. 1.3 Problems 45 (b) Starting with (13), show that in Cartesian coordinates, Eis E=λl 2π/epsilon1oµx x2+y2ix+y x2+y2iy¶ (a) (Standard Cartesian and cylindrical coordinates are deﬁned in Table I at the end of the text.) (c) Show that integration of /epsilon1oE·daover the part of the surface at x=a leads to the integral Z /epsilon1oE·da=λl 2πZ1 0Za −aa a2+y2dydz (b) (d) Finally, show that integration over the entire closed surface indeed gives λl. 1.3.2 Using the spherical symmetry and a spherical surface, the electric ﬁeld associated with the point charge qof Fig. 1.3.6 is found to be given by (12). Evaluation of the left-hand side of (1) over any other surface that encloses the point charge must also give q. Suppose that the closed surface Sis composed of a hemisphere of radius ain the upper half-plane, a hemisphere of radius bin the lower half-plane, and a washer-shaped ﬂat surface that joins the two. In spherical coordinates (deﬁned in Table I), these three parts of the closed surface Sare deﬁned by ( r=a,0< θ <1 2π,0≤φ < 2π),(r=b,1 2π < θ < π, 0≤φ < 2π), and ( θ=1 2π, b≤r≤a,0≤ φ <2π). For this surface, use (12) to evaluate the left-hand side of (1) and show that it results in q. 1.3.3∗A cylindrically symmetric charge conﬁguration extends to inﬁnity in the ±zdirections and has the same cross-section in any constant zplane. Inside the radius b, the charge density has a parabolic dependence on radius while over the range b < r < a outside that radius, the charge density is zero. ρ=½ ρo(r/b)2;r < b 0; b < r < a(a) There is no surface charge density at r=b. (a) Use the axial symmetry and Gauss’ integral law to show that Ein the two regions is E=½ (ρor3/4/epsilon1ob2)ir;r < b (ρob2/4/epsilon1or)ir;b < r < a(b) (b) Outside a shell at r=a,E= 0. Use (17) to show that the surface charge density at r=ais σs=−ρob2/4a (c) 46 Maxwell’s Integral Laws in Free Space Chapter 1 (c) Integrate this charge per unit area over the surface of the shell and show that the resulting charge per unit length on the shell is the negative of the charge per unit length inside. (d) Show that, in Cartesian coordinates, Eis E=ρo 4/epsilon1o½ [x(x2+y2)/b2]ix+ [y(x2+y2)/b2]iy;r < b b2x(x2+y2)−1ix+b2y(x2+y2)−1iy;b < r < a(d) Note that ( r=p x2+y2,cosφ=x/r, sinφ=y/r,ir=ixcosφ+ iysinφ) and the result takes the form E=Ex(x, y)ix+Ey(x, y)iy. (e) Now, imagine that the circular cylinder of charge in the region r < b is enclosed by a cylindrical surface of square cross-section with the z coordinate as its axis and unit length in the zdirection. The walls of this surface are at x=±c, y=±candz= 0 and z= 1. (To be sure that the cylinder of the charge distribution is entirely within the surface, b < r < a, b < c < a/√ 2.) Show that the surface integral on the left in (1) is I S/epsilon1oE·da=ρob2 4½Zc −c£c c2+y2−(−c) c2+y2¤ dy +Zc −c£c x2+c2−(−c) x2+c2¤ dx¾ (e) Without carrying out these integrations, what is the answer? 1.3.4 In a spherically symmetric conﬁguration, the region r < b has the uniform charge density ρband is surrounded by a region b < r < a having the uniform charge density ρa. At r=bthere is no surface charge density, while at r=athere is that surface charge density that assures E= 0 for a < r . (a) Determine Ein the two regions. (b) What is the surface charge density at r=a? (c) Now suppose that there is a surface charge density given at r=bof σs=σo. Determine Ein the two regions and σsatr=a. 1.3.5∗The region between the plane parallel sheets of surface charge density shown in Fig. 1.3.8 is ﬁlled with a charge density ρ= 2ρoz/s, where ρo is a given constant. Again, assume that the electric ﬁeld below the lower sheet is Eoizand show that between the sheets Ez=Eo−σo /epsilon1o+ρo /epsilon1os£ z2−(s/2)2¤ (a) 1.3.6 In a conﬁguration much like that of Fig. 1.3.8, there are three rather than two sheets of charge. One, in the plane z= 0, has the given surface charge density σo. The second and third, respectively located at z=s/2 and Sec. 1.4 Problems 47 z=−s/2, have unknown charge densities σaandσb. The electric ﬁeld outside the region −1 2s < z <1 2sis zero, and σa= 2σb. Determine σaand σb. 1.3.7 Particles having charges of the same sign are constrained in their positions by a plastic tube which is tilted with respect to the horizontal by the angle α, as shown in Fig. P1.3.7. Given that the lower particle has charge Qoand is ﬁxed, while the upper one (which has charge Qand mass M) is free to move without friction, at what relative position, ξ, can the upper particle be in a state of static equilibrium? Fig. P1.3.7 1.4 Amp` ere’s Integral Law 1.4.1∗A static Hﬁeld is produced by the cylindrically symmetric current density distribution J=Joexp(−r/a)iz, where Joandaare constants and ris the radial cylindrical coordinate. Use the integral form of Amp` ere’s law to show that Hφ=Joa2 r£ 1−e−r/a¡ 1 +r a¢¤ (a) 1.4.2∗In polar coordinates, a uniform current density Joizexists over the cross- section of a wire having radius b. This current is returned in the −zdirection as a uniform surface current at the radius r=a > b . (a) Show that the surface current density at r=ais K=−(Job2/2a)iz (a) (b) Use the integral form of Amp` ere’s law to show that Hin the regions 0< r < b andb < r < a is H=½(Jor/2)iφ; r < b (Job2/2r)iφ;b < r < a(b) (c) Use Amp` ere’s continuity condition, (16), to show that H= 0 for r > a . 48 Maxwell’s Integral Laws in Free Space Chapter 1 (d) Show that in Cartesian coordinates, His H=Jo 2½−yix+xiy; r < b −b2y(x2+y2)−1ix+b2x(x2+y2)−1iy;b < r < a(c) (e) Suppose that the inner cylinder is now enclosed by a contour Cthat encloses a square surface in a constant zplane with edges at x=±c andy=±c(so that Cis in the region b < r < a, b < c < a/√ 2). Show that the contour integral on the left in (1) is I CH·ds=Zc −cJob2 2µc c2+y2−(−c) c2+y2¶ dy +Zc −cJob2 2µc x2+c2−(−c) x2+c2¶ dx(d) Without carrying out the integrations, use Amp` ere’s integral law to deduce the result of evaluating (d). 1.4.3 In a conﬁguration having axial symmetry about the zaxis and extending to inﬁnity in the ±zdirections, a line current Iﬂows in the −zdirection along the zaxis. This current is returned uniformly in the + zdirection in the region b < r < a . There is no current density in the region 0 < r < b and there are no surface current densities. (a) In terms of I, what is the current density in the region b < r < a ? (b) Use the symmetry of the conﬁguration and the integral form of Amp` ere’s law to deduce Hin the regions 0 < r < b andb < r < a . (c) Express Hin each region in Cartesian coordinates. (d) Now, consider the evaluation of the left-hand side of (1) for a contour Cthat encloses a square surface Shaving sides of length 2 cand the z axis as a perpendicular. That is, Clies in a constant zplane and has sides x=±candy=±cwith c < a/√ 2). In Cartesian coordinates, set up the line integral on the left in (1). Without carrying out the integrations, what must the answer be? 1.4.4∗In a conﬁguration having axial symmetry about the zaxis, a line current I ﬂows in the −zdirection along the zaxis. This current is returned at the radii aandb, where there are uniform surface current densities Kzaand Kzb, respectively. The current density is zero in the regions 0 < r < b, b < r < a anda < r . (a) Given that Kza= 2Kzb, show that Kza=I/π(2a+b). (b) Show that His H=−I 2πiφ½ 1/r; 0 < r < b 2a/r(2a+b);b < r < a(a) Sec. 1.6 Problems 49 1.4.5 Uniform surface current densities K=±Koiyare in the planes z=±1 2s, respectively. In the region −1 2s < z <1 2s, the current density is J= 2Joz/siy. In the region z <−1 2s,H= 0. Determine Hfor−1 2s < z . 1.5 Charge Conservation in Integral Form 1.5.1∗In the region of space of interest, the charge density is uniform and a given function of time, ρ=ρo(t). Given that the system has spherical symmetry, with rthe distance from the center of symmetry, use the integral form of the law of charge conservation to show that the current density is J=−r 3dρo dtir (a) 1.5.2 In the region x > 0, the charge density is known to be uniform and the given function of time ρ=ρo(t). In the plane x= 0, the current density is zero. Given that it is xdirected and only dependent on xandt, what is J? 1.5.3∗In the region z >0, the current density J= 0. In the region z <0,J= Jo(x, y) cosωtiz, where Jois a given function of ( x, y). Given that when t= 0, the surface charge density σs= 0 over the plane z= 0, show that fort >0, the surface charge density in the plane z= 0 is σs(x, y, t ) = [Jo(x, y)/ω] sinωt. 1.5.4 In cylindrical coordinates, the current density J= 0 for r < R , and J= Jo(φ, z) sinωtirforr > R . The surface charge density on the surface at r=Risσs(φ, z, t ) = 0 when t= 0. What is σs(φ, z, t ) for t >0? 1.6 Faraday’s Integral Law 1.6.1∗Consider the calculation of the circulation of E, the left-hand side of (1), around a contour consisting of three segments enclosing a surface lying in thex−yplane: from ( x, y) = (0 ,0)→(g, s) along the line y=sx/g; from (x, y) = ( g, s)→(0, s) along y=sand from ( x, y) = (0 , s) to (0 ,0) along x= 0. (a) Show that along the ﬁrst leg, ds= [ix+ (s/g)iy]dx. (b) Given that E=Eoiywhere Eois a given constant, show that the line integral along the ﬁrst leg is sEoand that the circulation around the closed contour is zero. 1.6.2 The situation is the same as in Prob. 1.6.1 except that the ﬁrst segment of the closed contour is along the curve y=s(x/g)2. 50 Maxwell’s Integral Laws in Free Space Chapter 1 (a) Once again, show that for a uniform ﬁeld E=Eoiy, the circulation ofEis zero. (b) For E=Eo(x/g)iy, what is the circulation around this contour? 1.6.3∗TheEﬁeld of a line charge density uniformly distributed along the zaxis is given in cylindrical coordinates by (1.3.13). (a) Show that in Cartesian coordinates, with x=rcosφandy=rsinφ, E=λl 2π/epsilon1o·x x2+y2ix+y x2+y2iy¸ (a) (b) For the contour shown in Fig. P1.6.3, show that I CE·ds=λl 2π/epsilon1o·Zg k(1/x)dx+Zh 0y g2+y2dy −Zg kx x2+h2dx−Zh 0y k2+y2dy¸ (b) and complete the integrations to prove that the circulation is zero. Fig. P1.6.3 Fig. P1.6.4 1.6.4 A closed contour consisting of six segments is shown in Fig. P1.6.4. For the electric ﬁeld intensity of Prob. 1.6.3, calculate the line integral of E·ds on each of these segments and show that the integral around the closed contour is zero. 1.6.5∗The experiment in Fig. 1.6.4 is carried out with the coil positioned hori- zontally, as shown in Fig. 1.7.2. The left edge of the coil is directly below the wire, at a distance d, while the right edge is at the radial distance R from the wire, as shown. The area element daisydirected (the vertical direction). Sec. 1.7 Problems 51 (a) Show that, in Cartesian coordinates, the magnetic ﬁeld intensity due to the current iis H=i 2πµ−ixy x2+y2+iyx x2+y2¶ (a) (b) Use this ﬁeld to show that the magnetic ﬂux linking the coil is as given by (1.7.5). (c) What is the circulation of Earound the contour representing the coil? (d) Given that the coil has Nturns, what is the EMF measured at its terminals? 1.6.6 The magnetic ﬁeld intensity is given to be H=Ho(t)(ix+iy), where Ho(t) is a given function of time. What is the circulation of Earound the contour shown in Fig. P1.6.6? Fig. P1.6.6 1.6.7∗In the plane y= 0, there is a uniform surface charge density σs=σo. In the region y <0,E=E1ix+E2iywhere E1andE2are given constants. Use the continuity conditions of Gauss and Faraday, (1.3.17) and (12), to show that just above the plane y= 0, where y= 0+, the electric ﬁeld intensity isE=E1ix+ [E2+ (σo//epsilon1o)]iy. 1.6.8 Inside a circular cylindrical region having radius r=R, the electric ﬁeld intensity is E=Eoiy, where Eois a given constant. There is a surface charge density σocosφon the surface at r=R(the polar coordinate φis measured relative to the xaxis). What is Ejust outside the surface, where r=R+? 1.7 Integral Magnetic Flux Continuity Law 1.7.1∗A region is ﬁlled by a uniform magnetic ﬁeld intensity Ho(t)iz. (a) Show that in spherical coordinates (deﬁned in Fig. A.1.3 of Appendix 1),H=Ho(t)(ircosθ−iθsinθ). (b) A circular contour lies in the z= 0 plane and is at r=R. Using the enclosed surface in the plane z= 0 as the surface S, show that the circulation of Ein the φdirection around Cis−πR2µodHo/dt. 52 Maxwell’s Integral Laws in Free Space Chapter 1 (c) Now compute the same circulation using as a surface Senclosed by Cthe hemispherical surface at r=R,0≤θ <1 2π. 1.7.2 With Ho(t) a given function of time and da given constant, three distri- butions of Hare proposed. H=Ho(t)iy (a) H=Ho(t)(x/d)ix (b) H=Ho(t)(y/d)ix (c) Which one of these will not satisfy (1) for a surface Sas shown in Fig. 1.5.3? 1.7.3∗In the plane y= 0, there is a given surface current density K=Koix. In the region y <0,H=H1iy+H2iz. Use the continuity conditions of (1.4.16) and (6) to show that just above the current sheet, where y= 0+,H= (H1−Ko)iy+Hziz. 1.7.4 In the circular cylindrical surface r=R, there is a surface current density K=Koiz. Just inside this surface, where r=R,H=H1ir. What is H just outside the surface, where r=R+? 2 MAXWELL’S DIFFERENTIAL LAWS IN FREE SPACE 2.0 INTRODUCTION Maxwell’s integral laws encompass the laws of electrical circuits. The transition from ﬁelds to circuits is made by associating the relevant volumes, surfaces, and contours with electrodes, wires, and terminal pairs. Begun in an informal way in Chap. 1, this use of the integral laws will be formalized and examined as the following chapters unfold. Indeed, many of the empirical origins of the integral laws are in experiments involving electrodes, wires and the like. The remarkable fact is that the integral laws apply to any combination of volume and enclosing surface or surface and enclosing contour, whether associated with a circuit or not. This was implicit in our use of the integral laws for deducing ﬁeld distributions in Chap. l. Even though the integral laws can be used to determine the ﬁelds in highly symmetric conﬁgurations, they are not generally applicable to the analysis of re- alistic problems. Reasons for this lie beyond the geometric complexity of practical systems. Source distributions are not generally known, even when materials are idealized as insulators and “perfect” conductors. In actual materials, for example, those having ﬁnite conductivity, the self-consistent interplay of ﬁelds and sources, must be described. Because they apply to arbitrary volumes, surfaces, and contours, the integral laws also contain the diﬀerential laws that apply at each point in space. The dif- ferential laws derived in this chapter provide a more broadly applicable basis for predicting ﬁelds. As might be expected, the point relations must involve informa- tion about the shape of the ﬁelds in the neighborhood of the point. Thus it is that the integral laws are converted to point relations by introducing partial derivatives of the ﬁelds with respect to the spatial coordinates. The plan in this chapter is ﬁrst to write each of the integral laws in terms of one type of integral. For example, in the case of Gauss’ law, the surface integral is 1 2 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 converted to one over the volume Venclosed by the surface. Z Vdiv(/epsilon1oE)dv=I S/epsilon1oE·da (1) Here divis some combination of spatial derivatives of /epsilon1oEto be determined in the next section. With this mathematical theorem accepted for now, Gauss’ integral law, (1.3.1), can be written in terms of volume integrals. Z Vdiv(/epsilon1oE)dv=Z Vρdv (2) The desired diﬀerential form of Gauss’ law is obtained by equating the integrands in this expression. div(/epsilon1oE) =ρ (3) Is it true that if two integrals are equal, their integrands are as well? In general, the answer is no! For example, if x2is integrated from 0 to 1, the result is the same as for an integration of 2 x/3 over the same interval. However, x2is hardly equal to 2x/3 for every value of x. It is because the volume Vis arbitrary that we can equate the integrands in (1). For a one-dimensional integral, this is equivalent to having endpoints that are arbitrary. With the volume arbitrary (the endpoints arbitrary), the integrals can only be equal if the integrands are as well. The equality of the three-dimensional volume integration on the left in (1) and the two-dimensional surface integration on the right is analogous to the case of a one-dimensional integral being equal to the function evaluated at the integration endpoints. That is, suppose that the operator deroperates on f(x) in such a way that Zx2 x1der(f)dx=f(x2)−f(x1) (4) The integration on the left over the “volume” interval between x1andx2is reduced by this “theorem” to an evaluation on the “surface,” where x=x1andx=x2. The procedure for determining the operator derin (4) is analogous to that used to deduce the divergence and curl operators in Secs. 2.1 and 2.4, respectively. The point xat which deris to be evaluated is taken midway in the integration interval, as in Fig. 2.0.1. Then the interval is taken as incremental (∆ x=x2−x1) and for small ∆ x, (4) becomes Fig. 2.0.1 General function of xdeﬁned between endpoints x1andx2. [der(f)]∆x=f(x2)−f(x1) (5) Sec. 2.1 The Divergence Operator 3 Fig. 2.1.1 Incremental volume element for determination of divergence op- erator. It follows that der= lim ∆x→0·f¡ x+∆x 2¢ −f¡ x−∆x 2¢ ∆x¸ (6) Thus, as we knew to begin with, deris the derivative of fwith respect to x. Byproducts of the derivation of the divergence and curl operators in Secs. 2.1 and 2.4 are the integral theorems of Gauss and Stokes, derived in Secs. 2.2 and 2.5, respectively. A theorem is a mathematical relation and must be distinguished from a physical law, which establishes a physical relation among physical variables. The diﬀerential laws, together with the operators and theorems that are the point of this chapter, are summarized in Sec. 2.8. 2.1 THE DIVERGENCE OPERATOR If Gauss’ integral theorem, (1.3.1), is to be written with the surface integral replaced by a volume integral, then it is necessary that an operator be found such that Z VdivAdv=I SA·da (1) With the objective of ﬁnding this divergence operator ,div, (1) is applied to an incremental volume ∆ V. Because the volume is small, the volume integral on the left can be taken as the product of the integrand and the volume. Thus, the divergence of a vector Ais deﬁned in terms of the limit of a surface integral. divA≡lim ∆V→01 ∆VI SA·da (2) Once evaluated, it is a function of r. That is, in the limit, the volume shrinks to zero in such a way that all points on the surface approach the point r. With this condition satisﬁed, the actual shape of the volume element is arbitrary. In Cartesian coordinates, a convenient incremental volume is a rectangular parallelepiped ∆ x∆y∆zcentered at ( x, y, z ), as shown in Fig. 2.1.1. With the limit where ∆ x∆y∆z→0 in view, the right-hand side of (2) is approximated by 4 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 I SA·da/similarequal∆y∆z£ Ax¡ x+∆x 2, y, z¢ −Ax¡ x−∆x 2, y, z¢¤ + ∆z∆x£ Ay¡ x, y+∆y 2, z¢ −Ay¡ x, y−∆y 2, z¢¤ + ∆x∆y£ Az¡ x, y, z +∆z 2¢ −Az¡ x, y, z −∆z 2¢¤(3) With the above expression used to evaluate (2), along with ∆ V= ∆x∆y∆z, divA= lim ∆x→0" Ax¡ x+∆x 2, y, z¢ −Ax¡ x−∆x 2, y, z¢ ∆x# + lim ∆y→0" Ay¡ x, y+∆y 2, z¢ −Ay¡ x, y−∆y 2, z¢ ∆y# + lim ∆z→0" Az¡ x, y, z +∆z 2¢ −Az¡ x, y, z −∆z 2¢ ∆z#(4) It follows that in Cartesian coordinates, the divergence operator is divA=∂Ax ∂x+∂Ay ∂y+∂Az ∂z (5) This result suggests an alternative notation. The deloperator is deﬁned as ∇ ≡ix∂ ∂x+iy∂ ∂y+iz∂ ∂z (6) so that (5) can be written as divA=∇ ·A (7) Thedivnotation suggests that this combination of derivatives describes the outﬂow ofAfrom the neighborhood of the point of evaluation. The deﬁnition (2) is inde- pendent of the choice of a coordinate system. On the other hand, the delnotation suggests the mechanics of the operation in Cartesian coordinates. We will have it both ways by using the delnotation in writing equations in Cartesian coordinates, but using the name divergence in the text. Problems 2.1.4 and 2.1.6 lead to the divergence operator in cylindrical and spherical coordinates, respectively (summarized in Table I at the end of the text), and provide the opportunity to develop the connection between the general deﬁni- tion, (2), and speciﬁc representations. Sec. 2.2 Gauss’ Integral Theorem 5 Fig. 2.2.1 (a) Three mutually perpendicular slices deﬁne an incremental volume in the volume Vshown in cross-section. (b) Adjacent volume elements with common surface. 2.2 GAUSS’ INTEGRAL THEOREM The operator that is required for (2.1.1) to hold has been identiﬁed by considering an incremental volume element. But does the relation hold for volumes of ﬁnite size? The volume enclosed by the surface Scan be subdivided into diﬀerential elements, as shown in Fig. 2.2.1. Each of the elements has a surface of its own with thei-th being enclosed by the surface Si. We now prove that the surface integral of the vector Aover the surface Sis equal to the sum of the surface integrals over each surface S I SA·da=X i£Z SiA·da¤ (1) Note ﬁrst that the surface normals of two surfaces between adjacent volume el- ements are oppositely directed, while the vector Ahas the same value for both surfaces. Thus, as illustrated in Fig. 2.2.1, the ﬂuxes through surfaces separating two volume elements in the interior of Scancel. The only contributions to the summation in (1) which do not cancel are the ﬂuxes through the surfaces which do not separate one volume element from another, i.e., those surfaces that lie on S. But because these surfaces together form S, (1) follows. Finally, with the right-hand side rewritten, (1) is I SA·da=X i£R SiA·da ∆Vi¤ ∆Vi (2) where ∆ Viis the volume of the i-th element. Because these volume elements are diﬀerential, what is in brackets on the right in (2) can be represented using the deﬁnition of the divergence operator, (2.1.2). I SA·da=X i(∇ ·A)i∆Vi (3) Gauss’ integral theorem follows by replacing the summation over the diﬀerential volume elements by an integration over the volume. 6 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. 2.2.2 Volume between planes x=x1andx=x2having unit area in y−zplanes. I SA·da=Z V∇ ·Adv (4) Example 2.2.1. One-Dimensional Theorem If the vector Ais one-dimensional so that A=f(x)ix (5) what does Gauss’ integral theorem say about an integration over a volume Vbetween the planes x=x1andx=x2and of unit cross-section in any y−zplane between these planes? The volume Vand surface Sare as shown in Fig. 2.2.2. Because Aisxdirected, the only contributions are from the right and left surfaces. These respectively have da=ixdydz andda=−ixdydz. Hence, substitution into (4) gives the familiar form, f(x2)−f(x1) =Zx2 x1∂f ∂xdx (6) which is a reminder of the one-dimensional analogy discussed in the introduction. Gauss’ theorem extends into three dimensions the relationship that exists between the derivative and integral of a function. 2.3 GAUSS’ LAW, MAGNETIC FLUX CONTINUITY, AND CHARGE CONSERVATION Of the ﬁve integral laws summarized in Table 1.8.1, three involve integrations over closed surfaces. By Gauss’ theorem, (2.2.4), each of the surface integrals is now expressed as a volume integral. Because the volume is arbitrary, the integrands must vanish, and so the diﬀerential laws are obtained. Thediﬀerential form of Gauss’ law follows from (1.3.1) in that table. ∇ ·/epsilon1oE=ρ (1) Magnetic ﬂux continuity in diﬀerential form follows from (1.7.1). Sec. 2.4 The Curl Operator 7 ∇ ·µoH= 0 (2) In the integral charge conservation law, (1.5.2), there is a time derivative. Because the geometry of the integral we are considering is ﬁxed, the time derivative can be taken inside the integral. That is, the spatial integration can be carried out after the time derivative has been taken. But because ρis not only a function of t but of ( x, y, z ) as well, the time derivative is taken holding ( x, y, z ) constant. Thus, thediﬀerential charge conservation law is stated using a partial time derivative. ∇ ·J+∂ρ ∂t= 0(3) These three diﬀerential laws are summarized in Table 2.8.1. 2.4 THE CURL OPERATOR If the integral laws of Amp` ere and Faraday, (1.4.1) and (1.6.1), are to be written in terms of one type of integral, it is necessary to have an operator such that the contour integrals are converted to surface integrals. This operator is called the curl. Z ScurlA·da=I CA·ds (1) The operator is identiﬁed by making the surface an incremental one, ∆ a. At the particular point rwhere the operator is to be evaluated, pick a direction nand construct a plane normal to nthrough the point r. In this plane, choose a contour Caround rthat encloses the incremental area ∆ a. It follows from (1) that (curlA)n= lim ∆a→01 ∆aI CA·ds (2) The shape of the contour Cis arbitrary except that all its points are assumed to approach the point runder study in the limit ∆ a→0. Such an arbitrary elemental surface with its unit normal nis illustrated in Fig. 2.4.1a. The deﬁnition of the curl operator given by (2) is independent of the coordinate system. To express (2) in Cartesian coordinates, consider the incremental surface shown in Fig. 2.4.1b. The center of ∆ ais at the location ( x, y, z ), where the oper- ator is to be evaluated. The contour is composed of straight segments at y±∆y/2 andz±∆z/2. To ﬁrst order in ∆ yand ∆ z, it follows that the n=ixcomponent of (2) is (curlA)x= lim ∆y∆z→01 ∆y∆z(· Az¡ x, y+∆y 2, z¢ −Az¡ x, y−∆y 2, z¢¸ ∆z −· Ay¡ x, y, z +∆z 2¢ −Ay¡ x, y, z −∆z 2¢¸ ∆y) (3) 8 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. 2.4.1 (a) Incremental contour for evaluation of the component of the curl in the direction of n. (b) Incremental contour for evaluation of xcompo- nent of curl in Cartesian coordinates. Here the ﬁrst two terms represent integrations along the vertical segments, ﬁrst in the + zdirection and then in the −zdirection. Note that integration on this second leg results in a minus sign, because there, Ais oppositely directed to ds. In the limit, (3) becomes (curlA)x=∂Az ∂y−∂Ay ∂z(4) The same procedure, applied to elemental areas having normals in the yandz directions, result in three “components” for the curloperator. curlA=µ∂Az ∂y−∂Ay ∂z¶ ix+µ∂Ax ∂z−∂Az ∂x¶ iy +µ∂Ay ∂x−∂Ax ∂y¶ iz(5) In fact, we should be able to select the surface for evaluating (2) as having a unit normal nin any arbitrary direction. For (5) to be a vector, its dot product with n must give the same result as obtained for the direct evaluation of (2). This is shown to be true in Appendix 2. The result of cross-multiplying Aby the deloperator, deﬁned by (2.1.6), is the curl operator. This is the reason for the alternate notation for the curl operator. curlA=∇ ×A (6) Thus, in Cartesian coordinates ∇ ×A=¯¯¯¯¯ix iy iz ∂/∂x ∂/∂y ∂/∂z Ax Ay Az¯¯¯¯¯(7) The problems give the opportunity to derive expressions having similar forms in cylindrical and spherical coordinates. The results are summarized in Table I at the end of the text. Sec. 2.5 Stokes’ Integral Theorem 9 Fig. 2.5.1 Arbitrary surface enclosed by contour Cis subdivided into incre- mental elements, each enclosed by a contour having the same sense as C. 2.5 STOKES’ INTEGRAL THEOREM In Sec. 2.4, curlAwas identiﬁed as that vector function which had an integral over a surface Sthat could be reduced to an integral on Aover the enclosing contour C. This was done by applying (2.4.1) to an incremental surface. But does this relation hold for SandCof ﬁnite size and arbitrary shape? The generalization to an arbitrary surface begins by subdividing Sinto dif- ferential area elements, each enclosed by a contour C. As shown in Fig. 2.5.1, each diﬀerential contour coincides in direction with the positive sense of the original contour. We shall now prove that I CA·ds=X iI CiA·ds (1) where the sum is over all contours bounding the surface elements into which the surface Shas been subdivided. Because the segments are followed in opposite senses when evaluated for the adjacent area elements, line integrals along those segments of the contours which separate two adjacent surface elements add to zero in the sum of (1). Only those line integrals remain which pertain to the segments coinciding with the original contour. Hence, (1) is demonstrated. Next, (1) is written in the slightly diﬀerent form. I CA·ds=X i·1 ∆aiI CiA·ds¸ ∆ai (2) We can now appeal to the deﬁnition of the component of the curl in the direction of the normal to the surface element, (2.4.2), and replace the summation by an integration. I CA·ds=Z S(curlA)nda (3) Another way of writing this expression is to take advantage of the vector character of the curland the deﬁnition of a vector area element, da=nda: I CA·ds=Z S∇ ×A·da (4) 10 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 This is Stokes’ integral theorem . If a vector function can be written as the curl of a vector A, then the integral of that function over a surface Scan be reduced to an integral of Aon the enclosing contour C. 2.6 DIFFERENTIAL LAWS OF AMP `ERE AND FARADAY With the help of Stokes’ theorem, Amp` ere’s integral law (1.4.1) can now be stated asZ S∇ ×H·da=Z SJ·da+d dtZ S/epsilon1oE·da (1) That is, by virtue of (2.5.4), the contour integral in (1.4.1) is replaced by a surface integral. The surface Sis ﬁxed in time, so the time derivative in (1) can be taken inside the integral. Because Sis also arbitrary, the integrands in (1) must balance. ∇ ×H=J+∂/epsilon1oE ∂t (2) This is the diﬀerential form of Amp` ere’s law . In the last term, which is called the displacement current density, a partial time derivative is used to make it clear that the location ( x, y, z ) at which the expression is evaluated is held ﬁxed as the time derivative is taken. In Sec. 1.5, it was seen that the integral forms of Amp` ere’s and Gauss’ laws combined to give the integral form of the charge conservation law. Thus, we should expect that the diﬀerential forms of these laws would also combine to give the diﬀerential charge conservation law. To see this, we need the identity ∇·(∇×A) = 0 (Problem 2.4.5). Thus, the divergence of (2) gives 0 =∇ ·J+∂ ∂t(∇ ·/epsilon1oE) (3) Here the time and space derivatives have been interchanged in the last term. By Gauss’ diﬀerential law, (2.3.1), the time derivative is of the charge density, and so (3) becomes the diﬀerential form of charge conservation, (2.3.3). Note that we are taking a diﬀerential view of the interrelation between laws that parallels the integral developments of Sec. 1.5. Finally, Stokes’ theorem converts Faraday’s integral law (1.6.1) to integrations overSonly. It follows that the diﬀerential form of Faraday’s law is ∇ ×E=−∂µoH ∂t (4) The diﬀerential forms of Maxwell’s equations in free space are summarized in Table 2.8.1. Sec. 2.7 Visualization of Fields 11 Fig. 2.7.1 Construction of ﬁeld line. 2.7 VISUALIZATION OF FIELDS AND THE DIVERGENCE AND CURL A three-dimensional vector ﬁeld A(r) is speciﬁed by three components that are, individually, functions of position. It is diﬃcult enough to plot a single scalar func- tion in three dimensions; a plot of three is even more diﬃcult and hence less useful for visualization purposes. Field lines are one way of picturing a ﬁeld distribution. A ﬁeld line through a particular point ris constructed in the following way: At the point r, the vector ﬁeld has a particular direction. Proceed from the point rin the direction of the vector A(r) a diﬀerential distance dr. At the new point r+dr, the vector has a new direction A(r+dr). Proceed a diﬀerential distance dr/primealong this new (diﬀerentially diﬀerent) direction to a new point, and so forth as shown in Fig. 2.7.1. By this process, a ﬁeld line is traced out. The tangent to the ﬁeld line at any one of its points gives the direction of the vector ﬁeld A(r) at that point. The magnitude of A(r) can also be indicated in a somewhat rough way by means of the ﬁeld lines. The convention is used that the number of ﬁeld lines drawn through an area element perpendicular to the ﬁeld line at a point ris proportional to the magnitude of A(r) at that point. The ﬁeld might be represented in three dimensions by wires. If it has no divergence, a ﬁeld is said to be solenoidal . If it has no curl, it is irrotational . It is especially important to conceptualize solenoidal and irrotational ﬁelds. We will discuss the nature of irrotational ﬁelds in the following examples, but become especially in tune with their distributions in Chap. 4. Consider now the “wire-model” picture of the solenoidal ﬁeld. Single out a surface with sides formed of a continuum of adjacent ﬁeld lines, a “hose” of lines as shown in Fig. 2.7.2, with endfaces spanning across the ends of the hose. Then, because a solenoidal ﬁeld can have no net ﬂux out of this tube, the number of ﬁeld lines entering the hose through one endface must be equal to the number of lines leaving the hose through the other end. Because the hose is picked arbitrarily, we conclude that a solenoidal ﬁeld is represented by lines that are continuous; they do not appear or disappear within the region where they are solenoidal. The following examples begin to develop an appreciation for the attributes of the ﬁeld lines associated with the divergence and curl. Example 2.7.1. Fields with Divergence but No Curl (Irrotational but Not Solenoidal) 12 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. 2.7.2 Solenoidal ﬁeld lines form hoses within which the lines neither begin nor end. Fig. 2.7.3 Spherically symmetric ﬁeld that is irrotational. Volume elements VaandVcare used with Gauss’ theorem to show why ﬁeld is solenoidal outside the sphere but has a divergence inside. Surface elements CbandCdare used with Stokes’ theorem to show why ﬁelds are irrotational everywhere. The spherical region r < R supports a charge density ρ=ρor/R. The exterior region is free of charge. In Example 1.3.1, the radially symmetric electric ﬁeld intensity is found from the integral laws to be E=irρo 4/epsilon1o½r2 R;r < R R3 r2;r > R(1) In spherical coordinates, the divergence operator is (from Table I) ∇ ·E=1 r2∂ ∂r(r2Er) +1 rsinθ∂ ∂θ(sinθEθ) +1 rsinθ∂Eφ ∂φ(2) Thus, evaluation of Gauss’ diﬀerential law, (2.3.1), gives /epsilon1o∇ ·E=nρor R;r < R 0; r > R(3) which of course agrees with the charge distribution used in the original derivation. This exercise serves to emphasize that the diﬀerential laws apply point by point throughout the region. The ﬁeld lines can be sketched as in Fig. 2.7.3. The magnitude of the charge density is represented by the density of + (or −) symbols. Sec. 2.7 Visualization of Fields 13 Where in this plot does the ﬁeld have a divergence? Because the charge density has already been pictured, we already know the answer to this question. The ﬁeld has divergence only where there is a charge density. Thus, even though the ﬁeld lines are thinning out with increasing radius in the exterior region, at any given point in this region the ﬁeld has no divergence. The situation in this region is typiﬁed by the ﬂux of Ethrough the “hose” deﬁned by the volume Va. The ﬁeld does indeed decrease with radius, but the cross-sectional area of the hose increases so as to exactly compensate and maintain the net ﬂux constant. In the interior region, a volume element having the shape of a tube with sides parallel to the radial ﬁeld can also be considered, volume Vc. That the ﬁeld is not solenoidal is evident from the fact that its intensity is least over the cross-section of the tube having the least area. That there must be a net outward ﬂux is evidence of the net charge enclosed. Field lines originate inside the volume on the enclosed charges. Are the ﬁeld lines in Fig. 2.7.3 irrotational? In spherical coordinates, the curl is ∇ ×E=ir1 rsinθ· ∂ ∂θ(Eφsinθ)−∂Eθ ∂φ¸ +iθ· 1 rsinθ∂Er ∂φ−1 r∂ ∂r(rEφ)¸ +iφ· 1 r∂ ∂r(rEθ)−1 r∂Er ∂θ¸(4) and it follows from a substitution of (1) that there is no curl, either inside or outside. This result is corroborated by evaluating the circulation of Efor contours enclosing areas ∆ ahaving normals in any one of the coordinate directions. [Remember the deﬁnition of the curl, (2.4.2).] Examples are the contours enclosing the surfaces Sb andSdin Fig. 2.7.3. Contributions to the C/prime/primeandC/prime/prime/primesegments vanish because these are perpendicular to E, while (because Eis independent of φandθ) the contribution from one C/primesegment cancels that from the other. Example 2.7.2. Fields with Curl but No Divergence (Solenoidal but Not Irrotational) A wire having radius Rcarries an axial current density that increases linearly with radius. Amp` ere’s integral law was used in Example 1.4.1 to show that the associated magnetic ﬁeld intensity is H=iφJo 3nr2/R;r < R R2/r;r > R(5) Where does this ﬁeld have curl? The answer follows from Amp` ere’s law, (2.6.2), with the displacement current neglected. The curl is the current density, and hence restricted to the region r < R , where it tends to be concentrated at the periphery. Evaluation of the curl in cylindrical coordinates gives a result consistent with this expectation. ∇ ×H=ir¡1 r∂Hz ∂φ−∂Hφ ∂z¢ +iφ¡∂Hr ∂z−∂Hz ∂r¢ +iz¡1 r∂ ∂r(rHφ)−1 r∂Hr ∂φ¢ =nJor/Riz;r < R 0; r > R(6) 14 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. 2.7.4 Cylindrically symmetric ﬁeld that is solenoidal. Volume elements VaandVcare used with Gauss’ theorem to show why the ﬁeld has no divergence anywhere. Surface elements SbandSdare used with Stokes’ theorem to show that the ﬁeld is irrotational outside the cylinder but does have a curl inside. The current density and magnetic ﬁeld intensity are sketched in Fig. 2.7.4. In accordance with the “wire” representation, the spacing of the ﬁeld lines indicates their intensity. A similar convention applies to the current density. When seen “end- on,” a current density headed out of the paper is indicated by ⊙, while ⊗indicates the vector is headed into the paper. The suggestion is of the vector pictured as an arrow, with the symbols representing its tip and feathers, respectively. Can the azimuthally directed ﬁeld vary with r(a direction perpendicular to φ) and still have no curl in the outer region? The integration of Haround the contour Cbin Fig. 2.7.4 shows why it can. The contours C/prime bare arranged to make ds perpendicular to H, so that H·ds= 0 there. Integrations on the segments C/prime/prime/prime band C/prime/prime bcancel because the diﬀerence in the length of the segments just compensates the decrease in the ﬁeld with radius. In the interior region, a similar integration surely gives a ﬁnite result. On the contour Cd, the ﬁeld is larger on the outside leg where the contour length is larger, so it is clear that the curl must be ﬁnite. Of course, this ﬁeld shape simply reﬂects the presence of the current density. The ﬁeld is solenoidal everywhere. This can be checked by taking the diver- gence of (5) in each of the regions. In cylindrical coordinates, Table I gives ∇ ·H=1 r∂ ∂r(rHr) +1 r∂Hφ ∂φ+∂Hz ∂z(7) The ﬂux tubes deﬁned as incremental volumes VaandVcin Fig. 2.7.4, in the exterior and interior regions, respectively, clearly sustain no net ﬂux through their surfaces. That the ﬁeld lines circulate in tubes without originating or disappearing in certain regions is the hallmark of the solenoidal ﬁeld. It is important to distinguish between ﬁelds “in the large” (in terms of the integral laws written for volumes, surfaces, and contours of ﬁnite size) and “in the small” (in terms of diﬀerential laws). To this end, consider some questions that might be raised. Sec. 2.7 Visualization of Fields 15 Fig. 2.7.5 Volume element with sides tangential to ﬁeld lines is used to interpret divergence from ﬁeld coordinate system. Is it possible for a ﬁeld that has no divergence at each point on a closed surface Sto have a net ﬂux through that surface? Example 2.7.1 illustrates that the answer is yes. At each point on a surface Sthat encloses the charged interior region, the divergence of /epsilon1oEis zero. Yet integration of /epsilon1oE·daover such a surface gives a ﬁnite value, indeed, the net charge enclosed. The divergence can be viewed as a weighted derivative along the direction of the ﬁeld, or along the ﬁeld “hose.” With δadeﬁned as the cross-sectional area of such a tube having sides parallel to the ﬁeld /epsilon1oE, as shown in Fig. 2.7.5, it follows from (2.1.2) that the divergence is ∇ ·A= lim δa→0 δξ→01 δaµA·δa|ξ+∆ξ−A·δa|ξ δξ¶ (8) The minus sign in the second term results because daandδaare negatives on the left surface. Written in this form, the divergence is the derivative of eoE·δawith respect to a coordinate in the direction of E. Examples of such tubes are volumes VaandVcin Fig. 2.7.3. That the divergence is zero in the exterior region of that example is equivalent to having a radial derivative of the displacement ﬂux /epsilon1oE·δa that is zero. A further observation returns to the distinction between ﬁelds as they are described “in the large” by means of the integral laws and as they are represented “in the small” by the diﬀerential laws. Is it possible for a ﬁeld to have a circulation on some contour Cand yet be irrotational at each point on C? Example 2.7.2 shows that the answer is again yes. The exterior magnetic ﬁeld encircles the center current-carrying region. Therefore, it has a circulation on any contour that encloses the center region. Yet at all exterior points, the curlofHis zero. The cross-product of two vectors is perpendicular to both vectors. Is the curl of a vector necessarily perpendicular to that vector? Example 2.7.2 would seem to say yes. There the current density is the curl of Hand is in the zdirection, while His in the azimuthal direction. However, this time the answer is no. By deﬁnition we can add to Hany irrotational ﬁeld without altering the curl. If that irrotational ﬁeld has a component in the direction of the curl, then the curl of the combined ﬁelds is not perpendicular to the combined ﬁelds. 16 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. 2.7.6 Three surfaces, having orthogonal normal vectors, have geometry determined by the ﬁeld hose. Thus, the curl of the ﬁeld is interpreted in terms of a ﬁeld coordinate system. Illustration. A Vector Field Not Perpendicular to Its Curl In the interior of the conductor shown in Fig. 2.7.4, the magnetic ﬁeld intensity and its curl are H=Jo 3r2 Riφ;∇ ×H=J=Jor Riz (9) Suppose that we add to this Ha ﬁeld that is uniform and zdirected. H=Jor2 3Riφ+Hoiz (10) Then the new ﬁeld has a component in the zdirection and yet has the same z- directed curl as given by (9). Note that the new ﬁeld lines are helixes having in- creasingly tighter pitches as the radius is increased. The curl can also be viewed in terms of a ﬁeld hose. The deﬁnition, (2.4.2), is applied to any one of the three contours and associated surfaces shown in Fig. 2.7.6. Contours CξandCηare perpendicular and across the hose while ( Cζ) is around the hose. The former are illustrated by contours CbandCdin Fig. 2.7.4. The component of the curl in the ξdirection is the limit in which the area 2δrδlgoes to zero of the circulation around the contour Cξdivided by that area. The contributions to this line integration from the segments that are perpendicular to the ζaxis are by deﬁnition zero. Thus, for this component of the curl, transverse to the ﬁeld, (2.4.2) becomes (∇ ×H)ξ= lim δl→0 δξ→01 δlµδl·H|η+δη 2−δl·H|η−δη 2 δη¶ (11) The transverse components of the curlcan be regarded as derivatives with respect to transverse directions of the vector ﬁeld weighted by incremental line elements δl. Sec. 2.8 Summary of Maxwell’s Laws 17 At its center, the surface enclosed by the contour Cζhas its normal in the direction of the ﬁeld. It would seem that the curl in the ζdirection would therefore have to be zero. However, the previous discussion and illustration give a warning that the contour integral around Cζis not necessarily zero. Even though, to zero order in the diameter of the hose, the ﬁeld is perpendic- ular to the contour, to higher order it can have components parallel to the contour. This means that if the contour Cζwere actually perpendicular to the ﬁeld at each point, it would not close on itself. An equivalent contour, shown by the inset to Fig. 2.7.6, begins and terminates on the central ﬁeld line. With the exception of the segment in the ζdirection used to close this contour, each segment is now by deﬁnition perpendicular to ζ. The contribution to the circulation around the con- tour now comes from the ζ-directed segment. Remember that the length of this segment is determined by the shape of the ﬁeld lines. Thus, it is proportional to (δr)2, and therefore so also is the circulation. The limit deﬁned by (2.1.2) can result in a ﬁnite value in the ζdirection. The “cross-product” of an operator with a vector has properties that are not identical with the cross-product of two vectors. 2.8 SUMMARY OF MAXWELL’S DIFFERENTIAL LAWS AND INTEGRAL THEOREMS In this chapter, the divergence and curl operators have been introduced. A third, the gradient, is naturally deﬁned where it is put to use, in Chap. 4. A summary of these operators in the three standard coordinate systems is given in Table I at the end of the text. The problems for Secs. 2.1 and 2.4 outline the derivations of the gradient and curl operators in cylindrical and spherical coordinates. The integral theorems of Gauss and Stokes are two of three theorems sum- marized in Table II at the end of the text. Gauss’ theorem states how the volume integral of any scalar that can be represented as the divergence of a vector can be reduced to an integration of the normal component of that vector over the surface enclosing that volume. A volume integration is reduced to a surface integration. Similarly, Stokes’ theorem reduces the surface integration of any vector that can be represented as the curl of another vector to a contour integration of that second vector. A surface integral is reduced to a contour integral. These generally useful theorems are the basis for moving from the integral law point of view of Chap. 1 to a diﬀerential point of view. This transition from a global to a point-wise view of ﬁelds is summarized by the shift from the integral laws of Table 1.8.1 to the diﬀerential laws of Table 2.8.1. The aspects of a vector ﬁeld encapsulated in the divergence and curl can always be recalled by returning to the fundamental deﬁnitions, (2.1.2) and (2.4.2), respectively. The divergence is indeed deﬁned to represent the net outward ﬂux through a closed surface. But keep in mind that the surface is incremental, and that the divergence describes only the neighborhood of a given point. Similarly, the curl represents the circulation around an incremental contour, not around one that is of ﬁnite size. What should be committed to memory from this chapter? The theorems of Gauss and Stokes are the key to relating the integral and diﬀerential forms of Maxwell’s equations. Thus, with these theorems and the integral laws in mind, 18 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 TABLE 2.8.1 MAXWELL’S DIFFERENTIAL LAWS IN FREE SPACE NAME DIFFERENTIAL LAW EQ. NUMBER Gauss’ Law ∇ ·/epsilon1oE=ρ 2.3.1 Amp` ere’s Law ∇ ×H=J+ (∂/epsilon1oE)/(∂t) 2.6.2 Faraday’s Law ∇ ×E=−(∂µoH)/(∂t) 2.6.4 Magnetic Flux Continuity∇ ·µoH= 0 2.3.2 Charge Conservation∇ ·J+∂ρ ∂t= 0 2.3.3 it is easy to remember the diﬀerential laws. Applied to diﬀerential volumes and surfaces, the theorems also provide the deﬁnitions (and hence the signiﬁcances) of the divergence and curl operators independent of the coordinate system. Also, the evaluation in Cartesian coordinates of these operators should be remembered. Sec. 2.1 Problems 19 P R O B L E M S 2.1 The Divergence Operator 2.1.1∗In Cartesian coordinates, A= (Ao/d2)(x2ix+y2iy+z2iz), where Aoand dare constants. Show that divA= 2Ao(x+y+z)/d2. 2.1.2∗In Cartesian coordinates, three vector functions are A= (Ao/d)(yix+xiy) ( a) A= (Ao/d)(xix−yiy) ( b) A=Aoe−ky(coskxix−sinkxiy) ( c) where Ao, k, and dare constants. (a) Show that the divergence of each is zero. (b) Devise three vector functions that have a ﬁnite divergence and eval- uate their divergences. 2.1.3 In cylindrical coordinates, the divergence operator is given in Table I at the end of the text. Evaluate the divergence of the following vector functions. A= (Ao/d)(rcos 2φir−rsin 2φiφ) ( a) A=Ao(cosφir−sinφiφ) ( b) A= (Aor2/d2)ir (c) 2.1.4∗In cylindrical coordinates, unit vectors are as deﬁned in Fig. P2.1.4a. An incremental volume element having sides (∆ r, r∆φ,∆z) is as shown in Fig. P2.1.4b. Determine the divergence operator by evaluating (2), using steps analogous to those leading from (3) to (5). Show that the result is as given in Table I at the end of the text. (Hint: In carrying out the integra- tions over the surface elements in Fig. P2.1.4b having normals ±ir, note that not only is Arevaluated at r=r±1 2∆r, but so also is r. For this reason, it is most convenient to group Arandrtogether in manipulating the contributions from this surface.) 2.1.5 The divergence operator is given in spherical coordinates in Table I at the end of the text. Use that operator to evaluate the divergence of the following vector functions. A= (Ao/d3)r3ir (a) A= (Ao/d2)r2iφ (b) 20 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. P2.1.4 A=Ao(cosθir−sinθiθ) ( c) 2.1.6∗In spherical coordinates, an incremental volume element has sides ∆ r, r∆θ, rsinθ∆φ. Using steps analogous to those leading from (3) to (5), determine the divergence operator by evaluating (2.1.2). Show that the result is as given in Table I at the end of the text. 2.2 Gauss’ Integral Theorem 2.2.1∗Given a well-behaved vector function A, Gauss’ theorem shows that the same result will be obtained by integrating its divergence over a volume V or by integrating its normal component over the surface Sthat encloses that volume. The following steps exemplify this fact. Consider the particular vector function A= (Ao/d)(xix+yiy) and a cubical volume having surfaces in the planes x=±d, y=±d, and z=±d. (a) Show that the area elements on these surfaces are respectively da= ±ixdydz,±iydxdz, and±izdydx. (b) Show that evaluation of the left-hand side of (4) gives I SA·da=Ao d·Zd −dZd −d(d)dydz−Zd −dZd −d(−d)dydz +Zd −dZd −d(d)dxdz−Zd −dZd −d(−d)dxdz¸ = 16 Aod2 (c) Evaluate the divergence of Aand the right-hand side of (4) and show that it gives the same result. 2.2.2 With A= (Ao/d3)(xy2ix+x2yiy), carry out the steps in Prob. 2.2.1. Sec. 2.4 Problems 21 2.3 Diﬀerential Forms of Gauss’ Law, Magnetic Flux Continuity, and Charge Conservation 2.3.1∗For a line charge along the zaxis of Prob. 1.3.1, Ewas written in Cartesian coordinates as (a). (a) Use Gauss’ diﬀerential law in Cartesian coordinates to show that the charge density is indeed zero everywhere except along the zaxis. (b) Obtain the same result by evaluating Gauss’ law using Eas given by (1.3.13) and the divergence operator from Table I in cylindrical coordinates. 2.3.2∗Show that at each point r < a, Eandρas given respectively by (b) and (a) of Prob. 1.3.3 are consistent with Gauss’ diﬀerential law. 2.3.3∗For the ﬂux linkage λfto be independent of S, (2) must hold. Return to Prob. 1.6.6 and check to see that this condition was indeed satisﬁed by the magnetic ﬂux density. 2.3.4∗Using Hexpressed in cylindrical coordinates by (1.4.10), show that the magnetic ﬂux density of a line current is indeed solenoidal (has no diver- gence) everywhere except at r= 0. 2.3.5 Use the diﬀerential law of magnetic ﬂux continuity, (2), to answer Prob. 1.7.2. 2.3.6∗In Prob. 1.3.5, Eandρare found for a one-dimensional conﬁguration using the integral charge conservation law. Show that the diﬀerential form of this law is satisﬁed at each position −1 2s < z <1 2s. 2.3.7 ForJandρas found in Prob. 1.5.1, show that the diﬀerential form of charge conservation, (3), is satisﬁed. 2.4 The Curl Operator 2.4.1∗Show that the curls of the three vector functions given in Prob. 2.1.2 are zero. Devise three such functions that have ﬁnite curls (are rotational) and give their curls. 2.4.2 Vector functions are given in cylindrical coordinates in Prob. 2.1.3. Using the curl operator as given in cylindrical coordinates by Table I at the end of the text, show that all of these functions are irrotational. Devise three functions that are rotational and give their curls. 22 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. P2.4.3 2.4.3∗In cylindrical coordinates, deﬁne incremental surface elements having nor- mals in the r, φandzdirections, respectively, as shown in Fig. P2.4.3. Determine the r, φ, and zcomponents of the curl operator. Show that the result is as given in Table I at the end of the text. (Hint: In integrating in the ±φdirections on the outer and inner incremental contours of Fig. P2.4.3c, note that not only is Aφevaluated at r=r±1 2∆r, respectively, but so also isr. It is therefore convenient to treat Aφras a single function.) 2.4.4 In spherical coordinates, incremental surface elements have normals in the r, θ, and φdirections, respectively, as described in Appendix 1. Determine ther, θ, and φcomponents of the curl operator and compare to the result given in Table I at the end of the text. 2.4.5 The following is an identity. ∇ ·(∇ ×A) = 0 ( a) This can be shown in two ways. (a) Apply Stokes’ theorem to an arbitrary but closed surface S(one hav- ing no edge, so C= 0) and then Gauss’ theorem to argue the identity. (b) Write out the the divergence of the curl in Cartesian coordinates and show that it is indeed identically zero. 2.5 Stokes’ Integral Theorem 2.5.1∗To exemplify Stokes’ integral theorem, consider the evaluation of (4) for the vector function A= (Ao/d2)x2iyand a rectangular contour consisting of the segments at x=g+ ∆, y=h, x =g, and y= 0. The direction of the contour is such that da=izdxdy. Sec. 2.7 Problems 23 (a) Show that the left-hand side of (4) is hAo[(g+ ∆)2−g2]d2. (b) Verify (4) by obtaining the same result integrating curlAover the area enclosed by C. 2.5.2 For the vector function A= (Ao/d)(−ixy+iyx), evaluate the contour and surface integrals of (4) on CandSas prescribed in Prob. 2.5.1 and show that they are equal. 2.6 Diﬀerential Laws of Amp` ere and Faraday 2.6.1∗In Prob. 1.4.2, His given in Cartesian coordinates by (c). With ∂/epsilon1oE/∂t= 0, show that Amp` ere’s diﬀerential law is satisﬁed at each point r < a . 2.6.2∗For the HandJgiven in Prob. 1.4.1, show that Amp` ere’s diﬀerential law, (2), is satisﬁed with ∂/epsilon1oE/∂t= 0. 2.7 Visualization of Fields and the Divergence and Curl 2.7.1 Using the conventions exempliﬁed in Fig. 2.7.3, (a) Sketch the distributions of charge density ρand electric ﬁeld intensity Efor Prob. 1.3.5 and with Eo= 0 and σo= 0. (b) Verify that Eis irrotational. (c) From observation of the ﬁeld sketch, why would you suspect that E is indeed irrotational? 2.7.2 Using Fig. 2.7.4 as a model, sketch JandH (a) For Prob. 1.4.1. (b) For Prob. 1.4.4. (c) Verify that in each case, His solenoidal. (d) From observation of these ﬁeld sketches, why would you suspect that His indeed solenoidal? 2.7.3 Three two-dimensional vector ﬁelds are shown in Fig. P2.7.3. (a) Which of these is irrotational? (b) Which are solenoidal? 2.7.4 For the ﬁelds of Prob. 1.6.7, sketch Ejust above and just below the plane y= 0 and σsin the surface y= 0. Assume that E1=E2=σo//epsilon1o>0 and adhere to the convention that the ﬁeld intensity is represented by the spacing of the ﬁeld lines. 24 Maxwell’s Diﬀerential Laws In Free Space Chapter 2 Fig. P2.7.3 2.7.5 For the ﬁelds of Prob. 1.7.3, sketch Hjust above and just below the plane y= 0 and Kin the surface y= 0. Assume that H1=H2=Ko>0 and represent the intensity of Hby the spacing of the ﬁeld lines. 2.7.6 Field lines in the vicinity of the surface y= 0 are shown in Fig. P2.7.6. (a) If the ﬁeld lines represent E, there is a surface charge density σson the surface. Is σspositive or negative? (b) If the ﬁeld lines represent H, there is a surface current density K= Kzizon the surface. Is Kzpositive or negative? Fig. P2.7.6 3 INTRODUCTION TO ELECTROQUASISTATICS AND MAGNETOQUASISTATICS 3.0 INTRODUCTION The laws represented by Maxwell’s equations are remarkably general. Nevertheless, they are deceptively simple. In diﬀerential form they are ∇ ×E=−∂µoH ∂t(1) ∇ ×H=J+∂/epsilon1oE ∂t(2) ∇ ·/epsilon1oE=ρ (3) ∇ ·µoH= 0 (4) The sources of the electric and magnetic ﬁeld intensities, EandH, are the charge and current densities, ρandJ. If, at an initial instant, electric and magnetic ﬁelds are speciﬁed throughout all of a source-free space, then Maxwell’s equations in their diﬀerential form predict these ﬁelds as they subsequently evolve in space and time. Proof of this assertion is our starting point in Sec. 3.1. This makes it natural to attribute a physical signiﬁ- cance to the ﬁelds in their own right. Fields can exist in regions far removed from their sources because they can propagate as electromagnetic waves. An introduc- tion to such waves is given in Sec. 3.2. It is shown that the coupling between Eand Hproduced by the magnetic induction in Faraday’s law, the term on the right in (1) and the displacement current density in Amp` ere’s law, the time derivative term on the right in (2), gives rise to electromagnetic waves. Even though ﬁelds can propagate without sources, where they are initiated or detected they must be related to their sources or sinks. To do this, the Lorentz force law must be brought into play. In Sec. 3.1, this law is used to complete Newton’s law 1 2 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 and describe the evolution of a charge distribution. Generally, the Lorentz force law does not act so directly as it does in this example; nevertheless, it usually underlies a constitutive law for conduction that is added to Maxwell’s equations to relate the ﬁelds to the sources. The most commonly used constitutive law is Ohm’s law, which is not introduced until Chap. 7. However, in the intervening chapters we will often model electrodes and wires as being perfectly conducting in the sense that Lorentz’s law is responsible for making the charges move in just such a way that there is eﬀectively no electric ﬁeld intensity in the material. Maxwell’s equations describe the most intricate electromagnetic wave phe- nomena. Of course, the analysis of such ﬁelds is diﬃcult and not always necessary. Wave phenomena occur on short time scales or at high frequencies that are often of no practical concern. If this is the case, the ﬁelds may be described by truncated versions of Maxwell’s equations applied to relatively long time scales and low fre- quencies (quasistatics). The objective in Sec. 3.3 is to identify the two quasistatic approximations and rank the laws in order of importance in these approximations. In Sec. 3.4, we ﬁnd what turns out to be one typical condition that must be satisﬁed if either of these quasistatic approximations is to be justiﬁed. Thus, we will ﬁnd that a system composed of perfect conductors and free space is either electroquasistatic (EQS) or magnetoquasistatic (MQS) if an electromagnetic wave can propagate through a typical dimension of the system in a time that is shorter than times of interest. If fulﬁllment of the same condition justiﬁes either the EQS or MQS approxi- mation, how do we know which to use? We begin to form insights in this regard in Sec. 3.4. A formal justiﬁcation of the quasistatic approximations would be based on what might be termed a time-rate expansion. As time rates of change are increased, more terms are required in a series having its ﬁrst term predicted by the appropriate quasistatic laws. In Sec. 3.4, a speciﬁc example is used to illustrate this expansion and the error committed by omission of the higher-order terms. Whether they be electromagnetic, or perhaps thermal or mechanical, dynam- ical systems that proceed from one state to another as though they are static are commonly said to be quasistatic in their behavior. In this text, the quasistatic ﬁelds are indeed related to their sources as if they were truly static. That is, given the charge or current distribution, EorHare determined without regard for the dy- namics of electromagnetism. However, other dynamical processes can play a role in determining the source distributions. In the systems we are prepared to consider in this chapter, composed of free space and perfect conductors, the quasistatic source distributions within a given quasistatic subregion do not depend on time rates of change. Thus, for now, we will ﬁnd that geometry and spatial and temporal scales alone determine whether a subregion is magnetoquasistatic or electroquasistatic. Illustrated in Sec. 3.5 is the interconnection of such subsystems. In a way that is familiar from circuit theory, the resulting model for the total system has apportionments of sources in the subregions (charges in the EQS regions and currents in the MQS regions) that do depend on the time rates of change. After we have considered eﬀects of ﬁnite conductivity in Chaps. 7 and 10, it will be clear that there are many other situations where quasistatic models represent dynamical processes. Again, Sec. 3.6 provides an overview, this time not of the laws but rather of the parts of the physical world to which they pertain. The discussion is qualitative Sec. 3.1 Temporal Evolution of World 3 and the section is for “feet on the table” reading. Finally, Sec. 3.7 summarizes the electroquasistatic and magnetoquasistatic ﬁeld laws that, respectively, are the themes of Chaps. 4–7 and 8–10. We return to the subject of quasistatic approximations in Chap. 12, where electromagnetic waves are again considered. In Chap. 15 we will come to recog- nize that the concept of quasistatics promulgated in Chaps. 7 and 10 (where loss phenomena are considered) has made the classiﬁcation into electroquasistatic and magnetoquasistatic regions depend not only on geometry and spatial and temporal scales, but on material properties as well. 3.1 TEMPORAL EVOLUTION OF WORLD GOVERNED BY LAWS OF MAXWELL, LORENTZ, AND NEWTON If certain initial conditions are given, Maxwell’s equations, along with the Lorentz law and Newton’s law, describe the time evolution of EandH. This can be argued by expressing Maxwell’s equations, (1)–(4), with the time derivatives and charge density on the left. ∂H ∂t=−1 µo(∇ ×E) (1) ∂E ∂t=1 /epsilon1o(∇ ×H−J) (2) ρ=∇ ·/epsilon1oE (3) 0 =∇ ·µoH (4) The region of interest is vacuum, where particles having a mass mand charge qare subject only to the Lorentz force. Thus, Newton’s law (here used in its non- relativistic form), also written with the time derivative (of the particle velocity) on the left, links the charge distribution to the ﬁelds. mdv dt=q(E+v×µoH) (5) The Lorentz force on the right is given by (1.1.1). Suppose that at a particular instant, t=to, we are given the ﬁelds throughout the entire space of interest, E(r, to) and H(r, to). Suppose we are also given the velocity v(r, to) of all the charges when t=to. It follows from Gauss’ law, (3), that at this same instant, the distribution of charge density is known. ρ(r, to) =∇ ·/epsilon1oE(r, to) (6) Then the current density at the time t=tofollows as J(r, to) =ρ(r, to)v(r, to) (7) So that (4) is satisﬁed when t=to, we must require that the given distribution of Hbe solenoidal. 4 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 The curl operation involves only spatial derivatives, so the right-hand sides of the remaining laws, (1), (2), and (5), can now be evaluated. Thus, the time rates of change of the quantities, E,H, and v, given when t=to, are now known. This allows evaluation of these quantities an instant later, when t=to+∆t. For example, at this later time, E=E(r, to) + ∆ t∂E ∂t¯¯ (r,to)(8) Thus, when t=to+ ∆twe have the same three vector functions throughout all space we started with. This process can be repeated iteratively to determine the distributions at an arbitrary later time. Note that if the initial distribution of H is solenoidal, as required by (4), all subsequent distributions will be solenoidal as well. This follows by taking the divergence of Faraday’s law, (1), and noting that the divergence of the curl is zero. The left-hand side of (5) is written as a total derivative because it is required to represent the time derivative as measured by an observer moving with a given particle. The preceding argument shows that in free space, for given initial E,H, andv, the Lorentz law (here used with Newton’s law) and Maxwell’s equations determine the charge distributions and the associated ﬁelds for all later time. In this sense, Maxwell’s equations and the Lorentz law may be said to provide a complete description of electrodynamic interactions in free space. Commonly, more than one species of charge is involved and the charged particles respond to the ﬁeld in a manner more complex than simply represented by the laws of Newton and Lorentz. In that case, the role played by (5) is taken by a conduction constitutive law which nevertheless reﬂects the Lorentz force law. Another interesting property of Maxwell’s equations emerges from the preced- ing discussion. The electric and magnetic ﬁelds are coupled. The temporal evolution ofEis determined in part by the curlofH, (2), and, similarly, it is the curlofE that determines how fast His changing in time, (1). Example 3.1.1. Evolution of an Electromagnetic Wave The interplay of the magnetic induction and the electric displacement current is illustrated by considering ﬁelds that evolve in Cartesian coordinates from the initial distributions E=Eoixe−z2/2a2(9) H=p /epsilon1o/µoEoiye−z2/2a2(10) In this example, we let to= 0, so these are the ﬁelds when t= 0. Shown in Fig. 3.1.1, these ﬁelds are transverse, in that they have a direction perpendicular to the coordi- nate upon which they depend. Thus, they are both solenoidal, and Gauss’ law makes it clear that the physical situation we consider does not involve a charge density. It follows from (7) that the current density is also zero. With the initial ﬁelds given and J= 0, the right-hand sides of (1) and (2) can be evaluated to give the rates of change of HandE. µo∂H ∂t=−∇ × E=−iy∂Ex ∂z=−iyEod dze−z2/2a2(11) Sec. 3.1 Temporal Evolution of World 5 Fig. 3.1.1 A schematic representation of the EandHﬁelds of Exam- ple 3.2.1. The distributions move to the right with the speed of light, c. /epsilon1o∂E ∂t=∇ ×H=−ixp /epsilon1o/µoEod dze−z2/2a2(12) It follows from (11), Faraday’s law, that when t= ∆t, H=iyp /epsilon1o/µoEo¡ e−z2/2a2−c∆td dze−z2/2a2¢ (13) where c= 1/√/epsilon1oµo, and from (12), Amp` ere’s law, that the electric ﬁeld is E=Eoix¡ e−z2/2a2−c∆td dze−z2/2a2¢ (14) When t= ∆t, theEandHﬁelds are equal to the original Gaussian distribution minus c∆ttimes the spatial derivatives of these Gaussians. But these represent the original Gaussians shifted by c∆tin the + zdirection. Indeed, witness the relation applicable to any function f(z). f(z−∆z) =f(z)−∆zd f dz. (15) On the left, f(z−∆z) is the function f(z) shifted by ∆ z. The Taylor expansion on the right takes the same form as the ﬁelds when t= ∆t, (13) and (14). Thus, within ∆ t, theEandHﬁeld distributions have shifted by c∆tin the + zdirection. Iteration of this process shows that the ﬁeld distributions shown in Fig. 3.1.1 travel in the + zdirection without change of shape at the speed c, the speed of light. c=1√/epsilon1oµo∼=3×108m/sec (16) Note that the derivation would not have changed if we had substituted for the initial Gaussian functions any other continuous functions f(z). In retrospect, it should be recognized that the initial conditions were premed- itated so that they would result in a single wave propagating in the + zdirection. Also, the method of solution was really not numerical. If we were interested in pursu- ing the numerical approach, care would have to be taken to avoid the accumulation of errors. 6 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 The above example illustrated that the electromagnetic wave is caused by the interplay of the magnetic induction and the displacement current, the terms on the left in (1) and (2). Through Faraday’s law, (1), the curl of an initial Eimplies that an instant later, the initial His altered. Similarly, Amp` ere’s law requires that the curl of an initial Hleads to a change in E. In turn, the curls of the altered Eand Himply further changes in HandE, respectively. There are two main points in this section. First, Maxwell’s equations, aug- mented by laws describing the interaction of the ﬁelds with the sources, are suﬃcient to describe the evolution of electromagnetic ﬁelds. Second, in regions well removed from materials, electromagnetic ﬁelds evolve as electromagnetic waves. Typically, the time required for ﬁelds to propagate from one region to another, say over a distance L, is τem=L c(17) where cis the velocity of light. The origin of these waves is the coupling between the laws of Faraday and Amp` ere aﬀorded by the magnetic induction and the dis- placement current. If either one or the other of these terms is neglected, so too is any electromagnetic wave eﬀect. 3.2 QUASISTATIC LAWS The quasistatic laws are obtained from Maxwell’s equations by neglecting either the magnetic induction or the electric displacement current. ELECTROQUASTATIC MAGNETOQUASISTATIC ∇ ×E=−∂µoH ∂t/similarequal0 (1 a) ∇ ×E=−∂µoH ∂t(1b) ∇ ×H=∂/epsilon1oE ∂t+J (2a) ∇ ×H=∂/epsilon1oE ∂t+J/similarequalJ (2b) ∇ ·/epsilon1oE=ρ (3a) ∇ ·/epsilon1oE=ρ (3b) ∇ ·µoH= 0 (4 a) ∇ ·µoH= 0 (4 b) Sec. 3.2 Quasistatic Laws 7 The electromagnetic waves that result from the coupling of the magnetic in- duction and the displacement current are therefore neglected in either set of qua- sistatic laws. Before considering order of magnitude arguments in support of these approximate laws, we recognize their diﬀering orders of importance. In Chaps. 4 and 8 it will be shown that if the curl and divergence of a vector are speciﬁed, then that vector is determined. In the EQS approximation, (1a) re- quires that Eis essentially irrotational. It then follows from (3a) that if the charge density is given, both the curl and divergence of Eare speciﬁed. Thus, Gauss’ law and the EQS form of Fara- day’s law come ﬁrst.In the MQS approximation, the dis- placement current is negligible in (2b), while (4b) requires that His solenoidal. Thus, if the current den- sity is given, both the curl and di- vergence of Hare known. Thus, the MQS form of Amp` ere’s law and the ﬂux continuity condition come ﬁrst. ∇ ·/epsilon1oE=ρ (5a) ∇ ×H=J;∇ ·J= 0 (5 b−c) ∇ ×E= 0 (6 a) ∇ ·µoH= 0 (6 b) Implied by the approximate form of Amp` ere’s law is the continuity condition of J, given also by (5b). In these relations, there are no time derivatives. This does not mean that the sources, and hence the ﬁelds, are not functions of time. But given the sources at a certain instant, the ﬁelds at that same instant are determined without regard for what the sources of ﬁelds were an instant earlier. Figuratively, a snapshot of the source distribution determines the ﬁeld distribution at the same instant in time. Generally, the sources of the ﬁelds are not known. Rather, because of the Lorentz force law, which acts to set charges into motion, they are determined by the ﬁelds themselves. It is for this reason that time rates of change come into play. We now bring in the equation retaining a time derivative. Because His often not crucial to the EQS motion of charges, it is elimi- nated from the picture by taking the divergence of (2a).Faraday’s law makes it clear that a time varying Himplies an induced electric ﬁeld. ∇ ·J+∂ρ ∂t= 0 (7 a) ∇ ×E=−∂µoH ∂t(7b) 8 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 In the EQS approximation, His usu- ally a “leftover” quantity. In any case, onceEandJare determined, Hcan be found by solving (2a) and (4a).In the MQS approximation, the charge density is a “leftover” quantity, which can be found by applying Gauss’ law, (3b), to the previously deter- mined electric ﬁeld intensity. ∇ ×H=∂/epsilon1oE ∂t+J (8a)∇ ·/epsilon1oE=ρ (8b) ∇ ·µoH= 0 (9 a) In the EQS approximation, it is clear that with EandJdetermined from the “zero order” laws (5a)–(7a), the curl and divergence of Hare known [(8a) and (9a)]. Thus, Hcan be found in an “after the fact” way. Perhaps not so obvious is the fact that in the MQS approximation, the divergence and curl of Eare also determined without regard for ρ. The curl of Efollows from Faraday’s law, (7b), while the divergence is often speciﬁed by combining a conduction constitutive law with the continuity condition on J, (5b). The diﬀerential quasistatic laws are summarized in Table 3.6.1 at the end of the chapter. Because there is a direct correspondence between terms in the diﬀer- ential and integral laws, the quasistatic integral laws are as summarized in Table 3.6.2. The conditions under which these quasistatic approximations are valid are examined in the next section. 3.3 CONDITIONS FOR FIELDS TO BE QUASISTATIC An appreciation for the quasistatic approximations will come with a consideration of many case studies. Justiﬁcation of one or the other of the approximations hinges on using the quasistatic ﬁelds to estimate the “error” ﬁelds, which are then hopefully found to be small compared to the original quasistatic ﬁelds. In developing any mathematical “theory” for the description of some part of the physical world, approximations are made. Conclusions based on this “theory” should indeed be made with a concern for implicit approximations made out of ignorance or through oversight. But in making quasistatic approximations, we are fortunate in having available the “exact” laws. These can always be used to test the validity of a tentative approximation. Provided that the system of interest has dimensions that are all within a factor of two or so of each other, order of magnitude arguments easily illustrate how the error ﬁelds are related to the quasistatic ﬁelds. The examples shown in Fig. 3.3.1 are not to be considered in detail, but rather should be regarded as prototypes. The candidate for the EQS approximation in part (a) consists of metal spheres that are insulated from each other and driven by a source of EMF. In the case of part (b), which is proposed for the MQS approximation, a current source drives a current around a one-turn loop. The dimensions are “on the same order” if the diameter of one of the spheres, is within a factor of two or so of the spacing between spheres Sec. 3.3 Conditions for Quasistatics 9 Fig. 3.3.1 Prototype systems involving one typical length. (a) EQS system in which source of EMF drives a pair of perfectly conducting spheres having radius and spacing on the order of L. (b) MQS system consisting of perfectly conducting loop driven by current source. The radius of the loop and diameter of its cross-section are on the order of L. and if the diameter of the conductor forming the loop is within a similar factor of the diameter of the loop. If the system is pictured as made up of “perfect conductors” and “perfect insulators,” the decision as to whether a quasistatic ﬁeld ought to be classiﬁed as EQS or MQS can be made by a simple rule of thumb: Lower the time rate of change (frequency) of the driving source so that the ﬁelds become static. If the magnetic ﬁeld vanishes in this limit, then the ﬁeld is EQS; if the electric ﬁeld vanishes the ﬁeld is MQS. In reality, materials are not “perfect,” neither perfect conductors nor perfect insulators. Therefore, the usefulness of this rule depends on understanding under what circumstances materials tend to behave as “perfect” conductors, and insulators. Fortunately, nature provides us with metals that are extremely good conductors– and with gases, liquids, and solids that are very good insulators– so that this rule is a good intuitive starting point. Chapters 7, 10, and 15 will provide a more mature view of how to classify quasistatic systems. The quasistatic laws are now used in the order summarized by (3.2.5)-(3.2.9) to estimate the ﬁeld magnitudes. With only one typical length scale L, we can approximate spatial derivatives that make up the curl and divergence operators by 1/L. ELECTROQUASISTATIC MAGNETOQUASISTATIC Thus, it follows from Gauss’ law, (3.2.5a), that typical values of Eandρare re- lated byThus, it follows from Amp` ere’s law, (3.2.5b), that typical values of H andJare related by /epsilon1oE L=ρ⇒E=ρL /epsilon1o(1a)H L=J⇒H=JL (1b) As suggested by the integral forms of the laws so far used, these ﬁelds and their sources are sketched in Fig. 3.3.1. The EQS laws will predict Elines that originate on the positive charges on one electrode and terminate on the negative charges on the other. The MQS laws will predict lines of Hthat close around the circulating current. If the excitation were sinusoidal in time, the characteristic time τfor the 10 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 sinusoidal steady state response would be the reciprocal of the angular frequency ω. In any case, if the excitations are time varying, with a characteristic time τ, then the time varying charge implies a cur- rent, and this in turn induces an H. We could compute the current in the conductors from charge conservation, (3.2.7a), but because we are interested in the induced H, we use Amp` ere’s law, (3.2.8a), evaluated in the free space region. The electric ﬁeld is replaced in favor of the charge density in this ex- pression using (1a).the time-varying current implies an Hthat is time-varying. In accor- dance with Faraday’s law, (3.2.7b), the result is an induced E. The mag- netic ﬁeld intensity is replaced by J in this expression by making use of (1b). H L=/epsilon1oE τ⇒ H=/epsilon1oEL τ=L2ρ τ(2a)E L=µoH τ⇒ E=µoHL τ=µoJL2 τ(2b) What errors are committed by ignoring the magnetic induction and displace- ment current terms in the respective EQS and MQS laws? The electric ﬁeld induced by the qua- sistatic magnetic ﬁeld is estimated by using the Hﬁeld from (2a) to esti- mate the contribution of the induc- tion term in Faraday’s law. That is, the term originally neglected in (3.2.1a) is now estimated, and from this a curl of an error ﬁeld evaluated.The magnetic ﬁeld induced by the displacement current represents an error ﬁeld. It can be estimated from Amp` ere’s law, by using (2b) to eval- uate the displacement current that was originally neglected in (3.2.2b). Eerror L=µoρL2 τ2⇒ Eerror=µoρL3 τ2(3a)Herror L=/epsilon1oµoJL2 τ2⇒ Herror=/epsilon1oµoJL3 τ2(3b) Sec. 3.3 Conditions for Quasistatics 11 It follows from this expression and (1a) that the ratio of the error ﬁeld to the quasistatic ﬁeld isIt then follows from this and (1b) that the ratio of the error ﬁeld to the quasistatic ﬁeld is Eerror E=µo/epsilon1oL2 τ2(4a)Herror H=/epsilon1oµoL2 τ2(4b) For the approximations to be justiﬁed, these error ﬁelds must be small com- pared to the quasistatic ﬁelds. Note that whether (4a) is used to represent the EQS system or (4b) is used for the MQS system, the conditions on the spatial scale L and time τ(perhaps the reciprocal frequency) are the same. Both the EQS and MQS approximations are predicated on having suﬃciently slow time variations (low frequencies) and suﬃciently small dimensions so that µo/epsilon1oL2 τ2/lessmuch1⇒L c/lessmuchτ (5) where c= 1/√/epsilon1oµo. The ratio L/cis the time required for an electromagnetic wave to propagate at the velocity cover a length Lcharacterizing the system. Thus, either of the quasistatic approximations is valid if an electromagnetic wave can propagate a characteristic length of the system in a time that is short compared to times τof interest. If the conditions that must be fulﬁlled in order to justify the quasistatic ap- proximations are the same, how do we know which approximation to use? For systems modeled by free space and perfect conductors, such as we have considered here, the answer comes from considering the ﬁelds that are retained in the static limit (inﬁnite τor zero frequency ω). Recapitulating the rule expressed earlier, consider the pair of spheres shown in Fig. 3.3.1a. Excited by a constant source of EMF, they are charged, and the charges give rise to an electric ﬁeld. But in this static limit, there is no current and hence no magnetic ﬁeld. Thus, the static system is dominated by the electric ﬁeld, and it is natural to represent it as being EQS even if the excitation is time-varying. Excited by a dc source, the circulating current in Fig. 3.3.1b gives rise to a magnetic ﬁeld, but there are no charges with attendant electric ﬁelds. This time it is natural to use the MQS approximation when the excitation is time varying. Example 3.3.1. Estimate of Error Introduced by Electroquasistatic Approximation Consider a simple structure fed by a set of idealized sources of EMF as shown in Fig. 3.3.2. Two circular metal disks, of radius b, are spaced a distance dapart. A distribution of EMF generators is connected between the rims of the plates so that the complete system, plates and sources, is cylindrically symmetric. With the understanding that in subsequent chapters we will be examining the underlying physical processes, for now we assume that, because the plates are highly conducting, Emust be perpendicular to their surfaces. The electroquasistatic ﬁeld laws are represented by (3.2.5a) and (3.2.6a). A simple solution for the electric ﬁeld between the plates is E=E diz≡Eoiz (6) 12 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 Fig. 3.3.2 Plane parallel electrodes having no resistance, driven at their outer edges by a distribution of sources of EMF. Fig. 3.3.3 Parallel plates of Fig. 3.3.2, showing volume containing lower plate and radial surface current density at its periphery. where the sign deﬁnition of the EMF, E, is as indicated in Fig. 3.3.2. The ﬁeld of (6) satisﬁes (3.2.5a) and (3.2.6a) in the region between the plates because it is both irrotational and solenoidal (no charge is assumed to exist in the region between the plates). Further, the ﬁeld has no component tangential to the plates which is consistent with the assumption of plates with no resistance. Finally, Gauss’ jump condition, (1.3.17), can be used to ﬁnd the surface charges on the top and bottom plates. Because the ﬁelds above the upper plate and below the lower plate are assumed to be zero, the surface charge densities on the bottom of the top plate and on the top of the bottom plate are σs=n−/epsilon1oEz(z=d) =−/epsilon1oEo;z=d /epsilon1oEz(z= 0) = /epsilon1oEo; z= 0(7) There remains the question of how the electric ﬁeld in the neighborhood of the distributed source of EMF is constrained. We assume here that these sources are connected in such a way that they make the ﬁeld uniform right out to the outer edges of the plates. Thus, it is consistent to have a ﬁeld that is uniform throughout the entire region between the plates. Note that the surface charge density on the plates is also uniform out to r=b. At this point, (3.2.5a) and (3.2.6a) are satisﬁed between and on the plates. In the EQS order of laws, conservation of charge comes next. Rather than using the diﬀerential form, (3.2.7a), we use the integral form, (1.5.2). The volume Vis a cylinder of circular cross-section enclosing the lower plate, as shown in Fig. 3.3.3. Be- cause the radial surface current density in the plate is independent of φ, integration ofJ·daon the enclosing surface amounts to multiplying Krby the circumference, while the integration over the volume is carried out by multiplying σsby the surface area, because the surface charge density is uniform. Thus, Kr2πb+πb2/epsilon1odEo dt= 0⇒Kr¯¯ r=b=−b/epsilon1o 2dEo dt(8) In order to ﬁnd the magnetic ﬁeld, we make use of the “secondary” EQS laws, (3.2.8a) and (3.2.9a). Amp` ere’s law in integral form, (1.4.1), is convenient for the present case of high symmetry. The displacement current is zdirected, so the Sec. 3.3 Conditions for Quasistatics 13 Fig. 3.3.4 Cross-section of system shown in Fig. 3.3.2 showing surface and contour used in evaluating correction Eﬁeld. surface Sis taken as being in the free space region between the plates and having a z-directed normal. I CH·ds=Z S∂/epsilon1oE ∂t·izda (9) The symmetry of structure and source suggests that Hmust be φindependent. A centered circular contour of radius r, as in Fig. 3.3.2, with zin the range 0 < z < d , gives Hφ2πr=/epsilon1odEo dtπr2⇒Hφ=r 2/epsilon1odEo dt(10) Thus, for this speciﬁc conﬁguration, we are at a point in the analysis represented by (2a) in the order of magnitude arguments. Consider now “higher order” ﬁelds and speciﬁcally the error committed by neglecting the magnetic induction in the EQS approximation. The correct statement of Faraday’s law is (3.2.1a), with the magnetic induction retained. Now that the quasistatic Hhas been determined, we are in a position to compute the curl of E that it generates. Again, for this highly symmetric conﬁguration, it is best to use the integral law. Because Hisφdirected, the surface is chosen to have its normal in the φ direction, as shown in Fig. 3.3.4. Thus, Faraday’s integral law (1.6.1) becomes I CE·ds=−Z S∂µoHφ ∂tiφ·da (11) We use the contour shown in Fig. 3.3.4 and assume that the Einduced by the magnetic induction is independent of z. Because the tangential Eﬁeld is zero on the plates, the only contributions to the line integral on the left in (11) come from the vertical legs of the contour. The surface integral on the right is evaluated using (10). [Ez(b)−Ez(r)]d=µo/epsilon1od 2Zb rr/primedr/primed2Eo dt2 =µo/epsilon1od 4(b2−r2)d2Eo dt2(12) The ﬁeld at the outer edge is constrained by the EMF sources to be Eo, and so it follows from (12) that to this order of approximation the electric ﬁeld is Ez=Eo+/epsilon1oµo 4d2Eo dt2(r2−b2) (13) We have found that the electric ﬁeld at r/negationslash=bdiﬀers from the ﬁeld at the edge. How big is the diﬀerence? This depends on the time rate of variation of the electric ﬁeld. 14 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 For purposes of illustration, assume that the electric ﬁeld is sinusoidally varying with time. Eo(t) =Acosωt (14) Thus, the time characterizing the dynamics is 1 /ω. Introducing this expression into (13), and calling the second term the “error ﬁeld,” the ratio of the error ﬁeld and the ﬁeld at the rim, where r=b, is |Eerror| Eo=1 4ω2/epsilon1oµo(b2−r2) (15) The error ﬁeld will be negligible compared to the quasistatic ﬁeld if ω2/epsilon1oµob2 4/lessmuch1 (16) for all rbetween the plates. In terms of the free space wavelength λ, deﬁned as the distance an electromagnetic wave propagates at the velocity c= 1/√/epsilon1oµoin one cycle 2 π/ω λ c=2π ω: c≡1/√µo/epsilon1o (17) (16) becomes b2/lessmuch(λ/π)2(18) In free space and at a frequency of 1 MHz, the wavelength is 300 meters. Hence, if we build a circular disk capacitor and excite it at a frequency of 1 MHz, then the quasistatic laws will give a good approximation to the actual ﬁeld as long as the radius of the disk is much less than 300 meters. The correction ﬁeld for a MQS system is found by following steps that are analogous to those used in the previous example. Once the magnetic and electric ﬁelds have been determined using the MQS laws, the error magnetic ﬁeld induced by the displacement current can be found. 3.4 QUASISTATIC SYSTEMS1 Whether we ignore the magnetic induction and use the EQS approximation, or neglect the displacement current and make a MQS approximation, times of interest τmust be long compared to the time τemrequired for an electromagnetic wave to propagate at the velocity cover the largest length Lof the system. τem=L c/lessmuchτ (1) 1This section makes use of the integral laws at a level somewhat more advanced than neces- sary in preparation for the next chapter. It can be skipped without loss of continuity. Sec. 3.4 Quasistatic Systems 15 Fig. 3.4.1 Range of characteristic times over which quasistatic approxima- tion is valid. The transit time of an electromagnetic wave is τemwhile τ?is a time characterizing the dynamics of the quasistatic system. Fig. 3.4.2 (a) Quasistatic system showing (b) its EQS subsystem and (c) its MQS subsystem. This requirement is given a graphic representation in Fig. 3.4.1. For a given characteristic time (for example, a given reciprocal frequency), it is clear from (1) that the region described by the quasistatic laws is limited in size. Systems can often be divided into subregions that are small enough to be quasistatic but, by virtue of being interconnected through their boundaries, are dynamic in their behavior. With the elements regarded as the subregions, electric circuits are an example. In the physical world of perfect conductors and free space (to which we are presently limited), it is the topology of the conductors that determines whether these subregions are EQS or MQS. A system that is described by quasistatic laws but retains a dynamical be- havior exhibits one or more characteristic times. On the characteristic time axis in Fig. 3.4.1, τ?is one such time. The quasistatic system model provides a meaningful description provided that the one or more characteristic times τ?are long compared toτem. The following example illustrates this concept. Example 3.4.1. A Quasistatic System Exhibiting Resonance Shown in cross-section in Fig. 3.4.2 is a resonator used in connection with electron beam devices at microwave frequencies. The volume enclosed by its perfectly con- ducting boundaries can be broken into the two regions shown. The ﬁrst of these is bounded by a pair of circular plane parallel conductors having spacing dand radius b. This region is EQS and described in Example 3.3.1. The second region is bounded by coaxial, perfectly conducting cylinders which form an annular region having outside radius aand an inside radius bthat matches up to the outer edge of the lower plate of the EQS system. The coaxial cylinders are 16 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 Fig. 3.4.3 Surface Sand contour Cfor evaluating H-ﬁeld using Amp` ere’s law. shorted by a perfectly conducting plate at the bottom, where z= 0. A similar plate at the top, where z=h, connects the outer cylinder to the outer edge of the upper plate in the EQS subregion. For the moment, the subsystems are isolated from each other by driving the MQS system with a current source Ko(amps/meter) distributed around the periph- ery of the gap between conductors. This gives rise to axial surface current densities ofKoand−Ko(b/a) on the inner and outer cylindrical conductors and radial surface current densities contributing to J·dain the upper and lower plates, respectively. (Note that these satisfy the MQS current continuity requirement.) Because of the symmetry, the magnetic ﬁeld can be determined by using the integral MQS form of Amp` ere’s law. So that there is a contribution to the integration ofJ·da, a surface is selected with a normal in the axial direction. This surface is enclosed by a circular contour having the radius r, as shown in Fig. 3.4.3. Because of the axial symmetry, Hφis independent of φ, and the integrations on SandC amount to multiplications. I CH·ds=Z SJ·izda⇒2πrH φ= 2πbK o (2) Thus, in the annulus, Hφ=b rKo (3) In the regions outside the annulus, His zero. Note that this is consistent with Amp` ere’s jump condition, (1.4.16), evaluated on any of the boundaries using the already determined surface current densities. Also, we will ﬁnd in Chap. 10 that there can be no time-varying magnetic ﬂux density normal to a perfectly conducting boundary. The magnetic ﬁeld given in (3) satisﬁes this condition as well. In the hierarchy of MQS laws, we have now satisﬁed (3.2.5b) and (3.2.6b) and come next to Faraday’s law, (3.2.7b). For the present purposes, we are not interested in the details of the distribution of electric ﬁeld. Rather, we use the integral form of Faraday’s law, (1.6.1), integrated on the surface Sshown in Fig. 3.4.4. The integral ofE·dsalong the perfect conductor vanishes and we are left with Eab=Zb aE·ds=dλf dt(4) where the EMF across the gap is as deﬁned by (1.6.2), and the ﬂux linked by Cis consistent with (1.6.8). λf=hZa bµoHφdr=µobhK oZa bdr r=µohb ln¡a b¢ Ko (5) Sec. 3.4 Quasistatic Systems 17 Fig. 3.4.4 Surface Sand contour Cused to determine EMF using Faraday’s law. These last two expressions combine to give Eab=µohb ln¡a b¢dKo dt(6) Just as this expression serves to relate the EMF and surface current density at the gap of the MQS system, (3.3.8) relates the gap variables deﬁned in Fig. 3.4.2b for the EQS subsystem. The subsystems are now interconnected by replacing the distributed current source driving the MQS system with the peripheral surface current density of the EQS system. Kr+Ko= 0 (7) In addition, the EMF’s of the two subsystems are made to match where they join. −E=Eab (8) With (3.3.8) and (3.3.6), respectively, substituted for KrandEab, these expressions become two diﬀerential equations in the two variables EoandKodescribing the complete system. −b/epsilon1o 2dEo dt+Ko= 0 (9) −dEo=µobh ln¡a b¢dKo dt(10) Elimination of Kobetween these expressions gives d2Eo dt2+ω2 oEo= 0 (11) where ωois deﬁned as ω2 o=2d /epsilon1oµohb2ln¡a b¢ (12) and it follows that solutions are a linear combination of sin ωotand cos ωot. As might have been suspected from the outset, what we have found is a re- sponse to initial conditions that is oscillatory, with a natural frequency ωo. That is, the parallel plate capacitor that comprises the EQS subsystem, connected in parallel with the one-turn inductor that is the MQS subsystem, responds to initial values ofEoandKowith an oscillation that at one instant has Eoat its peak magnitude andKo= 0, and a quarter cycle later has Eo= 0 and Koat its peak magnitude. 18 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 Fig. 3.4.5 In terms of characteristic time τ, the dynamic regime in which the system of Fig. 3.4.2 is quasistatic but capable of being in a state of resonance. Remember that /epsilon1oEois the surface charge density on the lower plate in the EQS section. Thus, the oscillation is between the charges in the EQS subsystem and the currents in the MQS subsystem. The distribution of ﬁeld sources in the system as a whole is determined by a dynamical interaction between the two subsystems. If the system were driven by a current source having the frequency ω, it would display a resonance at the natural frequency ωo. Under what conditions can the system be in resonance and still be quasistatic? In this case, the characteristic time for the system dynamics is the reciprocal of the resonance frequency. The EQS subsystem is indeed EQS if b/c/lessmuchτ, while the annular subsystem is MQS if h/c/lessmuchτ. Thus, the resonance is correctly described by the quasistatic model if the times have the ordering shown in Fig. 3.4.5. Essentially, this is achieved by making the spacing din the EQS section very small. With the region of interest containing media, the appropriate quasistatic limit is often as much determined by the material properties as by the topology. In Chaps. 7 and 10, we will consider lossy materials where the distributions of ﬁeld sources depend on the time rates of change and a given region can be EQS or MQS depending on the electrical conductivity. We return to the subject of quasistatics in Chaps. 12 and 14. 3.5 OVERVIEW OF APPLICATIONS Electroquasistatics is the subject of Chaps. 4–7 and magnetoquasistatics the topic of Chaps. 8–10. Before embarking on these subjects, consider in this section some practical examples that fall in each category, and some that involve the electrody- namics of Chaps. 12–14. Our starting point is at location Aat the upper right in Fig. 3.5.1. With frequencies that range from 60-400 MHz, television signals propagate from remote locations to our homes as electromagnetic waves. If the frequency is f, the ﬁeld passes through one period in the time 1 /f. Setting this equal to the transit time, (3.1.l7) gives an expression for the wavelength, the distance the wave travels during one cycle. L≡λ=c f Thus, for channel 2 (60 MHz) the wavelength is about 5 m, while for channel 54 it is about 20 cm. The distance between antenna and receiver is many wavelengths, and hence the ﬁelds undergo many oscillations while traversing the space between the two. The dynamics is not quasistatic but rather intimately involves the electro- magnetic wave represented by inset Band described in Sec. 3.1. Sec. 3.5 Overview of Applications 19 Fig. 3.5.1 Quasistatic and electrodynamic ﬁelds in the physical world. The ﬁeld induces charges and currents in the antenna, and the resulting sig- nals are conveyed to the TV set by a transmission line. At TV frequencies, the line is likely to be many wavelengths long. Hence, the ﬁelds surrounding the line are also not quasistatic. But the radial distributions of current in the elements of the anten- nas and in the wires of the transmission line are governed by magnetoquasistatic (MQS) laws. As suggested by inset C, the current density tends to concentrate 20 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 adjacent to the conductor surfaces and this skin eﬀect is MQS. Inside the television set, in the transistors and picture tube that convert the signal to an image and sound, electroquasistatic (EQS) processes abound. Included are dynamic eﬀects in the transistors (E) that result from the time required for an electron or hole to migrate a ﬁnite distance through a semiconductor. Also included are the eﬀects of inertia as the electrons are accelerated by the electric ﬁeld in the picture tube (D). On the other hand, the speaker that transduces the electrical signals into sound is most likely MQS. Electromagnetic ﬁelds are far closer to the viewer than the television set. As is obvious to those who have had an electrocardiogram, the heart (F) is the source of a pulsating current. Are the distributions of these currents and the associated ﬁelds described by the EQS or MQS approximation? On the largest scales of the body, we will ﬁnd that it is MQS. Of course, there are many other sources of electrical currents in the body. Nerve conduction and other electrical activity in the brain occur on much smaller length scales and can involve regions of much less conductivity. These cases can be EQS. Electrical power systems provide diverse examples as well. The step-down transformer on a pole outside the home (G) is MQS, with dynamical processes including eddy currents and hysteresis. The energy in all these examples originates in the fuel burned in a power plant. Typically, a steam turbine drives a synchronous alternator (H). The ﬁelds within this generator of electrical power are MQS. However, most of the electronics in the control room (J) are described by the EQS approximation. In fact, much of the payoﬀ in making computer components smaller is gained by having them remain EQS even as the bit rate is increased. The electrostatic precipitator (I), used to remove ﬂyash from the combustion gases before they are vented from the stacks, seems to be an obvious candidate for the EQS approximation. Indeed, even though some modern precipitators use pulsed high voltage and all involve dynamic electrical discharges, they are governed by EQS laws. The power transmission system is at high voltage and therefore might nat- urally be regarded as EQS. Certainly, speciﬁcation of insulation performance (K) begins with EQS approximations. However, once electrical breakdown has occurred, enough current can be faulted to bring MQS considerations into play. Certainly, they are present in the operation of high-power switch gear. To be even a fraction of a wavelength at 60 Hz, a line must stretch the length of California. Thus, in so far as the power frequency ﬁelds are concerned, the system is quasistatic. But certain aspects of the power line itself are MQS, and others EQS, although when lightning strikes it is likely that neither approximation is appropriate. Not all ﬁelds in our bodies are of physiological origin. The man standing under the power line (L) ﬁnds himself in both electric and magnetic ﬁelds. How is it that our bodies can shield themselves from the electric ﬁeld while being essentially transparent to the magnetic ﬁeld without having obvious eﬀects on our hearts or nervous systems? We will ﬁnd that currents are indeed induced in the body by both the electric and magnetic ﬁelds, and that this coupling is best understood in terms of the quasistatic ﬁelds. By contrast, because the wavelength of an electromagnetic wave at TV frequencies is on the order of the dimensions of the body, the currents induced in the person standing in front of the TV antenna at Aare not quasistatic. Sec. 3.6 Summary 21 As we make our way through the topics outlined in Fig. 3.5.1, these and other physical situations will be taken up by the examples. 3.6 SUMMARY From a mathematical point of view, the summary of quasistatic laws given in Table 3.6.1 is an outline of the next seven chapters. An excursion down the left column and then down the right column of the outline represented by Fig. 1.0.1 carries us down the corresponding columns of the table. Gauss’ law and the requirement that Ebe irrotational, (3.2.5a) and (3.2.6a), are the subjects of Chaps. 4–5. In Chaps. 6 and 7, two types of charge density are distinguished and used to represent the eﬀects of macroscopic media on the electric ﬁeld. In Chap. 6, where polarization charge is used to represent insulating media, charge is automatically conserved. But in Chap. 7, where unpaired charges are created through conduction processes, the charge conservation law, (3.2.7a), comes into play on the same footing as (3.2.5a) and (3.2.6a). In stages, starting in Chap. 4, the ability to predict self-consistent distributions of Eandρis achieved in this last EQS chapter. Amp´ ere’s law and magnetic ﬂux continuity, (3.2.5b) and (3.2.6b), are featured in Chap. 8. First, the magnetic ﬁeld is determined for a given distribution of current density. Because current distributions are often controlled by means of wires, it is easy to think of practical situations where the MQS source, the current density, is known at the outset. But even more, the ﬁrst half of Chap. 7 was already devoted to determining distributions of “stationary” current densities. The MQS current density is always solenoidal, (3.2.5c), and the magnetic induction on the right in Faraday’s law, (3.2.7b), is sometimes negligible so that the electric ﬁeld can be essentially irrotational. Thus, the ﬁrst half of Chap. 7 actually starts the sequence of MQS topics. In the second half of Chap. 8, the magnetic ﬁeld is determined for systems of perfect conductors, where the source distribution is not known until the ﬁelds meet certain boundary conditions. The situation is analogous to that for EQS systems in Chap. 5. Chapters 9 and 10 distinguish between eﬀects of magnetization and conduction currents caused by macroscopic media. It is in Chap. 10 that Faraday’s law, (3.2.7b), comes into play in a ﬁeld theoretical sense. Again, in stages, in Chaps. 8–10, we attain the ability to describe a self-consistent ﬁeld and source evolution, this time of Hand its sources, J. The quasistatic approximations and ordering of laws can just as well be stated in terms of the integral laws. Thus, the diﬀerential laws summarized in Table 3.6.1 have the integral law counterparts listed in Table 3.6.2. 22 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 TABLE 3.6.1 SUMMARY OF QUASISTATIC DIFFERENTIAL LAWS IN FREE SPACE ELECTROQUASISTATIC MAGNETOQUASISTATIC Reference Eq. ∇ ·/epsilon1oE=ρ ∇ ×H=J;∇ ·J= 0 (3.2.5) ∇ ×E= 0 ∇ ·µoH= 0 (3.2.6) ∇ ·J+∂ρ ∂t= 0 ∇ ×E=−∂µoH ∂t(3.2.7) Secondary ∇ ×H=J+∂/epsilon1oE ∂t∇ ·/epsilon1oE=ρ (3.2.8) ∇ ·µoH= 0 (3.2.9) TABLE 3.6.2 SUMMARY OF QUASISTATIC INTEGRAL LAWS IN FREE SPACE (a) (b) ELECTROQUASISTATIC MAGNETOQUASISTATIC Eq. H S/epsilon1oE·da=R VρdvH CH·ds=R SJ·da;H SJ·da= 0 (1) H CE·ds= 0H SµoH·da= 0 (2) H SJ·da+d dtR VρdV= 0R CE·ds=−d dtR Sµ0H·da (3) Secondary H CH·ds=R SJ·da+d dtR S/epsilon1oE·daH S/epsilon1oE·da=R Vρdv (4) H SµoH·da= 0 (5) Sec. 3.2 Problems 23 P R O B L E M S 3.1 Temporal Evolution of World Governed by Laws of Maxwell, Lorentz, and Newton 3.1.1 In Example 3.1.1, it was shown that solutions to Maxwell’s equations can take the form E=Ex(z−ct)ixandH=Hy(z−ct)iyin a region where J= 0 and ρ= 0. (a) Given EandHby (9) and (10) when t= 0, what are these ﬁelds for t >0? (b) By substituting these expressions into (1)–(4), show that they are exact solutions to Maxwell’s equations. (c) Show that for an observer at z=ct+ constant, these ﬁelds are con- stant. 3.1.2∗Show that in a region where J= 0 and ρ= 0 and a solution to Maxwell’s equations E(r, t) and H(r, t) has been obtained, a second solution is ob- tained by replacing Hby−E,EbyH, /epsilon1byµandµby/epsilon1. 3.1.3 In Prob. 3.1.1, the initial conditions given by (9) and (10) were arranged so that for t >0, the ﬁelds took the form of a wave traveling in the + z direction. (a) How would you alter the magnetic ﬁeld intensity, (10), so that the ensuing ﬁeld took the form of a wave traveling in the −zdirection? (b) What would you make H, so that the result was a pair of electric ﬁeld intensity waves having the same shape, one traveling in the + z direction and the other traveling in the −zdirection? 3.1.4 When t= 0,E=Eoizcosβx, where Eoandβare given constants. When t= 0, what must Hbe to result in E=Eoizcosβ(x−ct) for t >0. 3.2 Quasistatic Laws 3.2.1 In Sec. 13.1, we will ﬁnd that ﬁelds of the type considered in Example 3.1.1 can exist between the plane parallel plates of Fig. P3.2.1. In the particular case where the plates are “open” at the right, where z= 0, it will be found that between the plates these ﬁelds are E=Eocosβz cosβlcosωtix (a) H=Eor/epsilon1o µosinβz cosβlsinωtiy (b) 24 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 Fig. P3.2.1 Fig. P3.2.2 where β=ω√µo/epsilon1oandEois a constant established by the voltage source at the left. (a) By substitution, show that in the free space region between the plates (where J= 0 and ρ= 0), (a) and (b) are exact solutions to Maxwell’s equations. (b) Use trigonometric identities to show that these ﬁelds can be decom- posed into sums of waves traveling in the ±zdirections. For example, Ex=E+(z−ct) +E−(z+ct), where cis deﬁned by (3.1.16) and E± are functions of z∓ct, respectively. (c) Show that if βl/lessmuch1, the time l/crequired for an electromagnetic wave to traverse the length of the electrodes is short compared to the time τ≡1/ωwithin which the driving voltage is changing. (d) Show that in the limit where this is true, (a) and (b) become E→Eocosωtix (c) H→Eo/epsilon1oωzsinωtiy (d) so that the electric ﬁeld between the plates is uniform. (e) With the frequency low enough so that (c) and (d) are good approx- imations to the ﬁelds, do these solutions satisfy the EQS or MQS laws? 3.2.2 In Sec. 13.1, it will be shown that the electric and magnetic ﬁelds between the plane parallel plates of Fig. P3.2.2 are E=rµo /epsilon1oHosinβz cosβlsinωtix (a) Sec. 3.3 Problems 25 H=Hocosβz cosβlcosωtiy (b) where β=ω√µo/epsilon1oandHois a constant determined by the current source at the left. Note that because the plates are “shorted” at z= 0, the electric ﬁeld intensity given by (a) is zero there. (a) Show that (a) and (b) are exact solutions to Maxwell’s equations in the region between the plates where J= 0 and ρ= 0. (b) Use trigonometric identities to show that these ﬁelds take the form of waves traveling in the ±zdirections with the velocity cdeﬁned by (3.1.16). (c) Show that the condition βl/lessmuch1 is equivalent to the condition that the wave transit time l/cis short compared to τ≡1/ω. (d) For the frequency ωlow enough so that the conditions of part (c) are satisﬁed, give approximate expressions for EandH. Describe the distribution of Hbetween the plates. (e) Are these approximate ﬁelds governed by the EQS or the MQS laws? 3.3 Conditions for Fields to be Quasistatic 3.3.1 Rather than being in the circular geometry of Example 3.3.1, the conﬁgu- ration considered here and shown in Fig. P3.3.1 consists of plane parallel rectangular electrodes of (inﬁnite) width win the ydirection, spacing din thexdirection and length 2 lin the zdirection. The region between these electrodes is free space. Voltage sources constrain the integral of Ebetween the electrode edges to be the same functions of time. v=Zd 0Ex(z=±l)dx (a) (a) Assume that the voltage sources are varying so slowly that the electric ﬁeld is essentially static (irrotational). Determine the electric ﬁeld between the electrodes in terms of vand the dimensions. What is the surface charge density on the inside surfaces of the electrodes? (These steps are very similar to those in Example 3.3.1.) (b) Use conservation of charge to determine the surface current density Kzon the electrodes. (c) Now use Amp` ere’s integral law and symmetry arguments to ﬁnd H. With this ﬁeld between the plates, use Amp` ere’s continuity condition, (1.4.16), to ﬁnd Kin the plates and show that it is consistent with the result of part (b). (d) Because of the Hfound in part (c), Eis not irrotational. Return to the integral form of Faraday’s law to ﬁnd a corrected electric ﬁeld intensity, using the magnetic ﬁeld of part (c). [Note that the electric ﬁeld found in part (a) already satisﬁes the conditions imposed by the voltage sources.] 26 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 Fig. P3.3.1 (e) If the driving voltage takes the form v=vocosωt, determine the ratio of the correction (error) ﬁeld to the quasistatic ﬁeld of part (a). 3.3.2 The conﬁguration shown in Fig. P3.3.2 is similar to that for Prob. 3.3.1 ex- cept that the sources distributed along the left and right edges are current rather than voltage sources and are of opposite rather than the same polar- ity. Thus, with the current sources varying slowly, a ( z-independent) surface current density K(t) circulates around a loop consisting of the sources and the electrodes. The roles of EandHare the reverse of what they were in Example 3.3.1 or Prob. 3.3.1. Because the electrodes are pictured as having no resistance, the low-frequency electric ﬁeld is zero while, even if the exci- tations are constant in time, there is an H. The following steps answer the question, Under what circumstances is the electric displacement current negligible compared to the magnetic induction? (a) Determine Hin the region between the electrodes in a manner consis- tent with there being no Houtside. (Amp` ere’s continuity condition relates HtoKat the electrodes. Like the Eﬁeld in Example 3.3.1 or Prob. 3.3.1, the His extremely simple.) (b) Use the integral form of Faraday’s law to determine Ebetween the electrodes. Note that symmetry requires that this ﬁeld be zero where z= 0. (c) Because of this time-varying E, there is a displacement current density between the electrodes in the xdirection. Use Amp` ere’s integral law to ﬁnd the correction (error) H. Note that the quasistatic ﬁeld already meets the conditions imposed by the current sources where z=±l. (d) Given that the driving currents are sinusoidal with angular frequency ω, determine the ratio of the “error” of Hto the MQS ﬁeld of part (a). 3.4 Quasistatic Systems Sec. 3.4 Problems 27 Fig. P3.3.2 3.4.1 The conﬁguration shown in cutaway view in Fig. P3.4.1 is essentially the outer region of the system shown in Fig. 3.4.2. The object here is to deter- mine the error associated with neglecting the displacement current density in this outer region. In this problem, the region of interest is pictured as bounded on three sides by material having no resistance, and on the fourth side by a distributed current source. The latter imposes a surface current density Koin the zdirection at the radius r=b. This current passes ra- dially outward through a plate in the z=hplane, axially downward in another conductor at the radius r=a, and radially inward in the plate at z= 0. (a) Use the MQS form of Amp` ere’s integral law to determine Hinside the “donut”-shaped region. This ﬁeld should be expressed in terms of Ko. (Hint: This step is essentially the same as for Example 3.4.1.) (b) There is no Houtside the structure. The interior ﬁeld is terminated on the boundaries by a surface current density in accordance with Amp` ere’s continuity condition. What is Kon each of the boundaries? (c) In general, the driving current is time varying, so Faraday’s law re- quires that there be an electric ﬁeld. Use the integral form of this law and the contour Cand surface S shown in Fig. P3.4.2 to determine E. Assume that Etangential to the zero-resistance boundaries is zero. Also, assume that Eiszdirected and independent of z. (d) Now determine the error in the MQS Hby using Amp` ere’s integral law. This time the displacement current density is not approximated as zero but rather as implied by the Efound in part (c). Note that the MQS Hﬁeld already satisﬁes the condition imposed by the current source at r=b. (e) With Ko=Kpcosωt, write the condition for the error ﬁeld to be small compared to the MQS ﬁeld in terms of ω, c, and l. 28 Introduction To Electroquasistatics and Magnetoquasistatics Chapter 3 Fig. P3.4.1 Fig. P3.4.2 4 ELECTROQUASISTATIC FIELDS: THE SUPERPOSITION INTEGRAL POINT OF VIEW 4.0 INTRODUCTION The reason for taking up electroquasistatic ﬁelds ﬁrst is the relative ease with which such a vector ﬁeld can be represented. The EQS form of Faraday’s law requires that the electric ﬁeld intensity Ebe irrotational. ∇ ×E= 0 (1) The electric ﬁeld intensity is related to the charge density ρby Gauss’ law. ∇ ·/epsilon1oE=ρ (2) Thus, the source of an electroquasistatic ﬁeld is a scalar, the charge density ρ. In free space, the source of a magnetoquasistatic ﬁeld is a vector, the current density. Scalar sources, are simpler than vector sources and this is why electroquasistatic ﬁelds are taken up ﬁrst. Most of this chapter is concerned with ﬁnding the distribution of Epredicted by these laws, given the distribution of ρ. But before the chapter ends, we will be ﬁnding ﬁelds in limited regions bounded by conductors. In these more practical situations, the distribution of charge on the boundary surfaces is not known until after the ﬁelds have been determined. Thus, this chapter sets the stage for the solving of boundary value problems in Chap. 5. We start by establishing the electric potential as a scalar function that uniquely represents an irrotational electric ﬁeld intensity. Byproducts of the derivation are the gradient operator and gradient theorem. The scalar form of Poisson’s equation then results from combining (1) and (2). This equation will be shown to be linear. It follows that the ﬁeld due to a superposition of charges is the superposition of the ﬁelds associated with the in- dividual charge components. The resulting superposition integral speciﬁes how the 1 2 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 potential, and hence the electric ﬁeld intensity, can be determined from the given charge distribution. Thus, by the end of Sec. 4.5, a general approach to ﬁnding solutions to (1) and (2) is achieved. The art of arranging the charge so that, in a restricted region, the resulting ﬁelds satisfy boundary conditions, is illustrated in Secs. 4.6 and 4.7. Finally, more general techniques for using the superposition integral to solve boundary value problems are illustrated in Sec. 4.8. For those having a background in circuit theory, it is helpful to recognize that the approaches used in this and the next chapter are familiar. The solution of (1) and (2) in three dimensions is like the solution of circuit equations, except that for the latter, there is only the one dimension of time. In the ﬁeld problem, the driving function is the charge density. One approach to ﬁnding a circuit response is based on ﬁrst ﬁnding the response to an impulse. Then the response to an arbitrary drive is determined by superim- posing responses to impulses, the superposition of which represents the drive. This response takes the form of a superposition integral, the convolution integral. The impulse response of Poisson’s equation that is our starting point is the ﬁeld of a point charge. Thus, the theme of this chapter is a convolution approach to solving (1) and (2). In the boundary value approach of the next chapter, concepts familiar from circuit theory are again exploited. There, solutions will be divided into a particular part, caused by the drive, and a homogeneous part, required to satisfy boundary conditions. It will be found that the superposition integral is one way of ﬁnding the particular solution. 4.1 IRROTATIONAL FIELD REPRESENTED BY SCALAR POTENTIAL: THE GRADIENT OPERATOR AND GRADIENT INTEGRAL THEOREM The integral of an irrotational electric ﬁeld from some reference point rrefto the position ris independent of the integration path. This follows from an integration of (1) over the surface Sspanning the contour deﬁned by alternative paths I and II, shown in Fig. 4.1.1. Stokes’ theorem, (2.5.4), gives Z S∇ ×E·da=I CE·ds= 0 (1) Stokes’ theorem employs a contour running around the surface in a single direction, whereas the line integrals of the electric ﬁeld from rtorref, from point ato point b, run along the contour in opposite directions. Taking the directions of the path increments into account, (1) is equivalent to I CE·ds=Zb apath IE·ds−Zb apath IIE·ds/prime= 0 (2) and thus, for an irrotational ﬁeld, the EMF between two points is independent of path.Zb apath IE·ds=Zb apath IIE·ds/prime(3) Sec. 4.1 Irrotational Field 3 Fig. 4.1.1 Paths I and II between positions randrrefare spanned by surface S. A ﬁeld that assigns a unique value of the line integral between two points independent of path of integration is said to be conservative . With the understanding that the reference point is kept ﬁxed, the integral is a scalar function of the integration endpoint r. We use the symbol Φ( r) to deﬁne this scalar function Φ(r)−Φ(rref) =Zrref rE·ds (4) and call Φ( r) the electric potential of the point rwith respect to the reference point. With the endpoints consisting of “nodes” where wires could be attached, the potential diﬀerence of (1) would be the voltage atrrelative to that at the reference. Typically, the latter would be the “ground” potential. Thus, for an irrotational ﬁeld, the EMF deﬁned in Sec. 1.6 becomes the voltage at the point arelative to point b. We shall show that speciﬁcation of the scalar function Φ( r) contains the same information as speciﬁcation of the ﬁeld E(r). This is a remarkable fact because a vector function of rrequires, in general, the speciﬁcation of three scalar functions ofr, say the three Cartesian components of the vector function. On the other hand, speciﬁcation of Φ( r) requires one scalar function of r. Note that the expression Φ( r) = constant represents a surface in three dimen- sions. A familiar example of such an expression describes a spherical surface having radius R. x2+y2+z2=R2(5) Surfaces of constant potential are called equipotentials . Shown in Fig. 4.1.2 are the cross-sections of two equipotential surfaces, one passing through the point r, the other through the point r+ ∆r. With ∆ rtaken as a diﬀerential vector, the potential at the point r+ ∆rdiﬀers by the diﬀerential 4 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.1.2 Two equipotential surfaces shown cut by a plane containing their normal, n. amount ∆Φ from that at r. The two equipotential surfaces cannot intersect. Indeed, if they intersected, both points randr+ ∆rwould have the same potential, which is contrary to our assumption. Illustrated in Fig. 4.1.2 is the shortest distance ∆ nfrom the point rto the equipotential at r+ ∆r. Because of the diﬀerential geometry assumed, the length element ∆ nis perpendicular to both equipotential surfaces. From Fig. 4.1.2, ∆ n= cosθ∆r, and we have ∆Φ =∆Φ ∆ncosθ∆r=∆Φ ∆nn·∆r (6) The vector ∆ rin (6) is of arbitrary direction. It is also of arbitrary diﬀerential length. Indeed, if we double the distance ∆ n, we double ∆Φ and ∆ r; ∆Φ /∆nre- mains unchanged and thus (6) holds for any ∆ r(of diﬀerential length). We conclude that (6) assigns to every diﬀerential vector length element ∆ r, originating from r, a scalar of magnitude proportional to the magnitude of ∆ rand to the cosine of the angle between ∆ rand the unit vector n. This assignment of a scalar to a vector is representable as the scalar product of the vector length element ∆ rwith a vector of magnitude ∆Φ /∆nand direction n. That is, (6) is equivalent to ∆Φ = grad Φ·∆r (7) where the gradient of the potential is deﬁned as grad Φ≡∆Φ ∆nn (8) Because it is independent of any particular coordinate system, (8) provides the best way to conceptualize the gradient operator. The same equation provides the algorithm for expressing grad Φ in any particular coordinate system. Consider, as an example, Cartesian coordinates. Thus, r=xix+yiy+ziz; ∆ r= ∆xix+ ∆yiy+ ∆ziz (9) and an alternative to (6) for expressing the diﬀerential change in Φ is ∆Φ = Φ( x+ ∆x, y+ ∆y, z+ ∆z)−Φ(x, y, z ) =∂Φ ∂x∆x+∂Φ ∂y∆y+∂Φ ∂z∆z.(10) Sec. 4.1 Irrotational Field 5 In view of (9), this expression is ∆Φ =µ ix∂Φ ∂x+iy∂Φ ∂y+iz∂Φ ∂z¶ ·∆r=∇Φ·∆r (11) and it follows that in Cartesian coordinates the gradient operation, as deﬁned by (7), is grad Φ≡ ∇Φ =∂Φ ∂xix+∂Φ ∂yiy+∂Φ ∂ziz (12) Here, the del operator deﬁned by (2.1.6) is introduced as an alternative way of writing the gradient operator. Problems at the end of this chapter serve to illustrate how the gradient is similarly determined in other coordinates, with results summarized in Table I at the end of the text. We are now ready to show that the potential function Φ( r) deﬁnes E(r) uniquely. According to (4), the potential changes from the point rto the point r+ ∆rby ∆Φ = Φ( r+ ∆r)−Φ(r) =−Zr+∆r rrefE·ds+Zr rrefE·ds =−Zr+∆r rE·ds(13) The ﬁrst two integrals in (13) follow from the deﬁnition of Φ, (4). By recognizing thatdsis ∆rand that ∆ ris of diﬀerential length, so that E(r) can be considered constant over the length of the vector ∆ r, it can be seen that the last integral in (13) becomes ∆Φ = −E·∆r (14) The vector element ∆ ris arbitrary. Therefore, comparison of (14) to (7) shows that E=−∇Φ (15) Given the potential function Φ( r), the associated electric ﬁeld intensity is the negative gradient of Φ. Note that we also obtained a useful integral theorem, for if (15) is substituted into (4), it follows that Zr rref∇Φ·ds= Φ(r)−Φ(rref) (16) That is, the line integration of the gradient of Φ is simply the diﬀerence in potential between the endpoints. Of course, Φ can be any scalar function. In retrospect, we can observe that the representation of Eby (15) guarantees that it is irrotational, for the vector identity holds ∇ ×(∇Φ) = 0 (17) 6 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 The curl of the gradient of a scalar potential Φ vanishes. Therefore, given an electric ﬁeld represented by a potential in accordance with (15), (4.0.1) is automatically satisﬁed. Because the preceding discussion shows that the potential Φ contains full information about the ﬁeld E, the replacement of Ebygrad (Φ) constitutes a gen- eral solution, or integral, of (4.0.1). Integration of a ﬁrst-order ordinary diﬀerential equation leads to one arbitrary integration constant. Integration of the ﬁrst-order vector diﬀerential equation curlE= 0 yields a scalar function of integration, Φ( r). Thus far, we have not made any speciﬁc assignment for the reference point rref. Provided that the potential behaves properly at inﬁnity, it is often convenient to let the reference point be at inﬁnity. There are some exceptional cases for which such a choice is not possible. All such cases involve problems with inﬁnite amounts of charge. One such example is the ﬁeld set up by a charge distribution that extends to inﬁnity in the ±zdirections, as in the second Illustration in Sec. 1.3. The ﬁeld decays like 1 /rwith radial distance rfrom the charged region. Thus, the line integral of E, (4), from a ﬁnite distance out to inﬁnity involves the diﬀerence of ln revaluated at the two endpoints and becomes inﬁnite if one endpoint moves to inﬁnity. In problems that extend to inﬁnity but are not of this singular nature, we shall assume that the reference is at inﬁnity. Example 4.1.1. Equipotential Surfaces Consider the potential function Φ( x, y), which is independent of z: Φ(x, y) =Voxy a2(18) Surfaces of constant potential can be represented by a cross-sectional view in any x−yplane in which they appear as lines, as shown in Fig. 4.1.3. For the potential given by (18), the equipotentials appear in the x−yplane as hyperbolae. The contours passing through the points ( a, a) and ( −a,−a) have the potential Vo, while those at ( a,−a) and ( −a, a) have potential −Vo. The magnitude of Eis proportional to the spatial rate of change of Φ in a direction perpendicular to the constant potential surface. Thus, if the surfaces of constant potential are sketched at equal increments in potential, as is done in Fig. 4.1.3, where the increments are Vo/4, the magnitude of Eis inversely propor- tional to the spacing between surfaces. The closer the spacing of potential lines, the higher the ﬁeld intensity. Field lines, sketched in Fig. 4.1.3, have arrows that point from high to low potentials. Note that because they are always perpendicular to the equipotentials, they naturally are most closely spaced where the ﬁeld intensity is largest. Example 4.1.2. Evaluation of Gradient and Line Integral Our objective is to exemplify by direct evaluation the fact that the line integration of an irrotational ﬁeld between two given points is independent of the integration path. In particular, consider the potential given by (18), which, in view of (12), implies the electric ﬁeld intensity E=−∇Φ =−Vo a2(yix+xiy) (19) Sec. 4.1 Irrotational Field 7 Fig. 4.1.3 Cross-sectional view of surfaces of constant potential for two-dimensional potential given by (18). We integrate this vector function along two paths, shown in Fig. 4.1.3, which join points (1) and (2). For the ﬁrst path, C1, yis held ﬁxed at y=aand hence ds=dxix. Thus, the integral becomes Z C1E·ds=Za −aEx(x, a)dx=−Za −aVo a2adx=−2Vo (20) For path C2, y−x2/a= 0 and in general, ds=dxix+dyiy, so the required integral isZ C2E·ds=Z C2(Exdx+Eydy) (21) However, for the path C2we have dy−(2x)dx/a = 0, and hence (21) becomes Z C2E·ds=Za −a¡ Ex+2x aEy¢ dx =Za −a−Vo a2µ x2 a+2x2 a¶ dx=−2Vo(22) Because Eis found by taking the negative gradient of Φ, and is therefore irrotational, it is no surprise that (20) and (22) give the same result. Example 4.1.3. Potential of Spherical Cloud of Charge 8 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 A uniform static charge distribution ρooccupies a spherical region of radius R. The remaining space is charge free (except, of course, for the balancing charge at inﬁn- ity). The following illustrates the determination of a piece-wise continuous potential function. The spherical symmetry of the charge distribution imposes a spherical symme- try on the electric ﬁeld that makes possible its determination from Gauss’ integral law. Following the approach used in Example 1.3.1, the ﬁeld is found to be Er=½rρo 3/epsilon1o;r < R R3ρo 3/epsilon1or2;r > R(23) The potential is obtained by evaluating the line integral of (4) with the reference point taken at inﬁnity, r=∞. The contour follows part of a straight line through the origin. In the exterior region, integration gives Φ(r) =Z∞ rErdr=4πR3 3ρo¡1 4π/epsilon1or¢ ; r > R (24) To ﬁnd Φ in the interior region, the integration is carried through the outer region, (which gives (24) evaluated at r=R) and then into the radius rin the interior region. Φ(r) =4πR3 3ρo¡1 4π/epsilon1oR¢ +ρo 6/epsilon1o(R2−r2) (25) Outside the charge distribution, where r≥R, the potential acquires the form of the coulomb potential of a point charge. Φ =q 4π/epsilon1or; q≡4πR3 3ρo (26) Note that qis the net charge of the distribution. Visualization of Two-Dimensional Irrotational Fields. In general, equipo- tentials are three-dimensional surfaces. Thus, any two-dimensional plot of the con- tours of constant potential is the intersection of these surfaces with some given plane. If the potential is two-dimensional in its dependence, then the equipotential surfaces have a cylindrical shape. For example, the two-dimensional potential of (18) has equipotential surfaces that are cylinders having the hyperbolic cross-sections shown in Fig. 4.1.3. We review these geometric concepts because we now introduce a diﬀerent point of view that is useful in picturing two-dimensional ﬁelds. A three-dimensional picture is now made in which the third dimension represents the amplitude of the potential Φ. Such a picture is shown in Fig. 4.1.4, where the potential of (18) is used as an example. The ﬂoor of the three-dimensional plot is the x−yplane, while the vertical dimension is the potential. Thus, contours of constant potential are represented by lines of constant altitude. The surface of Fig. 4.1.4 can be regarded as a membrane stretched between supports on the periphery of the region of interest that are elevated or depressed in proportion to the boundary potential. By the deﬁnition of the gradient, (8), the lines of electric ﬁeld intensity follow contours of steepest descent on this surface. Sec. 4.2 Poisson’s Equation 9 Fig. 4.1.4 Two-dimensional potential of (18) and Fig. 4.1.3 represented in three dimensions. The vertical coordinate, the potential, is analogous to the vertical deﬂection of a taut membrane. The equipotentials are then contours of constant altitude on the membrane surface. Potential surfaces have their greatest value in the mind’s eye, which pictures a two-dimensional potential as a contour map and the lines of electric ﬁeld intensity as the ﬂow lines of water streaming down the hill. 4.2 POISSON’S EQUATION Given that Eis irrotational, (4.0.1), and given the charge density in Gauss’ law, (4.0.2), what is the distribution of electric ﬁeld intensity? It was shown in Sec. 4.1 that we can satisfy the ﬁrst of these equations identically by representing the vector Eby the scalar electric potential Φ. E=−∇Φ (1) That is, with the introduction of this relation, (4.0.1) has been integrated. Having integrated (4.0.1), we now discard it and concentrate on the second equation of electroquasistatics, Gauss’ law. Introduction of (1) into Gauss’ law, (1.0.2), gives ∇ · ∇ Φ =−ρ /epsilon1o which is identically 10 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 ∇2Φ =−ρ /epsilon1o (2) Integration of this scalar Poisson’s equation , given the charge density on the right, is the objective in the remainder of this chapter. By analogy to the ordinary diﬀerential equations of circuit theory, the charge density on the right is a “driving function.” What is on the left is the operator ∇2, denoted by the second form of (2) and called the Laplacian of Φ. In Cartesian coor- dinates, it follows from the expressions for the divergence and gradient operators, (2.1.5) and (4.1.12), that ∂2Φ ∂x2+∂2Φ ∂y2+∂2Φ ∂z2=−ρ /epsilon1o(3) The Laplacian operator in cylindrical and spherical coordinates is determined in the problems and summarized in Table I at the end of the text. In Cartesian coordinates, the derivatives in this operator have constant coeﬃcients. In these other two coordinate systems, some of the coeﬃcients are space varying. Note that in (3), time does not appear explicitly as an independent variable. Hence, the mathematical problem of ﬁnding a quasistatic electric ﬁeld at the time tofor a time-varying charge distribution ρ(r, t) is the same as ﬁnding the static ﬁeld for the time-independent charge distribution ρ(r) equal to ρ(r, t=to), the charge distribution of the time-varying problem at the particular instant to. In problems where the charge distribution is given, the evaluation of a qua- sistatic ﬁeld is therefore equivalent to the evaluation of a succession of static ﬁelds, each with a diﬀerent charge distribution, at the time of interest. We emphasize this here to make it understood that the solution of a static electric ﬁeld has wider ap- plicability than one would at ﬁrst suppose: Every static ﬁeld solution can represent a “snapshot” at a particular instant of time. Having said that much, we shall not indicate the time dependence of the charge density and ﬁeld explicitly, but shall do so only when this is required for clarity. 4.3 SUPERPOSITION PRINCIPLE As illustrated in Cartesian coordinates by (4.2.3), Poisson’s equation is a linear second-order diﬀerential equation relating the potential Φ( r) to the charge distri- bution ρ(r). By “linear” we mean that the coeﬃcients of the derivatives in the diﬀerential equation are not functions of the dependent variable Φ. An important consequence of the linearity of Poisson’s equation is that Φ( r) obeys the superpo- sition principle. It is perhaps helpful to recognize the analogy to the superposition principle obeyed by solutions of the linear ordinary diﬀerential equations of circuit theory. Here the principle can be shown as follows. Consider two diﬀerent spatial distributions of charge density, ρa(r) and ρb(r). These might be relegated to diﬀerent regions, or occupy the same region. Suppose we have found the potentials Φ aand Φ bwhich satisfy Poisson’s equation, (4.2.3), Sec. 4.4 Fields of Charge Singularities 11 with the respective charge distributions ρaandρb. By deﬁnition, ∇2Φa(r) =−ρa(r) /epsilon1o(1) ∇2Φb(r) =−ρb(r) /epsilon1o(2) Adding these expressions, we obtain ∇2Φa(r) +∇2Φb(r) =−1 /epsilon1o[ρa(r) +ρb(r)] (3) Because the derivatives called for in the Laplacian operation– for example, the second derivatives of (4.2.3)– give the same result whether they operate on the potentials and then are summed or operate on the sum of the potentials, (3) can also be written as ∇2[Φa(r) + Φ b(r)] =−1 /epsilon1o[ρa(r) +ρb(r)] (4) The mathematical statement of the superposition principle follows from (1) and (2) and (4). That is, if ρa⇒Φa ρb⇒Φb (5) then ρa+ρb⇒Φa+ Φb (6) The potential distribution produced by the superposition of the charge distributions is the sum of the potentials associated with the individual distributions. 4.4 FIELDS ASSOCIATED WITH CHARGE SINGULARITIES At least three objectives are set in this section. First, the superposition concept from Sec. 4.3 is exempliﬁed. Second, we begin to deal with ﬁelds that are not highly symmetric. The potential proves invaluable in picturing such ﬁelds, and so we continue to develop ways of picturing the potential and ﬁeld distribution. Finally, the potential functions developed will reappear many times in the chapters that follow. Solutions to Poisson’s equation as pictured here ﬁlling all of space will turn out to be solutions to Laplace’s equation in subregions that are devoid of charge. Thus, they will be seen from a second point of view in Chap. 5, where Laplace’s equation is featured. First, consider the potential associated with a point charge at the origin of a spherical coordinate system. The electric ﬁeld was obtained using the integral form of Gauss’ law in Sec. 1.3, (1.3.12). It follows from the deﬁnition of the potential, (4.1.4), that the potential of a point charge qis Φ =q 4π/epsilon1or(1) 12 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.4.1 Point charges of equal magnitude and opposite sign on the zaxis. This “impulse response” for the three-dimensional Poisson’s equation is the starting point in derivations and problem solutions and is worth remembering. Consider next the ﬁeld associated with a positive and a negative charge, lo- cated on the zaxis at d/2 and −d/2, respectively. The conﬁguration is shown in Fig. 4.4.1. In (1), ris the scalar distance between the point of observation and the charge. With Pthe observation position, these distances are denoted in Fig. 4.4.1 byr+andr−. It follows from (1) and the superposition principle that the potential distribution for the two charges is Φ =q 4π/epsilon1oµ1 r+−1 r−¶ (2) To ﬁnd the electric ﬁeld intensity by taking the negative gradient of this function, it is necessary to express r+andr−in Cartesian coordinates. r+=r x2+y2+¡ z−d 2¢2; r−=r x2+y2+¡ z+d 2¢2(3) Thus, in these coordinates, the potential for the two charges given by (2) is Φ =q 4π/epsilon1oÃ 1q x2+y2+¡ z−d 2¢2−1q x2+y2+¡ z+d 2¢2! (4) Equation (2) shows that in the immediate vicinity of one or the other of the charges, the respective charge dominates the potential. Thus, close to the point charges the equipotentials are spheres enclosing the charge. Also, this expression makes it clear that the plane z= 0 is one of zero potential. One straightforward way to plot the equipotentials in detail is to program a calculator to evaluate (4) at a speciﬁed coordinate position. To this end, it is convenient to normalize the potential and the coordinates such that (4) is Φ=1q x2+y2+¡ z−1 2¢2−1q x2+y2+¡ z+1 2¢2(5) Sec. 4.4 Fields of Charge Singularities 13 where x=x d, y =y d, z =z d,Φ=Φ (q/4πd/epsilon1o) By evaluating Φ for various coordinate positions, it is possible to zero in on the co- ordinates of a given equipotential in an iterative fashion. The equipotentials shown in Fig. 4.4.2a were plotted in this way with x= 0. Of course, the equipotentials are actually three-dimensional surfaces obtained by rotating the curves shown about thezaxis. Because Eis the negative gradient of Φ, lines of electric ﬁeld intensity are perpendicular to the equipotentials. These can therefore be easily sketched and are shown as lines with arrows in Fig. 4.4.2a. Dipole at the Origin. An important limit of (2) corresponds to a view of the ﬁeld for an observer far from either of the charges. This is a very important limit because charge pairs of opposite sign are the model for polarized atoms or molecules. The dipole is therefore at center stage in Chap. 6, where we deal with polarizable matter. Formally, the dipole limit is taken by recognizing that rays joining the point of observation with the respective charges are essentially parallel to the rcoordinate when r/greatermuchd. The approximate geometry shown in Fig. 4.4.3 motivates the approximations. r+/similarequalr−d 2cosθ; r−/similarequalr+d 2cosθ (6) Because the ﬁrst terms in these expressions are very large compared to the second, powers of r+andr−can be expanded in a binomial expansion. (a+b)n=an+nan−1b+. . . (7) With n=−1, (2) becomes approximately Φ =q 4π/epsilon1o·¡1 r+d 2r2cosθ+. . .¢ −¡1 r−d 2r2cosθ+. . .¢¸ =qd 4π/epsilon1ocosθ r2(8) Remember, the potential is pictured in spherical coordinates. Suppose the equipotential is to be sketched that passes through the zaxis at some speciﬁed location. What is the shape of the potential as we move in the positive θdirection? On the left in (8) is a constant. With an increase in θ, the cosine function on the right decreases. Thus, to stay on the surface, the distance rfrom the origin must decrease. As the angle approaches π/2, the cosine decreases to zero, making it clear that the equipotential must approach the origin. The equipotentials and associated lines of Eare shown in Fig. 4.4.2b. 14 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.4.2 (a) Cross-section of equipotentials and lines of electric ﬁeld inten- sity for the two charges of Fig. 4.4.1. (b) Limit in which pair of charges form a dipole at the origin. (c) Limit of charges at inﬁnity. Sec. 4.4 Fields of Charge Singularities 15 Fig. 4.4.3 Far from the dipole, rays from the charges to the point of obser- vation are essentially parallel to rcoordinate. The dipole model is made mathematically exact by deﬁning it as the limit in which two charges of equal magnitude and opposite sign approach to within an inﬁnitesimal distance of each other while increasing in magnitude. Thus, with the dipole moment pdeﬁned as p= lim d→0 q→∞qd (9) the potential for the dipole, (8), becomes Φ =p 4π/epsilon1ocosθ r2(10) Another more general way of writing (10) with the dipole positioned at an arbitrary point r/primeand lying along a general axis is to introduce the dipole moment vector. This vector is deﬁned to be of magnitude pand directed along the axis of the two charges pointing from the −charge to the + charge. With the unit vector ir/primerdeﬁned as being directed from the point r/prime(where the dipole is located) to the point of observation at r, it follows from (10) that the generalized potential is Φ =p·ir/primer 4π/epsilon1o|r−r/prime|2(11) Pair of Charges at Inﬁnity Having Equal Magnitude and Opposite Sign. Con- sider next the appearance of the ﬁeld for an observer located between the charges of Fig. 4.4.2a, in the neighborhood of the origin. We now conﬁne interest to distances from the origin that are small compared to the charge spacing d. Eﬀectively, the charges are at inﬁnity in the + zand−zdirections, respectively. With the help of Fig. 4.4.4 and the three-dimensional Pythagorean theorem, the distances from the charges to the observer point are expressed in spherical coordinates as r+=r¡d 2−rcosθ¢2+ (rsinθ)2; r−=r¡d 2+rcosθ¢2+ (rsinθ)2(12) 16 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.4.4 Relative displacements with charges going to inﬁnity. In these expressions, dis large compared to r, so they can be expanded by again using (7) and keeping only linear terms in r. r−1 +/similarequal2 d+4r d2cosθ; r−1 −/similarequal2 d−4r d2cosθ (13) Introduction of these approximations into (2) results in the desired expression for the potential associated with charges that are at inﬁnity on the zaxis. Φ→2(q/d2) π/epsilon1orcosθ (14) Note that z=rcosθ, so what appears to be a complicated ﬁeld in spherical coor- dinates is simply Φ→2q/d2 π/epsilon1oz (15) Thezcoordinate can just as well be regarded as Cartesian, and the electric ﬁeld evaluated using the gradient operator in Cartesian coordinates. Thus, the surfaces of constant potential, shown in Fig. 4.4.2c, are horizontal planes. It follows that the electric ﬁeld intensity is uniform and downward directed. Note that the electric ﬁeld that follows from (15) is what is obtained by direct evaluation of (1.3.12) as the ﬁeld of point charges qat a distance d/2 above and below the point of interest. Other Charge Singularities. A two-dimensional dipole consists of a pair of oppositely charged parallel lines, rather than a pair of point charges. Pictured in a plane perpendicular to the lines, and in polar coordinates, the equipotentials ap- pear similar to those of Fig. 4.4.2b. However, in three dimensions the surfaces are cylinders of circular cross-section and not at all like the closed surfaces of revolu- tion that are the equipotentials for the three-dimensional dipole. Two-dimensional dipole ﬁelds are derived in Probs. 4.4.1 and 4.4.2, where the potentials are given for reference. Sec. 4.5 Solution of Poisson’s Equation 17 Fig. 4.5.1 An elementary volume of charge at r/primegives rise to a potential at the observer position r. There is an inﬁnite number of charge singularities. One of the “higher order” singularities is illustrated by the quadrupole ﬁelds developed in Probs. 4.4.3 and 4.4.4. We shall see these same potentials again in Chap. 5. 4.5 SOLUTION OF POISSON’S EQUATION FOR SPECIFIED CHARGE DISTRIBUTIONS The superposition principle is now used to ﬁnd the solution of Poisson’s equation for any given charge distribution ρ(r). The argument presented in the previous section for singular charge distributions suggests the approach. For the purpose of representing the arbitrary charge density distribution as a sum of “elementary” charge distributions, we subdivide the space occupied by the charge density into elementary volumes of size dx/primedy/primedz/prime. Each of these elements is denoted by the Cartesian coordinates ( x/prime, y/prime, z/prime), as shown in Fig. 4.5.1. The charge contained in one of these elementary volumes, the one with the coordinates (x/prime, y/prime, z/prime), is dq=ρ(r/prime)dx/primedy/primedz/prime=ρ(r/prime)dv/prime(1) We now express the total potential due to the charge density ρas the superpo- sition of the potentials dΦ due to the diﬀerential elements of charge, (1), positioned at the points r/prime. Note that each of these elementary charge distributions has zero charge density at all points outside of the volume element dv/primesituated at r/prime. Thus, they represent point charges of magnitudes dqgiven by (1). Provided that |r−r/prime| is taken as the distance between the point of observation rand the position of one incremental charge r/prime, the potential associated with this incremental charge is given by (4.4.1). dΦ(r,r/prime) =ρ(r/prime)dv/prime 4π/epsilon1o|r−r/prime|(2) where in Cartesian coordinates |r−r/prime|=p (x−x/prime)2+ (y−y/prime)2+ (z−z/prime)2 Note that (2) is a function of two sets of Cartesian coordinates: the (observer) coordinates ( x, y, z ) of the point rat which the potential is evaluated and the 18 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 (source) coordinates ( x/prime, y/prime, z/prime) of the point r/primeat which the incremental charge is positioned. According to the superposition principle, we obtain the total potential pro- duced by the sum of the diﬀerential charges by adding over all diﬀerential potentials, keeping the observation point ( x, y, z ) ﬁxed. The sum over the diﬀerential volume elements becomes a volume integral over the coordinates ( x/prime, y/prime, z/prime). Φ(r) =Z V/primeρ(r/prime)dv/prime 4π/epsilon1o|r−r/prime| (3) This is the superposition integral for the electroquasistatic potential. The evaluation of the potential requires that a triple integration be carried out. With the help of a computer, or even a programmable calculator, this is a straightforward process. There are few examples where the three successive inte- grations are carried out analytically without considerable diﬃculty. There are special representations of (3), appropriate in cases where the charge distribution is conﬁned to surfaces, lines, or where the distribution is two dimen- sional. For these, the number of integrations is reduced to two or even one, and the diﬃculties in obtaining analytical expressions are greatly reduced. Three-dimensional charge distributions can be represented as the superposi- tion of lines and sheets of charge and, by exploiting the potentials found analytically for these distributions, the numerical integration that might be required to deter- mine the potential for a three-dimensional charge distribution can be reduced to two or even one numerical integration. Superposition Integral for Surface Charge Density. If the charge density is conﬁned to regions that can be described by surfaces having a very small thickness ∆, then one of the three integrations of (3) can be carried out in general. The situation is as pictured in Fig. 4.5.2, where the distance to the observation point is large compared to the thickness over which the charge is distributed. As the integration of (3) is carried out over this thickness ∆, the distance between source and observer, |r−r/prime|, varies little. Thus, with ξused to denote a coordinate that is locally perpendicular to the surface, the general superposition integral, (3), reduces to Φ(r) =Z A/primeda/prime 4π/epsilon1o|r−r/prime|Z∆ 0ρ(r/prime)dξ (4) The integral on ξis by deﬁnition the surface charge density. Thus, (4) becomes a form of the superposition integral applicable where the charge distribution can be modeled as being on a surface. Φ(r) =Z A/primeσs(r/prime)da/prime 4π/epsilon1o|r−r/prime| (5) The following example illustrates the application of this integral. Sec. 4.5 Solution of Poisson’s Equation 19 Fig. 4.5.2 An element of surface charge at the location r/primegives rise to a potential at the observer point r. Fig. 4.5.3 A uniformly charged disk with coordinates for ﬁnding the potential along the zaxis. Example 4.5.1. Potential of a Uniformly Charged Disk The disk shown in Fig. 4.5.3 has a radius Rand carries a uniform surface charge density σo. The following steps lead to the potential and ﬁeld on the axis of the disk. The distance |r−r/prime|between the point r/primeat radius ρand angle φ(in cylindrical coordinates) and the point ron the axis of the disk (the zaxis) is given by |r−r/prime|=p ρ/prime2+z2 (6) It follows that (5) is expressible in terms of the following double integral Φ =σo 4π/epsilon1oZ2π 0ZR 0ρ/primedρ/primedφ/prime p ρ/prime2+z2 =σo 4π/epsilon1o2πZR 0ρ/primedρ/prime p ρ/prime2+z2 =σo 2/epsilon1o¡p R2+z2− |z|¢(7) where we have allowed for both positive z, the case illustrated in the ﬁgure, and negative z. Note that these are points on opposite sides of the disk. 20 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 The axial ﬁeld intensity Ezcan be found by taking the gradient of (7) in the zdirection. Ez=−∂Φ ∂z=−σo 2/epsilon1od dz¡p R2+z2− |z|¢ =−σo 2/epsilon1oµ z√ R2+z2∓1¶ (8) The upper sign applies to positive z, the lower sign to negative z. The potential distribution of (8) can be checked in two limiting cases for which answers are easily obtained by inspection: the potential at a distance |z| /greatermuchR, and the ﬁeld at |z| /lessmuchR. (a) At a very large distance |z|of the point of observation from the disk, the radius of the disk Ris small compared to |z|, and the potential of the disk must approach the potential of a point charge of magnitude equal to the total charge of the disk, σoπR2. The potential given by (7) can be expanded in powers of R/z p R2+z2− |z|=|z|µ 1 +1 2R2 z2¶ (9) to ﬁnd that Φ indeed approaches the potential function Φ/similarequalσo 4π/epsilon1oπR21 |z|(10) of a point charge at distance |z|from the observation point. (b) At |z| /lessmuchR, on either side of the disk, the ﬁeld of the disk must approach that of a charge sheet of very large (inﬁnite) extent. But that ﬁeld is ±σo/2/epsilon1o. We ﬁnd, indeed, that in the limit |z| →0, (8) yields this limiting result. Superposition Integral for Line Charge Density. Another special case of the general superposition integral, (3), pertains to ﬁelds from charge distributions that are conﬁned to the neighborhoods of lines. In practice, dimensions of interest are large compared to the cross-sectional dimensions of the area A/primeof the charge distribution. In that case, the situation is as depicted in Fig. 4.5.4, and in the integration over the cross-section the distance from source to observer is essentially constant. Thus, the superposition integral, (3), becomes Φ(r) =Z L/primedl/prime 4π/epsilon1o|r−r/prime|Z A/primeρ(r/prime)da/prime(11) In view of the deﬁnition of the line charge density, (1.3.10), this expression becomes Φ(r) =Z L/primeλl(r/prime)dl/prime 4π/epsilon1o|r−r/prime| (12) Example 4.5.2. Field of Collinear Line Charges of Opposite Polarity Sec. 4.5 Solution of Poisson’s Equation 21 Fig. 4.5.4 An element of line charge at the position r/primegives rise to a potential at the observer location r. Fig. 4.5.5 Collinear positive and negative line elements of charge sym- metrically located on the zaxis. A positive line charge density of magnitude λois uniformly distributed along the z axis between the points z=dandz= 3d. Negative charge of the same magnitude is distributed between z=−dandz=−3d. The axial symmetry suggests the use of the cylindrical coordinates deﬁned in Fig. 4.5.5. The distance from an element of charge λodz/primeto an arbitrary observer point (r, z) is |r−r/prime|=p r2+ (z−z/prime)2 (13) Thus, the line charge form of the superposition integral, (12), becomes Φ =λo 4π/epsilon1oµZ3d ddz/prime p (z−z/prime)2+r2−Z−d −3ddz/prime p (z−z/prime)2+r2¶ (14) These integrations are carried out to obtain the desired potential distribution Φ=ln¡ 3−z+p (3−z)2+r2¢¡ z+ 1 +p (z+ 1)2+r2¢ ¡ 1−z+p (1−z)2+r2¢¡ z+ 3 +p (z+ 3)2+r2¢ (15) 22 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.5.6 Cross-section of equipotential surfaces and lines of electric ﬁeld intensity for the conﬁguration of Fig. 4.5.5. Here, lengths have been normalized to d, so that z=z/dandr=r/d. Also, the potential has been normalized such that Φ≡Φ (λo/4π/epsilon1o)(16) A programmable calculator can be used to evaluate (15), given values of ( r, z). The equipotentials in Fig. 4.5.6 were, in fact, obtained in this way, making it possible to sketch the lines of ﬁeld intensity shown. Remember, the conﬁguration is axisym- metric, so the equipotentials are surfaces generated by rotating the cross-section shown about the zaxis. Two-Dimensional Charge and Field Distributions. In two-dimensional con- ﬁgurations, where the charge distribution uniformly extends from z=−∞ to z= +∞, one of the three integrations of the general superposition integral is carried out by representing the charge by a superposition of line charges, each ex- tending from z=−∞ toz= +∞. The fundamental element of charge, shown in Sec. 4.5 Solution of Poisson’s Equation 23 Fig. 4.5.7 For two-dimensional charge distributions, the elementary charge takes the form of a line charge of inﬁnite length. The observer and source position vectors, randr/prime, are two-dimensional vectors. Fig. 4.5.7, is not the point charge of (1) but rather an inﬁnitely long line charge. The associated potential is not that of a point charge but rather of a line charge. With the line charge distributed along the zaxis, the electric ﬁeld is given by (1.3.13) as Er=−∂Φ ∂r=λl 2π/epsilon1or(17) and integration of this expression gives the potential Φ =−λl 2π/epsilon1oln¡r ro¢ (18) where rois a reference radius brought in as a constant of integration. Thus, with da denoting an area element in the plane upon which the source and ﬁeld depend and randr/primethe vector positions of the observer and source respectively in that plane, the potential for the incremental line charge of Fig. 4.5.7 is written by making the identiﬁcations λl→ρ(r/prime)da/prime; r→ |r−r/prime| (19) Integration over the given two-dimensional source distribution then gives as the two-dimensional superposition integral Φ =−Z S/primeρ(r/prime)da/primeln|r−r/prime| 2π/epsilon1o (20) In dealing with charge distributions that extend to inﬁnity in the zdirection, the potential at inﬁnity can not be taken as a reference. The potential at an arbitrary ﬁnite position can be deﬁned as zero by adding an integration constant to (20). The following example leads to a result that will be found useful in solving boundary value problems in Sec. 4.8. Example 4.5.3. Two-Dimensional Potential of Uniformly Charged Sheet 24 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.5.8 Strip of uniformly charged material stretches to inﬁnity in the±zdirections, giving rise to two-dimensional potential distribution. A uniformly charged strip lying in the y= 0 plane between x=x2andx=x1 extends from z= +∞toz=−∞, as shown in Fig. 4.5.8. Because the thickness of the sheet in the ydirection is very small compared to other dimensions of interest, the integrand of (20) is essentially constant as the integration is carried out in the ydirection. Thus, the yintegration amounts to a multiplication by the thickness ∆ of the sheet ρ(r/prime)da/prime=ρ(r/prime)∆dx=σsdx (21) and (20) is written in terms of the surface charge density σsas Φ =−Z σs(x/prime)dx/primeln|r−r/prime| 2π/epsilon1o(22) If the distance between source and observer is written in terms of the Cartesian coordinates of Fig. 4.5.8, and it is recognized that the surface charge density is uniform so that σs=σois a constant, (22) becomes Φ =−σo 2π/epsilon1oZx1 x2lnp (x−x/prime)2+y2dx/prime(23) Introduction of the integration variable u=x−x/primeconverts this integral to an expression that is readily integrated. Φ =σo 2π/epsilon1oZx−x1 x−x2lnp u2+y2du =σo 2π/epsilon1o· (x−x1)lnp (x−x1)2+y2 −(x−x2)lnp (x−x2)2+y2+ytan−1¡x−x1 y¢ −ytan−1¡x−x2 y¢ + (x1−x2)¸(24) Two-dimensional distributions of surface charge can be piece-wise approximated by uniformly charged planar segments. The associated potentials are then represented by superpositions of the potential given by (24). Sec. 4.5 Solution of Poisson’s Equation 25 Potential of Uniform Dipole Layer. The potential produced by a dipole of charges ±qspaced a vector distance dapart has been found to be given by (4.4.11) Φ =p·ir/primer 4π/epsilon1o1 |r−r/prime|2(25) where p≡qd Adipole layer , shown in Fig. 4.5.9, consists of a pair of surface charge distributions ±σsspaced a distance dapart. An area element daof such a layer, with the direction of da(pointing from the negative charge density to the positive one), can be regarded as a diﬀerential dipole producing a (diﬀerential) potential dΦ dΦ =(σsd)da·ir/primer 4π/epsilon1o1 |r−r/prime|2(26) Denote the surface dipole density by πswhere πs≡σsd (27) and the potential produced by a surface dipole distribution over the surface Sis given by Φ =1 4π/epsilon1oZ Sπsir/primer |r−r/prime|2·da (28) This potential can be interpreted particularly simply if the dipole density is con- stant. Then πscan be pulled out from under the integral, and there Φ is equal to πs/(4π/epsilon1o) times the integral Ω≡Z Sir/primer·da/prime |r−r/prime|2(29) This integral is dimensionless and has a simple geometric interpretation. As shown in Fig. 4.5.9, ir/primer·dais the area element projected into the direction connecting the source point to the point of observation. Division by |r−r/prime|2reduces this projected area element onto the unit sphere. Thus, the integrand is the diﬀerential solid angle subtended by daas seen by an observer at r. The integral, (29), is equal to the solid angle subtended by the surface Swhen viewed from the point of observation r. In terms of this solid angle, Φ =πs 4π/epsilon1oΩ (30) Next consider the discontinuity of potential in passing through the surface Scontaining the dipole layer. Suppose that the surface Sis approached from the + side; then, from Fig. 4.5.10, the surface is viewed under the solid angle Ω o. 26 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.5.9 The diﬀerential solid angle subtended by dipole layer of area da. Fig. 4.5.10 The solid angle from opposite sides of dipole layer. Approached from the other side, the surface subtends the solid angle −(4π−Ωo). Thus, there is a discontinuity of potential across the surface of ∆Φ =πs 4π/epsilon1oΩo−πs 4π/epsilon1o(Ωs−4π) =πs /epsilon1o(31) Because the dipole layer contains an inﬁnite surface charge density σs, the ﬁeld within the layer is inﬁnite. The “fringing” ﬁeld, i.e., the external ﬁeld of the dipole layer, is ﬁnite and hence negligible in the evaluation of the internal ﬁeld of the dipole layer. Thus, the internal ﬁeld follows directly from Gauss’ law under the assumption that the ﬁeld exists solely between the two layers of opposite charge density (see Prob. 4.5.12). Because contributions to (28) are dominated by πsin the immediate vicinity of a point ras it approaches the surface, the discontinuity of potential is given by (31) even if πsis a function of position. In this case, the tangential Eis not continuous across the interface (Prob. 4.5.12). 4.6 ELECTROQUASISTATIC FIELDS IN THE PRESENCE OF PERFECT CONDUCTORS In most electroquasistatic situations, the surfaces of metals are equipotentials. In fact, if surrounded by insulators, the surfaces of many other conducting materials Sec. 4.6 Perfect Conductors 27 Fig. 4.6.1 Once the superposition principle has been used to determine the potential, the ﬁeld in a volume Vconﬁned by equipotentials is just as well induced by perfectly conducting electrodes having the shapes and potentials of the equipotentials they replace. also tend to form equipotential surfaces. The electrical properties and dynamical conditions required for representing a boundary surface of a material by an equipo- tential will be identiﬁed in Chap. 7. Consider the situation shown in Fig. 4.6.l, where three surfaces Si, i= 1,2,3 are held at the potentials Φ 1,Φ2,and Φ 3, respectively. These are presumably the surfaces of conducting electrodes. The ﬁeld in the volume Vsurrounding the sur- faces Siand extending to inﬁnity is not only due to the charge in that volume but due to charges outside that region as well. Fields normal to the boundaries terminate on surface charges. Thus, as far as the ﬁelds in the region of interest are concerned, the sources are the charge density in the volume V(if any) and the surface charges on the surrounding electrodes. The superposition integral, which is a solution to Poisson’s equation, gives the potential when the volume and surface charges are known. In the present statement of the problem, the volume charge densities are known in V, but the surface charge densities are not. The only fact known about the latter is that they must be so distributed as to make the Si’s into equipotential surfaces at the potentials Φ i. The determination of the charge distribution for the set of speciﬁed equipo- tential surfaces is not a simple matter and will occupy us in Chap. 5. But many interesting physical situations are uncovered by a diﬀerent approach. Suppose we are given a potential function Φ( r). Then any equipotential surface of that poten- tial can be replaced by an electrode at the corresponding potential. Some of the electrode conﬁgurations and associated ﬁelds obtained in this manner are of great practical interest. Suppose such a procedure has been followed. To determine the charge on the i-th electrode, it is necessary to integrate the surface charge density over the surface of the electrode. qi=Z Siσsda=Z Si/epsilon1oE·da (1) In the volume V, the contributions of the surface charges on the equipoten- tial surfaces are exactly equivalent to those of the charge distribution inside the regions enclosed by the surface Sicausing the original potential function. Thus, an alternative to the use of (1) for ﬁnding the total charge on the electrode is qi=Z Viρdv (2) 28 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.6.2 Pair of electrodes used to deﬁne capacitance. where Viis the volume enclosed by the surface Siandρis the charge density inside Siassociated with the original potential. Capacitance. Suppose the system consists of only two electrodes, as shown in Fig. 4.6.2. The charges on the surfaces of conductors (1) and (2) can be evaluated from the assumedly known solution by using (1). q1=I S1/epsilon1oE·da; q2=I S2/epsilon1oE·da (3) Further, there is a charge at inﬁnity of q∞=I S∞/epsilon1oE·da=−q1−q2 (4) The charge at inﬁnity is the negative of the sum of the charges on the two electrodes. This follows from the fact that the ﬁeld is divergence free, and all ﬁeld lines origi- nating from q1andq2must terminate at inﬁnity. Instead of the charges, one could specify the potentials of the two electrodes with respect to inﬁnity. If the charge on electrode 1 is brought to it by a voltage source (battery) that takes charge away from electrode 2 and deposits it on electrode 1, the normal process of charging up two electrodes, then q1=−q2. A capacitance Cbetween the two electrodes can be deﬁned as the ratio of charge on electrode 1 divided by the voltage between the two electrodes. In terms of the ﬁelds, this deﬁnition becomes C=H S1/epsilon1oE·da R(2) (1)E·ds(5) In order to relate this deﬁnition to the capacitance concept used in circuit theory, one further observation must be made. The capacitance relates the charge of one electrode to the voltage between the two electrodes. In general, there may also exist a voltage between electrode 1 and inﬁnity. In this case, capacitances must Sec. 4.6 Perfect Conductors 29 also be assigned to relate the voltage with regard to inﬁnity to the charges on the electrodes. If the electrodes are to behave as the single terminal-pair element of circuit theory, these capacitances must be negligible. Returning to (5), note that Cis independent of the magnitude of the ﬁeld variables. That is, if the magnitude of the charge distribution is doubled everywhere, it follows from the superposition integral that the potential doubles as well. Thus, the electric ﬁeld in the numerator and denominator of (3) is doubled everywhere. Each of the integrals therefore also doubles, their ratio remaining constant. Example 4.6.1. Capacitance of Isolated Spherical Electrodes A spherical electrode having radius a has a well-deﬁned capacitance Crelative to an electrode at inﬁnity. To determine C, note that the equipotentials of a point charge qat the origin Φ =q 4π/epsilon1or(6) are spherical. In fact, the equipotential having radius r=ahas a voltage with respect to inﬁnity of Φ =v=q 4π/epsilon1oa(7) The capacitance is deﬁned as the the net charge on the surface of the electrode per unit voltage, (5). But the net charge found by integrating the surface charge density over the surface of the sphere is simply q, and so the capacitance follows from (7) as C=q v= 4π/epsilon1oa (8) By way of illustrating the conditions necessary for the capacitance to be well deﬁned, consider a pair of spherical electrodes. Electrode (1) has radius a while electrode (2) has radius R. If these are separated by many times the larger of these radii, the potentials in their vicinities will again take the form of (6). Thus, with the voltages v1andv2deﬁned relative to inﬁnity, the charges on the respective spheres are q1= 4π/epsilon1oav1; q2= 4π/epsilon1oRv2 (9) With all of the charge on sphere (1) taken from sphere (2), q1=−q2⇒av1=−Rv2 (10) Under this condition, all of the ﬁeld lines from sphere (1) terminate on sphere (2). To determine the capacitance of the electrode pair, it is necessary to relate the charge q1to the voltage diﬀerence between the spheres. To this end, (9) is used to write q1 4π/epsilon1oa−q2 4π/epsilon1oR=v1−v2≡v (11) and because q1=−q2, it follows that q1=vC; C≡4π/epsilon1o¡1 a+1 R¢ (12) where Cis now the capacitance of one sphere relative to the other. 30 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.6.3 The Φ = 1 and Φ = 0 equipotentials of Fig. 4.5.6 are turned into perfectly conducting electrodes having the capacitance of (4.6.16). Note that in order to maintain no net charge on the two spheres, it follows from (9), (10), and (12) that the average of the voltages relative to inﬁnity must be retained at 1 2(v1+v2) =1 2µ q1 4π/epsilon1oa+q2 4π/epsilon1oR¶ =1 2v¡1 a−1 R¢ ¡1 a+1 R¢ (13) Thus, the average potential must be raised in proportion to the potential diﬀerence v. Example 4.6.2. Field and Capacitance of Shaped Electrodes The ﬁeld due to oppositely charged collinear line charges was found to be (4.5.15) in Example 4.5.2. The equipotential surfaces, shown in cross-section in Fig. 4.5.6, are melon shaped and tend to enclose one or the other of the line charge elements. Suppose that the surfaces on which the normalized potentials are equal to 1 and to 0, respectively, are turned into electrodes, as shown in Fig. 4.6.3. Now the ﬁeld lines originate on positive surface charges on the upper electrode and terminate on negative charges on the ground plane. By contrast with the original ﬁeld from the line charges, the ﬁeld in the region now inside the electrodes is zero. One way to determine the net charge on one of the electrodes requires that the electric ﬁeld be found by taking the gradient of the potential, that the unit normal vector to the surface of the electrode be determined, and hence that the surface charge be determined by evaluating /epsilon1oE·daon the electrode surface. Integration of this quantity over the electrode surface then gives the net charge. A far easier way to determine this net charge is to recognize that it is the same as the net charge enclosed by this surface for the original line charge conﬁguration. Thus, the net charge is simply 2 dλl, and if the potentials of the respective electrodes are taken as ±V, the capacitance is C≡q v=2dλl V(14) Sec. 4.6 Perfect Conductors 31 Fig. 4.6.4 Deﬁnition of coordinates for ﬁnding ﬁeld from line charges of opposite sign at x=±a. The displacement vectors are two dimen- sional and hence in the x−yplane. For the surface of the electrode in Fig. 4.6.3, V λl/4π/epsilon1o= 1⇒λl V= 4π/epsilon1o (15) It follows from these relations that the desired capacitance is simply C= 8π/epsilon1od (16) In these two examples, the charge density is zero everywhere between the electrodes. Thus, throughout the region of interest, Poisson’s equation reduces to Laplace’s equation. ∇2Φ = 0 (17) The solution to Poisson’s equation throughout all space is tantamount to solving Laplace’s equation in a limited region, subject to certain boundary conditions. A more direct approach to ﬁnding such solutions is taken in the next chapter. Even then, it is well to keep in mind that solutions to Laplace’s equation in a limited region are solutions to Poisson’s equation throughout the entire space, including those regions that contain the charges. The next example leads to an often-used result, the capacitance per unit length of a two-wire transmission line. Example 4.6.3. Potential of Two Oppositely Charged Conducting Cylinders The potential distribution between two equal and opposite parallel line charges has circular cylinders for its equipotential surfaces. Any pair of these cylinders can be replaced by perfectly conducting surfaces so as to obtain the solution to the potential set up between two perfectly conducting parallel cylinders of circular cross-section. We proceed in the following ways: (a) The potentials produced by two oppo- sitely charged parallel lines positioned at x= +aandx=−a, respectively, as shown in Fig. 4.6.4, are superimposed. (b) The intersections of the equipotential surfaces with the x−yplane are circles. The above results are used to ﬁnd the potential dis- tribution produced by two parallel circular cylinders of radius Rwith their centers spaced by a distance 2 l. (c) The cylinders carry a charge per unit length λland have a potential diﬀerence V, and so their capacitance per unit length is determined. (a) The potential associated with a single line charge on the zaxis is most easily obtained by integrating the electric ﬁeld, (1.3.13), found from Gauss’ integral 32 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.6.5 Cross-section of equipotentials and electric ﬁeld lines for line charges. law. It follows by superposition that the potential for two parallel line charges of charge per unit length + λland−λl, positioned at x= +aandx=−a, respectively, is Φ =−λl 2π/epsilon1oln r1+λl 2π/epsilon1oln r2=−λl 2π/epsilon1olnr1 r2(18) Here r1andr2are the distances of the ﬁeld point Pfrom the + and −line charges, respectively, as shown in Fig. 4.6.4. (b) On an equipotential surface, Φ = Uis a constant and the equation for that surface, (18), is r2 r1= exp¡2π/epsilon1oU λl¢ = const (19) where in Cartesian coordinates r2 2= (a+x)2+y2; r2 1= (a−x)2+y2 With the help of Fig. 4.6.4, (19) is seen to represent cylinders of circular cross-section with centers on the xaxis. This becomes apparent when the equation is expressed in Cartesian coordinates. The equipotential circles are shown in Fig. 4.6.5 for diﬀerent values of k≡expµ 2π/epsilon1oU λl¶ (20) (c) Given two conducting cylinders whose centers are a distance 2 lapart, as shown in Fig. 4.6.6, what is the location of the two line charges such that their ﬁeld Sec. 4.6 Perfect Conductors 33 Fig. 4.6.6 Cross-section of parallel circular cylinders with centers at x=±land line charges at x=±a, having equivalent ﬁeld. has equipotentials coincident with these two cylinders? In terms of kas deﬁned by (20), (19) becomes k2=(x+a)2+y2 (x−a)2+y2(21) This expression can be written as a quadratic function of xandy. x2−2xa(k2+ 1) (k2−1)+a2+y2= 0 (22) Equation (22) conﬁrms that the loci of constant potential in the x−yplane are indeed circles. In order to relate the radius and location of these circles to the parameters aandk, note that the expression for a circle having radius Rand center on the xaxis at x=lis (x−l)2+y2−R2= 0⇒x2−2xl+ (l2−R2) +y2= 0 (23) We can make (22) identical to this expression by setting −2l=−2a(k2+ 1) (k2−1)(24) and a2=l2−R2(25) Given the spacing 2 land radius Rof parallel conductors, this last expression can be used to locate the positions of the line charges. It also can be used to see that (l−a) =R2/(l+a), which can be used with (24) solved for k2to deduce that k=l+a R(26) Introduction of this expression into (20) then relates the potential of the cylinder on the right to the line charge density. The net charge per unit length that is actually on the surface of the right conductor is equal to the line charge density λl. With the voltage diﬀerence between the cylinders deﬁned as V= 2U, we can therefore solve for the capacitance per unit length. C=λl V=π/epsilon1o ln£l R+p (l/R)2−1¤ (27) 34 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.6.7 Cross-section of spherical electrode having radius Rand center at the origin of xaxis, showing charge qatx=X. Charge Q1atx=Dmakes spherical surface an equipotential, while Qoat origin makes the net charge on the sphere zero without disturbing the equipotential condition. Often, the cylinders are wires and it is appropriate to approximate this result for large ratios of l/R. l R+p (l/R)2−1 =l R£ 1 +p 1−(R/l)2¤ /similarequal2l R(28) Thus, the capacitance per unit length is approximately λl V≡C=π/epsilon1o ln2l R(29) This same result can be obtained directly from (18) by recognizing that when a/greatermuchl, the line charges are essentially at the center of the cylinders. Thus, evaluated on the surface of the right cylinder where the potential is V/2, r1/similarequalRandr2/similarequal2l, (18) gives (29). Example 4.6.4. Attraction of a Charged Particle to a Neutral Sphere A charged particle facing a conducting sphere induces a surface charge distribution on the sphere. This distribution adjusts itself so as to make the spherical surface an equipotential. In this problem, we take advantage of the fact that two charges of opposite sign produce a potential distribution, one equipotential surface of which is a sphere. First we ﬁnd the potential distribution set up by a perfectly conducting sphere of radius R, carrying a net charge Q, and a point charge qat a distance X(X≥R) from the center of the sphere. Then the result is used to determine the force on the charge qexerted by a neutral sphere ( Q= 0)! The conﬁguration is shown in Fig. 4.6.7. Consider ﬁrst the potential distribution set up by a point charge Q1and another point charge q. The construction of the potential is familiar from Sec. 4.4. Φ(r) =q 4π/epsilon1or2+Q1 4π/epsilon1or1(30) In general, the equipotentials are not spherical. However, the surface of zero potential Φ(r) = 0 =q 4π/epsilon1or2+Q1 4π/epsilon1or1(31) Sec. 4.7 Method of Images 35 is described byr2 r1=−q Q1(32) and if q/Q 1≤0, this represents a sphere. This can be proven by expressing (32) in Cartesian coordinates and noting that in the plane of the two charges, the result is the equation of a circle with its center on the axis intersecting the two charges [compare (19)]. Using this fact, we can apply (32) to the points AandBin Fig. 4.6.7 and eliminate q/Q 1. Taking Ras the radius of the sphere and Das the distance of the point charge Q1from the center of the sphere, it follows that R−D X−R=R+D X+R⇒D=R2 X(33) This speciﬁes the distance Dof the point charge Q1from the center of the equipo- tential sphere. Introduction of this result into (32) applied to point Agives the (ﬁctitious) charge Q1. −Q1=qR X(34) With this value for Q1located in accordance with (33), the surface of the sphere has zero potential. Without altering its equipotential character, the potential of the sphere can be shifted by positioning another ﬁctitious charge at its center. If the net charge of the spherical conductor is to be Q, then a charge Qo=Q−Q1is to be positioned at the center of the sphere. The net ﬁeld retains the sphere as an equipotential surface, now of nonzero potential. The ﬁeld outside the sphere is the sought-for solution. With r3deﬁned as the distance from the center of the sphere to the point of observation, the ﬁeld outside the sphere is Φ =q 4π/epsilon1or2+Q1 4π/epsilon1or1+Q−Q1 4π/epsilon1or3(35) With Q= 0, the force on the charge follows from an evaluation of the electric ﬁeld intensity directed along an axis passing through the center of the sphere and the charge q. The self-ﬁeld of the charge is omitted from this calculation. Thus, along thexaxis the potential due to the ﬁctitious charges within the sphere is Φ =Q1 4π/epsilon1o(x−D)−Q1 4π/epsilon1ox(36) Thexdirected electric ﬁeld intensity, and hence the required force, follows as fx=qEx=−q∂Φ ∂x=qQ1 4π/epsilon1o· 1 (x−D)2−1 x2¸ x=X(37) In view of (33) and (34), this can be written in terms of the actual physical quantities as fx=−q2R 4π/epsilon1oX3· 1£ 1−(R/X )2¤2−1¸ (38) The ﬁeld implied by (34) with Q= 0 is shown in Fig. 4.6.8. As the charge approaches the spherical conductor, images are induced on the nearest parts of the surface. To 36 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.6.8 Field of point charge in vicinity of neutral perfectly conducting spherical electrode. keep the net charge zero, charges of opposite sign must be induced on parts of the surface that are more remote from the point charge. The force of attraction results because the charges of opposite sign are closer to the point charge than those of the same sign. 4.7 METHOD OF IMAGES Given a charge distribution throughout all of space, the superposition integral can be used to determine the potential that satisﬁes Poisson’s equation. However, it is often the case that interest is conﬁned to a limited region, and the potential must satisfy a boundary condition on surfaces bounding this region. In the previous section, we recognized that any equipotential surface could be replaced by a physical electrode, and found solutions to boundary value problems in this way. The art of solving problems in this “backwards” fashion can be remarkably practical but hinges on having a good grasp of the relationship between ﬁelds and sources. Symmetry is often the basis for superimposing ﬁelds to satisfy boundary con- ditions. Consider for example the ﬁeld of a point charge a distance d/2 above a plane conductor, represented by an equipotential. As illustrated in Fig. 4.7.1a, the ﬁeldE+of the charge by itself has a component tangential to the boundary, and hence violates the boundary condition on the surface of the conductor. To satisfy this condition, forget the conductor and consider the ﬁeld of two charges of equal magnitude and opposite signs, spaced a distance 2 dapart. In the symmetry plane, the normal components add while the tangential components cancel. Thus, the composite ﬁeld is normal to the symmetry plane, as illustrated in the ﬁgure. In fact, the conﬁguration is the same as discussed in Sec. 4.4. The Sec. 4.7 Method of Images 37 Fig. 4.7.1 (a) Field of positive charge tangential to horizontal plane is can- celed by that of symmetrically located image charge of opposite sign. (b) Net ﬁeld of charge and its image. ﬁelds are as in Fig. 4.4.2a, where now the planar Φ = 0 surface is replaced by a conducting sheet. This method of satisfying the boundary conditions imposed on the ﬁeld of a point charge by a plane conductor by using an opposite charge at the mirror image position of the original charge, is called the method of images . The charge of opposite sign at the mirror-image position is the “image-charge.” Any superposition of charge pairs of opposite sign placed symmetrically on two sides of a plane results in a ﬁeld that is normal to the plane. An example is the ﬁeld of the pair of line charge elements shown in Fig. 4.5.6. With an electrode having the shape of the equipotential enclosing the upper line charge and a ground plane in the plane of symmetry, the ﬁeld is as shown in Fig. 4.6.3. This identiﬁcation of a physical situation to go with a known ﬁeld was used in the previous section. The method of images is only a special case involving planar equipotentials. To compare the replacement of the symmetry plane by a planar conductor, consider the following demonstration. Demonstration 4.7.1. Charge Induced in Ground Plane by Overhead Conductor The circular cylindrical conductor of Fig. 4.7.2, separated by a distance lfrom an equipotential (grounded) metal surface, has a voltage U=Uocosωt. The ﬁeld between the conductor and the ground plane is that of a line charge inside the con- ductor and its image below the ground plane. Thus, the potential is that determined in Example 4.6.3. In the Cartesian coordinates shown, (4.6.18), the deﬁnitions of r1 andr2with (4.6.19) and (4.6.25) (where U=V/2) provide the potential distribution Φ =−λl 2π/epsilon1olnp (a−x)2+y2 p (a+x)2+y2(1) The charge per unit length on the cylinder is [compare (4.6.27)] λl=CU; C=2π/epsilon1o ln· l R+q¡l R¢2−1¸ (2) 38 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.7.2 Charge induced on ground plane by overhead conductor is measured by probe. Distribution shown is predicted by (4.7.7). In the actual physical situation, images of this charge are induced on the surface of the ground plane. These can be measured by using a ﬂat probe that is connected through the cable to ground and insulated from the ground plane just below. The input resistance of the oscilloscope is low enough so that the probe surface is at essentially the same (zero) potential as the ground plane. What is the measured current, and hence voltage vo, as a function of the position Yof the probe? Given the potential, the surface charge is (1.3.17) σs=/epsilon1oEx(x= 0) = −/epsilon1o∂Φ ∂x¯¯¯¯ x=0(3) Evaluation of this expression using (1) gives σs=CU 2π· −(a−x) (a−x)2+y2−(a+x) (a+x)2+y2¸ x=0 =−CU πa a2+y2(4) Conservation of charge requires that the probe current be the time rate of change of the charge qon the probe surface. is=dq dt(5) Because the probe area is small, the integration of the surface charge over its surface is approximated by the product of the area and the surface charge evaluated at the position Yof its center. q=Z Aσsdydz/similarequalAσs (6) Sec. 4.8 Charge Simulation Approach to Boundary ValueProblems 39 Fig. 4.7.3 Image charges arranged to satisfy equipotential conditions in two planes. Thus, it follows from (4)–(6) that the induced voltage, vo=−Rsis, is vo=−Vosinωt1 1 + (Y/a)2; Vo≡RsACU oω aπ(7) This distribution of the induced signal with probe position is shown in Fig. 4.7.2. In the analysis, it is assumed that the plane x= 0, including the section of surface occupied by the probe, is constrained to zero potential. In ﬁrst computing the current to the probe using this assumption and then ﬁnding the probe voltage, we are clearly making an approximation that is valid only if the voltage is “small.” This can be insured by making the resistance Rssmall. The usual scope resistance is 1 MΩ. It may come as a surprise that such a resistance is treated here as a short. However, the voltage given by (7) is proportional to the frequency, so the value of acceptable resistance depends on the frequency. As the frequency is raised to the point where the voltage of the probe does begin to inﬂuence the ﬁeld distribution, some of the ﬁeld lines that originally terminated on the electrode are diverted to the grounded part of the plane. Also, charges of opposite polarity are induced on the other side of the probe. The result is an output signal that no longer increases with frequency. A frequency response of the probe voltage that does not increase linearly with frequency is therefore telltale evidence that the resistance is too large or the frequency too high. In the demonstration, where “desk-top” dimensions are typical, the frequency response is linear to about 100 Hz with a scope resistance of 1 MΩ. As the frequency is raised, the system becomes one with two excitations con- tributing to the potential distribution. The multiple terminal-pair systems treated in Sec. 5.1 start to model the full frequency response of the probe. Symmetry also motivates the use of image charges to satisfy boundary condi- tions on more than one planar surface. In Fig. 4.7.3, the objective is to ﬁnd the ﬁeld of the point charge in the ﬁrst quadrant with the planes x= 0 and y= 0 at zero potential. One image charge gives rise to a ﬁeld that satisﬁes one of the boundary conditions. The second is satisﬁed by introducing an image for the pairof charges. Once an image or a system of images has been found for a point charge, the same principle of images can be used for a continuous charge distribution. The charge density distributions have density distributions of image charges, and the total ﬁeld is again found using the superposition integral. Even where symmetry is not involved, charges located outside the region of interest to produce ﬁelds that satisfy boundary conditions are often referred to 40 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.8.1 (a) Surface of circular cylinder over a ground plane broken into planar segments, each having a uniform surface charge density. (b) Special case where boundaries are in planes y= constant. as image charges. Thus, the charge Q1located within the spherical electrode of Example 4.6.4 can be regarded as the image of q. 4.8 CHARGE SIMULATION APPROACH TO BOUNDARY VALUE PROBLEMS In solving a boundary value problem, we are in essence ﬁnding that distribution of charges external to the region of interest that makes the total ﬁeld meet the bound- ary conditions. Commonly, these external charges are actually on the surfaces of conductors bounding or embedded in the region of interest. By way of prepara- tion for the boundary value point of view taken in the next chapter, we consider in this section a direct approach to adjusting surface charges so that the ﬁelds meet prescribed boundary conditions on the potential. Analytically, the technique is cumbersome. However, with a computer, it becomes one of a class of powerful numerical techniques[1]for solving boundary value problems. Suppose that the ﬁelds are two dimensional, so that the region of interest can be “enclosed” by a surface that can be approximated by strip segments, as illustrated in Fig. 4.8.1a. This example becomes an approximation to the circular conductor over a ground plane (Example 4.7.1) if the magnitudes of the charges on the strips are adjusted to make the surfaces approximate appropriate equipotentials. With the surface charge density on each of these strips taken as uniform , a “stair-step” approximation to the actual distribution of charge is obtained. By increasing the number of segments, the approximation is reﬁned. For purposes of illustration, we conﬁne ourselves here to boundaries lying in planes of constant y, as shown in Fig. 4.8.1b. Then the potential associated with a single uniformly charged strip is as found in Example 4.5.3. Consider ﬁrst the potential due to a strip of width (a) lying in the plane y= 0 with its center at x= 0, as shown in Fig. 4.8.2a. This is a special case of the conﬁguration considered in Example 4.5.3. It follows from (4.5.24) with x1=a/2 andx2=−a/2 that the potential at the observer location ( x, y) is Φ(x, y) =σoS(x, y) (1) Sec. 4.8 Charge Simulation Approach 41 Fig. 4.8.2 (a) Charge strip of Fig. 4.5.8 centered at origin. (b) Charge strip translated so that its center is at ( X, Y). where S(x, y)≡·¡ x−a 2¢ lnr¡ x−a 2¢2+y2 −¡ x+a 2¢ lnr¡ x+a 2¢2+y2 +ytan−1¡x−a/2 y¢ −ytan−1¡x+a/2 y¢ +a¸ /2π/epsilon1o(2) With the strip located at ( x, y) = ( X, Y), as shown in Fig. 4.8.2b, this potential becomes Φ(x, y) =σoS(x−X, y−Y) (3) In turn, by superposition we can write the potential due to Nsuch strips, the one having the uniform surface charge density σibeing located at ( x, y) = (Xi, Yi). Φ(x, y) =NX i=1σiSi; Si≡S(x−Xi, y−Yi) (4) Given the surface charge densities, σi, the potential at any given location ( x, y) can be evaluated using this expression. We assume that the net charge on the strips is zero, so that their collective potential goes to zero at inﬁnity. With the strips representing surfaces that are constrained in potential (for example, perfectly conducting boundaries), the charge densities are adjusted to meet boundary conditions. Each strip represents part of an electrode surface. The potential Vjat the center of the j-th strip is set equal to the known voltage of the electrode to which it belongs. Evaluating (4) for the center of the j-th strip one obtains NX i=1σiSij=Vj; Sij≡S(xj−Xi, yj−Yi), j = 1, . . . N (5) 42 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. 4.8.3 Charge distribution on plane parallel electrodes approxi- mated by six uniformly charged strips. This statement can be made for each of the strips, so that it holds with j= 1, . . . N . These relations comprise Nequations that are linear in the Nunknowns σ1. . . σ N. ÃC11C12 . . . C21 . . . C NN!0 @σ1 ... σN1 A=0 @V1 ... VN1 A (6) The potentials V1. . . V Non the right are known, so these expressions can be solved for the surface charge densities. Thus, the potential that meets the approximate boundary conditions, (4), has been determined. We have found an approximation to the surface charge density needed to meet the potential boundary condition. Example 4.8.1. Fields of Finite Width Parallel Plate Capacitor In Fig. 4.8.3, the parallel plates of a capacitor are divided into six segments. The potentials at the centers of those in the top row are required to be V/2, while those in the lower row are −V/2. In this simple case of six segments, symmetry gives σ1=σ3=−σ4=−σ6, σ 2=−σ5 (7) and the six equations in six unknowns, (6) with N= 6, reduces to two equations in two unknowns. Thus, it is straightforward to write analytical expressions for the surface charge densities (See Prob. 4.8.1). The equipotentials and associated surface charge distributions are shown in Fig. 4.8.4 for increasing numbers of charge sheets. The ﬁrst is a reminder of the distribution of potential for uniformly charged sheets. Shown next are the equipo- tentials that result from using the six-segment approximation just evaluated. In the last case, 20 segments have been used and the inversion of (6) carried out by means of a computer. Sec. 4.8 Charge Simulation Approach 43 Fig. 4.8.4 Potential distributions using 2, 6, and 20 sheets to approxi- mate the ﬁelds of a plane parallel capacitor. Only the ﬁelds in the upper half-plane are shown. The distributions of surface charge density on the upper plate are shown to the right. Note that the approximate capacitance per unit length is C=1 VN/2X i=1b (N/2)σi (8) This section shows how the superposition integral point of view can be the basis for a numerical approach to solving boundary value problems. But as we 44 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 proceed to a more direct approach to boundary value problems, it is especially important to proﬁt from the physical insight inherent in the method used in this section. We have found a mathematical procedure for adjusting the distributions of surface charge so that boundaries are equipotentials. Conducting surfaces sur- rounded by insulating material tend to become equipotentials by similarly redis- tributing their surface charge. For example, consider how the surface charge redis- tributes itself on the parallel plates of Fig. 4.8.4. With the surface charge uniformly distributed, there is a strong electric ﬁeld tangential to the surface of the plate. In the upper plate, the charges move radially outward in response to this tangential ﬁeld. Thus, the charge redistributes itself as shown in the subsequent cases. The correct distribution of surface charge density is the one that makes this tangential electric ﬁeld approach zero, which it is when the surfaces become equipotentials. Thus, the surface charge density is higher near the edges of the plates than it is in the middle. The additional surface charges near the edges result in just that inward-directed electric ﬁeld which is needed to make the net ﬁeld perpendicular to the surfaces of the electrodes. We will ﬁnd in Sec. 8.6 that the solution to a class of two-dimensional MQS boundary value problems is completely analogous to that for EQS systems of perfect conductors. 4.9 SUMMARY The theme in this chapter is set by the two equations that determine E, given the charge density ρ. The ﬁrst of these, (4.0.1), requires that Ebe irrotational. Through the representation of Eas the negative gradient of the electric potential, Φ, it is eﬀectively integrated. E=−∇Φ (1) This gradient operator, determined in Cartesian coordinates in Sec. 4.1 and found in cylindrical and spherical coordinates in the problems of that section, is summarized in Table I. The associated gradient integral theorem, (4.1.16), is added for reference to the integral theorems of Gauss and Stokes in Table II. The substitution of (1) into Gauss’ law, the second of the two laws forming the theme of this chapter, gives Poisson’s equation. ∇2Φ =−ρ /epsilon1o(2) The Laplacian operator on the left, deﬁned as the divergence of the gradient of Φ, is summarized in the three standard coordinate systems in Table I. It follows from the linearity of (2) that the potential for the superposition of charge distributions is the superposition of potentials for the individual charge distributions. The potentials for dipoles and other singular charge distributions are therefore found by superimposing the potentials of point or line charges. The su- perposition integral formalizes the determination of the potential, given the distri- bution of charge. With the surface and line charges recognized as special (singular) volume charge densities, the second and third forms of the superposition integral Sec. 4.9 Summary 45 summarized in Table 4.9.1 follow directly from the ﬁrst. The fourth is convenient if the source and ﬁeld are two dimensional. Through Sec. 4.5, the charge density is regarded as given throughout all space. From Sec. 4.6 onward, a shift is made toward ﬁnding the ﬁeld in conﬁned regions of space bounded by surfaces of constant potential. At ﬁrst, the approach is oppor- tunistic. Given a solution, what problems have been solved? However, the numerical convolution method of Sec. 4.8 is a direct and practical approach to solving bound- ary value problems with arbitrary geometry. R E F E R E N C E S [1] R. F. Harrington, Field Computation by Moment Methods , MacMillan, NY (1968). 46 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 TABLE 4.9.1 SUPERPOSITION INTEGRALS FOR ELECTRIC POTENTIAL Volume Charge (4.5.3)Φ =Z V/primeρ(r/prime)dv/prime 4π/epsilon1o|r−r/prime| Surface Charge (4.5.5)Φ =I A/primeσs(r/prime)da/prime 4π/epsilon1o|r−r/prime| Line Charge (4.5.12)Φ =Z L/primeλl(r/prime)dl/prime 4π/epsilon1o|r−r/prime| Two-dimensional (4.5.20)Φ =−Z S/primeρ(r/prime)ln|r−r/prime|da/prime 2π/epsilon1o Double-layer (4.5.28)Φ =πs 4π/epsilon1oΩ Ω≡Z Sir/primer·da |r−r/prime|2 Sec. 4.1 Problems 47 P R O B L E M S 4.1 Irrotational Field Represented by Scalar Potential: The Gradient Operator and Gradient Integral Theorem 4.1.1 Surfaces of constant Φ that are spherical are given by Φ =Vo a2(x2+y2+z2) ( a) For example, the surface at radius ahas the potential Vo. (a) In Cartesian coordinates, what is grad(Φ)? (b) By the deﬁnition of the gradient operator, the unit normal nto an equipotential surface is n=∇Φ |∇Φ|(b) Evaluate nin Cartesian coordinates for the spherical equipotentials given by (a) and show that it is equal to ir, the unit vector in the radial direction in spherical coordinates. 4.1.2 For Example 4.1.1, carry out the integral of E·dsfrom the origin to ( x, y) = (a, a) along the line y=xand show that it is indeed equal to Φ(0 ,0)− Φ(a, a). 4.1.3 In Cartesian coordinates, three two-dimensional potential functions are Φ =Vox a(a) Φ =Voy a(b) Φ =Vo a2(x2−y2) ( c) (a) Determine Efor each potential. (b) For each function, make a sketch of Φ and Eusing the conventions of Fig. 4.1.3. (c) For each function, make a sketch using conventions of Fig. 4.1.4. 4.1.4∗A cylinder of rectangular cross-section is shown in Fig. P4.1.4. The electric potential inside this cylinder is Φ =ρo(t) /epsilon1o£¡π a¢2+¡π b¢2¤sinπ axsinπ by (a) 48 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. P4.1.4 where ρo(t) is a given function of time. (a) Show that the electric ﬁeld intensity is E=−ρo(t) /epsilon1o£¡π a¢2+¡π b¢2¤£π acosπ axsinπ byix +π bsinπ axcosπ byiy¤(b) (b) By direct evaluation, show that Eis irrotational. (c) Show that the charge density ρis ρ=ρo(t) sinπ axsinπ by (c) (d) Show that the tangential Eis zero on the boundaries. (e) Sketch the distributions of Φ , ρ, andEusing conventions of Figs. 2.7.3 and 4.1.3. (f) Compute the line integral of E·dsbetween the center and corner of the rectangular cross-section (points shown in Fig. P4.1.4) and show that it is equal to Φ( a/2, b/2, t). Why would you expect the integration to give the same result for any path joining the point (a) to any point on the wall? (g) Show that the net charge inside a length dof the cylinder in the z direction is Q=dρo4ab π2(d) ﬁrst by integrating the charge density over the volume and then by using Gauss’ integral law and integrating /epsilon1oE·daover the surface enclosing the volume. (h) Find the surface charge density on the electrode at y= 0 and use your result to show that the net charge on the electrode segment between x=a/4 and x= 3a/4 having depth dinto the paper is q=−√ 2a bdρo£¡π a¢2+¡π b¢2¤ (e) Sec. 4.1 Problems 49 (i) Show that the current, i(t), to this electrode segment is i=√ 2ad bdρo dt£¡π a¢2+¡π b¢2¤ (f) 4.1.5 Inside the cylinder of rectangular cross-section shown in Fig. P4.1.4, the potential is given as Φ =ρo(t) /epsilon1o£¡π a¢2+¡π b¢2¤cosπ axcosπ by (a) where ρo(t) is a given function of time. (a) Find E. (b) By evaluating the curl, show that Eis indeed irrotational. (c) Find ρ. (d) Show that Eis tangential to all of the boundaries. (e) Using the conventions of Figs. 2.7.3 and 4.1.3, sketch Φ , ρ, and E. (f) Use Eas found in part (a) to compute the integral of E·dsfrom (a) to (b) in Fig. P4.1.4. Check your answer by evaluating the potential diﬀerence between these points. (g) Evaluate the net charge in the volume by ﬁrst using Gauss’ integral law and integrating /epsilon1oE·daover the surface enclosing the volume and then by integrating ρover the volume. 4.1.6 Given the potential Φ =Asinhmxsinkyysinkzzsinωt (a) where A, m, and ωare given constants. (a) Find E. (b) By direct evaluation, show that Eis indeed irrotational. (c) Determine the charge density ρ. (d) Can you adjust mso that ρ= 0 throughout the volume? 4.1.7 The system, shown in cross-section in Fig. P4.1.7, extends to ±∞in the z direction. It consists of a cylinder having a square cross-section with sides which are resistive sheets (essentially many resistors in series). Thus, the voltage sources ±Vat the corners of the cylinder produce linear distribu- tions of potential along the sides. For example, the potential between the corners at ( a,0) and (0 , a) drops linearly from Vto−V. 50 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. P4.1.7 (a) Show that the potential inside the cylinder can match that on the walls of the cylinder if it takes the form A(x2−y2). What is A? (b) Determine Eand show that there is no volume charge density ρwithin the cylinder. (c) Sketch the equipotential surfaces and lines of electric ﬁeld intensity. 4.1.8 Figure P4.1.8 shows a cross-sectional view of a model for a “capacitance” probe designed to measure the depth hof penetration of a tool into a metallic groove. Both the “tool” and the groove can be considered con- stant potential surfaces having the potential diﬀerence v(t) as shown. An insulating segment at the tip of the tool is used as a probe to measure h. This is done by measuring the charge on the surface of the segment. In the following, we start with a ﬁeld distribution that can be made to ﬁt the problem, determine the charge and complete some instructive manipula- tions along the way. Fig. P4.1.8 (a) Given that the electric ﬁeld intensity between the groove and tool takes the form E=C[xix−yiy] ( a) show that Eis irrotational and evaluate the coeﬃcient Cby comput- ing the integral of E·dsbetween point (a) and the origin. Sec. 4.4 Problems 51 (b) Find the potential function consistent with (a) and evaluate Cby inspection. Check with part (a). (c) Using the conventions of Figs. 2.7.3 and 4.1.3, sketch lines of constant potential and electric ﬁeld Efor the region between the groove and the tool surfaces. (d) Determine the total charge on the insulated segment, given v(t). (Hint: Use the integral form of Gauss’ law with a convenient surface Senclosing the electrode.) 4.1.9∗In cylindrical coordinates, the incremental displacement vector, given in Cartesian coordinates by (9), is ∆r= ∆rir+r∆φiφ+ ∆ziz (a) Using arguments analogous to (7)–(12), show that the gradient operator in cylindrical coordinates is as given in Table I at the end of the text. 4.1.10∗Using arguments analogous to those of (7)–(12), show that the gradient operator in spherical coordinates is as given in Table I at the end of the text. 4.2 Poisson’s Equation 4.2.1∗In Prob. 4.1.4, the potential Φ is given by (a). Use Poisson’s equation to show that the associated charge density is as given by (c) of that problem. 4.2.2 In Prob. 4.1.5, Φ is given by (a). Use Poisson’s equation to ﬁnd the charge density. 4.2.3 Use the expressions for the divergence and gradient in cylindrical coor- dinates from Table I at the end of the text to show that the Laplacian operator is as summarized in that table. 4.2.4 Use the expressions from Table I at the end of the text for the divergence and gradient in spherical coordinates to show that the Laplacian operator is as summarized in that table. 4.3 Superposition Principle 4.3.1 A current source I(t) is connected in parallel with a capacitor Cand a resistor R. Write the ordinary diﬀerential equation that can be solved for the voltage v(t) across the three parallel elements. Follow steps analogous to those used in this section to show that if Ia(t)⇒va(t) and Ib(t)⇒vb(t), then Ia(t) +Ib(t)⇒va(t) +vb(t). 52 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 4.4 Fields Associated with Charge Singularities 4.4.1∗A two-dimensional ﬁeld results from parallel uniform distributions of line charge, + λlatx=d/2, y= 0 and −λlatx=−d/2, y= 0, as shown in Fig. P4.4.1. Thus, the potential distribution is independent of z. Fig. P4.4.1 (a) Start with the electric ﬁeld of a line charge, (1.3.13), and determine Φ. (b) Deﬁne the two-dimensional dipole moment as pλ=dλland show that in the limit where d→0 (while this moment remains constant), the electric potential is Φ =pλ 2π/epsilon1ocosφ r(a) 4.4.2∗For the conﬁguration of Prob. 4.4.1, consider the limit in which the line charge spacing dgoes to inﬁnity. Show that, in polar coordinates, the po- tential distribution is of the form Φ→Arcosφ (a) Express this in Cartesian coordinates and show that the associated Eis uniform. 4.4.3 A two-dimensional charge distribution is formed by pairs of positive and negative line charges running parallel to the zaxis. Shown in cross-section in Fig. P4.4.3, each line is at a distance d/2 from the origin. Show that in the limit where d/lessmuchr, this potential takes the form Acos 2φ/rn. What are the constants Aandn? 4.4.4 The charge distribution described in Prob. 4.4.3 is now at inﬁnity ( d/greatermuchr). (a) Show that the potential in the neighborhood of the origin takes the form A(x2−y2). (b) How would you position the line charges so that in the limit where they moved to inﬁnity, the potential would take the form of (4.1.18)? 4.5 Solution of Poisson’s Equation for Speciﬁed Charge Distributions Sec. 4.5 Problems 53 Fig. P4.4.3 Fig. P4.5.1 4.5.1 The only charge is restricted to a square patch centered at the origin and lying in the x−yplane, as shown in Fig. P4.5.1. (a) Assume that the patch is very thin in the zdirection compared to other dimensions of interest. Over its surface there is a given surface charge density σs(x, y). Express the potential Φ along the zaxis for z >0 in terms of a two-dimensional integral. (b) For the particular surface charge distribution σs=σo|xy|/a2where σoandaare constants, determine Φ along the positive zaxis. (c) What is Φ at the origin? (d) Show that Φ has a zdependence for z/greatermuchathat is the same as for a point charge at the origin. In this limit, what is the equivalent point charge for the patch? (e) What is Ealong the positive zaxis? 4.5.2∗The highly insulating spherical shell of Fig. P4.5.2 has radius Rand is “coated” with a surface charge density σs=σocosθ, where σois a given constant. 54 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. P4.5.2 (a) Show that the distribution of potential along the zaxis in the range z > R is Φ =σoR3 3/epsilon1oz2(a) [Hint: Remember that for the triangle shown in the ﬁgure, the law of cosines gives c= (b2+a2−2abcosα)1/2.] (b) Show that the potential distribution for the range z < R along the z axis inside the shell is Φ =σoz 3/epsilon1o(b) (c) Show that along the zaxis,Eis E=iz( 2σoR3 3/epsilon1oz3R < z −σo 3/epsilon1oR > z(c) (d) By comparing the zdependence of the potential to that of a dipole polarized in the zdirection, show that the equivalent dipole moment isqd= (4π/3)σoR3. 4.5.3 All of the charge is on the surface of a cylindrical shell having radius R and length 2 l, as shown in Fig. P4.5.3. Over the top half of this cylinder at r=Rthe surface charge density is σo(coulomb/m2), where σois a positive constant, while over the lower half it is −σo. (a) Find the potential distribution along the zaxis. (b) Determine Ealong the zaxis. (c) In the limit where z/greatermuchl, show that Φ becomes that of a dipole at the origin. What is the equivalent dipole moment? 4.5.4∗A uniform line charge of density λland length dis distributed parallel to the yaxis and centered at the point ( x, y, z ) = (a,0,0), as shown in Fig. P4.5.4. Use the superposition integral to show that the potential Φ( x, y, z ) is Φ =λl 4π/epsilon1oln·d 2−y+q (x−a)2+¡d 2−y¢2+z2 −d 2−y+q (x−a)2+¡d 2+y¢2+z2¸ (a) Sec. 4.5 Problems 55 Fig. P4.5.3 Fig. P4.5.4 Fig. P4.5.5 4.5.5 Charge is distributed with density λl=±λox/lcoulomb/m along the lines z=±a, y= 0, respectively, between the points x= 0 and x=l, as shown in Fig. P4.5.5. Take λoas a given charge per unit length and note that λlvaries from zero to λoover the lengths of the line charge distributions. Determine the distribution of Φ along the zaxis in the range 0 < z < a . 4.5.6 Charge is distributed along the zaxis such that the charge per unit length λl(z) is given by λl=½λoz a−a < z < a 0 z <−a;a < z(a) Determine Φ and Eat a position z > a on the zaxis. 56 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. P4.5.9 4.5.7∗A strip of charge lying in the x−zplane between x=−bandx=bextends to±∞in the zdirection. On this strip the surface charge density is σs=σo(d−b) (d−x)(a) where d > b . Show that at the location ( x, y) = (d,0), the potential is Φ(d,0) =σo 4π/epsilon1o(d−b){[ln(d−b)]2−[ln(d+b)]2} (b) 4.5.8 A pair of charge strips lying in the x−zplane and running from z= +∞to z=−∞are each of width 2 dwith their left and right edges, respectively, located on the zaxis. The one between the zaxis and ( x, y) = (2 d,0) has a uniform surface charge density σo, while the one between ( x, y) = (−2d,0) and the zaxis has σs=−σo. (Note that the symmetry makes the plane x= 0 one of zero potential.) What must be the value of σoif the potential at the center of the right strip, where ( x, y) = (d,0), is to be V? 4.5.9∗Distributions of line charge can be approximated by piecing together uni- formly charged segments. Especially if a computer is to be used to carry out the integration by summing over the ﬁelds due to the linear elements of line charge, this provides a convenient basis for calculating the electric potential for a given line distribution of charge. In the following, you de- termine the potential at an arbitrary observer coordinate rdue to a line charge that is uniformly distributed between the points r+bandr+c, as shown in Fig. P4.5.9a. The segment over which this charge (of line charge density λl) is distributed is denoted by the vector a, as shown in the ﬁgure. Viewed in the plane in which the position vectors a,b, and clie, a coordinate ξdenoting the position along the line charge is as shown in Fig. P4.5.9b. The origin of this coordinate is at the position on the line segment collinear with athat is nearest to the observer position r. Sec. 4.5 Problems 57 (a) Argue that in terms of ξ, the base and tip of the avector are as designated in Fig. P4.5.9b along the ξaxis. (b) Show that the superposition integral for the potential due to the seg- ment of line charge at r/primeis Φ =Zb·a/|a| c·a/|a|λldξ 4π/epsilon1o|r−r/prime|(a) where |r−r/prime|=s ξ2+|b×a|2 |a|2(b) (c) Finally, show that the potential is Φ =λ 4π/epsilon1oln¯¯¯¯b·a |a|+q¡b·a |a|¢2+|b×a|2 |a|2¯¯¯¯ ¯¯¯¯c·a |a|+s ¡c·a |a|¢2+|b×a|2 |a|2¯¯¯¯(c) (d) A straight segment of line charge has the uniform density λobetween the points ( x, y, z ) = (0 ,0, d) and ( x, y, z ) = (d, d, d ). Using (c), show that the potential φ(x, y, z ) is Φ =λo 4π/epsilon1oln¯¯¯¯2d−x−y+p 2[(d−x)2+ (d−y)2+ (d−z)2] −x−y+p 2[x2+y2+ (d−z)2]¯¯¯¯(d) 4.5.10∗Given the charge distribution, ρ(r), the potential Φ follows from (3). This expression has the disadvantage that to ﬁnd E, derivatives of Φ must be taken. Thus, it is not enough to know Φ at one location if Eis to be determined. Start with (3) and show that a superposition integral for the electric ﬁeld intensity is E=1 4π/epsilon1oZ V/primeρ(r/prime)ir/primerdv/prime |r−r/prime|2(a) where ir/primeris a unit vector directed from the source coordinate r/primeto the ob- server coordinate r. (Hint: Remember that when the gradient of Φ is taken to obtain E, the derivatives are with respect to the observer coordinates with the source coordinates held ﬁxed.) A similar derivation is given in Sec. 8.2, where an expression for the magnetic ﬁeld intensity His obtained from a superposition integral for the vector potential A. 4.5.11 For a better understanding of the concepts underlying the derivation of the superposition integral for Poisson’s equation, consider a hypothetical situation where a somewhat diﬀerent equation is to be solved. The charge 58 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 density is assumed in part to be a predetermined density s(x, y, z ), and in part to be induced at a given point ( x, y, z ) in proportion to the potential itself at that same point. That is, ρ=s−/epsilon1oκ2Φ ( a) (a) Show that the expression to be satisﬁed by Φ is then not Poisson’s equation but rather ∇2Φ−κ2Φ =−s /epsilon1o(b) where s(x, y, z ) now plays the role of ρ. (b) The ﬁrst step in the derivation of the superposition integral is to ﬁnd the response to a point source at the origin, deﬁned such that lim R→0ZR 0s4πr2dr=Q (c) Because the situation is then spherically symmetric, the desired re- sponse to this point source must be a function of ronly. Thus, for this response, (b) becomes 1 r2∂ ∂r¡ r2∂Φ ∂r¢ −κ2Φ =−s /epsilon1o(d) Show that for r/negationslash= 0, a solution is Φ =Ae−κr r(e) and use (c) to show that A=Q/4π/epsilon1o. (c) What is the superposition integral for Φ? 4.5.12∗Because there is a jump in potential across a dipole layer, given by (31), there is an inﬁnite electric ﬁeld within the layer. (a) With ndeﬁned as the unit normal to the interface, argue that this internal electric ﬁeld is Eint=−/epsilon1oσsn (a) (b) In deriving the continuity condition on E, (1.6.12), using (4.1.1), it was assumed that Ewas ﬁnite everywhere, even within the interface. With a dipole layer, this assumption cannot be made. For example, suppose that a nonuniform dipole layer πs(x) is in the plane y= 0. Show that there is a jump in tangential electric ﬁeld, Ex, given by Ea x−Eb x=−/epsilon1o∂πs ∂x(b) Sec. 4.6 Problems 59 Fig. P4.6.1 4.6 Electroquasistatic Fields in the Presence of Perfect Conductors 4.6.1∗A charge distribution is represented by a line charge between z=cand z=balong the zaxis, as shown in Fig. P4.6.1a. Between these points, the line charge density is given by λl=λo(a−z) (a−c)(a) and so it has the distribution shown in Fig. P4.6.1b. It varies linearly from the value λowhere z=ctoλo(a−b)/(a−c) where z=b. The only other charges in the system are at inﬁnity, where the potential is deﬁned as being zero. An equipotential surface for this charge distribution passes through the point z=aon the zaxis. [This is the same “ a” as appears in (a).] If this equipotential surface is replaced by a perfectly conducting electrode, show that the capacitance of the electrode relative to inﬁnity is C= 2π/epsilon1o(2a−c−b) ( b) 4.6.2 Charges at “inﬁnity” are used to impose a uniform ﬁeld E=Eoizon a region of free space. In addition to the charges that produce this ﬁeld, there are positive and negative charges, of magnitude q, atz= +d/2 and z=−d/2, respectively, as shown in Fig. P4.6.2. Spherical coordinates (r, θ, φ ) are deﬁned in the ﬁgure. (a) The potential, radial coordinate and charge are normalized such that Φ=Φ Eod; r=r d; q=q 4π/epsilon1oEod2(a) Show that the normalized electric potential Φ can be written as Φ=−rcosθ+q©£ r2+1 4−rcosθ¤−1/2−£ r2+1 4+rcosθ¤−1/2}(b) 60 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. P4.6.2 (b) There is an equipotential surface Φ = 0 that encloses these two charges. Thus, if a “perfectly conducting” object having a surface tak- ing the shape of this Φ = 0 surface is placed in the initially uniform electric ﬁeld, the result of part (a) is a solution to the boundary value problem representing the potential, and hence electric ﬁeld, around the object. The following establishes the shape of the object. Use (b) to ﬁnd an implicit expression for the radius rat which the surface intersects the zaxis. Use a graphical solution to show that there will always be such an intersection with r > d/ 2. For q= 2, ﬁnd this radius to two-place accuracy. (c) Make a plot of the surface Φ = 0 in a φ= constant plane. One way to do this is to use a programmable calculator to evaluate Φ given r andθ. It is then straightforward to pick a θand iterate on rto ﬁnd the location of the surface of zero potential. Make q= 2. (d) We expect Eto be largest at the poles of the object. Thus, it is in these regions that we expect electrical breakdown to ﬁrst occur. In terms of Eoand with q= 2, what is the electric ﬁeld at the north pole of the object? (e) In terms of Eoandd, what is the total charge on the northern half of the object. [Hint: A numerical calculation is notrequired.] 4.6.3∗For the disk of charge shown in Fig. 4.5.3, there is an equipotential surface that passes through the point z=don the zaxis and encloses the disk. Show that if this surface is replaced by a perfectly conducting electrode, the capacitance of this electrode relative to inﬁnity is C=2πR2/epsilon1o (√ R2+d2−d)(a) 4.6.4 The purpose of this problem is to get an estimate of the capacitance of, and the ﬁelds surrounding, the two conducting spheres of radius Rshown in Fig. P4.6.4, with the centers separated by a distance h. We construct Sec. 4.6 Problems 61 Fig. P4.6.4 an approximate ﬁeld solution for the ﬁeld produced by charges ±Qon the two spheres, as follows: (a) First we place the charges at the centers of the spheres. If R/lessmuchh, the two equipotentials surrounding the charges at r1≈Randr2≈R are almost spherical. If we assume that they arespherical, what is the potential diﬀerence between the two spherical conductors? Where does the maximum ﬁeld occur and how big is it? (b) We can obtain a better solution by noting that a spherical equipo- tential coincident with the top sphere is produced by a set of three charges. These are the charge −Qatz=−h/2 and the two charges inside the top sphere properly positioned according to (33) of appro- priate magnitude and total charge + Q. Next, we replace the charge −Qby two charges, just like we did for the charge + Q. The net ﬁeld is now due to four charges. Find the potential diﬀerence and capaci- tance for the new ﬁeld conﬁguration and compare with the previous result. Do you notice that you have obtained higher-order terms in R/h? You are in the process of obtaining a rapidly convergent series in powers of R/h. 4.6.5 This is a continuation of Prob. 4.5.4. The line distribution of charge given there is the only charge in the region 0 ≤x. However, the y−zplane is now a perfectly conducting surface, so that the electric ﬁeld is normal to the plane x= 0. (a) Determine the potential in the half-space 0 ≤x. (b) For the potential found in part (a), what is the equation for the equipotential surface passing through the point ( x, y, z ) = (a/2,0,0)? (c) For the remainder of this problem, assume that d= 4a. Make a sketch of this equipotential surface as it intersects the plane z= 0. In doing this, it is convenient to normalize xandytoaby deﬁning ξ=x/aand 62 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 η=y/a. A good way to make the plot is then to compute the potential using a programmable calculator. By iteration, you can quickly zero in on points of the desired potential. It is suﬃcient to show that in addition to the point of part (a), your curve passes through three well-deﬁned points that suggest its being a closed surface. (d) Suppose that this closed surface having potential Vis actually a metallic (perfect) conductor. Sketch the lines of electric ﬁeld intensity in the region between the electrode and the ground plane. (e) The capacitance of the electrode relative to the ground plane is de- ﬁned as C=q/V, where qis the total charge on the surface of the electrode having potential V. For the electrode of part (c), what is C? 4.7 Method of Images 4.7.1∗A point charge Qis located on the zaxis a distance dabove a perfect conductor in the plane z= 0. (a) Show that Φ above the plane is Φ =Q 4π/epsilon1o½1 [x2+y2+ (z−d)2]1/2 −1 [x2+y2+ (z+d)2]1/2¾ (a) (b) Show that the equation for the equipotential surface Φ = Vpassing through the point z=a < d is [x2+y2+ (z−d)2]−1/2−[x2+y2+ (z+d)2]−1/2 =2a d2−a2(b) (c) Use intuitive arguments to show that this surface encloses the point charge. In terms of a, d, and /epsilon1o, show that the capacitance relative to the ground plane of an electrode having the shape of this surface is C=2π/epsilon1o(d2−a2) a(c) 4.7.2 A positive uniform line charge is along the zaxis at the center of a perfectly conducting cylinder of square cross-section in the x−yplane. (a) Give the location and sign of the image line charges. (b) Sketch the equipotentials and Elines in the x−yplane. Sec. 4.7 Problems 63 Fig. P4.7.3 4.7.3 When a bird perches on a dc high-voltage power line and then ﬂies away, it does so carrying a net charge. (a) Why? (b) For the purpose of measuring this net charge Qcarried by the bird, we have the apparatus pictured in Fig. P4.7.3. Flush with the ground, a strip electrode having width wand length lis mounted so that it is insulated from ground. The resistance, R, connecting the electrode to ground is small enough so that the potential of the electrode (like that of the surrounding ground) can be approximated as zero. The bird ﬂies in the xdirection at a height habove the ground with a velocity U. Thus, its position is taken as y=handx=Ut. (c) Given that the bird has ﬂown at an altitude suﬃcient to make it appear as a point charge, what is the potential distribution? (d) Determine the surface charge density on the ground plane at y= 0. (e) At a given instant, what is the net charge, q, on the electrode? (As- sume that the width wis small compared to hso that in an integration over the electrode surface, the integration in the zdirection is simply a multiplication by w.) (f) Sketch the time dependence of the electrode charge. (g) The current through the resistor is dq/dt . Find an expression for the voltage, v, that would be measured across the resistance, R, and sketch its time dependence. 4.7.4∗Uniform line charge densities + λland−λlrun parallel to the zaxis at x=a, y= 0 and x=b, y= 0, respectively. There are no other charges in the half-space 0 < x. The y−zplane where x= 0 is composed of ﬁnely segmented electrodes. By connecting a voltage source to each segment, the potential in the x= 0 plane can be made whatever we want. Show that the potential distribution you would impose on these electrodes to insure that there is no normal component of Ein the x= 0 plane, Ex(0, y, z), is Φ(0, y, z) =−λl 2π/epsilon1oln(a2+y2) (b2+y2)(a) 64 Electroquasistatic Fields: The Superposition Integral Point of View Chapter 4 Fig. P4.7.5 4.7.5 The two-dimensional system shown in cross-section in Fig. P4.7.5 consists of a uniform line charge at x=d, y=dthat extends to inﬁnity in the ±z directions. The charge per unit length in the zdirection is the constant λ. Metal electrodes extend to inﬁnity in the x= 0 and y= 0 planes. These electrodes are grounded so that the potential in these planes is zero. (a) Determine the electric potential in the region x >0, y > 0. (b) An equipotential surface passes through the line x=a, y=a(a < d ). This surface is replaced by a metal electrode having the same shape. In terms of the given constants a, d, and /epsilon1o, what is the capacitance per unit length in the zdirection of this electrode relative to the ground planes? 4.7.6∗The disk of charge shown in Fig. 4.5.3 is located at z=srather than z= 0. The plane z= 0 consists of a perfectly conducting ground plane. (a) Show that for 0 < z, the electric potential along the zaxis is given by Φ =σo 2/epsilon1o·¡p R2+ (z−s)2− |z−s|¢ −¡p R2+ (z+s)2− |z+s|¢¸ (a) (b) Show that the capacitance relative to the ground plane of an electrode having the shape of the equipotential surface passing through the point z=d < s on the zaxis and enclosing the disk of charge is C=2πR2/epsilon1o£p R2+ (d−s)2−p R2+ (d+s)2+ 2d¤ (b) 4.7.7 The disk of charge shown in Fig. P4.7.7 has radius Rand height habove a perfectly conducting plane. It has a surface charge density σs=σor/R. A perfectly conducting electrode has the shape of an equipotential surface Sec. 4.8 Problems 65 Fig. P4.7.7 that passes through the point z=a < h on the zaxis and encloses the disk. What is the capacitance of this electrode relative to the plane z= 0? 4.7.8 A straight segment of line charge has the uniform density λobetween the points ( x, y, z ) = (0 ,0, d) and ( x, y, z ) = (d, d, d ). There is a perfectly con- ducting material in the plane z= 0. Determine the potential for z≥0. [See part (d) of Prob. 4.5.9.] 4.8 Charge Simulation Approach to Boundary Value Problems 4.8.1 For the six-segment approximation to the ﬁelds of the parallel plate ca- pacitor in Example 4.8.1, determine the respective strip charge densities in terms of the voltage Vand dimensions of the system. What is the approx- imate capacitance? 5 ELECTROQUASISTATIC FIELDS FROM THE BOUNDARY VALUE POINT OF VIEW 5.0 INTRODUCTION The electroquasistatic laws were discussed in Chap. 4. The electric ﬁeld intensity Eis irrotational and represented by the negative gradient of the electric potential. E=−∇Φ (1) Gauss’ law is then satisﬁed if the electric potential Φ is related to the charge density ρby Poisson’s equation ∇2Φ =−ρ /epsilon1o(2) In charge-free regions of space, Φ obeys Laplace’s equation, (2), with ρ= 0. The last part of Chap. 4 was devoted to an “opportunistic” approach to ﬁnding boundary value solutions. An exception was the numerical scheme described in Sec. 4.8 that led to the solution of a boundary value problem using the source- superposition approach. In this chapter, a more direct attack is made on solving boundary value problems without necessarily resorting to numerical methods. It is one that will be used extensively not only as eﬀects of polarization and conduction are added to the EQS laws, but in dealing with MQS systems as well. Once again, there is an analogy useful for those familiar with the description of linear circuit dynamics in terms of ordinary diﬀerential equations. With time as the independent variable, the response to a drive that is turned on when t= 0 can be determined in two ways. The ﬁrst represents the response as a superposition of impulse responses. The resulting convolution integral represents the response for all time, before and after t= 0 and even when t= 0. This is the analogue of the point of view taken in the ﬁrst part of Chap. 4. The second approach represents the history of the dynamics prior to when t= 0 in terms of initial conditions. With the understanding that interest is con- ﬁned to times subsequent to t= 0, the response is then divided into “particular” 1 2 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 and “homogeneous” parts. The particular solution to the diﬀerential equation rep- resenting the circuit is not unique, but insures that at each instant in the temporal range of interest, the diﬀerential equation is satisﬁed. This particular solution need not satisfy the initial conditions. In this chapter, the “drive” is the charge density, and the particular potential response guarantees that Poisson’s equation, (2), is satisﬁed everywhere in the spatial region of interest. In the circuit analogue, the homogeneous solution is used to satisfy the ini- tial conditions. In the ﬁeld problem, the homogeneous solution is used to satisfy boundary conditions. In a circuit, the homogeneous solution can be thought of as the response to drives that occurred prior to when t= 0 (outside the temporal range of interest). In the determination of the potential distribution, the homoge- neous response is one predicted by Laplace’s equation, (2), with ρ= 0, and can be regarded either as caused by ﬁctitious charges residing outside the region of interest or as caused by the surface charges induced on the boundaries. The development of these ideas in Secs. 5.1–5.3 is self-contained and does not depend on a familiarity with circuit theory. However, for those familiar with the solution of ordinary diﬀerential equations, it is satisfying to see that the approaches used here for dealing with partial diﬀerential equations are a natural extension of those used for ordinary diﬀerential equations. Although it can often be found more simply by other methods, a particu- lar solution always follows from the superposition integral. The main thrust of this chapter is therefore toward a determination of homogeneous solutions, of ﬁnd- ing solutions to Laplace’s equation. Many practical conﬁgurations have boundaries that are described by setting one of the coordinate variables in a three-dimensional coordinate system equal to a constant. For example, a box having rectangular cross- sections has walls described by setting one Cartesian coordinate equal to a constant to describe the boundary. Similarly, the boundaries of a circular cylinder are natu- rally described in cylindrical coordinates. So it is that there is great interest in hav- ing solutions to Laplace’s equation that naturally “ﬁt” these conﬁgurations. With many examples interwoven into the discussion, much of this chapter is devoted to cataloging these solutions. The results are used in this chapter for describing EQS ﬁelds in free space. However, as eﬀects of polarization and conduction are added to the EQS purview, and as MQS systems with magnetization and conduction are considered, the homogeneous solutions to Laplace’s equation established in this chapter will be a continual resource. A review of Chap. 4 will identify many solutions to Laplace’s equation. As long as the ﬁeld source is outside the region of interest, the resulting potential obeys Laplace’s equation. What is diﬀerent about the solutions established in this chapter? A hint comes from the numerical procedure used in Sec. 4.8 to satisfy arbitrary boundary conditions. There, a superposition of Nsolutions to Laplace’s equation was used to satisfy conditions at Npoints on the boundaries. Unfortunately, to determine the amplitudes of these Nsolutions, Nequations had to be solved for Nunknowns. The solutions to Laplace’s equation found in this chapter can also be used as the terms in an inﬁnite series that is made to satisfy arbitrary boundary conditions. But what is diﬀerent about the terms in this series is their orthogonality. This property of the solutions makes it possible to explicitly determine the individual amplitudes in the series. The notion of the orthogonality of functions may already Sec. 5.1 Particular and Homogeneous Solutions 3 Fig. 5.1.1 Volume of interest in which there can be a distribution of charge density. To illustrate bounding surfaces on which potential is constrained, n isolated surfaces and one enclosing surface are shown. be familiar through an exposure to Fourier analysis. In any case, the fundamental ideas involved are introduced in Sec. 5.5. 5.1 PARTICULAR AND HOMOGENEOUS SOLUTIONS TO POISSON’S AND LAPLACE’S EQUATIONS Suppose we want to analyze an electroquasistatic situation as shown in Fig. 5.1.1. A charge distribution ρ(r) is speciﬁed in the part of space of interest, designated by the volume V. This region is bounded by perfect conductors of speciﬁed shape and location. Known potentials are applied to these conductors and the enclosing surface, which may be at inﬁnity. In the space between the conductors, the potential function obeys Poisson’s equation, (5.0.2). A particular solution of this equation within the prescribed volume Vis given by the superposition integral, (4.5.3). Φp(r) =Z V/primeρ(r/prime)dv/prime 4π/epsilon1o|r−r/prime|(1) This potential obeys Poisson’s equation at each point within the volume V. Since we do not evaluate this equation outside the volume V, the integration over the sources called for in (1) need include no sources other than those within the volume V. This makes it clear that the particular solution is not unique, because the addition to the potential made by integrating over arbitrary charges outside the volume Vwill only give rise to a potential, the Laplacian derivative of which is zero within the volume V. Is (1) the complete solution? Because it is not unique, the answer must be, surely not. Further, it is clear that no information as to the position and shape of the conductors is built into this solution. Hence, the electric ﬁeld obtained as the negative gradient of the potential Φ pof (1) will, in general, possess a ﬁnite tangential component on the surfaces of the electrodes. On the other hand, the conductors 4 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 have surface charge distributions which adjust themselves so as to cause the net electric ﬁeld on the surfaces of the conductors to have vanishing tangential electric ﬁeld components. The distribution of these surface charges is not known at the outset and hence cannot be included in the integral (1). A way out of this dilemma is as follows: The potential distribution we seek within the space not occupied by the conductors is the result of two charge distri- butions. First is the prescribed volume charge distribution leading to the potential function Φ p, and second is the charge distributed on the conductor surfaces. The po- tential function produced by the surface charges must obey the source-free Poisson’s equation in the space Vof interest. Let us denote this solution to the homogeneous form of Poisson’s equation by the potential function Φ h. Then, in the volume V,Φh must satisfy Laplace’s equation. ∇2Φh= 0 (2) The superposition principle then makes it possible to write the total potential as Φ = Φ p+ Φh (3) The problem of ﬁnding the complete ﬁeld distribution now reduces to that of ﬁnding a solution such that the net potential Φ of (3) has the prescribed potentials vion the surfaces Si. Now Φ pis known and can be evaluated on the surface Si. Evaluation of (3) on Sigives vi= Φ p(Si) + Φ h(Si) (4) so that the homogeneous solution is prescribed on the boundaries Si. Φh(Si) =vi−Φp(Si) (5) Hence, the determination of an electroquasistatic ﬁeld with prescribed potentials on the boundaries is reduced to ﬁnding the solution to Laplace’s equation, (2), that satisﬁes the boundary condition given by (5). The approach which has been formalized in this section is another point of view applicable to the boundary value problems in the last part of Chap. 4. Cer- tainly, the abstract view of the boundary value situation provided by Fig. 5.1.1 is not diﬀerent from that of Fig. 4.6.1. In Example 4.6.4, the ﬁeld shown in Fig. 4.6.8 is determined for a point charge adjacent to an equipotential charge-neutral spher- ical electrode. In the volume Vof interest outside the electrode, the volume charge distribution is singular, the point charge q. The potential given by (4.6.35), in fact, takes the form of (3). The particular solution can be taken as the ﬁrst term, the potential of a point charge. The second and third terms, which are equivalent to the potentials caused by the ﬁctitious charges within the sphere, can be taken as the homogeneous solution. Superposition to Satisfy Boundary Conditions. In the following sections, superposition will often be used in another way to satisfy boundary conditions. Sec. 5.2 Uniqueness of Solutions 5 Suppose that there is no charge density in the volume V, and again the potentials on each of the nsurfaces Sjarevj. Then ∇2Φ = 0 (6) Φ =vjonSj, j= 1, . . . n (7) The solution is broken into a superposition of solutions Φ jthat meet the required condition on the j-th surface but are zero on all of the others. Φ =nX j=1Φj (8) Φj≡½ vjonSj 0 on S1. . . S j−1, Sj+1. . . S n(9) Each term is a solution to Laplace’s equation, (6), so the sum is as well. ∇2Φj= 0 (10) In Sec. 5.5, a method is developed for satisfying arbitrary boundary conditions on one of four surfaces enclosing a volume of interest. Capacitance Matrix. Suppose that in the nelectrode system the net charge on the i-th electrode is to be found. In view of (8), the integral of E·daover the surface Sienclosing this electrode then gives qi=−I Si/epsilon1o∇Φ·da=−I Si/epsilon1onX j=1∇Φj·da (11) Because of the linearity of Laplace’s equation, the potential Φ jis proportional to the voltage exciting that potential, vj. It follows that (11) can be written in terms of capacitance parameters that are independent of the excitations. That is, (11) becomes qi=nX j=1Cijvj (12) where the capacitance coeﬃcients are Cij=−H Si/epsilon1o∇Φj·da vj(13) The charge on the i-th electrode is a linear superposition of the contributions of allnvoltages. The coeﬃcient multiplying its own voltage, Cii, is called the self- capacitance , while the others, Cij, i/negationslash=j, are the mutual capacitances . 6 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.2.1 Field line originating on one part of bounding surface and termi- nating on another after passing through the point ro. 5.2 UNIQUENESS OF SOLUTIONS TO POISSON’S EQUATION We shall show in this section that a potential distribution obeying Poisson’s equa- tion is completely speciﬁed within a volume Vif the potential is speciﬁed over the surfaces bounding that volume. Such a uniqueness theorem is useful for two reasons: (a) It tells us that if we have found such a solution to Poisson’s equation, whether by mathematical analysis or physical insight, then we have found the only solution; and (b) it tells us what boundary conditions are appropriate to uniquely specify a solution. If there is no charge present in the volume of interest, then the theorem states the uniqueness of solutions to Laplace’s equation. Following the method “reductio ad absurdum”, we assume that the solution is not unique– that two solutions, Φ aand Φ b, exist, satisfying the same boundary conditions– and then show that this is impossible. The presumably diﬀerent solu- tions Φ aand Φ bmust satisfy Poisson’s equation with the same charge distribution and must satisfy the same boundary conditions. ∇2Φa=−ρ /epsilon1o; Φ a= Φ ionSi (1) ∇2Φb=−ρ /epsilon1o; Φ b= Φ ionSi (2) It follows that with Φ ddeﬁned as the diﬀerence in the two potentials, Φ d= Φ a−Φb, ∇2Φd≡ ∇ · (∇Φd) = 0; Φ d= 0 on Si (3) A simple argument now shows that the only way Φ dcan both satisfy Laplace’s equation and be zero on all of the bounding surfaces is for it to be zero. First, it is argued that Φ dcannot possess a maximum or minimum at any point within V. With the help of Fig. 5.2.1, visualize the negative of the gradient of Φ d, a ﬁeld line, as it passes through some point ro. Because the ﬁeld is solenoidal (divergence free), such a ﬁeld line cannot start or stop within V(Sec. 2.7). Further, the ﬁeld deﬁnes a potential (4.1.4). Hence, as one proceeds along the ﬁeld line in the direction of the negative gradient, the potential has to decrease until the ﬁeld line reaches one of the surfaces Sibounding V. Similarly, in the opposite direction, the potential has to increase until another one of the surfaces is reached. Accordingly, all maximum and minimum values of Φ d(r) have to be located on the surfaces. Sec. 5.3 Continuity Conditions 7 The diﬀerence potential at any interior point cannot assume a value larger than or smaller than the largest or smallest value of the potential on the surfaces. But the surfaces are themselves at zero potential. It follows that the diﬀerence potential is zero everywhere in Vand that Φ a= Φ b. Therefore, only one solution exists to the boundary value problem stated with (1). 5.3 CONTINUITY CONDITIONS At the surfaces of metal conductors, charge densities accumulate that are only a few atomic distances thick. In describing their ﬁelds, the details of the distribution within this thin layer are often not of interest. Thus, the charge is represented by a surface charge density (1.3.11) and the surface supporting the charge treated as a surface of discontinuity. In such cases, it is often convenient to divide a volume in which the ﬁeld is to be determined into regions separated by the surfaces of discontinuity, and to use piece-wise continuous functions to represent the ﬁelds. Continuity conditions are then needed to connect ﬁeld solutions in two regions separated by the discontinuity. These conditions are implied by the diﬀerential equations that apply throughout the region. They assure that the ﬁelds are consistent with the basic laws, even in passing through the discontinuity. Each of the four Maxwell’s equations implies a continuity condition. Because of the singular nature of the source distribution, these laws are used in integral form to relate the ﬁelds to either side of the surface of discontinuity. With the vector n deﬁned as the unit normal to the surface of discontinuity and pointing from region (b) to region (a), the continuity conditions were summarized in Table 1.8.3. In the EQS approximation, the laws of primary interest are Faraday’s law without the magnetic induction and Gauss’ law, the ﬁrst two equations of Chap. 4. Thus, the corresponding EQS continuity conditions are n×[Ea−Eb] = 0 (1) n·(/epsilon1oEa−/epsilon1oEb) =σs (2) Because the magnetic induction makes no contribution to Faraday’s continuity con- dition in any case, these conditions are the same as for the general electrodynamic laws. As a reminder, the contour enclosing the integration surface over which Fara- day’s law was integrated (Sec. 1.6) to obtain (1) is shown in Fig. 5.3.1a. The inte- gration volume used to obtain (2) from Gauss’ law (Sec. 1.3) is similarly shown in Fig. 5.3.1b. What are the continuity conditions on the electric potential? The potential Φ is continuous across a surface of discontinuity even if that surface carries a surface charge density. This will be the case when the Eﬁeld is ﬁnite (a dipole layer containing an inﬁnite ﬁeld causes a jump of potential), because then the line integral of the electric ﬁeld from one side of the surface to the other side is zero, the path- length being inﬁnitely small. Φa−Φb= 0 (3) To determine the jump condition representing Gauss’ law through the surface of discontinuity, it was integrated (Sec. 1.3) over the volume shown intersecting the 8 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.3.1 (a) Diﬀerential contour intersecting surface supporting surface charge density. (b) Diﬀerential volume enclosing surface charge on surface hav- ing normal n. surface in Fig. 5.3.1b. The resulting continuity condition, (2), is written in terms of the potential by recognizing that in the EQS approximation, E=−∇Φ. n·[(∇Φ)a−(∇Φ)b] =−σs /epsilon1o(4) At a surface of discontinuity that carries a surface charge density, the normal derivative of the potential is discontinuous . The continuity conditions become boundary conditions if they are made to represent physical constraints that go beyond those already implied by the laws that prevail in the volume. A familiar example is one where the surface is that of an electrode constrained in its potential. Then the continuity condition (3) requires that the potential in the volume adjacent to the electrode be the given potential of the electrode. This statement cannot be justiﬁed without invoking information about the physical nature of the electrode (that it is “inﬁnitely conducting,” for example) that is not represented in the volume laws and hence is not intrinsic to the continuity conditions. 5.4 SOLUTIONS TO LAPLACE’S EQUATION IN CARTESIAN COORDINATES Having investigated some general properties of solutions to Poisson’s equation, it is now appropriate to study speciﬁc methods of solution to Laplace’s equation subject to boundary conditions. Exempliﬁed by this and the next section are three standard steps often used in representing EQS ﬁelds. First, Laplace’s equation is set up in the coordinate system in which the boundary surfaces are coordinate surfaces. Then, the partial diﬀerential equation is reduced to a set of ordinary diﬀerential equations by separation of variables. In this way, an inﬁnite set of solutions is generated. Finally, the boundary conditions are satisﬁed by superimposing the solutions found by separation of variables. In this section, solutions are derived that are natural if boundary conditions are stated along coordinate surfaces of a Cartesian coordinate system. It is assumed that the ﬁelds depend on only two coordinates, xandy, so that Laplace’s equation Sec. 5.4 Solutions to Laplace’s Equation 9 is (Table I) ∂2Φ ∂x2+∂2Φ ∂y2= 0 (1) This is a partial diﬀerential equation in two independent variables. One time- honored method of mathematics is to reduce a new problem to a problem previously solved. Here the process of ﬁnding solutions to the partial diﬀerential equation is reduced to one of ﬁnding solutions to ordinary diﬀerential equations. This is accom- plished by the method of separation of variables . It consists of assuming solutions with the special space dependence Φ(x, y) =X(x)Y(y) (2) In (2), Xis assumed to be a function of xalone and Yis a function of yalone. If need be, a general space dependence is then recovered by superposition of these special solutions. Substitution of (2) into (1) and division by Φ then gives 1 X(x)d2X(x) dx2=−1 Y(y)d2Y(y) dy2(3) Total derivative symbols are used because the respective functions XandYare by deﬁnition only functions of xandy. In (3) we now have on the left-hand side a function of xalone, on the right- hand side a function of yalone. The equation can be satisﬁed independent of xand yonly if each of these expressions is constant. We denote this “separation” constant byk2, and it follows that d2X dx2=−k2X (4) and d2Y dy2=k2Y (5) These equations have the solutions X∼coskx or sin kx (6) Y∼coshky or sinh ky (7) Ifk= 0, the solutions degenerate into X∼constant or x (8) Y∼constant or y (9) The product solutions, (2), are summarized in the ﬁrst four rows of Table 5.4.1. Those in the right-hand column are simply those of the middle column with the roles of xandyinterchanged. Generally, we will leave the prime oﬀ the k/primein writing these solutions. Exponentials are also solutions to (7). These, sometimes more convenient, solutions are summarized in the last four rows of the table. 10 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 The solutions summarized in this table can be used to gain insight into the nature of EQS ﬁelds. A good investment is therefore made if they are now visualized. The ﬁelds represented by the potentials in the left-hand column of Table 5.4.1 are all familiar. Those that are linear in xandyrepresent uniform ﬁelds, in the xandydirections, respectively. The potential xyis familiar from Fig. 4.1.3. We will use similar conventions to represent the potentials of the second column, but it is helpful to have in mind the three-dimensional portrayal exempliﬁed for the potential xyin Fig. 4.1.4. In the more complicated ﬁeld maps to follow, the sketch is visualized as a contour map of the potential Φ with peaks of positive potential and valleys of negative potential. On the top and left peripheries of Fig. 5.4.1 are sketched the functions cos kx and cosh ky, respectively, the product of which is the ﬁrst of the potentials in the middle column of Table 5.4.1. If we start out from the origin in either the + yor−y directions (north or south), we climb a potential hill. If we instead proceed in the +xor−xdirections (east or west), we move downhill. An easterly path begun on the potential hill to the north of the origin corresponds to a decrease in the cos kx factor. To follow a path of equal elevation, the cosh kyfactor must increase, and this implies that the path must turn northward. A good starting point in making these ﬁeld sketches is the identiﬁcation of the contours of zero potential. In the plot of the second potential in the middle column of Table 5.4.1, shown in Fig. 5.4.2, these are the yaxis and the lines kx= +π/2,+3π/2, etc. The dependence on yis now odd rather than even, as it was for the plot of Fig. 5.4.1. Thus, the origin is now on the side of a potential hill that slopes downward from north to south. The solutions in the third and fourth rows of the second column possess the same ﬁeld patterns as those just discussed provided those patterns are respectively shifted in the xdirection. In the last four rows of Table 5.4.1 are four additional possible solutions which are linear combinations of the previous four in that column. Because these decay exponentially in either the + yor−ydirections, they are useful for representing solutions in problems where an inﬁnite half-space is considered. The solutions in Table 5.4.1 are nonsingular throughout the entire x−yplane. This means that Laplace’s equation is obeyed everywhere within the ﬁnite x−y plane, and hence the ﬁeld lines are continuous; they do not appear or disappear. The sketches show that the ﬁelds become stronger and stronger as one proceeds in the positive and negative ydirections. The lines of electric ﬁeld originate on positive charges and terminate on negative charges at y→ ±∞ . Thus, for the plots shown in Figs. 5.4.1 and 5.4.2, the charge distributions at inﬁnity must consist of alternating distributions of positive and negative charges of inﬁnite amplitude. Two ﬁnal observations serve to further develop an appreciation for the nature of solutions to Laplace’s equation. First, the third dimension can be used to repre- sent the potential in the manner of Fig. 4.1.4, so that the potential surface has the shape of a membrane stretched from boundaries that are elevated in proportion to their potentials. Laplace’s equation, (1), requires that the sum of quantities that reﬂect the curvatures in the xandydirections vanish. If the second derivative of a function is positive, it is curved upward; and if it is negative, it is curved downward. If the curvature is positive in the xdirection, it must be negative in the ydirection. Thus, at the origin in Fig. 5.4.1, the potential is cupped downward for excursions in the Sec. 5.5 Modal Expansion 11 Fig. 5.4.1 Equipotentials for Φ = cos( kx) cosh( ky) and ﬁeld lines. As an aid to visualizing the potential, the separate factors cos( kx) and cosh( ky) are, respectively, displayed at the top and to the left. xdirection, and so it must be cupped upward for variations in the ydirection. A similar deduction must apply at every point in the x−yplane. Second, because the kthat appears in the periodic functions of the second column in Table 5.4.1 is the same as that in the exponential and hyperbolic func- tions, it is clear that the more rapid the periodic variation, the more rapid is the decay or apparent growth. 5.5 MODAL EXPANSION TO SATISFY BOUNDARY CONDITIONS Each of the solutions obtained in the preceding section by separation of variables could be produced by an appropriate potential applied to pairs of parallel surfaces 12 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.4.2 Equipotentials for Φ = cos( kx) sinh( ky) and ﬁeld lines. As an aid to visualizing the potential, the separate factors cos( kx) and sinh( ky) are, respectively, displayed at the top and to the left. in the planes x= constant and y= constant. Consider, for example, the fourth solution in the column k2≥0 of Table 5.4.1, which with a constant multiplier is Φ =Asinkxsinhky (1) This solution has Φ = 0 in the plane y= 0 and in the planes x=nπ/k , where nis an integer. Suppose that we set k=nπ/a so that Φ = 0 in the plane y=aas well. Then at y=b, the potential of (1) Φ(x, b) =Asinhnπ absinnπ ax (2) Sec. 5.5 Modal Expansion 13 TABLE 5.4.1 TWO-DIMENSIONAL CARTESIAN SOLUTIONS OF LAPLACE’S EQUATION k= 0 k2≥0 k2≤0 (k→jk/prime) Constant coskxcosh ky cosh k/primexcosk/primey y coskxsinhky cosh k/primexsink/primey x sinkxcosh ky sinhk/primexcosk/primey xy sinkxsinhky sinhk/primexsink/primey coskx ekyek/primexcosk/primey coskx e−kye−k/primexcosk/primey sinkx ekyek/primexsink/primey sinkx e−kye−k/primexsink/primey Fig. 5.5.1 Two of the inﬁnite number of potential functions having the form of (1) that will ﬁt the boundary conditions Φ = 0 at y= 0 and at x= 0 and x=a. has a sinusoidal dependence on x. If a potential of the form of (2) were applied along the surface at y=b, and the surfaces at x= 0, x=a, and y= 0 were held at zero potential (by, say, planar conductors held at zero potential), then the potential, (1), would exist within the space 0 < x < a, 0< y < b . Segmented electrodes having each segment constrained to the appropriate potential could be used to approximate the distribution at y=b. The potential and ﬁeld plots for n= 1 and n= 2 are given in Fig. 5.5.1. Note that the theorem of Sec. 5.2 insures 14 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.5.2 Cross-section of zero-potential rectangular slot with an electrode having the potential vinserted at the top. that the speciﬁed potential is unique. But what can be done to describe the ﬁeld if the wall potentials are not con- strained to ﬁt neatly the solution obtained by separation of variables? For example, suppose that the ﬁelds are desired in the same region of rectangular cross-section, but with an electrode at y=bconstrained to have a potential vthat is independent ofx. The conﬁguration is now as shown in Fig. 5.5.2. A line of attack is suggested by the inﬁnite number of solutions, having the form of (1), that meet the boundary condition on three of the four walls. The superposition principle makes it clear that any linear combination of these is also a solution, so if we let Anbe arbitrary coeﬃcients, a more general solution is Φ =∞X n=1Ansinhnπ aysinnπ ax (3) Note that khas been assigned values such that the sine function is zero in the planes x= 0 and x=a. Now how can we adjust the coeﬃcients so that the boundary condition at the driven electrode, at y=b, is met? One approach that we will not have to use is suggested by the numerical method described in Sec. 4.8. The electrode could be divided into Nsegments and (3) evaluated at the center point of each of the segments. If the inﬁnite series were truncated at Nterms, the result would be Nequations that were linear in the Nunknowns An. This system of equations could be inverted to determine the An’s. Substitution of these into (3) would then comprise a solution to the boundary value problem. Unfortunately, to achieve reasonable accuracy, large values of Nwould be required and a computer would be needed. The power of the approach of variable separation is that it results in solutions that are orthogonal in a sense that makes it possible to determine explicitly the coeﬃcients An. The evaluation of the coeﬃcients is remarkably simple. First, (3) is evaluated on the surface of the electrode where the potential is known. Φ(x, b) =∞X n=1Ansinhnπb asinnπ ax (4) Sec. 5.5 Modal Expansion 15 On the right is the inﬁnite series of sinusoidal functions with coeﬃcients that are to be determined. On the left is a given function of x. We multiply both sides of the expression by sin( mπx/a ), where mis one integer, and then both sides of the expression are integrated over the width of the system. Za 0Φ(x, b) sinmπ axdx=∞X n=1Ansinhnπb aZa 0sinmπ axsinnπ axdx (5) The functions sin( nπx/a ) and sin( mπx/a ) are orthogonal in the sense that the integral of their product over the speciﬁed interval is zero, unless m=n. Za 0sinmπ axsinnπ axdx=½0, n/negationslash=m a 2, n =m(6) Thus, all the terms on the right in (5) vanish, except the one having n=m. Of course, mcan be any integer, so we can solve (5) for the m-th amplitude and then replace mbyn. An=2 asinhnπb aZa 0Φ(x, b) sinnπ axdx (7) Given any distribution of potential on the surface y=b, this integral can be carried out and hence the coeﬃcients determined. In this speciﬁc problem, the potential is vat each point on the electrode surface. Thus, (7) is evaluated to give An=2v(t) nπ(1−cosπn) sinh¡nπb a¢=(0; neven 4v nπ1 sinh¡ nπb a¢;nodd (8) Finally, substitution of these coeﬃcients into (3) gives the desired potential. Φ =∞X n=1 odd4v(t) π1 nsinh¡nπ ay¢ sinh¡nπb a¢sinnπ ax (9) Each product term in this inﬁnite series satisﬁes Laplace’s equation and the zero potential condition on three of the surfaces enclosing the region of interest. The sumsatisﬁes the potential condition on the “last” boundary. Note that the sum is not itself in the form of the product of a function of xalone and a function of y alone. The modal expansion is applicable with an arbitrary distribution of potential on the “last” boundary. But what if we have an arbitrary distribution of potential on all four of the planes enclosing the region of interest? The superposition principle justiﬁes using the sum of four solutions of the type illustrated here. Added to the series solution already found are three more, each analogous to the previous one, but rotated by 90 degrees. Because each of the four series has a ﬁnite potential only on the part of the boundary to which its series applies, the sum of the four satisfy all boundary conditions. The potential given by (9) is illustrated in Fig. 5.5.3. In the three-dimensional portrayal, it is especially clear that the ﬁeld is inﬁnitely large in the corners where 16 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.5.3 Potential and ﬁeld lines for the conﬁguration of Fig. 5.5.2, (9), shown using vertical coordinate to display the potential and shown in x−y plane. the driven electrode meets the grounded walls. Where the electric ﬁeld emanates from the driven electrode, there is surface charge, so at the corners there is an inﬁnite surface charge density. In practice, of course, the spacing is not inﬁnitesimal and the ﬁelds are not inﬁnite. Demonstration 5.5.1. Capacitance Attenuator Because neither of the ﬁeld laws in this chapter involve time derivatives, the ﬁeld that has been determined is correct for v=v(t), an arbitrary function of time. As a consequence, the coeﬃcients Anare also functions of time. Thus, the charges induced on the walls of the box are time varying, as can be seen if the wall at y= 0 is isolated from the grounded side walls and connected to ground through a resistor. The conﬁguration is shown in cross-section by Fig. 5.5.4. The resistance Ris small enough so that the potential vois small compared with v. The charge induced on this output electrode is found by applying Gauss’ integral law with an integration surface enclosing the electrode. The width of the electrode in the zdirection is w, so q=I S/epsilon1oE·da=/epsilon1owZa 0Ey(x,0)dx=−/epsilon1owZa 0∂Φ ∂y(x,0)dx (10) This expression is evaluated using (9). q=−Cmv; Cm≡8/epsilon1ow π∞X n=1 odd1 nsinh¡nπb a¢ (11) Conservation of charge requires that the current through the resistance be the rate of change of this charge with respect to time. Thus, the output voltage is vo=−Rdq dt=RCmdv dt(12) Sec. 5.5 Modal Expansion 17 Fig. 5.5.4 The bottom of the slot is replaced by an insulating electrode connected to ground through a low resistance so that the induced current can be measured. and if v=Vsinωt, then vo=RCmωVcosωt≡Vocosωt (13) The experiment shown in Fig. 5.5.5 is designed to demonstrate the dependence of the output voltage on the spacing bbetween the input and output electrodes. It follows from (13) and (11) that this voltage can be written in normalized form as Vo U=∞X n=1 odd1 2nsinh¡nπb a¢; U≡16/epsilon1owωR πV (14) Thus, the natural log of the normalized voltage has the dependence on the electrode spacing shown in Fig. 5.5.5. Note that with increasing b/athe function quickly becomes a straight line. In the limit of large b/a, the hyperbolic sine can be approximated by exp( nπb/a )/2 and the series can be approximated by one term. Thus, the dependence of the output voltage on the electrode spacing becomes simply ln¡Vo U¢ =ln e−(πb/a )=−πb a(15) and so the asymptotic slope of the curve is −π. Charges induced on the input electrode have their images either on the side walls of the box or on the output electrode. If b/ais small, almost all of these images are on the output electrode, but as it is withdrawn, more and more of the images are on the side walls and fewer are on the output electrode. In retrospect, there are several matters that deserve further discussion. First, the potential used as a starting point in this section, (1), is one from a list of four in Table 5.4.1. What type of procedure can be used to select the appropriate form? In general, the solution used to satisfy the zero potential boundary condition on the 18 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.5.5 Demonstration of electroquasistatic attenuator in which normalized output voltage is measured as a function of the distance be- tween input and output electrodes normalized to the smaller dimension of the box. The normalizing voltage Uis deﬁned by (14). The output electrode is positioned by means of the attached insulating rod. In op- eration, a metal lid covers the side of the box. “ﬁrst” three surfaces is a linear combination of the four possible solutions. Thus, with the A’s denoting undetermined coeﬃcients, the general form of the solution is Φ =A1coskxcoshky+A2coskxsinhky +A3sinkxcoshky+A4sinkxsinhky(16) Formally, (1) was selected by eliminating three of these four coeﬃcients. The ﬁrst two must vanish because the function must be zero at x= 0. The third is excluded because the potential must be zero at y= 0. Thus, we are led to the last term, which, if A4=A, is (1). The methodical elimination of solutions is necessary. Because the origin of the coordinates is arbitrary, setting up a simple expression for the potential is a matter of choosing the origin of coordinates properly so that as many of the solutions (16) are eliminated as possible. We purposely choose the origin so that a single term from the four in (16) meets the boundary condition at x= 0 and y= 0. The selection of product solutions from the list should interplay with the choice of coordinates. Some combinations are much more convenient than others. This will be exempliﬁed in this and the following chapters. The remainder of this section is devoted to a more detailed discussion of the expansion in sinusoids represented by (9). In the plane y=b, the potential distribution is of the form Φ(x, b) =∞X n=1Vnsinnπ ax (17) Sec. 5.5 Modal Expansion 19 Fig. 5.5.6 Fourier series approximation to square wave given by (17) and (18), successively showing one, two, and three terms. Higher-order terms tend to ﬁll in the sharp discontinuity at x= 0 and x=a. Outside the range of interest, the series represents an odd function of xhaving a periodicity length 2a. where the procedure for determining the coeﬃcients has led to (8), written here in terms of the coeﬃcients Vnof (17) as Vn=½0, n even 4v nπ, n odd(18) The approximation to the potential vthat is uniform over the span of the driving electrode is shown in Fig. 5.5.6. Equation (17) represents a square wave of period 2 a extending over all x,−∞< x < +∞. One half of a period appears as shown in the ﬁgure. It is possible to represent this distribution in terms of sinusoids alone because it is odd in x. In general, a periodic function is represented by a Fourier series of both sines and cosines. In the present problem, cosines were missing because the potential had to be zero at x= 0 and x=a. Study of a Fourier series shows that the series converges to the actual function in the sense that in the limit of an inﬁnite number of terms,Za 0[Φ2(x)−F2(x)]dx= 0 (19) where Φ( x) is the actual potential distribution and F(x) is the Fourier series ap- proximation. To see the generality of the approach exempliﬁed here, we show that the orthogonality property of the functions X(x) results from the diﬀerential equation and boundary conditions. Thus, it should not be surprising that the solutions in other coordinate systems also have an orthogonality property. In all cases, the orthogonality property is associated with any one of the factors in a product solution. For the Cartesian problem considered here, it is X(x) that satisﬁes boundary conditions at two points in space. This is assured by adjusting 20 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 the eigenvalue kn=nπ/a so that the eigenfunction or mode, sin( nπx/a ), is zero at x= 0 and x=a. This function satisﬁes (5.4.4) and the boundary conditions. d2Xm dx2+k2 mXm= 0; Xm= 0 at x= 0, a (20) The subscript mis used to recognize that there is an inﬁnite number of solutions to this problem. Another solution, say the n-th, must also satisfy this equation and the boundary conditions. d2Xn dx2+k2 nXn= 0; Xn= 0 at x= 0, a (21) The orthogonality property for these modes, exploited in evaluating the coeﬃcients of the series expansion, is Za 0XmXndx= 0, n /negationslash=m (22) To prove this condition in general, we multiply (20) by Xnand integrate between the points where the boundary conditions apply. Za 0Xnd dx¡dXm dx¢ dx+Za 0k2 mXmXndx= 0 (23) By identifying u=Xnandv=dXm/dx, the ﬁrst term is integrated by parts to obtainZa 0Xnd dx¡dXm dx¢ dx=XndXm dx¯¯¯¯a 0−Za 0dXn dxdXm dxdx (24) The ﬁrst term on the right vanishes because of the boundary conditions. Thus, (23) becomes −Za 0dXm dxdXn dxdx+k2 mZa 0XmXndx= 0 (25) If these same steps are completed with nandminterchanged, the result is (25) with nandminterchanged. Because the ﬁrst term in (25) is the same as its counterpart in this second equation, subtraction of the two expressions yields (k2 m−k2 n)Za 0XmXndx= 0 (26) Thus, the functions are orthogonal provided that kn/negationslash=km. For this speciﬁc problem, the eigenfunctions are Xn= sin( nπ/a ) and the eigenvalues are kn=nπ/a . But in general we can expect that our product solutions to Laplace’s equation in other coordinate systems will result in a set of functions having similar orthogonality properties. Sec. 5.6 Solutions to Poisson’s Equation 21 Fig. 5.6.1 Cross-section of layer of charge that is periodic in the x direction and bounded from above and below by zero potential plates. With this charge translating to the right, an insulated electrode inserted in the lower equipotential is used to detect the motion. 5.6 SOLUTIONS TO POISSON’S EQUATION WITH BOUNDARY CONDITIONS An approach to solving Poisson’s equation in a region bounded by surfaces of known potential was outlined in Sec. 5.1. The potential was divided into a particular part, the Laplacian of which balances −ρ//epsilon1othroughout the region of interest, and a homogeneous part that makes the sum of the two potentials satisfy the boundary conditions. In short, Φ = Φ p+ Φh (1) ∇2Φp=−ρ /epsilon1o(2) ∇2Φh= 0 (3) and on the enclosing surfaces, Φh= Φ−Φp onS (4) The following examples illustrate this approach. At the same time they demon- strate the use of the Cartesian coordinate solutions to Laplace’s equation and the idea that the ﬁelds described can be time varying. Example 5.6.1. Field of Traveling Wave of Space Charge between Equipotential Surfaces The cross-section of a two-dimensional system that stretches to inﬁnity in the x andzdirections is shown in Fig. 5.6.1. Conductors in the planes y=aandy=−a bound the region of interest. Between these planes the charge density is periodic in thexdirection and uniformly distributed in the ydirection. ρ=ρocosβx (5) The parameters ρoandβare given constants. For now, the segment connected to ground through the resistor in the lower electrode can be regarded as being at the same zero potential as the remainder of the electrode in the plane x=−aand the electrode in the plane y=a. First we ask for the ﬁeld distribution. 22 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Remember that any particular solution to (2) will do. Because the charge density is independent of y, it is natural to look for a particular solution with the same property. Then, on the left in (2) is a second derivative with respect to x, and the equation can be integrated twice to obtain Φp=ρo /epsilon1oβ2cosβx (6) This particular solution is independent of y. Note that it is not the potential that would be obtained by evaluating the superposition integral over the charge between the grounded planes. Viewed over all space, that charge distribution is not indepen- dent of y. In fact, the potential of (6) is associated with a charge distribution as given by (5) that extends to inﬁnity in the + yand−ydirections. The homogeneous solution must make up for the fact that (6) does not satisfy the boundary conditions. That is, at the boundaries, Φ = 0 in (1), so the homoge- neous and particular solutions must balance there. Φh¯¯ y=±a=−Φp¯¯ y=±a=−ρo /epsilon1oβ2cosβx (7) Thus, we are looking for a solution to Laplace’s equation, (3), that satisﬁes these boundary conditions. Because the potential has the same value on the boundaries, and the origin of the yaxis has been chosen to be midway between, it is clear that the potential must be an even function of y. Further, it must have a periodicity in thexdirection that matches that of (7). Thus, from the list of solutions to Laplace’s equation in Cartesian coordinates in the middle column of Table 5.4.1, k=β, the sinkxterms are eliminated in favor of the cos kxsolutions, and the cosh kysolution is selected because it is even in y. Φh=Acoshβycosβx (8) The coeﬃcient Ais now adjusted so that the boundary conditions are satisﬁed by substituting (8) into (7). Acoshβacosβx=−ρo /epsilon1oβ2cosβx→A=−ρo /epsilon1oβ2coshβa(9) Superposition of the particular solution, (7), and the homogeneous solution given by substituting the coeﬃcient of (9) into (8), results in the desired potential distribution. Φ =ρo /epsilon1oβ2µ 1−coshβy coshβa¶ cosβx (10) The mathematical solutions used in deriving (10) are illustrated in Fig. 5.6.2. The particular solution describes an electric ﬁeld that originates in regions of positive charge density and terminates in regions of negative charge density. It is purely x directed and is therefore tangential to the equipotential boundary. The homogeneous solution that is added to this ﬁeld is entirely due to surface charges. These give rise to a ﬁeld that bucks out the tangential ﬁeld at the walls, rendering them surfaces of constant potential. Thus, the sum of the solutions (also shown in the ﬁgure), satisﬁes Gauss’ law and the boundary conditions. With this static view of the ﬁelds ﬁrmly in mind, suppose that the charge distribution is moving in the xdirection with the velocity v. ρ=ρocosβ(x−vt) (11) Sec. 5.6 Solutions to Poisson’s Equation 23 Fig. 5.6.2 Equipotentials and ﬁeld lines for conﬁguration of Fig. 5.6.1 showing graphically the superposition of particular and homogeneous parts that gives the required potential. The variable xin (5) has been replaced by x−vt. With this moving charge distri- bution, the ﬁeld also moves. Thus, (10) becomes Φ =ρo /epsilon1oβ2µ 1−coshβy coshβa¶ cosβ(x−vt) (12) Note that the homogeneous solution is now a linear combination of the ﬁrst and third solutions in the middle column of Table 5.4.1. As the space charge wave moves by, the charges induced on the perfectly conducting walls follow along in synchronism. The current that accompanies the redistribution of surface charges is detected if a section of the wall is insulated from the rest and connected to ground through a resistor, as shown in Fig. 5.6.1. Under the assumption that the resistance is small enough so that the segment remains at essentially zero potential, what is the output voltage vo? The current through the resistor is found by invoking charge conservation for the segment to ﬁnd the current that is the time rate of change of the net charge on the segment. The latter follows from Gauss’ integral law and (12) as q=wZl/2 −l/2/epsilon1oEy¯¯¯¯ y=−adx =−wρo β2tanhβa£ sinβ¡l 2−vt¢ + sin β¡l 2+vt¢¤(13) It follows that the dynamics of the traveling wave of space charge is reﬂected in a measured voltage of vo=−Rdq dt=−2Rwρ ov βtanhβasinβl 2sinβvt (14) 24 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.6.3 Cross-section of sheet beam of charge between plane par- allel equipotential plates. Beam is modeled by surface charge density having dc and ac parts. In writing this expression, the double-angle formulas have been invoked. Several predictions should be consistent with intuition. The output voltage varies sinusoidally with time at a frequency that is proportional to the velocity and inversely proportional to the wavelength, 2 π/β. The higher the velocity, the greater the voltage. Finally, if the detection electrode is a multiple of the wavelength 2 π/β, the voltage is zero. If the charge density is concentrated in surface-like regions that are thin com- pared to other dimensions of interest, it is possible to solve Poisson’s equation with boundary conditions using a procedure that has the appearance of solving Laplace’s equation rather than Poisson’s equation. The potential is typically bro- ken into piece-wise continuous functions, and the eﬀect of the charge density is brought in by Gauss’ continuity condition, which is used to splice the functions at the surface occupied by the charge density. The following example illustrates this procedure. What is accomplished is a solution to Poisson’s equation in the entire region, including the charge-carrying surface. Example 5.6.2. Thin Bunched Charged-Particle Beam between Conducting Plates In microwave ampliﬁers and oscillators of the electron beam type, a basic problem is the evaluation of the electric ﬁeld produced by a bunched electron beam. The cross-section of the beam is usually small compared with a free space wavelength of an electromagnetic wave, in which case the electroquasistatic approximation applies. We consider a strip electron beam having a charge density that is uniform over its cross-section δ. The beam moves with the velocity vin the xdirection between two planar perfect conductors situated at y=±aand held at zero potential. The conﬁguration is shown in cross-section in Fig. 5.6.3. In addition to the uniform charge density, there is a “ripple” of charge density, so that the net charge density is ρ=8 < :0 a > y >δ 2 ρo+ρ1cos£2π Λ(x−vt)¤δ 2> y > −δ 2 0 −δ 2> y > −a(15) where ρo, ρ1, and Λ are constants. The system can be idealized to be of inﬁnite extent in the xandydirections. The thickness δof the beam is much smaller than the wavelength of the periodic charge density ripple, and much smaller than the spacing 2 aof the planar conductors. Thus, the beam is treated as a sheet of surface charge with a density σs=σo+σ1cos£2π Λ(x−vt)¤ (16) Sec. 5.6 Solutions to Poisson’s Equation 25 where σo=ρoδandσ1=ρ1δ. In regions (a) and (b), respectively, above and below the beam, the poten- tial obeys Laplace’s equation. Superscripts (a) and (b) are now used to designate variables evaluated in these regions. To guarantee that the fundamental laws are satisﬁed within the sheet, these potentials must satisfy the jump conditions implied by the laws of Faraday and Gauss, (5.3.4) and (5.3.5). That is, at y= 0 Φa= Φb(17) −/epsilon1oµ ∂Φa ∂y−∂Φb ∂y¶ =σo+σ1cos· 2π Λ(x−vt)¸ (18) To complete the speciﬁcation of the ﬁeld in the region between the plates, boundary conditions are, at y=a, Φa= 0 (19) and at y=−a, Φb= 0 (20) In the respective regions, the potential is split into dc and ac parts, respectively, produced by the uniform and ripple parts of the charge density. Φ = Φ o+ Φ 1 (21) By deﬁnition, Φ oand Φ 1satisfy Laplace’s equation and (17), (19), and (20). The dc part, Φ o, satisﬁes (18) with only the ﬁrst term on the right, while the ac part, Φ 1, satisﬁes (18) with only the second term. The dc surface charge density is independent of x, so it is natural to look for potentials that are also independent of x. From the ﬁrst column in Table 5.4.1, such solutions are Φa=A1y+A2 (22) Φb=B1y+B2 (23) The four coeﬃcients in these expressions are determined from (17)–(20), if need be, by substitution of these expressions and formal solution for the coeﬃcients. More attractive is the solution by inspection that recognizes that the system is symmetric with respect to y, that the uniform surface charge gives rise to uniform electric ﬁelds that are directed upward and downward in the two regions, and that the associated linear potential must be zero at the two boundaries. Φa o=σo 2/epsilon1o(a−y) (24) Φb o=σo 2/epsilon1o(a+y) (25) Now consider the ac part of the potential. The xdependence is suggested by (18), which makes it clear that for product solutions, the xdependence of the potential must be the cosine function moving with time. Neither the sinh nor the cosh functions vanish at the boundaries, so we will have to take a linear combination of these to satisfy the boundary conditions at y= +a. This is eﬀectively done by inspection if it is recognized that the origin of the yaxis used in writing the 26 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.6.4 Equipotentials and ﬁeld lines caused by ac part of sheet charge in the conﬁguration of Fig. 5.6.3. solutions is arbitrary. The solutions to Laplace’s equation that satisfy the boundary conditions, (19) and (20), are Φa 1=A3sinh2π Λ(y−a) cos£2π Λ(x−vt)¤ (26) Φb 1=B3sinh2π Λ(y+a) cos£2π Λ(x−vt)¤ (27) These potentials must match at y= 0, as required by (17), so we might just as well have written them with the coeﬃcients adjusted accordingly. Φa 1=−Csinh2π Λ(y−a) cos£2π Λ(x−vt)¤ (28) Φb 1=Csinh2π Λ(y+a) cos£2π Λ(x−vt)¤ (29) The one remaining coeﬃcient is determined by substituting these expressions into (18) (with σoomitted). C=σ1 2/epsilon1oΛ 2π/cosh¡2πa Λ¢ (30) We have found the potential as a piece-wise continuous function. In region (a), it is the superposition of (24) and (28), while in region (b), it is (25) and (29). In both expressions, Cis provided by (30). Φa=σo 2/epsilon1o(a−y)−σ1 2/epsilon1oΛ 2πsinh£2π Λ(y−a)¤ cosh¡2π Λa¢cos£2π Λ(x−vt)¤ (31) Φb=σo 2/epsilon1o(a+y) +σ1 2/epsilon1oΛ 2πsinh£2π Λ(y+a)¤ cosh¡2π Λa¢cos£2π Λ(x−vt)¤ (32) When t= 0, the ac part of this potential distribution is as shown by Fig. 5.6.4. With increasing time, the ﬁeld distribution translates to the right with the velocity v. Note that some lines of electric ﬁeld intensity that originate on the beam terminate elsewhere on the beam, while others terminate on the equipotential walls. If the walls are even a wavelength away from the beam ( a= Λ), almost all the ﬁeld lines terminate elsewhere on the beam. That is, coupling to the wall is signiﬁcant only if the wavelength is on the order of or larger than a. The nature of solutions to Laplace’s equation is in evidence. Two-dimensional potentials that vary rapidly in one direction must decay equally rapidly in a perpendicular direction. Sec. 5.7 Laplace’s Eq. in Polar Coordinates 27 Fig. 5.7.1 Polar coordinate system. A comparison of the ﬁelds from the sheet beam shown in Fig. 5.6.4 and the periodic distribution of volume charge density shown in Fig. 5.6.2 is a reminder of the similarity of the two physical situations. Even though Laplace’s equation applies in the subregions of the conﬁguration considered in this section, it is really Poisson’s equation that is solved “in the large,” as in the previous example. 5.7 SOLUTIONS TO LAPLACE’S EQUATION IN POLAR COORDINATES In electroquasistatic ﬁeld problems in which the boundary conditions are speciﬁed on circular cylinders or on planes of constant φ, it is convenient to match these conditions with solutions to Laplace’s equation in polar coordinates (cylindrical coordinates with no zdependence). The approach adopted is entirely analogous to the one used in Sec. 5.4 in the case of Cartesian coordinates. As a reminder, the polar coordinates are deﬁned in Fig. 5.7.1. In these coordi- nates and with the understanding that there is no zdependence, Laplace’s equation, Table I, (8), is 1 r∂ ∂r¡ r∂Φ ∂r¢ +1 r2∂2Φ ∂φ2= 0 (1) One diﬀerence between this equation and Laplace’s equation written in Cartesian coordinates is immediately apparent: In polar coordinates, the equation contains coeﬃcients which not only depend on the independent variable rbut become sin- gular at the origin. This singular behavior of the diﬀerential equation will aﬀect the type of solutions we now obtain. In order to reduce the solution of the partial diﬀerential equation to the sim- pler problem of solving total diﬀerential equations, we look for solutions which can be written as products of functions of ralone and of φalone. Φ =R(r)F(φ) (2) When this assumed form of φis introduced into (1), and the result divided by φ and multiplied by r, we obtain r Rd dr¡ rdR dr¢ =−1 Fd2F dφ2(3) 28 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 We ﬁnd on the left-hand side of (3) a function of ralone and on the right-hand side a function of φalone. The two sides of the equation can balance if and only if the function of φand the function of rare both equal to the same constant. For this “separation constant” we introduce the symbol −m2. d2F dφ2=−m2F (4) rd dr¡ rdR dr¢ =m2R (5) Form2>0, the solutions to the diﬀerential equation for Fare conveniently written as F∼cosmφ or sin mφ (6) Because of the space-varying coeﬃcients, the solutions to (5) are not exponentials or linear combinations of exponentials as has so far been the case. Fortunately, the solutions are nevertheless simple. Substitution of a solution having the form rninto (5) shows that the equation is satisﬁed provided that n=±m. Thus, R∼rmor r−m(7) In the special case of a zero separation constant, the limiting solutions are F∼constant or φ (8) and R∼constant or ln r (9) The product solutions shown in the ﬁrst two columns of Table 5.7.1, constructed by taking all possible combinations of these solutions, are those most often used in polar coordinates. But what are the solutions if m2<0? In Cartesian coordinates, changing the sign of the separation constant k2 amounts to interchanging the roles of the xandycoordinates. Solutions that are periodic in the xdirection become exponential in character, while the exponential decay and growth in the ydirection becomes periodic. Here the geometry is such that the randφcoordinates are not interchangeable, but the new solutions resulting from replacing m2by−p2, where pis a real number, essentially make the oscillating dependence radial instead of azimuthal, and the exponential dependence azimuthal rather than radial. To see this, let m2=−p2, orm=jp, and the solutions given by (7) become R∼rjpor r−jp(10) These take a more familiar appearance if it is recognized that rcan be written identically as r≡elnr(11) Introduction of this identity into (10) then gives the more familiar complex expo- nential, which can be split into its real and imaginary parts using Euler’s formula. R∼r±jp=e±jp ln r= cos( p ln r )±jsin(p ln r ) (12) Sec. 5.7 Laplace’s Eq. in Polar Coordinates 29 Thus, two independent solutions for R(r) are the cosine and sine functions of p ln r . Theφdependence is now either represented by exp ±pφor the hyperbolic functions that are linear combinations of these exponentials. These solutions are summarized in the right-hand column of Table 5.7.1. In principle, the solution to a given problem can be approached by the me- thodical elimination of solutions from the catalogue given in Table 5.7.1. In fact, most problems are best approached by attributing to each solution some physical meaning. This makes it possible to deﬁne coordinates so that the ﬁeld representa- tion is kept as simple as possible. With that objective, consider ﬁrst the solutions appearing in the ﬁrst column of Table 5.7.1. The constant potential is an obvious solution and need not be considered further. We have a solution in row two for which the potential is proportional to the angle. The equipotential lines and the ﬁeld lines are illustrated in Fig. 5.7.2a. Evaluation of the ﬁeld by taking the gradient of the potential in polar coordinates (the gradient operator given in Table I) shows that it becomes inﬁnitely large as the origin is reached. The potential increases from zero to 2 πas the angle φis increased from zero to 2 π. If the potential is to be single valued, then we cannot allow that φincrease further without leaving the region of validity of the solution. This observation identiﬁes the solution with a physical ﬁeld observed when two semi-inﬁnite conducting plates are held at diﬀerent potentials and the distance between the conducting plates at their junction is assumed to be negligible. In this case, shown in Fig. 5.7.2, the outside ﬁeld between the plates is properly represented by a potential proportional to φ. With the plates separated by an angle of 90 degrees rather than 360 degrees, the potential that is proportional to φis seen in the corners of the conﬁguration shown in Fig. 5.5.3. The m2= 0 solution in the third row is familiar from Sec. 1.3, for it is the potential of a line charge. The fourth m2= 0 solution is sketched in Fig. 5.7.3. In order to sketch the potentials corresponding to the solutions in the second column of Table 5.7.1, the separation constant must be speciﬁed. For the time being, let us assume that mis an integer. For m= 1, the solutions rcosφand rsinφrepresent familiar potentials. Observe that the polar coordinates are related to the Cartesian ones deﬁned in Fig. 5.7.1 by rcosφ=x rsinφ=y (13) The ﬁelds that go with these potentials are best found by taking the gradient in Cartesian coordinates. This makes it clear that they can be used to represent uni- form ﬁelds having the xandydirections, respectively. To emphasize the simplicity of these solutions, which are made complicated by the polar representation, the second function of (13) is shown in Fig. 5.7.4a. Figure 5.7.4b shows the potential r−1sinφ. To stay on a contour of constant potential in the ﬁrst quadrant of this ﬁgure as φis increased toward π/2, it is necessary to ﬁrst increase r, and then as the sine function decreases in the second quadrant, to decrease r. The potential is singular at the origin of r; as the origin is approached from above, it is large and positive; while from below it is large and negative. Thus, the ﬁeld lines emerge from the origin within 0 < φ < π and converge toward the origin in the lower half-plane. There must be a source at 30 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.7.2 Equipotentials and ﬁeld lines for (a) Φ = φ, (b) region exterior to planar electrodes having potential diﬀerence V. Fig. 5.7.3 Equipotentials and ﬁeld lines for Φ = φ ln(r). the origin composed of equal and opposite charges on the two sides of the plane rsinφ= 0. The source, which is uniform and of inﬁnite extent in the zdirection, is a line dipole. This conclusion is conﬁrmed by direct evaluation of the potential produced by two line charges, the charge −λlsituated at the origin, the charge + λlat a very small distance away from the origin at r=d, φ=π/2. The potential follows from Sec. 5.7 Laplace’s Eq. in Polar Coordinates 31 Fig. 5.7.4 Equipotentials and ﬁeld lines for (a) Φ = rsin(φ), (b) Φ = r−1sin(φ). 32 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.7.5 Equipotentials and ﬁeld lines for (a) Φ = r2sin(2φ), (b) Φ = r−2sin(2φ). steps paralleling those used for the three-dimensional dipole in Sec. 4.4. Φ = lim d→0 λl→∞· −λl 2π/epsilon1oln(r−dsinφ) +λl 2π/epsilon1oln r¸ =pλ 2π/epsilon1osinφ r(14) The spatial dependence of the potential is indeed sin φ/r. In an analogy with the three-dimensional dipole of Sec. 4.4, pλ≡λldis deﬁned as the line dipole moment. In Example 4.6.3, it is shown that the equipotentials for parallel line charges are circular cylinders. Because this result is independent of spacing between the line charges, it is no surprise that the equipotentials of Fig. 5.7.4b are circular. In summary, the m= 1 solutions can be thought of as the ﬁelds of dipoles at inﬁnity and at the origin. For the sine dependencies, the dipoles are ydirected, while for the cosine dependencies they are xdirected. The solution of Fig. 5.7.5a, φ∝r2sin 2φ, has been met before in Carte- sian coordinates. Either from a comparison of the equipotential plots or by direct transformation of the Cartesian coordinates into polar coordinates, the potential is recognized as xy. Them= 2 solution that is singular at the origin is shown in Fig. 5.7.5b. Field lines emerge from the origin and return to it twice as φranges from 0 to 2 π. This observation identiﬁes four line charges of equal magnitude, alternating in sign as the source of the ﬁeld. Thus, the m= 2 solutions can be regarded as those of quadrupoles at inﬁnity and at the origin. It is perhaps a bit surprising that we have obtained from Laplace’s equation solutions that are singular at the origin and hence associated with sources at the origin. The singularity of one of the two independent solutions to (5) can be traced to the singularity in the coeﬃcients of this diﬀerential equation. From the foregoing, it is seen that increasing mintroduces a more rapid variation of the ﬁeld with respect to the angular coordinate. In problems where Sec. 5.8 Examples in Polar Coordinates 33 TABLE 5.7.1 SOLUTIONS TO LAPLACE’S EQUATION IN POLAR COORDINATES m= 0 m2≥0 m2≤0 (m→jp) Constant cos[p ln(r)] cosh pφ φ cos[p ln(r)] sinh pφ ln r sin [p ln(r)] cosh pφ φ ln r sin [p ln(r)] sinh pφ rmcosmφ cos [p ln(r)]epφ rmsinmφ cos [p ln(r)]e−pφ r−mcosmφ sin [p ln(r)]epφ r−msinmφ sin [p ln(r)]e−pφ the region of interest includes all values of φ, m must be an integer to make the ﬁeld return to the same value after one revolution. But, mdoes not have to be an integer. If the region of interest is pie shaped, mcan be selected so that the potential passes through one cycle over an arbitrary interval of φ. For example, the periodicity angle can be made φoby making mφo=nπorm=nπ/φ o, where n can have any integer value. The solutions for m2<0, the right-hand column of Table 5.7.1, are illustrated in Fig. 5.7.6 using as an example essentially the fourth solution. Note that the radial phase has been shifted by subtracting p ln(b) from the argument of the sine. Thus, the potential shown is Φ = sin£ p ln(r/b)¤ sinhpφ (15) and it automatically passes through zero at the radius r=b. The distances between radii of zero potential are not equal. Nevertheless, the potential distribution is qual- itatively similar to that in Cartesian coordinates shown in Fig. 5.4.2. The exponen- tial dependence is azimuthal; that direction is thus analogous to yin Fig. 5.4.2. In essence, the potentials for m2<0 are similar to those in Cartesian coordinates but wrapped around the zaxis. 5.8 EXAMPLES IN POLAR COORDINATES With the objective of attaching physical insight to the polar coordinate solutions to Laplace’s equation, two types of examples are of interest. First are certain classic 34 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.7.6 Equipotentials and ﬁeld lines representative of solutions in right- hand column of Table 5.7.1. Potential shown is given by (15). Fig. 5.8.1 Natural boundaries in polar coordinates enclose region V. problems that have simple solutions. Second are examples that require the generally applicable modal approach that makes it possible to satisfy arbitrary boundary conditions. The equipotential cylinder in a uniform applied electric ﬁeld considered in the ﬁrst example is in the ﬁrst category. While an important addition to our resource of case studies, the example is also of practical value because it allows estimates to be made in complex engineering systems, perhaps of the degree to which an applied ﬁeld will tend to concentrate on a cylindrical object. In the most general problem in the second category, arbitrary potentials are imposed on the polar coordinate boundaries enclosing a region V, as shown in Fig. 5.8.1. The potential is the superposition of four solutions, each meeting the potential constraint on one of the boundaries while being zero on the other three. In Cartesian coordinates, the approach used to ﬁnd one of these four solutions, the modal approach of Sec. 5.5, applies directly to the other three. That is, in writing the solutions, the roles of xandycan be interchanged. On the other hand, in polar coordinates the set of solutions needed to represent a potential imposed on the boundaries at r=aorr=bis diﬀerent from that appropriate for potential constraints on the boundaries at φ= 0 or φ=φo. Examples 5.8.2 and 5.8.3 illustrate the two types of solutions needed to determine the ﬁelds in the most general case. In the second of these, the potential is expanded in a set of orthogonal functions that are not sines or cosines. This gives the opportunity to form an appreciation for an orthogonality property of the product solutions to Laplace’s equation that prevails in many other coordinate systems. Sec. 5.8 Examples in Polar Coordinates 35 Simple Solutions. The example considered now is the ﬁrst in a series of “cylinder” case studies built on the same m= 1 solutions. In the next chapter, the cylinder will become a polarizable dielectric. In Chap. 7, it will have ﬁnite conductivity and provide the basis for establishing just how “perfect” a conductor must be to justify the equipotential model used here. In Chaps. 8–10, the ﬁeld will be magnetic and the cylinder ﬁrst perfectly conducting, then magnetizable, and ﬁnally a shell of ﬁnite conductivity. Because of the simplicity of the dipole solutions used in this series of examples, in each case it is possible to focus on the physics without becoming distracted by mathematical details. Example 5.8.1. Equipotential Cylinder in a Uniform Electric Field A uniform electric ﬁeld Eais applied in a direction perpendicular to the axis of a (perfectly) conducting cylinder. Thus, the surface of the conductor, which is at r=R, is an equipotential. The objective is to determine the ﬁeld distribution as modiﬁed by the presence of the cylinder. Because the boundary condition is stated on a circular cylindrical surface, it is natural to use polar coordinates. The ﬁeld excitation comes from “inﬁnity,” where the ﬁeld is known to be uniform, of magnitude Ea, and xdirected. Because our solution must approach this uniform ﬁeld far from the cylinder, it is important to recognize at the outset that its potential, which in Cartesian coordinates is −Eax, is Φ(r→ ∞ )→ −Earcosφ (1) To this must be added the potential produced by the charges induced on the surface of the conductor so that the surface is maintained an equipotential. Because the solutions have to hold over the entire range 0 < φ < 2π, only integer values of the separation constant mare allowed, i.e., only solutions that are periodic in φ. If we are to add a function to (1) that makes the potential zero at r=R, it must cancel the value given by (1) at each point on the surface of the cylinder. There are two solutions in Table 5.7.1 that have the same cos φdependence as (1). We pick the 1/rdependence because it decays to zero as r→ ∞ and hence does not disturb the potential at inﬁnity already given by (1). With Aan arbitrary coeﬃcient, the solution is therefore Φ =−Earcosφ+A rcosφ (2) Because Φ = 0 at r=R, evaluation of this expression shows that the boundary condition is satisﬁed at every angle φif A=EaR2(3) and the potential is therefore Φ =−EaR· r R−R r¸ cosφ (4) The equipotentials given by this expression are shown in Fig. 5.8.2. Note that the x= 0 plane has been taken as having zero potential by omitting an additive constant in (1). The ﬁeld lines shown in this ﬁgure follow from taking the gradient of (4). E=irEa· 1 +¡R r¢2¸ cosφ−iφEa· 1−¡R r¢2¸ sinφ (5) 36 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.8.2 Equipotentials and ﬁeld lines for perfectly conducting cylin- der in initially uniform electric ﬁeld. Field lines tend to concentrate on the surface where φ= 0 and φ=π. At these locations, the ﬁeld is maximum and twice the applied ﬁeld. Now that the boundary value problem has been solved, the surface charge on the cylindrical conductor fol- lows from Gauss’ jump condition, (5.3.2), and the fact that there is no ﬁeld inside the cylinder. σs=n·/epsilon1oE=/epsilon1oEr¯¯ r=R= 2/epsilon1oEacosφ (6) In retrospect, the boundary condition on the circular cylindrical surface has been satisﬁed by adding to the uniform potential that of an xdirected line dipole. Its moment is that necessary to create a ﬁeld that cancels the tangential ﬁeld on the surface caused by the imposed ﬁeld. Azimuthal Modes. The preceding example considered a situation in which Laplace’s equation is obeyed in the entire range 0 < φ < 2π. The next two examples Sec. 5.8 Examples in Polar Coordinates 37 Fig. 5.8.3 Region of interest with zero potential boundaries at φ= 0,Φ =φo, and r=band electrode at r=ahaving potential v. illustrate how the polar coordinate solutions are adapted to meeting conditions on polar coordinate boundaries that have arbitrary locations as pictured in Fig. 5.8.1. Example 5.8.2. Modal Analysis in φ: Fields in and around Corners The conﬁguration shown in Fig. 5.8.3, where the potential is zero on the walls of the region Vatr=band at φ= 0 and φ=φo, but is von a curved electrode at r=a, is the polar coordinate analogue of that considered in Sec. 5.5. What solutions from Table 5.7.1 are pertinent? The region within which Laplace’s equation is to be obeyed does not occupy a full circle, and hence there is no requirement that the potential be a single-valued function of φ. The separation constant mcan assume noninteger values. We shall attempt to satisfy the boundary conditions on the three zero-potential boundaries using individual solutions from Table 5.7.1. Because the potential is zero atφ= 0, the cosine and ln(r) terms are eliminated. The requirement that the potential also be zero at φ=φoeliminates the functions φandφln(r). Moreover, the fact that the remaining sine functions must be zero at φ=φotells us that mφo=nπ. Solutions in the last column are not appropriate because they do not pass through zero more than once as a function of φ. Thus, we are led to the two solutions in the second column that are proportional to sin( nπφ/φ o). Φ =∞X n=1· An¡r b¢nπ/φ o+Bn¡r b¢−nπ/φ o¸ sinµ nπφ φo¶ (7) In writing these solutions, the r’s have been normalized to b, because it is then clear by inspection how the coeﬃcients AnandBnare related to make the potential zero at r=b, A n=−Bn. Φ =∞X n=1An·¡r b¢nπ/φ o−¡r b¢−nπ/φ o¸ sinµ nπφ φo¶ (8) Each term in this inﬁnite series satisﬁes the conditions on the three boundaries that are constrained to zero potential. All of the terms are now used to meet the condition at the “last” boundary, where r=a. There we must represent a potential which jumps abruptly from zero to vatφ= 0, stays at the same vup to φ=φo, and then jumps abruptly from vback to zero. The determination of the coeﬃcients in (8) that make the series of sine functions meet this boundary condition is the same as for (5.5.4) in the Cartesian analogue considered in Sec. 5.5. The parameter nπ(x/a) 38 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.8.4 Pie-shaped region with zero potential boundaries at φ= 0 andφ=φoand electrode having potential vatr=a. (a) With included angle less than 180 degrees, ﬁelds are shielded from region near origin. (b) With angle greater than 180 degrees, ﬁelds tend to concentrate at origin. of Sec. 5.5 is now to be identiﬁed with nπ(φ/φo). With the potential given by (8) evaluated at r=a, the coeﬃcients must be as in (5.5.17) and (5.5.18). Thus, to meet the “last” boundary condition, (8) becomes the desired potential distribution. Φ =∞X n=1 odd4v nπ·¡r b¢nπ/φ o−¡r b¢−nπ/φ o¸ ·¡a b¢nπ/φ o−¡a b¢−nπ/φ o¸sinµ nπ φoφ¶ (9) The distribution of potential and ﬁeld intensity implied by this result is much like that for the region of rectangular cross-section depicted in Fig. 5.5.3. See Fig. 5.8.3. In the limit where b→0, the potential given by (9) becomes Φ =∞X n=1 odd4v nπ¡r a¢nπ/φ osinnπ φoφ (10) and describes the conﬁgurations shown in Fig. 5.8.4. Although the wedge-shaped region is a reasonable “distortion” of its Cartesian analogue, the ﬁeld in a region with an outside corner ( π/φo<1) is also represented by (10). As long as the leading term has the exponent π/φo>1, the leading term in the gradient [with the exponent ( π/φo)−1] approaches zero at the origin. This means that the ﬁeld in a wedge with φo< πapproaches zero at its apex. However, if π/φo<1, which is true for π < φ o<2πas illustrated in Fig. 5.8.4b, the leading term in the gradient of Φ has the exponent ( π/Φo)−1<0, and hence the ﬁeld approaches inﬁnity as r→0. We conclude that the ﬁeld in the neighborhood of a sharp edge is inﬁnite. This observation teaches a lesson for the design of conductor shapes so as to avoid electrical breakdown. Avoid sharp edges! Radial Modes. The modes illustrated so far possessed sinusoidal φdepen- dencies, and hence their superposition has taken the form of a Fourier series. To satisfy boundary conditions imposed on constant φplanes, it is again necessary to have an inﬁnite set of solutions to Laplace’s equation. These illustrate how the Sec. 5.8 Examples in Polar Coordinates 39 Fig. 5.8.5 Radial distribution of ﬁrst three modes given by (13) for a/b= 2. Then= 3 mode is the radial dependence for the potential shown in Fig. 5.7.6. product solutions to Laplace’s equation can be used to provide orthogonal modes that are not Fourier series. To satisfy zero potential boundary conditions at r=bandr=a, it is neces- sary that the function pass through zero at least twice. This makes it clear that the solutions must be chosen from the last column in Table 5.7.1. The functions that are proportional to the sine and cosine functions can just as well be proportional to the sine function shifted in phase (a linear combination of the sine and cosine). This phase shift is adjusted to make the function zero where r=b, so that the radial dependence is expressed as R(r) = sin[ p ln(r)−p ln(b)] = sin[ p ln(r/b)] (11) and the function made to be zero at r=aby setting p ln(a/b) =nπ⇒p=nπ ln(a/b)(12) where nis an integer. The solutions that have now been deﬁned can be superimposed to form a series analogous to the Fourier series. S(r) =∞X n=1SnRn(r); Rn≡sin· nπln(r/b) ln(a/b)¸ (13) Fora/b= 2, the ﬁrst three terms in the series are illustrated in Fig. 5.8.5. They have similarity to sinusoids but reﬂect the polar geometry by having peaks and zero crossings skewed toward low values of r. With a weighting function g(r) =r−1, these modes are orthogonal in the sense that Za b1 rsin· nπln(r/b) ln(a/b)¸ sin· mπln(r/b) ln(a/b)¸ dr=½1 2ln(a/b), m =n 0, m /negationslash=n(14) It can be shown from the diﬀerential equation deﬁning R(r), (5.7.5), and the boundary conditions, that the integration gives zero if the integration is over 40 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.8.6 Region with zero potential boundaries at r=a, r=b, and φ= 0. Electrode at φ=φohas potential v. the product of diﬀerent modes. The proof is analogous to that given in Cartesian coordinates in Sec. 5.5. Consider now an example in which these modes are used to satisfy a speciﬁc boundary condition. Example 5.8.3. Modal Analysis in r The region of interest is of the same shape as in the previous example. However, as shown in Fig. 5.8.6, the zero potential boundary conditions are at r=aandr=b and at φ= 0. The “last” boundary is now at φ=φo, where an electrode connected to a voltage source imposes a uniform potential v. The radial boundary conditions are satisﬁed by using the functions described by (13) for the radial dependence. Because the potential is zero where φ= 0, it is then convenient to use the hyperbolic sine to represent the φdependence. Thus, from the solutions in the last column of Table 5.7.1, we take a linear combination of the second and fourth. Φ =∞X n=1Ansin· nπln(r/b) ln(a/b)¸ sinh· nπ ln(a/b)φ¸ (15) Using an approach that is analogous to that for evaluating the Fourier coeﬃcients in Sec. 5.5, we now use (15) on the “last” boundary, where φ=φoand Φ = v, multiply both sides by the mode Rmdeﬁned with (13) and by the weighting factor 1 /r, and integrate over the radial span of the region. Za b1 rΦ(r, φo) sin· mπln(r/b) ln(a/b)¸ dr=∞X n=1Za bAn rsinh· nπ ln(a/b)φo¸ ·sin· nπln(r/b) ln(a/b)¸ sin· mπln(r/b) ln(a/b)¸ dr(16) Out of the inﬁnite series on the right, the orthogonality condition, (14), picks only them-th term. Thus, the equation can be solved for Amandm→n. With the substitution u=mπln (r/b)/ln(a/b), the integrals can be carried out in closed form. An=( 4v nπsinh£ nπ ln(a/b)φo¤, n odd 0, n even(17) A picture of the potential and ﬁeld intensity distributions represented by (15) and its negative gradient is visualized by “bending” the rectangular region shown by Fig. 5.5.3 into the curved region of Fig. 5.8.6. The role of yis now played by φ. Sec. 5.9 Laplace’s Eq. in Spherical Coordinates 41 Fig. 5.9.1 Spherical coordinate system. 5.9 THREE SOLUTIONS TO LAPLACE’S EQUATION IN SPHERICAL COORDINATES The method employed to solve Laplace’s equation in Cartesian coordinates can be repeated to solve the same equation in the spherical coordinates of Fig. 5.9.1. We have so far considered solutions that depend on only two independent variables. In spherical coordinates, these are commonly randθ. These two-dimensional solutions therefore satisfy boundary conditions on spheres and cones. Rather than embark on an exploration of product solutions in spherical co- ordinates, attention is directed in this section to three such solutions to Laplace’s equation that are already familiar and that are remarkably useful. These will be used to explore physical processes ranging from polarization and charge relaxation dynamics to the induction of magnetization and eddy currents. Under the assumption that there is no φdependence, Laplace’s equation in spherical coordinates is (Table I) 1 r2∂ ∂r¡ r2∂Φ ∂r¢ +1 r2sinθ∂ ∂θ¡ sinθ∂Φ ∂θ¢ = 0 (1) The ﬁrst of the three solutions to this equation is independent of θand is the potential of a point charge. Φ∼1 r(2) If there is any doubt, substitution shows that Laplace’s equation is indeed satisﬁed. Of course, it is not satisﬁed at the origin where the point charge is located. Another of the solutions found before is the three-dimensional dipole, (4.4.10). Φ∼cosθ r2(3) This solution factors into a function of ralone and of θalone, and hence would have to turn up in developing the product solutions to Laplace’s equation in spherical coordinates. Substitution shows that it too is a solution of (1). The third solution represents a uniform z-directed electric ﬁeld in spherical coordinates. Such a ﬁeld has a potential that is linear in z, and in spherical coordi- nates, z=rcosθ. Thus, the potential is Φ∼rcosθ (4) 42 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 These last two solutions, for the three-dimensional dipole at the origin and a ﬁeld due to ±charges at z→ ±∞ , are similar to those for dipoles in two dimensions, them= 1 solutions that are proportional to cos φfrom the second column of Table 5.7.1. However, note that the two-dimensional dipole potential varies as r−1, while the three dimensional dipole potential has an r−2dependence. Also note that whereas the polar coordinate dipole can have an arbitrary orientation (can be a sine as well as a cosine function of φ, or any linear combination of these), the three- dimensional dipole is zdirected. That is, do not replace the cosine function in (3) by a sine function and expect that the potential will satisfy Laplace’s equation in spherical coordinates. Example 5.9.1. Equipotential Sphere in a Uniform Electrical Field Consider a raindrop in an electric ﬁeld. If in the absence of the drop, that ﬁeld is uniform over many drop radii R, the ﬁeld in the vicinity of the drop can be computed by taking the ﬁeld as being uniform “far from the sphere.” The ﬁeld is zdirected and has a magnitude Ea. Thus, on the scale of the drop, the potential must approach that of the uniform ﬁeld (4) as r→ ∞ . Φ(r→ ∞ )→ −Earcosθ (5) We will see in Chap. 7 that it takes only microseconds for a water drop in air to become an equipotential. The condition that the potential be zero at r=Rand yet approach the potential of (5) as r→ ∞ is met by adding to (5) the potential of a dipole at the origin, an adjustable coeﬃcient times (3). By writing the rdependencies normalized to the drop radius R, it is possible to see directly what this coeﬃcient must be. That is, the proposed solution is Φ =−EaRcosθ£r R+A¡R r¢2¤ (6) and it is clear that to make this function zero at r=R, A =−1. Φ =−EaRcosθ£r R−¡R r¢2¤ (7) Note that even though the conﬁguration of a perfectly conducting rod in a uniform transverse electric ﬁeld (as considered in Example 5.8.1) is very diﬀerent from the perfectly conducting sphere in a uniform electric ﬁeld, the potentials are deduced from very similar arguments, and indeed the potentials appear similar. In cross- section, the distribution of potential and ﬁeld intensity is similar to that for the cylinder shown in Fig. 5.8.2. Of course, their appearance in three-dimensional space is very diﬀerent. For the polar coordinate conﬁguration, the equipotentials shown are the cross-sections of cylinders, while for the spherical drop they are cross-sections of surfaces of revolution. In both cases, the potential acquired (by the sphere or the rod) is that of the symmetry plane normal to the applied ﬁeld. The surface charge on the spherical surface follows from (7). σs=−/epsilon1on· ∇Φ¯¯ r=R=/epsilon1oEr¯¯ r=R= 3/epsilon1oEacosθ (8) Thus, for Ea>0, the north pole is capped by positive surface charge while the south pole has negative charge. Although we think of the second solution in (7) as being Sec. 5.9 Laplace’s Eq. in Spherical Coordinates 43 due to a ﬁctitious dipole located at the sphere’s center, it actually represents the ﬁeld of these surface charges. By contrast with the rod, where the maximum ﬁeld is twice the uniform ﬁeld, it follows from (8) that the ﬁeld intensiﬁes by a factor of three at the poles of the sphere. In making practical use of the solution found here, the “uniform ﬁeld at inﬁnity Ea” is that of a ﬁeld that is slowly varying over dimensions on the order of the drop radius R. To demonstrate this idea in speciﬁc terms, suppose that the imposed ﬁeld is due to a distant point charge. This is the situation considered in Example 4.6.4, where the ﬁeld produced by a point charge and a conducting sphere is considered. If the point charge is very far away from the sphere, its ﬁeld at the position of the sphere is essentially uniform over the region occupied by the sphere. (To relate the directions of the ﬁelds in Example 4.6.4 to the present case, mount the θ= 0 axis from the center of the sphere pointing towards the point charge. Also, to make the ﬁeld in the vicinity of the sphere positive, make the point charge negative, q→ −q.) At the sphere center, the magnitude of the ﬁeld intensity due to the point charge is Ea=q 4π/epsilon1oX2(9) The magnitude of the image charge, given by (4.6.34), is Q1=|q|R X(10) and it is positioned at the distance D=R2/Xfrom the center of the sphere. If the sphere is to be charge free, a charge of strength −Q1has to be mounted at its center. If Xis very large compared to R, the distance Dbecomes small enough so that this charge and the charge given by (10) form a dipole of strength p=Q1R2 X=|q|R3 X2(11) The potential resulting from this dipole moment is given by (4.4.10), with pevaluated using this moment. With the aid of (9), the dipole ﬁeld induced by the point charge is recognized as Φ =p 4π/epsilon1or2cosθ=EaR3 r2cosθ (12) As witnessed by (7), this potential is identical to the one we have found necessary to add to the potential of the uniform ﬁeld in order to match the boundary conditions on the sphere. Of the three spherical coordinate solutions to Laplace’s equation given in this section, only two were required in the previous example. The next makes use of all three. Example 5.9.2. Charged Equipotential Sphere in a Uniform Electric Field Suppose that the highly conducting sphere from Example 5.9.1 carries a net charge qwhile immersed in a uniform applied electric ﬁeld Ea. Thunderstorm electriﬁcation is evidence that raindrops are often charged, and Eacould be the ﬁeld they generate collectively. 44 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 In the absence of this net charge, the potential is given by (7). On the boundary atr=R, this potential remains uniform if we add the potential of a point charge at the origin of magnitude q. Φ =−EaRcosθ£r R−¡R r¢2¤ +q 4π/epsilon1or(13) The surface potential has been raised from zero to q/4π/epsilon1oR, but this potential is independent of φand so the tangential electric ﬁeld remains zero. The point charge is, of course, ﬁctitious. The actual charge is distributed over the surface and is found from (13) to be σs=−/epsilon1o∂Φ ∂r¯¯ r=R= 3/epsilon1oEa¡ cosθ+q qc¢ ; qc≡12π/epsilon1oEaR2(14) The surface charge density switches sign when the term in parentheses vanishes, when q/qc<1 and −cosθc=q qc(15) Figure 5.9.2a is a graphical solution of this equation. For Eaandqpositive, the positive surface charge capping the sphere extends into the southern hemisphere. The potential and electric ﬁeld distributions implied by (13) are illustrated in Fig. 5.9.2b. Ifqexceeds qc≡12π/epsilon1oEaR2, the entire surface of the sphere is covered with positive surface charge density and Eis directed outward over the entire surface. 5.10 THREE-DIMENSIONAL SOLUTIONS TO LAPLACE’S EQUATION Natural boundaries enclosing volumes in which Poisson’s equation is to be satisﬁed are shown in Fig. 5.10.1 for the three standard coordinate systems. In general, the distribution of potential is desired within the volume with an arbitrary potential distribution on the bounding surfaces. Considered ﬁrst in this section is the extension of the Cartesian coordinate two-dimensional product solutions and modal expansions introduced in Secs. 5.4 and 5.5 to three dimensions. Given an arbitrary potential distribution over one of the six surfaces of the box shown in Fig. 5.10.1, and given that the other ﬁve surfaces are at zero potential, what is the solution to Laplace’s equation within? If need be, a superposition of six such solutions can be used to satisfy arbitrary conditions on all six boundaries. To use the same modal approach in conﬁgurations where the boundaries are natural to other than Cartesian coordinate systems, for example the cylindrical and spherical ones shown in Fig. 5.10.1, essentially the same extension of the basic ideas already illustrated is used. However, the product solutions involve less familiar functions. For those who understand the two-dimensional solutions, how they are used to meet arbitrary boundary conditions and how they are extended to three- dimensional Cartesian coordinate conﬁgurations, the literature cited in this section should provide ready access to what is needed to exploit solutions in new coordinate systems. In addition to the three standard coordinate systems, there are many Sec. 5.10 Three Solutions 45 Fig. 5.9.2 (a) Graphical solution of (15) for angle θcat which electric ﬁeld switches from being outward to being inward directed on surface of sphere. (b) Equipotentials and ﬁeld lines for perfectly conducting sphere having net charge qin an initially uniform electric ﬁeld. others in which Laplace’s equation admits product solutions. The latter part of this section is intended as an introduction to these coordinate systems and associated product solutions. 46 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.10.1 Volumes deﬁned by natural boundaries in (a) Cartesian, (b) cylindrical, and (c) spherical coordinates. Cartesian Coordinate Product Solutions. In three-dimensions, Laplace’s equation is ∂2Φ ∂x2+∂2Φ ∂y2+∂2Φ ∂z2= 0 (1) We look for solutions that are expressible as products of a function of xalone, X(x), a function of yalone, Y(y), and a function of zalone, Z(z). Φ =X(x)Y(y)Z(z) (2) Introducing (2) into (1) and dividing by Φ, we obtain 1 Xd2X dx2+1 Yd2Y dy2+1 Zd2Z dz2= 0 (3) A function of xalone, added to one of yalone and one of of zalone, gives zero. Because x, y, and zare independent variables, the zero sum is possible only if each of these three “functions” is in fact equal to a constant. The sum of these constants must then be zero. 1 Xd2X dx2=−k2 x;1 Yd2Y dy2=k2 y;1 Zd2Z dz2=−k2 z (4) −k2 x+k2 y−k2 z= 0 (5) Note that if two of these three separation constants are positive, it is then necessary that the third be negative. We anticipated this by writing (4) accordingly. The solutions of (4) are X∼coskxxor sin kxx Y∼coshkyyor sinh kyy (6) Z∼coskzzor sin kzz where k2 y=k2 x+k2 z. Sec. 5.10 Three Solutions 47 Of course, the roles of the coordinates can be interchanged, so either the xor zdirections could be taken as having the exponential dependence. From these solutions it is evident that the potential cannot be periodic or be exponential in its dependencies on all three coordinates and still be a solution to Laplace’s equation. In writing (6) we have anticipated satisfying potential constraints on planes of constant yby taking XandZas periodic. Modal Expansion in Cartesian Coordinates. It is possible to choose the constants and the solutions from (6) so that zero potential boundary conditions are met on ﬁve of the six boundaries. With coordinates as shown in Fig. 5.10.1a, the sine functions are used for XandZto insure a zero potential in the planes x= 0 andz= 0. To make the potential zero in planes x=aandz=w, it is necessary that sinkxa= 0; sin kzw= 0 (7) Solution of these eigenvalue equations gives kx=mπ/a, k z=nπ/w , and hence XZ∼sinmπ axsinnπ wz (8) where mandnare integers. To make the potential zero on the ﬁfth boundary, say where y= 0, the hyperbolic sine function is used to represent the ydependence. Thus, a set of solutions, each meeting a zero potential condition on ﬁve boundaries, is Φ∼sinmπ axsinnπ wzsinhkmny (9) where in view of (5) kmn≡p (mπ/a )2+ (nπ/w )2 These can be used to satisfy an arbitrary potential constraint on the “last” boundary, where y=b. The following example, which extends Sec. 5.5, illustrates this concept. Example 5.10.1. Capacitive Attenuator in Three Dimensions In the attenuator of Example 5.5.1, the two-dimensional ﬁeld distribution is a good approximation because one cross-sectional dimension is small compared to the other. In Fig. 5.5.5, a/lessmuchw. If the cross-sectional dimensions aandware comparable, as shown in Fig. 5.10.2, the ﬁeld can be represented by the modal superposition given by (9). Φ =∞X m=1∞X n=1Amnsinmπ axsinnπ wzsinhkmny (10) In the ﬁve planes x= 0, x=a, y= 0, z= 0, and z=wthe potential is zero. In the plane y=b, it is constrained to be vby an electrode connected to a voltage source. 48 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.10.2 Region bounded by zero potentials at x= 0, x=a, z= 0, z=w, and y= 0. Electrode constrains plane y=bto have potential v. Evaluation of (10) at the electrode surface must give v. v=∞X m=1∞X n=1Amnsinhkmnbsinmπ axsinnπ wz (11) The coeﬃcients Amnare determined by exploiting the orthogonality of the eigen- functions. That is, Za 0XmXidx=n0, m /negationslash=i a 2, m =i;Zw 0ZnZjdz=n0, n/negationslash=j w 2, n =j(12) where Xm≡sinmπ ax; Zn≡sinnπ wz. The steps that now lead to an expression for any given coeﬃcient Amnare a nat- ural extension of those used in Sec. 5.5. Both sides of (11) are multiplied by the eigenfunction XiZjand then both sides are integrated over the surface at y=b. Za 0Zw 0vXiZjdxdz =∞X m=1∞X n=1Amnsinh(kmnb) Za 0Zw 0XmXiZnZjdxdz(13) Because of the product form of each term, the integrations can be carried out on x andzseparately. In view of the orthogonality conditions, (12), the only none-zero term on the right comes in the summation with m=iandn=j. This makes it possible to solve the equation for the coeﬃcient Aij. Then, by replacing i→mand j→n, we obtain Amn=Ra 0Rw 0vsinmπ axsinnπ wzdxdz aw 4sinh(kmnb)(14) Sec. 5.10 Three Solutions 49 The integral can be carried out for any given distribution of potential. In this par- ticular situation, the potential of the surface at y=bis uniform. Thus, integration gives Amn=n16v mnπ21 sinh(kmnb)formandnboth odd 0 for either morneven(15) The desired potential, satisfying the boundary conditions on all six surfaces, is given by (10) and (15). Note that the ﬁrst term in the solution we have found is not the same as the ﬁrst term in the two-dimensional ﬁeld representation, (5.5.9). No matter what the ratio of atow, the ﬁrst term in the three-dimensional solution has a sinusoidal dependence on z, while the two-dimensional one has no dependence on z. For the capacitive attenuator of Fig. 5.5.5, what output signal is predicted by this three dimensional representation? From (10) and (15), the charge on the output electrode is q=Za 0Zw 0£ −/epsilon1o∂Φ ∂y¤ y=0dxdz≡ −CMv (16) where CM=64 π4/epsilon1oaw∞X m=1 odd∞X n=1 oddkmn m2n2sinh(kmnb) With v=Vsinωt, we ﬁnd that vo=Vocosωtwhere Vo=RCnωV (17) Using (16), it follows that the amplitude of the output voltage is Vo U/prime=∞X m=1 odd∞X n=1 oddakmn 2πm2n2sinh£ (kmna)b a¤ (18) where the voltage is normalized to U/prime=128/epsilon1owRωV π3 and kmna=p (nπ)2+ (mπ)2(a/w)2 This expression can be used to replace the plot of Fig. 5.5.5. Here we compare the two-dimensional and three-dimensional predictions of output voltage by considering (18) in the limit where b/greatermucha. In this limit, the hyperbolic sine is dominated by one of its exponentials, and the ﬁrst term in the series gives ln¡Vo U/prime¢ →lnp 1 + (a/w)2−πp 1 + (a/w)2b a(19) In the limit a/w/lessmuch1, the dependence on spacing between input and output elec- trodes expressed by the right hand side becomes identical to that for the two- dimensional model, (5.5.15). However, U/prime= (8/π2)Uregardless of a/w. This three-dimensional Cartesian coordinate example illustrates how the or- thogonality property of the product solution is exploited to provide a potential 50 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. 5.10.3 Two-dimensional square wave function used to represent elec- trode potential for system of Fig. 5.10.2 in plane y=b. that is zero on ﬁve of the boundaries while assuming any desired distribution on the sixth boundary. On this sixth surface, the potential takes the form Φ =∞X m=1 odd∞X n=1 oddVmnFmn (20) where Fmn≡XmZn≡sinmπ axsinnπ wz The two-dimensional functions Fmnhave been used to represent the “last” bound- ary condition. This two-dimensional Fourier series replaces the one-dimensional Fourier series of Sec. 5.5 (5.5.17). In the example, it represents the two-dimensional square wave function shown in Fig. 5.10.3. Note that this function goes to zero along x= 0, x=aandz= 0, z=w, as it should. It changes sign as it passes through any one of these “nodal” lines, but the range outside the original rectangle is of no physical interest, and hence the behavior outside that range does not aﬀect the validity of the solution applied to the example. Because the function represented is odd in both xandy, it can be represented by sine functions only. Our foray into three-dimensional modal expansions extends the notion of or- thogonality of functions with respect to a one-dimensional interval to orthogonality of functions with respect to a two-dimensional section of a plane. We are able to determine the coeﬃcients Vmnin (20) as it is made to ﬁt the potential prescribed on the “sixth” surface because the terms in the series are orthogonal in the sense that Za 0Zw 0FmnFijdxdz =½0 m/negationslash=iorn/negationslash=j aw 4m=iandn=j(21) In other coordinate systems, a similar orthogonality relation will hold for the prod- uct solutions evaluated on one of the surfaces deﬁned by a constant natural coordi- nate. In general, a weighting function multiplies the eigenfunctions in the integrand of the surface integral that is analogous to (21). Except for some special cases, this is as far as we will go in considering three- dimensional product solutions to Laplace’s equation. In the remainder of this sec- tion, references to the literature are given for solutions in cylindrical, spherical, and other coordinate systems. Sec. 5.11 Summary 51 Modal Expansion in Other Coordinates. A general volume having natural boundaries in cylindrical coordinates is shown in Fig. 5.10.1b. Product solutions to Laplace’s equation take the form Φ =R(r)F(φ)Z(z) (22) The polar coordinates of Sec. 5.7 are a special case where Z(z) is a constant. The ordinary diﬀerential equations, analogous to (4) and (5), that determine F(φ) and Z(z), have constant coeﬃcients, and hence the solutions are sines and cosines of mφandkz, respectively. The radial dependence is predicted by an or- dinary diﬀerential equation that, like (5.7.5), has space-varying coeﬃcients. Un- fortunately, with the zdependence, solutions are not simply polynomials. Rather, they are Bessel’s functions of order mand argument kr. As applied to product solutions to Laplace’s equation, these functions are described in standard ﬁelds texts[1−4]. Bessel’s and associated functions are developed in mathematics texts and treatises[5−8]. As has been illustrated in two- and now three-dimensions, the solution to an arbitrary potential distribution on the boundaries can be written as the super- position of solutions each having the desired potential on one boundary and zero potential on the others. Summarized in Table 5.10.1 are the forms taken by the product solution, (22), in representing the potential for an arbitrary distribution on the speciﬁed surface. For example, if the potential is imposed on a surface of constant r, the radial dependence is given by Bessel’s functions of real order and imaginary argument. What is needed to represent Φ in the constant rsurface are functions that are periodic in φandz, so we expect that these Bessel’s functions have an exponential-like dependence on r. In spherical coordinates, product solutions take the form Φ =R(r)/circleminus(θ)F(φ) (23) From the cylindrical coordinate solutions, it might be guessed that new functions are required to describe R(r). In fact, these turn out to be simple polynomials. The φdependence is predicted by a constant coeﬃcient equation, and hence represented by familiar trigonometric functions. But the θdependence is described by Legendre functions. By contrast with the Bessel’s functions, which are described by inﬁnite polynomial series, the Legendre functions are ﬁnite polynomials in cos( θ). In con- nection with Laplace’s equation, the solutions are summarized in ﬁelds texts[1−4]. As solutions to ordinary diﬀerential equations, the Legendre polynomials are pre- sented in mathematics texts[5,7]. The names of other coordinate systems suggest the surfaces generated by set- ting one of the variables equal to a constant: Elliptic-cylinder coordinates and pro- late spheroidal coordinates are examples in which Laplace’s equation is separable[2]. The ﬁrst step in exploiting these new systems is to write the Laplacian and other diﬀerential operators in terms of those coordinates. This is also described in the given references. 5.11 SUMMARY There are two themes in this chapter. First is the division of a solution to a partial diﬀerential equation into a particular part, designed to balance the “drive” in the 52 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 TABLE 5.10.1 FORM OF SOLUTIONS TO LAPLACE’S EQUATION IN CYLINDRICAL COORDINATES WHEN POTENTIAL IS CONSTRAINED ON GIVEN SURFACE AND OTHERS ARE AT ZERO POTENTIAL Surface of ConstantR(r) F(φ) Z(z) r Bessel’s functions of real order and imaginary argument (modiﬁed Bessel’s functions)trigonometric func- tions of real argu- menttrigonometric func- tions of real argu- ment φ Bessel’s functions of imaginary order and imaginary argu- menttrigonometric func- tions of imaginary argumenttrigonometric func- tions of real argu- ment z Bessel’s functions of real order and real argumenttrigonometric func- tions of real argu- menttrigonometric func- tions of imaginary argument diﬀerential equation, and a homogeneous part, used to make the total solution satisfy the boundary conditions. This chapter solves Poisson’s equation; the “drive” is due to the volumetric charge density and the boundary conditions are stated in terms of prescribed potentials. In the following chapters, the approach used here will be applied to boundary value problems representing many diﬀerent physical situations. Diﬀerential equations and boundary conditions will be diﬀerent, but because they will be linear, the same approach can be used. Second is the theme of product solutions to Laplace’s equation which by virtue of their orthogonality can be superimposed to satisfy arbitrary boundary conditions. The thrust of this statement can be appreciated by the end of Sec. 5.5. In the conﬁguration considered in that section, the potential is zero on all but one of the natural Cartesian boundaries of an enclosed region. It is shown that the product solutions can be superimposed to satisfy an arbitrary potential condition on the “last” boundary. By making the “last” boundary any one of the boundaries and, if need be, superimposing as many series solutions as there are boundaries, it is then possible to meet arbitrary conditions on all of the boundaries. The section on polar coordinates gives the opportunity to extend these ideas to systems where the coordinates are not interchangeable, while the section on three-dimensional Cartesian solutions indicates a typical generalization to three dimensions. In the chapters that follow, there will be a frequent need for solving Laplace’s equation. To this end, three classes of solutions will often be exploited: the Carte- sian solutions of Table 5.4.1, the polar coordinate ones of Table 5.7.1, and the three Sec. 5.11 Summary 53 spherical coordinate solutions of Sec. 5.9. In Chap. 10, where magnetic diﬀusion phenomena are introduced and in Chap. 13, where electromagnetic waves are de- scribed, the application of these ideas to the diﬀusion and the Helmholtz equations is illustrated. R E F E R E N C E S [1] M. Zahn, Electromagnetic Field Theory: A Problem Solving Approach , John Wiley and Sons, N.Y. (1979). [2] P. Moon and D. E. Spencer, Field Theory for Engineers , Van Nostrand, Princeton, N.J. (1961). [3] S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Com- munication Electronics , John Wiley and Sons, N.Y. (1967). [4] J. R. Melcher, Continuum Electromechanics , M.I.T. Press, Cambridge, Mass. (1981). [5] F. B. Hildebrand, Advanced Calculus for Applications , Prentice-Hall, Inc, Englewood Cliﬀs, N.J. (1962). [6] G. N. Watson, A Treatise on the Theory of Bessel Functions , Cambridge University Press, London E.C.4. (1944). [7] P. M. Morse and H. Feshbach, Methods of Theoretical Physics , McGraw-Hill Book Co., N.Y. (1953). [8] N. W. McLachlan, Bessel Functions for Engineers , Oxford University Press, London E.C.4 (1941). 54 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 P R O B L E M S 5.1 Particular and Homogeneous Solutions to Poisson’s and Laplace’s Equations 5.1.1 In Problem 4.7.1, the potential of a point charge over a perfectly conducting plane (where z > 0) was found to be Eq. (a) of that problem. Identify particular and homogeneous parts of this solution. 5.1.2 A solution for the potential in the region −a < y < a , where there is a charge density ρ, satisﬁes the boundary conditions Φ = 0 in the planes y= +aandy=−a. Φ =ρo /epsilon1oβ2µ 1−coshβy coshβa¶ cosβx (a) (a) What is ρin this region? (b) Identify Φ pand Φ h. What boundary conditions are satisﬁed by Φ hat y= +aandy=−a? (c) Illustrate another combination of Φ pand Φ hthat could just as well be used and give the boundary conditions that apply for Φ hin that case. 5.1.3∗The charge density between the planes x= 0 and x=ddepends only on x. ρ=4ρo(x−d)2 d2(a) Boundary conditions are that Φ( x= 0) = 0 and Φ( x=d) =V, so Φ = Φ(x) is independent of yandz. (a) Show that Poisson’s equation therefore reduces to ∂2Φ ∂x2=−4ρo d2/epsilon1o(x−d)2(b) (b) Integrate this expression twice and use the boundary conditions to show that the potential distribution is Φ =−ρo 3d2/epsilon1o(x−d)4+¡V d−ρod 3/epsilon1o¢ x+ρod2 3/epsilon1o(c) (c) Argue that the ﬁrst term in (c) can be Φ p, with the remaining terms then Φ h. (d) Show that in that case, the boundary conditions satisﬁed by Φ hare Φh(0) =ρod2 3/epsilon1o; Φ h(d) =V (d) Sec. 5.3 Problems 55 5.1.4 With the charge density given as ρ=ρosinπx d(a) carry out the steps in Prob. 5.1.3. Fig. P5.1.5 5.1.5∗A frequently used model for a capacitor is shown in Fig. P5.1.5, where two plane parallel electrodes have a spacing that is small compared to either of their planar dimensions. The potential diﬀerence between the electrodes is v, and so over most of the region between the electrodes, the electric ﬁeld is uniform. (a) Show that in the region well removed from the edges of the electrodes, the ﬁeld E=−(v/d)izsatisﬁes Laplace’s equation and the boundary conditions on the electrode surfaces. (b) Show that the surface charge density on the lower surface of the upper electrode is σs=/epsilon1ov/d. (c) For a single pair of electrodes, the capacitance Cis deﬁned such that q=Cv(13). Show that for the plane parallel capacitor of Fig. P5.1.5, C=A/epsilon1o/d, where Ais the area of one of the electrodes. (d) Use the integral form of charge conservation, (1.5.2), to show that i=dq/dt =Cdv/dt . 5.1.6∗In the three-electrode system of Fig. P5.1.6, the bottom electrode is taken as having the reference potential. The upper and middle electrodes then have potentials v1andv2, respectively. The spacings between electrodes, 2 d andd, are small enough relative to the planar dimensions of the electrodes so that the ﬁelds between can be approximated as being uniform. (a) Show that the ﬁelds denoted in the ﬁgure are then approximately E1=v1/2d, E 2=v2/dandEm= (v1−v2)/d. (b) Show that the net charges q1andq2on the top and middle electrodes, respectively, are related to the voltages by the capacitance matrix [in the form of (12)] · q1 q2¸ =· /epsilon1ow(L+l)/2d−/epsilon1owl/d −/epsilon1owl/d 2/epsilon1owl/d¸· v1 v2¸ (a) 56 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. P5.1.6 5.3 Continuity Conditions 5.3.1∗The electric potentials Φaand Φbabove and below the plane y= 0 are Φa=Vcosβxexp(−βy); y >0 Φb=Vcosβxexp(βy); y <0(a) (a) Show that (4) holds. (The potential is continuous at y= 0.) (b) Evaluate Etangential to the surface y= 0 and show that it too is continuous. [Equation (1) is then automatically satisﬁed at y= 0.] (c) Use (5) to show that in the plane y= 0, the surface charge density, σs= 2/epsilon1oβVcosβx, accounts for the discontinuity in the derivative of Φ normal to the plane y= 0. 5.3.2 By way of appreciating how the continuity of Φ guarantees the continuity of tangential E[(4) implies that (1) is satisﬁed], suppose that the potential is given in the plane y= 0: Φ = Φ( x,0, z). (a) Which components of Ecan be determined from this information alone? (b) For example, if Φ( x,0, z) =Vsin(βx) sin(βz), what are those compo- nents of E? 5.4 Solutions to Laplace’s Equation in Cartesian Coordinates 5.4.1∗A region that extends to ±∞in the zdirection has the square cross-section of dimensions as shown in Fig. P5.4.1. The walls at x= 0 and y= 0 are at zero potential, while those at x=aandy=ahave the linear distributions shown. The interior region is free of charge density. (a) Show that the potential inside is Φ =Voxy a2(a) Sec. 5.4 Problems 57 Fig. P5.4.1 Fig. P5.4.2 (b) Show that plots of Φ and Eare as shown in the ﬁrst quadrant of Fig. 4.1.3. 5.4.2 One way to constrain a boundary so that it has a potential distribution that is a linear function of position is shown in Fig. P5.4.2a. A uniformly resistive sheet having a length 2 ais driven by a voltage source V. For the coordinate xshown, the resulting potential distribution is the linear function of xshown. The constant Cis determined by the deﬁnition of where the potential is zero. In the case shown in Fig. 5.4.2a, if Φ is zero at 58 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 x= 0, then C= 0. (a) Suppose a cylindrical region having a square cross-section of length 2aon a side, as shown in Fig. 5.4.2b, is constrained in potential by resistive sheets and voltage sources, as shown. Note that the potential is deﬁned to be zero at the lower right-hand corner, where ( x, y) = (a,−a). Inside the cylinder, what must the potential be in the planes x=±aandy=±a? (b) Find the linear combination of the potentials from the ﬁrst column of Table 5.4.1 that satisﬁes the conditions on the potentials required by the resistive sheets. That is, if Φ takes the form Φ =Ax+By+C+Dxy (a) so that it satisﬁes Laplace’s equation inside the cylinder, what are the coeﬃcients A,B,C, and D? (c) Determine Efor this potential. (d) Sketch Φ and E. (e) Now the potential on the walls of the square cylinder is constrained as shown in Fig. 5.4.2c. This time the potential is zero at the location (x, y) = (0 ,0). Adjust the coeﬃcients in (a) so that the potential satisﬁes these conditions. Determine Eand sketch the equipotentials and ﬁeld lines. 5.4.3∗Shown in cross-section in Fig. P5.4.3 is a cylindrical system that extends to inﬁnity in the ±zdirections. There is no charge density inside the cylinder, and the potentials on the boundaries are Φ =Vocosπ axaty=±b (a) Φ = 0 at x=±a 2(b) (a) Show that the potential inside the cylinder is Φ =Vocosπx acoshπy a/coshπb a(c) (b) Show that a plot of Φ and Eis as given by the part of Fig. 5.4.1 where −π/2< kx < π/ 2. 5.4.4 The square cross-section of a cylindrical region that extends to inﬁnity in the±zdirections is shown in Fig. P5.4.4. The potentials on the boundaries are as shown. (a) Inside the cylindrical space, there is no charge density. Find Φ. (b) What is Ein this region? Sec. 5.4 Problems 59 Fig. P5.4.4 Fig. P5.4.5 (c) Sketch Φ and E. 5.4.5∗The cross-section of an electrode structure which is symmetric about the x= 0 plane is shown in Fig. P5.4.5. Above this plane are electrodes that alternately either have the potential v(t) or the potential −v(t). The system has depth d(into the paper) which is very long compared to such dimen- sions as aorl. So that the current i(t) can be measured, one of the upper electrodes has a segment which is insulated from the rest of the electrode, but driven by the same potential. The geometry of the upper electrodes is speciﬁed by giving their altitudes above the x= 0 plane. For example, the upper electrode between y=−bandy=bhas the shape η=1 ksinh−1¡sinhka coskx¢ ; k=π 2b(a) where ηis as shown in Fig. P5.4.5. (a) Show that the potential in the region between the electrodes is Φ =v(t) coskxsinhky/sinhka (b) 60 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 (b) Show that Ein this region is E=v(t) sinh¡πa 2b¢¡π 2b¢· sin¡πx 2b¢ sinh¡πy 2b¢ ix −cos¡πx 2b¢ cosh¡πy 2b¢ iy¸ (c) (c) Show that plots of Φ and Eare as shown in Fig. 5.4.2. (d) Show that the net charge on the upper electrode segment between y=−landy=lis q=2/epsilon1od sinhkasinkl· 1 +¡sinhka coskl¢2¸1/2 v(t) =Cv (d) (Because the surface Sin Gauss’ integral law is arbitrary, it can be chosen so that it both encloses this electrode and is convenient for integration.) (e) Given that v(t) =Vosinωt, where Voandωare constants, show that the current to the electrode segment i(t), as deﬁned in Fig. P5.4.5, is i=dq dt=Cdv dt=CωV ocosωt (e) 5.4.6 In Prob. 5.4.5, the polarities of all of the voltage sources driving the lower electrodes are reversed. (a) Find Φ in the region between the electrodes. (b) Determine E. (c) Sketch Φ and E. (d) Find the charge qon the electrode segment in the upper middle elec- trode. (e) Given that v(t) =Vocosωt, what is i(t)? 5.5 Modal Expansion to Satisfy Boundary Conditions 5.5.1∗The system shown in Fig. P5.5.1a is composed of a pair of perfectly con- ducting parallel plates in the planes x= 0 and x=athat are shorted in the plane y=b. Along the left edge, the potential is imposed and so has a given distribution Φ d(x). The plates and short have zero potential. (a) Show that, in terms of Φ d(x), the potential distribution for 0 < y < b,0< x < a is Φ =∞X n=1Ansin¡nπx a¢ sinh£nπ a(y−b)¤ (a) Sec. 5.5 Problems 61 Fig. P5.5.1 where An=2 asinh¡ −nπb a¢Za 0Φd(x) sin¡nπx a¢ dx (b) (At this stage, the coeﬃcients in a modal expansion for the ﬁeld are left expressed as integrals over the yet to be speciﬁed potential distribution.) (b) In particular, if the imposed potential is as shown in Fig. P5.5.1b, show that Anis An=−4V1 nπcos¡nπ 4¢ sinh¡nπb a¢ (c) 5.5.2∗The walls of a rectangular cylinder are constrained in potential as shown in Fig. P5.5.2. The walls at x=aandy=bhave zero potential, while those at y= 0 and x= 0 have the potential distributions V1(x) and V2(y), respectively. In particular, suppose that these distributions of potential are uni- form, so that V1(x) =VaandV2(y) =Vb, with VaandVbdeﬁned to be independent of xandy. (a) The region inside the cylinder is free space. Show that the potential distribution there is Φ =∞X n=1 odd· −4Vb nπsinhnπ b(x−a) sinhnπa bsinnπy b −4Va nπsinhnπ a(y−b) sinhnπb asinnπx a¸(a) (b) Show that the distribution of surface charge density along the wall at x=ais σs=∞X n=1 odd· −4/epsilon1oVb bsinnπ by sinhnπa b+4/epsilon1oVa asinhnπ a(y−b) sinhnπb a¸ (b) 62 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. P5.5.2 Fig. P5.5.3 5.5.3 In the conﬁguration described in Prob. 5.5.2, the distributions of potentials on the walls at x= 0 and y= 0 are as shown in Fig. P5.5.3, where the peak voltages VaandVbare given functions of time. (a) Determine the potential in the free space region inside the cylinder. (b) Find the surface charge distribution on the wall at y=b. 5.5.4∗The cross-section of a system that extends to “inﬁnity” out of the paper is shown in Fig. P5.5.4. An electrode in the plane y=dhas the potential V. A second electrode has the shape of an “L.” One of its sides is in the plane y= 0, while the other is in the plane x= 0, extending from y= 0 almost toy=d. This electrode is at zero potential. (a) The electrodes extend to inﬁnity in the −xdirection. Show that, far to the left, the potential between the electrodes tends to Φ =V y d(a) (b) Using this result as a part of the solution, Φ a, the potential between the plates is written as Φ = Φ a+ Φb. Show that the boundary condi- tions that must be satisﬁed by Φ bare Φb= 0 at y= 0 and y=d (b) Φb→0 as x→ −∞ (c) Φb=−V y datx= 0 ( d) Sec. 5.5 Problems 63 Fig. P5.5.4 Fig. P5.5.5 (c) Show that the potential between the electrodes is Φ =V y d+∞X n=12V nπ(−1)nsinnπy dexp¡nπx d¢ (e) (d) Show that a plot of Φ and Eappears as shown in Fig. 6.6.9c, turned upside down. 5.5.5 In the two-dimensional system shown in cross-section in Fig. P5.5.5, plane parallel plates extend to inﬁnity in the −ydirection. The potentials of the upper and lower plates are, respectively, −Vo/2 and Vo/2. The potential over the plane y= 0 terminating the plates at the right is speciﬁed to be Φd(x). (a) What is the potential distribution between the plates far to the left? (b) If Φ is taken as the potential Φ athat assumes the correct distribution asy→ −∞ , plus a potential Φ b, what boundary conditions must be satisﬁed by Φ b? (c) What is the potential distribution between the plates? 5.5.6 As an alternative (and in this case much more complicated) way of ex- pressing the potential in Prob. 5.4.1, use a modal approach to express the potential in the interior region of Fig. P5.4.1. 5.5.7∗Take an approach to ﬁnding the potential in the conﬁguration of Fig. 5.5.2 that is an alternative to that used in the text. Let Φ = ( V y/b ) + Φ 1. (a) Show that the boundary conditions that must be satisﬁed by Φ 1are that Φ 1=−V y/b atx= 0 and at x=a, and Φ 1= 0 at y= 0 and y=b. 64 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. P5.6.1 (b) Show that the potential is Φ =V y b+∞X n=1Ancoshnπ b¡ x−a 2¢ sinnπy b(a) where An=2V(−1)n cosh¡nπa 2b¢ (b) (It is convenient to exploit the symmetry of the conﬁguration about the plane x=a/2.) 5.6 Solutions to Poisson’s Equation with Boundary Conditions 5.6.1∗The potential distribution is to be determined in a region bounded by the planes y= 0 and y=dand extending to inﬁnity in the xandzdirections, as shown in Fig. P5.6.1. In this region, there is a uniform charge density ρo. On the upper boundary, the potential is Φ( x, d, z ) =Vasin(βx). On the lower boundary, Φ( x,0, z) =Vbsin(αx). Show that Φ( x, y, z ) throughout the region 0 < y < d is Φ =Vasinβxsinhβy sinhβd−Vbsinαxsinhα(y−d) sinhαd −ρo /epsilon1o¡y2 2−yd 2¢(a) 5.6.2 For the conﬁguration of Fig. P5.6.1, the charge is again uniform in the region between the boundaries, with density ρo, but the potential at y=d is Φ = Φ osin(kx), while that at y= 0 is zero (Φ oandkare given constants). Find Φ in the region where 0 < y < d , between the boundaries. 5.6.3∗In the region between the boundaries at y=±d/2 in Fig. P5.6.3, the charge density is ρ=ρocosk(x−δ); −d 2< y <d 2(a) Sec. 5.6 Problems 65 Fig. P5.6.3 where ρoandδare given constants. Electrodes at y=±d/2 constrain the tangential electric ﬁeld there to be Ex=Eocoskx (b) The charge density might represent a traveling wave of space charge on a modulated particle beam, and the walls represent the traveling-wave structure which interacts with the beam. Thus, in a practical device, such as a traveling-wave ampliﬁer designed to convert the kinetic energy of the moving charge to ac electrical energy available at the electrodes, the charge and potential distributions move to the right with the same velocity. This does not concern us, because we consider the interaction at one instant in time. (a) Show that a particular solution is Φp=ρo /epsilon1ok2cosk(x−δ) ( c) (b) Show that the total potential is the sum of this solution and that solution to Laplace’s equation that makes the total solution satisfy the boundary conditions. Φ = Φ p−coshky cosh¡kd 2¢·Eo ksinkx+ρo /epsilon1ok2cosk(x−δ)¸ (d) (c) The force density (force per unit volume) acting on the charge is ρE. Show that the force fxacting on a section of the charge of length in thexdirection λ= 2π/kspanning the region −d/2< y < d/ 2 and unit length in the zdirection is fx=2πρoEo k2tanh¡kd 2¢ coskδ (e) 5.6.4 In the region 0 < y < d shown in cross-section in Fig. P5.6.4, the charge density is ρ=ρocosk(x−δ); 0 < y < d (a) where ρoandδare constants. Electrodes at y=dconstrain the potential there to be Φ( x, d) = Vocos(kx) (Voandkgiven constants), while an electrode at y= 0 makes Φ( x,0) = 0. 66 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. P5.6.4 Fig. P5.6.5 (a) Find a particular solution that satisﬁes Poisson’s equation everywhere between the electrodes. (b) What boundary conditions must the homogeneous solution satisfy at y=dandy= 0? (c) Find Φ in the region 0 < y < d . (d) The force density (force per unit volume) acting on the charge is ρE. Find the total force fxacting on a section of the charge spanning the system from y= 0 to y=d, of unit length in the zdirection and of length λ= 2π/kin the xdirection. 5.6.5∗A region that extends to inﬁnity in the ±zdirections has a rectangular cross-section of dimensions 2 aandb, as shown in Fig. P5.6.5. The bound- aries are at zero potential while the region inside has the distribution of charge density ρ=ρosin¡πy b¢ (a) where ρois a given constant. Show that the potential in this region is Φ =ρo /epsilon1o(b/π)2sin¡πy b¢£ 1−coshπx b/coshπa b¤ (b) 5.6.6 The cross-section of a two-dimensional conﬁguration is shown in Fig. P5.6.6. The potential distribution is to be determined inside the boundaries, which are all at zero potential. (a) Given that a particular solution inside the boundaries is Φp=Vsin¡πy b¢ sinβx (a) Sec. 5.6 Problems 67 Fig. P5.6.6 Fig. P5.6.7 where Vandβare given constants, what is the charge density in that region? (b) What is Φ? 5.6.7 The cross-section of a metal box that is very long in the zdirection is shown in Fig. P5.6.7. It is ﬁlled by the charge density ρox/l. Determine Φ inside the box, given that Φ = 0 on the walls. 5.6.8∗In region (b), where y <0, the charge density is ρ=ρocos(βx)eαy, where ρo, β, and αare positive constants. In region (a), where 0 < y, ρ = 0. (a) Show that a particular solution in the region y <0 is Φp=ρo /epsilon1o(β2−α2)cos(βx) exp( αy) ( a) (b) There is no surface charge density in the plane y= 0. Show that the potential is Φ =−ρocosβx /epsilon1o(β2−α2)2(¡α β−1¢ exp(−βy); 0 < y −2 exp( αy) +¡α β+ 1¢ exp(βy);y <0(b) 5.6.9 A sheet of charge having the surface charge density σs=σosinβ(x−xo) is in the plane y= 0, as shown in Fig. 5.6.3. At a distance aabove and below the sheet, electrode structures are used to constrain the potential to be Φ =Vcosβx. The system extends to inﬁnity in the xandzdirections. The regions above and below the sheet are designated (a) and (b), respectively. (a) Find Φ aand Φ bin terms of the constants V, β, σ o, and xo. 68 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 (b) Given that the force per unit area acting on the charge sheet is σsEx(x,0), what is the force acting on a section of the sheet hav- ing length din the zdirection and one wavelength 2 π/β in the x direction? (c) Now, the potential on the wall is made a traveling wave having a given angular frequency ω,Φ(x,±a, t) =Vcos(βx−ωt), and the charge moves to the right with a velocity U, so that σs=σosinβ(x−Ut−xo), where U=ω/β. Thus, the wall potentials and surface charge density move in synchronism. Building on the results from parts (a)–(b), what is the potential distribution and hence total force on the section of charged sheet? (d) What you have developed is a primitive model for an electron beam device used to convert the kinetic energy of the electrons (accelerated to the velocity vby a dc voltage) to high-frequency electrical power output. Because the system is free of dissipation, the electrical power output (through the electrode structure) is equal to the mechanical power input. Based on the force found in part (c), what is the electrical power output produced by one period 2 π/β of the charge sheet of width w? (e) For what values of xowould the device act as a generator of electrical power? 5.7 Solutions to Laplace’s Equation in Polar Coordinates 5.7.1∗A circular cylindrical surface r=ahas the potential Φ = Vsin 5φ. The regions r < a anda < r are free of charge density. Show that the potential is Φ =V½ (r/a)5sin 5φ r < a (a/r)5sin 5φ a < r(a) 5.7.2 Thex−zplane is one of zero potential. Thus, the yaxis is perpendicular to a zero potential plane. With φmeasured relative to the xaxis and zthe third coordinate axis, the potential on the surface at r=Ris constrained by segmented electrodes there to be Φ = Vsinφ. (a) If ρ= 0 in the region r < R , what is Φ in that region? (b) Over the range r < R , what is the surface charge density on the surface at y= 0? 5.7.3∗An annular region b < r < a where ρ= 0 is bounded from outside at r=a by a surface having the potential Φ = Vacos 3φand from the inside at r=b by a surface having the potential Φ = Vbsinφ. Show that Φ in the annulus can be written as the sum of two terms, each a combination of solutions to Laplace’s equation designed to have the correct value at one radius while Sec. 5.8 Problems 69 being zero at the other. Φ =Vacos 3φ[(r/b)3−(b/r)3] [(a/b)3−(b/a)3]+Vbsinφ[(r/a)−(a/r)] [(b/a)−(a/b)](a) 5.7.4 In the region b < r < a, 0< φ < α, ρ = 0. On the boundaries of this region at r=a, atφ= 0 and φ=α,Φ = 0. At r=b,Φ =Vbsin(πφ/α ). Determine Φ in this region. 5.7.5∗In the region b < r < a, 0< φ < α, ρ = 0. On the boundaries of this region at r=a, r =band at φ= 0,Φ = 0. At φ=α, the potential is Φ =Vsin[3πln(r/a)/ln(b/a)]. Show that within the region, Φ =Vsinh·3πφ ln(b/a)¸ sin· 3πln(r/a) ln(b/a)¸± sinh·3πα ln(b/a)¸ (a) 5.7.6 The plane φ= 0 is at potential Φ = V, while that at φ= 3π/2 is at zero potential. The system extends to inﬁnity in the ±zandrdirections. Determine and sketch Φ and Ein the range 0 < φ < 3π/2. 5.8 Examples in Polar Coordinates 5.8.1∗Show that Φ and Eas given by (4) and (5), respectively, describe the potential and electric ﬁeld intensity around a perfectly conducting half- cylinder at r=Ron a perfectly conducting plane at x= 0 with a uniform ﬁeld Eaixapplied at x→ ∞ . Show that the maximum ﬁeld intensity is twice that of the applied ﬁeld, regardless of the radius of the half-cylinder. 5.8.2 Coaxial circular cylindrical surfaces bound an annular region of free space where b < r < a . On the inner surface, where r=b, Φ = Vb>0. On the outer surface, where r=a,Φ =Va>0. (a) What is Φ in the annular region? (b) How large must Vbbe to insure that all lines of Eare outward directed from the inner cylinder? (c) What is the net charge per unit length on the inner cylinder under the conditions of (b)? 5.8.3∗A device proposed for using the voltage voto measure the angular velocity Ω of a shaft is shown in Fig. P5.8.3a. A cylindrical grounded electrode has radius R. (The resistance Rois “small.”) Outside and concentric at r=a is a rotating shell supporting the surface charge density distribution shown in Fig. P5.8.3b. 70 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 Fig. P5.8.3 (a) Given θoandσo, show that in regions (a) and (b), respectively, outside and inside the rotating shell, Φ =2σoa π/epsilon1o∞X m=1 odd1 m2· ½ [(a/R)m−(R/a)m](R/r)msinm(φ−θo);a < r (R/a)m[(r/R)m−(R/r)m] sinm(φ−θo);R < r < a(a) (b) Show that the charge on the segment of the inner electrode attached to the resistor is q=∞X m=1 oddQm[cosmθo−cosm(α−θo)]; Qm≡4wσoa m2π(R/a)m(b) where wis the length in the zdirection. (c) Given that θo= Ωt, show that the output voltage is related to Ω by vo(t) =∞X m=1 oddQmmΩRo[sinm(α−Ωt) + sin mΩt] ( c) so that its amplitude can be used to measure Ω. 5.8.4 Complete the steps of Prob. 5.8.3 with the conﬁguration of Fig. P5.8.3 altered so that the rotating shell is inside rather than outside the grounded electrode. Thus, the radius aof the rotating shell is less than the radius R, and region (a) is a < r < R , while region (b) is r < a . 5.8.5∗A pair of perfectly conducting zero potential electrodes form a wedge, one in the plane φ= 0 and the other in the plane φ=α. They essentially extend to inﬁnity in the ±zdirections. Closing the region between the electrodes atr=Ris an electrode having potential V. Show that the potential inside the region bounded by these three surfaces is Φ =∞X m=1 odd4V mπ(r/R)mπ/αsin¡mπφ α¢ (a) Sec. 5.9 Problems 71 Fig. P5.8.7 5.8.6 In a two-dimensional system, the region of interest is bounded in the φ= 0 plane by a grounded electrode and in the φ=αplane by one that has Φ =V. The region extends to inﬁnity in the rdirection. At r=R,Φ =V. Determine Φ. 5.8.7 Figure P5.8.7 shows a circular cylindrical wall having potential Vorelative to a grounded ﬁn in the plane φ= 0 that reaches from the wall to the center. The gaps between the cylinder and the ﬁn are very small. (a) Find all solutions in polar coordinates that satisfy the boundary con- ditions at φ= 0 and φ= 2π. Note that you cannot accept solutions for Φ of negative powers in r. (b) Match the boundary condition at r=R. (c) One of the terms in this solution has an electric ﬁeld intensity that is inﬁnite at the tip of the ﬁn, where r= 0. Sketch Φ and Ein the neighborhood of the tip. What is the σson the ﬁn associated with this term as a function of r? What is the net charge associated with this term? (d) Sketch the potential and ﬁeld intensity throughout the region. 5.8.8 A two-dimensional system has the same cross-sectional geometry as that shown in Fig. 5.8.6 except that the wall at φ= 0 has the potential v. The wall at φ=φois grounded. Determine the interior potential. 5.8.9 Use arguments analogous to those used in going from (5.5.22) to (5.5.26) to show the orthogonality (14) of the radial modes Rndeﬁned by (13). [Note the comment following (14).] 5.9 Three Solutions to Laplace’s Equation in Spherical Coordinates 5.9.1 On the surface of a spherical shell having radius r=a, the potential is Φ =Vcosθ. (a) With no charge density either outside or inside this shell, what is Φ forr < a and for r > a ? (b) Sketch Φ and E. 72 Electroquasistatic Fields from the Boundary Value Point of View Chapter 5 5.9.2∗A spherical shell having radius asupports the surface charge density σocosθ. (a) Show that if this is the only charge in the volume of interest, the potential is Φ =σoa 3/epsilon1o½ (a/r)2cosθ r≥a (r/a) cosθ r ≤a(a) (b) Show that a plot of Φ and Eappears as shown in Fig. 6.3.1. 5.9.3∗A spherical shell having zero potential has radius a. Inside, the charge density is ρ=ρocosθ. Show that the potential there is Φ =a2ρo 4/epsilon1o[(r/a)−(r/a)2] cosθ (a) 5.9.4 The volume of a spherical region is ﬁlled with the charge density ρ= ρo(r/a)mcosθ, where ρoandmare given constants. If the potential Φ = 0 atr=a, what is Φ for r < a ? 5.10 Three-Dimensional Solutions to Laplace’s Equation 5.10.1∗In the conﬁguration of Fig. 5.10.2, all surfaces have zero potential except those at x= 0 and x=a, which have Φ = v. Show that Φ =∞X m=1 odd∞X n=1 oddAmnsin¡mπy b¢ sin¡nπz w¢ coshkmn¡ x−a 2¢ (a) and Amn=16v mnπ2/coshkmna 2(b) 5.10.2 In the conﬁguration of Fig. 5.10.2, all surfaces have zero potential. In the plane y=a/2, there is the surface charge density σs=σosin(πx/a ) sin(πz/w ). Find the potentials Φaand Φbabove and below this surface, respectively. 5.10.3 The conﬁguration is the same as shown in Fig. 5.10.2 except that all of the walls are at zero potential and the volume is ﬁlled by the uniform charge density ρ=ρo. Write four essentially diﬀerent expressions for the potential distribution. 6 POLARIZATION 6.0 INTRODUCTION The previous chapters postulated surface charge densities that appear and disap- pear as required by the boundary conditions obeyed by surfaces of conductors. Thus, the idea that the distribution of the charge density may be linked to the ﬁeld it induces is not new. Thus far, however, no consideration has been given in any detail to the physical laws which determine the occurrence and behavior of charge densities in matter. To set the stage for this and the next chapter, consider two possible pictures that could be used to explain why an object distorts an initially uniform electric ﬁeld. In Fig. 6.0.1a, the sphere is composed of a metallic conductor, and therefore composed of atoms having electrons that are free to move from one atomic site to another. Suppose, to begin with, that there are equal numbers of positive sites and negative electrons. In the absence of an applied ﬁeld and on a scale that is large compared to the distance between atoms (that is, on a macroscopic scale), there is therefore no charge density at any point within the material. When this object is placed in an initially uniform electric ﬁeld, the electrons are subject to forces that tend to make them concentrate on the south pole of the sphere. This requires only that the electrons migrate downward slightly (on the average, less than an interatomic distance). Because the interior of the sphere must be ﬁeld free in the ﬁnal equilibrium (steady) state, the charge density remains zero at each point within the volume of the material. However, to preserve a zero net charge, the positive atomic sites on the north pole of the sphere are uncovered. After a time, the net result is the distribution of surface charge density shown in Fig. 6.0.1b. [In fact, provided the electrodes are well-removed from the sphere, this is the distribution found in Example 5.9.1.] Now consider an alternative picture of the physics that can lead to a very sim- ilar result. As shown in Fig. 6.0.1c, the material is composed of atoms, molecules, 1 2 Polarization Chapter 6 Fig. 6.0.1 In the left-hand sequence, the sphere is conducting, while on the right, it is polarizable and not conducting. or groups of molecules (domains) in which the electric ﬁeld induces dipole mo- ments. For example, suppose that the dipole moments are of an atomic scale and, in the absence of an electric ﬁeld, do not exist; the moments are induced because atoms contain positively charged nuclei and electrons orbiting around the nuclei. According to quantum theory, electrons orbiting the nuclei are not to be viewed as localized at any particular instant of time. It is more appropriate to think of the electrons as “clouds” of charge surrounding the nuclei. Because the charge of the orbiting electrons is equal and opposite to the charge of the nuclei, a neutral atom has no net charge. An atom with no permanent dipole moment has the further Sec. 6.0 Introduction 3 Fig. 6.0.2 Nucleus with surrounding electronic charge cloud displaced by applied electric ﬁeld. property that the center of the negative charge of the electron “clouds” coincides with the center of the positive charge of the nuclei. In the presence of an electric ﬁeld, the center of positive charge is pulled in the direction of the ﬁeld while the center of negative charge is pushed in the opposite direction. At the atomic level, this relative displacement of charge centers is as sketched in Fig. 6.0.2. Because the two centers of charge no longer coincide, the particle acquires a dipole moment. We can represent each atom by a pair of charges of equal magnitude and opposite sign separated by a distance d. On the macroscopic scale of the sphere and in an applied ﬁeld, the dipoles then appear somewhat as shown in Fig. 6.0.1d. In the interior of the sphere, the polarization leaves each positive charge in the vicinity of a negative one, and hence there is no net charge density. However, at the north pole there are no negative charges to neutralize the positive ones, and at the south pole no positive ones to pair up with the negative ones. The result is a distribution of surface charge density that does not diﬀer qualitatively from that for the metal sphere. How can we distinguish between these two very diﬀerent situations? Suppose that the two spheres make contact with the lower electrode, as shown in parts (e) and (f) of the ﬁgure. By this we mean that in the case of the metal sphere, electrons are now free to pass between the sphere and the electrode. Once again, electrons move slightly downward, leaving positive sites exposed at the top of the sphere. However, some of those at the bottom ﬂow into the lower electrode, thus reducing the amount of negative surface charge on the lower side of the metal sphere. At the top, the polarized sphere shown by Fig. 6.0.1f has a similar distribution of positive surface charge density. But one very important diﬀerence between the two situations is apparent. On an atomic scale in the ideal dielectric, the orbiting electrons are paired with the parent atom, and hence the sphere must remain neu- tral. Thus, the metallic sphere now has a net charge, while the one made up of dipoles does not. Experimental evidence that a metallic sphere had indeed acquired a net charge could be gained in a number of diﬀerent ways. Two are clear from demonstrations in Chap. 1. A pair of spheres, each charged by “induction” in this fashion, would repel each other, and this could be demonstrated by the experiment in Fig. 1.3.10. The charge could also be measured by charge conservation, as in Demonstration 1.5.1. Presumably, the same experiments carried out using insulating spheres would demonstrate the existence of no net charge. Because charge accumulations occur via displacements of paired charges (po- larization) as well as of charges that can move far away from their partners of opposite sign, it is often appropriate to distinguish between these by separating the total charge density ρinto parts ρuandρp, respectively, produced by unpaired and 4 Polarization Chapter 6 paired charges. ρ=ρu+ρp (1) In this chapter, we consider insulating materials and therefore focus on the eﬀects of the paired or polarization charge density. Additional eﬀects of unpaired charges are taken up in the next chapter. Our ﬁrst step, in Sec. 6.1, is to relate the polarization charge density to the density of dipoles– to the polarization density. We do this because it is the polar- ization density that can be most easily speciﬁed. Sections 6.2 and 6.3 then focus on the ﬁrst of two general classes of polarization. In these sections, the polariza- tion density is permanent and therefore speciﬁed without regard for the electric ﬁeld. In Sec. 6.4, we discuss simple constitutive laws expressing the action of the ﬁeld upon the polarization. This ﬁeld-induced atomic polarization just described is typical of physical situations. The ﬁeld action on the atom, molecule, or domain is accompanied by a reaction of the dipoles on the ﬁeld that must be considered simultaneously. That is, within such a polarizable body placed into an electric ﬁeld, a polarization charge density is produced which, in turn, modiﬁes the electric ﬁeld. In Secs. 6.5–6.7, we shall study methods by which self-consistent solutions to such problems are obtained. 6.1 POLARIZATION DENSITY The following development is applicable to polarization phenomena having diverse microscopic origins. Whether representative of atoms, molecules, groups of ordered atoms or molecules (domains), or even macroscopic particles, the dipoles are pic- tured as opposite charges ±qseparated by a vector distance ddirected from the negative to the positive charge. Thus, the individual dipoles, represented as in Sec. 4.4, have moments pdeﬁned as p=qd (1) Because dis generally smaller in magnitude than the size of the atom, molecule, or other particle, it is small compared with any macroscopic dimension of interest. Now consider a medium consisting of Nsuch polarized particles per unit volume. What is the net charge qcontained within an arbitrary volume Venclosed by a surface S? Clearly, if the particles of the medium within Vwere unpolarized, the net charge in Vwould be zero. However, now that they are polarized, some charge centers that were contained in Vin their unpolarized state have moved out of the surface Sand left behind unneutralized centers of charge. To determine the net unneutralized charge left behind in V, we will assume (without loss of generality) that the negative centers of charge are stationary and that only the positive centers of charge are mobile during the polarization process. Consider the particles in the neighborhood of an element of area daon the surface S, as shown in Fig. 6.1.1. All positive centers of charge now outside Swithin the volume dV=d·dahave left behind negative charge centers. These contribute a net negative charge to V. Because there are Nd·dasuch negative centers of charge indV, the net charge left behind in Vis Sec. 6.1 Polarization Density 5 Fig. 6.1.1 Volume element containing positive charges which have left neg- ative charges on the other side of surface S. Q=−I S(qNd)·da (2) Note that the integrand can be either positive or negative depending on whether positive centers of charge are leaving or entering Vthrough the surface element da. Which of these possibilities occurs is reﬂected by the relative orientation of d andda. Ifdhas a component parallel (anti-parallel) to da, then positive centers of charge are leaving (entering) Vthrough da. The integrand of (1) has the dimensions of dipole moment per unit volume and will therefore be deﬁned as the polarization density . P≡Nqd (3) Also by deﬁnition, the net charge in Vcan be determined by integrating the polar- ization charge density over its volume. Q=Z VρpdV (4) Thus, we have two ways of calculating the net charge, the ﬁrst by using the polar- ization density from (3) in the surface integral of (2). Q=−I SP·da=−Z V∇ ·PdV (5) Here Gauss’ theorem has been used to convert the surface integral to one over the enclosed volume. The charge found from this volume integral must be the same as given by the second way of calculating the net charge, by (4). Because the volume under consideration is arbitrary, the integrands of the volume integrals in (4) and (5) must be identical. ρp=−∇ ·P(6) In this way, the polarization charge density ρphas been related to the polarization density P. 6 Polarization Chapter 6 Fig. 6.1.2 Polarization surface charge due to uniform polarization of right cylinder. It may seem that little has been accomplished in this development because, instead of the unknown ρp, the new unknown Pappeared. In some instances, P is known. But even in the more common cases where the polarization density and hence the polarization charge density is not known a priori but is induced by the ﬁeld, it is easier to directly link PwithEthan ρpwithE. In Fig. 6.0.1, the polarized sphere could acquire no net charge. Our repre- sentation of the polarization charge density in terms of the polarization density guarantees that this is true. To see this, suppose Vis interpreted as the volume containing the entire polarized body so that the surface Senclosing the volume V falls outside the body. Because Pvanishes on S, the surface integral in (5) must vanish. Any distribution of charge density related to the polarization density by (6) cannot contribute a net charge to an isolated body. We will often ﬁnd it necessary to represent the polarization density by a discontinuous function. For example, in a material surrounded by free space, such as the sphere in Fig. 6.0.1, the polarization density can fall from a ﬁnite value to zero at the interface. In such regions, there can be a surface polarization charge density. With the objective of determining this density from P, (6) can be integrated over a pillbox enclosing an incremental area of an interface. With the substitution −P→/epsilon1oEandρp→ρ, (6) takes the same form as Gauss’ law, so the proof is identical to that leading from (1.3.1) to (1.3.17). We conclude that where there is a jump in the normal component of P, there is a surface polarization charge density σsp=−n·(Pa−Pb)(7) Just as (6) tells us how to determine the polarization charge density for a given distribution of Pin the volume of a material, this expression serves to evaluate the singularity in polarization charge density (the surface polarization charge density) at an interface. Note that according to (6), Poriginates on negative polarization charge and terminates on positive charge. This contrasts with the relationship between Eand the charge density. For example, according to (6) and (7), the uniformly polarized cylinder of material shown in Fig. 6.1.2 with Ppointing upward has positive σsp on the top and negative on the bottom. Sec. 6.2 Laws and Continuity 7 6.2 LAWS AND CONTINUITY CONDITIONS WITH POLARIZATION With the unpaired and polarization charge densities distinguished, Gauss’ law becomes ∇ ·/epsilon1oE=ρu+ρp (1) where (6.1.6) relates ρptoP. ρp=−∇ ·P (2) Because Pis an “averaged” polarization per unit volume, it is a “smooth” vector function of position on an atomic scale. In this sense, it is a macroscopic variable. The negative of its divergence, the polarization charge density, is also a macroscopic quantity that does not reﬂect the “graininess” of the microscopic charge distribu- tion. Thus, as it appears in (1), the electric ﬁeld intensity is also a macroscopic variable. Integration of (1) over an incremental volume enclosing a section of the inter- face, as carried out in obtaining (1.3.7), results in n·/epsilon1o(Ea−Eb) =σsu+σsp (3) where (6.1.7) relates σsptoP. σsp=−n·(Pa−Pb) (4) These last two equations, respectively, give expression to the continuity con- dition of Gauss’ law, (1), at a surface of discontinuity. Polarization Current Density and Amp` ere’s Law. Gauss’ law is not the only one aﬀected by polarization. If the polarization density varies with time, then the ﬂow of charge across the surface Sdescribed in Sec. 6.1 comprises an electrical current. Thus, we need to investigate charge conservation, and more generally the eﬀect of a time-varying polarization density on Amp´ ere’s law. To this end, the following steps lead to the polarization current density implied by a time-varying polarization density. According to the deﬁnition of Pevolved in Sec. 6.1, the process of polarization transfers an amount of charge dQ dQ=P·da (5) through a surface area element da. This is perhaps envisioned in terms of the volume d·dashown in Fig. 6.2.1. If the polarization density Pvaries with time, then according to this equation, charge is passed through the area element at a ﬁnite rate. For a change in qNd, orP, of ∆ P, the amount of charge that has passed through the incremental area element dais ∆(dQ) = ∆ P·da (6) 8 Polarization Chapter 6 Fig. 6.2.1 Charges passing through area element daresult in polarization current density. Note that we have two indicators of diﬀerentials in this expression. The d refers to the fact that Qis diﬀerential because dais a diﬀerential. The rate of change with time of dQ,∆(dQ)/∆t, can be identiﬁed with a current dipthrough da, from side (b) to side (a). dip=∆(dQ) ∆t=∂P ∂t·da (7) The partial diﬀerentiation symbol is used to distinguish the diﬀerentiation with respect to tfrom the space dependence of P. A current dipthrough an area element dais usually written as a current density dot-multiplied by da dip=Jp·da (8) Hence, we compare these last two equations and deduce that the polarization cur- rent density is Jp=∂P ∂t(9) Note that Jpandρp, via (2) and (9), automatically obey a continuity law having the same form as the charge conservation equation, (2.3.3). ∇ ·Jp+∂ρp ∂t= 0 (10) Hence, we can think of a rate of charge transport in a material medium as consisting of a current density of unpaired charges Juand a polarization current density Jp, each obeying its own conservation law. This is also implied by Amp` ere’s law, as now generalized to include the eﬀects of polarization. In the EQS approximation, the magnetic ﬁeld intensity is not usually of in- terest, and so Amp` ere’s law is of secondary importance. But if Hwere to be deter- mined, Jpwould make a contribution. That is, Amp` ere’s law as given by (2.6.2) is now written with the current density divided into paired and unpaired parts. With the latter given by (9), Amp` ere’s diﬀerential law, generalized to include polariza- tion, is ∇ ×H=Ju+∂ ∂t(/epsilon1oE+P) (11) Sec. 6.3 Permanent Polarization 9 This law is valid whether quasistatic approximations are to be made or not. How- ever, it is its implication for charge conservation that is usually of interest in the EQS approximation. Thus, the divergence of (11) gives zero on the left and, in view of (1), (2), and (9), the expression becomes ∇ ·Ju+∂ρu ∂t+∇ ·Jp+∂ρp ∂t= 0 (12) Thus, with the addition of the polarization current density to (11), the divergence of Amp` ere’s law gives the sum of the conservation equations for polarization charges, (10), and unpaired charges ∇ ·Ju+∂ρu ∂t= 0 (13) In the remainder of this chapter, it will be assumed that in the polarized material, ρuis usually zero. Thus, (13) will not come into play until Chap. 7. Displacement Flux Density. Primarily in dealing with ﬁeld-dependent polar- ization phenomena, it is customary to deﬁne a combination of quantities appearing in Gauss’ law and Amp` ere’s law as the displacement ﬂux density D. D≡/epsilon1oE+P (14) We regard Pas representing the material and Eas a ﬁeld quantity induced by the external sources and the sources within the material. This suggests that Dbe considered a “hybrid” quantity. Not all texts on electromagnetism take this point of view. Our separation of all quantities appearing in Maxwell’s equations into ﬁeld and material quantities aids in the construction of models for the interaction of ﬁelds with matter. With ρpreplaced by (2), Gauss’ law (1) can be written in terms of Ddeﬁned by (14), ∇ ·D=ρu (15) while the associated continuity condition, (3) with σspreplaced by (4), becomes n·(Da−Db) =σsu (16) The divergence of Dand the jump in normal Ddetermine the unpaired charge densities. Equations (15) and (16) hold, unchanged in form, both in free space and matter. To adapt the laws to free space, simply set D=/epsilon1oE. Amp` ere’s law is also conveniently written in terms of D. Substitution of (14) into (11) gives ∇ ×H=Ju+∂D ∂t (17) 10 Polarization Chapter 6 Now the displacement current density ∂D/∂tincludes the polarization current den- sity. 6.3 PERMANENT POLARIZATION Usually, the polarization depends on the electric ﬁeld intensity. However, in some materials a permanent polarization is “frozen” into the material. Ideally, this means thatP(r, t) is prescribed, independent of E. Electrets, used to make microphones and telephone speakers, are often modeled in this way. With Pa given function of space, and perhaps of time, the polarization charge density and surface charge density follow from (6.2.2) and (6.2.4) respectively. If the unpaired charge density is also given throughout the material, the total charge density in Gauss’ law and surface charge density in the continuity condition for Gauss’ law are known. [The right-hand sides of (6.2.1) and (6.2.3) are known.] Thus, a description of permanent polarization problems follows the same format as used in Chaps. 4 and 5. Examples in this section are intended to develop an appreciation for the re- lationship between the polarization density P, the polarization charge density ρp, and the electric ﬁeld intensity E. It should be recognized that once ρpis determined from the given P, the methods of Chaps. 4 and 5 are directly applicable. The distinction between paired and unpaired charges is sometimes academic. By subjecting an insulating material to an extremely large ﬁeld, especially at an elevated temperature, it is possible to coerce molecules or domains of molecules into a polarization state that is retained for some period of time at lower ﬁelds and temperatures. It is natural to take this as a state of permanent polarization. But, if ions are made to impact the surface of the material, they can form sites of permanent charge. Certainly, the origin of these ions suggests that they be regarded as unpaired. Yet if the material attracts other charges to become neutral, as it tends to do, these permanent charges could also be regarded as due to polarization and represented by a permanent polarization charge density. In this section, the EQS laws prevail. Thus, with the understanding that throughout the region of interest (exclusive of enclosing boundaries) the charge densities are given, E=−∇Φ (1) ∇2Φ =−1 /epsilon1o(ρu+ρp) (2) The example now considered is akin to that pictured qualitatively in Fig. 6.1.2. By making the uniformly polarized material spherical, it is possible to obtain a simple solution for the ﬁeld distribution. Example 6.3.1. A Permanently Polarized Sphere A sphere of material having radius Ris uniformly polarized along the zaxis, P=Poiz (3) Sec. 6.3 Permanent Polarization 11 Given that the surrounding region is free space with no additional ﬁeld sources, what is the electric ﬁeld intensity Eproduced by this permanent polarization? The ﬁrst step is to establish the distribution of ρp, in the material volume and on its surfaces. In the volume, the negative divergence of Pis zero, so there is no volumetric polarization charge density (6.2.2). This is obvious with Pwritten in Cartesian coordinates. It is less obvious when Pis expressed in its spherical coordinate components. P=Pocosθir−Posinθiθ (4) Abrupt changes of the normal component of Pentail polarization surface charge densities. These follow from using (4) to evaluate the continuity condition of (6.2.4) applied at r=R, where the normal component is irand region (a) is outside the sphere. σsp=Pocosθ (5) This surface charge density gives rise to E. Now that the ﬁeld sources have been identiﬁed, the situation reverts to one much like that illustrated by Problem 5.9.2. Both within the sphere and in the surrounding free space, the potential must satisfy Laplace’s equation, (2), with ρu+ ρp= 0. In terms of Φ the continuity conditions at r=Rimplied by (1) and (2) [(5.3.3) and (6.2.3)] with the latter evaluated using (5) are Φo−Φi= 0 (6) −/epsilon1o∂Φo ∂r+/epsilon1o∂Φi ∂r=Pocosθ (7) where ( o) and ( i) denote the regions outside and inside the sphere. The source of the Eﬁeld represented by this potential is a surface polarization charge density that varies cosinusoidally with θ. It is possible to fulﬁll the boundary conditions, (6) and (7), with the two spherical coordinate solutions to Laplace’s equation (from Sec. 5.9) having the θdependence cos θ. Because there are no sources in the region outside the sphere, the potential must go to zero as r→ ∞ . Of the two possible solutions having the cos θdependence, the dipole ﬁeld is used outside the sphere. Φo=Acosθ r2(8) Inside the sphere, the potential must be ﬁnite, so this solution is excluded. The solution is Φi=Brcosθ (9) which is that of a uniform electric ﬁeld intensity. Substitution of these expressions into the continuity conditions, (6) and (7), gives expressions from which cos θcan be factored. Thus, the boundary conditions are satisﬁed at every point on the surface ifA R2−BR= 0 (10) 2/epsilon1oA R3+/epsilon1oB=Po (11) These expressions can be solved for AandB, which are introduced into (8) and (9) to give the potential distribution Φo=PoR3 3/epsilon1ocosθ r2(12) 12 Polarization Chapter 6 Fig. 6.3.1 Equipotentials and lines of electric ﬁeld intensity of perma- nently polarized sphere having uniform polarization density. Inset shows polarization density and associated surface polarization charge density. Φi=Po 3/epsilon1orcosθ (13) Finally, the desired distribution of electric ﬁeld is obtained by taking the negative gradient of this potential. Eo=PoR3 3/epsilon1or3(2 cos θir+ sin θiθ) (14) Ei=Po 3/epsilon1o(−cosθir+ sin θiθ) (15) With the distribution of polarization density shown in the inset, Fig. 6.3.1 shows this electric ﬁeld intensity. It comes as no surprise that the Elines originate on the positive charge and terminate on the negative. The polarization density originates on negative polarization charge and terminates on positive polarization charge. The resulting electric ﬁeld is classic because outside it is exactly that of a dipole at the origin, while inside it is uniform. What would be the moment of the dipole at the origin giving rise to the same external ﬁeld as the uniformly polarized sphere? This can be seen from a comparison of (12) and (4.4.10). |P|=4 3πR3Po (16) The moment is simply the volume multiplied by the uniform polarization density. There are two new ingredients in the next example. First, the region of interest has boundaries upon which the potential is constrained. Second, the given polar- ization density represents a volumetric distribution of polarization charge density rather than a surface distribution. Example 6.3.2. Fields Due to Volume Polarization Charge with Boundary Conditions Sec. 6.3 Permanent Polarization 13 Fig. 6.3.2 Periodic distribution of polarization density and associated polarization charge density ( ρo<0) gives rise to potential and ﬁeld shown in Fig. 5.6.2. Fig. 6.3.3 Cross-section of electret microphone. Plane parallel electrodes, in the planes y=±a, are constrained to zero potential. In the planar region between, the polarization density is the spatially periodic function P=−ixρo βsinβx (17) We wish to determine the ﬁeld distribution. First, the distribution of polarization charge density is determined by taking the negative divergence of (17) [(17) is substituted into (6.1.6)]. ρp=ρocosβx (18) The distribution of polarization density and polarization charge density which has been found is shown in Fig. 6.3.2 ( ρo<0). Now the situation reverts to solving Poisson’s equation, given this source dis- tribution and subject to the zero potential conditions on the boundaries at y=±a. The problem is identical to that considered in Example 5.6.1. The potential and ﬁeld are the superposition of particular and homogeneous parts depicted in Fig. 5.6.2. The next example illustrates how a permanent polarization can conspire with a mechanical deformation to produce a useful electrical signal. Example 6.3.3. An Electret Microphone Shown in cross-section in Fig. 6.3.3 is a thin sheet of permanently polarized material having thickness d. It is bounded from below by a ﬁxed electrode having the potential vand from above by an air gap. On the other side of this gap is a conducting grounded diaphragm which serves as the movable element of a microphone. It is mounted so that it can undergo displacements. Thus, the spacing h=h(t). Given h(t), what is the voltage developed across a load resistance R? In the sheet, the polarization density is uniform, with magnitude Po, and di- rected from the lower electrode toward the upper one. This vector has no divergence, 14 Polarization Chapter 6 Fig. 6.3.4 (a) Distribution of polarization density and surface charge density in electret microphone. (b) Electric ﬁeld intensity and surface polarization and unpaired charges. and so evaluation of (6.1.6) shows that the polarization charge density is zero in the volume of the sheet. The polarization surface charge density on the electret air gap interface follows from (6.1.7) as σsp=−n·(Pa−Pb) =Po (19) Because σspis uniform and the equipotential boundaries are plane and parallel, the electric ﬁeld in the air gap [region (a)] and in the electret [region (b)] are taken as uniform. E=ixnEa;d < x < h Eb; 0< x < d(20) Formally, we have just solved Laplace’s equation in each of the bulk regions. The ﬁelds EaandEbmust satisfy two conditions. First, the potential diﬀerence between the electrodes is v, so v=Zh 0Exdx=dEb+ (h−d)Ea (21) Second, Gauss’ jump condition at the electret air gap interface, (6.2.3), requires that /epsilon1oEa−/epsilon1oEb=Po (22) Simultaneous solution of these last two expressions evaluates the electric ﬁelds in terms of vandh. Ea=v h+d hPo /epsilon1o(24a) Eb=v h−(h−d) hPo /epsilon1o(24b) What has been found is illustrated in Fig. 6.3.4. The uniform Pand associated σspshown in part (a) combine with the unpaired charges on the lower electrode and upper diaphragm to produce the ﬁelds shown in part (b). In this picture, it is assumed that vis positive and ( h−d)Po//epsilon1o> v. In the air gap, the ﬁeld due to the unpaired charges on the electrodes reinforces that due to σsp, while in the electret, it opposes the downward-directed ﬁeld due to σsp. To compute the current i, deﬁned in Fig. 6.3.3, the lower electrode and the electret are enclosed by a surface S, and Gauss’ law is used to evaluate the enclosed unpaired charge. ∇ ·(/epsilon1oE+P) =ρu⇒q=I S(/epsilon1oE+P)·nda (25) Sec. 6.3 Permanent Polarization 15 Just how the surface Scuts through the system does not matter. Here we take the surface as enclosing the lower electrode by passing through the air gap. It follows from (24) that the unpaired charge is q=A/epsilon1oEa=A/epsilon1o hµ v+dPo /epsilon1o¶ (26) where Ais the area of the electrode. Conservation of unpaired charge requires that the current be the rate of change of the total unpaired charge on the lower electrode. i=dq dt(27) With the resistor attached to the terminals (the input resistance of an ampliﬁer driven by the microphone), the voltage and current must also satisfy Ohm’s law. v=−iR (28) These last three relations combine to give an expression for v(t), given h(t). −v R=−A/epsilon1o h2µ v+dPo /epsilon1o¶ dh dt+A/epsilon1o hdv dt(29) This diﬀerential equation has time-varying coeﬃcients. Not only is this equa- tion diﬃcult to solve, but also the predicted voltage response cannot be a good replica of h(t), as required for a good microphone, if all terms are of equal impor- tance. That situation can be remedied if the deﬂections h1are kept small compared with the equilibrium position, ho/greatermuchh1. In the absence of a time variation of h1, it is clear from (29) that vis zero. By making h1small, we can make vsmall. Expanding the right-hand side of (29) to ﬁrst order in h1, dh 1/dt, v , and dv/dt , we obtain Codv dt+v R=Co ho¡dPo /epsilon1o¢dh1 dt(30) where Co=A/epsilon1o/ho. We could solve this equation for its response to a sinusoidal drive. Alter- natively, the resulting frequency response can be determined, with more physical insight, by considering two limits. First, suppose that time rate of change is so slow (frequencies so low) that the ﬁrst term on the left is negligible compared to the second. Then the output voltage is v=CoR ho¡dPo /epsilon1o¢dh1 dt; ωRC o/lessmuch1 (31) In this limit, the resistor acts as a short. The charge can be determined by the diaphragm displacement with the contribution of vignored (i.e., the charge required to produce vby charging the capacitance Cois ignored). The small but ﬁnite voltage is then obtained as the time rate of change of the charge multiplied by −R. Second, suppose that time rates of change are so rapid that the second term is negligible compared to the ﬁrst. Within an integration constant, v=dPo /epsilon1oh1 ho; ωRC o/greatermuch1 (32) 16 Polarization Chapter 6 Fig. 6.3.5 Frequency response of electret microphone for imposed di- aphragm displacement. In this limit, the electrode charge is essentially constant. The voltage is obtained from (26) with qset equal to its equilibrium value, ( A/epsilon1o/ho)(dPo//epsilon1o). The frequency response gleaned from these asymptotic responses is in Fig. 6.3.5. Because its displacement was taken as known, we have been able to ignore the dynamical equations of the diaphragm. If the mass and damping of the diaphragm are ignored, the displacement indeed reﬂects the pressure of a sound wave. In this limit, a linear distortion-free response of the microphone to pressure is assured at frequencies ω > 1/RC. However, in predicting the response to a sound wave, it is usually necessary to include the detailed dynamics of the diaphragm. In a practical microphone, subjecting the electret sheet to an electric ﬁeld would induce some polarization over and beyond the permanent component Po. Thus, a more realistic model would incorporate features of the linear dielectrics introduced in Sec. 6.4. 6.4 CONSTITUTIVE LAWS OF POLARIZATION Dipole formation, or orientation of dipolar particles, usually depends on the local ﬁeld in which the particles are situated. This local microscopic ﬁeld is not necessarily equal to the macroscopic Eﬁeld. Yet certain relationships between the macroscopic quantities EandPcan be established without a knowledge of the relations between the local microscopic ﬁelds and the macroscopic Eﬁelds. Usually, these relations, called constitutive laws, originate in experimental observations characteristic of the material being investigated. First, the permanent polarization model developed in the previous section is one constitutive law. In such a medium, P(r) is prescribed independent of E. There are media, and these are much more common, in which the polarization depends on E. Consider an isotropic medium, which, in the absence of an electric ﬁeld has no preferred orientation. Amorphous media such as glass are isotropic. Crystalline media, made up of randomly oriented microscopic crystals, also behave as isotropic media on a macroscopic scale. If we assume that the polarization Pin an isotropic medium depends on the instantaneous ﬁeld and not on its past history, thenPis a function of E P=P(E) (1) where PandEare parallel to each other. Indeed, if Pwere not parallel to E, then a preferred direction diﬀerent from the direction of Ewould need to exist in the medium, which contradicts the assumption of isotropy. A possible relation between Sec. 6.5 Fields in Linear Dielectrics 17 Fig. 6.4.1 Polarization characteristic for nonlinear isotropic material. the magnitudes of EandPis shown in Fig. 6.4.1 and represents an “electrically nonlinear” medium for which P“saturates” for large values of E. If the medium is electrically linear, in addition to being isotropic, then a linear relationship exists between EandP P=/epsilon1oχeE (2) where χeis the dielectric susceptibility . Typical values are given in Table 6.4.1. All isotropic media behave as linear media and obey (2) if the applied Eﬁeld is suﬃciently small. As long as Eis small enough, any continuous function P(E) can be expanded in a Taylor series of Eand broken oﬀ with the ﬁrst term in E. (An isotropic medium cannot have a term in the Taylor expansion independent of E.) For a linear isotropic material, where (2) is obeyed, it follows that DandE are related by D=/epsilon1E (3) where /epsilon1≡/epsilon1o(1 +χe) (4) is the permittivity ordielectric constant . The permittivity normalized to /epsilon1o,(1+χe), is the relative dielectric constant . In our discussion, it has been assumed that the state of polarization depends only on the instantaneous electric ﬁeld intensity. There are materials in which the polarization depends not only on the current electric ﬁeld intensity but on the sequence of preceding states as well (hysteresis). Because we will ﬁnd magnetiza- tion phenomena analogous in many ways to polarization phenomena, we will defer consideration of hysteretic phenomena to Chap. 9. Many types of transducers exploit the dependence of polarization on variables other than the electric ﬁeld. In pyroelectric materials, polarization is a function of temperature. Pyroelectrics are used for optical detectors of high-power infrared ra- diation. Piezoelectric materials have a polarization which is a function of strain (deformation). Such media are suited to low-power electromechanical energy con- version. 18 Polarization Chapter 6 TABLE 6.4.1 MATERIAL DIELECTRIC SUSCEPTIBILITIES Gasesχe Air, 0◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.00059 40 atmospheres . . . . . . . . . . . . . . . . . . . . . . . . . 0.0218 80 atmospheres . . . . . . . . . . . . . . . . . . . . . . . . . 0.0439 Carbon dioxide, 0◦C. . . . . . . . . . . . . . . . . . . . . . . . . 0.000985 Hydrogen, 0◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.000264 Water vapor, 145◦C. . . . . . . . . . . . . . . . . . . . . . . . . 0.00705 Liquidsχe Acetone, 0◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.6 Air, -191◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.43 Alcohol amyl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.0 ethyl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.8 methyl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 Benzene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.29 Glycerine, 15◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55.2 Oils, castor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.67 linseed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.35 corn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Water, distilled . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79.1 Solidsχe Diamond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Glass, ﬂint, density 4.5 . . . . . . . . . . . . . . . . . . . . . . . . 8.90 ﬂint, density 2.87 . . . . . . . . . . . . . . . . . . . . . . . 5.61 lead, density 3.0-3.5 . . . . . . . . . . . . . . . . . . . . . 4.4-7.0 Mica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6-5.0 Paper (cable insulation) . . . . . . . . . . . . . . . . . . . . . 1.0-1.5 Paraﬃn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Porcelain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Quartz, 1 to axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.69 11 to axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.06 Rubber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3-3.0 Shellac . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Sec. 6.5 Fields in Linear Dielectrics 19 Fig. 6.5.1 Field region ﬁlled by (a) uniform dielectric, (b) piece-wise uniform dielectric and (c) smoothly varying dielectric. 6.5 FIELDS IN THE PRESENCE OF ELECTRICALLY LINEAR DIELECTRICS In Secs. 6.2 and 6.3, the polarization density was given independently of the electric ﬁeld intensity. In this and the next two sections, the polarization is induced by the electric ﬁeld. Not only does the electric ﬁeld give rise to the polarization, but in return, the polarization modiﬁes the ﬁeld. The polarization feeds back on the electric ﬁeld intensity. This “feedback” is described by the constitutive law for a linear dielectric. Thus, (6.4.3) and Gauss’ law, (6.2.15), combine to give ∇ ·/epsilon1E=ρu (1) and the electroquasistatic form of Faraday’s law requires that ∇ ×E= 0⇒E=−∇Φ (2) The continuity conditions implied by these two laws across an interface separating media having diﬀerent permittivities are (6.2.16) expressed in terms of the consti- tutive law and either (5.3.1) or (5.3.4). These are n·(/epsilon1aEa−/epsilon1bEb) =σsu (3) n×(Ea−Eb) = 0⇒Φa−Φb= 0 (4) Figure 6.5.1 illustrates three classes of situations involving linear dielectrics. In the ﬁrst, the entire region of interest is ﬁlled with a uniform dielectric. In the second, the region of interest can be broken into uniform subregions within which 20 Polarization Chapter 6 the permittivity is constant. The continuity conditions are needed to insure that the basic laws are satisﬁed through the interfaces between these regions. Systems of this type are said to be composed of piece-wise uniform dielectrics. Finally, the dielectric material may vary in its permittivity over dimensions that are on the same order as those of interest. Such a smoothly inhomogeneous dielectric is illustrated in Fig. 6.5.1c. The remainder of this section makes some observations that are generally applicable provided that ρu= 0 throughout the volume of the region of interest. Section 6.6 is devoted to systems having uniform and piece-wise uniform dielectrics, while Sec. 6.7 illustrates ﬁelds in smoothly inhomogeneous dielectrics. Capacitance. How does the presence of a dielectric alter the capacitance? To answer this question, recognize that conservation of unpaired charge, as expressed by (6.2.13), still requires that the current imeasured at terminals connected to a pair of electrodes is the time rate of change of the unpaired charge on the electrode. In view of Gauss’ law, with the eﬀects of polarization included, (6.2.15), the net unpaired charge on an electrode enclosed by a surface Sis q=Z VρudV=Z V∇ ·DdV=I SD·nda (5) Here, Gauss’ theorem has been used to convert the volume integral to a surface integral. We conclude that the capacitance of an electrode (a) relative to a reference electrode (b) is C=H SD·nda Rb aC/primeE·ds=H SD·nda v(6) Note that this is the same as for electrodes in free space except that /epsilon1oE→D. Because there is no unpaired charge density in the region between the electrodes, Sis any surface that encloses the electrode (a). As before, with no polarization, E is irrotational, and therefore C/primeis any contour connecting the electrode (a) to the reference (b). In an electrically linear dielectric, where D=/epsilon1E, both the numerator and denominator of (6) are proportional to the voltage, and as a result, the capacitance Cis independent of the voltage. However, with the introduction of an electrically nonlinear material, perhaps having the polarization constitutive law of Fig. 6.4.1, the numerator of (6) is not a linear function of the voltage. As deﬁned by (6), the capacitance is then a function of the applied voltage. Induced Polarization Charge. Stated as (1)–(4), the laws and continuity conditions for ﬁelds in a linear dielectric put the polarization charge out of view. Yet it is this charge that contains the eﬀect of the dielectric on the ﬁeld. Where does the polarization charge accumulate? Again, assuming that ρuis zero, a vector identity casts Gauss’ law as given by (1) into the form /epsilon1∇ ·E+E· ∇/epsilon1= 0 (7) Sec. 6.6 Piece-Wise Uniform Electrically Linear Dielectrics 21 Multiplied by /epsilon1oand divided by /epsilon1, this expression can be written as ∇ ·/epsilon1oE=−/epsilon1o /epsilon1E· ∇/epsilon1 (8) Comparison of this expression to Gauss’ law written in terms of ρp, (6.2.1), shows that the polarization charge density is ρp=−/epsilon1o /epsilon1E· ∇/epsilon1 (9) This equation makes it clear that polarization charge will be induced only where there are gradients in /epsilon1. A special case is where there is an abrupt disconti- nuity in /epsilon1. Then the gradient in (9) is singular and represents a polarization surface charge density (the gradient represents the spatial derivative of a step function, which is an impulse). This surface charge density can best be determined by mak- ing use of the polarization charge density continuity condition, (6.1.7). Substitution of the constitutive law P= (/epsilon1−/epsilon1o)Ethen gives σsp=−n·[(/epsilon1a−/epsilon1o)Ea−(/epsilon1b−/epsilon1o)Eb] (10) Because σsu= 0, it follows from the jump condition for n·D, (3), that σsp=n·/epsilon1oEa¡ 1−/epsilon1a /epsilon1b¢ (11) Remember that nis directed from region (b) to region (a). Because Dis solenoidal, we can construct tubes of Dcontaining constant ﬂux. Lines of Dmust therefore begin and terminate on the boundaries. The constitutive law,D=/epsilon1E, requires that Dis proportional to E. Thus, although Ecan intensify or rarify as it passes through a ﬂux tube, it can not reverse direction. Therefore, if we follow a bundle of electric ﬁeld lines from the boundary point of high potential to the one of low potential, the polarization charge encountered [in accordance with (9) and (11)] is positive at points where /epsilon1is decreasing, negative where it is increasing. Consider the examples in Fig. 6.5.1. In the case of the uniform dielectric, Fig. 6.5.1a, the typical ﬂux tube shown passes through no variations in /epsilon1, and it follows from (8) that there is no volume polarization charge density. Thus, it will come as no surprise that the ﬁeld distribution in this case is predicted by Laplace’s equation. In the piece-wise uniform dielectrics, there is no polarization charge density in a ﬂux tube except where it passes through an interface. For the ﬂux tube shown, (11) shows that if the upper region has the greater permittivity ( /epsilon1a> /epsilon1b), then there is an accumulation of negative surface charge density at the interface. Thus, the ﬁeld originating on positive charges at the lower electrode is in part terminated by negative polarization surface charge at the interface, and the ﬁeld in the upper region tends to be weakened relative to that below. In the smoothly inhomogeneous dielectric of Fig. 6.5.1c, the typical ﬂux tube shown passes through a region where /epsilon1increases with ξ. It follows from (8) that negative polarization charge density is induced in the volume of the material. Here 22 Polarization Chapter 6 again, the electric ﬁeld associated with positive charge on the lower electrode is in part terminated on the polarization charge density induced in the volume. As a result, the dielectric tends to make the electric ﬁeld weaken with increasing ξ. The next two sections give the opportunity to solve for the ﬁelds in simple conﬁgurations and then see that the results are consistent with the physical picture that has been found here. 6.6 PIECE-WISE UNIFORM ELECTRICALLY LINEAR DIELECTRICS In a region where the permittivity is uniform and where there is no unpaired charge, the electric potential obeys Laplace’s equation. ∇2Φ = 0 (1) This follows from (6.5.1) and (6.5.2). Uniform Dielectrics. If all of the region of interest is ﬁlled by a uniform dielectric, it is clear from the foregoing that all equations developed for ﬁelds in free space are now valid in the presence of the uniform dielectric. The only alteration is the replacement of the permittivity of free space /epsilon1oby that of the uniform dielectric. In every problem from Chaps. 4 and 5 where Φ and Ewere determined in a region of free space bounded by equipotentials, that region could just as well be ﬁlled with a uniform dielectric, and for the same potentials the electric ﬁeld intensity would be unaltered. However, the surface charge density σsuon the boundaries would then be increased by the ratio /epsilon1//epsilon1o. Illustration. Capacitance of a Sphere A sphere having radius Rhas a potential vrelative to inﬁnity. Formally, the po- tential, and hence the electric ﬁeld, follow from (1). Φ =vR r⇒E=vR r2(2) Evaluation of the capacitance, (6.5.6), then gives C≡q v=4πR2 v/epsilon1Er|r=R= 4πR/epsilon1 (3) The dielectric has increased the capacitance in the ratio of the dielectric constant of the material to the dielectric constant of free space. The susceptibilities listed in Table 6.4.1 illustrate the increase in capacitance that would be observed if vacuum were replaced by one of the materials. In gases, atoms or molecules are so dilute that the increase in capacitance is usually negligi- ble. With solids and liquids, the increase is of practical importance. Some, having Sec. 6.6 Piece-Wise Uniform Dielectrics 23 Fig. 6.6.1 (a) Plane parallel capacitor with region between electrodes occupied by a dielectric. (b) Artiﬁcial dielectric composed of cubic array of perfectly conducting spheres having radius Rand spacing s. molecules of large permanent dipole moments that are aligned by the ﬁeld, increase the capacitance dramatically. The following example is intended to provide an appreciation for why the polarized dielectric increases the capacitance. Example 6.6.1. An Artiﬁcial Dielectric In the plane parallel capacitor of Fig. 6.6.1, the electric ﬁeld intensity is ( v/d)iz. Thus, the unpaired charge density on the lower electrode is Dz=/epsilon1v/d, and if the electrode area is A, the capacitance is C≡q v=A vDz|z=0=A/epsilon1 d(4) Here we assume that dis much less than either of the electrode dimensions, so the fringing ﬁelds can be ignored. Now consider the plane parallel capacitor of Fig. 6.6.1b. The dielectric is com- posed of “molecules” that are actually perfectly conducting spheres. These have radius Rand are in a cubic array with spacing s >> R . With the application of a voltage, the spheres acquire the positive and negative surface charges on their northern and southern poles required to make their surfaces equipotentials. In so far as the ﬁeld outside the spheres is concerned, the system is modeled as an array of dipoles, each induced by the applied ﬁeld. If there are many of the spheres, the change in capacitance caused by inserting the array between the plates can be determined by treating it as a continuum. This we will do under the assumption that s >> R . In that case, the ﬁeld in regions removed several radii from the sphere centers is essentially uniform, and taken as Ez=v/d. The resulting ﬁeld in the vicinity of a sphere is then as determined in Example 5.9.1. The dipole moment of each sphere follows from a comparison of the potential for the perfectly conducting sphere in a uniform electric ﬁeld, (5.9.7), with that of a dipole, (4.4.10). p= 4π/epsilon1oR3Ea (5) The polarization density is the moment/dipole multiplied by the number of dipoles per unit volume, the number density N. Pz=/epsilon1o(4πR3N)Ea (6) For the cubic array, a unit volume contains 1 /s3spheres, and so N=1 s3(7) 24 Polarization Chapter 6 Fig. 6.6.2 From the microscopic point of view, the increase in capaci- tance results because the dipoles adjacent to the electrode induce image charges on the electrode in addition to those from the unpaired charges on the opposite electrode. From (6) and (7) it follows that P=/epsilon1o£ 4π¡R s¢3¤ E (8) Thus, the polarization density is a linear function of E. The susceptibility follows from a comparison of (8) with (6.4.2) and, in turn, the permittivity is given by (6.4.4). χe= 4π¡R s¢3⇒/epsilon1=£ 1 + 4 π¡R s¢3¤ /epsilon1o (9) Of course, this expression is accurate only if the interaction between spheres is negligible. As the array of spheres is inserted between the electrodes, surface charges are induced, as shown in Fig. 6.6.2. Within the array, each cap of positive surface charge on the north pole of a sphere is compensated by an opposite charge on the south pole of a neighboring sphere. Thus, on a scale large compared to the spacing s, there is no charge density in the volume of the array. Nevertheless, the average ﬁeld at the electrode is larger than the applied ﬁeld Ea. This is caused by surface charges on the last layers of spheres which have their images in unpaired charges on the electrodes. For a given applied voltage, the ﬁeld between the top and bottom layers of spheres and the adjacent electrodes is increased, with an attendant increase in observed capacitance. Demonstration 6.6.1. Artiﬁcial Dielectric In Fig. 6.6.3, the artiﬁcial dielectric is composed of an array of ping-pong balls with conducting coatings. The parallel plate capacitor is in one leg of a bridge, as shown in the circuit pictured in Fig. 6.6.4. The resistors shunt the input terminals of balanced ampliﬁers so that the oscilloscope displays vo. With the array removed, capacitor C2is adjusted to null the output voltage vo. The output voltage resulting from the the insertion of the array is a measure of the change in capacitance. To simplify the interpretation of this voltage, the resistances Rsare made small compared to the impedance of the parallel plate capacitor. Thus, almost all of the applied voltage V appears across the lower legs of the bridge. With the introduction of the array, the change in current through the parallel plate capacitor is Sec. 6.6 Piece-Wise Uniform Dielectrics 25 Fig. 6.6.3 Demonstration in which change in capacitance is used to measure the equivalent dielectric constant of an artiﬁcial dielectric. Fig. 6.6.4 Balanced ampliﬁers of oscilloscope, balancing capacitors, and demonstration capacitor shown in Fig. 6.6.4 comprise the elements in the bridge circuit. The driving voltage comes from the transformer, while vois the oscilloscope voltage. |∆i|=ω(∆C)|V| (10) Thus, there is a change of current through the resistance in the right leg and hence a change of voltage across that resistance given by vo=Rsω(∆C)V (11) Because the current through the left leg has remained the same, this change in voltage is the measured output voltage. Typical experimental values are R= 1.87 cm, s= 8 cm, A= (0.40)2m2, d= 0.15 m, ω= 2π(250 Hz), Rs= 100 kΩ and V= 566 v peak with a measured voltage of vo= 0.15 V peak. From (4), (9), and (11), the output voltage is predicted to be 0.135 V peak. Piece-Wise Uniform Dielectrics. So far we have only considered systems ﬁlled with uniform dielectrics, as in Fig. 6.5.1a. We turn now to the description of ﬁelds in piece-wise uniform dielectrics, as exempliﬁed by Fig. 6.5.1b. 26 Polarization Chapter 6 Fig. 6.6.5 Insulating rod having uniform permittivity /epsilon1bsurrounded by material of uniform permittivity /epsilon1a. Uniform electric ﬁeld is imposed by electrodes that are at “inﬁnity.” In each of the regions of constant permittivity, the ﬁeld distribution is de- scribed by Laplace’s equation, (1). The ﬁeld problem is attacked by solving this equation in each of the regions and then using the jump conditions to match these solutions at the surfaces of discontinuity between the dielectrics. The following ex- ample has a relatively simple solution that helps form further insights. Example 6.6.2. Dielectric Rod in Uniform Transverse Field A uniform electric ﬁeld Eoix, perhaps produced by means of a parallel plate ca- pacitor, exists in a dielectric having permittivity /epsilon1a. With its axis perpendicular to this ﬁeld, a circular cylindrical dielectric rod having permittivity /epsilon1band radius Ris introduced, as shown in Fig. 6.6.5. With the understanding that the electrodes are suﬃciently far from the rod so that the ﬁeld at “inﬁnity” is essentially uniform, our objective is to determine and then interpret the electric ﬁeld inside and outside the rod. The shape of the circular cylindrical boundary suggests that we use polar coordinates. In these coordinates, x=rcosφ, and so the potential far from the cylinder is Φ(r→ ∞ )→ −Eorcosφ (12) Because this potential varies like the cosine of the angle, it is reasonable to attempt satisfying the jump conditions with solutions of Laplace’s equation having the same φdependence. Thus, outside the cylinder, the potential is assumed to take the form Φa=−Eorcosφ+AR rcosφ (13) Here the dipole ﬁeld is multiplied by an adjustable coeﬃcient A, but the uniform ﬁeld has a magnitude set to match the potential at large r, (12). Inside the cylinder, the solution with a 1 /rdependence cannot be accepted because it becomes singular at the origin. Thus, the only solution having the cosine dependence on φis a uniform ﬁeld, with the potential Φb=Br Rcosφ (14) Can the coeﬃcients AandBbe adjusted to satisfy the two jump conditions implied by the laws of Gauss and Faraday, (6.5.3) and (6.5.4), at r=R? /epsilon1aEa r−/epsilon1bEb r= 0 (15) Sec. 6.6 Piece-Wise Uniform Dielectrics 27 Fig. 6.6.6 Electric ﬁeld intensity in and around dielectric rod of Fig. 6.6.5 for (a) /epsilon1b> /epsilon1aand (b) /epsilon1b≤/epsilon1a. Φa−Φb= 0 (16) Substitution of (13) and (14) into these conditions shows that the answer is yes. Continuity of potential, (16), requires that (−EoR+A) cosφ=Bcosφ (17) while continuity of normal D, (15), is satisﬁed if ¡ −/epsilon1aEo−/epsilon1aA R¢ cosφ=/epsilon1bB Rcosφ (18) Note that these conditions contain the cos φdependence on both sides, and so can be satisﬁed at each angle φ. This conﬁrms the correctness of the originally assumed φdependence of our solutions. Simultaneous solution of (17) and (18) for AandB gives A=/epsilon1b−/epsilon1a /epsilon1b+/epsilon1aEoR (19) B=−2/epsilon1a /epsilon1b+/epsilon1aEoR (20) Introducing these values of the coeﬃcients into the potentials, (13) and (14), gives Φa=−REocosφ·¡r R¢ −¡R r¢(/epsilon1b−/epsilon1a) (/epsilon1b+/epsilon1a)¸ (21) Φb=−2/epsilon1a /epsilon1b+/epsilon1aEorcosφ (22) The electric ﬁeld is obtained as the gradient of this potential. Ea=Eo( ircosφ· 1 +¡R r¢2(/epsilon1b−/epsilon1a) (/epsilon1b+/epsilon1a)¸ −iφsinφ· 1−¡R r¢2(/epsilon1b−/epsilon1a) /epsilon1b+/epsilon1a¸) (23) Eb=2/epsilon1a /epsilon1b+/epsilon1aEo(ircosφ−iφsinφ) (24) 28 Polarization Chapter 6 Fig. 6.6.7 Surface polarization charge density responsible for distortion of ﬁelds as shown in Fig. 6.6.6. (a) /epsilon1b> /epsilon1a, (b) /epsilon1a> /epsilon1b. The electric ﬁeld intensity given by these expressions is shown in Fig. 6.6.6. If the cylinder has the higher dielectric constant, as would be the case for a dielectric rod in air, the lines of electric ﬁeld intensity tend to concentrate in the rod. In the opposite case– for example, representing a cylindrical void in a dielectric– the ﬁeld lines tend to skirt the cylinder. With an understanding of the relationship between the electric ﬁeld intensity and the induced polarization charge comes the ability to see in advance how di- electrics distort the electric ﬁeld. The circular cylindrical dielectric rod introduced into a uniform tranverse electric ﬁeld in Example 6.6.2 serves as an illustration. Without carrying out the detailed analysis which led to (23) and (24), could we see in advance that the electric ﬁeld has the distribution illustrated in Fig. 6.6.6? The induced polarization charge provides the sources for the ﬁeld induced by polarized material. For piece-wise uniform dielectrics, this is a polarization surface charge, given by (6.5.11). σsp=n·/epsilon1oEa¡ 1−/epsilon1a /epsilon1b¢ (25) The electric ﬁeld intensity in the cylindrical rod example is generally directed to the right. It follows from (25) that the distribution of surface polarization charge at the cylindrical interface is as illustrated in Fig. 6.6.7. With the rod having the higher permittivity, Fig. 6.6.7a, the induced positive polarization surface charge density is at the right and the negative surface charge is at the left. These charges give rise to ﬁelds that generally originate at the positive charge and terminate at the negative. Thus, it is clear without any analysis that if /epsilon1b> /epsilon1a, the induced ﬁeld inside tends to cancel the imposed ﬁeld. In this case, the interior ﬁeld is decreased or “depolarized.” In the exterior region, vector addition of the induced ﬁeld to the right-directed imposed ﬁeld shows that incoming ﬁeld lines at the left must be deﬂected inward, while outgoing ones at the right are deﬂected outward. These same ideas, applied to the case where /epsilon1a> /epsilon1b, show that the interior ﬁeld is increased while the exterior one tends to be ducted around the cylinder. The circular cylinder is one of a series of examples having exact solutions. These give the opportunity to highlight the physical phenomena without encum- bering mathematics. If it is actually necessary to account for detailed geometry, Sec. 6.6 Piece-Wise Uniform Dielectrics 29 Fig. 6.6.8 Grounded upper electrode and lower electrode extending from x= 0 to x→ ∞ form plane parallel capacitor with fringing ﬁeld that extends into the region 0 < xbetween grounded electrodes. then some of the approaches introduced in Chaps. 4 and 5 can be used. The fol- lowing example illustrates the use of the orthogonal modes approach introduced in Sec. 5.5. Example 6.6.3. Fringing Field of Dielectric Filled Parallel Plate Capacitor Fields are to be determined in the planar region between a grounded conductor in the plane y=aand a pair of conductors in the plane y= 0, shown in Fig. 6.6.8. To the right of x= 0 in the y= 0 plane is a second grounded conductor. To the left of x= 0 in this same plane is an electrode at the potential V. The regions to the right and left of the plane x= 0 are, respectively, ﬁlled with uniform dielectrics having permittivities /epsilon1aand/epsilon1b. Under the assumption that the system extends to inﬁnity in the ±xand±zdirections, we now determine the fringing ﬁelds in the vicinity of the interface between dielectrics. Our approach is to write solutions to Laplace’s equation in the respective regions that satisfy the boundary conditions in the planes y= 0 and y=aand asx→ ±∞ . These are then matched up by the jump conditions at the interface between dielectrics. Consider ﬁrst the region to the right, where Φ = 0 in the planes y= 0 and y=aand goes to zero as x→ ∞ . From Table 5.4.1, we select the inﬁnite set of solutions Φa=∞X n=1Ane−nπ axsinnπ ay (26) Here we have set k=nπ/a so that the sine functions are zero at each of the boundaries. In the region to the left, the ﬁeld is uniform in the limit x→ −∞ . This suggests writing the solution as the sum of a “particular” part meeting the “inhomogeneous part” of the boundary condition and a homogeneous part that is zero on each of the boundaries. Φb=−V¡y a−1¢ +∞X n=1Bnenπ axsinnπ ay (27) The coeﬃcients AnandBnmust now be adjusted so that the jump conditions are met at the interface between the dielectrics, where x= 0. First, consider the jump condition on the potential, (6.5.4). Evaluated at x= 0, (26) and (27) must give the same potential regardless of y. Φa¯¯ x=0= Φb¯¯ x=0⇒∞X n=1Ansinnπ ay=−V¡y a−1¢ +∞X n=1Bnsinnπ ay (28) 30 Polarization Chapter 6 To satisfy this relation at each value of y, expand the linear potential distribution on the right in a series of the same form as the other two terms. −V¡y a−1¢ =∞X n=1Vnsinnπ ay (29) Multiplication of both sides by sin( mπy/a ) and integration from y= 0 to y=a gives only one term on the right and an integral that can be carried out on the left. Hence, we can solve for the coeﬃcients Vnin (29). Za 0−V¡y a−1¢ sinmπ aydy=aVm 2⇒Vn=2V nπ(30) Thus, the series provided by (29) and (30) can be substituted into (28) to obtain an expression with each term a sum over the same type of series. ∞X n=1Ansinnπ ay=∞X n=12V nπsinnπ ay+∞X n=1Bnsinnπ ay (31) This expression is satisﬁed if the coeﬃcients of the like terms are equal. Thus, we have An=2V nπ+Bn (32) To make the normal component of Dcontinuous at the interface, −/epsilon1a∂Φa ∂x¯¯ x=0=−/epsilon1b∂Φb ∂x¯¯ x=0⇒∞X n=1/epsilon1anπ aAnsinnπ ay =−∞X n=1/epsilon1bnπ aBnsinnπ ay(33) and a second relation between the coeﬃcients results. /epsilon1aAn=−/epsilon1bBn (34) The coeﬃcients AnandBnare now determined by simultaneously solving (32) and (34). These are substituted into the original expressions for the potential, (26) and (27), to give the desired potential distribution. Φa=∞X n=12V nπ¡ 1 +/epsilon1a /epsilon1b¢e−nπ axsinnπ ay (35) Φb=−V¡y a−1¢ −∞X n=12 nπ/epsilon1a /epsilon1bV¡ 1 +/epsilon1a /epsilon1b¢enπ axsinnπ ay (36) These potential distributions, and sketches of the associated ﬁelds, are illus- trated in Fig. 6.6.9. Shown ﬁrst is the uniform dielectric. Laplace’s equation prevails throughout, even at the “interface.” Far to the left, we know that the potential is Sec. 6.7 Inhomogeneous Dielectrics 31 Fig. 6.6.9 Equipotentials and ﬁeld lines for conﬁguration of Fig. 6.6.8. (a) Fringing for uniform dielectric. (b) With high permittivity material between capacitor plates, ﬁeld inside tends to become tangential to the interface and uniform throughout the region to the left. (c) With high permittivity material outside the region between the capacitor plates, the ﬁeld inside tends to be perpendicular to the interface. linear in y, and hence represented by the equally spaced parallel straight lines. These lines must end at other points on the bounding surface having the same potential. The only place where this is possible is in the singular region at the origin where the potential makes an abrupt change from Vto 0. These observations provide a starting point in sketching the ﬁeld lines. Shown next is the ﬁeld distribution in the limit where the permittivity between the capacitor plates (to the left) is very large compared to that outside. As is clear by taking the limit /epsilon1a//epsilon1b→0 in (36), the ﬁeld inside the capacitor tends to be uniform right up to the edge of the capacitor. The dielectric eﬀectively ducts the electric ﬁeld. As far as the ﬁeld inside the capacitor is concerned, there tends to be no normal component of E. In the opposite extreme, where the region to the right has a high permittivity compared to that between the capacitor plates, the electric ﬁeld inside the capaci- tor tends to approach the interface normally. As far as the potential to the left is concerned, the interface is an equipotential. In Chap. 9, we ﬁnd that magnetization and polarization phenomena are analo- gous. There we delve further into approximations on magnetic ﬁeld distributions in the presence of magnetizable materials that can just as well be used to understand systems of piece-wise uniform dielectrics. 32 Polarization Chapter 6 6.7 SMOOTHLY INHOMOGENEOUS ELECTRICALLY LINEAR DIELECTRICS The potential distribution in a dielectric that is free of unpaired charge and which has a space-varying permittivity is governed by ∇ ·/epsilon1∇Φ = 0 (1) This is (6.5.1) combined with (6.5.2) and with ρu= 0. The contribution of the spatially varying permittivity is emphasized by using the vector identity for the divergence of a scalar ( /epsilon1) times a vector ( ∇Φ). ∇2Φ +∇Φ·∇/epsilon1 /epsilon1= 0 (2) With a spatially varying permittivity, polarization charge is induced in proportion to the component of Ethat is in the direction of the gradient in /epsilon1. Thus, in general, the potential is not a solution to Laplace’s equation. Equation (2) gives a diﬀerent perspective to the approach taken in dealing with piece-wise uniform systems. In Sec. 6.6, the polarization charge density represented by the ∇/epsilon1term in (2) is conﬁned to interfaces and accounted for by jump conditions. Thus, the section was a variation on the theme of Laplace’s equation. The theme of this section broadens the developments of Sec. 6.6. It is the objective in this section to demonstrate how familiar methods are adapted to dealing with unfamiliar laws. In general, (2) has spatially varying coef- ﬁcients. Thus, even though it is linear, we are not guaranteed simple closed-form solutions. However, if the spatial dependence of /epsilon1is exponential, the equation does have constant coeﬃcients and simple solutions. Our example exploits this fact. Example 6.7.1. Fields in an Exponentially Varying Dielectric A dielectric has a permittivity that varies exponentially in the ydirection, as illustrated in Fig. 6.7.1a. /epsilon1=/epsilon1(y) =/epsilon1pe−βy(3) Here /epsilon1pandβare given constants. In this example, the dielectric ﬁlls the rectangular region shown in Fig. 6.7.1b. This conﬁguration is familiar from Sec. 5.5. The ﬁelds are two dimensional, Φ = 0 atx= 0 and x=aandy= 0. The potential on the “last” surface, where y=b, is v(t). It follows from (3) that ∇Φ·∇/epsilon1 /epsilon1=−β∂Φ ∂y(4) and (2) becomes ∂2Φ ∂x2+∂2Φ ∂y2−β∂Φ ∂y= 0 (5) Sec. 6.7 Inhomogeneous Dielectrics 33 Fig. 6.7.1 (a) Smooth permittivity distribution of material enclosed by (b) zero potential boundaries at x= 0, x=a, and y= 0, and electrode at potential vaty=b. The dielectric ﬁlls a region having boundaries that are natural in Cartesian coordinates. Thus, we look for product solutions having the form Φ = X(x)Y(y). Substitution into (5) gives 1 Yµ d2Y dy2−1 βdY dy¶ +1 Xd2X dx2= 0 (6) The ﬁrst term, a function of yalone, must sum with the function of xalone to give zero. Thus, the ﬁrst is set equal to the separation coeﬃcient k2and the second equal to−k2. d2X dx2+k2X= 0 (7) d2Y dy2−βdY dy−k2Y= 0 (8) This assignment of sign for the separation coeﬃcient is motivated by the requirement that Φ = 0 at two locations. This results in periodic solutions for (7). X=nsinkx coskx(9) Because it also has constant coeﬃcients, the solutions to (8) are exponentials. Sub- stitution of exp( py) shows that p=β 2±r¡β 2¢2+k2 (10) and it follows that solutions are linear combinations of two exponentials. Y=eβ 2y2 4coshq¡β 2¢2+k2y sinhq¡β 2¢2+k2y3 5 (11) 34 Polarization Chapter 6 For the speciﬁc problem at hand, we look for the products of these sets of solutions that satisfy the homogeneous boundary conditions. Those at x= 0 and x=aare met by making k=nπ/a , with nan integer. The origin of the yaxis was made to coincide with the third zero potential boundary so that the hyperbolic sine function could be used. Thus, we arrive at an inﬁnite series of solutions, each satisfying the homogeneous boundary conditions. Φ =∞X n=1Aneβ 2ysinhr¡β 2¢2+¡nπ a¢2ysin¡nπ ax¢ (12) The assignment of the coeﬃcients so that the potential constraint at y=bis met follows the procedure familiar from Sec. 5.5. Φ =∞X n=1 odd4v nπeβ 2(y−b)sinhq¡β 2¢2+¡nπ a¢2y sinhq¡β 2¢2+¡nπ a¢2bsin¡nπ ax¢ (13) For interpretation of (13), suppose that βis positive so that /epsilon1decreases with y, as illustrated in Fig. 6.7.1a. Without the analysis, we know that the lines of Doriginate on the electrode at y=band terminate on the zero potential walls. This means that Elines either terminate on the grounded walls or on polarization charges induced in the volume. If v >0, we can see from (6.5.9) that because E·∇/epsilon1 is positive, the induced polarization charge density must be negative. Thus, some of theElines terminate on this negative charge density and it comes as no surprise that we have found a potential that decays away from the excitation electrode at y=b at a rate that is faster than if the potential were governed by Laplace’s equation. The electric ﬁeld is eﬀectively shielded out of the lower region of higher permittivity by the induced polarization charge. One approach to determining ﬁelds in spatially varying dielectrics is suggested in Fig. 6.7.2. The smooth distribution has been approximated by “stair steps.” Physically, the equivalent system consists of uniform layers. Thus, the ﬁelds re- vert to the solutions of Laplace’s equation matched to each other at the interfaces by the jump conditions. According to (6.5.11), Elines originating at y=band passing downward through these interfaces will induce positive surface polarization charge. Thus, replacing the smoothly varying dielectric with the layers of uniform dielectric is equivalent to representing the volume polarization charge density by a distribution of surface polarization charges. 6.8 SUMMARY Table 6.8.1 is useful both as an outline of this chapter and as a reference. Gauss’ theorem is the basis for deriving the surface relations in the right-hand column from the respective volume relations in the left-hand column. By remembering the volume relations, one is able to recall the surface relations. Our ﬁrst task, in Sec. 6.1, was to introduce the polarization density as a way of representing the polarization charge density. The ﬁrst volume and surface Sec. 6.8 Summary 35 Fig. 6.7.2 Stair-step distribution of permittivity approximating smooth dis- tribution. relations resulted. These are deceptively similar in appearance to Gauss’ law and the associated jump condition. However, they are not electric ﬁeld laws. Rather, they simply relate the volume and surface sources representing the material to the polarization density. Next we considered the ﬁelds due to permanently polarized materials. The polarization density was given. For this purpose, Gauss’ law and the associated jump condition were conveniently written as (6.2.2) and (6.2.3), respectively. With the polarization induced by the ﬁeld itself, it was convenient to intro- duce the displacement ﬂux density Dand write Gauss’ law and the jump condition as (6.2.15) and (6.2.16). In particular, for linear polarization, the equivalent consti- tutive laws of (6.4.2) and (6.4.3) were introduced. The theme of this chapter has been the determination of EQS ﬁelds when the polarization charge density makes a contribution. In cases where the polarization density is given, this is easy to keep in mind, because the ﬁrst step in formulating a problem is to evaluate ρpfrom the given P. However, when ρpis induced, variables such as Dare used and we must be reminded that when all is said and done, ρp (or its surface counterpart, σsp) is still responsible for the eﬀect of the material on the ﬁeld. The expressions for ρpandσspgiven by the last two relations in the table are useful not only for interpreting the distributions of ﬁelds after they have been found but for forming an impression of the ﬁelds in complex systems where it would not be worthwhile to ﬁnd an analytic solution. Remember that these relations hold only in regions where there is no unpaired charge density. In Chap. 9, we will ﬁnd that most of this chapter is directly applicable to the description of magnetization. There we will continue to develop insights that will be equally applicable to the polarization phenomena of this chapter. 36 Polarization Chapter 6 TABLE 6.8.1 SUMMARY OF POLARIZATION RELATIONS AND LAWS Polarization Charge Density and Polarization Density ρp=−∇ ·P (6.1.6) σsp=−n·(Pa−Pb) 6.1.7) Gauss’ Law with Polarization ∇ ·/epsilon1oE=ρp+ρu (6.2.1) n·/epsilon1o(Ea−Eb) =σsp+σsu (6.2.3) ∇ ·D=ρu (6.2.15) n·(Da−Db) =σsu (6.2.16) where D≡/epsilon1oE+P (6.2.14) Electrically Linear Polarization Constitutive Law P=/epsilon1oχeE= (/epsilon1−/epsilon1o)E (6.4.2) D=/epsilon1E (6.4.3) Source Distribution, ρu= 0 ρp=−/epsilon1o /epsilon1E· ∇/epsilon1 (6.5.9) σsp=n·/epsilon1oEa¡ 1−/epsilon1a /epsilon1b¢ (6.5.11) P R O B L E M S 6.1 Polarization Density 6.1.1 The layer of polarized material shown in cross-section in Fig. P6.1.1, having thickness dand surfaces in the planes y=dandy= 0, has the polarization density P=Pocosβx(ix+iy). (a) Determine the polarization charge density throughout the slab. (b) What is the surface polarization charge density on the layer surfaces? Sec. 6.3 Problems 37 Fig. P6.1.1 6.2 Laws and Continuity Conditions with Polarization 6.2.1 For the polarization density given in Prob. 6.1.1, with Po(t) =Pocosωt: (a) Determine the polarization current density and polarization charge density. (b) Using Jpandρp, show that the diﬀerential charge conservation law, (10), is indeed satisﬁed. 6.3 Permanent Polarization 6.3.1∗A layer of permanently polarized material is sandwiched between plane parallel perfectly conducting electrodes in the planes x= 0 and x=a, respectively, having potentials Φ = 0 and Φ = −V. The system extends to inﬁnity in the ±yand±zdirections. (a) Given that P=Pocosβxix, show that the potential between the electrodes is Φ =Po β/epsilon1o(sinβx−x asinβa)−V x a(a) (b) Given that P=Pocosβyiy, show that the potential between the electrodes is Φ =Po β/epsilon1osinβy· 1−coshβ(x−a/2) cosh( βa/2)¸ −V x a(b) 6.3.2 The cross-section of a conﬁguration that extends to inﬁnity in the ±zdi- rections is shown in Fig. P6.3.2. What is the potential distribution inside the cylinder of rectangular cross-section? 6.3.3∗A polarization density is given in the semi-inﬁnite half-space y <0 to be P=Pocos[(2 π/Λ)x]iy. There are no other ﬁeld sources in the system and Poand Λ are given constants. (a) Show that ρp= 0 and σsp=Pocos(2 πx/Λ). 38 Polarization Chapter 6 Fig. P6.3.2 Fig. P6.3.5 (b) Show that Φ =PoΛ 4/epsilon1oπcos(2 πx/Λ) exp( ∓2πy/Λ); y><0 ( a) 6.3.4 A layer in the region −a < y < 0 has the polarization density P= Poiysinβ(x−xo). In the planes y=±a, the potential is constrained to be Φ =Vcosβx, where Po, βandVare given constants. The region 0 < y < a is free space and the system extends to inﬁnity in the ±xand±zdirections. Find the potential in regions (a) and (b) in the free space and polarized regions, respectively. (If you have already solved Prob. 5.6.12, you can solve this problem by inspection.) 6.3.5∗Figure P6.3.5 shows a material having the uniform polarization density P=Poiz, with a spherical cavity having radius R. On the surface of the cavity is a uniform distribution of unpaired charge having density σsu=σo. The interior of the cavity is free space, and Poandσoare given constants. The potential far from the cavity is zero. Show that the electric potential is Φ =( −Po 3/epsilon1orcosθ+σoR /epsilon1o;r≤R −PoR3 3/epsilon1or2cosθ+σoR2 /epsilon1or;r≥R(a) 6.3.6 The cross-section of a groove (shaped like a half-cylinder having radius R) cut from a uniformly polarized material is shown in Fig. P6.3.6. The Sec. 6.3 Problems 39 Fig. P6.3.6 Fig. P6.3.7 Fig. P6.3.8 material rests on a grounded perfectly conducting electrode at y= 0, and Pois a given constant. Assume that the conﬁguration extends to inﬁnity in the ydirection and ﬁnd Φ in regions (a) and (b), respectively, outside and inside the groove. 6.3.7 The system shown in cross-section in Fig. P6.3.7 extends to inﬁnity in the ±xand±zdirections. The electrodes at y= 0 and y=a+bare shorted. Given Poand the dimensions, what is Ein regions (a) and (b)? 6.3.8∗In the two-dimensional conﬁguration shown in Fig. P6.3.8, a perfectly con- ducting circular cylindrical electrode at r=ais grounded. It is coax- ial with a rotor of radius bwhich supports the polarization density P= ∇[Porcos(φ−α)]. (a) Show that the polarization charge density is zero inside the rotor. (b) Show that the potential functions ΦIand ΦIIrespectively in the regions outside and inside the rotor are ΦI=Pob2 2/epsilon1o¡1 r−r a2¢ cos(φ−α) ( a) 40 Polarization Chapter 6 Fig. P6.3.9 ΦII=Po(a2−b2) 2/epsilon1oa2rcos(φ−α) ( b) (c) Show that if α= Ωt, where Ω is an angular velocity, the ﬁeld rotates in the φdirection with this angular velocity. 6.3.9 A circular cylindrical material having radius bhas the polarization density P=∇[Po(rm+1/bm) cosmφ], where mis a given positive integer. The region b < r < a , shown in Fig. P6.3.9, is free space. (a) Determine the volume and surface polarization charge densities for the circular cylinder. (b) Find the potential in regions (a) and (b). (c) Now the cylinder rotates with the constant angular velocity Ω. Argue that the resulting potential is obtained by replacing φ→(φ−Ωt). (d) A section of the outer cylinder is electrically isolated and connected to ground through a resistance R. This resistance is low enough so that, as far as the potential in the gap is concerned, the potential of the segment can still be taken as zero. However, as the rotor rotates, the charge induced on the segment is time varying. As a result, there is a current through the resistor and hence an output signal vo. Assume that the segment subtends an angle π/m and has length lin the z direction, and ﬁnd vo. 6.3.10∗Plane parallel electrodes having zero potential extend to inﬁnity in the x−z planes at y= 0 and y=d. (a) In a ﬁrst conﬁguration, the region between the electrodes is free space, except for a segmented electrode in the plane x= 0 which constrains the potential there to be V(y). Given V(y), what is the potential distribution in the regions 0 < x andx < 0, regions (a) and (b), respectively? (b) Now the segmented electrode is removed and the region x <0 is ﬁlled with a permanently polarized material having P=Poix, where Pois a given constant. What continuity conditions must the potential satisfy in the x= 0 plane? Sec. 6.5 Problems 41 (c) Show that the potential is given by Φ =dPo /epsilon1o∞X n=1[1−(−1)n] (nπ)2sinnπ dyexp¡ ∓nπ dx¢ ; x><0 ( a) (The method used here to represent Φ is used in Example 6.6.3.) 6.3.11 In Prob. 6.1.1, there is a perfect conductor in the plane y= 0 and the region d < y is free space. What are the potentials in regions (a) and (b), the regions where d < y and 0 < y < d , respectively? 6.4 Polarization Constitutive Laws 6.4.1 Suppose that a solid or liquid has a mass density of ρ= 103kg/m3and a molecular weight of Mo= 18 (typical of water). [The number of molecules per unit mass is Avogadro’s number ( Ao= 6.023×1026molecules/kg-mole) divided by Mo.] This material has a permittivity /epsilon1= 2/epsilon1oand is subject to an electric ﬁeld intensity E= 107v/m (approaching the highest ﬁeld strength that can be sustained without breakdown on scales of a centimeters in liquids and solids). Assume that each molecule has a polarization qdwhere q=e= 1.6×10−19C, the charge of an electron). What is |d|? 6.5 Fields in the Presence of Electrically Linear Dielectrics 6.5.1∗The plane parallel electrode conﬁgurations of Fig. P6.5.1 have in common the fact that the linear dielectrics have dielectric “constants” that are func- tions of x, /epsilon1=/epsilon1(x). The systems have depth cin the zdirection. (a) Show that regardless of the speciﬁc functional dependence on x,Eis uniform and simply iyv/d. (b) For the system of Fig. P6.5.1a, where the dielectric is composed of uniform regions having permittivities /epsilon1aand/epsilon1b, show that the capac- itance is C=c d(/epsilon1bb+/epsilon1aa) ( a) (c) For the smoothly inhomogeneous capacitor of Fig. P6.5.1b, /epsilon1=/epsilon1o(1+ x/l). Show that C=3/epsilon1ocl 2d(b) 6.5.2 In the conﬁguration shown in Fig. P6.5.1b, what is the capacitance Cif /epsilon1=/epsilon1a(1 +αcosβx), where 0 < α < 1 and βare given constants? 42 Polarization Chapter 6 Fig. P6.5.1 Fig. P6.5.3 6.5.3∗The region of Fig. P6.5.3 between plane parallel perfectly conducting elec- trodes in the planes y= 0 and y=lis ﬁlled by a uniformly inhomogeneous dielectric having permittivity /epsilon1=/epsilon1o[1+χa(1+y/l)]. The electrode at y= 0 has potential vrelative to that at y=l. The electrode separation lis much smaller than the dimensions of the system in the xandzdirections, so the ﬁelds can be regarded as not depending on xorz. (a) Show that Dyis independent of y. (b) With the electrodes having area A, show that the capacitance is C=/epsilon1oA lχa/lnh1 + 2 χa 1 +χai (a) 6.5.4 The dielectric in the system of Prob. 6.5.3 is replaced by one having per- mittivity /epsilon1=/epsilon1pexp(−y/d), where /epsilon1pis constant. What is the capacitance C? 6.5.5 In the two conﬁgurations shown in cross-section in Fig. P6.5.5, circular cylindrical conductors are used to make coaxial capacitors. In Fig. P6.5.5a, the linear dielectric has a wedge shape with interfaces with the free space region that are surfaces of constant φ. In Fig. P6.5.5b, the interface is at r=R. (a) Determine E(r) in regions (1) and (2) in each conﬁguration, showing that simple ﬁelds satisfy all boundary conditions on the electrode surfaces and at the interfaces between dielectric and free space. (b) For lengths lin the zdirection, what are the capacitances? 6.5.6∗For the conﬁguration of Fig. P6.5.5a, the wedge-shaped dielectric is re- placed by one that ﬁlls the gap (over all φas well as over the radius Sec. 6.6 Problems 43 Fig. P6.5.5 Fig. P6.6.1 b < r < a ) with material having the permittivity /epsilon1=/epsilon1a+/epsilon1bcos2φ, where /epsilon1aand/epsilon1bare constants. Show that the capacitance is C= (2/epsilon1a+/epsilon1b)πl/ln (a/b) ( a) 6.6 Piece-Wise Uniform Electrically Linear Dielectrics 6.6.1∗An insulating sphere having radius Rand uniform permittivity /epsilon1sis sur- rounded by free space, as shown in Fig. P6.6.1. It is immersed in an electric ﬁeldEo(t)izthat, in the absence of the sphere, is uniform. (a) Show that the potential is Φ =Eo(t)½ −rcosθ+R3Acosθ r2;R < r Brcosθ; r < R(a) where A= (/epsilon1s−/epsilon1o)/(/epsilon1s+ 2/epsilon1o) and B=−3/epsilon1o/(/epsilon1s+ 2/epsilon1o). (b) Show that, in the limit where /epsilon1s→ ∞ , the electric ﬁeld intensity tangential to the surface of the sphere goes to zero. Thus, the surface becomes an equipotential. (c) Show that the same solution is obtained for the potential outside the sphere as in the limit /epsilon1s→ ∞ if this boundary condition is used at the outset. 44 Polarization Chapter 6 Fig. P6.6.2 6.6.2 An electric dipole having a z-directed moment pis situated at the origin, as shown in Fig. P6.6.2. Surrounding it is a spherical cavity of free space having radius a. Outside of the radius ais a linearly polarizable dielectric having permittivity /epsilon1. (a) Determine Φ and Ein regions (a) and (b) outside and inside the cavity. (b) Show that in the limit where /epsilon1→ ∞ , the electric ﬁeld intensity tan- gential to the interface of the dielectric goes to zero. That is, in this limit, the eﬀect of the dielectric on the interior ﬁelds is the same as if the dielectric were a perfect conductor. (c) Show that the same interior potential is obtained as in the limit /epsilon1→ ∞if this boundary condition is used at the outset. 6.6.3∗In Example 6.6.1, an artiﬁcial dielectric is made from an array of perfectly conducting spheres. Here, an artiﬁcial dielectric is constructed using an array of rods, each having a circular cross-section with radius R. The rods run parallel to the capacitor plates and hence perpendicular to the imposed electric ﬁeld intensity. The spacing between rod centers is s, and they are in a square array. Show that, for slarge enough so that the ﬁelds induced by the rods do not interact, the equivalent electric susceptibility is χc= 2π(R/s)2. 6.6.4 Each of the conducting spheres in the artiﬁcial dielectric of Example 6.6.1 is replaced by the dielectric sphere of Prob. 6.6.1. Again, with the un- derstanding that the spacing between spheres is large enough to justify ignoring their interaction, what is the equivalent susceptibility of the ar- ray? 6.6.5∗A point charge ﬁnds itself at a height habove an inﬁnite half-space of dielectric material. The charge has magnitude q, the dielectric has a uniform permittivity /epsilon1, and there are no unpaired charges in the volume of the dielectric or on its surface. The Cartesian coordinates xandzare in the plane of the dielectric interface, while yis directed perpendicular to the interface and into the free space region. Thus, the charge is at y=h. The ﬁeld in the free space region can be taken as the superposition of a particular solution due to the point charge and a homogeneous solution due to a charge qbaty=−hbelow the interface. The ﬁeld in the dielectric can be taken as that of a charge qaaty=h. Sec. 6.6 Problems 45 (a) Show that the potential is given by Φ =1 4π/epsilon1o½ (q/r+)−(qb/r−); 0 < y qa/r+; y <0 where r±=p x2+ (y∓h)2+z2and the magnitudes of the charges turn out to be qa=2q¡/epsilon1 /epsilon1o+ 1¢; qb=q¡/epsilon1 /epsilon1o−1¢ ¡/epsilon1 /epsilon1o+ 1¢ (b) (b) Show that the charge is attracted to the dielectric with the force f=qqb 16π/epsilon1oh2(c) 6.6.6 The half-space y >0 is ﬁlled by a dielectric having uniform permittivity /epsilon1a, while the remaining region 0 > yis ﬁlled by a dielectric having the uniform permittivity /epsilon1b. Running parallel to the interface between these dielectrics along the line where x= 0 and y=his a uniform line charge of density λ. Determine the potentials in regions (a) and (b), respectively. 6.6.7∗If the permittivities are nearly the same, so that (1 −/epsilon1a//epsilon1b)≡κis small, the qualitative approach to determining the ﬁeld distribution given in con- nection with Fig. 6.6.7 can be made quantitative. That is, if κis small, the polarization charge induced by the imposed ﬁeld can be determined to a good approximation and that charge, in turn, used to ﬁnd the change in the applied ﬁeld. Consider the following approximate approach to ﬁnding the ﬁelds in and around the dielectric cylinder of Example 6.6.2. (a) In the limit where κis zero, the ﬁeld is equal to the applied ﬁeld, both inside and outside the cylinder. Write this ﬁeld in polar coordinates. (b) Show that this ﬁeld gives rise to σsp=/epsilon1bEoκcosφat the surface of the cylinder. (c) Find the ﬁeld due to this induced polarization surface charge and add it to the imposed ﬁeld to show that, with the ﬁrst-order contribution of the induced polarization surface charge, the ﬁeld is Φ =−REo½¡r R−κR 2r¢ cosφ;r > R r R¡ 1−κ 2¢ cosφ;r < R(a) (d) Expand the exact ﬁelds given by (21) and (22) to ﬁrst order in κand show that they are in agreement with this result. 6.6.8 As an illustration of how identiﬁcation of the induced polarization charge can be used in a qualitative determination of the ﬁelds, consider the ﬁelds 46 Polarization Chapter 6 Fig. P6.6.8 Fig. P6.6.9 between the plane parallel electrodes of Fig. P6.6.8. In Fig. P6.6.8a, there are two layers of dielectric. (a) In the limit where κ= (1−/epsilon1a//epsilon1b) is zero, what is the imposed E? (b) What is the σspinduced by this ﬁeld at the interface between the dielectrics. (c) For /epsilon1a> /epsilon1b, sketch the ﬁeld lines in the two regions. (You should be able to see, from the superposition of the ﬁelds induced by this σsp and that imposed, which of the ﬁelds is the greater.) (d) Now consider the more complicated geometry of Fig. 6.6.8b and carry out the same steps. Based on your deductions, draw a sketch of σsp andEfor the case where /epsilon1b> /epsilon1a. 6.6.9 The conﬁguration of perfectly conducting electrodes and perfectly insulat- ing dielectrics shown in Fig. P6.6.9 is similar to that shown in Fig. 6.6.8 except that at the left and right, the electrodes are “shorted” together and the top electrode is also divided at the middle. Thus, the ⊃shaped electrode is grounded while the ⊂shaped one is at potential V. (a) Determine Φ in regions (a) and (b). (b) With the permittivities equal, sketch Φ and E. (Use physical reason- ing rather than the mathematical result.) (c) Assuming that the permittivities are nearly equal, use the result of (b) to deduce σspon the interface between dielectrics in the case where /epsilon1a//epsilon1bis somewhat greater than and then somewhat less than 1. Sketch Ededuced as the sum of the ﬁelds induced by these surface charges and the imposed ﬁeld. (d) With /epsilon1amuch greater that /epsilon1b, draw a sketch of Φ and Ein region (b). (e) With /epsilon1amuch less than /epsilon1b, sketch Φ and Ein both regions. 6.7 Smoothly Inhomogeneous Electrically Linear Dielectrics Sec. 6.7 Problems 47 Fig. P6.7.1 6.7.1∗For the two-dimensional system shown in Fig. P6.7.1, show that the po- tential in the smoothly inhomogeneous dielectric is Φ =V x a+∞X n=1¡2V nπ¢ eβy/2 exp· −p (β/2)2+ (nπ/a )2y¸ sin¡nπ ax¢(a) 6.7.2 In Example 6.6.3, the dielectrics to right and left, respectively, have the per- mittivities /epsilon1a=/epsilon1pexp(−βx) and /epsilon1b=/epsilon1pexp(βx). Determine the potential throughout the dielectric regions. 6.7.3 A linear dielectric has the permittivity /epsilon1=/epsilon1a{1 +χpexp[−(x2+y2+z2)/a2]} (a) An electric ﬁeld that is uniform far from the origin (where it is equal toEoiy) is imposed. (a) Assume that /epsilon1//epsilon1ois not much diﬀerent from unity and ﬁnd ρp. (b) With this induced polarization charge as a guide, sketch E. 7 CONDUCTION AND ELECTROQUASISTATIC CHARGE RELAXATION 7.0 INTRODUCTION This is the last in the sequence of chapters concerned largely with electrostatic and electroquasistatic ﬁelds. The electric ﬁeld Eis still irrotational and can therefore be represented in terms of the electric potential Φ. ∇ ×E= 0⇔E=−∇Φ (1) The source of Eis the charge density. In Chap. 4, we began our exploration of EQS ﬁelds by treating the distribution of this source as prescribed. By the end of Chap. 4, we identiﬁed solutions to boundary value problems, where equipotential surfaces were replaced by perfectly conducting metallic electrodes. There, and throughout Chap. 5, the sources residing on the surfaces of electrodes as surface charge densities were made self-consistent with the ﬁeld. However, in the volume, the charge density was still prescribed. In Chap. 6, the ﬁrst of two steps were taken toward a self-consistent description of the charge density in the volume. In relating Eto its sources through Gauss’ law, we recognized the existence of two types of charge densities, ρuandρp, which, respectively, represented unpaired and paired charges. The paired charges were related to the polarization density Pwith the result that Gauss’ law could be written as (6.2.15) ∇ ·D=ρu (2) where D≡/epsilon1oE+P. Throughout Chap. 6, the volume was assumed to be perfectly insulating. Thus, ρpwas either zero or a given distribution. 1 2 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.0.1 EQS distributions of potential and current density are analogous to those of voltage and current in a network of resistors and capacitors. (a) Systems of perfect dielectrics and perfect conductors are analogous to capaci- tive networks. (b) Conduction eﬀects considered in this chapter are analogous to those introduced by adding resistors to the network. The second step toward a self-consistent description of the volume charge density is taken by adding to (1) and (2) an equation expressing conservation of the unpaired charges, (2.3.3). ∇ ·Ju+∂ρu ∂t= 0(3) That the charge appearing in this equation is indeed the unpaired charge den- sity follows by taking the divergence of Amp` ere’s law expressed with polarization, (6.2.17), and using Gauss’ law as given by (2) to eliminate D. To make use of these three diﬀerential laws, it is necessary to specify Pand J. In Chap. 6, we learned that the former was usually accomplished by either specifying the polarization density Por by introducing a polarization constitutive law relating PtoE. In this chapter, we will almost always be concerned with linear dielectrics, where D=/epsilon1E. A new constitutive law is required to relate Juto the electric ﬁeld intensity. The ﬁrst of the following sections is therefore devoted to the constitutive law of conduction. With the completion of Sec. 7.1, we have before us the diﬀerential laws that are the theme of this chapter. To anticipate the developments that follow, it is helpful to make an analogy to circuit theory. If the previous two chapters are regarded as describing circuits consisting of interconnected capacitors, as shown in Fig. 7.0.1a, then this chapter adds resistors to the circuit, as in Fig. 7.0.1b. Suppose that the voltage source is a step function. As the circuit is composed of resistors and capacitors, the distribution of currents and voltages in the circuit is ﬁnally determined by the resistors alone. That is, as t→ ∞ , the capacitors cease charging and are equivalent to open circuits. The distribution of voltages is then determined by the steady ﬂow of current through the resistors. In this long-time limit, the charge on the capacitors is determined from the voltages already speciﬁed by the resistive network. The steady current ﬂow is analogous to the ﬁeld situation where ∂ρu/∂t→0 in the conservation of charge expression, (3). We will ﬁnd that (1) and (3), the latter written with Jurepresented by the conduction constitutive law, then fully determine the distribution of potential, of E, and hence of Ju. Just as the charges Sec. 7.1 Conduction Constitutive Laws 3 on the capacitors in the circuit of Fig. 7.0.1b are then speciﬁed by the already determined voltage distribution, the charge distribution can be found in an after- the-fact fashion from the already determined ﬁeld distribution by using Gauss’ law, (2). After considering the physical basis for common conduction constitutive laws in Sec. 7.1, Secs. 7.2–7.6 are devoted to steady conduction phenomena. In the circuit of Fig. 7.0.1b, the distribution of voltages an instant after the voltage step is applied is determined by the capacitors without regard for the re- sistors. From a ﬁeld theory point of view, this is the physical situation described in Chaps. 4 and 5. It is the objective of Secs. 7.7–7.9 to form an appreciation for how this initial distribution of the ﬁelds and sources relaxes to the steady condition, already studied in Secs. 7.2–7.6, that prevails when t→ ∞ . In Chaps. 3–5 we invoked the “perfect conductivity” model for a conductor. For electroquasistatic systems, we will conclude this chapter with an answer to the question, “Under what circumstances can a conductor be regarded as perfect?” Finally, if the ﬁelds and currents are essentially static, there is no distinction between EQS and MQS laws. That is, if ∂B/∂tis negligible in an MQS system, Faraday’s law again reduces to (1). Thus, the ﬁrst half of this chapter provides an understanding of steady conduction in some MQS as well as EQS systems. In Chap. 8, we determine the magnetic ﬁeld intensity from a given distribution of current density. Provided that rates of change are slow enough so that eﬀects of magnetic induction can be ignored, the solution to the steady conduction problem as addressed in Secs. 7.2–7.6 provides the distribution of the magnetic ﬁeld source, the current density, needed to begin Chap. 8. Just how fast can the ﬁelds vary without producing eﬀects of magnetic in- duction? For EQS systems, the answer to this question comes in Secs. 7.7–7.9. The EQS eﬀects of ﬁnite conductivity and ﬁnite rates of change are in sharp contrast to their MQS counterparts, studied in the last half of Chap. 10. 7.1 CONDUCTION CONSTITUTIVE LAWS In the presence of materials, ﬁelds vary in space over at least two length scales. Themicroscopic scale is typically the distance between atoms or molecules while the much larger macroscopic scale is typically the dimension of an object made from the material. As developed in the previous chapter, ﬁelds in polarized media are averages over the microscopic scale of the dipoles. In eﬀect, the experimental determination of the polarization constitutive law relating the macroscopic Pand E(Sec. 6.4) does not deal with the microscopic ﬁeld. With the understanding that experimentally measured values will again be used to evaluate macroscopic parameters, we assume that the average force acting on an unpaired or free charge, q, within matter is of the same form as the Lorentz force, (1.1.1). f=q(E+v×µoH) (1) By contrast with a polarization charge, a free charge is not bound to the atoms and molecules, of which matter is constituted, but under the inﬂuence of the electric and magnetic ﬁelds can travel over distances that are large compared to interatomic or intermolecular distances. In general, the charged particles collide with the atomic 4 Conduction and Electroquasistatic Charge Relaxation Chapter 7 or molecular constituents, and so the force given by (1) does not lead to uniform acceleration, as it would for a charged particle in free space. In fact, in the conven- tional conduction process, a particle experiences so many collisions on time scales of interest that the average velocity it acquires is quite low. This phenomenon gives rise to two consequences. First, inertial eﬀects can be disregarded in the time aver- age balance of forces on the particle. Second, the velocity is so low that the forces due to magnetic ﬁelds are usually negligible. (The magnetic force term leads to the Hall eﬀect, which is small and very diﬃcult to observe in metallic conductors, but because of the relatively larger translational velocities reached by the charge carriers in semiconductors, more easily observed in these.) With the driving force ascribed solely to the electric ﬁeld and counterbalanced by a “viscous” force, proportional to the average translational velocity vof the charged particle, the force equation becomes f=±|q±|E=ν±v (2) where the upper and lower signs correspond to particles of positive and negative charge, respectively. The coeﬃcients ν±are positive constants representing the time average “drag” resulting from collisions of the carriers with the ﬁxed atoms or molecules through which they move. Written in terms of the mobilities ,µ±, the velocities of the positive and neg- ative particles follow from (2) as v±=±µ±E (3) where µ±=|q±|/ν±. The mobility is deﬁned as positive. The positive and negative particles move with and against the electric ﬁeld intensity, respectively. Now suppose that there are two types of charged particles, one positive and the other negative. These might be the positive sodium and negative chlorine ions resulting when salt is dissolved in water. In a metal, the positive charges represent the (zero mobility) atomic sites, while the negative particles are electrons. Then, with N+andN−, respectively, deﬁned as the number of these charged particles per unit volume, the current density is Ju=N+|q+|v+−N−|q−|v− (4) A ﬂux of negative particles comprises an electrical current that is in a direction opposite to that of the particle motion. Thus, the second term in (4) appears with a negative sign. The velocities in this expression are related to Eby (3), so it follows that the current density is Ju= (N+|q+|µ++N−|q−|µ−)E (5) In terms of the same variables, the unpaired charge density is ρu=N+|q+| −N−|q−| (6) Ohmic Conduction. In general, the distributions of particle densities N+and N−are determined by the electric ﬁeld. However, in many materials, the quantity in brackets in (5) is a property of the material, called the electrical conductivity σ. Sec. 7.2 Steady Ohmic Conduction 5 Ju=σE; σ≡(N+|q+|µ++N−|q−|µ−) (7) The MKS units of σare (ohm - m)−1≡Siemens/m = S/m. In these materials, the charge densities N+q+andN−q−keep each other in (approximate) balance so that there is little eﬀect of the applied ﬁeld on their sum. Thus, the conductivity σ(r) is speciﬁed as a function of position in nonuniform media by the distribution N±in the material and by the local mobilities, which can also be functions of r. The conduction constitutive law given by (7) is Ohm’s law generalized in a ﬁeld-theoretical sense. Values of the conductivity for some common materials are given in Table 7.1.1. It is important to keep in mind that any constitutive law is of restricted use, and Ohm’s law is no exception. For metals and semiconductors, it is usually a good model on a suﬃciently large scale. It is also widely used in dealing with electrolytes. However, as materials become semi-insulators, it can be of questionable validity. Unipolar Conduction. To form an appreciation for the implications of Ohm’s law, it will be helpful to contrast it with the law for unipolar conduction . In that case, charged particles of only one sign move in a neutral background, so that the expressions for the current density and charge density that replace (5) and (6) are Ju=|ρ|µE (8) ρu=ρ (9) where the charge density ρnow carries its own sign. Typical of situations described by these relations is the passage of ions through air. Note that a current density exists in unipolar conduction only if there is a net charge density. By contrast, for Ohmic conduction, where the current density and the charge density are given by (7) and (6), respectively, there can be a current density at a location where there is no netcharge density. For example, in a metal, negative electrons move through a background of ﬁxed positively charged atoms. Thus, in (7), µ+= 0 and the conductivity is due solely to the electrons. But it follows from (6) that the positive charges do have an important eﬀect, in that they can nullify the charge density of the electrons. We will often ﬁnd that in an Ohmic conductor there is a current density where there is no net unpaired charge density. 7.2 STEADY OHMIC CONDUCTION To set the stage for the next two sections, consider the ﬁelds in a material that has a linear polarizability and is described by Ohm’s law, (7.1.7). J=σ(r)E;D=/epsilon1(r)E (1) 6 Conduction and Electroquasistatic Charge Relaxation Chapter 7 TABLE 7.1.1 CONDUCTIVITY OF VARIOUS MATERIALS Metals and Alloys in Solid State σ−mhos/m at 20◦C Aluminum, commercial hard drawn . . . . . . . . . . . . . . . . . . . . . . . . . . 3.54 x 107 Copper, annealed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.80 x 107 Copper, hard drawn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.65 x 107 Gold, pure drawn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.10 x 107 Iron, 99.98% . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.0 x 107 Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.5–1.0 x 107 Lead . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.48 x 107 Magnesium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17 x 107 Nichrome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.10 x 107 Nickel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.28 x 107 Silver, 99.98% . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 x 107 Tungsten . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.81 x 107 Semi-insulating and Dielectric Solids Bakelite (average range)* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10−8−1010 Celluloid* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10−8 Glass, ordinary* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10−12 Hard rubber* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10−14−10−16 Mica* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10−11−10−15 Paraﬃn* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10−14−10−16 Quartz, fused* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . less than 10−17 Sulfur* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . less than 10−16 Teﬂon* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . less than 10−16 Liquids Mercury . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.10 x 107 Alcohol, ethyl, 15◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 x 10−4 Water, Distilled, 18◦C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 x 10−4 Corn Oil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 x 10−11 *For highly insulating materials. Ohm’s law is of dubious validity and conductivity values are only useful for making estimates. In general, these properties are functions of position, r. Typically, electrodes are used to constrain the potential over some of the surface enclosing this material, as suggested by Fig. 7.2.1. In this section, we suppose that the excitations are essentially constant in Sec. 7.2 Steady Ohmic Conduction 7 Fig. 7.2.1 Conﬁguration having volume enclosed by surfaces S/prime, upon which the potential is constrained, and S/prime/prime, upon which its normal derivative is con- strained. time, in the sense that the rate of accumulation of charge at any given location has a negligible inﬂuence on the distribution of the current density. Thus, the time derivative of the unpaired charge density in the charge conservation law, (7.0.3), is negligible. This implies that the current density is solenoidal. ∇ ·σE= 0 (2) Of course, in the EQS approximation, the electric ﬁeld is also irrotational. ∇ ×E= 0⇔E=−∇Φ (3) Combining (2) and (3) gives a second-order diﬀerential equation for the potential distribution. ∇ ·σ∇Φ = 0 (4) In regions of uniform conductivity ( σ= constant), it assumes a familiar form. ∇2Φ = 0 (5) In a uniform conductor, the potential distribution satisﬁes Laplace’s equation. It is important to realize that the physical reasons for obtaining Laplace’s equation for the potential distribution in a uniform conductor are quite diﬀerent from those that led to Laplace’s equation in the electroquasistatic cases of Chaps. 4 and 5. With steady conduction, the governing requirement is that the divergence of the current density vanish. The unpaired charge density does not inﬂuence the current distribution, but is rather determined by it. In a uniform conductor, the continuity constraint on Jhappens to imply that there is no unpaired charge density. 8 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.2.2 Boundary between region (a) that is insulating relative to region (b). In a nonuniform conductor, (4) shows that there is an accumulation of un- paired charge. Indeed, with σa function of position, (2) becomes σ∇ ·E+E· ∇σ= 0 (6) Once the potential distribution has been found, Gauss’ law can be used to determine the distribution of unpaired charge density. ρu=/epsilon1∇ ·E+E· ∇/epsilon1 (7) Equation (6) can be solved for divEand that quantity substituted into (7) to obtain ρu=−/epsilon1 σE· ∇σ+E· ∇/epsilon1(8) Even though the distribution of /epsilon1plays no part in determining E, through Gauss’ law, it does inﬂuence the distribution of unpaired charge density. Continuity Conditions. Where the conductivity changes abruptly, the con- tinuity conditions follow from (2) and (3). The condition n·(σaEa−σbEb) = 0 (9) is derived from (2), just as (1.3.17) followed from Gauss’ law. The continuity con- ditions implied by (3) are familiar from Sec. 5.3. n×(Ea−Eb) = 0⇔Φa−Φb= 0 (10) Illustration. Boundary Condition at an Insulating Surface Insulated wires and ordinary resistors are examples where a conducting medium is bounded by one that is essentially insulating. What boundary condition should be used to determine the current distribution inside the conducting material? Sec. 7.2 Steady Ohmic Conduction 9 In Fig. 7.2.2, region (a) is relatively insulating compared to region (b), σa/lessmuch σb. It follows from (9) that the normal electric ﬁeld in region (a) is much greater than in region (b), Ea n/greatermuchEb n. According to (10), the tangential components of Eare equal, Ea t=Eb t. With the assumption that the normal and tangential components of Eare of the same order of magnitude in the insulating region, these two statements establish the relative magnitudes of the normal and tangential components of E, respectively, sketched in Fig. 7.2.2. We conclude that in the relatively conducting region (b), the normal component of Eis essentially zero compared to the tangential component. Thus, to determine the ﬁelds in the relatively conducting region, the boundary condition used at an insulating surface is n·J= 0⇒n· ∇Φ = 0 (11) At an insulating boundary, inside the conductor, the normal derivative of the potential is zero, while the boundary potential adjusts itself to make this true. Current lines are diverted so that they remain tangential to the insulating boundary, as sketched in Fig. 7.2.2. Just as Gauss’ law embodied in (8) is used to ﬁnd the unpaired volume charge density ex post facto , Gauss’ continuity condition (6.5.3) serves to evaluate the unpaired surface charge density. Combined with the current continuity condition, (9), it becomes σsu=n·/epsilon1aEaµ 1−/epsilon1b /epsilon1aσa σb¶ (12) Conductance. If there are only two electrodes contacting the conductor of Fig. 7.2.1 and hence one voltage v1=vand current i1=i, the voltage-current relation for the terminal pair is of the form i=Gv (13) where Gis the conductance. To relate Gto ﬁeld quantities, (2) is integrated over a volume Venclosed by a surface S, and Gauss’ theorem is used to convert the volume integral to one of the current σE·daover the surface S. This integral law is then applied to the surface shown in Fig. 7.2.1 enclosing the electrode that is connected to the positive terminal. Where it intersects the wire, the contribution is−i, so that the integral over the closed surface becomes −i+Z S1σE·da= 0 (14) where S1is the surface where the perfectly conducting electrode having potential v1interfaces with the Ohmic conductor. Division of (14) by the terminal voltage vgives an expression for the conduc- tance deﬁned by (13). 10 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.2.3 Typical conﬁgurations involving a conducting material and per- fectly conducting electrodes. (a) Region of interest is ﬁlled by material having uniform conductivity. (b) Region composed of diﬀerent materials, each having uniform conductivity. Conductivity is discontinuous at interfaces. (c) Conduc- tivity is smoothly varying. G=i v=R S1σE·da v (15) Note that the linearity of the equation governing the potential distribution, (4), assures that iis proportional to v. Hence, (15) is independent of vand, indeed, a parameter characterizing the system independent of the excitation. A comparison of (15) for the conductance with (6.5.6) for the capacitance suggests an analogy that will be developed in Sec. 7.5. Qualitative View of Fields in Conductors. Three classes of steady conduction conﬁgurations are typiﬁed in Fig. 7.2.3. In the ﬁrst, the region of interest is one of uniform conductivity bounded either by surfaces with constrained potentials or by perfect insulators. In the second, the conductivity varies abruptly but by a ﬁnite amount at interfaces, while in the third, it varies smoothly. Because Gauss’ law plays no role in determining the potential distribution, the permittivity distributions in these three classes of conﬁgurations are arbitrary. Of course, they do have a strong inﬂuence on the resulting distributions of unpaired charge density. A qualitative picture of the electric ﬁeld distribution within conductors emerges from arguments similar to those used in Sec. 6.5 for linear dielectrics. Because Jis solenoidal and has the same direction as E, it passes from the high-potential to the low-potential electrodes through tubes within which lines of Jneither terminate nor originate. The Elines form the same tubes but either terminate or originate on Sec. 7.2 Steady Ohmic Conduction 11 the sum of unpaired and polarization charges. The sum of these charge densities is div /epsilon1 oE, which can be determined from (6). ρu+ρp=∇ ·/epsilon1oE=−/epsilon1oE·∇σ σ=−/epsilon1oJ·∇σ σ2(16) At an abrupt discontinuity, the sum of the surface charges determines the discon- tinuity of normal E. In view of (9), σsu+σsp=n·(/epsilon1oEa−/epsilon1oEb) =n·/epsilon1oEa¡ 1−σa σb¢ (17) Note that the distribution of /epsilon1plays no part in shaping the Elines. In following a typical current tube from high potential to low in the uniform conductor of Fig. 7.2.3a, no conductivity gradients are encountered, so (16) tells us there is no source of E. Thus, it is no surprise that Φ satisﬁes Laplace’s equation throughout the uniform conductor. In following the current tube through the discontinuity of Fig. 7.2.3b, from low to high conductivity, (17) shows that there is a negative surface source of E. Thus, Etends to be excluded from the more conducting region and intensiﬁed in the less conducting region. With the conductivity increasing smoothly in the direction of E, as illustrated in Fig. 7.2.3c, E· ∇σis positive. Thus, the source of Eis negative and the Elines attenuate along the ﬂux tube. Uniform and piece-wise uniform conductors are commonly encountered, and examples in this category are taken up in Secs. 7.4 and 7.5. Examples where the conductivity is smoothly distributed are analogous to the smoothly varying permit- tivity conﬁgurations exempliﬁed in Sec. 6.7. In a simple one-dimensional conﬁgu- ration, the following example illustrates all three categories. Example 7.2.1. One-Dimensional Resistors The resistor shown in Fig. 7.2.4 has a uniform cross-section of area Ain any x−z plane. Over its length dit has a conductivity σ(y). Perfectly conducting electrodes constrain the potential to be vaty= 0 and to be zero at y=d. The cylindrical conductor is surrounded by a perfect insulator. The potential is assumed to depend only on y. Thus, the electric ﬁeld and cur- rent density are ydirected, and the condition that there be no component of Enor- mal to the insulating boundaries is automatically satisﬁed. For the one-dimensional ﬁeld, (4) reduces to d dy¡ σdΦ dy¢ = 0 (18) The quantity in parentheses, the negative of the current density, is conserved over the length of the resistor. Thus, with Jodeﬁned as constant, σdΦ dy=−Jo (19) This expression is now integrated from the lower electrode to an arbitrary location y.ZΦ vdΦ =−Zy 0Jo σdy⇒Φ =v−Zy 0Jo σdy (20) 12 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.2.4 Cylindrical resistor having conductivity that is a function of position ybetween the electrodes. The material surrounding the con- ductor is insulating. Evaluation of this expression where y=dand Φ = 0 relates the current density to the terminal voltage. v=Zd 0Jo σdy⇒Jo=v/Zd 0dy σ(21) Introduction of this expression into (20) then gives the potential distribution. Φ =v· 1−Zy 0dy σ/Zd 0dy σ¸ (22) The conductance, deﬁned by (15), follows from (21). G=AJo v=A/Zd 0dy σ(23) These relations hold for any one-dimensional distribution of σ. Of course, there is no dependence on /epsilon1, which could have any distribution. The permittivity could even depend on xandz. In terms of the circuit analogy suggested in the introduction, the resistors determine the distribution of voltages regardless of the interconnected capacitors. Three special cases conform to the three categories of conﬁgurations illustrated in Fig. 7.2.3. Uniform Conductivity. Ifσis uniform, evaluation of (22) and (23) gives Φ =v¡ 1−y d¢ (24) G=Aσ d(25) Sec. 7.2 Steady Ohmic Conduction 13 Fig. 7.2.5 Conductivity, potential, charge density, and ﬁeld distribu- tions in special cases for the conﬁguration of Fig. 7.2.4. (a) Uniform conductivity. (b) Layers of uniform but diﬀerent conductivities. (c) Ex- ponentially varying conductivity. The potential and electric ﬁeld are the same as they would be between plane parallel electrodes in free space in a uniform perfect dielectric. However, because of the insulating walls, the conduction ﬁeld remains uniform regardless of the length of the resistor compared to its transverse dimensions. It is clear from (16) that there is no volume charge density, and this is consis- tent with the uniform ﬁeld that has been found. These distributions of σ,Φ, and E are shown in Fig. 7.2.5a. Piece-Wise Uniform Conductivity. With the resistor composed of uni- formly conducting layers in series, as shown in Fig. 7.2.5b, the potential and con- ductance follow from (22) and (23) as Φ =8 >>< >>:v½ 1−G Ay σb¾ 0< y < b v½ 1−G A[(b/σb) + (y−b)/σa]¾ b < y < a +b(26) G=A [(b/σb) + (a/σa)](27) Again, there are no sources to distort the electric ﬁeld in the uniformly conducting regions. However, at the discontinuity in conductivity, (17) shows that there is sur- face charge. For σb> σa, this surface charge is positive, tending to account for the more intense ﬁeld shown in Fig. 7.2.5b in the upper region. Smoothly Varying Conductivity. With the exponential variation σ= σoexp(−y/d), (22) and (23) become Φ =v· 1−(ey/d−1) (e−1)¸ (28) 14 Conduction and Electroquasistatic Charge Relaxation Chapter 7 G=Aσo d(e−1)(29) Here the charge density that accounts for the distribution of Efollows from (16). ρu+ρp=/epsilon1oJo σodey/d(30) Thus, the ﬁeld is shielded from the lower region by an exponentially increasing volume charge density. 7.3 DISTRIBUTED CURRENT SOURCES AND ASSOCIATED FIELDS Under steady conditions, conservation of charge requires that the current density be solenoidal. Thus, Jlines do not originate or terminate. We have so far thought of current tubes as originating outside the region of interest, on the boundaries. It is sometimes convenient to introduce a volume distribution of current sources, s(r, t) A/m3, deﬁned so that the steady charge conservation equation becomes I SJ·da=Z Vsdv⇔ ∇ · J=s (1) The motivation for introducing a distributed source of current becomes clear as we now deﬁne singular sources and think about how these can be realized physically. Distributed Current Source Singularities. The analogy between (1) and Gauss’ law begs for the deﬁnition of point, line, and surface current sources, as depicted in Fig. 7.3.1. In returning to Sec. 1.3 where the analogous singular charge distributions were deﬁned, it should be kept in mind that we are now considering a source of current density, not of electric ﬂux. A point source of current gives rise to a net current ipout of a volume Vthat shrinks to zero while always enveloping the source. I SJ·da=ip ip≡lims→∞ V→0Z Vsdv (2) Such a source might be used to represent the current distribution around a small electrode introduced into a conducting material. As shown in Fig. 7.3.1d, the electrode is connected to a source of current ipthrough an insulated wire. At least under steady conditions, the wire and its insulation can be made ﬁne enough so that the current distribution in the surrounding conductor is not disturbed. Note that if the wire and its insulation are considered, the current density remains solenoidal. A surface surrounding the spherical electrode is pierced by the Sec. 7.3 Distributed Current Sources 15 Fig. 7.3.1 Singular current source distributions represented conceptually by the top row, suggesting how these might be realized physically by the bottom row by electrodes fed through insulated wires. wire. The contribution to the integral of J·dafrom this part of the surface integral is equal and opposite to that of the remainder of the surface surrounding the electrode. The point source is, in this case, an artiﬁce for ignoring the eﬀect of the insulated wire on the current distribution. The tubular volume having a cross-sectional area Aused to deﬁne a line charge density in Sec. 1.3 (Fig. 1.3.4) is equally applicable here to deﬁning a line current density. Kl≡lims→∞ A→0Z Asda (3) In general, Klis a function of position along the line, as shown in Fig. 7.3.1b. If this is the case, a physical realization would require a bundle of insulated wires, each terminated in an electrode segment delivering its current to the surrounding medium, as shown in Fig. 7.3.1e. Most often, the line source is used with two- dimensional ﬂows and describes a uniform wire electrode driven at one end by a current source. The surface current source of Figs. 7.3.1c and 7.3.1f is deﬁned using the same incremental control volume enclosing the surface source as shown in Fig. 1.3.5. Js≡lims→∞ h→0Zξ+h 2 ξ−h 2sdξ (4) Note that Jsis the netcurrent density entering the surrounding material at a given location. 16 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.3.2 For a small spherical electrode, the conductance relative to a large conductor at “inﬁnity” is given by (7). Fields Associated with Current Source Singularities. In the immediate vicinity of a point current source immersed in a uniform conductor, the current distribution is spherically symmetric. Thus, with J=σE, the integral current continuity law, (1), requires that 4πr2σEr=ip (5) From this, the electric ﬁeld intensity and potential of a point source follow as Er=ip 4πσr2⇒Φ =ip 4πσr(6) Example 7.3.1. Conductance of an Isolated Spherical Electrode A simple way to measure the conductivity of a liquid is based on using a small spherical electrode of radius a, as shown in Fig. 7.3.2. The electrode, connected to an insulated wire, is immersed in the liquid of uniform conductivity σ. The liquid is in a container with a second electrode having a large area compared to that of the sphere, and located many radii afrom the sphere. Thus, the potential drop associated with a current ithat passes from the spherical electrode to the large electrode is largely in the vicinity of the sphere. By deﬁnition the potential at the surface of the sphere is v, so evaluation of the potential for a point source, (6), at r=agives v=i 4πσa⇒G≡i v= 4πσa (7) This conductance is analogous to the capacitance of an isolated spherical electrode, as given by (4.6.8). Here, a ﬁne insulated wire connected to the sphere would have little eﬀect on the current distribution. The conductance associated with a contact on a conducting material is often approximated by picturing the contact as a hemispherical electrode, as shown in Fig. 7.3.3. The region above the surface is an insulator. Thus, there is no current density and hence no electric ﬁeld intensity normal to this surface. Note that this condition Sec. 7.4 Superposition and Uniqueness ofSteady Conduction Solutions 17 Fig. 7.3.3 Hemispherical electrode provides contact with inﬁnite half- space of material with conductance given by (8). is satisﬁed by the ﬁeld associated with a point source positioned on the conductor- insulator interface. An additional requirement is that the potential on the surface of the electrode be v. Because current is carried by only half of the spherical surface, it follows from reevaluation of (6a) that the conductance of the hemispherical surface contact is G= 2πσa (8) The ﬁelds associated with uniform line and surface sources are analogous to those discussed for line and surface charges in Sec. 1.3. The superposition principle, as discussed for Poisson’s equation in Sec. 4.3, is equally applicable here. Thus, the ﬁelds associated with higher-order source sin- gularities can again be found by superimposing those of the basic singular sources already deﬁned. Because it can be used to model a battery imbedded in a conductor, the dipole source is of particular importance. Example 7.3.2. Dipole Current Source in Spherical Coordinates A positive point current source of magnitude ipis located at z=d, just above a negative source (a sink) of equal magnitude at the origin. The source-sink pair, shown in Fig. 7.3.4, gives rise to ﬁelds analogous to those of Fig. 4.4.2. In the limit where the spacing dgoes to zero while the product of the source strength and this spacing remains ﬁnite, this pair of sources forms a dipole. Starting with the potential as given for a source at the origin by (6), the limiting process is the same as leading to (4.4.8). The charge dipole moment qdis replaced by the current dipole moment ipdand/epsilon1o→σ, qd→ipd. Thus, the potential of the dipole current source is Φ =ipd 4πσcosθ r2(9) The potential of a polar dipole current source is found in Prob. 7.3.3. Method of Images. With the new boundary conditions describing steady current distributions come additional opportunities to exploit symmetry, as dis- cussed in Sec. 4.7. Figure 7.3.5 shows a pair of equal magnitude point current sources located at equal distances to the right and left of a planar surface. By con- trast with the point charges of Fig. 4.7.1, these sources are of the same sign. Thus, 18 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.3.4 Three-dimensional dipole current source has potential given by (9). Fig. 7.3.5 Point current source and its image representing an insulating boundary. the electric ﬁeld normal to the surface is zero rather than the tangential ﬁeld. The ﬁeld and current distribution in the right half is the same as if that region were ﬁlled by a uniform conductor and bounded by an insulator on its left. 7.4 SUPERPOSITION AND UNIQUENESS OF STEADY CONDUCTION SOLUTIONS The physical laws and boundary conditions are diﬀerent, but the approach in this section is similar to that of Secs. 5.1 and 5.2 treating Poisson’s equation. In a material having the conductivity distribution σ(r) and source distribution s(r), a steady potential distribution Φ must satisfy (7.2.4) with a source density −son the right. Typically, the conﬁgurations of interest are as in Fig. 7.2.1, except that we now include the possibility of a distribution of current source density in the volume V. Electrodes are used to constrain this potential over some of the surface enclosing the volume Voccupied by this material. This part of the surface, where the material contacts the electrodes, will be called S/prime. We will assume here that on the remainder of the enclosing surface, denoted by S/prime/prime, the normal current density is speciﬁed. Depicted in Fig. 7.2.1 is the special case where the boundary S/prime/primeis insulating and hence where the normal current density is zero. Thus, according to Sec. 7.4 Superposition and Uniqueness 19 (7.2.1), (7.2.3), and (7.3.1), the desired EandJare found from a solution Φ to ∇ ·σ∇Φ =−s (1) where Φ = Φ ionS/prime i −n·σ∇Φ =JionS/prime/prime i Except for the possibility that part of the boundary is a surface S/prime/primewhere the normal current density rather than the potential is speciﬁed, the situation here is analogous to that in Sec. 5.1. The solution can be divided into a particular part [that satisﬁes the diﬀerential equation of (1) at each point in the volume, but not the boundary conditions] and a homogeneous part. The latter is then adjusted to make the sum of the two satisfy the boundary conditions. Superposition to Satisfy Boundary Conditions. Suppose that a system is composed of a source-free conductor ( s= 0) contacted by one reference electrode at ground potential and nelectrodes, respectively, at the potentials vj, j= 1, . . . n . The contacting surfaces of these electrodes comprise the surface S/prime. As shown in Fig. 7.2.1, there may be other parts of the surface enclosing the material that are insulating ( Ji= 0) and denoted by S/prime/prime. The solution can be represented as the sum of the potential distributions associated with each of the electrodes of speciﬁed potential while the others are grounded. Φ =nX j=1Φj (2) where ∇ ·σ∇Φj= 0 Φj=½ vjonS/prime i, j =i 0 on S/prime i, j/negationslash=i Each Φ jsatisﬁes (1) with s= 0 and the boundary condition on S/prime/prime iwith Ji= 0. This decomposition of the solution is familiar from Sec. 5.1. However, the boundary condition on the insulating surface S/prime/primerequires a somewhat broadened view of what is meant by the respective terms in (2). As the following example illustrates, modes that have zero derivatives rather than zero amplitude at boundaries are now useful for satisfying the insulating boundary condition. Example 7.4.1. Modal Solution with an Insulating Boundary In the two-dimensional conﬁguration of Fig. 7.4.1, a uniformly conducting material is grounded along its left edge, bounded by insulating material along its right edge, 20 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.4.1 (a) Two terminal pairs attached to conducting material having one wall at zero potential and another that is insulating. (b) Field solution is broken into part due to potential v1and (c) potential v2. (d) The boundary condition at the insulating wall is satisﬁed by using the symmetry of an equivalent problem with all of the walls constrained in potential. and driven by electrodes having the potentials v1andv2at the top and bottom, respectively. Decomposition of the potential, as called for by (2), amounts to the superpo- sition of the potentials for the two problems of (b) and (c) in the ﬁgure. Note that for each of these, the normal derivative of the potential must be zero at the right boundary. Pictured in part (d) of Fig. 7.4.1 is a conﬁguration familiar from Sec. 5.5. The potential distribution for the conﬁguration of Fig. 5.5.2, (5.5.9), is equally applicable to that of Fig. 7.4.1. This is so because the symmetry requires that there be no x- directed electric ﬁeld along the surface x=a/2. In turn, the potential distribution for part (c) is readily determined from this one by replacing v1→v2andy→b−y. Thus, the total potential is Φ =∞X n=1 odd4 π½ v1 nsinh¡nπ ay¢ sinh¡nπ ab¢sinnπ ax +v2 nsinh£nπ a(b−y)¤ sinh¡nπb a¢sinnπ ax¾(3) If we were to solve this problem without reference to Sec. 5.5, the modes used to expand the electrode potential would be zero at x= 0 and have zero derivative at the insulating boundary (at x=a/2). Sec. 7.5 Piece-Wise Uniform Conductors 21 The Conductance Matrix. With S/prime ideﬁned as the surface over which the i-th electrode contacts the conducting material, the current emerging from that electrode is ii=Z Siσ∇Φ·da (4) [See Fig. 7.2.1 for deﬁnition of direction of da.] In terms of the potential decompo- sition represented by (2), this expression becomes ii=nX j=1Z S/prime iσ∇Φj·da=nX j=1Gijvj (5) where the conductances are Gij=R S/prime iσ∇Φj·da vj (6) Because Φ jis by deﬁnition proportional to vj, these parameters are independent of the excitations. They depend only on the physical properties and geometry of the conﬁguration. Example 7.4.2. Two Terminal Pair Conductance Matrix For the system of Fig. 7.4.1, (5) becomes hi1 i2i =hG11G12 G21G22ihv1 v2i (7) With the potential given by (3), the self-conductances G11andG22and the mutual conductances G12andG21follow by evaluation of (5). This potential is singular in the left-hand corners, so the self-conductances determined in this way are represented by a series that does not converge. However, the mutual conductances are determined by integrating the current density over an electrode that is at the same potential as the grounded wall, so they are well represented. For example, with cdeﬁned as the length of the conducting block in the zdirection, G12=σc v2Za/2 0∂Φ2 ∂y¯¯¯ y=bdx=4 πσc∞X n=1 odd1 nsinh¡nπb a¢ (8) Uniqueness. With Φ i, Ji, σ(r), and s(r) given, a steady current distribution is uniquely speciﬁed by the diﬀerential equation and boundary conditions of (1). As in Sec. 5.2, a proof that a second solution must be the same as the ﬁrst hinges on deﬁning a diﬀerence potential Φ d= Φ a−Φband showing that, because Φ d= 0 on S/prime iandn·σ∇Φd= 0 on S/prime/prime iin Fig. 7.2.1, Φ dmust be zero. 22 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.5.1 Conducting circular rod is immersed in a conducting mate- rial supporting a current density that would be uniform in the absence of the rod. 7.5 STEADY CURRENTS IN PIECE-WISE UNIFORM CONDUCTORS Conductor conﬁgurations are often made up from materials that are uniformly conducting. The conductivity is then uniform in the subregions occupied by the diﬀerent materials but undergoes step discontinuities at interfaces between regions. In the uniformly conducting regions, the potential obeys Laplace’s equation, (7.2.5), ∇2Φ = 0 (1) while at the interfaces between regions, the continuity conditions require that the normal current density and tangential electric ﬁeld intensity be continuous, (7.2.9) and (7.2.10). n·(σaEa−σbEb) = 0 (2) Φa−Φb= 0 (3) Analogy to Fields in Linear Dielectrics. If the conductivity is replaced by the permittivity, these laws are identical to those underlying the examples of Sec. 6.6. The role played by Dis now taken by J. Thus, the analysis for the following example has already been carried out in Sec. 6.6. Example 7.5.1. Conducting Circular Rod in Uniform Transverse Field A rod of radius Rand conductivity σbis immersed in a material of conductivity σa, as shown in Fig. 7.5.1. Perhaps imposed by means of plane parallel electrodes far to the right and left, there is a uniform current density far from the cylinder. The potential distribution is deduced using the same steps as in Example 6.6.2, with /epsilon1a→σaand/epsilon1b→σb. Thus, it follows from (6.6.21) and (6.6.22) as Φa=−REocosφ·¡r R¢ −¡R r¢(σb−σa) (σb+σa)¸ (4) Φb=−2σa σa+σbEorcosφ (5) and the lines of electric ﬁeld intensity are as shown in Fig. 6.6.6. Note that although the lines of EandJare in the same direction and have the same pattern in each of the Sec. 7.5 Piece-Wise Uniform Conductors 23 Fig. 7.5.2 Distribution of current density in and around the rod of Fig. 7.5.1. (a) σb≥σa. (b) σa≥σb. regions, they have very diﬀerent behaviors where the conductivity is discontinuous. In fact, the normal component of the current density is continuous at the interface, and the spacing between lines of Jmust be preserved across the interface. Thus, in the distribution of current density shown in Fig. 7.5.2, the lines are continuous. Note that the current tends to concentrate on the rod if it is more conducting, but is diverted around the rod if it is more insulating. A surface charge density resides at the interface between the conducting media of diﬀerent conductivities. This surface charge density acts as the source of Eon the cylindrical surface and is identiﬁed by (7.2.17). Inside-Outside Approximations. In exploiting the formal analogy between ﬁelds in linear dielectrics and in Ohmic conductors, it is important to keep in mind the very diﬀerent physical phenomena being described. For example, there is no conduction analog to the free space permittivity /epsilon1o. There is no minimum value of the conductivity, and although /epsilon1can vary between a minimum of /epsilon1oin free space and 1000 /epsilon1oor more in special solids, the electrical conductivity is even more widely varying. The ratio of the conductivity of a copper wire to that of its insulation exceeds 1021. Because some materials are very good conductors while others are very good insulators, steady conduction problems can exemplify the determination of ﬁelds for large ratios of physical parameters. In Sec. 6.6, we examined ﬁeld distributions in cases where the ratios of permittivities were very large or very small. The “inside- outside” viewpoint is applicable not only to approximating ﬁelds in dielectrics but to ﬁnding the ﬁelds in the transient EQS systems in the latter part of this chapter and in MQS systems with magnetization and conduction. Before attempting a more general approach, consider the following example, where the ﬁelds in and around a resistor are described. Example 7.5.2. Fields in and around a Conductor The circular cylindrical conductor of Fig. 7.5.3, having radius band length L, is surrounded by a perfectly conducting circular cylindrical “can” having inside 24 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.5.3 Circular cylindrical conductor surroun- ded by coaxial perfectly conducting “can” that is connected to the right end by a perfectly conducting “short” in the plane z= 0. The left end is at potential vrelative to right end and surrounding wall and is connected to that wall at z=−Lby a washer-shaped resistive material. Fig. 7.5.4 Distribution of potential and electric ﬁeld intensity for the conﬁguration of Fig. 7.5.3. radius a. With respect to the surrounding perfectly conducting shield, a dc voltage source applies a voltage vto the perfectly conducting disk. A washer-shaped material of thickness δand also having conductivity σis connected between the perfectly conducting disk and the outer can. What are the distributions of Φ and Ein the conductors and in the annular free space region? Note that the ﬁelds within each of the conductors are fully speciﬁed without regard for the shape of the can. The surfaces of the circular cylindrical conductor are either constrained in potential or bounded by free space. On the latter, the normal component of J, and hence of E, is zero. Thus, in the language of Sec. 7.4, the potential is constrained on S/primewhile the normal derivative of Φ is constrained on the insulating surfaces S/prime/prime. For the center conductor, S/primeis at z= 0 and z=−Lwhile S/prime/primeis at r=b. For the washer-shaped conductor, S/primeis at r=bandr=aandS/prime/prime is at z=−Landz=−(L+δ). The theorem of Sec. 7.4 shows that the potential inside each of the conductors is uniquely speciﬁed. Note that this is true regardless of the arrangement outside the conductors. In the cylindrical conductor, the solution for the potential that satisﬁes Laplace’s equation and all these boundary conditions is simply a linear function of z. Φb=−v Lz (6) Thus, the electric ﬁeld intensity is uniform and zdirected. Eb=v Liz (7) These equipotentials and Elines are sketched in Fig. 7.5.4. By way of reinforcing what is new about the insulating surface boundary condition, note that (6) and (7) apply to the cylindrical conductor regardless of its cross-section geometry and its length. However, the longer it is, the more stringent is the requirement that the annular region be insulating compared to the central region. Sec. 7.5 Piece-Wise Uniform Conductors 25 In the washer-shaped conductor, the axial symmetry requires that the poten- tial not depend on z. If it depends only on the radius, the boundary conditions on the insulating surfaces are automatically satsﬁed. Two solutions to Laplace’s equa- tion are required to meet the potential constraints at r=aandr=b. Thus, the solution is assumed to be of the form Φc=Alnr +B (8) The coeﬃcients AandBare determined from the radial boundary conditions, and it follows that the potential within the washer-shaped conductor is Φc=vln¡r a¢ ln¡b a¢ (9) The “inside” ﬁelds can now be used to determine those in the insulating annular “outside” region. The potential is determined on all of the surface surrounding this region. In addition to being zero on the surfaces r=aandz= 0, the potential is given by (6) at r=band by (9) at z=−L. So, in turn, the potential in this annular region is uniquely determined. This is one of the few problems in this book where solutions to Laplace’s equation that have both an rand a zdependence are considered. Because there is noφdependence, Laplace’s equation requires that µ ∂2 ∂z2+1 r∂ ∂rr∂ ∂r¶ Φ = 0 (10) The linear dependence on zof the potential at r=bsuggests that solutions to Laplace’s equation take the product form R(r)z. Substitution into (10) then shows that the rdependence is the same as given by (9). With the coeﬃcients adjusted to make the potential Φ a(a,−L) = 0 and Φ a(b,−L) =v, it follows that in the outside insulating region Φa=v ln¡a b¢ln¡r a¢z L(11) To sketch this potential and the associated Elines in Fig. 7.5.4, observe that the equipotentials join points of the given potential on the central conductor with those of the same potential on the washer-shaped conductor. Of course, the zero potential surface is at r=aand at z= 0. The lines of electric ﬁeld intensity that originate on the surfaces of the conductors are perpendicular to these equipoten- tials and have tangential components that match those of the inside ﬁelds. Thus, at the surfaces of the ﬁnite conductors, the electric ﬁeld in region (a) is neither perpendicular nor tangential to the boundary. For a positive potential v, it is clear that there must be positive surface charge on the surfaces of the conductors bounding the annular insulating region. Remember that the normal component of Eon the conductor sides of these surfaces is zero. Thus, there is a surface charge that is proportional to the normal component of E on the insulating side of the surfaces. σs(r=b) =/epsilon1oEa r(r=b) =−/epsilon1ov b ln(a/b)z L(12) The order in which we have determined the ﬁelds makes it clear that this surface charge is the one required to accommodate the ﬁeld conﬁguration outside 26 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.5.5 Demonstration of the absence of volume charge density and existence of a surface charge density for a uniform conductor. (a) A slightly conducting oil is contained by a box constructed from a pair of electrodes to the left and right and with insulating walls on the other two sides and the bottom. The top surface of the conducting oil is free to move. The resulting surface force density sets up a circulating motion of the liquid, as shown. (b) With an insulating sheet resting on the interface, the circulating motion is absent. the conducting regions. A change in the shield geometry changes Φabut does not alter the current distribution within the conductors. In terms of the circuit analogy used in Sec. 7.0, the potential distributions have been completely determined by the rod-shaped and washer-shaped resistors. The charge distribution is then determined ex post facto by the “distributed capacitors” surrounding the resistors. The following demonstration shows that the unpaired charge density is zero in the volume of a uniformly conducting material and that charges do indeed tend to accumulate at discontinuities of conductivity. Demonstration 7.5.1. Distribution of Unpaired Charge A box is constructed so that two of its sides and its bottom are plexiglas, the top is open, and the sides shown to left and right in Fig. 7.5.5 are highly conducting. It is ﬁlled with corn oil so that the region between the vertical electrodes in Fig. 7.5.5 is semi-insulating. The region above the free surface is air and insulating compared to the corn oil. Thus, the corn oil plays a role analogous to that of the cylindrical rod in Example 7.5.2. Consistent with its insulating transverse boundaries and the potential constraints to left and right is an “inside” electric ﬁeld that is uniform. The electric ﬁeld in the outside region (a) determines the distribution of charge on the interface. Since we have determined that the inside ﬁeld is uniform, the potential of the interface varies linearly from vat the right electrode to zero at the left electrode. Thus, the equipotentials are evenly spaced along the interface. The equipotentials in the outside region (a) are planes joining the inside equipotentials and extending to inﬁnity, parallel to the canted electrodes. Note that this ﬁeld satisﬁes the boundary conditions on the slanted electrodes and matches the potential on the liquid interface. The electric ﬁeld intensity is uniform, originating on the upper electrode and terminating either on the interface or on the lower slanted electrode. Because both the spacing and the potential diﬀerence vary linearly with horizontal distance, the negative surface charge induced on the interface is uniform. Sec. 7.5 Piece-Wise Uniform Conductors 27 Wherever there is an unpaired charge density, the corn oil is subject to an electrical force. There is unpaired charge in the immediate vicinity of the interface in the form of a surface charge, but not in the volume of the conductor. Consistent with this prediction is the observation that with the application of about 20 kV to electrodes having 20 cm spacing, the liquid is set into a circulating motion. The liquid moves rapidly to the right at the interface and recirculates in the region below. Note that the force at the interface is indeed to the right because it is proportional to the product of a negative charge and a negative electric ﬁeld intensity. The ﬂuid moves as though each part of the interface is being pulled to the right. But how can we be sure that the circulation is not due to forces on unpaired charges in the ﬂuid volume? An alteration to the same experiment answers this question. With a plexiglas sheet placed on the interface, it is mechanically pinned down. That is, the electrical force acting on the unpaired charges in the immediate vicinity of the interface is countered by viscous forces tending to prevent the ﬂuid from moving tangential to the solid boundary. Yet because the sheet is insulating, the ﬁeld distribution within the conductor is presumably unaltered from what it was before. With the plexiglas sheet in place, the circulations of the ﬁrst experiment are no longer observed. This is consistent with a model that represents the corn-oil as a uniform Ohmic conductor1. (For a mathematical analysis, see Prob. 7.5.3.) In general, there is a two-way coupling between the ﬁelds in adjacent uniformly conducting regions. If the ratio of conductivities is either very large or very small, it is possible to calculate the ﬁelds in an “inside” region ignoring the eﬀect of “outside” regions, and then to ﬁnd the ﬁelds in the “outside” region. The region in which the ﬁeld is ﬁrst found, the “inside” region, is usually the one to which the excitation is applied, as illustrated in Example 7.5.2. This will be further illustrated in the following example, which pursues an approximate treatment of Example 7.5.1. The exact solutions found there can then be compared to the approximate ones. Example 7.5.3. Approximate Current Distribution around Relatively Insulating and Conducting Rods Consider ﬁrst the ﬁeld distribution around and then in a circular rod that has a small conductivity relative to its surroundings. Thus, in Fig. 7.5.1, σa/greatermuchσb. Electrodes far to the left and right are used to apply a uniform ﬁeld and current density to region (a). It is therefore in this inside region outside the cylinder that the ﬁelds are ﬁrst approximated. With the rod relatively insulating, it imposes on region (a) the approximate boundary condition that the normal current density, and hence the radial derivative of the potential, be zero at the rod surface, where r=R. n·Ja≈0⇒∂Φa ∂r≈0 at r=R (13) Given that the ﬁeld at inﬁnity must be uniform, the potential distribution in region (a) is now uniquely speciﬁed. A solution to Laplace’s equation that satisﬁes this condition at inﬁnity and includes an arbitrary coeﬃcient for hopefully satisfying the 1See ﬁlm Electric Fields and Moving Media , produced by the National Committee for Electri- cal Engineering Films and distributed by Education Development Center, 39 Chapel St., Newton, Mass. 02160. 28 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.5.6 Distributions of electric ﬁeld intensity around conducting rod immersed in conducting medium: (a) σa/greatermuchσb; (b) σb/greatermuchσa. Com- pare these to distributions of current density shown in Fig. 7.5.2. ﬁrst condition is Φa=−Eorcosφ+Acosφ r(14) With Aadjusted to satisfy (13), the approximate potential in region (a) is Φa=−Eo¡ r+R2 r¢ cosφ (15) This is the potential in the exterior region, implying the ﬁeld lines shown in Fig. 7.5.6a. Now that we have obtained the approximate potential at r=R,Φb= −2EoRcos(φ), we can in turn approximate the potential in region (b). Φb=Brcosφ=−2Eorcosφ (16) The ﬁeld lines associated with this potential are also shown in Fig. 7.5.6a. Note that if we take the limits of (4) and (5) where σa/σb/greatermuch1, we obtain these potentials. Contrast these steps with those that are appropriate in the opposite extreme, where σa/σb/lessmuch1. There the rod tends to behave as an equipotential and the bound- ary condition at r=Ris Φa= constant = 0. This condition is now used to evaluate the coeﬃcient Ain (14) to obtain Φa=−Eo¡ r−R2 r¢ cosφ (17) This potential implies that there is a current density at the rod surface given by Ja r(r=R) =−σa∂Φa ∂r(r=R) = 2 σaEocosφ (18) The normal current density at the inside surface of the rod must be the same, so the coeﬃcient Bin (16) can be evaluated. Φb=−2σa σbEorcosφ (19) Sec. 7.5 Piece-Wise Uniform Conductors 29 Fig. 7.5.7 Rotor of insulating material is immersed in somewhat con- ducting corn oil. Plane parallel electrodes are used to impose constant electric ﬁeld, so from the top, the distribution of electric ﬁeld should be that of Fig. 7.5.6a, at least until the rotor begins to rotate spontaneously in either direction. Now the ﬁeld lines are as shown in Fig. 7.5.6b. Again, the approximate potential distributions given by (17) and (19), respec- tively, are consistent with what is obtained from the exact solutions, (4) and (5), in the limit σa/σb/lessmuch1. In the following demonstration, a surprising electromechanical response has its origins in the charge distribution implied by the potential distributions found in Example 7.5.3. Demonstration 7.5.2. Rotation of an Insulating Rod in a Steady Current In the apparatus shown in Fig. 7.5.7, a teﬂon rod is mounted at its ends on bearings so that it is free to rotate. It, and a pair of plane parallel electrodes, are immersed in corn oil. Thus, from the top, the conﬁguration is as shown in Fig. 7.5.1. The applied ﬁeldEo=v/d, where vis the voltage applied between the electrodes and dis their spacing. In the experiment, R= 1.27 cm , d= 11.8 cm, and the applied voltage is 10–20 kV. As the voltage is raised, there is a threshold at which the rod begins to rotate. With the voltage held ﬁxed at a level above the threshold, the ensuing rotation is continuous and in either direction. [See footnote 1.] To explain this “motor,” note that even though the corn oil used in the ex- periment has a conductivity of σa= 5×10−11S/m, that is still much greater than the conductivity σbof the rod. Thus, the potential around and in the rod is given by (15) and (16) and the Eﬁeld distribution is as shown in Fig. 7.5.6a. Also shown in this ﬁgure is the distribution of unpaired surface charge, which can be evaluated using (16). σs(r=R) =n·(/epsilon1aEa r−/epsilon1bEb r) =/epsilon1b∂Φb ∂r(r=R) =−2/epsilon1bEocosφ (20) Positive charges on the left electrode induce charges of the same sign on the nearer side of the rod, as do the negative charges on the electrode to the right. Thus, when static, the rod is in a posture analogous to that of a compass needle oriented 30 Conduction and Electroquasistatic Charge Relaxation Chapter 7 backwards in a magnetic ﬁeld. Its static state is unstable and it attempts to reorient itself in the ﬁeld. The continuous rotation results because once it begins to rotate, additional ﬁelds are generated that allow the charge to leak oﬀ the cylinder through currents in the surrounding oil. Note that if the rod were much more conducting than its surroundings, charges on the electrodes would induce charges of opposite sign on the nearer surfaces of the rod. This more familiar situation is the one shown in Fig. 7.5.6b. The condition requiring that there be no normal current density at an insu- lating boundary can have a dramatic eﬀect on fringing ﬁelds. This has already been illustrated by Example 7.5.2, where the ﬁeld was uniform in the central conductor no matter what its length relative to its radius. Whenever we take the resistance of a wire having length L, cross-sectional area A, and conductivity σas being L/σA , we exploit this boundary condition. The conduction analogue of Example 6.6.3 gives a further illustration of how an insulating boundary ducts the electric ﬁeld intensity. With /epsilon1a→σaand/epsilon1b→σb, the conﬁguration of Fig. 6.6.8 becomes the edge of a plane parallel resistor ﬁlled out to the edge of the electrodes by a material having conductivity σb. The fringing ﬁeld then depends on the conductivity σaof the surrounding material. The fringing ﬁeld that would result if the entire region were ﬁlled by a ma- terial having a uniform conductivity is shown in Fig. 6.6.9a. By contrast, the ﬁeld distribution with the conducting material extending only to the edge of the elec- trode is shown in Fig. 6.6.9b. The ﬁeld inside is exactly uniform and independent of the geometry of what is outside. Of course, there is always a fringing ﬁeld outside that does depend on the outside geometry. But because there is little associated current density, the resistance is unaﬀected by this part of the ﬁeld. 7.6 CONDUCTION ANALOGS The potential distribution for steady conduction is determined by solving (7.4.1) ∇ ·σ∇Φc=−s (1) in a volume Vhaving conductivity σ(r) and current source distribution s(r), re- spectively. On the other hand, if the volume is ﬁlled by a perfect dielectric having permit- tivity /epsilon1(r) and unpaired charge density distribution ρu(r), respectively, the potential distribution is determined by the combination of (6.5.1) and (6.5.2). ∇ ·/epsilon1∇Φe=−ρu (2) It is clear that solutions pertaining to one of these physical situations are solutions for the other, provided that the boundary conditions are also analogous. We have been exploiting this analogy in Sec. 7.5 for piece-wise continuous systems. There, solutions for the ﬁelds in dielectrics were applied to conduction problems. Of course, measurements made on dielectrics can also be used to predict steady conduction phemonena. Sec. 7.6 Conduction Analogs 31 Conversely, ﬁelds found either theoretically or by experimentation in a steady conduction situation can be used to describe those in perfect dielectrics. When measurements are used, the latter procedure is a particularly useful one, because conduction processes are conveniently simulated and comparatively easy to mea- sure. It is more diﬃcult to measure the potential in free space than in a conductor, and to measure a capacitance than a resistance. Formally, a quantitative analogy is established by introducing the constant ratios for the magnitudes of the properties, sources, and potentials, respectively, in the two systems throughout the volumes and on the boundaries. With k1andk2 deﬁned as scaling constants, /epsilon1 σ=k1,Φc Φe=k2,k2 k1=s ρu(3) substitution of the conduction variables into (2) converts it into (1). The boundary conditions on surfaces S/primewhere the potential is constrained are analogous, provided the boundary potentials also have the constant ratio k2given by (3). Most often, interest is in systems where there are no volume source distribu- tions. Thus, suppose that the capacitance of a pair of electrodes is to be determined by measuring the conductance of analogously shaped electrodes immersed in a con- ducting material. The ratio of the measured capacitance to conductance, the ratio of (6.5.6) to (7.2.15), follows from substituting /epsilon1=k1σ, (3a), C G=R S1/epsilon1E·da/vR S1σE·da/v=k1R S1σE·da/vR S1σE·da/v=k1=/epsilon1 σ(4) In multiple terminal pair systems, the capacitance matrix deﬁned by (5.1.12) and (5.1.13) is similarly deduced from measurement of a conductance matrix, deﬁned in (7.4.6). Demonstration 7.6.1. Electrolyte-Tank Measurements If great accuracy is required, ﬁelds in complex geometries are most easily determined numerically. However, especially if the capacitance is sought– and not a detailed ﬁeld mapping– a conduction analog can prove convenient. A simple experiment to determine the capacitance of a pair of electrodes is shown in Fig. 7.6.1, where they are mounted on insulated rods, contacted through insulated wires, and immersed in tap water. To avoid electrolysis, where the conductors contact the water, low-frequency ac is used. Care should be taken to insure that boundary conditions imposed by the tank wall are either analogous or inconsequential. Often, to motivate or justify approximations used in analytical modeling of complex systems, it is helpful to probe the potential distribution using such an experiment. The probe consists of a small metal tip, mounted and wired like the electrodes, but connected to a divider. By setting the probe potential to the desired rms value, it is possible to trace out equipotential surfaces by moving the probe in such a way as to keep the probe current nulled. Commercial equipment is automated with a feedback system to perform such measurements with great precision. However, given the alternative of numerical simulation, it is more likely that such approaches are appropriate in establishing rough approximations. 32 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.6.1 Electrolytic conduction analog tank for determining poten- tial distributions in complex conﬁgurations. Fig. 7.6.2 In two dimensions, equipotential and ﬁeld lines predicted by Laplace’s equation form a grid of curvilinear squares. Mapping Fields that Satisfy Laplace’s Equation. Laplace’s equation deter- mines the potential distribution in a volume ﬁlled with a material of uniform con- ductivity that is source free. Especially for two-dimensional ﬁelds, the conduction analog then also gives the opportunity to reﬁne the art of sketching the equipoten- tials of solutions to Laplace’s equation and the associated ﬁeld lines. Before considering how a sheet of conducting paper provides the medium for determining two-dimensional ﬁelds, it is worthwhile to identify the properties of a ﬁeld sketch that indeed represents a two-dimensional solution to Laplace’s equation. A review of the many two-dimensional plots of equipotentials and ﬁelds given in Chaps. 4 and 5 shows that they form a grid of curvilinear rectangles. In terms of variables deﬁned for the ﬁeld sketch of Fig. 7.6.2, where the distance between equipotentials is denoted by ∆ nand the distance between Elines is ∆ s, the ratio ∆n/∆stends to be constant, as we shall now show. Sec. 7.6 Conduction Analogs 33 The condition that the ﬁeld be irrotational gives E=−∇Φ⇒ |E| ≈|∆Φ| |∆n|(5) while the steady charge conservation law implies that along a ﬂux tube, ∇ ·σE= 0⇒σ|E|∆s= constant ≡∆K (6) Thus, along a ﬂux tube, σ∆Φ ∆n∆s= ∆K⇒∆s ∆n=∆K σ∆Φ= constant (7) If each of the ﬂux tubes carries the same current, and if the equipotential lines are drawn for equal increments of ∆Φ, then the ratio ∆ s/∆nmust be constant throughout the mapping. The sides of the curvilinear rectangles are commonly made equal, so that the equipotentials and ﬁeld lines form a grid of curvilinear squares. The faithfulness to Laplace’s equation of a map of equipotentials at equal increments in potential can be checked by sketching in the perpendicular ﬁeld lines. With the ﬁeld lines forming curvilinear squares in the starting region, a correct distribution of the equipotentials is achieved when a grid of squares is maintained throughout the region. With some practice, it is possible to iterate between re- ﬁnements of the equipotentials and the ﬁeld lines until a satisfactory map of the solution is sketched. Demonstration 7.6.2. Two-Dimensional Solution to Laplace’s Equation by Means of Teledeltos Paper For the mapping of two-dimensional ﬁelds, the conduction analog has the advantage that it is not necessary to make the electrodes and conductor “inﬁnitely” long in the third dimension. Two-dimensional current distributions will result even in a thin- sheet conductor, provided that it has a conductivity that is large compared to its surroundings. Here again we exploit the boundary condition applying to the surfaces of the paper. As far as the ﬁelds inside the paper are concerned, a two-dimensional current distribution automatically meets the requirement that there be no current density normal to those parts of the paper bounded by air. A typical ﬁeld mapping apparatus is as simple as that shown in Fig. 7.6.3. The paper has the thickness ∆ and a conductivity σ. The electrodes take the form of silver paint or copper tape put on the upper surface of the paper, with a shape simulating the electrodes of the actual system. Because the paper is so thin compared to dimensions of interest in the plane of the paper surface, the currents from the electrodes quickly assume an essentially uniform proﬁle over the cross-section of the paper, much as suggested by the inset to Fig. 7.6.3. In using the paper, it is usual to deal in terms of a surface resistance 1 /∆σ. The conductance of the plane parallel electrode system shown in Fig. 7.6.4 can be used to establish this parameter. i v=w∆σ S≡Gp⇒∆σ=GpS w(8) 34 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.6.3 Conducting paper with attached electrodes can be used to determine two-dimensional potential distributions. Fig. 7.6.4 Apparatus for determining surface conductivity ∆ σof pa- per used in experiment shown in Fig. 7.6.3. The units are simply ohms, and 1 /∆σis the resistance of a square of the material having any sidelength. Thus, the units are commonly denoted as “ohms/square.” To associate a conductance as measured at the terminals of the experiment shown in Fig. 7.6.3 with the capacitance of a pair of electrodes having length lin the third dimension, note that the surface integrations used to deﬁne CandGreduce to C=l vI C/epsilon1E·ds; G=∆ vI CσE·ds (9) where the surface integrals have been reduced to line integrals by carrying out the integration in the third dimension. The ratio of these quantities follows in terms of the surface conductance ∆ σas C G=lk1 ∆=l/epsilon1 ∆σ(10) Here Gis the conductance as actually measured using the conducting paper, and C is the capacitance of the two-dimensional capacitor it simulates. In Chap. 9, we will ﬁnd that magnetic ﬁeld distributions as well can often be found by using the conduction analog. Sec. 7.7 Charge Relaxation 35 TABLE 7.7.1 CHARGE RELAXATION TIMES OF TYPICAL MATERIALS σ−S/m /epsilon1//epsilon1o τe−s Copper 5.8×1071 1.5×10−19 Water, distilled 2×10−481 3.6×10−6 Corn oil 5×10−113.1 0.55 Mica 10−11−10−155.8 5.1−5.1×104 7.7 CHARGE RELAXATION IN UNIFORM CONDUCTORS In a region that has uniform conductivity and permittivity, charge conservation and Gauss’ law determine the unpaired charge density throughout the volume of the material, without regard for the boundary conditions. To see this, Ohm’s law (7.1.7) is substituted for the current density in the charge conservation law, (7.0.3), ∇ ·σE+∂ρu ∂t= 0 (1) and Gauss’ law (6.2.15) is written using the linear polarization constitutive law, (6.4.3). ∇ ·/epsilon1E=ρu (2) In a region where σand/epsilon1are uniform, these parameters can be pulled outside the divergence operators in these equations. Substitution of divEfound from (2) into (1) then gives the charge relaxation equation for ρu. ∂ρu ∂t+ρu τe= 0; τe≡/epsilon1 σ (3) Note that it has not been assumed that Eis irrotational, so the unpaired charge obeys this equation whether the ﬁelds are EQS or not . The solution to (3) takes on the same appearance as if it were an ordinary diﬀerential equation, say predicting the voltage of an RC circuit. ρu=ρi(x, y, z )e−t/τe(4) However, (3) is a partial diﬀerential equation, and so the coeﬃcient of the exponen- tial in (4) is an arbitrary function of the spatial coordinates. The relaxation time τehas the typical values illustrated in Table 7.7.1. The function ρi(x, y, z ) is the unpaired charge density when t= 0. Given any initial distribution, the subsequent distribution of ρuis given by (4). Once the 36 Conduction and Electroquasistatic Charge Relaxation Chapter 7 unpaired charge density has decayed to zero at a given point, it will remain zero. This is true regardless of the constraints on the surface bounding the region of uniform σand/epsilon1.Except for a transient that can only be initiated from very special initial conditions, the unpaired charge density in a material of uniform conductivity and permittivity is zero. This is true even if the system is not EQS. The following example is intended to help emphasize these implications of (3) and (4). Example 7.7.1. Charge Relaxation in Region of Uniform σand/epsilon1 In the region of uniform σand/epsilon1shown in Fig. 7.7.1, the initial distribution of unpaired charge density is ρi=nρo;r < a 0; a < r(5) where ρois a constant. It follows from (4) that the subsequent distribution is ρu=½ ρoe−t/τe;r < a 0; a < r As pictured in Fig. 7.7.1, the charge density in the spherical region r < a remains uniform as it decays to zero with the time constant τe. The charge density in the surrounding region is initially zero and remains so throughout the transient. Charge conservation implies that there must be a current density in the ma- terial surrounding the initially charged spherical region. Yet, according to the laws used here, there is never a net unpaired charge density in that region. This is pos- sible because in Ohmic conduction, there are at least two types of charges involved. In the uniformly conducting material, one or both of these migrate in the electric ﬁeld caused by the net charge [in accordance with (7.1.5)] while exactly neutralizing each other so that ρu= 0 (7.1.6). Net Charge on Bodies Immersed in Uniform Materials2.The integral charge relaxation law, (1.5.2), applies to the net charge within any volume con- taining a medium of constant /epsilon1andσ. If an initially charged particle ﬁnds itself suspended in a ﬂuid having uniform σand/epsilon1, this charge must decay with the charge relaxation time constant τe. Demonstration 7.7.1. Relaxation of Charge on Particle in Ohmic Conductor The pair of plane parallel electrodes shown in Fig. 7.7.2 is immersed in a semi- insulating liquid, such as corn oil, having a relaxation time on the order of a second. Initially, a metal particle rests on the lower electrode. Because this particle makes electrical contact with the lower electrode, application of a potential diﬀerence re- sults in charge being induced not only on the surfaces of the electrodes but on the surface of the particle as well. At the outset, the particle is an extension of the lower 2This subsection is not essential to the material that follows. Sec. 7.7 Charge Relaxation 37 Fig. 7.7.1 Within a material having uniform conductivity and permittivity, initially there is a uniform charge density ρuin a spherical region, having radius a. In the surrounding region the charge density is given to be initially zero and found to be always zero. Within the spherical region, the charge density is found to decay exponentially while retaining its uniform distribution. Fig. 7.7.2 The region between plane parallel electrodes is ﬁlled by a semi-insulating liquid. With the application of a constant potential diﬀerence, a metal particle resting on the lower plate makes upward excursions into the ﬂuid. [See footnote 1.] electrode. Thus, there is an electrical force on the particle that is upward. Note that changing the polarity of the voltage changes the sign of both the particle charge and the ﬁeld, so the force is always upward. As the voltage is raised, the electrical force outweighs the net gravitational force on the particle and it lifts oﬀ. As it separates from the lower electrode, it does so with a net charge suﬃcient to cause the electrical force to start it on its way toward charges of the opposite sign on the upper electrode. However, if the liquid is an Ohmic conductor with a relaxation time shorter than that required for the particle to reach the upper electrode, the net charge on the particle decays, and the upward electrical force falls below that of the downward gravitational force. In this case, the particle falls back to the lower electrode without reaching the upper one. Upon contacting the lower electrode, its charge is renewed and so it again lifts oﬀ. Thus, the particle appears to bounce on the lower electrode. By contrast, if the oil has a relaxation time long enough so that the particle can reach the upper electrode before a signiﬁcant fraction of its charge is lost, then the particle makes rapid excursions between the electrodes. Contact with the upper electrode results in a charge reversal and hence a reversal in the electrical force as well. The experiment demonstrates that as long as a particle is electrically isolated 38 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.7.3 Particle immersed in an initially uniform electric ﬁeld is charged by unipolar current of positive ions following ﬁeld lines to its surface. As the particle charges, the “window” over which it can collect ions becomes closed. in an Ohmic conductor, its charge will decay to zero and will do so with a time constant that is the relaxation time /epsilon1/σ. According to the Ohmic model, once the particle is surrounded by a uniformly conducting material, it cannot be given a net charge by any manipulation of the potentials on electrodes bounding the Ohmic conductor. The charge can only change upon contact with one of the electrodes. We have found that a particle immersed in an Ohmic conductor can only dis- charge. This is true even if it ﬁnds itself in a region where there is an externally imposed conduction current. By contrast, the next example illustrates how a unipo- larconduction process can be used to charge a particle. The ion-impact charging (or ﬁeld charging) process is put to work in electrophotography and air pollution control. Example 7.7.2. Ion-Impact Charging of Macroscopic Particles The particle shown in Fig. 7.7.3 is itself perfectly conducting. In its absence, the surrounding region is ﬁlled by an un-ionized gas such as air permeated by a uniform z-directed electric ﬁeld. Positive ions introduced at z→ −∞ then give rise to a unipolar current having a density given by the unipolar conduction law, (7.1.8). With the introduction of the particle, some of the lines of electric ﬁeld intensity can terminate on the particle. These carry ions to the particle. Other lines originate on the particle and it is assumed that there is no mechanism for the particle surface to initiate ions that would then carry charge away from the particle along these lines. Thus, as the particle intercepts some of the ion current, it charges up. Here the particle-charging process is described as a sequence of steady states. The charge conservation equation (7.0.3) obtained by using the unipolar conduction law (7.1.8) then requires that ∇ ·(µρE) = 0 (6) Thus, the “ﬁeld” ρE(consisting of the product of the charge density and the electric ﬁeld intensity) forms ﬂux tubes. These have walls tangential to Eand incremental Sec. 7.7 Charge Relaxation 39 cross-sectional areas δa, as illustrated in Figs.7.7.3 and 2.7.5, such that ρE·δa remains constant. As a second approximation, it is assumed that the dominant sources for the electric ﬁeld are on the boundaries, either on the surface of the particle or at inﬁnity. Thus, the ions in the volume of the gas are low enough in concentration so that their volume charge density makes a negligible contribution to the electric ﬁeld intensity. At each point in the volume of the gas, ∇ ·/epsilon1oE≈0 (7) From this statement of Gauss’ law, it follows that the Elines also form ﬂux tubes along which E·δais conserved. Because both E·δaandρE·δaare constant along a given Eline, it is necessary that the charge density ρbe constant along these lines. This fact will now be used to calculate the current of ions to the particle. At a given instant in the charging process, the particle has a net charge q. Its surface is an equipotential and it ﬁnds itself in an electric ﬁeld that is uniform at inﬁnity. The distribution of electric ﬁeld for this situation was found in Example 5.9.2. Lines of electric ﬁeld intensity terminate on the southern end of the sphere over the range π≥θ≥θc, where θcis shown in Figs. 7.7.3 and 5.9.2. In view of the unipolar conduction law, these lines carry with them a current density. Thus, there is a net current into the particle given by i=Zπ θc−µρE r(r=R, θ)(2πRsinθRdθ ) (8) Because ρis constant along an electric ﬁeld line and ρis uniform far from the charge-collecting particles, it is a constant over the surface of integration. It follows from (5.9.13) that the normal electric ﬁeld needed to evaluate (8) is Er=−∂Φ ∂r¯¯ r=R= 3Eacosθ+q 4π/epsilon1oR2(9) Substitution of (9) into (8) gives i=−µρ6πR2EaZπ θc¡ cosθ+q qc¢ sinθdθ (10) where, as in Example 5.9.2, qc= 12π/epsilon1oR2Eaand −cosθc=q qc(11) Remember, θcis the angle at which the radial electric ﬁeld switches from being outward to inward. Thus, it is a function of the amount of charge on the particle. Substitution of (11) into (10) and some manipulation gives the net current to the particle as i=qc τi¡ 1−q qc¢2(12) where τi= 4/epsilon1o/µρ. From (10) it is clear that the current depends on the particle charge. As charge accumulates on the particle, the angle θcincreases and so the southern surface over 40 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.7.4 Normalized particle charge as a function of normalized time. The saturation charge qcand charging time τare given after (10) and (12), respectively. which electric ﬁeld lines terminate decreases. By the time q=qc, the collection surface is zero and, as implied by (12), the current goes to zero. If the charging process is slow enough to be viewed as a sequence of stationary states, the current given by (12) is equal to the rate of increase of the particle charge. dq dt=i⇒d(q/qc) d(t/τi)=¡ 1−q qc¢2(13) Divided by what is on the right and multiplied by the denominator on the left, this expression can be integrated. Zq/qc 0d¡q/prime qc¢ ¡ 1−q/prime qc¢2=Zt/τi 0d¡t τi¢ (14) The result is a charging law that is not exponential but rather q qc=t/τi 1 +t/τi(15) This charging transient is shown in Fig. 7.7.4. By contrast with a particle placed in a conduction current that is Ohmic, a particle subjected to a unipolar current will charge up to the saturation charge qc. Note that the charging time, τi= 4/epsilon1o/µρ, again takes the form of /epsilon1divided by a “conductivity.” Demonstration 7.7.2. Electrostatic Precipitation Once dust, smoke, or fume particles are charged, they can be subjected to an electric ﬁeld and pulled out of the gas in which they are interspersed. In large precipitators used to ﬁlter combustion gases before they are released from a stack, the charging and precipitation processes are carried out in one region. The apparatus of Fig. 7.7.5 illustrates this process. A ﬁne wire is stretched along the axis of a grounded conducting cylinder having a radius of 5–10 cm. With the wire at a voltage of 10–30 kv, a hissing sound gives Sec. 7.8 Electroquasistatic Conduction 41 Fig. 7.7.5 Electrostatic precipitator consisting of ﬁne wire at high voltage relative to surrounding conducting transparent coaxial cylinder. Ions created in corona discharge in the immediate vicinity of the wire follow ﬁeld lines toward outer wall, some terminating on smoke particles. Once charged by the mechanism described in Example 7.7.2, the smoke particles are precipitated on the outer wall. evidence of ionization of the air in the immediate vicinity of the wire. This corona discharge provides positive and negative ion pairs adjacent to the wire. If the wire is positive, some of the positive ions are drawn out of this region and migrate to the cylindrical outer wall. Thus, outside the corona discharge region there is a unipolar conduction current of the type postulated in Example 7.7.2. The ion mobility is typically (1 →2)×10−4(m/s)/(v/m), while the ﬁeld is on the order of 5 ×105v/m, so the ion velocity (7.1.3) is in the range of 50 −100 m/s. Smoke particles, mixed with air rising through the cylinder, can be seen to be removed from the gas within a second or so. Large polyethylene particles dropped in from the top can be more readily seen to collect on the walls. In a practical precipitator, the collection electrodes are periodically rapped so that chunks of the collected material drop into a hopper below. Most of the time required to clear the air of smoke is spent by the particle in migrating to the wall after it has been charged. The charging time constant τiis typically only a few milliseconds. This demonstration further emphasizes the contrast between the behavior of a macroscopic particle when immersed in an Ohmic conductor, as in the previous demonstration, and when subjected to unipolar conduction. A particle immersed in a unipolar “conductor” becomes charged. In a uniform Ohmic conductor, it can only discharge. 7.8 ELECTROQUASISTATIC CONDUCTION LAWS FOR INHOMOGENEOUS MATERIALS In this section, we extend the discussion of transients to situations in which the electrical permittivity and Ohmic conductivity are arbitrary functions of space. /epsilon1=/epsilon1(r), σ =σ(r) (1) 42 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Distributions of these parameters, as exempliﬁed in Figs. 6.5.1 and 7.2.3, might be uniform, piece-wise uniform, or smoothly nonuniform. The speciﬁc examples falling into these categories answer three questions. (a) Where does the unpaired charge density, found in Sec. 7.7, tend to accumulate when it disappears from a region having uniform properties. (b) With the unpaired charge density determined by the self-consistent EQS laws, what is the equation governing the potential distribution throughout the vol- ume of interest? (c) What boundary and initial conditions make the solutions to this equation unique? The laws studied in this section and exempliﬁed in the next describe both the perfectly insulating limit of Chap. 6 and the conduction dominated limit of Secs. 7.1–7.6. More important, as suggested in Sec. 7.0, they describe how these limiting situations are related in EQS systems. Evolution of Unpaired Charge Density. With a nonuniform conductivity distribution, the statement of charge conservation and Ohm’s law expressed by (7.7.1) becomes σ∇ ·E+E· ∇σ+∂ρu ∂t= 0 (2) Similarly, with a nonuniform permittivity, Gauss’ law as given by (7.7.2) becomes /epsilon1∇ ·E+E· ∇/epsilon1=ρu (3) Elimination of ∇ ·Ebetween these equations gives an expression that is the gener- alization of the charge relaxation equation, (7.7.3). ∂ρu ∂t+ρu (/epsilon1/σ)=−E· ∇σ+σ /epsilon1E· ∇/epsilon1 (4) Wherever the electric ﬁeld has a component in the direction of a gradient of σor /epsilon1, the unpaired charge density can be present and can be temporally increasing or decreasing. If a steady state has been established, in the sense that time rates of change are negligible, the charge distribution is given by (4), because then, ∂ρu/∂t= 0. Note that this is the distribution of (7.2.8) that prevails for steady conduction. We can therefore expect that the charge density found to disappear from a region of uniform properties in Sec. 7.7 will reappear at surfaces of discontinuity of σand /epsilon1or in regions where /epsilon1andσvary smoothly. Electroquasistatic Potential Distribution. To evaluate (4), the self-consistent electric ﬁeld intensity is required. With the objective of determining that ﬁeld, Gauss’ law, (7.7.2), is used to eliminate ρufrom the charge conservation statement, (7.7.1). ∇ ·σE+∂ ∂t(∇ ·/epsilon1E) = 0 (5) Sec. 7.8 Electroquasistatic Conduction 43 For the ﬁrst time in the analysis of charge relaxation, we now introduce the elec- troquasistatic approximation ∇ ×E/similarequal0⇒E=−∇Φ (6) and (5) becomes the desired expression governing the evolution of the electric po- tential. ∇ ·¡ σ∇Φ +∂ ∂t/epsilon1∇Φ¢ = 0(7) Uniqueness. Consider now the initial and boundary conditions that make solutions to (7) unique. Suppose that throughout the volume V, the initial charge distribution is given as ρu(r, t= 0) = ρi(r) (8) and that on the surface Senclosing this volume, the potential is a given function of time Φ = Φ i(r, t) onSfort≥0. (9) Thus, when t= 0, the initial distribution of electric ﬁeld intensity satisﬁes Gauss’ law. The initial potential distribution satisﬁes the same law as for regions occupied by perfect dielectrics. ∇ ·/epsilon1∇Φi=−ρi (10) Given the boundary condition of (9) when t= 0, it follows from Sec. 5.2 that the initial distribution of potential is uniquely determined. Is the subsequent evolution of the ﬁeld uniquely determined by (7) and the initial and boundary conditions? To answer this question, we will take a somewhat more formal approach than used in Sec. 5.2 but nevertheless use the same reasoning. Supose that there are two solutions, Φ = Φ aand Φ = Φ b, that satisfy (7) and the same initial and boundary conditions. Equation (7) is written ﬁrst with Φ = Φ aand then with Φ = Φ b. With Φd≡Φa−Φb, the diﬀerence between these two equations becomes ∇ ·£ σ∇Φd+∂ ∂t(/epsilon1∇Φd)¤ = 0 (11) Multiplication of (11) by Φ dand integration over the volume Vgives Z VΦd∇ ·£ σ∇Φd+∂ ∂t(/epsilon1∇Φd)¤ dv= 0 (12) The objective in the following manipulation is to turn this integration either into one over positive deﬁnite quantities or into an integration over the surface S, where the boundary conditions determine the potential. The latter is achieved if the integrand can be expressed as a divergence. Thus, the vector identity ∇ ·ψA=ψ∇ ·A+A· ∇ψ (13) 44 Conduction and Electroquasistatic Charge Relaxation Chapter 7 is used to write (12) as Z V∇ ·£ Φd¡ σ∇Φd+∂ ∂t/epsilon1∇Φd¢¤ dv −Z V¡ σ∇Φd+∂ ∂t/epsilon1∇Φd¢ · ∇Φddv= 0(14) and then Gauss’ theorem converts the ﬁrst integral to one over the surface Sen- closing V.I SΦd¡ σ∇Φd+∂ ∂t/epsilon1∇Φd¢ ·da −Z V£ σ|∇Φd|2+∂ ∂t¡1 2/epsilon1|∇Φd|2¢¤ dv= 0(15) The conversion of (12) to (15) is an example of a three-dimensional integration by parts. The surface integral is analogous to an evaluation at the endpoints of a one-dimensional integral. If both Φ aand Φ bsatisfy the same condition on S, namely (9), then the diﬀerence potential is zero on Sfor all 0 ≤t. Thus, the surface integral in (15) vanishes. We are left with the requirement that for 0 ≤t, d dtZ V1 2/epsilon1|∇Φd|2dv=−Z Vσ|∇Φd|2dv (16) Because both Φ aand Φ bsatisfy the same initial conditions, Φ dmust initially be zero. Thus, for ∇Φdto change to a nonzero value from zero, the derivative on the left must be positive. However, the integral on the right can only be zero or negative. Thus, Φ dmust stay zero for all time. We conclude that the ﬁelds found using (7), the initial condition of (8), and boundary conditions of (9) are unique. 7.9 CHARGE RELAXATION IN UNIFORM AND PIECE-WISE UNIFORM SYSTEMS Conﬁgurations composed of subregions where the material has uniform properties are already familiar from Secs. 6.6 and 7.5. The conductivity and permittivity are then step functions of position, and the terms on the right in (7.8.4) are spatial impulses. Thus, the charge density tends to accumulate at interfaces between regions and is represented by a surface charge density. We consider ﬁrst the evolution of the potential distribution in a region hav- ing uniform properties. With the inhomogeneities represented by the continuity conditions, the discussion is then extended to piece-wise uniform conﬁgurations. Fields in Regions Having Uniform Properties. Where /epsilon1andσare uniform, (7.8.7) becomes ∇2·∂Φ ∂t+Φ (/epsilon1/σ)¸ = 0 (1) Sec. 7.9 Piece-Wise Uniform Systems 45 This expression is satisﬁed either if the potential obeys the relaxation equation ∂Φp ∂t+Φp (/epsilon1/σ)= 0 (2) or if it satisﬁes Laplace’s equation ∇2Φh= 0 (3) In general, the potential is a linear combination of these solutions. Φ = Φ p+ Φh (4) The potential satisfying (2) is that associated with the relaxation of the charge density initially distributed in the volume of the material. We can think of this as being a particular solution, because the divergence of the associated electric displacement D=/epsilon1E=−/epsilon1∇Φpgives the unpaired charge density, (7.7.4), at each point in the volume Vfort >0. The solutions Φ hto Laplace’s equation can then be used to make the sum of the two solutions satisfy the boundary conditions. Given that the initial charge density throughout the volume is ρi(r), the subsequent distribution is given by (7.7.4). One particular solution for the potential that then satisﬁes Poisson’s equation throughout the volume follows from evaluating the superposition integral [(4.5.3) with /epsilon1o→/epsilon1] over that volume. Φp=Z V/primeρi(r/prime) 4π/epsilon1|r−r/prime|dv/primee−t/(/epsilon1/σ)(5) Note that this potential indeed satisﬁes (2) and the initial conditions on the charge density in the volume. Of course, the integral could be extended to charges outside the volume V, and the particular solution would be equally valid. The solutions to Laplace’s equation make it possible to make the total poten- tial satisfy boundary conditions. Because an initial distribution of volume charge density cannot be initiated by means of boundary electrodes, the decay of an initial charge density is not usually of interest. The volume potential is most often simply a solution to Laplace’s equation. Before delving into these more common examples, consider one that illustrates the more general situation. Example 7.9.1. Potential Associated with Relaxation of Volume Charge In Example 7.7.1, the decay of charge having a spherical distribution in space was described. This could be done without regard for boundary constraints. To determine the associated potential, we stipulate the nature of the boundary surrounding the uniform material in which the charge is initially embedded. The uniform material ﬁlls the upper half-space and is bounded in the plane z= 0 by a perfect conductor constrained to zero potential. As shown in Fig. 7.9.1, when t= 0, there is an initial distribution of charge density that is uniform and of 46 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.9.1 Inﬁnite half-space of material having uniform conductivity and permittivity is bounded from below by a perfectly conducting plate. When t= 0, there is a uniform charge density in a spherical region. density ρothroughout a spherical region of radius acentered at z=hon the zaxis, where h > a . In terms of a spherical coordinate system centered on the zaxis at z=h, a particular solution for the potential follows from the integral form of Gauss’ law, much as in Example 1.3.1. With r+denoting the radial distance from the center of the spherical region, Φp=( 3a2−r2 + 6/epsilon1ρoe−t/τ;r+< a a3ρo 3/epsilon1r+e−t/τ; a < r +(6) where r+= [x2+y2+ (z−h)2]1/2andτ≡/epsilon1/σ. Note that this potential satisﬁes (2) and the initial condition but does not satisfy the zero potential condition at z= 0. To satisfy the latter, we add a potential that is a solution to Laplace’s equation, (3), everywhere in the upper half-space. This is the potential associated with an image charge density −ρoexp(−t/τ) distributed uniformly over a spherical region of radius acentered at z=−h. Φh=−a3ρo 3/epsilon1r−e−t/τ(7) where r−= [x2+y2+ (z+h)2]1/2, z > 0. Thus, the total potential Φ = Φ p+Φhthat satisﬁes both the initial conditions and boundary conditions for 0 < tis Φ =(3a2−r2 + 6/epsilon1ρoe−t/τ−a3ρo 3/epsilon1r−e−t/τ;r+< a a3ρo 3/epsilon1¡1 r+−1 r−¢ e−t/τ; a < r +(8) At each instant in time, the potential distribution is the same as if the charge and its image were static. As the charge relaxes, so does its image. Note that the charge relaxes to the boundary without producing a net charge density anywhere outside the spherical region where the charge was initiated. Continuity Conditions in Piece-Wise Uniform Systems. Where the material properties undergo step discontinuities, the diﬀerential equations are represented by continuity conditions. The one representing the condition that the ﬁeld be irro- tational, (7.8.6), is the same as that in Sec. 5.3. n×(Ea−Eb) = 0⇔Φa−Φb= 0 (9) Sec. 7.9 Piece-Wise Uniform Systems 47 Fig. 7.9.2 Incremental volume for writing charge conservation boundary condition. The continuity condition representing Gauss’ law, (7.7.2), is also familiar (6.2.16). σsu=n·(/epsilon1aEa−/epsilon1bEb) (10) The continuity condition representing charge conservation, (7.7.1), is (1.5.12). With the current density expressed in terms of Ohm’s law, this continuity condition becomes n·(σaEa−σbEb) +∂ ∂tσsu= 0(11) For the incremental volume of Fig. 7.9.2, this continuity condition requires that if the conduction current entering the volume from region (b) exceeds that leaving to region (a), there must be an increasing surface charge density within the volume. The fact that we are solving a second-order diﬀerential equation, (7.8.7), sug- gests that there are really only two continuity conditions. Thus, Gauss’ continuity condition only serves to relate the ﬁeld to the unknown surface charge density, and the combination of (10) and (11) comprise one continuity condition. n·(σaEa−σbEb) +∂ ∂tn·(/epsilon1aEa−/epsilon1bEb) = 0 (12) This continuity condition and the one on the tangential ﬁeld or potential, (9), are needed to splice together solutions representing ﬁelds in piece-wise uniform conﬁgurations. The following example illustrates how the time dependence of the continuity condition allows the ﬁelds and charge distribution to evolve from the distributions for perfect dielectrics described in the latter part of Chap. 6 to the steady conduction distributions discussed in the ﬁrst part of this chapter. Example 7.9.2. Maxwell’s Capacitor A conﬁguration that brings out the roles of polarization and conduction in the ﬁeld evolution while avoiding geometric complications is shown in Fig. 7.9.3. The space 48 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.9.3 Maxwell’s capacitor. between perfectly conducting parallel plates is ﬁlled by layers of material. The one above has thickness a, permittivity /epsilon1a, and conductivity σa, while for the one below, these parameters are b, /epsilon1b, and σb, respectively. When t= 0, a switch is closed and the potential Vof a battery is applied across the two electrodes. Initially, there is no unpaired charge between the electrodes either in the volume or on the interface. The electrodes are assumed long enough so that the fringing can be neglected and the ﬁelds in each of the materials taken as uniform. E=ixnEa(t); 0 < x < a Eb(t);−b < x < 0(13) The linear potential associated with this distribution satisﬁes Laplace’s equation, (3). Because there is no initial charge density in the volumes of the layers, the particular part of the potential, the solution to (2), is zero. The voltage source imposes the condition that the line integral of the electric ﬁeld between the plates must be equal to v(t). Za −bExdx=v(t) =aEa+bEb (14) Because the layers are conducting, they respond to the application of the voltage with conduction currents. Since the currents diﬀer, they cause a time rate of change of unpaired surface charge density at the interface between the layers, as expressed by (12). (σaEa−σbEb) +d dt(/epsilon1aEa−/epsilon1bEb) = 0 (15) Note that the boundary conditions on tangential Eat the electrode surfaces and at the interface are automatically satisﬁed. Given the driving voltage, these last two expressions comprise two equations in the two unknowns EaandEb. Thus, the solution to (14) for Eband substitution into (15) gives a ﬁrst-order diﬀerential equation for the ﬁeld response in the upper layer. (b/epsilon1a+a/epsilon1b)dEa dt+ (bσa+aσb)Ea=σbv+/epsilon1bdv dt(16) In particular, consider the response to a step in voltage, v=V u−1(t). The drive on the right in (16) then consists of a step and an impulse. The impulse must be matched by an impulse on the left. That is, the ﬁeld Eaalso undergoes a step change when t= 0. To identify the magnitude of this step, integrate (16) from 0−to 0+. (b/epsilon1a+a/epsilon1b)Z0+ 0−dEa dtdt+ (bσa+aσb)Z0+ 0−Eadt =σbZ0+ 0−vdt+/epsilon1bZ0+ 0−dv dtdt(17) Sec. 7.9 Piece-Wise Uniform Systems 49 The result is a relationship between the jumps in voltage and in ﬁeld. ¡ /epsilon1a+a b/epsilon1b¢ [Ea(0+)−Ea(0−)] =/epsilon1b b[v(0+)−v(0−)] (18) Because v(0−) = 0 and Ea(0−) = 0, it follows that Ea(0+) =/epsilon1bV b/epsilon1a+a/epsilon1b(19) Fort >0, the particular plus homogeneous solution to (16) is Ea=σbV bσa+aσb+Ae−t/τ(20) where τ≡b/epsilon1a+a/epsilon1b bσa+aσb. The coeﬃcient Ais adjusted to make Eameet the initial condition given by (19). Thus, the ﬁeld transient in the upper layer is found to be Ea=σbV (bσa+aσb)(1−e−t/τ) +/epsilon1bV (b/epsilon1a+a/epsilon1b)e−t/τ(21) It follows from (14) that the ﬁeld in the lower layer is then Eb=V b−a bEa (22) The unpaired surface charge density, (10), follows from these ﬁelds. σsu=V(σb/epsilon1a−σa/epsilon1b) (bσa+aσb)(1−e−t/τ) (23) The ﬁeld and unpaired surface charge density transients are shown in Fig. 7.9.4. The curves are drawn to depict a lower layer that has a somewhat greater permittivity and a much greater conductivity than the upper layer. Just after the step in voltage, when t= 0+, the surface charge density remains zero. Thus, the electric ﬁelds are at ﬁrst what they would be if the layers were regarded as perfectly insulating dielectrics. As the surface charge accumulates, these ﬁelds approach values consistent with steady conduction. The limiting surface charge density approaches a saturation value that could be found by ﬁrst evaluating the steady conduction ﬁelds and then ﬁnding σsu. Note that this surface charge can be positive or negative. With the lower region much more conducting than the upper one ( σb/epsilon1a/greatermuchσa/epsilon1b) the surface charge is positive. In this case, the ﬁeld ends up tending to be shielded out of the lower layer. Piece-wise continuous conﬁgurations can often be represented by capacitor- resistor networks. An exact circuit representation of Maxwell’s capacitor is shown in Fig. 7.9.5. The voltages across the capacitors are simply va=Eaaandvb=Ebb. In the circuit, the surface charge density given by (23) is the sum of the net charge per unit area on the lower plate of the top capacitor and that on the upper plate of the lower capacitor. 50 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.9.4 With a step in voltage applied to the plane parallel conﬁg- uration of Fig. 7.9.3, the electric ﬁeld intensity above and below the in- terface responds as shown on the left, while the unpaired surface charge density has the time dependence shown on the right. Fig. 7.9.5 Maxwell’s capacitor, Fig. 7.9.3, is exactly equivalent to the circuit shown. Nonuniform Fields in Piece-Wise Uniform Systems. We continue now to consider examples with no initial charge density in the regions having uniform conductivity and dielectric constant. Since it is not possible to establish a charge density in these regions by means of boundary constraints, this is almost always the situation in practice. The ﬁeld distributions in the uniform subregions have potentials that satisfy Laplace’s equation, (3). These are “spliced” together at the interfaces between regions and constrained at boundaries by conditions that vary with time. The continuity conditions vary with time to account for the accumulation of unpaired charge at the interfaces between regions. Maxwell’s capacitor, Example 7.9.2, illustrates most features of the surface charge relaxation process. The response to a step function of voltage across an electrode pair is at ﬁrst the ﬁeld distribution of a system of perfect dielectrics, as developed in Chap. 6. After many charge relaxation times, steady conduction prevails, and the ﬁelds are as described in Sec. 7.5. In the remainder of this section, conﬁgurations will be considered that, by contrast to Maxwell’s capacitor, have ﬁelds that change their shape as the relaxation process evolves. The interplay of polarization and conduction processes is also evident in the Sec. 7.9 Piece-Wise Uniform Systems 51 Fig. 7.9.6 A spherical material with conductivity σband permittiv- ity/epsilon1bis surrounded by a material with conductivity and permittivity (σa, /epsilon1a). An electric ﬁeld E(t) that is uniform far from the sphere is applied. sinusoidal steady state response of a system. Just as the Maxwell capacitor has short-time and long-time responses dominated by the “capacitors” and “resistors,” respectively, the high-frequency and low-frequency responses are dominated by po- larization and conduction, respectively. This too will now be illustrated. Example 7.9.3. Spherical Semi-insulating Material Embedded in a Second Material Stressed by Uniform Electric Field An electric ﬁeld intensity E(t) is imposed on a material having permittivity and conductivity ( /epsilon1a, σa), perhaps by means of plane parallel electrodes. At the origin of a spherical coordinate system embedded in this material is a spherical region having permittivity and conductivity ( /epsilon1b, σb) and radius R, as shown in Fig. 7.9.6. Limiting cases include a conducting sphere surrounded by free space ( /epsilon1a=/epsilon1o, σa= 0) or an insulating spherical cavity surrounded by a conducting material ( σb= 0). In each of the regions, the potential must satisfy Laplace’s equation. From our experience with the potentials for perfect dielectric and for steady conduction conﬁgurations, we can expect that the boundary conditions can be satisﬁed using combinations of uniform and dipole ﬁelds. With the understanding that the coeﬃ- cients A(t) and B(t) are functions of time, the solutions to Laplace’s equation are therefore postulated to take the form Φ =½ −E(t)rcosθ+A(t)cosθ r2;r > R B(t)rcosθ; r < R(24) Note that the uniform part of the exterior ﬁeld has been matched at r→ ∞ to the given driving ﬁeld. Continuity of the tangential electric ﬁeld at r=R, (9), requires that these potential functions match at r=R. Φa(r=R) = Φb(r=R) (25) Conservation of charge, with the surface charge density represented using Gauss’ law, (12), makes the further requirement that (σaEa r−σbEb r) +∂ ∂t(/epsilon1aEa r−/epsilon1bEb r) = 0 (26) In substituting the potentials of (24) into these two conditions, no derivatives with respect to θare taken, so each term has the θdependence cos( θ). It is for this 52 Conduction and Electroquasistatic Charge Relaxation Chapter 7 reason that such a simple solution can be used to satisfy the continuity conditions. Substitution into (25) relates the coeﬃcients −ER+A R2=BR⇒B=−E+A R3(27) and with this relation used to eliminate B, substitution into (26) results in a diﬀer- ential equation for A(t), with E(t) as a driving function. (2/epsilon1a+/epsilon1b)dA dt+ (2σa+σb)A= (σb−σa)R3E(t) + (/epsilon1b−/epsilon1a)R3dE dt(28) Step Response. Note that expression (28) has the same form as that for Maxwell’s capacitor, (16). The procedure leading to the ﬁeld response to a step function of applied ﬁeld, E=Eou−1(t), is therefore identical to that illustrated in Example 7.9.2. In fact, comparison of these equations makes it clear that the required solution, given that there were no initial ﬁelds (when t= 0−), is A=EoR3· σb−σa 2σa+σb(1−e−t/τ) +/epsilon1b−/epsilon1a 2/epsilon1a+/epsilon1be−t/τ¸ (29) where the relaxation time τ= (2/epsilon1a+/epsilon1b)/(2σa+σb). The coeﬃcient Bfollows from (27). Thus, the potential of (24) is determined for t≥0. Φ =−EoRcosθ8 >>< >>:r R+· σa−σb 2σa+σb(1−e−t/τ) +/epsilon1a−/epsilon1b 2/epsilon1a+/epsilon1be−t/τ¸ (R r)2;R < r r R· 1 +σa−σb 2σa+σb(1−e−t/τ) +/epsilon1a−/epsilon1b 2/epsilon1a+/epsilon1be−t/τ¸ ; r < R(30) The accumulation of unpaired surface charge at r=Raccounts for the redistribution of potential with time. It follows from (10) that σsu=/epsilon1aEa r−/epsilon1bEb r= 3Eo(/epsilon1aσb−/epsilon1bσa) (2σa+σb)(1−e−t/τ) cosθ (31) Thus, the unpaired surface charge density accumulates at the poles of the sphere, exponentially approaching a saturation value at a rate determined by the relaxation time τ.Just after the ﬁeld is turned on, this surface charge density is zero and the ﬁeld distribution should be that for a uniform ﬁeld applied to perfect dielectrics . Indeed, evaluated when t= 0, (30) gives the potential for perfect dielectrics. In the opposite extreme, where many relaxation times have passed so that the exponentials in (30) are negligible, the potential assumes the distribution for steady conduction . A graphical portrayal of this ﬁeld transient is given in Fig. 7.9.7. The case shown was chosen because it involves a drastic redistribution of the ﬁeld as time progresses. The spherical region is highly conducting compared to its surroundings, but the exterior material is highly polarizable compared to the spherical region. Thus, just after the switch is closed, the ﬁeld lines tend to be trapped in the outer region. As time progresses and conduction rules, these lines tend to pass through Sec. 7.9 Piece-Wise Uniform Systems 53 Fig. 7.9.7 Evolution of the displacement ﬂux density Din and around the sphere of Fig. 7.9.6 and of σsuin response to the application of a step in applied ﬁeld. The sphere is more conducting than its surroundings (σa/σb= 0.2), while the outer region has a greater permittivity than the inner one, /epsilon1a//epsilon1b= 5. Thus, when the distribution of Dis determined by the polarization just after the ﬁeld is applied, the ﬁeld lines tend to be trapped in the outer region. By the time t= 0.5τ, enough σsuhas been induced to cancel the ﬁeld associated with σsp, and the electric ﬁeld intensity is essentially uniform. In the ﬁnal state, conduction alone determines the distribution of E. However, it is Dthat is shown in the ﬁgure, so, in fact, the permittivities do contribute to the ﬁnal relative intensities. the highly conducting sphere. The temporal scale of the transient is determined by the relaxation time τ. Sinusoidal Steady State Response. Consider now the sinusoidal steady state that results from applying the uniform ﬁeld E(t) =Epcosωt=ReE pejωt(32) 54 Conduction and Electroquasistatic Charge Relaxation Chapter 7 As in dealing with ac circuits, where the currents and voltages are also solutions to constant coeﬃcient ordinary diﬀerential equations, the response is now assumed to have the same frequency ωas the drive but to have a yet to be determined amplitude and phase represented by the complex coeﬃcients AandB. A(t) =ReˆAejωt; B(t) =ReˆBejωt(33) Substitution of (32) and (33a) into (28) gives an expression that can be solved for ˆAin terms of the drive, Ep. ˆA=[(σb−σa) +jω(/epsilon1b−/epsilon1a)] (2σa+σb) +jω(2/epsilon1a+/epsilon1b)R3Ep (34) In turn, the complex amplitude Bfollows from this result and (27). ˆB=−Ep+ˆA R3=−3Ep(σa+jω/epsilon1a) (2σa+σb) +jω(2/epsilon1a+/epsilon1b)(35) Now, with the amplitudes in (31) and (32) given by these expressions, the sinusoidal steady state ﬁelds postulated with (24) are determined. Φ =−Re E pRcosθejωt8 < :r R+· (σa−σb)+jω(/epsilon1a−/epsilon1b) (2σa+σb)+jω(2/epsilon1a+/epsilon1b)¸ (R r)2;r > R 3r R(σa+jω/epsilon1a) (2σa+σb)+jω(2/epsilon1a+/epsilon1b); R > r(36) The surface charge density associated with these ﬁelds is then σsu=Re3Ep(σb/epsilon1a−σa/epsilon1b) (2σa+σb) +jω(2/epsilon1a+/epsilon1b)cosθejωt(37) With the frequency rather than the time as the parameter, these expressions can be interpreted analogously to the step function response, (30) and (31). In the high- frequency limit, where ω(2/epsilon1a+/epsilon1b) 2σa+σb≡ωτ/greatermuch1; ω· (/epsilon1a−/epsilon1b) (σa−σb)¸ /greatermuch1 (38) the conductivity terms become negligible in (36), the coeﬃcients ˆAandˆBbecome independent of frequency and real. Thus, the ﬁelds are in temporal phase with the applied ﬁeld and sinusoidally varying versions of what would be found if the materials were assumed to be perfect dielectrics. If the frequency is high compared to the reciprocal charge relaxation times, the ﬁeld distributions are the same as they would be just after a step in applied ﬁeld [when t= 0+in (30)]. With the inequalities of (38) reversed, the terms involving the permittivity in (36) are negligible, the coeﬃcients ˆAandˆBare again real and hence the ﬁelds are just as they would be for stationary conduction except that they vary sinusoidally with time. Thus, in the low frequency limit, the ﬁelds are sinusoidally varying versions of the steady conduction ﬁelds that prevail long after a step in applied ﬁeld [(30) in the limit t→ ∞ ]. Sec. 7.9 Piece-Wise Uniform Systems 55 These high- and low-frequency limits are consistent with the frequency de- pendence of the unpaired surface charge density, given by (37). At low frequencies, this surface charge density varies sinusoidally in or out of phase with the applied ﬁeld and with an amplitude consistent with steady conduction. As the frequency is made to greatly exceed the reciprocal relaxation time, the magnitude of this charge falls to zero. In this high-frequency limit, there is insuﬃcient time during one cycle for signiﬁcant charge to relax to the spherical interface. Thus, at high frequencies the ﬁelds become the same as if the unpaired charge density were ignored and the dielectrics assumed to be perfectly insulating. In the two demonstrations that close this section, an obvious objective is the association of the previous example with practical situations. The approximations used to rederive the relevant ﬁelds cast further light on the physical processes at work. Demonstration 7.9.1. Capacitively Induced Fields in a Person in the Vicinity of a High-Voltage Power Line A person standing under a conventional power line, as in Fig. 7.9.8a, is subject to a 60 Hz alternating electric ﬁeld intensity that is typically 5 ×104v/m. In response to this ﬁeld, body currents are induced. Common experience suggests that these are not large enough to create discomfort, but are the currents appreciable enough to be of long-term medical concern? In the bare-handed maintenance of power lines, a person is brought to within arms length of the line by an insulated hoist, as shown in Fig. 7.9.8b. Without shielding, the body is in this case subjected to much more intense ﬁelds, perhaps 5×105v/m. For the ﬁrst person proving out this technique, the estimation of ﬁelds and currents within the body was of considerable interest. To the layman, these imposed ﬁelds seem to imply that a body one meter in length would be subject to a voltage diﬀerence of 50 kV at the ground and 500 kV near the line. However, as we will now illustrate, surrounded by air, the body does an excellent job of shielding out the electric ﬁeld. The hemispherical conductor resting on a ground plane, shown in Fig. 7.9.9, is a model for an individual on (and in electrical contact with) the ground. In the experiment, the hemisphere is jello, molded to have the radius Rand having a conductivity essentially that of the salt water used in its making. (To obtain the physiological conductivity of 0.2 S/m, unﬂavored gelatine is made using 0.02 M NaCl, a solution of 1.12 grams/liter.) Presumably, the potential in and around the hemisphere is given by (30). Thez= 0 plane is at zero potential for the spherical region described, and so the potential applies equally well to the hemisphere on the ground plane. Parameters are (/epsilon1a, σa) = ( /epsilon1o,0) in the air and ( /epsilon1b, σb) = ( /epsilon1, σ) in the hemisphere. A conductivity typical of physiological tissue is σ=.2 S/m. As a result, the charge relaxation time based on the permittivity of the body ( /epsilon1b= 81/epsilon1o) and the conductivity of the body is extremely short, τ= 4×10−9s. This makes it possible to approximate the potential distribution using the two simple steps that follow. First, because the charge can relax to the surface in a time that is far shorter than 1 /ω, and because the hemisphere is surrounded by material that has far less conductivity, as far as the ﬁeld in the air is concerned, its surface is an equipotential. Φa(r=R)/similarequal0 (39) 56 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.9.8 (a) Person in vicinity of power line terminates lines of elec- tric ﬁeld intensity and hence is subject to currents associated with in- duced charge. The electric ﬁeld intensity at the ground is as much as 5×104V/m. (b) Worker carrying out “bare-handed maintenance” is subject to ﬁeld that depends greatly on shielding provided, but can be 5×105V/m or more. (c) Hemispherical model for person on ground in (a). (d) Spherical model for person near line without shielding, (b). Thus, the potential distribution can be written by inspection [or by recourse to (5.9.7)] as Φa/similarequal −EpRcosωt·¡r R¢ −¡R r¢2¸ cosθ (40) Because of the short relaxation time and high conductivity for the sphere relative to the air, the surface charge density is essentially determined by the exterior ﬁeld. Sec. 7.9 Piece-Wise Uniform Systems 57 Fig. 7.9.9 Demonstration of currents induced in ﬂesh-simulating hemi- sphere by ﬁeld applied in surrounding air. Thus, the conservation of charge continuity condition, (12), is approximately σEb r(r=R)/similarequal∂ ∂t[/epsilon1oEa r(r=R)] (41) The rate of change of the surface charge density on the right in this expression has already been determined, so the expression serves to evaluate the normal conduction current density just inside the hemispherical surface. Eb r(r=R) =−3ω/epsilon1oEp σsinωtcosθ (42) In the interior region, the potential is uniform and thus takes the form Brcos(θ). Evaluation of the coeﬃcient Bby using (42) then gives the approximate potential distribution within the hemisphere. Φb/similarequal3ω/epsilon1o σEprcosθsinωt=3ω/epsilon1o σEpzsinωt (43) In retrospect, note that the potentials given by (40) and (43) are obtained by taking the appropriate limit of the potential obtained without making approxima- tions, (36). Inside the hemisphere, the conditions for essentially steady conduction prevail. Thus, the potential predicted by (43) is probed by means of metal spheres (Ag/AgCl electrodes) embedded in the jello and connected to an oscilloscope through insulated wires. Inside the hemisphere, surface charge stored on the surfaces of the insulated wires has a minor eﬀect on the current distribution. Typical experimental values for a 250 Hz excitation are R= 3.8 cm, s= 12.7 cm,v= 565 V peak, and σ= 0.2 S/m. With the probes located at z= 2.86 cm and z= 0.95 cm, the measured potentials are 25 µV peak and 10 µV peak, respectively. With the given parameters, (43) gives 26.5 µV peak and 8.8 µV peak, respectively. What are the typical current densities that would be induced in a person in the vicinity of a power line? According to (41), for the person on the ground in a 58 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.9.10 Conﬁguration for an electrocardiogram, including volt- ages typically generated at body periphery by the heart. ﬁeld of 5 ×104V/m (Fig. 7.9.8a), the current density is Jz=σEz= 0.05µA/cm2. For the person doing bare-handed maintenance where the ﬁeld is perhaps 5 ×105 V/m (Fig. 7.9.8b), the model is a sphere in a uniform ﬁeld (Fig. 7.9.8d). The current density is again given by (43), Jz=σEz= 0.5µA/cm2. Of course, the geometry of a person is not spherical. Thus, it can be expected that the ﬁeld will concentrate more in the actual situation than for the hemispherical or spherical models. The approximations introduced in this demonstration would greatly simplify the development of a numerical model. Have we found estimates of current densities suggesting danger, especially for the maintenance worker? Physiological systems are far too complex for there to be a simple answer to this question. However, matters are placed in some perspective by recognizing that currents of diverse origins exist in the body so long as it lives. In the next demonstration, electrocardiogram potentials are used to estimate current densities that result from the muscular contractions of the heart. The magnitude of the current density found there will lend some perspective to that determined here. The approximate analysis introduced in support of the previous demonstra- tion is an example of the “inside-outside” viewpoint introduced in Sec. 7.5. The exterior insulating region, where the ﬁeld was applied, was “inside,” while the inte- rior conducting region was “outside.” The following demonstration continues this theme with a contrasting example, where the excitation is in the conducting region. Demonstration 7.9.2. Currents Induced by the Heart The conﬁguration for taking an electrocardiogram is typically as shown in Fig. 7.9.10. With care taken to balance out 60 Hz signals induced in each of the elec- trodes by external ﬁelds, the electrical signals induced by the muscle contractions in the heart are easily measured using a conventional oscilloscope. In practice, many electrodes are used so that detailed information on the distribution of the muscle contractions can be discerned. Here we simply represent the heart by a dipole source of current at the center of a conducting sphere, somewhat as depicted in Figs. 7.9.10 and 7.9.11. Relatively little current is induced in the limbs, so that potentials measured at the extremities roughly reﬂect the potentials on the surface of the equivalent sphere. Given that typical potential diﬀerences are on the order of millivolts, what current dipole mo- ment can we attribute to the heart, and what are the typical current densities in its neighborhood? Sec. 7.9 Piece-Wise Uniform Systems 59 Fig. 7.9.11 Body and heart modeled by spherical conductor and dipole current. With the heart represented by a current source of dipole moment ipdat the center of the spherical “torso,” the electric potential at the origin approaches that for the dipole current source, (7.3.9). Φb(r→0)→ipd 4πσcosθ r2(44) At the surface r=R, the spherical body is being surrounded by an insulator. Thus, again using Fig. 7.9.11, any normal conduction current must be accounted for by the accumulation of surface charge. Because the relaxation time is so short compared to the 1 speriod typical of the heart, the current density associated with the buildup of surface charge is extremely small. As a result, the current distribution inside the sphere is as though the normal current density at r=Rwere zero. ∂Φb ∂r(r=R)/similarequal0 (45) Thus, the potential within the body is fully determined without regard for con- straints from the surrounding region. The solution to Laplace’s equation that satis- ﬁes these last two conditions is Φb/similarequalipd 4πσR2·¡R r¢2+ 2¡r R¢¸ cosθ (46) Because the potential is continuous at r=R, the potential on the surface of the “torso” follows from evaluation of this expression at r=R. Φa(r=R) = Φ b(r=R) =3(ipd) 4πσR2cosθ (47) Thus, given that the potential diﬀerence between θ= 45 degrees and θ= 135 degrees is 1 mV, that R= 25 cm, and that σ= 0.2 S/m, it follows from (47) that the peak current dipole moment of the heart is 3 .7×10−5A - m. Typical current densities can now be found using (46) to evaluate the electric ﬁeld intensity. For example, the current density at the radius R/2 just above the dipole source is Jz=σEz¡ r=R 2, θ= 0¢ =7(ipd) 2πR3 = 2.6×10−3A/m2= 0.26µA/cm2(48) 60 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Note that at the particular position selected the current density exceeds with some margin that to which the maintenance worker is subjected in the previous demon- stration. To begin to correlate the state and function of the heart with electrocardio- grams, it is necessary to represent the heart by a current dipole that not only has a special temporal signature but rotates with time as well[1,2]. Unfortunately, much of the medical literature on the subject takes the analogy between electric dipoles (Sec.4.4) and current dipoles (Sec. 7.3) literally. The heart is described as an electric dipole[2], which it certainly is not. If it were, its ﬁelds would be shielded out by the surrounding conducting ﬂesh. R E F E R E N C E S [1] R. Plonsey, Bioelectric Phenomena , McGraw-Hill Book Co., N.Y. (1969), p. 205. [2] A. C. Burton, Physiology and Biophysics of the Circulation , Year Book Medical Pub., Inc., Chicago, Ill. 2nd ed., pp. 125-138. 7.10 SUMMARY This chapter can be divided into three parts. In the ﬁrst, Sec. 7.1, conduction constitutive laws are related to the average motions of microscopic charge carriers. Ohm’s law, as it relates the current density Juto the electric ﬁeld intensity E Ju=σE (1) is found to describe conduction in certain materials which are constituted of at least one positive and one negative species of charge carrier. As a reminder that the current density can be related to ﬁeld variables in many ways other than Ohm’s law, the unipolar conduction law is also derived in Sec. 7.1, (7.1.8). But in this chapter and those to follow, the conduction law (1) is used almost exclusively. The second part of this chapter, Secs. 7.2–7.6, is concerned with “steady” con- duction. A summary of the diﬀerential laws and corresponding continuity conditions is given in Table 7.10.1. Under steady conditions, the unpaired charge density is determined from the last expressions in the table after the ﬁrst two have been used to determine the electric potential and ﬁeld intensity. In the third part of this chapter, Secs. 7.7–7.9, the dynamics of EQS systems is developed and exempliﬁed. The laws used to determine the electric potential and ﬁeld intensity, given by the ﬁrst two lines in Table 7.10.2, are valid for frequencies and characteristic times that are arbitrary relative to electrical relaxation times, provided those times are themselves long compared to times required for an elec- tromagnetic wave to propagate through the system. The last expressions identify how the unpaired charge density is relaxing under dynamic conditions. In EQS systems, the magnetic induction makes a negligible contribution and the electric ﬁeld intensity is essentially irrotational. Thus, Eis represented by Sec. 7.10 Summary 61 TABLE 7.10.1 SUMMARY OF LAWS FOR STEADY STATE OHMIC CONDUCTION Diﬀerential Law Eq. No. Continuity Condition Eq. No. Faraday’s Law∇ ×E/similarequal0⇔E=−∇Φ (7.0.1) Φa−Φb= 0 (7.2.10) Charge conservation∇ ·σE=s(7.2.2) (7.3.1)n·(σaEa−σbEb) =Js(7.2.9) (7.3.4) Unpaired charge distributionρu=−/epsilon1 σE· ∇σ+E· ∇/epsilon1(7.2.8) σsu=n·/epsilon1aEa¡ 1−/epsilon1bσa /epsilon1aσb¢ (7.2.12) TABLE 7.10.2 SUMMARY OF EQS LAWS FOR INHOMOGENEOUS OHMIC MEDIA Diﬀerential Law Eq. No. Continuity Condition Eq. No. Faraday’s lawE=−∇Φ (7.0.1) Φa−Φb= 0 (7.2.10) Charge conservation, Ohm’s law, and Gauss’ law∇ ·£ σE+∂ ∂t(/epsilon1E)¤ =s (7.8.5) (7.3.2)n·(σaEa−σbEb) +∂ ∂tn·(/epsilon1aEa−/epsilon1bEb) =Js(7.9.12) (7.3.4) Relaxation of unpaired charge density∂ρu ∂t+ρu /epsilon1/σ=−E· ∇σ +σ /epsilon1E· ∇/epsilon1(7.8.4)∂σsu ∂t+n·(σaEa−σbEb) = 0(7.9.11) −grad (Φ) in both Table 7.10.1 and Table 7.10.2. In the EQS approximation, ne- glecting the magnetic induction is tantamount to ignoring the ﬁnite transit time eﬀects of electromagnetic waves. This we saw in Chap. 3 and will see again in Chaps. 14 and 15. 62 Conduction and Electroquasistatic Charge Relaxation Chapter 7 In MQS systems, ﬁelds may be varying so slowly that the eﬀect of magnetic induction on the current ﬂow is again ignorable. In that case, the laws of Table 7.10.1 are once again applicable. So it is that the second part of this chapter is a logical base from which to begin the next chapter. At least under steady conditions we already know how to predict the distribution of the current density, the source of the magnetic ﬁeld intensity. How rapidly can MQS ﬁelds vary without having the magnetic induction come into play? We will answer this question in Chap. 10. Sec. 7.2 Problems 63 P R O B L E M S 7.1 Conduction Constitutive Laws 7.1.1 In a metal such as copper, where each atom contributes approximately one conduction electron, typical current densities are the result of electrons moving at a surprisingly low velocity. To estimate this velocity, assume that each atom contributes one conduction electron and that the material is copper, where the molecular weight Mo= 63.5 and the mass density is ρ= 8.9×103kg/m3. Thus, the density of electrons is approximately ( Ao/Mo)ρ, where Ao= 6.023×1026molecules/kg-mole is Avogadro’s number. Given σfrom Table 7.1.1, what is the mobility of the electrons in copper? What electric ﬁeld intensity is required to drive a current density of l amp/cm2? What is the electron velocity? 7.2 Steady Ohmic Conduction 7.2.1∗The circular disk of uniformly conducting material shown in Fig. P7.2.1 has a dc voltage vapplied to its surfaces at r=aandr=bby means of perfectly conducting electrodes. The other boundaries are interfaces with free space. Show that the resistance R=ln(a/b)/2πσd. Fig. P7.2.1 7.2.2 In a spherical version of the resistor shown in Fig. P7.2.1, a uniformly conducting material is connected to a voltage source vthrough spherical perfectly conducting electrodes at r=aandr=b. What is the resistance? 7.2.3∗By replacing /epsilon1→σ, resistors are made to have the same geometry as shown in Fig. P6.5.1. In general, the region between the plane parallel perfectly conducting electrodes is ﬁlled by a material of conductivity σ=σ(x). The boundaries of the conductor that interface with the surrounding free space have normals that are either in the xor the zdirection. (a) Show that even if dis large compared to landc,Ebetween the plates is (v/d)iy. 64 Conduction and Electroquasistatic Charge Relaxation Chapter 7 (b) If the conductor is piece-wise uniform, with sections having conduc- tivities σaandσbof width aandb, respectively, as shown in Fig. P6.5.1a, show that the conductance G=c(σbb+σaa)/d. (c) If σ=σa(1 +x/l), show that G= 3σacl/2d. 7.2.4 A pair of uniform conductors form a resistor having the shape of a circular cylindrical half-shell, as shown in Fig. P7.2.4. The boundaries at r=a andr=b, and in planes parallel to the paper, interface with free space. Show that for steady conduction, all boundary conditions are satisﬁed by a simple piece-wise continuous potential that is an exact solution to Laplace’s equation. Determine the resistance. Fig. P7.2.4 7.2.5∗The region between the planar electrodes of Fig. 7.2.4 is ﬁlled with a ma- terial having conductivity σ=σo/(1 +y/a), where σoandaare constants. The permittivity /epsilon1is uniform. (a) Show that G=Aσo/d(1 +d/2a). (b) Show that ρu=/epsilon1Gv/Aσ oa. 7.2.6 The region between the planar electrodes of Fig. 7.2.4 is ﬁlled with a uni- formly conducting material having permittivity /epsilon1=/epsilon1a/(1 +y/a). (a) What is G? (b) What is ρuin the conductor? 7.2.7∗A section of a spherical shell of conducting material with inner radius b and outer radius ais shown in Fig. P7.2.7. Show that if σ=σo(r/a)2, the conductance G= 6π(1−cosα/2)ab3σo/(a3−b3). Sec. 7.3 Problems 65 Fig. P7.2.7 7.2.8 In a cylindrical version of the geometry shown in Fig. P7.2.7, the mate- rial between circular cylindrical outer and inner electrodes of radii aand b, respectively, has conductivity σ=σo(a/r). The boundaries parallel to the page interface free space and are a distance dapart. Determine the conductance G. 7.3 Distributed Current Sources and Associated Fields 7.3.1∗An inﬁnite half-space of uniformly conducting material in the region y >0 has an interface with free space in the plane y= 0. There is a point current source of Iamps located at ( x, y, z ) = (0 , h,0) on the yaxis. Using an approach analogous to that used in Prob. 6.6.5, show that the potential inside the conductor is Φa=I 4πσp x2+ (y−h)2+z2+I 4πσp x2+ (y+h)2+z2. (a) Now that the potential of the interface is known, show that the po- tential in the free space region outside the conductor, where y <0, is Φb=2I 4πσp x2+ (y−h)2+z2(b) 7.3.2 The half-space y >0 is of uniform conductivity while the remaining space is insulating. A uniform line current source of density Kl(A/m) runs parallel to the plane y= 0 along the line x= 0, y=h. (a) Determine Φ in the conductor. (b) In turn, what is Φ in the insulating half-space? 7.3.3∗A two-dimensional dipole current source consists of uniform line current sources ±Klhave the spacing d. The cross-sectional view is as shown in 66 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. 7.3.4, with θ→φ. Show that the associated potential is Φ =Kld 2πσcosφ r(a) in the limit Kl→ ∞ , d→0, Kldﬁnite. 7.4 Superposition and Uniqueness of Steady Conduction Solutions 7.4.1∗A material of uniform conductivity has a spherical insulating cavity of radius bat its center. It is surrounded by segmented electrodes that are driven by current sources in such a way that at the spherical outer surface r=a, the radial current density is Jr=−Jocosθ, where Jois a given constant. (a) Show that inside the conducting material, the potential is Φ =Job σ£ (r/b) +1 2(b/r)2¤ [1−(b/a)3]cosθ; b < r < a. (a) (b) Evaluated at r=b, this gives the potential on the surface bounding the insulating cavity. Show that the potential in the cavity is Φ =3Jo 2σrcosθ [1−(b/a)3]; r < b (b) 7.4.2 A uniformly conducting material has a spherical interface at r=a, with a surrounding insulating material and a spherical boundary at r=b(b < a ), where the radial current density is Jr=Jocosθ, essentially independent of time. (a) What is Φ in the conductor? (b) What is Φ in the insulating region surrounding the conductor? 7.4.3 In a system that stretches to inﬁnity in the ±xand±zdirections, there is a layer of uniformly conducting material having boundaries in the planes y= 0 and y=−a. The region y > 0 is free space, while a potential Φ =Vcosβxis imposed on the boundary at y=−a. (a) Determine Φ in the conducting layer. (b) What is Φ in the region y >0? 7.4.4∗The uniformly conducting material shown in cross-section in Fig. P7.4.4 extends to inﬁnity in the ±zdirections and has the shape of a 90-degree section from a circular cylindrical annulus. At φ= 0 and φ=π/2, it is in contact with grounded electrodes. The boundary at r=ainterfaces free Sec. 7.5 Problems 67 Fig. P7.4.4 Fig. P7.4.5 space, while at r=b, an electrode constrains the potential to be v. Show that the potential in the conductor is Φ =∞X m=1 odd4V mπ[(r/b)2m+ (a/b)4m(b/r)2m] [1 + ( a/b)4m]sin 2mφ (a) 7.4.5 The cross-section of a uniformly conducting material that extends to inﬁn- ity in the ±zdirections is shown in Fig. P7.4.5. The boundaries at r=b, atφ= 0, and at φ=αinterface insulating material. At r=a, voltage sources constrain Φ = −v/2 over the range 0 < φ < α/ 2, and Φ = v/2 over the range α/2< φ < α . (a) Find an inﬁnite set of solutions for Φ that satisfy the boundary con- ditions at the three insulating surfaces. (b) Determine Φ in the conductor. 7.4.6 The system of Fig. P7.4.4 is altered so that there is an electrode on the boundary at r=a. Determine the mutual conductance between this elec- trode and the one at r=b. 7.5 Steady Currents in Piece-Wise Uniform Conductors 7.5.1∗A sphere having uniform conductivity σbis surrounded by material having the uniform conductivity σa. As shown in Fig. P7.5.1, electrodes at “inﬁn- 68 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. P7.5.1 ity” to the right and left impose a uniform current density Joat inﬁnity. Steady conduction prevails. Show that Φ =−JoR σa8 >>< >>:· r R+µ σa−σb 2σa+σb¶¡R r¢2¸ cosθ;R < r µ 3σa σb+2σa¶¡r R¢ cosθ; r < R(a) 7.5.2 Assume at the outset that the sphere of Prob. 7.5.1 is much more highly conducting than its surroundings. (a) As far as the ﬁelds in region (a) are concerned, what is the boundary condition at r=R? (b) Determine the approximate potential in region (a) and compare to the appropriate limiting potential from Prob. 7.5.1. (c) Based on this potential in region (a), determine the approximate po- tential in the sphere and compare to the appropriate limit of Φ as found in Prob. 7.5.1. (d) Now, assume that the sphere is much more insulating than its sur- roundings. Repeat the steps of parts (a)–(c). 7.5.3∗A rectangular box having depth b, length land width much larger than b has an insulating bottom and metallic ends which serve as electrodes. In Fig. P7.5.3a, the right electrode is extended upward and then back over the box. The box is ﬁlled to a depth bwith a liquid having uniform conductivity. The region above is air. The voltage source can be regarded as imposing a potential in the plane z=−lbetween the left and top electrodes that is linear. (a) Show that the potential in the conductor is Φ = −vz/l. (b) In turn, show that in the region above the conductor, Φ = v(z/l)(x− a)/a. (c) What are the distributions of ρuandσu? Sec. 7.5 Problems 69 Fig. P7.5.3 Fig. P7.5.4 (d) Now suppose that the upper electrode is slanted, as shown in Fig. P7.5.3b. Show that Φ in the conductor is unaltered but in the region between the conductor and the slanted plate, Φ = v[(z/l) + (x/a)]. 7.5.4 The structure shown in Fig. P7.5.4 is inﬁnite in the ±zdirections. Each leg has the same uniform conductivity, and conduction is stationary. The walls in the xand in the yplanes are perfectly conducting. (a) Determine Φ ,E, and Jin the conductors. (b) What are Φ and Ein the free space region? (c) Sketch Φ and Ein this region and in the conductors. 7.5.5 The system shown in cross-section by Fig. P7.5.6a extends to inﬁnity in the±xand±zdirections. The material of uniform conductivity σato the right is bounded at y= 0 and y=aby electrodes at zero potential. The material of uniform conductivity σbto the left is bounded in these planes by electrodes each at the potential v. The approach to ﬁnding the ﬁelds is similar to that used in Example 6.6.3. (a) What is Φaasx→ ∞ and Φbasx→ −∞ ? (b) Add to each of these solutions an inﬁnite set such that the boundary conditions are satisﬁed in the planes y= 0 and y=aand as x→ ±∞ . (c) What two boundary conditions relate Φato Φbin the plane x= 0? (d) Use these conditions to determine the coeﬃcients in the inﬁnite series, and hence ﬁnd Φ throughout the region between the electrodes. 70 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. P7.5.5 (e) In the limits σb/greatermuchσaandσb=σa, sketch Φ and E. (A numerical evaluation of the expressions for Φ is not required.) (f) Shown in Fig. P7.5.6b is a similar system but with the conductors bounded from above by free space. Repeat the steps (a) through (e) for the ﬁelds in the conducting layer. 7.6 Conduction Analogs 7.6.1∗In deducing (4) relating the capacitance of electrodes in an insulating mate- rial to the conductance of electrodes having the same shape in a conducting material, it is assumed that not only are the ratios of all dimensions in one situation the same as in the other (the systems are geometrically similar), but that the actual size of the two physical situations is the same. Show that if the systems are again geometrically similar but the length scale of the capacitor is l/epsilon1while that of the conduction cell is lσ, RC = (/epsilon1/σ)(l/epsilon1/lσ). 7.7 Charge Relaxation in Uniform Conductors 7.7.1∗In the two-dimensional conﬁguration of Prob. 4.1.4, consider the ﬁeld tran- sient that results if the region within the cylinder of rectangular cross- section is ﬁlled by a material having uniform conductivity σand permit- tivity /epsilon1. (a) With the initial potential given by (a) of Prob. 4.1.4, with /epsilon1o→/epsilon1and ρoa given constant, show that ρu(x, y, t = 0) is given by (c) of Prob. 4.1.4. (b) Show that for t >0, ρis given by (c) of Prob. 4.1.4 multiplied by exp(−t/τ), where τ=/epsilon1/σ. (c) Show that for t > 0, the potential is given by (a) of Prob. 4.1.4 multiplied by exp( −t/τ). (d) Show that for t >0, the current i(t) from the electrode segment is (f) of Prob. 4.1.4 7.7.2 When t= 0, the only net charge in a material having uniform σand/epsilon1is the line charge of Prob. 4.5.4. As a function of time for t >0, determine Sec. 7.9 Problems 71 the (a) line charge density, (b) charge density elsewhere in the medium, and (c) the potential Φ( x, y, z, t ). 7.7.3∗When t= 0, the charged particle of Example 7.7.2 has a charge q=qo< −qc. (a) Show that, as long as qremains less than −qc, the net current to the particle is i=−µρ /epsilon1q. (b) Show that, as long as q <−qc, q=qoexp(−t/τ1) where τ1=/epsilon1/µρ. 7.7.4 Relative to the potential at inﬁnity on a plane passing through the equator of the particle in Example 7.7.2, what is the potential of the particle when its charge reaches q=qc? 7.8 Electroquasistatic Conduction Laws for Inhomogeneous Materials 7.8.1∗Use an approach similar to that illustrated in this section to show unique- ness of the solution to Poisson’s equation for a given initial distribution ofρand a given potential Φ = Φ Σon the surface S/prime, and a given current density −(σ∇Φ +∂/epsilon1∇Φ/∂t)·n=JΣonS/prime/primewhere S/prime+S/prime/primeencloses the volume of interest V. 7.9 Charge Relaxation in Uniform and Piece-Wise Uniform Systems 7.9.1∗We return to the coaxial circular cylindrical electrode conﬁgurations of Prob. 6.5.5. Now the material in region (2) of each has not only a uniform permittivity /epsilon1but a uniform conductivity σas well. Given that V(t) = ReˆVexp(jωt), (a) show that Ein the ﬁrst conﬁguration of Fig. P6.5.5 is irv/rln (a/b), (b) while in the second conﬁguration, E=ir rReˆv Detnjω/epsilon1o; R < r < a σ+jω/epsilon1;b < r < R(a) where Det = [σ ln(a/R)] +jω[/epsilon1oln(R/b) +/epsilon1ln(a/R)]. (c) Show that in the ﬁrst conﬁguration a length l(into the paper) is equivalent to a conductance Gin parallel with a capacitance Cwhere G=[σα]l ln(a/b); C=[/epsilon1o(2π−α) +/epsilon1α]l ln(a/b)(b) 72 Conduction and Electroquasistatic Charge Relaxation Chapter 7 Fig. P7.9.4 while in the second, it is equivalent to the circuit of Fig. 7.9.5 with Ga= 0; Gb=2πσl ln(R/b) Ca=2π/epsilon1ol ln(a/R); Cb=2π/epsilon1l ln(R/b)(c) 7.9.2 Interpret the conﬁgurations shown in Fig. P6.5.5 as spherical. An outer spherically shaped electrode has inside radius a, while an inner electrode positioned on the same center has radius b. Region (1) is free space while (2) has uniform /epsilon1andσ. (a) For V=Vocos(ωt), determine Ein each region. (b) What are the elements in the equivalent circuit for each? 7.9.3∗Show that the hemispherical electrode of Fig. 7.3.3 is equivalent to a circuit having a conductance G= 2πσain parallel with a capacitance C= 2π/epsilon1a. 7.9.4 The circular cylinder of Fig. P7.9.4a has /epsilon1bandσband is surrounded by material having /epsilon1aandσa. The electric ﬁeld E(t)ixis applied at x=±∞. (a) Find the potential in and around the cylinder and the surface charge density that result from applying a step in ﬁeld to a system that initially is free of charge. (b) Find these quantities for the sinusoidal steady state response. (c) Argue that these ﬁelds are equally applicable to the description of the conﬁguration shown in Fig. P7.9.4b with the cylinder replaced by a half-cylinder on a perfectly conducting ground plane. In the limit where the exterior region is free space while the half-cylinder is so conducting that its charge relaxation time is short compared to times characterizing the applied ﬁeld (1 /ωin the sinusoidal steady state case), what are the approximate ﬁelds in the exterior and in the interior regions? (See Prob. 7.9.5 for a direct calculation of these approximate ﬁelds.) Sec. 7.9 Problems 73 7.9.5∗The half-cylinder of Fig. P7.9.4b has a relaxation time that is short com- pared to times characterizing the applied ﬁeld E(t). The surrounding region is free space ( σa= 0). (a) Show that in the exterior region, the potential is approximately Φa/similarequal −aE(t)£r a−a r¤ cosφ (a) (b) In turn, show that the ﬁeld inside the half-cylinder is approximately Φb/similarequal −2/epsilon1o σdE dtrcosφ (b) 7.9.6 An electric dipole having a z-directed moment p(t) is situated at the origin and at the center of a spherical cavity of free space having a radius ain a material having uniform /epsilon1andσ. When t <0, p= 0 and there is no charge anywhere. The dipole is a step function of time, instantaneously assuming a moment powhen t= 0. (a) An instant after the dipole is established, what is the distribution of Φ inside and outside the cavity? (b) Long after the electric dipole is turned on and the ﬁelds have reached a steady state, what is the distribution of Φ? (c) Determine Φ( r, θ, t ). 7.9.7∗A planar layer of semi-insulating material has thickness d, uniform permit- tivity /epsilon1, and uniform conductivity σ, as shown in Fig. P7.9.7. From below it is bounded by contacting electrode segments that impose the potential Φ =Vcosβx. The system extends to inﬁnity in the ±xand±zdirections. (a) The potential has been applied for a long time. Show that at y= 0, σsu=/epsilon1oV βcosβx/coshβd. (b) When t= 0, the applied potential is turned oﬀ. Show that this un- paired surface charge density decays exponentially from the initial value from part (a) with the time constant τ= (/epsilon1otanhβd+/epsilon1)/σ. Fig. P7.9.7 7.9.8∗Region (b), where y <0, has uniform permittivity /epsilon1and conductivity σ, while region (a), where 0 < y, is free space. Before t= 0 there are no 74 Conduction and Electroquasistatic Charge Relaxation Chapter 7 charges. When t= 0, a point charge Qis suddenly “turned on” at the location ( x, y, z ) = (0 , h,0). (a) Show that just after t= 0, Φa=Q 4π/epsilon1op x2+ (y−h)2+z2−qb 4π/epsilon1op x2+ (y+h)2+z2(a) Φb=qa 4π/epsilon1op x2+ (y−h)2+z2(b) where qb→Q[(/epsilon1//epsilon1o)−1]/[(/epsilon1//epsilon1o) + 1] and qa→2Q/[(/epsilon1//epsilon1o) + 1]. (b) Show that as t→ ∞ , qb→Qand the ﬁeld in region (b) goes to zero. (c) Show that the transient is described by (a) and (b) with qb=Q· 1−µ2/epsilon1o /epsilon1+/epsilon1o¶ exp(−t/τ)¸ (c) qa=Q·2/epsilon1o (/epsilon1+/epsilon1o)¸ exp(−t/τ) ( d) where τ= (/epsilon1o+/epsilon1)/σ. 7.9.9∗The cross-section of a two-dimensional system is shown in Fig. P7.9.9. The parallel plate capacitor to the left of the plane x= 0 extends to x=−∞, with the lower electrode at potential v(t) and the upper one grounded. This upper electrode extends to the right to the plane x=b, where it is bent downward to y= 0 and inward to the plane x= 0 along the surface y= 0. Region (a) is free space while region (b) to the left of the plane x= 0 has uniform permittivity /epsilon1and conductivity σ. The applied voltage v(t) is a step function of magnitude Vo. (a) The voltage has been on for a long-time. What are the ﬁeld and potential distributions in region (b)? Having determined Φb, what is the potential in region (a)? (b) Now, Φ is to be found for t >0. Example 6.6.3 illustrates the approach that can be used. Show that in the limit t→ ∞ ,Φ becomes the result of part (a). (c) In the special case where /epsilon1=/epsilon1o, sketch the evolution of the ﬁeld from the time just after the voltage is applied to the long-time limit of part (a). Fig. P7.9.9 8 MAGNETOQUASISTATIC FIELDS: SUPERPOSITION INTEGRAL AND BOUNDARY VALUE POINTS OF VIEW 8.0 INTRODUCTION MQS Fields: Superposition Integral and Boundary Value Views We now follow the study of electroquasistatics with that of magnetoquasistat- ics. In terms of the ﬂow of ideas summarized in Fig. 1.0.1, we have completed the EQS column to the left. Starting from the top of the MQS column on the right, recall from Chap. 3 that the laws of primary interest are Amp` ere’s law (with the displacement current density neglected) and the magnetic ﬂux continuity law (Table 3.6.1). ∇ ×H=J (1) ∇ ·µoH= 0 (2) These laws have associated with them continuity conditions at interfaces. If the in- terface carries a surface current density K, then the continuity condition associated with (1) is (1.4.16) n×(Ha−Hb) =K (3) and the continuity condition associated with (2) is (1.7.6). n·(µoHa−µoHb) = 0 (4) In the absence of magnetizable materials, these laws determine the magnetic ﬁeld intensity Hgiven its source, the current density J. By contrast with the elec- troquasistatic ﬁeld intensity E,His not everywhere irrotational. However, it is solenoidal everywhere. 1 2Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 The similarities and contrasts between the primary EQS and MQS laws are the topic of this and the next two chapters. The similarities will streamline the development, while the contrasts will deepen the understanding of both MQS and EQS systems. Ideas already developed in Chaps. 4 and 5 will also be applicable here. Thus, this chapter alone plays the role for MQS systems taken by these two earlier chapters for EQS systems. Chapter 4 began by expressing the irrotational Ein terms of a scalar poten- tial. Here His not generally irrotational, although it may be in certain source-free regions. On the other hand, even with the eﬀects of magnetization that are in- troduced in Chap. 9, the generalization of the magnetic ﬂux density µoHhas no divergence anywhere. Therefore, Sec. 8.1 focuses on the solenoidal character of µoH and develops a vector form of Poisson’s equation satisﬁed by the vector potential, from which the Hﬁeld may be obtained. In Chap. 4, where the electric potential was used to represent an irrotational electric ﬁeld, we paused to develop insights into the nature of the scalar potential. Similarly, here we could delve into the way in which the vector potential represents the ﬂux of a solenoidal ﬁeld. For two reasons, we delay developing this interpretation of the vector potential for Sec. 8.6. First, as we see in Sec. 8.2, the superposition integral approach is often used to directly relate the source, the current density, to the magnetic ﬁeld intensity without the intetermediary of a potential. Second, many situations of interest involving current-carrying coils can be idealized by represent- ing the coil wires as surface currents. In this idealization, all of space is current free except for some surfaces within which surface currents ﬂow. But, because His irrotational everywhere except through these surfaces, this means that the Hﬁeld may be expressed as the gradient of a scalar potential. Further, since the magnetic ﬁeld is divergence free (at least as treated in this chapter, which does not deal with magnetizable materials), the scalar potential obeys Laplace’s equation. Thus, most methods developed for EQS systems using solutions to Laplace’s equation can be applied to the solution to MQS problems as well. In this way, we ﬁnd “dual” situations to those solved already in earlier chapters. The method extends to time- varying quasistatic magnetic ﬁelds in the presence of perfect conductors in Sec. 8.4. Eventually, in Chap. 9, we shall extend the approach to problems involving piece-wise uniform and linear magnetizable materials. Vector Field Uniquely Speciﬁed. A vector ﬁeld is uniquely speciﬁed by its curl and divergence. This fact, used in the next sections, follows from a slight modiﬁcation to the uniqueness theorem discussed in Sec. 5.2. Suppose that the vector and scalar functions C(r) and D(r) are given and represent the curl and divergence, respectively, of a vector function F. ∇ ×F=C(r) (5) ∇ ·F=D(r) (6) The same arguments used in this earlier uniqueness proof then shows that Fis uniquely speciﬁed provided the functions C(r) and D(r) are given everywhere and have distributions consistent with Fgoing to zero at inﬁnity. Suppose that Fa andFbare two diﬀerent solutions of (5) and (6). Then the diﬀerence solution Sec. 8.1 Vector Potential 3 Fd=Fa−Fbis both irrotational and solenoidal. ∇ ×Fd= 0 (7) ∇ ·Fd= 0 (8) The diﬀerence solution is governed by the same equations as in Sec. 5.2. With Fdtaken to be the gradient of a Laplacian potential, the remaining steps in the uniqueness argument are equally applicable here. The uniqueness proof shows the importance played by the two diﬀerential vector operations, curl and divergence. Among the many possible combinations of the partial derivatives of the vector components of F, these two particular combi- nations have the remarkable property that their speciﬁcation gives full information about F. In Chap. 4, we determined a vector ﬁeld F=Egiven that the vector source C= 0 and the scalar source D=ρ//epsilon1o. In Secs. 8.1 we ﬁnd the vector ﬁeld F=H, given that the scalar source D= 0 and that the vector source is C=J. The strategy in this chapter parallels that for Chaps. 4 and 5. We can again think of dividing the ﬁelds into two parts, a particular part due to the current density, and a homogeneous part that is needed to satisfy boundary conditions. Thus, with the understanding that the superposition principle makes it possible to take the ﬁelds as the sum of particular and homogeneous solutions, (1) and (2) become ∇ ×Hp=J (9) ∇ ·µoHp= 0 (10) ∇ ×Hh= 0 (11) ∇ ·µoHh= 0 (12) In sections 8.1–8.3, it is presumed that the current density is given everywhere. The resulting vector and scalar superposition integrals provide solutions to (9) and (10) while (11) and (12) are not relevant. In Sec. 8.4, where the ﬁelds are found in free-space regions bounded by perfect conductors, (11) and (12) are solved and boundary conditions are met without the use of particular solutions. In Sec. 8.5, where currents are imposed but conﬁned to surfaces, a boundary value approach is taken to ﬁnd a particular solution. Finally, Sec. 8.6 concludes with an example in which the region of interest includes a volume current density (which gives rise to a particular ﬁeld solution) bounded by a perfect conductor (in which surface currents are induced that introduce a homogeneous solution). 8.1 THE VECTOR POTENTIAL AND THE VECTOR POISSON EQUATION A general solution to (8.0.2) is µoH=∇ ×A (1) 4Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 where Ais the vector potential . Just as E=−gradΦ is the “integral” of the EQS equation curlE= 0, so too is (1) the “integral” of (8.0.2). Remember that we could add an arbitrary constant to Φ without aﬀecting E. In the case of the vector potential, we can add the gradient of an arbitrary scalar function to Awithout aﬀecting H. Indeed, because ∇ ×(∇ψ) = 0, we can replace AbyA/prime=A+∇ψ. The curl of Ais the same as of A/prime. We can interpret (1) as the speciﬁcation of Ain terms of the assumedly known physical Hﬁeld. But as pointed out in the introduction, to uniquely specify a vector ﬁeld, both its curl and divergence must be given. In order to specify Auniquely, we must also give its divergence. Just what we specify here is a matter of convenience and will vary in accordance with the application. In MQS systems, we shall ﬁnd it convenient to make the vector potential solenoidal ∇ ·A= 0 (2) Speciﬁcation of the potential in this way is sometimes called setting the gauge, and with (2) we have established the Coulomb gauge . We turn now to the evaluation of A, and hence H, from the MQS Amp` ere’s law and magnetic ﬂux continuity law, (8.0.1) and (8.0.2). The latter is automatically satisﬁed by letting the magnetic ﬂux density be represented in terms of the vector potential, (1). Substituting (1) into Amp` ere’s law (8.0.1) then gives ∇ ×(∇ ×A) =µoJ (3) The following identity holds. ∇ ×(∇ ×A) =∇(∇ ·A)− ∇2A (4) The reason for deﬁning Aas solenoidal was to eliminate the ∇ ·Aterm in this expression and to reduce (3) to the vector Poisson’s equation . ∇2A=−µoJ (5) The vector Laplacian on the left in this expression is deﬁned in Cartesian co- ordinates as having components that are the scalar Laplacian operating on the respective components of A. Thus, (5) is equivalent to three scalar Poisson’s equa- tions, one for each Cartesian component of the vector equation. For example, the zcomponent is ∇2Az=−µoJz (6) With the identiﬁcation of Az→Φ and µoJz→ρ//epsilon1o, this expression becomes the scalar Poisson’s equation of Chap. 4, (4.2.2). The integral of this latter equation is the superposition integral, (4.5.3). Thus, identiﬁcation of variables gives as the integral of (6) Az=µo 4πZ V/primeJz(r/prime) |r−r/prime|dv/prime(7) and two similar equations for the other two components of A. Reconstructing the vector Aby multiplying (7) by izand adding the corresponding xandycompo- nents, we obtain the superposition integral for the vector potential . Sec. 8.1 Vector Potential 5 A(r) =µo 4πZ V/primeJ(r/prime) |r−r/prime|dv/prime (8) Remember, r/primeis the coordinate of the current density source, while ris the coor- dinate of the point at which Ais evaluated, the observer coordinate. Given the current density everywhere, this integration provides the vector potential. Hence, in principle, the ﬂux density µoHis determined by carrying out the integration and then taking the curl in accordance with (1). The theorem at the end of Sec. 8.0 makes it clear that the solution provided by (8) is indeed unique when the current density is given everywhere. In order that ∇ ×Abe a physical ﬂux density, J(r) cannot be an arbitrary vector ﬁeld. Because div(curl) of any vector is identically equal to zero, the diver- gence of the quasistatic Amp` ere’s law, (8.0.1), gives ∇ ·(∇ ×H) = 0 = ∇ ·Jand thus ∇ ·J= 0 (9) The current distributions of magnetoquasistatics must be solenoidal. Of course, we know from the discussion of uniqueness given in Sec. 8.0 that (9) does not uniquely specify the current distribution. In an Ohmic conductor, sta- tionary current distributions satisfying (9) were determined in Secs. 7.1–7.5. Thus, any of these distributions can be used in (8). Even under dynamic conditions, (9) remains valid for MQS systems. However, in Secs. 8.4–8.6 and as will be discussed in detail in Chap. 10, if time rates of change become too rapid, Faraday’s law de- mands a rotational electric ﬁeld which plays a role in determining the distribution of current density. For now, we assume that the current distribution is that for steady Ohmic conduction. Two-Dimensional Current and Vector Potential Distributions. Suppose a current distribution J=izJz(x, y) exists through all of space. Then the vector potential is zdirected, according to (8), and its zcomponent obeys the scalar Poisson equation Az=µo 4πZJz(x/prime, y/prime)dv/prime |r−r/prime|(10) But this is formally the same expression, (4.5.3), as that of the scalar potential produced by a charge distribution ρ(x/prime, y/prime). Φ =1 4π/epsilon1oZρ(x/prime, y/prime)dv/prime |r−r/prime|(11) It was inconvenient to integrate the above equation directly. Instead, we determined the ﬁeld of a line charge from symmetry and Gauss’ law and integrated the resulting expression to obtain the potential (4.5.18) Φ =−λl 2π/epsilon1oln¡r ro¢ (12) 6Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 where ris the distance from the line charge r=p (x−x/prime)2+ (y−y/prime)2andro is the reference radius. The scalar potential can thus be evaluated from the two- dimensional integral Φ =−1 2π/epsilon1oZ Z ρ(x/prime, y/prime)lnµp (x−x/prime)2+ (y−y/prime)2/ro¶ dx/primedy/prime(13) The vector potential of a two-dimensional z-directed current distribution obeys the same equation and thus has a solution by analogy, after a proper interchange of parameters. Az=−µo 2πZ Jz(x/prime, y/prime)lnµp (x−x/prime)2+ (y−y/prime)2/ro¶ dx/primedy/prime(14) Two important consequences emerge from this derivation. (a) Every two-dimensional EQS potential Φ( x, y) produced by a given charge distribution ρ(x, y), has an MQS analog vector potential Az(x, y) caused by a current density Jz(x, y) with the same spatial distribution as ρ(x, y). The magnetic ﬁeld follows from (1) and thus µoH=∇ ×A=µ ix∂ ∂x+iy∂ ∂y¶ ×izAz =−iz×µ ix∂Az ∂x+iy∂Az ∂y¶ =−iz× ∇Az(15) Therefore the lines of magnetic ﬂux density are perpendicular to the gradient ofAz. A plot of ﬁeld lines and equipotential lines of the EQS problem is trans- formed into a plot of an MQS ﬁeld problem by interpreting the equipotential lines as the lines of magnetic ﬂux density. Lines of constant Azare lines of magnetic ﬂux. (b) The vector potential of a line current of magnitude ialong the zdirection is given by analogy with (12), Az=−µo 2πi ln(r/ro) (16) which is consistent with the magnetic ﬁeld H=iφ(i/2πr) given by (1.4.10), if one makes use of the curl expression in polar coordinates, µoH=1 r∂Az ∂φir−∂Az ∂riφ (17) The following illustrates the integration called for in (8). The ﬁelds associated with singular current distributions will be used in later sections and chapters. Example 8.1.1. Field Associated with a Current Sheet Sec. 8.1 Vector Potential 7 Fig. 8.1.1 Cross-section of surfaces of constant Azand lines of mag- netic ﬂux density for the uniform sheet of current shown. Az-directed current density is uniformly distributed over a strip located between x2 andx1as shown in Fig. 8.1.1. The thickness of the sheet, ∆, is very small compared to other dimensions of interest. So, the integration of (14) in the ydirection amounts to a multiplication of the current density by ∆. The vector potential is therefore determined by completing the integration on x/prime Az=−µoKo 2πZx1 x2lnµp (x−x/prime)2+y2/ro¶ dx/prime(18) where Ko≡Jz∆. This integral is carried out in Example 4.5.3, where the two dimensional elec- tric potential of a charged strip was determined. Thus, with σo//epsilon1o→µoKo, (4.5.24) becomes the desired vector potential. The proﬁles of surfaces of constant Azare shown in Fig. 8.1.1. Remember, these are also the lines of magnetic ﬂux density, µoH. Example 8.1.2. Two-Dimensional Magnetic Dipole Field A pair of closely spaced conductors carrying oppositely directed currents of mag- nitude iis shown in Fig. 8.1.2. The currents extend to + and −inﬁnity in the z direction, so the resulting ﬁelds are two-dimensional and can be represented by Az. In polar coordinates, the distance from the right conductor, which is at a distance dfrom the zaxis, to the observer location is essentially r−dcosφ. The Azfor each wire takes the form of (16), with rthe distance from the wire to the point of observation. Thus, superposition of the vector potentials due to the two wires gives Az=−µoi 2π[ln(r−dcosφ)−lnr] =−µoi 2πln¡ 1−d rcosφ) (19) In the limit d/lessmuchr, this expression becomes Az=µoid 2πcosφ r(20) 8Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.1.2 A pair of wires having the spacing dcarry the current iin opposite directions parallel to the zaxis. The two-dimensional dipole ﬁeld is shown in Fig. 8.1.3. Fig. 8.1.3 Cross-sections of surfaces of constant Azand hence lines of magnetic ﬂux density for conﬁguration of Fig. 8.1.2. Thus, the surfaces of constant Azhave intersections with planes of constant zthat are circular, as shown in Fig. 8.1.3. These are also the lines of magnetic ﬂux density, which follow from (17). µoH=µoid 2πµ −sinφ r2ir+cosφ r2iφ¶ (21) If the line currents are replaced by line charges, the resulting equipotential lines (intersections of the equipotential surfaces with the x−yplane) coincide with the magnetic ﬁeld lines shown in Fig. 8.1.3. Thus, the lines of electric ﬁeld intensity for the electric dual of the magnetic conﬁguration shown in Fig. 8.1.3 originate on the positive line charge on the right and terminate on the negative line charge at the left, following lines that are perpendicular to those shown. 8.2 THE BIOT-SAVART SUPERPOSITION INTEGRAL Once the vector potential has been determined from the superposition integral of Sec. 8.1, the magnetic ﬂux density follows from an evaluation of curlA. However, in certain ﬁeld evaluations, it is best to have a superposition integral for the ﬁeld itself. For example, in numerical calculations, numerical derivatives should be avoided. Sec. 8.2 The Biot-Savart Integral 9 The ﬁeld superposition integral follows by operating on the vector potential as given by (8.1.8) before the integration has been carried out. H=1 µo∇ ×A=1 4π∇ ×Z V/prime·J(r/prime) |r−r/prime|¸ dv/prime(1) The integration is with respect to the source coordinates denoted by r/prime, while the curloperation involves taking derivatives with respect to the observer coordinates r. Thus, the curl operation can be carried out before the integral is completed, and (1) becomes H=1 4πZ V/prime∇ ×·J(r/prime) |r−r/prime|¸ dv/prime(2) Thecurl operation required to evaluate the integrand in this expression can be carried out without regard for the particular dependence of the current density because the derivatives are with respect to r, not r/prime. To make this evaluation, observe that the curl operates on the product of the vector Jand the scalar ψ= |r−r/prime|−1, and that operation obeys the vector identity ∇ ×(ψJ) =ψ∇ ×J+∇ψ×J (3) Because Jis independent of r, the ﬁrst term on the right is zero. Thus, (2) becomes H=1 4πZ V/prime∇µ1 |r−r/prime|¶ ×Jdv/prime(4) To evaluate the gradient in this expression, consider the special case when r/prime is at the origin in a spherical coordinate system, as shown in Fig. 8.2.1. Then ∇(1/r) =−1 r2ir (5) where iris the unit vector directed from the source coordinate at the origin to the observer coordinate at ( r, θ, φ ). We now move the source coordinate from the origin to the arbitrary location r/prime. Then the distance rin (5) is replaced by the distance |r−r/prime|. To replace the unit vector ir, the source-observer unit vector ir/primeris deﬁned as being directed from an arbitrary source coordinate to the observer coordinate P. In terms of this source- observer unit vector, illustrated in Fig. 8.2.2, (5) becomes ∇µ1 r−r/prime¶ =−ir/primer |r−r/prime|2(6) Substitution of this expression into (4) gives the Biot-Savart Law for the magnetic ﬁeld intensity. H=1 4πZ V/primeJ(r/prime)×ir/primer |r−r/prime|2dv/prime (7) 10Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.2.1 Spherical coordinate system with r/primelocated at origin. Fig. 8.2.2 Source coordinate r/primeand observer coordinate rshowing unit vec- torir/primerdirected from r/primetor. In evaluating the integrand, the cross-product is evaluated at the source coordinate r/prime. The integrand represents the contribution of the current density at r/primeto the ﬁeld atr. The following examples illustrate the Biot-Savart law. Example 8.2.1. On Axis Field of Circular Cylindrical Solenoid The cross-section of an N-turn solenoid of axial length dand radius ais shown in Fig. 8.2.3. There are many turns, so the current ipassing through each is essentially φdirected. To keep the integration simple, we conﬁne ourselves to ﬁnding Hon the zaxis, which is the axis of symmetry. In cylindrical coordinates, the source coordinate incremental volume element isdv/prime=r/primedφ/primedr/primedz/prime. For many windings uniformly distributed over a thickness ∆, the current density is essentially the total number of turns multiplied by the current per turn and divided by the area through which the current ﬂows. J∼=iφNi ∆d(8) The superposition integral, (7), is carried out ﬁrst on r/prime. This extends from r/prime=a tor/prime=a+ ∆ over the radial thickness of the winding. Because ∆ /lessmucha, the source- observer distance and direction remain essentially constant over this interval, and so the integration amounts to a multiplication by ∆. The axial symmetry requires thatHon the zaxis be zdirected. The integration over z/primeandφ/primeis Hz=1 4πZ−d/2 d/2Z2π 0¡Ni d¢(iφ×ir/primer)z |r−r/prime|2adφ/primedz/prime(9) In terms of the angle αshown in Fig. 8.2.3 and its inset, the source-observer unit vector is ir/primer=−irsinα−izcosα (10) Sec. 8.2 The Biot-Savart Integral 11 Fig. 8.2.3 A solenoid consists of Nturns uniformly wound over a length d, each turn carrying a current i. The ﬁeld is calculated along thezaxis, so the observer coordinate is at ron the zaxis. so that (iφ×ir/primer)z= sin α=ap a2+ (z/prime−z)2;|r−r/prime|2=a2+ (z/prime−z)2(11) The integrand in (9) is φ/primeindependent, and the integration over φ/primeamounts to multiplication by 2 π. Hz=Ni 2dZd/2 −d/2a2dz/prime [a2+ (z/prime−z)2]3/2(12) With the substitution z/prime/prime=z/prime−z, it follows that Hz=Ni 2dz/prime/prime p a2+z/prime/prime2]d 2−z −d 2−z =Ni 2d·d 2a−z aq 1 +¡d 2a−z a¢2+d 2a+z aq 1 +¡d 2a+z a¢2¸(13) In the limit where d/2a/lessmuch1, the solenoid becomes a circular coil with Nturns concentrated at r=ain the plane z= 0. The ﬁeld intensity at the center of this coil follows from (13) as the amp-turns divided by the loop diameter. Hz→Ni 2a(14) Thus, a 100-turn circular loop having a radius a= 5 cm (that is large compared to its axial length d) and carrying a current of i= 1 A would have a ﬁeld intensity of 12Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.2.4 Experiment for documenting the axial Hpredicted in Ex- ample 8.2.1. Proﬁle of normalized Hzis for d/2a= 2.58. 1000 A/m at its center. The ﬂux density measured by a magnetometer would then beBz=µoHz= 4π×10−7(1000) tesla = 4 πgauss. Further implications of this ﬁnding are discussed in the following demonstra- tion. Demonstration 8.2.1. Fields of a Circular Cylindrical Solenoid The solenoid shown in Fig. 8.2.4 has N= 141 turns, an axial length d= 70 .5 cm, and a radius a= 13.6 cm. A Hall-type magnetometer measures the magnitude and direction of Hin and around the coil. The on-axis distribution of Hzpredicted by (13) for the experimental length-to-diameter ratio d/2a= 2.58 is shown in Fig. 8.2.4. With i= 1 amp, the ﬂux density at the center approaches 2.5 gauss. The accuracy with which theory and experiment agree is likely to be limited only by such matters as the care with which the probe can be mounted and the calibration of the magnetometer. Care must also be taken that there are no magnetizable materials, such as iron, in the vicinity of the coil. To avoid contributions from the earth’s magnetic ﬁeld (which is on the order of a gauss), ac ﬁelds should be used. If ac is used, there should be no large conducting objects near by in which eddy currents might be induced. (Magnetization and eddy currents, respectively, are taken up in the next two chapters.) The inﬁnitely long solenoid can be regarded as the analog for MQS systems of the “plane parallel plate capacitor.” Just as the capacitor can be constructed to create a uniform electric ﬁeld between the plates with zero ﬁeld outside the region bounded by the plates, so too the long solenoid gives rise to a uniform magnetic ﬁeld throughout the interior region and an exterior ﬁeld that is zero. This can be seen by probing the ﬁeld not only as a function of axial position but of radius as well. For the ﬁnite length solenoid, the on-axis interior ﬁeld designated by H∞in Fig. 8.2.4 is given by (13) for locations on the zaxis where d/2/greatermuchz. Hz→H∞≡· d/2aq 1 +¡d 2a¢2¸ Ni d(15) In the limit where the solenoid is also very long compared to its radius, where d/2a/greatermuch1, this expression becomes H∞→Ni d(16) Sec. 8.2 The Biot-Savart Integral 13 Fig. 8.2.5 A line current iis uniformly distributed over the length of the vector aoriginating at r+band terminating at r+c. The resulting magnetic ﬁeld intensity is determined at the observer position r. Probing of the ﬁeld shows the ﬁeld maintains the value and direction of (16) over the interior cross-section as well. It also shows that the magnetic ﬁeld intensity just outside the windings at an axial location that is several radii afrom the coil ends is relatively small. Continuity of magnetic ﬂux requires that the total ﬂux passing through the solenoid in the zdirection must be returned in the −zdirection outside the solenoid. How, then, can the exterior ﬁeld of a long solenoid be negligible compared to that inside? The outside ﬂux returns in the −zdirection through a much larger exte- rior area than the area πa2through which the interior ﬂux passes. In fact, as the coil becomes inﬁnitely long, this return ﬂux spreads out over an exterior area that stretches to inﬁnity in the xandydirections. The ﬁeld intensity just outside the winding tends to zero as the coil is made very long. Stick Model for Computing Fields of Electromagnet. The Biot-Savart su- perposition integral can be completed analytically for relatively few conﬁgurations. Nevertheless, its evaluation amounts to no more than a summation of the ﬁeld con- tributions from each of the current elements. Thus, on the computer, its evaluation is a straightforward matter. Many practical current distributions are, or can be approximated by, con- nected straight-line current segments, or current “sticks.” We will now use the Biot-Savart law to ﬁnd the ﬁeld at an arbitrary observer position rassociated with a current stick having an arbitrary location. The result is a practical resource, be- cause a numerical summation over diﬀerential volume current elements can then be replaced by one over the sticks. The current stick, shown in Fig. 8.2.5, is represented by a vector a. Thus, the current is uniformly distributed between the base of this vector at r+band the tip of the vector at r+c. The source coordinate r/primeis located along the current stick. The objective in the following paragraphs is to carry out an integration over the length of the current stick and obtain an expression for H(r). Because the current stick does not represent a solenoidal current density at its ends, the ﬁeld derived is of physical signiﬁcance only if used in conjunction with other current sticks that together represent a continuous current distribution. The detailed view of the current stick, Fig. 8.2.6, shows the source coordinate ξdenoting the position along the stick. The origin of this coordinate is at the point 14Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.2.6 View of current element from Fig. 8.2.5 in plane containing band c, and hence a. on a line through the stick that is closest to the observer coordinate. The projection of bonto a vector aisξb=a·b/|a|. Thus, the current stick begins at this distance from ξ= 0, as shown in Fig. 8.2.6, and terminates at ξc, the projection of conto the axis of a, as also shown. The cross-product c×a/|a|is perpendicular to the plane of Fig. 8.2.6 and equal in magnitude to the projection of conto a vector that is perpendicular to aand in the plane of Fig. 8.2.6. Thus, the shortest distance between the observer position and the axis of the current stick is ro=|c×a|/|a|. It follows from this fact and the deﬁnition of the cross-product that ds×ir/primer=dξ£c×a |a|¤ |r−r/prime|(17) where dsis the diﬀerential along the line current and |r−r/prime|= (ξ2+r2 o)1/2 Integration of the Biot-Savart law, (7), is ﬁrst performed over the cross-section of the stick. The cross-sectional dimensions are small, so during this integration, the integrand remains essentially constant. Thus, the current density is replaced by the total current and the integral reduced to one on the axial coordinate ξof the stick. H=i 4πZξc ξbds×ir/primer |r−r/prime|2(18) In view of (17), this integral is expressed in terms of the source coordinate integra- tion variable ξas H=i 4πZξc ξbc×adξ |a|(ξ2+r2o)3/2(19) Sec. 8.2 The Biot-Savart Integral 15 Fig. 8.2.7 A pair of square N-turn coils produce a ﬁeld at Pon the zaxis that is the superposition of the ﬁelds Hzdue to the eight linear elements comprising the coils. The coils are centered on the zaxis. This integral is carried out to obtain H=i 4πc×a |a|·ξ r2o[ξ2+r2o]1/2¸ξc ξb(20) In evaluating this expression at the integration endpoints, note that by deﬁnition, (ξ2 c+r2 o)1/2=|c|; ( ξ2 b+r2 o)1/2=|b| (21) so that (20) becomes an expression for the ﬁeld intensity at the observer location expressed in terms of vectors a,b, and cthat serve to deﬁne the relative location of the current stick.1 H=i 4πc×a |c×a|2µa·c |c|−a·b |b|¶ (22) The following illustrates how this expression can be used repetitively to determine the ﬁeld induced by currents represented in a piece-wise fashion by current sticks. Expressed in Cartesian coordinates, the vectors are a convenient way to specify the sticks making up a complex winding. On the computer, the evaluation of (22) is then conveniently carried out by a subroutine that is used many times. Example 8.2.2. Axial Field of a Pair of Square Coils Shown in Fig. 8.2.7 is a pair of coils, each having Nturns carrying a current iin such a direction that the ﬁelds induced by each coil reinforce along the zaxis. The four linear sections of the two coils comprise the sides of a cube, centered at the origin and with dimensions 2 d. 1Private communication, Mr. John G. Aspinall. 16Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.2.8 Demonstration of axial ﬁeld generated by pair of square coils having spacing equal to the side lengths. We conﬁne ourselves to ﬁnding Halong the zaxis where, by symmetry, it has only a zcomponent. Thus, for an observer at (0 ,0, z), the vectors specifying element (1) of the right-hand coil in Fig. 8.2.7 are a= 2dix b=−dix+diy+ (d−z)iz (23) c=dix+diy+ (d−z)iz Evaluation of the zcomponent of (22) then gives the part of Hzdue to element (1). Because of the axial symmetry, the ﬁeld induced by elements (2), (3), and (4) in the same coil are the same as already found for element (1). The ﬁeld induced by element (5) in the second coil is similarly found starting from vectors that are the same as in (23), except that d→ −din the zcomponents of bandc. Here too, the other three elements each contribute the same ﬁeld as already found. Thus, the axial ﬁeld intensity, the sum of the contributions from the individual coils, is Hz=−2iN πd½ 1£¡ 1−z d¢2+ 1¤£ 2 +¡ 1−z d¢2¤1/2 +1£¡ 1 +z d¢2+ 1¤£ 2 +¡ 1 +z d¢2¤1/2¾ (24) This distribution is plotted on the inset to Fig. 8.2.8. Because the ﬁelds induced by the separate coils reinforce, the pair can be used to produce a relatively uniform ﬁeld in the midregion. Sec. 8.3 Scalar Magnetic Potential 17 Demonstration 8.2.2. Field of Square Pair of Coils In the experiment of Fig. 8.2.8, the axial ﬁeld is probed by means of a Hall magne- tometer. The output is connected to the vertical trace of a high persistence scope. The probe is mounted on a carriage that is attached to a potentiometer in such a way that there is an output voltage proportional to the horizontal position of the probe. This is used to control the horizontal scope deﬂection. The result is a trace that follows the predicted contour. The plot is shown in terms of normalized coordi- nates that can be used to compare theory to experiment using any size of coils and any level of current. 8.3 THE SCALAR MAGNETIC POTENTIAL The vector potential Adescribes magnetic ﬁelds that possess curl wherever there is a current density J(r). In the space free of current, ∇ ×H= 0 (1) and thus Hought to be derivable there from the gradient of a potential. H=−∇Ψ (2) Because ∇ ·µoH= 0 (3) we further have ∇2Ψ = 0 (4) The potential obeys Laplace’s equation. Example 8.3.1. The Scalar Potential of a Line Current A line current is a source singularity (at the origin of a polar coordinate system if it is placed along its zaxis). From Amp` ere’s integral law applied to the contour C of Fig. 1.4.4, we have I CH·ds= 2πrH φ=Z SJ·da=i (5) and thus Hφ=i 2πr(6) It follows that the potential Ψ that has Hφof (6) as the negative of its gradient is Ψ =−i 2πφ (7) 18Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.3.1 Surface spanning loop, contour following loop, and contour for H H·ds. Note that the potential is multiple valued as the origin is encircled more than once. This property reﬂects the fact that strictly, His not curl free in all of space. As the origin is encircled, Amp` ere’s integral law identiﬁes Jas the source of the curl of H. Because Ψ is a solution to Laplace’s equation, it must possess an EQS analog. The electroquasistatic potential Φ =−V 2πφ; 0 < φ < 2π (8) describes the fringing ﬁeld of a capacitor of semi-inﬁnite extent, extending from x= 0 to x= +∞, with a voltage Vacross the plates, in the limit as the spacing between the plates is negligible (Fig. 5.7.2 with Vreversed in sign). It can also be interpreted as the ﬁeld of a semi-inﬁnite dipole layer with the dipole density πs=σsd=/epsilon1oVdeﬁned by (4.5.27), where dis the spacing between the surface charge densities, ±σs, on the outside surfaces of the semi-inﬁnite plates (Fig. 5.7.2 with the signs of the charges reversed). We now have further opportunity to relate Hﬁelds of current-carrying wires to EQS analogs involving dipole layers. The Scalar Potential of a Current Loop. A current loop carrying a current ihas a magnetic ﬁeld that is curl free everywhere except at the location of the wire. We shall now determine the scalar potential produced by the current loop. The line integralH H·dsenclosing the current does not give zero, and hence paths that enclose the current in the loop are not allowed, if the potential is to be single valued. Suppose that we mount over the loop a surface Sspanning the loop which is not crossed by any path of integration. The actual shape of the surface is arbitrary, but the contour Clis deﬁned by the wire which is its edge. The potential is then made single valued. The discontinuity of potential across the surface follows from Amp` ere’s law Z CH·ds=i (9) where the broken circle on the integral sign is to indicate a path as shown in Fig. 8.3.1 that goes from one side of the surface to a point on the opposite side. Thus, the potential Ψ of a current loop has the discontinuity Z H·ds=Z (−∇Ψ)·ds= ∆Ψ = i (10) Sec. 8.3 Scalar Magnetic Potential 19 Fig. 8.3.2 Solid angle for observer at rdue to current loop at r/prime. We have found in electroquasistatics that a uniform dipole layer of magnitude πson a surface Sproduces a potential that experiences a constant potential jump πs//epsilon1oacross the surface, (4.5.31). Its potential was (4.5.30) Φ(r) =πs 4π/epsilon1oΩ (11) where Ω is the solid angle subtended by the rim of the surface as seen by an observer at the point r. Thus, we conclude that the scalar potential Ψ, a solution to Laplace’s equation with a constant jump iacross the surface Sspanning the wire loop, must have a potential jump πs//epsilon1o→i, and hence the solution Ψ(r) =i 4πΩ(12) where again the solid angle is that subtended by the contour along the wire as seen by an observer at the point ras shown by Fig. 8.3.2. In the example of a dipole layer, the surface Sspeciﬁed the physical distribution of the dipole layer. In the present case, Sis arbitrary as long as it spans the contour Cof the wire. This is consistent with the fact that the solid angle Ω is invariant with respect to changes of the surface Sand depends only on the geometry of the rim. Example 8.3.2. TheHField of Small Loop Consider a small loop of area aat the origin of a spherical coordinate system with the normal to the surface parallel to the zaxis. According to (12), the scalar potential of the loop is then Ψ =i 4πir·iza r2=ia 4πcosθ r2(13) 20Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 This is the potential of a dipole. The Hﬁeld follows from using (2) H=−∇Ψ =ia 4πr3[2 cos θir+ sin θiθ] (14) As far as its ﬁeld around and far from the loop is concerned, the current loop can be viewed as if it were a “magnetic” dipole, consisting of two equal and opposite magnetic charges ±qmspaced a distance dapart (Fig. 4.4.1 with q→qm). The magnetic charges (monopoles) are sources of divergence of the magnetic ﬂux µoH analogous to electric charges as sources of divergence of the displacement ﬂux density /epsilon1oE. Thus, if Maxwell’s equations are modiﬁed to include the action of a magnetic charge density ρm= lim ∆V→0qm ∆V in units of voltsec/m4, then the new magnetic Gauss’ law must be ∇ ·µoH=ρm (15) in analogy with ∇ ·/epsilon1oE=ρ (16) Now, magnetic monopoles have been postulated by Dirac, and recent searches for the existence of such monopoles have been apparently successful2. Because the search is so diﬃcult, it is apparent that, if they exist at all, they are very rare in nature. Here the introduction of magnetic charge is a matter of convenience so that the ﬁeld produced by a small current loop can be pictured as the ﬁeld of a magnetic dipole . This can serve as a mnemonic for the reconstruction of the ﬁeld. Thus, if it is remembered that the potential of the electric dipole is Φ =p·ir/primer 4π/epsilon1o|r−r/prime|2(17) the potential of a magnetic dipole can be easily recalled as Ψ =pm·ir/primer 4πµo|r−r/prime|2(18) where pm≡qmd=µoia=µom (19) The magnetic dipole moment is deﬁned as the product of the magnetic charge, qm, and the separation, d, or by µotimes the current times the area of the current loop. Another symbol is used commonly for the “dipole moment” of a current loop, m≡ia, the product of the current times the area of the loop without the factor µo. The reader must gather from the context whether the words dipole moment refer topmorm=pm/µo. The magnetic ﬁeld intensity Hof a magnetic dipole at the origin, (14), is H=−∇Ψ =m 4πr3(2 cos θir+ sin θiθ) (20) Of course, the details of the ﬁeld produced by the current loop and the magnetic charge-dipole diﬀer in the near ﬁeld. One has ∇ ·µoH/negationslash= 0, and the other has a solenoidal Hﬁeld. 2Science Vol. 216, (June 4, 1982). Sec. 8.4 Perfect Conductors 21 8.4 MAGNETOQUASISTATIC FIELDS IN THE PRESENCE OF PERFECT CONDUCTORS There are physical situations in which the current distribution is not prespeciﬁed but is given by some equivalent information. Thus, for example, a perfectly conduct- ingbody in a time-varying magnetic ﬁeld supports surface currents that shield the Hﬁeld from the interior of the body. The eﬀect of the conductor on the magnetic ﬁeld is reminiscent of the EQS situations of Sec. 4.6, where charges distributed themselves on the surface of a conductor in such a way as to shield the electric ﬁeld out of the material. We found in Chap. 7 that the EQS model of a perfect conductor described the low-frequency response of systems in the sinusoidal steady state, or the long-time response to a step function drive. We will ﬁnd in Chap. 10 that the MQS model of a perfect conductor represents the high-frequency sinusoidal steady state response or the short-time response to a step drive. Usually, we use the model of perfect conductivity to describe bodies of high but ﬁnite conductivity. The value of conductivity which justiﬁes use of the perfect conductor model depends on the frequency (or time scale in the case of a transient) as well as the geometry and size, as will be seen in Chap. 10. When the material is cooled to the point where it becomes superconducting, a type I superconductor (for example lead) expels any mangetic ﬁeld that might have originally been within its interior, while showing zero resistance to currrent ﬂow. Thus, even for dc, the material acts on the magnetic ﬁeld like a perfect conductor. However, type I mate- rials also act to exclude the ﬂux from the material, so they should be regarded as perfect conductors in which ﬂux cannot be trapped. The newer “high temperature ceramic superconductors,” such as Y 1Ba2Cu3O7, show a type II regime. In this class of superconductors, there can be trapped ﬂux if the material is cooled in a dc ﬁeld. “High temperature superconductors” are those that show a zero resistance at temperatures above that of liquid nitrogen, 77 degrees Kelvin. As for EQS systems, Faraday’s continuity condition, (1.6.12), requires that the tangential Ebe continuous at a boundary between free space and a conductor. By deﬁnition, a stationary perfect conductor cannot have an electric ﬁeld in its interior. Thus, in MQS as well as EQS systems, there can be no tangential Eat the surface of a perfect conductor. But the primary laws determining Hin the free space region, Amp` ere’s law with J= 0 and the ﬂux continuity condition, do not involve the electric ﬁeld. Rather, they involve the magnetic ﬁeld, or perhaps the vector or scalar potential. Thus, it is desirable to also state the boundary condition in terms of Hor Ψ. Boundary Conditions and Evaluation of Induced Surface Current Den- sity. To identify the boundary condition on the magnetic ﬁeld at the surface of a perfect conductor, observe ﬁrst that the magnetic ﬂux continuity condition requires that if there is a time-varying ﬂux density n·µoHnormal to the surface on the free space side, then there must be the same ﬂux density on the conductor side. But this means that there is then a time-varying ﬂux density in the volume of the perfect conductor. Faraday’s law, in turn, requires that there be a curlofEin the conductor. For this to be true, Emust be ﬁnite there, a contradiction of our deﬁni- 22Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.4.1 Perfectly conducting circular cylinder of radius Rin a mag- netic ﬁeld that is ydirected and of magnitude Hofar from the cylinder. tion of the perfect conductor. We conclude that there can be no normal component of a time-varying magnetic ﬂux density at a perfectly conducting surface . n·µoH= 0 (1) Correspondingly, if the Hﬁeld is the gradient of the scalar potential Ψ, we ﬁnd that∂Ψ ∂n= 0 (2) on the surface of a perfect conductor. This should be contrasted with the boundary condition for an EQS potential Φ which must be constant on the surface of a perfect conductor. This boundary condition can be used to determine the magnetic ﬁeld distribution in the neighborhood of a perfect conductor. Once this has been done, Amp` ere’s continuity condition, (1.4.16), can be used to ﬁnd the surface current density that has been induced by the time-varying magnetic ﬁeld. With ndirected from the perfect conductor into the region of free space, K=n×H (3) Because there is no time-varying magnetic ﬁeld in the conductor, only the tangential ﬁeld intensity on the free space side of the surface is required in this evaluation of the surface current density. Example 8.4.1. Perfectly Conducting Cylinder in a Uniform Magnetic Field A perfectly conducting cylinder having radius Rand extending to z=±∞ is immersed in a uniform time-varying magnetic ﬁeld. This ﬁeld is ydirected and has intensity Hoat inﬁnity, as shown in Fig. 8.4.1. What is the distribution of Hin the neighborhood of the cylinder? In the free space region around the cylinder, there is no current density. Thus, the ﬁeld can be written as the gradient of a scalar potential (in two dimensions) H=−∇Ψ (4) The far ﬁeld has the potential Ψ =−Hoy=−Horsinφ; r→ ∞ (5) Sec. 8.4 Perfect Conductors 23 Fig. 8.4.2 Lines of magnetic ﬁeld intensity for perfectly conducting cylinder in transverse magnetic ﬁeld. The condition ∂Ψ/∂n= 0 on the surface of the cylinder suggests that the boundary condition at r=Rcan be satisﬁed by adding to (5) a dipole solution proportional to sin φ/r. By inspection, Ψ =−HosinφR¡r R+R r¢ (6) has the property ∂Ψ/∂r= 0 at r=R. The magnetic ﬁeld follows from (6) by taking its negative gradient H=−∇Ψ =Hosinφir¡ 1−R2 r2¢ +Hocosφiφ¡ 1 +R2 r2¢ (7) The current density induced on the surface of the cylinder, and responsible for generating the magnetic ﬁeld that excludes the ﬁeld from the interior of the cylinder, is found by evaluating (3) at r=R. K=n×H=izHφ(r=R) =iz2Hocosφ (8) The ﬁeld intensity of (7) and this surface current density are shown in Fig. 8.4.2. Note that the polarity of Kis such that it gives rise to a magnetic dipole ﬁeld that tends to buck out the imposed ﬁeld. Comparison of (7) and the ﬁeld of a two-dimensional dipole, (8.1.21), shows that the induced moment is id= 2πHoR2. 24Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.4.3 A coil having terminals at (a) and (b) links ﬂux through surface enclosed by a contour composed of C1adjacent to the perfectly conducting material and C2completing the circuit between the terminals. The direction of positive ﬂux is that of da, deﬁned with respect to dsby the right-hand rule (Fig. 1.4.1). For the eﬀect of magnetic induction to be negligible in the neighborhood of the terminals, the coil should have many turns, as shown by the inset. There is an analogy to steady conduction ( H↔J) in the neighborhood of an insulating rod immersed in a conductor carrying a uniform current density. In Demonstration 7.5.2, an electric dipole ﬁeld also bucked out an imposed uniform ﬁeld ( J) in such a way that there was no normal ﬁeld on the surface of a cylinder. Voltage at the Terminals of a Perfectly Conducting Coil. Faraday’s law was the underlying reason for the vanishing of the ﬂux density normal to a perfect conductor. By stating this boundary condition in terms of the magnetic ﬁeld alone, we have been able to formulate the magnetic ﬁeld of perfect conductors without explicitly solving for the distribution of electric ﬁeld intensity. It would seem that for the determination of the voltage induced by a time-varying magnetic ﬁeld at the terminals of the coil, knowledge of the Eﬁeld would be necessary. In fact, as we now take care to deﬁne the circumstances required to make the terminal voltage of a coil a well-deﬁned variable, we shall see that we can put oﬀ the detailed determination ofEfor Chap. 10. The EMF at point (a) relative to that at point (b) was deﬁned in Sec. 1.6 as the line integral of E·dsfrom (a) to (b). In Sec. 4.1, where the electric ﬁeld was irrotational, this integral was then deﬁned as the voltage at point (a) relative to (b). We shall continue to use this terminology, which is consistent with that used in circuit theory. If the voltage is to be a well-deﬁned quantity, independent of the layout of the connecting wires, the terminals of the coil shown in Fig. 8.4.3 must be in a region where the magnetic induction is negligible compared to that in other regions and where, as a result, the electric ﬁeld is irrotational. To determine the voltage, the integral form of Faraday’s law, (1.6.1), is applied to the closed line integral C shown in Fig. 8.4.3.I CE·ds=−d dtZ SµoH·da (9) Sec. 8.4 Perfect Conductors 25 The contour goes from the terminal at (a) to that at (b) along the coil wire and closes through a path outside the coil. However, we know that Eis zero along the perfectly conducting wire. Hence, the entire contribution to the line integral comes from the short path between the terminals. Thus, the left side of (9) reduces toZ C1+C2E·ds=Za bC2E·ds=−Za bC2∇Φ·ds =−(Φa−Φb) =−v(10) It follows from Faraday’s law, (9), that the terminal voltage is v=dλ dt (11) where λis the ﬂux linkage3 λ≡Z SµoH·da (12) By deﬁnition, the surface Sspans the closed contour C. Thus, as shown in Fig. 8.4.3, it has as its edge the perfectly conducting coil, C1, and the contour used to close the circuit in the region where the terminals are located, C2. If the magnetic induction is negligible in the latter region, the electric ﬁeld is irrotational. In that case, the speciﬁc contour, C2, is arbitrary, and the EMF between the terminals becomes the voltage of circuit theory. Our discussion has emphasized the importance of having the terminals in a region where the magnetic induction, ∂µoH/∂t, is negligible. If a time-varying magnetic ﬁeld is signiﬁcant in this region, then diﬀerent arrangements of the leads connecting the terminals to the voltmeter will result in diﬀerent voltmeter readings. (We will emphasize this point in Sec. 10.1, where we develop an appreciation for the electric ﬁeld implied by Faraday’s law throughout the free space region sur- rounding the perfect conductors.) However, there remains the task of identifying conﬁgurations in which the ﬂux linkage is not appreciably aﬀected by the layout of leads connected to the terminals. In the absence of magnetizable materials, this is generally realized by making coils with many turns that are connected to the outside world through leads arranged to link a minimum of ﬂux. The inset to Fig. 8.4.3 shows an example. The large number of turns assures a magnetic ﬁeld within the coil that is much larger than that associated with the wires that connect the coil to the terminals. By intertwining these wires, or at least having them close together, the terminal voltage becomes independent of the detailed wire layout. Demonstration 8.4.1. Surface used to Deﬁne the Flux Linkage 3We drop the subscript fon the symbol λfor ﬂux linkage where there is no chance to mistake it for line charge density. 26Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.4.4 To visualize the surface enclosed by the contour C1+C2of Fig. 8.4.3, imagine ﬁlling it in with yarn strung on a frame representing the contour. The surface Sused to deﬁne λin (12) is often geometrically complex. It is helpful to picture the surface in terms of a model. Shown in Fig. 8.4.4 is a three-turn coil. The surface is ﬁlled in by stringing yarn between a vertical rod joining the terminals in the external region and points on the wire. The surface is ﬁlled in by connecting points of decreasing altitude on the rod to points of increasing distance along the wire. Note from Fig. 8.4.3 that daanddsare related by the right-hand rule, where the latter is directed along the contour from the positive terminal to the negative one. Another way of demonstrating the relationship of the surface to the coil ge- ometry takes advantage of the phenomenon familiar from blowing bubbles. A small coil, closed along the external segment between the terminals, can be dipped into materials like soap solution to form a continuous ﬁlm having the wire as one contin- uous edge. In fact, if the ﬁlm is formed from a material that hardens into a plastic sheet, a permanent model for the surface is obtained. Inductance. When the ﬂux linked by the perfectly conducting coil of Fig. 8.4.3 is due entirely to a current iin the coil itself, λis proportional to i, λ=Li. Thus, the inductance L, deﬁned as L≡λ i=R SµoH·da i (13) becomes a parameter that is only a function of geometric variables and µo. In this case, the terminal voltage given by (11) assumes a form familiar from circuit theory. v=Ldi dt (14) The following example illustrates this rule. Example 8.4.2. Inductance of a Long Solenoid Sec. 8.4 Perfect Conductors 27 In Demonstration 8.2.1, we examined the ﬁeld of a long N-turn solenoid and found that in the limit where the length dbecomes very large, the ﬁeld intensity along the axis is Hz=Ni d(15) where iis the current in each turn. For an inﬁnitely long solenoid this is not only the ﬁeld on the axis of symmetry but everywhere inside the solenoid. To see this, observe that a uniform magnetic ﬁeld intensity satisﬁes both Amp` ere’s law and the ﬂux continuity condition throughout the free space interior region. (A uniform ﬁeld is irrotational and solenoidal.) Further, with the ﬁeld given by (15) inside the coil and taken as zero outside, Amp` ere’s continuity condition (1.4.16) is satisﬁed at the surface of the coil where Kφ=Ni/d . The normal ﬂux continuity condition is automatically satisﬁed, since there is no ﬂux density normal to the coil surface. Because the ﬁeld is uniform over the circular cylindrical cross-section, the magnetic ﬂux Φ λ4passing through one turn of the solenoid is simply the cross- sectional area Aof the solenoid multiplied by the ﬂux density µoH. Φλ=µoHzA=µoAN di (16) The ﬂux linkage, deﬁned by (12), is obtained by summing the contributions of all the turns. λ=X turnsΦλ=µoN2A di (17) Thus, from (13), L=λ i=µoN2A d(18) For the circular cylindrical solenoid of radius a, A =πa2. The same arguments used to see that the interior ﬁeld of a solenoid of circular cross-section is given by (15) show that the solenoid can have an arbitrary cross-sectional geometry and the ﬁeld will still be given by (15) everywhere inside and be zero outside. Thus, (18) is applicable to a solenoid of arbitrary cross-section. Example 8.4.3. Dipole Moment Induced in Perfectly Conducting Sphere by Imposed Uniform Magnetic Field If a highly conducting material is immersed in a magnetic ﬁeld, it will modify the ﬁeld in its vicinity via a surface current that cancels the ﬁeld in its interior. If the material is spherical, we can superimpose the ﬁeld of a dipole and the uniform ﬁeld to exactly satisfy the boundary condition on the conducting surface. For a sphere having radius Rin an imposed ﬁeld Hoiz, as shown in Fig. 8.4.5, what is the equivalent dipole moment m? The imposed ﬁeld is conveniently analyzed into radial and azimuthal compo- nents. Then the irrotational and solenoidal ﬁeld proposed to satisfy the boundary conditions is the sum of that uniform ﬁeld and the ﬁeld of a dipole at the origin, as given by (8.3.14) together with the deﬁnition (8.3.19). H=Ho¡ cosθir−sinθiθ¢ +m 4π¡2 cosθ r3ir+sinθ r3iθ¢ (19) 4We use the symbol Φ λfor the ﬂux through one turn of a coil or a loop. 28Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.4.5 Immersed in a uniform magnetic ﬁeld, a perfectly conduct- ing sphere has the same eﬀect as an oppositely directed magnetic dipole. Fig. 8.4.6 One-turn solenoid. By design, this ﬁeld already approaches the uniform ﬁeld at inﬁnity. To satisfy the condition that n·µoH= 0 at r=R, µoHr(r=R) = 0⇒2m 4πR3cosθ+Hocosθ= 0 (20) It follows that the equivalent dipole moment is m=−2πHoR3(21) The surface currents induced in the sphere which buck out the imposed magnetic ﬂux are responsible for the dipole moment, as illustrated in Fig. 8.4.5. Example 8.4.4. One-Turn “Solenoid” The structure of perfectly conducting sheets shown in Fig. 8.4.6 has width wmuch greater than aand is excited by a uniform (in the zdirection) current per unit length Katy=−b. The H-ﬁeld solution that satisﬁes the boundary condition n·H= 0 and n×H=Kon the perfect conductor is Hz=−K (22) Sec. 8.5 Piece-Wise Magnetic Fields 29 What is the voltage that appears across the current generator? From (11) and (12) we conclude v=dλ dt(23) with λ=Z µoH·da=µoKab =µoab wi where iis the total current supplied by the generator. The voltage is thus v=Ldi dt(24) where L=µoab w 8.5 PIECE-WISE MAGNETIC FIELDS In a typical physical situation to which the scalar potential is applicable, layers of wire are used to make a winding that is thin compared to other dimensions of interest. Currents are then conﬁned to surfaces that separate the regions where H is irrotational. Thus, the sources of the magnetic ﬁeld intensity can be represented as surface currents. The ﬁeld produced by these currents is then found by choosing source-free solutions in the space surrounding the current-carrying surfaces and “connecting” these solutions across the surfaces by the proper boundary conditions. This procedure is analogous to ﬁnding EQS potentials produced by charge sheets in Chap. 5. Solutions to Laplace’s equation were set up on the two sides of a charge sheet and the jump in normal /epsilon1oEadjusted to equal the surface charge density. In the MQS situation, the Hﬁeld obeys Amp` ere’s continuity condition, (1.4.16). n×(Ha−Hb) =K (1) At this same surface, the magnetic ﬂux continuity condition, (1.7.6), also applies. n·(µoHa−µoHb) = 0 (2) Remember that in Chap. 5, continuity of tangential Ewas implied by making the electric potential continuous. By contrast, according to (1), where there is a surface current density, the tangential His discontinuous and this implies that the magnetic scalar potential Ψis not generally continuous . To see this, consider the application of Amp` ere’s integral law to an incremental surface that is pierced by the surface current density, as shown in Fig. 8.5.1. If His ﬁnite, then in the limit where the width wgoes to zero, the contributions to the line integral from the segments B→B/primeandA/prime→Avanish, and so I CH·ds=Z SJ·da⇒ −(∇Ψa− ∇Ψb)·is=K·in (3) 30Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.5.1 Contour enclosing surface current density Kon surface having normal n. Integration of Amp` ere’s law on surface enclosed by the contour shows that the magnetic scalar potential is, in general, discontinuous across the surface. where the unit vectors isandinare deﬁned in Fig. 8.5.1. Multiplication of (3) by the incremental line element dsand integration over the length of the incremental surface gives −ZB A(∇Ψa− ∇Ψb)·isds=ZB AK·inds (4) In view of the gradient integral theorem, (4.1.16), the integrals on the left can be carried out to obtain (ΨB−ΨA)−(ΨB/prime−ΨA/prime) =−ZB AK·inds (5) Now think of A−A/primeas a ﬁxed reference position on the surface, where Ψ Ais deﬁned as being equal to Ψ A/prime. It then follows that the discontinuity in Ψ at the location B−B/primeis a measure of the net current passing normal to the strip joining A−A/prime toB−B/prime. A further contrast with the electric ﬁeld comes from the normal ﬁeld continuity condition, (2). At a surface carrying a surface current density in free space, the normal derivative of Ψis continuous. The following example shows how to ﬁnd Ψ, and hence H, when a surface current distribution is given. Example 8.5.1. The Spherical Coil The magnetic ﬁeld intensity produced inside a properly wound spherical coil has the important property that it is uniform. This should be contrasted with the ﬁeld of a long solenoid that is uniform only to the extent that the fringing ﬁeld can be neglected. The coil is wound of thin wire so that the turns density is sinusoidally dis- tributed between the north and south poles of a sphere. To the extent that we can disregard the slight pitch in the coil needed to connect the loops with each other, loops of appropriately varying diameter, spaced evenly as projected onto the zaxis, Sec. 8.5 Piece-Wise Magnetic Fields 31 Fig. 8.5.2 Cross-section of “ﬂux ball” consisting of sphere with wind- ing on its surface that is of uniform turns density with respect to the z axis. automatically simulate such a distribution. The coil, with a radius Rand a wire carrying the current i, is shown in Fig. 8.5.2. To deduce the surface current density representing this winding, note that the density of turns on the surface is the total number, N, divided by the total length, 2 R, and so the number of turns in the incremental length dzis (N/2R)dz. Because z=rcosθ, a diﬀerential length dzcorresponds to an angular increment dθ: dz=−sinθRdθ . Therefore, the number of turns in the diﬀerential length Rdθ as measured along the periphery of the sphere is ( N/2R) sinθ. With each turn carrying the current i, the surface current density is K=iφN 2Risinθ (6) In the spaces interior and exterior to the surface of the sphere, His both irrotational and solenoidal. Hence, it is represented by scalar magnetic potentials. Theφcomponent of (1) is the link between the surface current density and the induced ﬁeld. Ha θ−Hb θ=N 2Risinθ (7) To obtain Hθ, the derivative of Ψ with respect to θmust be taken, and this suggests that the θdependence of Ψ be taken as cos θ. The ﬁeld is ﬁnite at the origin and zero at inﬁnity, so, from the three solutions to Laplace’s equation given in Sec. 5.9, we select Ψ =C(r/R) cosθ; r < R (8) Ψ =A(R/r)2cosθ; r > R (9) The continuity conditions, used now to determine the coeﬃcients AandC, are in terms of the ﬁeld intensity. Thus, (8) and (9) are used to write Hin the two regions as H=−C R(ircosθ−iθsinθ); r < R (10) H=A R(R/r)3(ir2 cosθ+iθsinθ); r > R (11) Substitution of the appropriate components into the continuity conditions, (2) and (7), gives −C R=2A R(12) 32Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.5.3 Magnetic ﬁeld intensity of “ﬂux-ball” shown in Fig. 8.5.2. A R−C R=Ni 2R(13) Thus, the magnetic ﬁeld intensity of (10) and (11) is evaluated by setting C= −2A=−Ni/3. H=Ni 3R(ircosθ−iθsinθ); r < R (14) H=Ni 6R(R/r)3(ir2 cosθ+iθsinθ); r > R (15) The exterior lines of magnetic ﬁeld intensity are those of a dipole, while the interior ﬁeld is uniform. Thus, the total picture, shown in Fig. 8.5.3, is one of ﬁeld lines circulating from south to north inside the sphere and back from north to south on the outside around currents that follow lines of equilatitude around the sphere. The magnetic potential follows by substituting C=−2A=−Ni/3 for Cand Ain (8) and (9). Ψ =−Ni 3r Rcosθ; r < R (16) Ψ =Ni 6(R/r)2cosθ; r > R (17) Note that these potentials are equal at the equator of the sphere and become increasingly disparate as the poles are approached. With the vertical dimension used to denote Ψ, a sketch of Ψ evaluated in a plane of ﬁxed φwould appear as shown in Fig. 8.5.4. Inside, Ψ slopes linearly from its highest value at the south pole to its lowest at the north. Outside, Ψ has its highest value at the north pole and lowest at Sec. 8.5 Piece-Wise Magnetic Fields 33 Fig. 8.5.4 Magnetic scalar potential for “ﬂux ball” of Fig. 8.5.2. The vertical axis is Ψ. A line of Hcloses on itself as it circulates around surface current, going down the potential “hills” inside and outside the sphere and recovering its altitude at the surfaces of discontinuity at r=R, containing the surface current density. the south. This is consistent with the picture aﬀorded by Fig. 8.5.1 and (5). Even though it closes on itself, the line of Hshown goes continuously “down hill.” The potential Ψ regains its altitude in the region of discontinuity. Finally, we illustrate the computation of the inductance of a coil modeled by a surface current and represented in terms of the magnetic scalar potential. To compute the total ﬂux linked by the winding, ﬁrst consider the ﬂux linked by one turn at the location r=Randθ=θ/prime. Using the ﬂat surface at z/prime=Rcosθ/primethat is enclosed by this circular turn, the ﬂux is Φλ=ZRsinθ/prime 0µoHz2πrdr =π(Rsinθ/prime)2µoHz (18) In this particular problem, Hzis uniform inside the sphere, so this integration amounts to multiplying the area enclosed by the turn by the normal ﬂux density. The turns density multiplied by Rdθ gives the number of turns linking this ﬂux in an increment of peripheral length. Thus, the total ﬂux is obtained by carrying out a second integration over all of the turns. λ=Zπ 0ΦλN 2Rsinθ/primeRdθ/prime=Zπ 0iπN2Rµo 6sin3θ/primedθ/prime=Li (19) L≡2 9πN2µoR (20) Demonstration 8.5.1. Field and Inductance of a Spherical Coil In the experiment shown in Fig. 8.5.5, the “ﬂux ball” has 64 turns and a radius of R= 5 cm. The turns are wound on a plastic sphere that essentially has the magnetic properties of free space. 34Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.5.5 Demonstration of ﬁelds surrounding the magnetic “ﬂux ball.” The Hall magnetometer makes it possible to probe the magnitude and direc- tion of the ﬁeld outside the coil. For example, at the north pole, where the magnetic ﬂux density is perpendicular to the sphere surface, the ﬂux density is vertical and fori= 1 A predicted by either (14) or (15) to be µoNi/3R= 5.36×10−4T= 5.36 gauss. The inductance is determined by measuring the voltage and current, varying the frequency to determine that it is high enough to assure that the resistance of the coil plays a negligible role in the terminal impedance (the impedance should be of magnitude ωL, and hence vary linearly with frequency). The inductance predicted by (20) is 180 µH, and the value measured using the oscilloscope is typically within 10 percent. 8.6 VECTOR POTENTIAL AND THE BOUNDARY VALUE POINT OF VIEW We have found that many interesting MQS cases can be treated by the use of the scalar potential obeying Laplace’s equation. The vector potential, deﬁned by (8.1.1), is necessary when analyzing ﬁelds with nonzero curl. There are other cases as well in which its use may be advantageous. The vector potential is the natural variable for evaluating the ﬂux passing through a surface. In view of (8.1.1), integration of the ﬂux density over the open surface Sof Fig. 8.6.1 gives λ=Z SµoH·da=Z S∇ ×A·da (1) and it follows from Stokes’ theorem that this ﬂux is equal to the line integral of A·dsaround the contour enclosing the surface. λ=I CA·ds (2) Sec. 8.6 Vector Potential 35 Fig. 8.6.1 Open surface Shaving area element daenclosed by contour C having directed diﬀerential length ds. Fig. 8.6.2 Surface Swith sides of length lparallel to the zaxis at locations (a) and (b). The contour direction is consistent with the ﬂux being positive, as shown. In certain important cases, Ahas only one component and a vector ﬁeld is again represented in terms of one scalar function. Two such cases are identiﬁed in the following subsections. Vector Potential for Two-Dimensional Fields. Suppose that the ﬂux density is parallel to the x−yplane and is independent of z. It can then be represented by a vector potential having only a zcomponent. A=Az(x, y)iz (3) Note that the divergence of this Ais automatically zero and that in Cartesian coordinates, the components of the ﬂux density are given in terms of Azby µoH=∇ ×A=∂Az ∂yix−∂Az ∂xiy (4) Consider now the evaluation of the net ﬂux of magnetic ﬂux density through a surface Sthat has length lin the zdirection, as shown in Fig. 8.6.2. The points (a) and (b) denote the coordinates of the corners of the contour enclosing S. The contour consists of a pair of parallel straight segments of length lparallel to the z axis, one at the location (a) in the x−yplane and the other at (b), and contours joining (a) and (b) in x−yplanes. Contributions to the contour integral, (2), from these latter segments of Care zero, because Ais perpendicular to ds. Integration along the z-directed segments amounts to multiplication of Azevaluated at (a) or (b) by the length of the segment. Thus, (1) becomes λ=l(Aa z−Ab z) (5) 36Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.6.3 Diﬀerence between axisymmetric stream function Λ sevaluated at (a) and (b) is net ﬂux through surface enclosed by the contour shown. The vector potential at (a) relative to (b) is the net magnetic ﬂux per unit length passing through a surface of unit length in the zdirection subtended between the two points and a corresponding pair at unity distance along the zaxis. Note that the ﬂux has a sign, relative to the direction of the contour integration, governed by the right-hand rule (Fig. 1.4.1). Vector Potential for Axisymmetric Fields in Spherical Coordinates. If the magnetic ﬂux density is invariant with respect to rotation around the zaxis, having components in the randθdirections only, the vector potential again has a single component. A=Aφ(r, θ)iφ (6) The net ﬂux through the annular surface “spanned” over the contour shown in Fig. 8.6.3, having constant outer and inner radii denoted by (a) and (b), respectively, is given by the contributions to (2) of the azimuthal segments, Aφmultiplied by the circumferences. The contour is closed by adjacent oppositely directed segments joining points (a) and (b) in a plane of constant φ. Thus, the contributions to the line integral of (2) from these segments cancel, even if Ahad components in the direction of dson these segments. Thus, the net ﬂux through the annulus is simply theaxisymmetric stream function Λ at (a) relative to that at (b).5 λ= Λa s−Λb s (7) where Λs≡2πrsinθAφ (8) Lines of ﬂux density are tangential to the axisymmetric surfaces of constant Λs. Just as Azprovides a ready visualization of the ﬂux lines in two dimensions, Λsportrays the axisymmetric ﬂux lines. 5With Aused to represent the velocity distribution of an incompressible ﬂuid, Λ s(or Λ s/2π) is called Stokes’ stream function. Sec. 8.6 Vector Potential 37 Fig. 8.6.4 Surfaces of constant Azand hence lines of magnetic ﬁeld intensity for ﬁeld trapped between perfectly conducting electrodes. Boundary Value Solution by “Inspection”. In two-dimensional conﬁgura- tions, any surface of constant Azcan be replaced by the surface of a perfect conduc- tor. Moreover, in the free space region between conductors, Azsatisﬁes Laplace’s equation. Thus, any two-dimensional conﬁguration from Chaps. 4 and 5 can be replaced by one where the potential lines are ﬁeld lines. The equipotential (con- stant Φ) surfaces of the EQS perfect conductors become the perfectly conducting (constant Az) surfaces of an MQS system. Illustration. Field Trapped between Hyperbolic Perfect Conductors The two-dimensional potential distribution of Example 4.1.1 suggests the vector potential Az= Λoxy/a2. The lines of magnetic ﬁeld intensity, which are the surfaces of constant Az, are shown in Fig. 8.6.4. Here, the surfaces Az=±Λoare taken as being the surfaces of perfect conductors. Thus, the current density on the surfaces of these conductors are, given by using (4) to determine Hand, in turn, (8.4.3) to ﬁnd Kz. These currents shield the ﬁelds from the volume of the perfect conductors. The net ﬂux per unit length passing downward between the upper pair of conductors is [in view of (7)] simply 2Λ o. This solution is the superposition of the ﬁelds of four line currents. Two di- rected in the + zdirection are at inﬁnity in the ﬁrst and third quadrants, while two in the −zdirection are in the second and fourth quadrants. Example 8.6.1. Field and Inductance of Oppositely Directed Currents in Parallel Perfectly Conducting Cylinders The cross-section of a pair of parallel perfectly conducting cylinders that extend to±∞ in the zdirection is shown in Fig. 8.6.5. The conductors have the same geometry as in the EQS case considered in Example 4.6.3. However, they should be regarded as shorted at one end and driven by a current source iat the other. Thus, current in the + zdirection in the right conductor is returned in the left conductor. 38Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.6.5 Cross-section of perfectly conducting parallel conductors having radius Rand spacing 2 l. Fields of oppositely directed line cur- rents having spacing 2 aare shown to satisfy normal ﬂux boundary con- dition on circular cylindrical surfaces of conductors. Although the net current in each conductor is given, its distribution on the surface of the conductors is to be determined. Example 4.6.3 suggests our strategy. Instead of superimposing the potentials Φ of a pair of line charges of opposite sign, we superimpose the Azof oppositely directed line currents. With r1andr2the distances from the observer coordinate to the source coordinates, deﬁned in Fig. 8.6.5, it follows from the vector potential for a line current given by (8.1.16) that Az=−µoi 2π(ln r1−ln r2) (9) With the identiﬁcation of variables Az→Φ;µoi 2π→λl 2π/epsilon1o(10) this expression is identical to that for the antidual EQS conﬁguration, (4.6.18). We can conclude that the line currents should be located at a= (l2−R2)1/2, and that the constant kused in that deduction (4.6.20) is identiﬁed using (10). k≡expµ 2πΛ µoi¶ =l+a R(11) Here, the potential Uin (4.6.20) is replaced by the ﬂux per unit length Λ. Thus, the surfaces of constant Azare circular cylinders and represent the ﬁeld lines shown in Fig. 8.6.6. The inductance per unit length Lis now deduced from (11). L≡2Λ i=µo πln¡l+a R¢ =µo πln¯¯¯¯l R+r¡l R¢2−1¯¯¯¯(12) In the limit where the conductors represent wires that are thin compared to their spacing, the inductance per unit length of (12) is approximated using (4.6.28). L≈µo πln¡2l R¢ (13) Once the vector potential has been determined, it is possible to evaluate the distribution of current density on the conductors. Note that the currents tend to con- centrate on the inside surfaces of the conductors, where the magnetic ﬁeld intensity is more intense. Sec. 8.6 Vector Potential 39 Fig. 8.6.6 Surfaces of constant Azand hence lines of magnetic ﬁeld intensity for the parallel conductor conﬁguration shown in the same cross-sectional view by Fig. 8.6.5. We are one step short of a general relationship between the capacitance per unit length and inductance per unit length of a pair of parallel perfect conductors, regardless of the cross-sectional geometry. With Φ and Azdeﬁned as zero on one of the conductors, evaluated on the other conductor they represent the voltage and the ﬂux linkage per unit length, respectively. Thus, with the understanding that Φ andAzare evaluated on the second conductor, L=Az/i, and C=λl/Φ, (4.6.5). Here, iandλl, respectively, are the line current and line charge density that give rise to the same ﬁelds as do those sources actually on the surfaces of the conductors. These quantitities are related by (10), so we can conclude that regardless of the cross-sectional geometry, the product of the inductance per unit length and the capacitance per unit length is LC=Azλl iΦ=µo/epsilon1o=1 c2(14) where cis the velocity of light (3.1.16). Note that inductance per unit length of parallel circular conductors given by (12) and the capacitance per unit length for the same conductors under “open circuit” conditions (4.6.27) satisfy the general relation (14). Method of Images. In the presence of a planar perfect conductor, the zero normal ﬂux condition can be satisﬁed by symmetrically mounting source distribu- tions on both sides of the plane. This approach is familiar from Sec. 4.7, where the boundary condition required a plane of symmetry on which the tangential electric ﬁeld was zero. Here we require that the ﬁeld intensity be tangential to the bound- ary. For two-dimensional conﬁgurations, the analogy between the electric potential andAzmakes the image method of Sec. 4.7 directly applicable here. In both cases, the symmetry plane is one of constant potential (Φ or Az). 40Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.6.7 With the frequency high enough so that the currents dis- tribute themselves with a negligible normal ﬂux density on the conduc- tors, the ﬁeld intensity tangential to the conducting plane is that pre- dicted by (16) and shown by the graph. At low frequencies, the current tends to be uniformly distributed in the planar conductor. The most obvious example is an inﬁnitely long line current at a distance d/2 from a perfectly conducting plane. If Fig. 4.7.1 were a picture of line charges rather than point charges, this would be the dual situation. The appropriate image is then an oppositely directed line current located at a distance d/2 to the other side of the perfectly conducting plane. By making a pair of symmetrically located line currents the image for this pair of currents, the boundary condition on yet another plane can be satisﬁed, the analog to the conﬁguration of Fig. 4.7.3. The following demonstration is intended to emphasize that the perfectly con- ducting symmetry plane carries a surface current that terminates the ﬁeld in the region of interest. Demonstration 8.6.1. Surface Currents Induced in Ground Plane by Over- head Conductor The metal cylinder mounted over a metal ground plane shown in Fig. 8.6.7 is familiar from Demonstration 4.7.1. Rather than being insulated from the ground plane and driven by a voltage source, this cylinder is shorted to the ground plane at one end and driven by a current source at the other. The height lis small compared to the length, so that the two-dimensional model describes the ﬁeld distribution in the midregion. A probe is used to measure the magnetic ﬂux density tangential to the metal ground plane. The distribution of this ﬁeld, and hence of the surface current density in the adjacent metal, can be determined by recognizing that the ground plane boundary condition of no normal ﬂux density is met by symmetrically mounting a distribution of oppositely directed currents below the metal sheet. This is just what was done in determining the ﬁelds for the pair of cylindrical conductors, Fig. 8.6.5. Sec. 8.6 Vector Potential 41 Thus, (9) is the image solution for the region x≥0. In terms of xandy, Az=−µoi 2πlnp (a−x)2+y2 p (a+x)2+y2(15) The ﬂux density tangential to the ground plane at the location y=Yis µoHy(x= 0) = −∂Az ∂x(x= 0) = −µoi πa· 1 1 +¡Y a¢2¸ (16) Normalized to Ho=i/πa, this distribution is shown as a function of the probe position, Y, in the inset to Fig. 8.6.7. The role of the surface current density implied by this tangential ﬁeld is demon- strated by the same probe measurement of the magnetic ﬂux density normal to the conducting sheet. Provided that the frequency is high enough so that the sheet does indeed behave as a perfect conductor, this ﬂux density is small compared to that tangential to the sheet. This is also true at the surface of the cylindrical conductor. To appreciate the physical origins of this distribution, a dc current source is used in place of the ac source. The distribution of current in the sheet is then dictated by the rules of steady conduction, as enunciated in the ﬁrst half of Chap. 7. If the sheet is long enough compared to its width, the current is uniformly distributed over the sheet and over the cross-section of the cylinder. By contrast with the high- frequency ac case, where the ﬁeld is terminated by surface currents in the sheet, the magnetic ﬁeld now extends below the sheet. The method of images is not restricted to the two-dimensional situations where there is a convenient analogy between Φ and Az. In the following example, involving a three-dimensional ﬁeld, the symmetry conditions are viewed without the aid of the vector potential. Example 8.6.2. Current Loop above a Perfectly Conducting Plane A current loop with time-varying current iis mounted a distance habove a perfectly conducting plane, as shown in Fig. 8.6.8. Its axis is inclined at an angle θwith respect to the normal to the plane. What is the net ﬁeld produced by the current loop and the currents it induces in the plane? To satisfy the boundary condition in the plane of the perfectly conducting sheet, an image loop is mounted as shown in Fig. 8.6.9. For each current segment in the actual loop, there is a segment in the image loop giving rise to an oppositely directed vertical component of H. Thus, the net normal ﬂux density in the plane of the perfect conductor is zero. Two-Dimensional Boundary Value Problems. The vector potential of a two-dimensional ﬁeld parallel to the x−yplane is zdirected and thus only one scalar function describes fully the associated ﬁeld, as already pointed out earlier. In problems in which currents are conﬁned to the boundaries, the scalar potential can be used as eﬀectively as the vector potential. The lines of steepest descent of the scalar potential are the lines of constant height of the vector potential. When the 42Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.6.8 Current loop at distance habove a perfectly conducting plane. Fig. 8.6.9 Cross-section of conﬁguration of Fig. 8.6.8, showing image dipole giving rise to ﬁeld that cancels the ﬂux density normal to the planar perfect conductor. region of interest contains current distributions, then use of the vector potential is required. We shall consider both situations in the examples to follow. Example 8.6.3. Inductive Attenuator The cross-section of two conducting electrodes that extend to inﬁnity in the ±z directions is shown in Fig. 8.6.10. The time-varying current in the + zdirection in the electrode at y=bis returned in the −zdirection through the /unionsq-shaped electrode. This current is so rapidly varying that the electrodes behave as though they were perfectly conducting. The gaps of width ∆ insulating the electrodes from each other are small compared to the other dimensions of interest. The magnetic ﬂux (per unit length in the zdirection) passing through these gaps in the directions shown is deﬁned as Λ( t). The magnetic ﬁelds are two dimensional and there are no sources in the region of interest. Thus, µoHcan be represented in terms of Az, which satisﬁes ∇2Az= 0 (17) The walls are perfectly conducting in the sense that they are modeled as having no normal µoH. This means that Azis constant on these walls. We deﬁne Azto be zero on the vertical and bottom walls. Thus, Azmust be equal to Λ on the upper Sec. 8.6 Vector Potential 43 Fig. 8.6.10 Cross-section of inductive attenuator. electrode, so that the ﬂux per unit length in the zdirection through the gaps is Λ. Az(0, y) = 0 , A z(a, y) = 0 , A z(x,0) = 0 , A z(x, b) = Λ (18) The boundary value problem is now formally identical to the EQS capacitive atten- uator that was the theme of Sec. 5.5, with the identiﬁcation of variables Φ→Az, V →Λ (19) Thus, it follows from (5.5.9) that Az=∞X n=1 odd4Λ(t) nπsinh¡nπ ay¢ sinh¡nπb a¢sin¡nπ ax¢ (20) The lines of magnetic ﬂux density are the lines of constant Az. They are the equipo- tential “lines” of Fig. 5.5.3, shown in Fig. 8.6.10 with arrows added to indicate the ﬁeld direction. Remember, there is a z-directed surface current density that is pro- portional to the tangential ﬁeld intensity. For the ﬂux lines shown, Kzis out of the page in the upper electrode and returned into the page on the side walls and (to an extent determined by brelative to a) on the bottom wall as well. From the cross-sectional view given by Fig. 8.6.10, the provision for the current through the driven plate at the top to recirculate through the side and bottom plates is not shown. The following demonstration emphasizes the implied current paths at the ends of the conﬁguration. Demonstration 8.6.2. Inductive Attenuator One conﬁguration described by Example 8.6.3 is shown in Fig. 8.6.11. Here the upper plate is shorted to the adjacent walls at the near end and driven at the far end through a step-down transformer by a 20 kHz oscillator. The driving voltage v(t) at the far end of the upper plate is measured by means of an oscilloscope. The lower 44Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 plate is shorted to the side walls at the far end and also connected to these walls at the near end, but in such a way that the induced current i(t) can be measured by means of a current probe. The walls and upper and lower plates are made from brass or copper. To insure that the resistances of the plate terminations are negligible, they are made from heavy copper wire with the connections soldered. (To make it possible to adjust the spacing b, braided wire is used for the shorts on the lower electrode.) If the length wof the plates in the zdirection is large compared to aandb,H within the volume follows from (20). The surface current density Kzin the lower plate then follows from evaluation of the tangential Hon its surface. In turn, the total current follows from integration of Kzover the width, a, of the plate. i=−1 µo∞X n=1 odd16Λ 2nπ1 sinh¡nπb a¢ (21) With the objective of relating this current to the driving voltage, note that (8.4.11) gives v=wdΛ dt(22) so that with the driving voltage a sinusoid of magnitude V, v=Vcos(ωt)⇒Λ =V wωsin(ωt) (23) Thus, in terms of the driving voltage, the output current is iosin(ωt), where it follows from (21) and (23) that io=−I∞X n=1 odd1 2nsinh¡nπb a¢; I≡16V πwωµ o(24) We have found that the output current, normalized to I, has the dependence on spacing between upper and lower plates shown by the inset to Fig. 8.6.11. With the spacing bsmall compared to a, almost all of the current through the upper plate is returned in the lower one, and the ﬁeld between is essentially uniform. As the spacing bbecomes comparable to the distance abetween the side walls, most of the current through the upper electrode is returned in these side walls. Thus, for large b/a, the normalized output current of Fig. 8.6.11 reﬂects the exponential decay in the−ydirection of the ﬁeld. Value is added to this demonstration if it is compared to its EQS antidual, Demonstration 5.5.1. For the EQS conﬁguration, the lower plate was properly con- strained to essentially the same potential as the walls by connecting it to these side walls through a resistance (which was then used to measure the induced current). Up to frequencies above 100 Hz in the EQS case, this resistance could be as high as that of the oscilloscope (say 1 MΩ) and still constrain the lower plate to essentially the same zero potential as the walls. In the MQS case, we did not use a resistance to connect the lower plate to the side walls (and hence provide a means of measuring the output current), because that resistance would have had to be extremely low, even at 20 kHz, to prevent ﬂux from leaking through the gaps between the lower plate and the side walls. We used the current probe instead. The eﬀects of ﬁnite conductivity in MQS systems are the subject of Chap. 10. Sec. 8.6 Vector Potential 45 Fig. 8.6.11 Inductive attenuator demonstration. In a ﬁnal example, we exemplify how the particular and homogeneous solu- tions are combined to satisfy boundary conditions while also illustrating how the inductance of a distributed winding is determined. Example 8.6.4. Field and Inductance of Distributed Winding Bounded by Perfect Conductor The cross-section of a distributed winding of radius ais shown in Fig. 8.6.12. It consists of turns carrying current iin the + zdirection at a location ( r, φ) and returning the current at ( r,−φ) in the −zdirection. The density of turns, each carrying the current iin the + zdirection for 0 ≤φ≤πand in the −zdirection for π < φ < 2π, is n=no|sinφ| (25) The total number of wires Nin the left-hand half of the coil is N=Za 0Zπ 0nosinφrdrdφ =noa2(26) so that the current density is J=izinosinφ=iziN a2sinφ (27) The windings are very long in the zdirection so that eﬀects of the end turns are ignored and the ﬁelds taken as independent of z. 46Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. 8.6.12 Cross-section of two-dimensional distributed winding sur- rounded by perfectly conducting material. A typical coil consists of wires carrying current in the + zdirection at ( r, φ) somewhere to the right (0< φ < π ), and returning it in the −zdirection at ( r,−φ) to the left. The coil is bounded at r=aby a perfect conductor. With the following steps we determine the ﬁeld distribution throughout the winding and ﬁnally, its inductance. The vector potential is zindependent and must satisfy Poisson’s equation (8.1.6). In polar coordinates, 1 r∂ ∂rµ r∂Az ∂r¶ +1 r2∂2Az ∂φ2=−µoJz (28) First we look for a particular solution. If it is to take a product form, inspection shows that sin φis the appropriate φdependence. Substitution of an rdependence rnshows that the equation can be satisﬁed if n= 2. Thus, we have “guessed” a particular solution. Azp=−µoNi 3r2 a2sinφ (29) The magnetic ﬂux density normal to the perfectly conducting surface at r=a must be zero, so the total vector potential must be constant there. It follows that one must add a vector potential with no associated current density in the region r < a , a homogeneous solution Azh. Atr=a, the homogeneous solution, Azh, must be the negative of the particular solution, Azp. [Azp+Azh]r=a= 0⇒Azh(r=a) =µoNi 3sinφ (30) A linear combination of the two solutions to Laplace’s equation that have the same φdependence as this condition is Azh=Crsinφ+D rsinφ (31) The coeﬃcient Dmust be zero so that the solution is ﬁnite at the origin. The coeﬃcient Cis then adjusted to make (31) satisfy the condition of (30). Hence, the sum of the particular and homogeneous solutions is Az=−µoNi 3·¡r a¢2−r a¸ sinφ (32) Sec. 8.7 Summary 47 Fig. 8.6.13 Graphical representation of the surfaces of constant Az for the system of Fig. 8.6.12 as the sum of particular and homogeneous solutions. A graphical representation of what has been accomplished is given in Fig. 8.6.13, where the surfaces of constant Az(and hence the lines of ﬁeld intensity) are shown for the particular, homogeneous, and total solutions. Each turn of the coil links a diﬀerent magnetic ﬂux. Thus, to determine the total ﬂux linked by the distribution of turns, it is necessary to carry out an inte- gration. To do this, ﬁrst observe that the ﬂux linked by the turns with their right legs within the area rdφdr in the neighborhood of ( r, φ) and their left legs within a similar area in the neighborhood of ( r,−φ) is Φλ=l[Az(r, φ)−Az(r,−φ)]nosinφrdφdr (33) Here, lis the length of the system in the zdirection. The total ﬂux linked by all of the turns is obtained by integrating over all of the turns. λ=lnoZa 0Zπ 0[Az(r, φ)−Az(r,−φ)]rsinφdφdr (34) Substitution for Azfrom (32) and use of (26) then gives λ=Li with L≡π 36lµoN2(35) where Lwill be recognized as the inductance. 8.7 SUMMARY Just as Chap. 4 was initiated with the representation of an irrotational vector ﬁeld E, this chapter began by focusing on the solenoidal character of the magnetic ﬂux density. Thus, µoHwas portrayed as the curl of another vector, the vector potential A. The determination of the magnetic ﬁeld intensity, given the current density everywhere, was pursued ﬁrst using the vector potential. The integration of the 48Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 vector Poisson’s equation for Awas the ﬁrst of many exploitations of analogies between EQS and MQS descriptions. In Cartesian coordinates, the superposition integral for A, (8.1.8) in Table 8.7.1, has components that are analogous to the scalar potential superposition integral, (4.5.3), from Table 4.9.1. Similarly, the two- dimensional superposition integral, (8.1.14), has as its analog (4.5.20) from Table 4.9.l. Especially if a computer is to be used, it is often most practical to work directly with the magnetic ﬁeld intensity. The Biot-Savart law, (8.2.7) in Table 8.7.1, gives Hdirectly as an integration over the given distribution of current density. In many applications, the current distribution can be approximated by piece- wise continuous straight-line segments. In this case, the total ﬁeld is conveniently represented by the superposition of contributions given by (8.2.22) in Table 8.7.1 due to the individual “sticks.” In regions free of current density, His not only solenoidal, but also irrotational. Thus, like the electric ﬁeld intensity of Chap. 4, it can be represented by a scalar potential Ψ ,H=−∇Ψ. The magnetic scalar potential is, in general, discontinuous across a surface carrying a surface current density. It is its normal derivative that is continuous. The scalar potential provides an elegant representation of the ﬁelds in free space regions surrounding current loops. The superposition integral, (8.3.12) in Table 8.7.1, is written in terms of the solid angle Ω. Through the combined eﬀects of Faraday’s law, ﬂux continuity, and Ohm’s law, currents are induced in a conductor by a time-varying magnetic ﬁeld. In a perfect conductor, these currents are on the surface, distributed in such a way as to shield the magnetic ﬁeld out of the conductor. As a result, the normal component of the magnetic ﬂux density must be zero on the surface of a perfect conductor. Although useful for representing any solenoidal ﬁeld, the vector potential is especially useful in the situations summarized by Table 8.7.2. It is especially con- venient for describing systems with perfectly conducting boundaries. In two di- mensions, the boundary condition on a perfect conductor is satisﬁed by making the vector potential constant on the boundary. The approaches of Chaps. 4 and 5 apply equally well to solving MQS boundary value problems involving perfect conductors. In fact, the two-dimensional EQS and MQS conﬁgurations of perfect conductors in free space, exempliﬁed by the conﬁgurations of Figs. 4.7.2 and 8.6.7, were found to be duals. Formally, the solution for Hfollows from that for Eby identifying Φ →Az, ρ//epsilon1 o→µoJz. However, while the electric ﬁeld intensity E is perpendicular to the surfaces of constant Φ ,His tangential to the surfaces of constant Az. The boundary conditions obeyed by the vector potential at surfaces of discon- tinuity (containing surface currents) reﬂect the discontinuity in tangential Hﬁeld and the continuity of the normal ﬂux density. The vector potential itself must be continuous (a discontinuity of Awould imply an inﬁnite Hin the surface) (Aa−Ab) = 0 (1) where Amp` ere’s continuity condition n×[(∇ ×A)a−(∇ ×A)b] =µoK (2) requires that curlAhave discontinuous tangential components. The condition that Abe continuous, (1), guarantees the continuity of the normal ﬂux density. [Accord- ing to (1), the integral of A·dsaround an incremental closed contour lying on one Sec. 8.7 Summary 49 TABLE 8.7.1 side of the surface is equal to that on the other. Thus, the normal ﬂux which each of these integrals represents, is the same as well.] In ﬂuid mechanics, the scalar Azwould be called a “stream-function”, because in two dimensions, lines of constant vector potential constitute the ﬂux lines. In axisymmetric conﬁgurations, the ﬂux lines are lines of constant Λ s, as deﬁned in Table 8.7.2. Of course, a similar representation can be used for any solenoidal vector. For example, an expression for the two-dimensional lines of electric ﬁeld intensity in a region free of charge density could be obtained by ﬁnding a vector potential representation of E. Thus, in these special cases, the vector potential is convenient for plotting any solenoidal ﬁeld. The electric potential Φ of EQS systems, evaluated on the surface of a perfectly 50Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 TABLE 8.7.2 conducting capacitor electrode, can be used to evaluate the terminal voltage. The vector potential is similarly related to the terminal characteristics of a lumped parameter element, this time an inductor. Indeed, we found in Sec. 8.6 that the ﬂux per unit length linked by a pair of conductors in two dimensions was simply the diﬀerence of vector potentials evaluated on the two conductors. In Sec. 8.4, we found that the terminal voltage is the time rate of change of this ﬂux linkage. The division of the ﬁeld into particular and homogeneous parts makes possible a number of diﬀerent approaches to obtaining the total ﬁeld. The particular part can be obtained using the vector potential, using the Biot-Savart law, or by super- imposing the ﬁelds of thin coils represented in terms the scalar magnetic potential. The homogeneous solution is both irrotational and solenoidal, so it is possible to use either the vector or the scalar potential to represent this part of the ﬁeld everywhere. The vector potential helps determine the net ﬂux, as required for calculating the inductance, but is of limited usefulness for three-dimensional conﬁgurations. The scalar potential does not directly portray the net ﬂux, but does generally apply to three-dimensional conﬁgurations. Sec. 8.2 Problems 51 P R O B L E M S 8.1 The Vector Potential and the Vector Poisson Equation 8.1.1 A solenoid has radius a, length d, and turns N, as shown in Fig. 8.2.3. The length dis much greater than a, so it can be regarded as being inﬁnite. It is driven by a current i. (a) Show that Amp` ere’s diﬀerential law and the magnetic ﬂux continuity law [(8.0.1) and (8.0.2)], as well as the associated continuity condi- tions [(8.0.3) and (8.0.4)], are satisﬁed by an interior magnetic ﬁeld intensity that is uniform and an exterior one that is zero. (b) What is the interior ﬁeld? (c)Ais continuous at r=abecause otherwise the Hﬁeld would have a singularity. Determine A. 8.1.2∗A two-dimensional magnetic quadrupole is composed of four line currents of magnitudes i, two in the positive zdirection at x= 0,y=±d/2 and two in the negative zdirection at x=±d/2, y= 0. (With the line charges repre- senting line currents, the cross-section is the same as shown in Fig. P4.4.3.) Show that in the limit where r/greatermuchd, A z=−(µoid2/4π)(r−2) cos 2 φ. (Note that distances must be approximated accurately to order d2.) 8.1.3 A two-dimensional coil, shown in cross-section in Fig. P8.1.3, is composed ofNturns of length lin the zdirection that is much greater than the width wor spacing d. The thickness of the windings in the ydirection is much less than wandd. Each turn carries the current i. Determine A. Fig. P8.1.3 8.2 The Biot-Savart Superposition Integral 8.2.1∗The washer-shaped coil shown in Fig. P8.2.1 has a thickness ∆ that is much less than the inner radius band outer radius a. It supports a current density J=Joiφ. Show that along the zaxis, H=∆Joiz 2·b√ b2+z2−a√ a2+z2+ln(a+√ a2+z2) (b+√ b2+z2)¸ (a) 52Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. P8.2.1 Fig. P8.2.5 8.2.2∗A coil is wound so that the wire forms a spherical shell of radius Rwith the wire essentially running in the φdirection. With the wire driven by a current source, the resulting current distribution is a surface current at r= Rhaving the density K=Kosinθiφ, where Kois a given constant. There are no other currents. Show that at the center of the coil, H= (2Ko/3)iz. 8.2.3 In the conﬁguration of Prob. 8.2.2, the surface current density is uniformly distributed, so that K=Koiφ, where Kois again a constant. Find Hat the center of the coil. 8.2.4 Within a spherical region of radius R, the current density is J=Joiφ, where Jois a given constant. Outside this region is free space and no other sources of H. Determine Hat the origin. 8.2.5∗A current icirculates around a loop having the shape of an equilateral triangle having sides of length d, as shown in Fig. P8.2.5. The loop is in thez= 0 plane. Show that along the zaxis, H=ip 3/4d2iz 4π¡ z2+d2 12¢−1¡d2 3+z2¢−1/2(a) 8.2.6 For the two-dimensional coil of Prob. 8.1.3, use the Biot-Savart superposi- tion integral to ﬁnd Halong the xaxis. Sec. 8.4 Problems 53 8.2.7∗Show that Ainduced at point Pby the current stick of Figs. 8.2.5 and 8.2.6 is A=µoi 4πa |a|ln"c·a |a|+|c| b·a |a|+|b|# (a) 8.3 The Scalar Magnetic Potential 8.3.1 Evaluate the Hﬁeld on the axis of a circular loop of radius Rcarrying a current i. Show that your result is consistent with the result of Example 8.3.2 at distances from the loop much greater than R. 8.3.2 Determine Ψ for two inﬁnitely long parallel thin wires carrying currents iin opposite directions parallel to the zaxis of a Cartesian coordinate system and located along x=±a. Show that the lines Ψ = const in the x−yplane are circles. 8.3.3 Find the scalar potential on the axis of a stack of circular loops (a coil) of Nturns and length lusing 8.3.12 for an individual turn, integrating over all the turns. Find Hon the axis. 8.4 Magnetoquasistatic Fields in the Presence of Perfect Conductors 8.4.1∗A current loop of radius Ris at the center of a conducting spherical shell having radius b. Assume that R/lessmuchband that i(t) is so rapidly varying that the shell can be taken as perfectly conducting. Show that in spherical coordinates, where R/lessmuchr < b H=iπR2 4π· 2 cosθ¡1 r3−1 b3¢ ir+ sin θ¡1 r3+2 b3¢ iθ¸ (a) 8.4.2 The two-dimensional magnetic dipole of Example 8.1.2 is at the center of a conducting shell having radius a/greatermuchd. The current i(t) is so rapidly varying that the shell can be regarded as perfectly conducting. What are Ψ and H in the region d/lessmuchr < a ? 8.4.3∗The cross-section of a two-dimensional system is shown in Fig. P8.4.3. A magnetic ﬂux per unit length sµoHois trapped between perfectly conduct- ing plane parallel plates that extend to inﬁnity to the left and right. At the origin on the lower plate is a perfectly conducting half-cylinder of radius R. (a) Show that if s/greatermuchR, then Ψ =HoR¡r R+R r¢ cosφ (a) 54Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. P8.4.3 Fig. P8.4.6 (b) Show that a plot of Hwould appear as in the left half of Fig. 8.4.2 turned on its side. 8.4.4 In a three-dimensional version of that shown in Fig. P8.4.3, a perfectly conducting hemispherical bump of radius s/greatermuchRis attached to the lower of two perfectly conducting plane parallel plates. The hemisphere is centered at the origin of a spherical coordinate system such as in Fig. P8.4.3, with φ→θ. The magnetic ﬁeld intensity is uniform far from the hemisphere. Determine Ψ and H. 8.4.5∗Running from z=−∞toz= +∞at (x, y) = (0 ,−h) is a wire. The wire is parallel to a perfectly conducting plane at y= 0. When t= 0, a current stepi=Iu−1(t) is applied in the + zdirection to the wire. (a) Show that in the region y <0, H=i 2π½−(y+h)ix+xiy [x2+ (y+h)2]+(y−h)ix−xiy [x2+ (y−h)2]¾ for t >0 ( a) (b) Show that the surface current density at y= 0 is Kz=−ih/π(x2+ h2). 8.4.6 The cross-section of a system that extends to inﬁnity in the ±zdirections is shown in Fig. P8.4.6. Surrounded by free space, a sheet of current has Sec. 8.5 Problems 55 Fig. P8.5.1 Fig. P8.5.2 the surface current density Koizuniformly distributed between x=band x=a. The plane x= 0 is perfectly conducting. (a) Determine Ψ in the region 0 < x. (b) Find Kin the plane x= 0. 8.5 Piece-Wise Magnetic Fields 8.5.1∗The cross-section of a cylindrical winding is shown in Fig. P8.5.1. As pro- jected onto the y= 0 plane, the number of turns per unit length is constant and equal to N/2R. The cylinder can be modeled as inﬁnitely long in the axial direction. (a) Given that the winding carries a current i, show that Ψ =Ni 4½ (R/r) cosφ;R < r −(r/R) cosφ;r < R(a) 56Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 and that therefore H=Ni 4R½ (R/r)2[cosφir+ sin φiφ];R < r [cosφir−sinφiφ]; r < R(b) (b) Show that the inductance per unit length of the winding is L= πµoN2/8. 8.5.2 The cross-section of a rotor, coaxial with a perfectly conducting “magnetic shield,” is shown in Fig. P8.5.2. Windings consisting of Nturns per unit peripheral length are distributed uniformly at r=bso that at a given instant in time, the surface current distribution is as shown. At r=a, there is the inner surface of a perfect conductor. The system is very long in the zdirection. (a) What are the continuity conditions on Ψ at r=band the boundary condition at r=a? (b) Find Ψ, and hence H, in regions (a) and (b) outside and inside the winding, respectively. (c) With the understanding that the rotor is wound using one wire, so that each turn is in series with the next and a wire carrying the current in the + zdirection at φreturns the current in the −zdirection at −φ, what is the inductance of the rotor coil? Why is it independent of the rotor position φo? 8.6 Vector Potential 8.6.1∗In Example 1.4.1, the magnetic ﬁeld intensity is determined to be that given by (1.4.7). Deﬁne Azto be zero at the origin. (a) Show that if Hφis to be ﬁnite in the neighborhood of r=R,Azmust be continuous there. (b) Show that Ais given by A=−izµoJoR2 3½1 3(r/R)3; r < R ln(r/R) +1 3;r > R(a) (c) The loop designated by C/primein Fig. 1.4.2 has a length lin the zdirec- tion, an inner leg at r= 0, and an outer leg at r=a > R . Use Ato show that the ﬂux linked is λ=−lAz(a) =µoJoR2l 3£ ln(a/R) +1 3¤ (b) 8.6.2 For the conﬁguration of Prob. 1.4.2, deﬁne Azas being zero at the origin. (a) Determine Azin the regions r < b andb < r < a . Sec. 8.6 Problems 57 Fig. P8.6.5 (b) Use Ato determine the ﬂux linked by a closed rectangular loop having length lin the zdirection and each of its four sides in a plane of constant φ. Two of the sides are parallel to the zaxis, one at radius r=cand the other at r= 0. The other two, respectively, join the ends of these segments, running radially from r= 0 to r=c. 8.6.3∗In cylindrical coordinates, µoH=µo[Hr(r, z)ir+Hz(r, z)iz]. That is, the magnetic ﬂux density is axially symmetric and does not have a φcompo- nent. (a) Show that A= [Λ c(r, z)/r]iφ (a) (b) Show that the ﬂux passing between contours at r=aandr=bis λ= 2π[Λc(a)−Λc(b)] ( b) 8.6.4∗For the inductive attenuator considered in Example 8.6.3 and Demonstra- tion 8.6.2: (a) derive the vector potential, (20), without identifying this MQS prob- lem with its EQS counterpart. (b) Show that the current is as given by (21). (c) In the limit where b/a/greatermuch1, show that the response has the depen- dence on b/ashown in the plot of Fig. 8.6.11. (d) Show that in the opposite limit, where b/a/lessmuch1, the total current in the lower plate (21) is consistent with a magnetic ﬁeld intensity between the upper and lower plates that is uniform (with respect to y) and hence equal to (Λ /bµo)ix. Note that ∞X n=1 odd1 n2=π2 8(a) 8.6.5 Perfectly conducting electrodes are composed of sheets bent into the shape of/unionsq’s, as shown in Fig. P8.6.5. The length of the system in the zdirection is very large compared to the length 2 aor height d, so the ﬁelds can be 58Magnetoquasistatic Fields: Superposition Integral and Boundary Value Points of View Chapter 8 Fig. P8.6.6 regarded as two dimensional. The insulating gaps have a width ∆ that is small compared to all dimensions. Passing through these gaps is a magnetic ﬂux (per unit length in the zdirection) Λ( t). One method of solution is suggested by Example 6.6.3. (a) Find Ain regions (a) and (b) to the right and left, respectively, of the plane x= 0. (b) Sketch H. 8.6.6∗The wires comprising the winding shown in cross-section by Fig. P8.6.6 carry current in the −zdirection over the range 0 < x < a and return this current over the range −a < x < 0. These windings extend uniformly over the range 0 < y < b . Thus, the current density in the region of interest is J=−inosin(πx/a )iz, where iis the current carried by each wire and |nosin(πx/a )|is the number of turns per unit area. This region is surrounded by perfectly conducting walls at y= 0 and y=band at x=−aandx=a. The length lin the zdirection is much greater than either aorb. (a) Show that A=izµoino(a/π)2sin¡πx a¢·coshπ a¡ y−b 2¢ cosh¡πb 2a¢−1¸ (a) (b) Show that the inductance of the winding is L= 2µon2 ola4 π3·¡πb 2a¢ −tanh¡πb 2a¢¸ (b) (c) Sketch H. 8.6.7 In the conﬁguration of Prob. 8.6.6, the rectangular region is uniformly ﬁlled with wires that all carry their current in the zdirection. There are noof these wires per unit area. The current carried by each wire is returned in the perfectly conducting walls. (a) Determine A. (b) Assume that all the wires are connected to the wall by a terminating plate at z=land that each is driven by a current source i(t) in the plane z= 0. Note that it has been assumed that each of these current Sec. 8.6 Problems 59 sources is the same function of time. What is the voltage v(x, y, t ) of these sources? 8.6.8 In the conﬁguration of Prob. 8.6.6, the turns are uniformly distributed. Thus, nois a constant representing the number of wires per unit area carrying current in the −zdirection in the region 0 < x. Assume that the wire carrying current in the −zdirection at the location ( x, y) returns the current at ( −x, y). (a) Determine A. (b) Find the inductance L. 9 MAGNETIZATION 9.0 INTRODUCTION The sources of the magnetic ﬁelds considered in Chap. 8 were conduction currents associated with the motion of unpaired charge carriers through materials. Typically, the current was in a metal and the carriers were conduction electrons. In this chapter, we recognize that materials provide still other magnetic ﬁeld sources. These account for the ﬁelds of permanent magnets and for the increase in inductance produced in a coil by insertion of a magnetizable material. Magnetization eﬀects are due to the propensity of the atomic constituents of matter to behave as magnetic dipoles. It is natural to think of electrons circulating around a nucleus as comprising a circulating current, and hence giving rise to a magnetic moment similar to that for a current loop, as discussed in Example 8.3.2. More surprising is the magnetic dipole moment found for individual electrons. This moment, associated with the electronic property of spin, is deﬁned as the Bohr magneton me=±e m1 2¯h (1) where e/m is the electronic charge-to-mass ratio, 1 .76×1011coulomb/kg, and 2 π¯h is Planck’s constant, ¯ h= 1.05×10−34joule-sec so that mehas the units A−m2. The quantum mechanics of atoms and molecules dictates that, whether due to the orbits or to the spins, the electronic contributions to their net dipole moments tend to cancel. Those that do make a contribution are typically in unﬁlled shells. An estimate of the moment that would result if each atom or molecule of a material contributed only one Bohr magneton shows that the orbital and spin contributions from all the electrons comprising a typical solid had better tend to cancel or the resulting ﬁeld eﬀects would be prodigious indeed. Even if each atom or molecule is made to contribute only one Bohr magneton of magnetic moment, a 1 2 Magnetization Chapter 9 magnetic ﬁeld results comparable to that produced by extremely large conduction currents. To make this apparent, compare the magnetic ﬁeld induced by a current loop having a radius Rand carrying a current i(Fig. 9.0.la) to that from a spherical collection of dipoles (Fig. 9.0.1b), each having the magnetic moment of only one electron. Fig. 9.0.1 (a) Current iin loop of radius Rgives dipole moment m. (b) Spherical material of radius Rhas dipole moment approximated as the sum of atomic dipole moments. In the case of the spherical material, we consider the net dipole moment to be simply the moment meof a single molecule multiplied by the number of molecules. The number of molecules per unit mass is Avogadro’s number ( A0= 6.023×1026 molecules/kg-mole) divided by the molecular weight, Mo. The mass is the volume multiplied by the mass density ρ(kg/m3). Thus, for a sphere having radius R, the sum of the dipole moments is m=me¡4 3πR3ρ¢¡Ao Mo¢ (2) Suppose that the current loop shown in Fig. 9.0.1a has the same radius Ras the sphere. What current iwould give rise to a magnetic moment equal to that from the sphere of hypothetical material? If the moment of the loop, given by (8.3.19) as being m=iπR2, is set equal to that of the sphere, (2), it follows that imust be i=me4 3RρAo Mo(3) Hence, for iron (where ρ= 7.86×103andMo= 56) and a radius of 10 cm, the current required to produce the same magnetic moment is 105A. Material magnetization can either be permanent or be induced by the appli- cation of a ﬁeld, much as for the polarizable materials considered in Chap. 6. In most materials, the average moment per molecule that can be brought into play is much less than one Bohr magneton. However, highly magnetizable materials can produce net magnetic moments comparable to that estimated in (2). The development of magnetization in this chapter parallels that for polariza- tion in Chap. 6. Just as the polarization density was used in Sec. 6.1 to represent the eﬀect of electric dipoles on the electric ﬁeld intensity, the magnetization density introduced in Sec. 9.1 will account for the contributions of magnetic dipoles to the magnetic ﬁeld intensity. The MQS laws and continuity conditions then collected in Sec. 9.2 are the basis for the remaining sections, and for Chap. 10 as well. Because permanent magnets are so common, the permanent magnetization ﬁelds considered in Sec. 9.3 are more familiar than the permanent polarization electric ﬁelds of Sec. 6.3. Similarly, the force experienced as a piece of iron is brought Sec. 9.1 Magnetization Density 3 into a magnetic ﬁeld is common evidence of the induced magnetization described by the constitutive laws of Sec. 9.4. The extensive analogy between polarization and magnetization makes most of the examples from Chap. 6 analogous to magnetization examples. This is especially true in Secs. 9.5 and 9.6, where materials are considered that have a magnetization that is linearly related to the magnetic ﬁeld intensity. Thus, these sections not only build on the insights gained in the earlier sections on polarization, but give the opportunity to expand on both topics as well. The magnetic circuits considered in Sec. 9.7 are of great practical interest and exemplify an approximate way for the evaluation of ﬁelds in the presence of strongly magnetized materials. The saturation of magnetizable materials is of primary practical concern. The problems for Secs. 9.6 and 9.7 are an introduction to ﬁelds in materials that are magnetically nonlinear. We generalize Faraday’s law in Sec. 9.2 so that it can be used in this chapter to predict the voltage at the terminals of coils in systems that include magnetization. This generalization is used to determine terminal relations that include magneti- zation in Sec. 9.5. The examples in the subsequent sections study the implications of Faraday’s law with magnetization included. As in Chap. 8, we conﬁne ourselves in this chapter to examples that can be modeled using the terminal variables of perfectly conducting circuits. The MQS laws, generalized in Sec. 9.2 to include magnetization, form the basis for the discussion of electric ﬁelds in MQS systems that is the theme of Chap. 10. 9.1 MAGNETIZATION DENSITY The sources of magnetic ﬁeld in matter are the (more or less) aligned magnetic dipoles of individual electrons or currents caused by circulating electrons.1We now describe the eﬀect on the magnetic ﬁeld of a distribution of magnetic dipoles rep- resenting the material. In Sec. 8.3, we deﬁned the magnitude of the magnetic moment mof a cir- culating current loop of current iand area aasm=ia. The moment vector, m, was deﬁned as normal to the surface spanning the contour of the loop and pointing in the direction determined by the right-hand rule. In Sec. 8.3, where the moment was in the zdirection in spherical coordinates, the loop was found to produce the magnetic ﬁeld intensity H=µom 4πµor3[2 cos θir+ sin θiθ] (1) This ﬁeld is analogous to the electric ﬁeld associated with a dipole having the moment p. With pdirected along the zaxis, the electric dipole ﬁeld is given by taking the gradient of (4.4.10). E=p 4π/epsilon1or3[2 cos θir+ sin θiθ] (2) 1Magnetic monopoles, which would play a role with respect to magnetic ﬁelds analogous to that of the charge with respect to electric ﬁelds, may in fact exist, but are certainly not of engineering signiﬁcance. See Science , Research News, “In search of magnetic monopoles,” Vol. 216, p. 1086 (June 4, 1982). 4 Magnetization Chapter 9 Thus, the dipole ﬁelds are obtained from each other by making the identiﬁcations p↔µom (3) In Sec. 6.1, a spatial distribution of electric dipoles is represented by the polarization density P=Np, where Nis the number density of dipoles. Similarly, here we deﬁne amagnetization density as M=Nm (4) where again Nis the number of dipoles per unit volume. Note that just as the analog of the dipole moment pisµom, the analog of the polarization density Pis µoM. 9.2 LAWS AND CONTINUITY CONDITIONS WITH MAGNETIZATION Recall that the eﬀect of a spatial distribution of electric dipoles upon the electric ﬁeld is described by a generalization of Gauss’ law for electric ﬁelds, (6.2.1) and (6.2.2), ∇ ·/epsilon1oE=−∇ ·P+ρu (1) The eﬀect of the spatial distribution of magnetic dipoles upon the magnetic ﬁeld intensity is now similarly taken into account by generalizing the magnetic ﬂux continuity law. ∇ ·µoH=−∇ · µoM (2) In this law, there is no analog to an unpaired electric charge density. The continuity condition found by integrating (2) over an incremental volume enclosing a section of an interface having a normal nis n·µo(Ha−Hb) =−n·µo(Ma−Mb) (3) Suggested by the analogy to the description of polarization is the deﬁnition of the quantities on the right in (2) and (3), respectively, as the magnetic charge density ρmand the magnetic surface charge density σsm. ρm≡ −∇ · µoM (4) σsm≡ −n·µo(Ma−Mb) (5) Sec. 9.2 Laws and Continuity 5 Faraday’s Law Including Magnetization. The modiﬁcation of the magnetic ﬂux continuity law implies that another of Maxwell’s equations must be generalized. In introducing the ﬂux continuity law in Sec. 1.7, we observed that it was almost inherent in Faraday’s law. Because the divergence of the curl is zero, the divergence of the free space form of Faraday’s law reduces to ∇ ·(∇ ×E) = 0 = −∂ ∂t∇ ·µoH (6) Thus, in free space, µoHmust have a divergence that is at least constant in time. The magnetic ﬂux continuity law adds the information that this constant is zero. In the presence of magnetizable material, (2) shows that the quantity µo(H+M) is solenoidal. To make Faraday’s law consistent with this requirement, the law is now written as ∇ ×E=−∂ ∂tµo(H+M)(7) Magnetic Flux Density. The grouping of HandMin Faraday’s law and the ﬂux continuity law makes it natural to deﬁne a new variable, the magnetic ﬂux density B. B≡µo(H+M) (8) This quantity plays a role that is analogous to that of the electric displacement ﬂux density Ddeﬁned by (6.2.14). Because there are no macroscopic quantities of monopoles of magnetic charge, its divergence is zero. That is, the ﬂux continuity law, (2), becomes simply ∇ ·B= 0 (9) and the corresponding continuity condition, (3), becomes simply n·(Ba−Bb) = 0 (10) A similar simpliﬁcation is obtained by writing Faraday’s law in terms of the magnetic ﬂux density. Equation (7) becomes ∇ ×E=−∂B ∂t (11) If the magnetization is speciﬁed independent of H, it is usually best to have it entered explicitly in the formulation by not introducing B. However, if Mis given 6 Magnetization Chapter 9 as a function of H, especially if it is linear in H, it is most convenient to remove Mfrom the formulation by using Bas a variable. Terminal Voltage with Magnetization. In Sec. 8.4, where we discussed the terminal voltage of a perfectly conducting coil, there was no magnetization. The generalization of Faraday’s law to include magnetization requires a generalization of the terminal relation. The starting point in deriving the terminal relation was Faraday’s integral law, (8.4.9). This law is generalized to included magnetization eﬀects by replacing µoHwithB. Otherwise, the derivation of the terminal relation, (8.4.11), is the same as before. Thus, the terminal voltage is again v=dλ dt (12) but now the ﬂux linkage is λ≡Z SB·da (13) In Sec. 9.4 we will see that Faraday’s law of induction, as reﬂected in these last two relations, is the basis for measuring B. 9.3 PERMANENT MAGNETIZATION As the modern-day versions of the lodestone, which made the existence of magnetic ﬁelds apparent in ancient times, permanent magnets are now so cheaply manufac- tured that they are used at home to pin notes on the refrigerator and so reliable that they are at the heart of motors, transducers, and information storage systems. To a ﬁrst approximation, a permanent magnet can be modeled by a material hav- ing a speciﬁed distribution of magnetization density M. Thus, in this section we consider the magnetic ﬁeld intensity generated by prescribed distributions of M. In a region where there is no current density J, Amp` ere’s law requires that H be irrotational. It is then often convenient to represent the magnetic ﬁeld intensity in terms of the scalar magnetic potential Ψ introduced in Sec. 8.3. H=−∇Ψ (1) From the ﬂux continuity law, (9.2.2), it then follows that Ψ satisﬁes Poisson’s equation. ∇2Ψ =−ρm µo;ρm≡ −∇ · µoM (2) A speciﬁed magnetization density leads to a prescribed magnetic charge density ρm. The situation is analogous to that considered in Sec. 6.3, where the polarization density was prescribed and, as a result, where ρpwas known. Sec. 9.3 Permanent Magnetization 7 Fig. 9.3.1 (a) Cylinder of circular cross-section uniformly magnetized in the direction of its axis. (b) Axial distribution of scalar magnetic potential and (c) axial magnetic ﬁeld intensity. For these distributions, the cylinder length is assumed to be equal to its diameter. Of course, the net magnetic charge of a magnetizable body is always zero, because Z Vρmdv=I SµoH·da= 0 (3) if the integral is taken over the entire volume containing the body. Techniques for solving Poisson’s equation for a prescribed charge distribution developed in Chaps. 4 and 5 are directly applicable here. For example, if the magnetization is given throughout all space and there are no other sources, the magnetic scalar potential is given by a superposition integral. Just as the integral of (4.2.2) is (4.5.3), so the integral of (2) is Ψ =Z V/primeρm(r/prime)dv 4πµo|r−r/prime| (4) If the region of interest is bounded by material on which boundary conditions are speciﬁed, (4) provides the particular solution. Example 9.3.1. Magnetic Field Intensity of a Uniformly Magnetized Cylinder The cylinder shown in Fig. 9.3.1 is uniformly magnetized in the zdirection, M= Moiz. The ﬁrst step toward ﬁnding the resulting Hwithin the cylinder and in the surrounding free space is an evaluation of the distribution of magnetic charge density. The uniform Mhas no divergence, so ρm= 0 throughout the volume. Thus, the source of His on the surfaces where Moriginates and terminates. In view of (9.2.3), it takes the form of the surface charge density σsm=−n·µo(Ma−Mb) =±µoMo (5) The upper and lower signs refer to the upper and lower surfaces. 8 Magnetization Chapter 9 In principle, we could use the superposition integral to ﬁnd the potential ev- erywhere. To keep the integration simple, we conﬁne ourselves here to ﬁnding it on thezaxis. The integration of (4) then reduces to integrations over the endfaces of the cylinder. Ψ =ZR 0µoMo2πρ/primedρ/prime 4πµoq ρ/prime2+¡ z−d 2¢2−ZR 0µoMo2πρ/primedρ/prime 4πµoq ρ/prime2+¡ z+d 2¢2(6) With absolute magnitudes used to make the expressions valid regardless of position along the zaxis, these integrals become Ψ =dMo 2·r¡R d¢2+¡z d−1 2¢2−¯¯z d−1 2¯¯ −r¡R d¢2+¡z d+1 2¢2+¯¯z d+1 2¯¯¸ (7) The ﬁeld intensity follows from (1) Hz=−dMo 2· ¡z d−1 2¢ q¡R d¢2+¡z d−1 2¢2−¡z d+1 2¢ q¡R d¢2+¡z d+1 2¢2+u¸ (8) where u≡0 for|z|> d/2 and u≡2 for−d/2< z < d/ 2. Here, from top to bottom, respectively, the signs correspond to evaluating the ﬁeld above the upper surface, within the magnet, and below the bottom surface. The axial distributions of Ψ and Hzshown in Fig. 9.3.1 are consistent with a three-dimensional picture of a ﬁeld that originates on the top face of the magnet and terminates on the bottom face. As for the spherical magnet (the analogue of the permanently polarized sphere shown in Fig. 6.3.1), the magnetic ﬁeld intensity inside the magnet has a direction opposite to that of M. In practice, Mwould most likely be determined by making measurements of the external ﬁeld and then deducing Mfrom this ﬁeld. If the magnetic ﬁeld intensity is generated by a combination of prescribed currents and permanent magnetization, it can be evaluated by superimposing the ﬁeld due to the current and the magnetization. For example, suppose that the uniformly magnetized circular cylinder of Fig. 9.3.1 were surrounded by the N- turn solenoid of Fig. 8.2.3. Then the axial ﬁeld intensity would be the sum of that for the current [predicted by the Biot-Savart law, (8.2.7)], and for the magnetization [predicted by the negative gradient of (4)]. Example 9.3.2. Retrieval of Signals Stored on Magnetizable Tape Permanent magnetization is used for a permanent record in the tape recorder. Currents in an electromagnet are used to induce the permanent magnetization, ex- ploiting the hysteresis in the magnetization of certain materials, as will be discussed Sec. 9.3 Permanent Magnetization 9 Fig. 9.3.2 Permanently magnetized tape has distribution of Mrep- resenting a Fourier component of a recorded signal. From a frame of reference attached to the tape, the magnetization is static. Fig. 9.3.3 From the frame of reference of a sensing coil, the tape is seen to move in the x/primedirection with the velocity U. in Sec. 9.4. Here we look at a model of perpendicular magnetization, an actively pur- sued research ﬁeld. The conventional recording is done by producing magnetization Mparallel to the tape. In a thin tape at rest, the magnetization density shown in Fig. 9.3.2 is assumed to be uniform over the thickness and to be of the simple form M=Mocosβxiy (9) The magnetic ﬁeld is ﬁrst determined in a frame of reference attached to the tape, denoted by ( x, y, z ) as deﬁned in Fig. 9.3.2. The tape moves with a velocity Uwith respect to a ﬁxed sensing “head,” and so our second step will be to represent this ﬁeld in terms of ﬁxed coordinates. With Fig. 9.3.3 in view, it is clear that these coordinates, denoted by ( x/prime, y/prime, z/prime), are related to the moving coordinates by x/prime=x+Ut→x=x/prime−Ut; y=y/prime(10) Thus, from the ﬁxed reference frame, the magnetization takes the form of a traveling wave. M=Mocosβ(x/prime−Ut)iy (11) IfMis observed at a ﬁxed location x/prime, it has a sinusoidal temporal variation with the frequency ω=βU. This relationship between the ﬁxed frame frequency and the spatial periodicity suggests how the distribution of magnetization is established by “recording” a signal having the frequency ω. The magnetization density has no divergence in the volume of the tape, so the ﬁeld source is a surface charge density. With upper and lower signs denoting the upper and lower tape surfaces, it follows that σm=±µoMocosβx (12) The continuity conditions to be satisﬁed at the upper and lower surfaces represent the continuity of magnetic ﬂux (9.2.3) µoHa y−µoHo y=µoMocosβx aty=d 2 µoHo y−µoHb y=−µoMocosβx aty=−d 2(13) 10 Magnetization Chapter 9 and the continuity of tangential H Ψa= Ψ oaty=d 2 Ψo= Ψ baty=−d 2(14) In addition, the ﬁeld should go to zero as y→ ±∞ . Because the ﬁeld sources are conﬁned to surfaces, the magnetic scalar potential must satisfy Laplace’s equation, (2) with ρm= 0, in the bulk regions delimited by the interfaces. Motivated by the “odd” symmetry of the source with respect to the y= 0 plane and its periodicity in x, we pick solutions to Laplace’s equation for the magnetic potential above (a), inside ( o), and below (b) the tape that also satisfy the odd symmetry condition of having Ψ( y) =−Ψ(−y). ψa=A e−βycosβx ψo=Csinhβycosβx (15) ψb=−A eβycosβx Subject to the requirement that β >0, the exterior potentials go to zero at y=±∞. The interior function is made an odd function of yby excluding the cosh( βy) cos( βx) solution to Laplace’s equation, while the exterior functions are made odd by making the coeﬃcients equal in magnitude and opposite in sign. Thus, only two coeﬃcients remain to be determined. These follow from substituting the assumed solution into either of (13) and either of (14), and then solving the two equations to obtain A=Mo βeβd/2¡ 1 + cothβd 2¢−1 C=Mo β£¡ 1 + cothβd 2¢ sinhβd 2¤−1(16) The conditions at one interface are automatically satisﬁed if those at the other are met. This is a proof that the assumed solutions have indeed been correct. Our fore- sight in deﬁning the origin of the yaxis to be at the symmetry plane and exploiting the resulting odd dependence of Ψ on yhas reduced the number of undetermined coeﬃcients from four to two. This ﬁeld is now expressed in the ﬁxed frame coordinates. With Adeﬁned by (16a) and xandygiven in terms of the ﬁxed frame coordinates by (10), the magnetic potential above the tape has been determined to be Ψa=Mo βe−β(y/prime−d 2) ¡ 1 + cothβd 2¢cosβ(x/prime−Ut) (17) Next, we determine the output voltage of a ﬁxed coil, positioned at a height habove the tape, as shown in Fig. 9.3.3. This detecting “head” has Nturns, a length lin the x/primedirection, and width win the zdirection. With the objective of ﬁnding the ﬂux linkage, we use (17) to determine the y-directed ﬂux density in the neighborhood of the coil. By=−µo∂Ψa ∂y/prime=µoMoe−β(y/prime−d 2) ¡ 1 + cothβd 2¢cosβ(x/prime−Ut) (18) Sec. 9.3 Permanent Magnetization 11 Fig. 9.3.4 Magnitude of sensing coil output voltage as a function of βl= 2πl/Λ, where Λ is the wavelength of the magnetization. If the mag- netization is produced by a ﬁxed coil driven at the angular frequency ω, the horizontal axis, which is then ωl/U , is proportional to the recording frequency. The ﬂux linkage follows by multiplying the number of turns Ntimes Byintegrated over the surface in the plane y=h+1 2dspanned by the coil. λ=wNZl/2 −l/2By¡ y/prime=h+d 2¢ dx/prime =µoMowNe−βh β¡ 1 + cothβd 2¢£ sinβ¡l 2−Ut¢ + sin β¡l 2+Ut¢¤(19) The dependence on lis clariﬁed by using a trigonometric identity to simplify the last term in this expression. λ=2µoMowNe−βh β¡ 1 + cothβd 2¢sinβl 2cosβUt (20) Finally, the output voltage follows from (9.2.12). vo=dλ dt=−2µoMowUN¡ 1 + cothβd 2¢e−βhsinβl 2sinβUt (21) The strong dependence of this expression on the wavelength of the magnetization, 2π/β, reﬂects the nature of ﬁelds predicted using Laplace’s equation. It follows from (21) that the output voltage has the angular frequency ω=βU. Thus, (21) can also be regarded as giving the frequency response of the sensor. The magnitude of vohas the dependence on either the normalized βorωshown in Fig. 9.3.4. Two phenomena underlie the voltage response. The periodic dependence re- ﬂects the relationship between the length lof the coil and the wavelength 2 π/βof the magnetization. When the coil length is equal to the wavelength, there is as much positive as negative ﬂux linking the coil at a given instant, and the signal falls to zero. This is also the condition when lis any multiple of a wavelength and accounts for the sin(1 2βl) term in (21). 12 Magnetization Chapter 9 Fig. 9.4.1 Toroidal coil with donut-shaped magnetizable core. The strong decay of the envelope of the output signal as the frequency is increased, and hence the wavelength decreased, reﬂects a property of Laplace’s equation that frequently comes into play in engineering electromagnetic ﬁelds. The shorter the wavelength, the more rapid the decay of the ﬁeld in the direction per- pendicular to the tape. With the sensing coil at a ﬁxed height above the tape, this means that once the wavelength is on the order of 2 πh, there is an essentially expo- nential decrease in signal with increasing frequency. Thus, there is a strong incentive to place the coil as close to the tape as possible. We should expect that if the tape is very thin compared to the wavelength, the ﬁeld induced by magnetic surface charges on the top surface would tend to be canceled by those of opposite sign on the surface just below. This eﬀect is accounted for by the term [1 + coth(1 2βd)] in the denominator of (21). In a practical recording device, the sensing head of the previous example would incorporate magnetizable materials. To predict how these aﬀect the ﬁelds, we need a law relating the ﬁeld to the magnetization it induces. This is the subject of the next section. 9.4 MAGNETIZATION CONSTITUTIVE LAWS The permanent magnetization model of Sec. 9.3 is a somewhat artiﬁcial example of the magnetization density Mspeciﬁed, independent of the magnetic ﬁeld intensity. Even in the best of permanent magnets, there is actually some dependence of M onH. Constitutive laws relate the magnetization density Mor the magnetic ﬂux density Bto the macroscopic Hwithin a material. Before discussing some of the more common relations and their underlying physics, it is well to have in view an experiment giving direct evidence of the constitutive law of magnetization. The objective is to observe the establishment of Hby a current in accordance with Amp` ere’s law, and deduce Bfrom the voltage it induces in accordance with Fara- day’s law. Example 9.4.1. Toroidal Coil A coil of toroidal geometry is shown in Fig. 9.4.1. It consists of a donut-shaped core ﬁlled with magnetizable material with N1turns tightly wound on its periphery. By means of a source driving its terminals, this coil carries a current i. The resulting Sec. 9.4 Magnetization Constitutive Laws 13 Fig. 9.4.2 Surface Senclosed by contour Cused with Amp` ere’s inte- gral law to determine Hin the coil shown in Fig. 9.4.1. current distribution can be assumed to be so smooth that the ﬁne structure of the ﬁeld, caused by the ﬁnite size of the wires, can be disregarded. We will ignore the slight pitch of the coil and the associated small current component circulating around the axis of the toroid. Because of the toroidal geometry, the Hﬁeld in the magnetizable material is determined by Amp` ere’s law and symmetry considerations. Symmetry about the toroidal axis suggests that Hisφdirected. The integral MQS form of Amp` ere’s law is written for a contour Ccirculating about the toroidal axis within the core and at a radius r. Because the major radius Rof the torus is large compared to the minor radius1 2w, we will ignore the variation of rover the cross-section of the torus and approximate rby an average radius R. The surface Sspanned by this contour and shown in Fig. 9.4.2 is pierced N1times by the current i, giving a total current of N1i. Thus, the azimuthal ﬁeld inside the core is essentially 2πrH φ=N1i→Hφ≡H=N1i 2πr/similarequalN1i 2πR(1) Note that the same argument shows that the magnetic ﬁeld intensity outside the core is zero. In general, if we are given the current distribution and wish to determine H, recourse must be made not only to Amp` ere’s law but to the ﬂux continuity condition as well. In the idealized toroidal geometry, where the ﬂux lines automatically close on themselves without leaving the magnetized material, the ﬂux continuity condition is automatically satisﬁed. Thus, in the toroidal conﬁguration, the Himposed on the core is determined by a measurement of the current iand the geometry. How can we measure the magnetic ﬂux density in the core? Because Bappears in Faraday’s law of induction, the measurement of the terminal voltage of an addi- tional coil, having N2turns also wound on the donut-shaped core, gives information onB. The terminals of this coil are terminated in a high enough impedance so that there is a negligible current in this second winding. Thus, the Hﬁeld established by the current iremains unaltered. The ﬂux linked by each turn of the sensing coil is essentially the ﬂux density multiplied by the cross-sectional area πw2/4 of the core. Thus, the ﬂux linked by the terminals of the sensing coil is λ2=πw2 4N2B (2) and ﬂux density in the core material is directly reﬂected in the terminal ﬂux-linkage. The following demonstration shows how (1) and (2) can be used to infer the magnetization characteristic of the core material from measurement of the terminal current and voltage of the ﬁrst and second coils. Demonstration 9.4.1. Measurement of B−HCharacteristic 14 Magnetization Chapter 9 Fig. 9.4.3 Demonstration in which the B−Hcurve is traced out in the sinusoidal steady state. The experiment shown in Fig. 9.4.3 displays the magnetization characteristic on the oscilloscope. The magnetizable material is in the donut-shaped toroidal conﬁguration of Example 9.4.1 with the N1-turn coil driven by a current ifrom a Variac. The voltage across a series resistance then gives a horizontal deﬂection of the oscilloscope proportional to H, in accordance with (1). The terminals of the N2turn-coil are connected through an integrating net- work to the vertical deﬂection terminals of the oscilloscope. Thus, the vertical deﬂec- tion is proportional to the integral of the terminal voltage, to λ, and hence through (2), to B. In the discussions of magnetization characteristics which follow, it is helpful to think of the material as comprising the core of the torus in this experiment. Then the magnetic ﬁeld intensity His proportional to the current i, while the magnetic ﬂux density Bis reﬂected in the voltage induced in a coil linking this ﬂux. Many materials are magnetically linear in the sense that M=χmH (3) Here χmis the magnetic susceptibility . More commonly, the constitutive law for a magnetically linear material is written in terms of the magnetic ﬂux density, deﬁned by (9.2.8). B=µH; µ≡µo(1 +χm) (4) According to this law, the magnetization is taken into account by replacing the permeability of free space µoby the permeability µof the material. For purposes of comparing the magnetizability of materials, the relative permeability µ/µois often used. Typical susceptibilities for certain elements, compounds, and common materi- als are given in Table 9.4.1. Most common materials are only slightly magnetizable. Some substances that are readily polarized, such as water, are not easily magne- tized. Note that the magnetic susceptibility can be either positive or negative and that there are some materials, notably iron and its compounds, in which it can be enormous. In establishing an appreciation for the degree of magnetizability that can be expected of a material, it is helpful to have a qualitative picture of its mi- Sec. 9.4 Magnetization Constitutive Laws 15 TABLE 9.4.1 RELATIVE SUSCEPTIBILITIES OF COMMON MATERIALS Material χm PARAMAGNETIC Mg 1.2×10−5 Al 2.2×10−5 Pt 3.6×10−4 air 3.6×10−7 O2 2.1×10−6 DIAMAGNETIC Na −0.24×10−5 Cu −1.0×10−5 diamond −2.2×10−5 Hg −3.2×10−5 H2O −0.9×10−5 FERROMAGNETIC Fe (dynamo sheets) 5.5×103 Fe (lab specimens) 8.8×104 Fe (crystals) 1.4×106 Si-Fe transformer sheets 7×104 Si-Fe crystals 3.8×106 µ-metal 105 FERRIMAGNETIC Fe3O4 100 ferrites 5000 croscopic origins, beginning at the atomic level but including the collective eﬀects of groups of atoms or molecules that result when they become as densely packed as they are in solids. These latter eﬀects are prominent in the most easily magnetized materials. The magnetic moment of an atom (or molecule) is the sum of the orbital and spin contributions. Especially in a gas, where the atoms are dilute, the magnetic susceptibility results from the (partial) alignment of the individual magnetic mo- ments by a magnetic ﬁeld. Although the spin contributions to the moment tend to cancel, many atoms have net moments of one or more Bohr magnetons. At room temperature, the orientations of the moments are mostly randomized by thermal agitation, even under the most intense ﬁelds. As a result, an applied ﬁeld can give rise to a signiﬁcant magnetization only at very low temperatures. A paramagnetic material displays an appreciable susceptibility only at low temperatures. If, in the absence of an applied ﬁeld, the spin contributions to the moment of an atom very nearly cancel, the material can be diamagnetic , in the sense that it displays a slightly negative susceptibility. With the application of a ﬁeld, the 16 Magnetization Chapter 9 Fig. 9.4.4 Typical magnetization curve without hysteresis. For typical fer- romagnetic solids, the saturation ﬂux density is in the range of 1–2 Tesla. For ferromagnetic domains suspended in a liquid, it is .02–.04 Tesla. orbiting electrons are slightly altered in their circulations, giving rise to changes in moment in a direction opposite to that of the applied ﬁeld. Again, thermal energy tends to disorient these moments. At room temperature, this eﬀect is even smaller than that for paramagnetic materials. At very low temperatures, it is possible to raise the applied ﬁeld to such a level that essentially all the moments are aligned. This is reﬂected in the saturation of the ﬂux density B, as shown in Fig. 9.4.4. At low ﬁeld intensity, the slope of the magnetization curve is µ, while at high ﬁeld strengths, there are no more moments to be aligned and the slope is µo. As long as the ﬁeld is raised and lowered at a rate slow enough so that there is time for the thermal energy to reach an equilibrium with the magnetic ﬁeld, the B-Hcurve is single valued in the sense that the same curve is followed whether the magnetic ﬁeld is increasing or decreasing, and regardless of its rate of change. Until now, we have been considering the magnetization of materials that are suﬃciently dilute so that the atomic moments do not interact with each other. In solids, atoms can be so closely spaced that the magnetic ﬁeld due to the moment of one atom can have a signiﬁcant eﬀect on the orientation of another. In ferromagnetic materials, this mutual interaction is all important. To appreciate what makes certain materials ferromagnetic rather than simply paramagnetic, we need to remember that the electrons which surround the nuclei of atoms are assigned by quantum mechanical principles to layers or “shells.” Each shell has a particular maximum number of electrons. The electron behaves as if it possessed a net angular momentum, or spin, and hence a magnetic moment. A ﬁlled shell always contains an even number of electrons which are distributed spatially in such a manner that the total spin, and likewise the magnetic moment, is zero. For the majority of atoms, the outermost shell is unﬁlled, and so it is the outer- most electrons that play the major role in determining the net magnetic moment of the atom. This picture of the atom is consistent with paramagnetic and diamagnetic behavior. However, the transition elements form a special class. They have unﬁlled inner shells, so that the electrons responsible for the net moment of the atom are surrounded by the electrons that interact most intimately with the electrons of a neighboring atom. When such atoms are as closely packed as they are in solids, the combination of the interaction between magnetic moments and of electrostatic coupling results in the spontaneous alignment of dipoles, or ferromagnetism . The underlying interaction between atoms is both magnetic and electrostatic, and can be understood only by invoking quantum mechanical arguments. In a ferromagnetic material, atoms naturally establish an array of moments that reinforce. Nevertheless, on a macroscopic scale, ferromagnetic materials are Sec. 9.4 Magnetization Constitutive Laws 17 Fig. 9.4.5 Polycrystalline ferromagnetic material viewed at the domain level. In the absence of an applied magnetic ﬁeld, the domain moments tend to cancel. (This presumes that the material has not been left in a magnetized state by a previously applied ﬁeld.) As a ﬁeld is applied, the domain walls shift, giving rise to a net magnetization. In ideal materials, saturation results as all of the domains combine into one. In materials used for bulk fabrication of transformers, imperfections prevent the realization of this state. not necessarily permanently magnetized. The spontaneous alignment of dipoles is commonly conﬁned to microscopic regions, called domains. The moments of the individual domains are randomly oriented and cancel on a macroscopic scale. Macroscopic magnetization occurs when a ﬁeld is applied to a solid, because those domains that have a magnetic dipole moment nearly aligned with the applied ﬁeld grow at the expense of domains whose magnetic dipole moments are less aligned with the applied ﬁeld. The shift in domain structure caused by raising the applied ﬁeld from one level to another is illustrated in Fig. 9.4.5. The domain walls encounter a resistance to propagation that balances the eﬀect of the ﬁeld. A typical trajectory traced out in the B−Hplane as the ﬁeld is applied to a typical ferromagnetic material is shown in Fig. 9.4.6. If the magnetization is zero at the outset, the initial trajectory followed as the ﬁeld is turned up starts at the origin. If the ﬁeld is then turned down, the domains require a certain degree of coercion to reduce their average magnetization. In fact, with the applied ﬁeld turned oﬀ, there generally remains a ﬂux density, and the ﬁeld must be reversed to reduce the ﬂux density to zero. The trajectory traced out if the applied ﬁeld is slowly cycled between positive and negative values many times is the one shown in the ﬁgure, with the remanence ﬂux density Brwhen H= 0 and a coercive ﬁeld intensity Hcrequired to make the ﬂux density zero. Some values of these parameters, for materials used to make permanent magnets, are given in Table 9.4.2. In the toroidal geometry of Example 9.4.1, His proportional to the terminal current i. Thus, imposition of a sinusoidally varying current results in a sinusoidally varying H, as illustrated in Fig. 9.4.6b. As the iand hence Hincreases, the trajec- tory in the B−Hplane is the one of increasing H. With decreasing H, a diﬀerent trajectory is followed. In general, it is not possible to specify Bsimply by giving H(or even the time derivatives of H). When the magnetization state reﬂects the previous states of magnetization, the material is said to be hysteretic . The B−H 18 Magnetization Chapter 9 TABLE 9.4.2 MAGNETIZATION PARAMETERS FOR PERMANENT MAGNET From American Institute of Physics Handbook, McGraw-Hill, p. 5–188. Material Hc(A/m) Br(Tesla) Carbon steel 4000 1.00 Alnico 2 43,000 0.72 Alnico 7 83,500 0.70 Ferroxdur 2 143,000 .34 Fig. 9.4.6 Magnetization characteristic for material showing hysteresis with typical values of BrandHcgiven in Table 9.4.2. The curve is obtained after many cycles of sinusoidal excitation in apparatus such as that of Fig. 9.4.3. The trajectory is traced out in response to a sinusoidal current, as shown by the inset. trajectory representing the response to a sinusoidal His then called the hysteresis loop. Hysteresis can be both harmful and useful. Permanent magnetization is one result of hysteresis, and as we illustrated in Example 9.3.2, this can be the basis for the storage of information on tapes. When we develop a picture of energy dissipation in Chap. 11, it will be clear that hysteresis also implies the generation of heat, and this can impose limits on the use of magnetizable materials. Liquids having signiﬁcant magnetizabilities have been synthesized by perma- nently suspending macroscopic particles composed of single ferromagnetic domains. Sec. 9.5 Fields in Linear Materials 19 Here also the relatively high magnetizability comes from the ferromagnetic charac- ter of the individual domains. However, the very diﬀerent way in which the domains interact with each other helps in gaining an appreciation for the magnetization of ferromagnetic polycrystalline solids. In the absence of a ﬁeld imposed on the synthesized liquid, the thermal molec- ular energy randomizes the dipole moments and there is no residual magnetization. With the application of a low frequency Hﬁeld, the suspended particles assume an average alignment with the ﬁeld and a single-valued B−Hcurve is traced out, typically as shown in Fig. 9.4.4. However, as the frequency is raised, the reorien- tation of the domains lags behind the applied ﬁeld, and the B−Hcurve shows hysteresis, much as for solids. Although both the solid and the liquid can show hysteresis, the two diﬀer in an important way. In the solid, the magnetization shows hysteresis even in the limit of zero frequency. In the liquid, hysteresis results only if there is a ﬁnite rate of change of the applied ﬁeld. Ferromagnetic materials such as iron are metallic solids and hence tend to be relatively good electrical conductors. As we will see in Chap. 10, this means that unless care is taken to interrupt conduction paths in the material, currents will be induced by a time-varying magnetic ﬂux density. Often, these eddy currents are un- desired. With the objective of obtaining a highly magnetizable insulating material, iron atoms can be combined into an oxide crystal. Although the spontaneous inter- action between molecules that characterizes ferromagnetism is indeed observed, the alignment of neighbors is antiparallel rather than parallel. As a result, such pure oxides do not show strong magnetic properties. However, a mixed-oxide material like Fe 3O4(magnetite) is composed of sublattice oxides of diﬀering moments. The spontaneous antiparallel alignment results in a net moment. The class of relatively magnetizable but electrically insulating materials are called ferrimagnetic . Our discussion of the origins of magnetization began at the atomic level, where electronic orbits and spins are fundamental. However, it ends with a discussion of constitutive laws that can only be explained by bringing in additional eﬀects that occur on scales much greater than atomic or molecular. Thus, the macroscopic B andHused to describe magnetizable materials can represent averages with respect to scales of domains or of macroscopic particles. In Sec. 9.5 we will make an artiﬁcial diamagnetic material from a matrix of “perfectly” conducting particles. In a time- varying magnetic ﬁeld, a magnetic moment is induced in each particle that tends to cancel that being imposed, as was shown in Example 8.4.3. In fact, the currents induced in the particles and responsible for this induced moment are analogous to the induced changes in electronic orbits responsible on the atomic scale for diamagnetism[1]. 9.5 FIELDS IN THE PRESENCE OF MAGNETICALLY LINEAR INSULATING MATERIALS In this and the next two sections, we study materials with the linear magnetization characteristic of (9.4.4). With the understanding that µis a prescribed function of position, B=µH, the MQS forms of Amp` ere’s law and the ﬂux continuity law are 20 Magnetization Chapter 9 ∇ ×H=J (1) ∇ ·µH= 0 (2) In this chapter, we assume that the current density Jis conﬁned to perfect conduc- tors. We will ﬁnd in Chap. 10 that a time-varying magnetic ﬂux implies an electric ﬁeld. Thus, wherever a conducting material ﬁnds itself in a time-varying ﬁeld, there is the possibility that eddy currents will be induced. It is for this reason that the magnetizable materials considered in this and the next sections are presumed to be insulating. If the ﬁelds of interest vary slowly enough, these induced currents can be negligible. Ferromagnetic materials are often metallic, and hence also conductors. How- ever, materials can be made both readily magnetizable and insulating by breaking up the conduction paths. By engineering at the molecular or domain scale, or even introducing laminations of magnetizable materials, the material is rendered essen- tially free of a current density J. The considerations that determine the thickness of laminations used in transformers to prevent eddy currents will be taken up in Chap. 10. In the regions outside the perfect conductors carrying the current Jof (1), His irrotational and Bis solenoidal. Thus, we have a choice of representations. Either, as in Sec. 8.3, we can use the scalar magnetic potential and let H=−∇Ψ, or we can follow the lead from Sec. 8.6 and use the vector potential to represent the ﬂux density by letting B=∇ ×A. Where there are discontinuities in the permeability and/or thin coils modeled by surface currents, the continuity conditions associated with Amp` ere’s law and the ﬂux continuity law are used. With Bexpressed using the linear magnetization constitutive law, (1.4.16) and (9.2.10) become n×(Ha−Hb) =K (3) n·(µaHa−µbHb) = 0 (4) The classiﬁcation of physical conﬁgurations given in Sec. 6.5 for linearly polariz- able materials is equally useful here. In the ﬁrst of these, the region of interest is of uniform permeability. The laws summarized by (1) and (2) are the same as for free space except that µois replaced by µ, so the results of Chap. 6 apply directly. Conﬁgurations made up of materials having essentially uniform permeabilities are of the greatest practical interest by far. Thus, piece-wise uniform systems are the theme of Secs. 9.6 and 9.7. The smoothly inhomogeneous systems that are the last category in Fig. 9.5.1 are of limited practical interest. However, it is sometimes use- ful, perhaps in numerical simulations, to regard the uniform and piece-wise uniform systems as special cases of the smoothly nonuniform systems. Sec. 9.5 Fields in Linear Materials 21 Fig. 9.5.1 (a) Uniform permeability, (b) piece-wise uniform permeability, and (c) smoothly inhomogeneous conﬁgurations involving linearly magnetiz- able material. Inductance in the Presence of Linearly Magnetizable Materials. In the presence of linearly magnetizable materials, the magnetic ﬂux density is again pro- portional to the excitation currents. If ﬁelds are produced by a single perfectly conducting coil, its inductance is the generalization of that introduced with (8.4.13). L≡λ i=R SµH·da i(5) The surface Sspanning a contour deﬁned by the perfectly conducting wire is the same as that shown in Figs. 8.4.3 and 8.4.4. The eﬀect of having magnetizable material is, of course, represented in (5) by the eﬀect of this material on the intensity, direction, and distribution of B=µH. For systems in the ﬁrst category of Fig. 9.5.1, where the entire region occupied by the ﬁeld is ﬁlled by a material of uniform permeability µ, the eﬀect of the magnetization on the inductance is clear. The solutions to (1) and (2) for Hare not altered in the presence of the permeable material. It then follows from (5) that the inductance is simply proportional to µ. Because it imposes a magnetic ﬁeld intensity that never leaves the core mate- rial, the toroid of Example 9.4.1 is a special case of a piece-wise uniform magnetic material that acts as if all of space were ﬁlled with the magnetizable material. As shown by the following example, the inductance of the toroid is therefore also proportional to µ. Example 9.5.1. Inductance of a Toroid If the toroidal core of the winding shown in Fig. 9.4.1 and used in the experiment of Fig. 9.4.3 were made a linearly magnetizable material, what would be the voltage needed to supply the driving current i? If we deﬁne the ﬂux linkage of the driving coil as λ1, v=dλ1 dt(6) 22 Magnetization Chapter 9 Fig. 9.5.2 (a) Solenoid of length dand radius aﬁlled with material of uniform permeability µ. (b) Solenoid of (a) ﬁlled with artiﬁcial dia- magnetic material composed of an array of metal spheres having radius Rand spacing s. We now ﬁnd the inductance L, where λ1=Li, and hence determine the required input voltage. The ﬂux linked by one turn of the driving coil is essentially the cross-sectional area of the toroid multiplied by the ﬂux density. The total ﬂux linked is this quantity multiplied by the total turns N1. λ1=N1¡1 4πw2¢ B (7) According to the linear constitutive law, the ﬂux density follows from the ﬁeld intensity as B=µH. For the toroid, His related to the driving current iby (9.4.1), so B=µH=µ¡N1 2πR¢ i (8) The desired relation is the combination of these last two expressions. λ1=Li; L≡1 8µw2 RN2 1 (9) As predicted, the inductance is proportional to µ. Although inductances are gen- erally increased by bringing paramagnetic and especially ferromagnetic materials into their ﬁelds, the eﬀect of introducing ferromagnetic materials into coils can be less dramatic than in the toroidal geometry for reasons discussed in Sec. 9.6. The dependence of the inductance on the square of the turns results because not only is the ﬁeld induced by the current iproportional to the number of turns, but so too is the amount of the resulting ﬂux that is linked by the coil. Example 9.5.2. An Artiﬁcial Diamagnetic Material The cross-section of a long (ideally “inﬁnite”) solenoid ﬁlled with material of uniform permeability is shown in Fig. 9.5.2a. The azimuthal surface current Kφresults in an axial magnetic ﬁeld intensity Hz=Kφ. We presume that the axial length dis very large compared to the radius aof the coil. Thus, the ﬁeld inside the coil is uniform while that outside is zero. To see that this simple ﬁeld solution is indeed correct, note that it is both irrotational and solenoidal everywhere except at the surface r=a, and that there the boundary conditions, (3) and (4), are satisﬁed. For an n-turn coil carrying a current i, the surface current density Kφ=ni/d. Thus, the magnetic ﬁeld intensity is related to the terminal current by Hz=ni d(10) Sec. 9.5 Fields in Linear Materials 23 Fig. 9.5.3 Inductance of the coil in Fig. 9.5.2b is decreased because perfectly conducting spheres tend to reduce its eﬀective cross-sectional area. In the linearly magnetized core region, the ﬂux density is Bz=µHz, and so it is also uniform. As a result, the ﬂux linked by each turn is simply πa2Bzand the total ﬂux linked by the coil is λ=nπa2µHz (11) Substitution from (1) then gives λ=Li, L ≡πµa2n2 d(12) where Lis the inductance of the coil. Because the coil is assumed to be very long, its inductance is increased by a factor µ/µoover that of a coil in free space, much as for the toroid of Example 9.5.1. Now suppose that the permeable material is actually a cubic array of metal spheres, each having a radius R, as shown in Fig. 9.5.2b. The frequency of the current iis presumably high enough so that each sphere can be regarded as perfectly conducting in the MQS sense discussed in Sec. 8.4. The spacing sof the spheres is large compared to their radius, so that the ﬁeld of one sphere does not produce an appreciable ﬁeld at the positions of its neighbors. Each sphere ﬁnds itself in an essentially uniform magnetic ﬁeld. The dipole moment of the currents induced in a sphere by a magnetic ﬁeld that is uniform at inﬁnity was calculated in Example 8.4.3, (8.4.21). m=−2πHoR3(13) Because the induced currents must produce a ﬁeld that bucks out the imposed ﬁeld, a negative moment is induced by a positive ﬁeld. By deﬁnition, the magnetization density is the number of magnetic moments per unit volume. For a cubic array with spacing sbetween the sphere centers, the number per unit volume is s−3. Thus, the magnetization density is simply M=Nm=−2πHo¡R s¢3(14) Comparison of this expression to (9.4.3), which deﬁnes the susceptibility χm, shows that χm=−2π¡R s¢3(15) As we might have expected from the antiparallel moment induced in a sphere by an imposed ﬁeld, the susceptibility is negative. The permeability, related to χmby (9.4.4), is therefore less than 1. µ=µo(1 +χm) =µo£ 1−2π¡R s¢3¤ (16) The perfectly conducting spheres eﬀectively reduce the cross-sectional area of the ﬂux, as suggested by Fig. 9.5.3, and hence reduce the inductance. With the introduction of the array of metal spheres, the inductance goes from a value given by (12) with µ=µoto one with µgiven by (16). 24 Magnetization Chapter 9 Fig. 9.5.4 Experiment to measure the decrease of inductance that results when the artiﬁcial diamagnetic array of Fig. 9.5.2b is inserted into a solenoid. Faraday’s law of induction is also responsible for diamagnetism due to atomic moments. Instead of inducing circulating conduction currents in a metal sphere, as in this example, the time-varying ﬁeld induces changes in the orbits of electrons about the nucleus that, on the average, contribute an antiparallel magnetic moment to the atom. The following demonstration is the MQS analog of the EQS Demonstration 6.6.1. In the latter, a measurement was made of the change in capacitance caused by inserting an artiﬁcial dielectric between capacitor plates. Here the change in in- ductance is observed as an artiﬁcial diamagnetic material is inserted into a solenoid. Although the spheres are modeled as perfectly conducting in both demonstrations, we will ﬁnd in Chap. 10 that the requirements to justify this assumption in this MQS example are very diﬀerent from those for its EQS counterpart. Demonstration 9.5.1. Artiﬁcial Diamagnetic Material The experiment shown in Fig. 9.5.4 measures the change in solenoid inductance when an array of conducting spheres is inserted. The coil is driven at the angular frequency ωby an oscillator-ampliﬁer. Over the length dshown in the ﬁgure, the ﬁeld tends to be uniform. The circuit shown schematically in Fig. 9.5.5 takes the form of a bridge with the inductive reactance of L2used to balance the reactance of the central part of the empty solenoid. The input resistances of the oscilloscope’s balanced ampliﬁers, represented by Rs, are large compared to the inductor reactances. These branches dominate over the inductive reactances in determining the current through the inductors and, as a result, the inductor currents remain essentially constant as the inductances are varied. With the reactance of the inductor L2balancing that of the empty solenoid, these currents are equal and the balanced ampliﬁer voltage vo= 0. When the array of spheres is inserted into the solenoid, the currents through both legs remain essentially constant. Thus, the resulting voltage vois the change in voltage across the solenoid Sec. 9.5 Fields in Linear Materials 25 Fig. 9.5.5 Bridge used to measure the change in inductance in the experiment of Fig. 9.5.4. caused by its change in inductance ∆ L. vo= (∆ L)di dt→ |ˆvo|=ω(∆L)|ˆi| (17) In the latter expression, the current and voltage indicated by a circumﬂex are either peak or rms sinusoidal steady state amplitudes. In view of (12), this expression becomes |ˆvo|=ω(µ−µo)πa2n2 d|ˆi| (18) In terms of the sphere radius and spacing, the change in permeability is given by (16), so the voltage measured by the balanced ampliﬁers is |ˆvo|=2π2ωa2n2 d¡R s¢3|ˆi| (19) To evaluate this expression, we need only the frequency and amplitude of the coil current, the number of turns in the length d, and other dimensions of the system. Induced Magnetic Charge: Demagnetization. The complete analogy be- tween linearly polarized and linearly magnetized materials is proﬁtably carried yet another step. Magnetic charge is induced where µis spatially varying, and hence the magnetizable material can introduce sources that revise the free space ﬁeld dis- tribution. In the linearly magnetizable material, the distribution of these sources is not known until after the ﬁelds have been determined. However, it is often helpful in qualitatively predicting the ﬁeld eﬀects of magnetizable materials to picture the distribution of induced magnetic charges. Using a vector identity, (2) can be written µ∇ ·H+H· ∇µ= 0 (20) Rearrangement of this expression shows that the source of µoH, the magnetic charge density, is ∇ ·µoH=−µo µH· ∇µ≡ρm (21) 26 Magnetization Chapter 9 Most often we deal with piece-wise uniform systems where variations in µare con- ﬁned to interfaces. In that case, it is appropriate to write the continuity of ﬂux density condition in the form n·µo(Ha−Hb) =n·µoHa¡ 1−µa µb¢ ≡σsm (22) where σsmis the magnetic surface charge density. The following illustrates the use of this relation. Illustration. The Demagnetization Field A sphere of material having uniform permeability µis placed in an initially uniform upward-directed ﬁeld. It is clear from (21) that there are no distortions of the uniform ﬁeld from magnetic charge induced in the volume of the material. Rather, the sources of induced ﬁeld are located on the surface where the imposed ﬁeld has a component normal to the permeability discontinuity. It follows from (22) that positive and negative magnetic surface charges are induced on the top and bottom parts of the surface, respectively. TheHﬁeld caused by the induced magnetic surface charges originates at the positive charge at the top and terminates on the negative charge at the bottom. This is illustrated by the magnetization analog of the permanently polarized sphere, considered in Example 6.3.1. Our point here is that the ﬁeld resulting from these induced magnetic surface charges tends to cancel the one imposed. Thus, the ﬁeld intensity available to magnetize the material is reduced. The remarks following (6.5.11) apply equally well here. The roles of E,D, and /epsilon1are taken by H,B, and µ. In regions of uniform permeability, (1) and (2) are the same laws considered in Chap. 8, and where the current density is zero, Laplace’s equation governs. As we now consider piece-wise nonuniform systems, the eﬀect of the material is accounted for by the continuity conditions. 9.6 FIELDS IN PIECE-WISE UNIFORM MAGNETICALLY LINEAR MATERIALS Whether we choose to represent the magnetic ﬁeld in terms of the magnetic scalar potential Ψ or the vector potential A, in a current-free region having uniform permeability it assumes a distribution governed by Laplace’s equation. That is, where µis constant and J= 0, (9.5.1) and (9.5.2) require that His both solenoidal and irrotational. If we let H=−∇Ψ, the ﬁeld is automatically irrotational and ∇2Ψ = 0 (1) is the condition that it be solenoidal. If we let µH=∇×A, the ﬁeld is automatically solenoidal. The condition that it also be irrotational (together with the requirement thatAbe solenoidal) is then2 2∇ × ∇ × A=∇(∇ ·A)− ∇2A Sec. 9.6 Piece-Wise Uniform Materials 27 ∇2A= 0 (2) Thus, in Cartesian coordinates, each component of Asatisﬁes the same equation as does Ψ. The methods illustrated for representing piece-wise uniform dielectrics in Sec. 6.6 are applicable here as well. The major diﬀerence is that here, currents are used to excite the ﬁeld whereas there, unpaired charges were responsible for inducing the polarization. The sources are now the current density and surface current density rather than unpaired volume and surface charges. Thus, the external excitations drive the curl of the ﬁeld, in accordance with (9.5.1) and (9.5.3), rather than its divergence. The boundary conditions needed at interfaces between magnetically linear materials are n·(µaHa−µbHb) = 0 (3) for the normal component of the magnetic ﬁeld intensity, and n×(Ha−Hb) =K (4) for the tangential component, in the presence of a surface current. As before, we shall ﬁnd it convenient to represent windings by equivalent surface currents. Example 9.6.1. The Spherical Coil with a Permeable Core The spherical coil developed in Example 8.5.1 is now ﬁlled with a uniform core having the permeability µ. With the ﬁeld intensity again represented in terms of the magnetic scalar potential, H=−∇Ψ, the analysis diﬀers only slightly from that already carried out. Laplace’s equation, (1), again prevails inside and outside the coil. At the coil surface, the tangential Hagain suﬀers a discontinuity equal to the surface current density in accordance with Amp` ere’s continuity condition, (4). The eﬀect of the permeable material is only felt through the ﬂux continuity condition, (3), which requires that µoHa r−µHb r= 0 (5) Thus, the normal ﬂux continuity condition of (8.5.12) is generalized to include the eﬀect of the permeable material by −µC R=2µoA R(6) and it follows that the coeﬃcients needed to evaluate Ψ, and hence H, are now A=Ni 2¡ 1 +2µo µ¢; C=−µo µNi¡ 1 +2µo µ¢ (7) 28 Magnetization Chapter 9 Substitution of these coeﬃcients into (8.5.10) and (8.5.11) gives the ﬁeld inside and outside the spherical coil. H=8 < :µo µNi¡ 1+2µo µ¢ R(ircosθ−iθsinθ) =µo µ+2µoNi Riz;r < R Ni 2¡ 1+2µo µ¢ R¡R r¢3(ir2 cosθ+iθsinθ); r > R(8) If the coil is highly permeable, these expressions show that the ﬁeld intensity inside is much less than that outside. In the limit of “inﬁnite permeability,” where µo/µ→0, the ﬁeld inside is zero while that outside becomes Hθ(r=R) =Ni 2Rsinθ (9) This is the surface current density, (8.5.6). A surface current density backed by a highly permeable material terminates the tangential magnetic ﬁeld. Thus, Amp` ere’s continuity condition relating the ﬁelds to each side of the surface is replaced by a boundary condition on the ﬁeld on the low permeability side of the interface. Using this boundary condition, that Ha θbe equal to the given Kθ, (8.5.6), the solution for the exterior Ψ and Hcan be written by inspection in the limit when µ→ ∞ . Ψa=Ni 2¡R r¢2cosθ;H=Ni 2R¡R r¢3(ir2 cosθ+iθsinθ) (10) The interior magnetic ﬂux density can in turn be approximated by using this exterior ﬁeld to compute the ﬂux density normal to the surface. Because this ﬂux density must be the same inside, ﬁnding the interior ﬁeld reduces to solving Laplace’s equa- tion for Ψ subject to the boundary condition that −µ∂Ψb ∂r(r=R) =µoNi Rcosθ (11) Again, the solution represents a uniform ﬁeld and can be written by inspection. Ψb=−µo µNir Rcosθ (12) TheHﬁeld, the gradient of the above expression, is indeed that given by (8a) in the limit where µo/µis small. Note that the interior Hgoes to zero as the permeability goes to inﬁnity, but the interior ﬂux density Bremains ﬁnite. This fact makes it clear that the inductance of the coil must remain ﬁnite, even in the limit where µ→ ∞ . To determine an expression for the inductance that is valid regardless of the core permeability, (8a) can be used to evaluate (8.5.18). Note that the internal ﬂux density Bthat replaces µoHzis 3µ/[µ+2µo] times larger than the ﬂux density in the absence of the magnetic material. This enhancement factor increases monotonically with the ratio µ/µobut reaches a maximum of only 3 in the limit where this ratio goes to inﬁnity. Once again, we have evidence of the core demagnetization caused by the surface magnetic charge induced on the surface of the sphere. With the uniformity of the ﬁeld inside the sphere known in advance, a much simpler derivation of (8a) gives further insight into the role of the magnetization. Sec. 9.6 Piece-Wise Uniform Materials 29 Fig. 9.6.1 Sphere of material having uniform permeability with N- turn coil of radius Rat its center. Because R/lessmuchb, the coil can be modeled as a dipole. The surrounding region has permeability µa. Thus, in the core, the H-ﬁeld is the superposition of two ﬁelds. The ﬁrst is caused by the surface current, and given by (8a) with µ=µo. Hi=Ni 3Riz (13) The second is due to the uniform magnetization M=Miz, which is given by the magnetization analog to (6.3.15) ( E→H,P→µoM, /epsilon1o→µo). HM=−Mo 3iz (14) The net internal magnetic ﬁeld intensity is the sum of these. H=¡Ni 3R−Mo 3¢ iz (15) Only now do we introduce the constitutive law relating MotoHz, Mo=χmHz. [In Sec. 9.8 we will exploit the fact that the relation could be nonlinear.] If this law is introduced into (15), and that expression solved for Hz, a result is obtained that is familiar from from (8a). Hz=Ni/3R 1 +1 3χm=µo µNi/R¡ 1 +2µo µ¢ (16) This last calculation again demonstrates how the ﬁeld Ni/3Ris reduced by the magnetization through the “feedback factor” 1 /[1 + ( χm/3)]. Magnetic circuit models, introduced in the next section, exploit the capacity of highly permeable materials to guide the magnetic ﬂux. The example considered next uses familiar solutions to Laplace’s equation to illustrate how this guiding takes place. We will make reference to this case study when the subject of magnetic circuits is initiated. Example 9.6.2. Field Model for a Magnetic Circuit A small coil with Nturns and excited by a current iis used to make a magnetic ﬁeld in a spherically shaped material of permeability µb. As shown in Fig. 9.6.1, the coil has radius R, while the µsphere has radius band is surrounded by a magnetic medium of permeability µa. 30 Magnetization Chapter 9 Because the coil radius is small compared to that of the sphere, it will be modeled as a dipole having its moment m=πR2iin the zdirection. It follows from (8.3.13) that the magnetic scalar potential for this dipole is Ψdipole =R2Ni 4cosθ r2(17) No surface current density exists at the surface of the sphere. Thus, Amp` ere’s con- tinuity law requires that Ha θ−Hb θ= 0→Ψa= Ψ batr=b (18) Also, at the interface, the ﬂux continuity condition is µaHa r−µbHb r= 0 at r=b (19) Finally, the only excitation of the ﬁeld is the coil at the origin, so we require that the ﬁeld decay to zero far from the sphere. Ψa→0 as r→ ∞ (20) Given that the scalar potential has the θdependence cos( θ), we look for solu- tions having this same θdependence. In the exterior region, the solution representing a uniform ﬁeld is ruled out because there is no ﬁeld at inﬁnity. In the neighborhood of the origin, we know that Ψ must approach the dipole ﬁeld. These two conditions are implicit in the assumed solutions Ψa=Acosθ r2; Ψ b=R2Ni 4cosθ r2+Crcosθ (21) while the coeﬃcients AandCare available to satisfy the two remaining continuity conditions, (18) and (19). Substitution gives two expressions which are linear in A andCand which can be solved to give A=3 4µbNiR2 (µb+ 2µa); C=Ni b3R2(µb−µa) 2(µb+ 2µa)(22) We thus conclude that the scalar magnetic potential outside the sphere is that of a dipole Ψa=3 4µbNi (µb+ 2µa)¡R r¢2cosθ (23) while inside it is that of a dipole plus that of a uniform ﬁeld. Ψb=Ni 4·¡R r¢2cosθ+2(µb−µa) (µb+ 2µa)¡R b¢2r bcosθ¸ (24) For increasing values of the relative permeability, the equipotentials and ﬁeld lines are shown in Fig. 9.6.2. With µb/µa= 1, the ﬁeld is simply that of the dipole at the origin. In the opposite extreme, where the ratio of permeabilities is 100, it has Sec. 9.6 Piece-Wise Uniform Materials 31 Fig. 9.6.2 Magnetic potential and lines of ﬁeld intensity in and around the magnetizable sphere of Fig. 9.6.1. (a) With the ratio of permeabilities equal to 1, the dipole ﬁeld extends into the surrounding free space region without modiﬁcation. (b) With µb/µa= 3, ﬁeld lines tend to be more conﬁned to the sphere. (c) With µb/µa= 100, the ﬁeld lines (and hence the ﬂux lines) tend to remain inside the sphere. become clear that the interior ﬁeld lines tend to become tangential to the spherical surface. The results of Fig. 9.6.2 can be elaborated by taking the limit of µb/µagoing to inﬁnity. In this limit, the scalar potentials are Ψa=3 4Ni¡R r¢2cosθ (25) Ψb=Ni r¡R b¢2£¡b r¢2+ 2¡r b¢¤ cosθ (26) In the limit of a large permeability of the medium in which the coil is imbedded relative to that of the surrounding medium, guidance of the magnetic ﬂux occurs by the highly permeable medium. Indeed, in this limit, the ﬂux produced by the coil goes to inﬁnity, whereas the ﬂux of the ﬁeldR H·daescaping from the sphere (the so-called “fringing”) stays ﬁnite, because the exterior potential stays ﬁnite. The magnetic ﬂuxR B·dais guided within the sphere, and practically no magnetic ﬂux escapes. The ﬂux lines on the inside surface of the highly permeable sphere can be practically tangential as indeed predicted by (26). Another limit of interest is when the outside medium is highly permeable and the coil is situated in a medium of low permeability (like free space). In this limit, one obtains Ψa= 0 (27) Ψb=Ni 4¡R b¢2£¡b r¢2−r b¤ cosθ (28) The surface at r=bbecomes an equipotential of Ψ. The magnetic ﬁeld is perpen- dicular to the surface. The highly permeable medium behaves in a way analogous to a perfect conductor in the electroquasistatic case. 32 Magnetization Chapter 9 Fig. 9.6.3 Graphical representation of the relations between components of Hat an interface between a medium of permeability µaand a material having permeability µb. In order to gain physical insight, two types of approximate boundary condi- tions have been illustrated in the previous example. These apply when one region is of much greater permeability than another. In the limit of inﬁnite permeability of one of the regions, the two continuity conditions at the interface between these regions reduce to one boundary condition on the ﬁelds in one of the regions. We conclude this section with a summary of these boundary conditions. At a boundary between regions (a) and (b), having permeabilities µaandµb, respectively, the normal ﬂux density µHnis continuous. If there is no surface current density, the tangential components Htare also continuous. Thus, the magnetic ﬁeld intensity to either side of the interface is as shown in Fig. 9.6.3. With the angles between Hand the normal on each side of the interface denoted by αandβ, respectively, tanα=Ha t Han; tan β=Hb t Hbn(29) The continuity conditions can be used to express tan( α) in terms of the ﬁelds on the (b) side of the interface, so it follows that tanα tanβ=µa µb(30) In the limit where µa/µb→0, there are therefore two possibilities. Either tan( α)→ 0, so that α→0 and Hin region (a) becomes perpendicular to the boundary, or tan(β)→ ∞ so that β→90 degrees and Hin region (b) becomes tangential to the boundary. Which of these two possibilities pertains depends on the excitation conﬁguration. Excitation in Region of High Permeability. In these conﬁgurations, a closed contour can be found within the highly permeable material that encircles current- carrying wires. For the coil at the center of the highly permeable sphere considered in Example 9.6.2, such a contour is as shown in Fig. 9.6.4. As µb→ ∞ , the ﬂux density Balso goes to inﬁnity. In this limit, the ﬂux escaping from the body can be ignored compared to that guided by the body. The boundary is therefore one at which the interior ﬂux density is essentially tangential. n·B= 0 (31) Sec. 9.7 Magnetic Circuits 33 Fig. 9.6.4 Typical contour in conﬁguration of Fig. 9.6.1 encircling current without leaving highly permeable material. Fig. 9.6.5 (a) With coil in the low permeability region, the contour encircling the current must pass through low permeability material. (b) With coil on the surface between regions, contours encircling current must still leave highly permeable region. Once the ﬁeld has been determined in the inﬁnitely permeable material, continuity of tangential His used to provide a boundary condition on the free space side of the interface. Excitation in Region of Low Permeability. In this second class of con- ﬁgurations, there is no closed contour within the highly permeable material that encircles a current-carrying wire. If the current-carrying wires are within the free space region, as in Fig. 9.6.5a, a contour must leave the highly permeable material to encircle the wire. In the limit where µb→ ∞ , the magnetic ﬁeld intensity in the highly permeable material approaches zero, and thus Hon the interior side of the interface becomes perpendicular to the boundary. n×H= 0 (32) With wires on the interface between regions comprising a surface current den- sity, as illustrated in Fig. 9.6.5b, it is still not possible to encircle the current without following a contour that leaves the highly permeable material. Thus, the case of a surface current is also in this second category. The tangential His terminated by the surface current density. Thus, the boundary condition on Hon the interior side of the interface carrying the surface current Kis n×H=K (33) This boundary condition was illustrated in Example 9.6.1. Once the ﬁelds in the interior region have been found, continuity of normal ﬂux density provides a boundary condition for determining the ﬂux distribution in the highly permeable region. 34 Magnetization Chapter 9 Fig. 9.7.1 Highly magnetizable core in which ﬂux induced by winding can circulate in two paths. Fig. 9.7.2 Cross-section of highly permeable core showing contour C1spanned by surface S1, used with Amp´ ere’s integral law, and closed surface S2, used with the integral ﬂux continuity law. 9.7 MAGNETIC CIRCUITS The availability of relatively inexpensive magnetic materials, with magnetic suscep- tibilities of the order of 1000 or more, allows the production of high magnetic ﬂux densities with relatively small currents. Devices designed to exploit these materials include compact inductors, transformers, and rotating machines. Many of these are modeled as the magnetic circuits that are the theme of this section. A magnetic circuit typical of transformer cores is shown in Fig. 9.7.1. A core of high permeability material has a pair of rectangular windows cut through its center. Wires passing through these windows are wrapped around the central column. The ﬂux generated by this coil tends to be guided by the magnetizable material. It passes upward through the center leg of the material, and splits into parts that circulate through the legs to left and right. Example 9.6.2, with its highly permeable sphere excited by a small coil, oﬀered the opportunity to study the trapping of magnetic ﬂux. Here, as in that case with µb/µa/greatermuch1, the ﬂux density inside the core tends to be tangential to the surface. Thus, the magnetic ﬂux density is guided by the material and the ﬁeld distribution within the core tends to be independent of the exterior conﬁguration. In situations of this type, where the ducting of the magnetic ﬂux makes it possible to approximate the distribution of magnetic ﬁeld, the MQS integral laws serve much the same purpose as do Kirchhoﬀ’s laws for electrical circuits. Sec. 9.7 Magnetic Circuits 35 Fig. 9.7.3 Cross-section of magnetic circuit used to produce a mag- netic ﬁeld intensity Hgin an air gap. The MQS form of Amp` ere’s integral law applies to a contour, such as C1in Fig. 9.7.2, following a path of circulating magnetic ﬂux. I C1H·ds=Z S1J·da (1) The surface enclosed by this contour in Fig. 9.7.2 is pierced Ntimes by the current carried by the wire, so the surface integral of the current density on the right in (1) is, in this case, Ni. The same equation could be written for a contour circulating through the left leg, or for one circulating around through the outer legs. Note that the latter would enclose a surface Sthrough which the net current would be zero. If Amp` ere’s integral law plays a role analogous to Kirchhoﬀ’s voltage law, then the integral law expressing continuity of magnetic ﬂux is analogous to Kirchhoﬀ’s current law. It requires that through a closed surface, such as S2in Fig. 9.7.2, the net magnetic ﬂux is zero. I S2B·da= 0 (2) As a result, the ﬂux entering the closed surface S2in Fig. 9.7.2 through the central leg must be equal to that leaving to left and right through the upper legs of the magnetic circuit. We will return to this particular magnetic circuit when we discuss transformers. Example 9.7.1. The Air Gap Field of an Electromagnet The magnetic circuit of Fig. 9.7.3 might be used to produce a high magnetic ﬁeld intensity in the narrow air gap. An N-turn coil is wrapped around the left leg of the highly permeable core. Provided that the length gof the air gap is not too large, the ﬂux resulting from the current iin this winding is largely guided along the magnetizable material. By approximating the ﬁelds in sections of the circuit as being essentially uni- form, it is possible to use the integral laws to determine the ﬁeld intensity in the gap. In the left leg, the ﬁeld is approximated by the constant H1over the length l1and cross-sectional area A1. Similarly, over the lengths l2, which have the cross- sectional areas A2, the ﬁeld intensity is approximated by H2. Finally, under the assumption that the gap width gis small compared to the cross-sectional dimen- sions of the gap, the ﬁeld in the gap is represented by the constant Hg. The line 36 Magnetization Chapter 9 integral of Hin Amp` ere’s integral law, (1), is then applied to the contour Cthat follows the magnetic ﬁeld intensity around the circuit to obtain the left-hand side of the expression H1ll+ 2H2l2+gHg=Ni (3) The right-hand side of this equation represents the surface integral of J·dafor a surface Shaving this contour as its edge. The total current through the surface is simply the current through one wire multiplied by the number of times it pierces the surface S. We presume that the magnetizable material is operated under conditions of magnetic linearity. The constitutive law then relates the ﬂux density and ﬁeld in- tensity in each of the regions. B1=µH1;B2=µH2;Bg=µoHg (4) Continuity of magnetic ﬂux, (2), requires that the total ﬂux through each section of the circuit be the same. With the ﬂux densities expressed using (4), this requires that A1µH1=A2µH2=A2µoHg (5) Our objective is to determine Hg. To that end, (5) is used to write H2=µo µHg; H1=µo µA2 A1Hg (6) and these relations used to eliminate H1andH2in favor of Hgin (3). From the resulting expression, it follows that Hg=Ni¡µo µA2 A1l1+2µo µl2+g¢ (7) Note that in the limit of inﬁnite core permeability, the gap ﬁeld intensity is simply Ni/g . If the magnetic circuit can be broken into sections in which the ﬁeld intensity is essentially uniform, then the ﬁelds may be determined from the integral laws. The previous example is a case in point. A more general approach is required if the core is of complex geometry or if a more accurate model is required. We presume throughout this chapter that the magnetizable material is suf- ﬁciently insulating so that even if the ﬁelds are time varying, there is no current density in the core. As a result, the magnetic ﬁeld intensity in the core can be represented in terms of the scalar magnetic potential introduced in Sec. 8.3. H=−∇Ψ (8) According to Amp` ere’s integral law, (1), integration of H·dsaround a closed contour must be equal to the “Amp` ere turns” Nipassing through the surface spanning the contour. With Hexpressed in terms of Ψ, integration from (a) to (b) around a contour such as Cin Fig. 9.7.4, which encircles a net current equal to the product of the turns Nand the current per turn i, gives Ψ a−Ψb≡∆Ψ = Ni. With (a) and (b) adjacent to each other, it is clear that Ψ is multiple-valued. To specify the principal value of this multiple-valued function we must introduce a Sec. 9.7 Magnetic Circuits 37 Fig. 9.7.4 Typical magnetic circuit conﬁguration in which the magnetic scalar potential is ﬁrst determined inside the highly magnetizable material. The principal value of the multivalued scalar potential inside the core is taken by not crossing the surface Sd. discontinuity in Ψ somewhere along the contour. In the circuit of Fig. 9.7.4, this discontinuity is deﬁned to occur across the surface Sd. To make the line integral of H·dsfrom any point just above the surface Sdaround the circuit to a point just below the surface equal to Ni, the potential is required to suﬀer a discontinuity ∆Ψ = Niacross Sd. Everywhere inside the magnetic material, Ψ satisﬁes Laplace’s equation. If, in addition, the normal ﬂux density on the walls of the magnetizable material is required to vanish, the distribu- tion of Ψ within the core is uniquely determined. Note that only the discontinuity in Ψ is speciﬁed on the surface Sd. The magnitude of Ψ on one side or the other is not speciﬁed. Also, the normal derivative of Ψ, which is proportional to the normal component of H, must be continuous across Sd. The following simple example shows how the scalar magnetic potential can be used to determine the ﬁeld inside a magnetic circuit. Example 9.7.2. The Magnetic Potential inside a Magnetizable Core The core of the magnetic circuit shown in Fig. 9.7.5 has outer and inner radii aandb, respectively, and a length din the zdirection that is large compared to a. A current iis carried in the zdirection through the center hole and returned on the outer periphery by Nturns. Thus, the integral of H·dsover a contour circulating around the magnetic circuit must be Ni, and a surface of discontinuity Sdis arbitrarily introduced as shown in Fig. 9.7.5. With the boundary condition of no ﬂux leakage, ∂Ψ/∂r= 0 at r=aand at r=b, the solution to Laplace’s equation within the core is uniquely speciﬁed. In principle, the boundary value problem can be solved even if the geometry is complicated. For the conﬁguration shown in Fig. 9.7.5, the requirement of no radial derivative suggests that Ψ is independent of r. Thus, with Aan arbitrary coeﬃcient, a reasonable guess is Ψ =Aφ=−Ni¡φ 2π¢ (9) The coeﬃcient Ahas been selected so that there is indeed a discontinuity Niin Ψ between φ= 2πandφ= 0. The magnetic ﬁeld intensity given by substituting (9) into (8) is H=A riφ=Ni 2πriφ (10) Note that His continuous, as it should be. Now that the inside ﬁeld has been determined, it is possible, in turn, to ﬁnd the ﬁelds in the surrounding free space regions. The solution for the inside ﬁeld, together 38 Magnetization Chapter 9 Fig. 9.7.5 Magnetic circuit consisting of a core having the shape of a circular cylindrical annulus with an N-turn winding wrapped around half of its circumferential length. The length of the system into the paper is very long compared to the outer radius a. with the given surface current distribution at the boundary between regions, provides the tangential ﬁeld at the boundaries of the outside regions. Within an arbitrary constant, a boundary condition on Ψ is therefore speciﬁed. In the outside regions, there is no closed contour that both stays within the region and encircles current. In these regions, Ψ is continuous. Thus, the problem of ﬁnding the “leakage” ﬁelds is reduced to ﬁnding the boundary value solution to Laplace’s equation. This inside-outside approach gives an approximate ﬁeld distribution that is justiﬁed only if the relative permeability of the core is very large. Once the outside ﬁeld is approximated in this way, it can be used to predict how much ﬂux has left the magnetic circuit and hence how much error there is in the calculation. Generally, the error will be found to depend not only on the relative permeability but also on the geometry. If the magnetic circuit is composed of legs that are long and thin, then we would expect the leakage of ﬂux to be large and the approximation of the inside-outside approach to become invalid. Electrical Terminal Relations and Characteristics. Practical inductors (chokes) often take the form of magnetic circuits. With more than one winding on the same magnetic circuit, the magnetic circuit serves as the core of a transformer. Figure 9.7.6 gives the schematic representation of a transformer. Each winding is modeled as perfectly conducting, so its terminal voltage is given by (9.2.12). v1=dλ1 dt; v2=dλ2 dt(11) However, the ﬂux linked by one winding is due to two currents. If the core is magnetically linear, we have a ﬂux linked by the ﬁrst coil that is the sum of a ﬂux linkage L11i1due to its own current and a ﬂux linkage L12due to the current in the second winding. The situation for the second coil is similar. Thus, the ﬂux linkages are related to the terminal currents by an inductance matrix .· λ1 λ2¸ =· L11L12 L21L22¸· i1 i2¸ (12) Sec. 9.7 Magnetic Circuits 39 Fig. 9.7.6 Circuit representation of a transformer as deﬁned by the terminal relations of (12) or of an ideal transformer as deﬁned by (13). The coeﬃcients Lijare functions of the core and coil geometries and properties of the material, with L11andL22the familiar self-inductances andL12andL21the mutual inductances . The word “transformer” is commonly used in two ways, each often represented schematically, as in Fig. 9.7.6. In the ﬁrst, the implication is only that the terminal relations are as summarized by (12). In the second usage, where the device is said to be an ideal transformer , the terminal relations are given as voltage and current ratios. For an ideal transformer, i2 i1=−N1 N2;v2 v1=N2 N1(13) Presumably, such a device can serve to step up the voltage while stepping down the current. The relationships between terminal voltages and between terminal currents is linear, so that such a device is “ideal” for processing signals. The magnetic circuit developed in the next example is that of a typical trans- former. We have two objectives. First, we determine the inductances needed to complete (12). Second, we deﬁne the conditions under which such a transformer operates as an ideal transformer. Example 9.7.3. A Transformer The core shown in Fig. 9.7.7 is familiar from the introduction to this section, Fig. 9.7.1. The “windows” have been ﬁlled up by a pair of windings, having the turns N1 andN2, respectively. They share the center leg of the magnetic circuit as a common core and generate a ﬂux that circulates through the branches to either side. The relation between the terminal voltages for an ideal transformer depends only on unity coupling between the two windings. That is, if we call Φ λthe magnetic ﬂux through the center leg, the ﬂux linking the respective coils is λ1=N1Φλ; λ2=N2Φλ (14) These statements presume that there is no leakage ﬂux which would link one coil but bypass the other. In terms of the magnetic ﬂux through the center leg, the terminal voltages follow from (14) as v1=N1dΦλ dt; v2=N2dΦλ dt(15) From these expressions, without further restrictions on the mode of operation, fol- lows the relation between the terminal voltages of (13). 40 Magnetization Chapter 9 Fig. 9.7.7 In a typical transformer, coupling is optimized by wrapping the primary and secondary on the same core. The inset shows how full use is made of the magnetizable material in the core manufacture. We now use the integral laws to determine the ﬂux linkages in terms of the currents. Because it is desirable to minimize the peak magnetic ﬂux density at each point throughout the core, and because the ﬂux through the center leg divides evenly between the two circuits, the cross-sectional areas of the return legs are made half as large as that of the center leg.3As a result, the magnitude of B, and hence H, can be approximated as constant throughout the core. [Note that we have now used the ﬂux continuity condition of (2).] With the average length of a circulating magnetic ﬁeld line taken as l, Amp` ere’s integral law, (1), gives Hl=N1i1+N2i2 (16) In view of the presumed magnetic linearity of the core, the ﬂux through the cross- sectional area Aof the center leg is Φλ=AB=AµH (17) and it follows from these last two expressions that Φλ=AµN 1 li1+AµN 2 li2. (18) Multiplication by the turns N1and then N2, respectively, gives the ﬂux linkages λ1 andλ2. λ1=µ AµN2 1 l¶ i1+µ AµN 1N2 l¶ i2 λ2=µ AµN 1N2 l¶ i1+µ AµN2 2 l¶ i2 (19) 3To optimize the usage of core material, the relative dimensions are often taken as in the inset to Fig. 9.7.7. Two cores are cut from rectangular sections measuring 6 h×8h. Once the windows have been removed, the rectangle is cut in two, forming two “ E” cores which can then be combined with the “ I’s” to form two complete cores. To reduce eddy currents, the core is often made from varnished laminations. This will be discussed in Chap. 10. Sec. 9.7 Magnetic Circuits 41 Fig. 9.7.8 Transformer with a load resistance Rthat includes the internal resistance of the secondary winding. Comparison of this expression with (12) identiﬁes the self- and mutual inductances as L11=AµN2 1 l;L22=AµN2 2 l;L12=L21=AµN 1N2 l(20) Note that the mutual inductances are equal. In Sec. 11.7, we shall see that this is a consequence of energy conservation. Also, the self-inductances are related to either mutual inductance by√ L11L22=L12 (21) Under what conditions do the terminal currents obey the relations for an “ideal transformer”? Suppose that the (1) terminals are selected as the “primary” terminals of the transformer and driven by a current source I(t), and that the terminals of the (2) winding, the “secondary,” are connected to a resistive load R. To recognize that the winding in fact has an internal resistance, this load includes the winding resistance as well. The electrical circuit is as shown in Fig. 9.7.8. The secondary circuit equation is −i2R=dλ2 dt(22) and using (12) with i1=I, it follows that the secondary current i2is governed by L22di2 dt+i2R=−L21dI dt(23) For purposes of illustration, consider the response to a drive that is in the sinusoidal steady state. With the drive angular frequency equal to ω, the response has the same time dependence in the steady state. I=ReˆIejωt⇒i2=Reˆi2ejωt(24) Substitution into (23) then shows that the complex amplitude of the response is ˆi2=−jωL 21ˆI jωL 22+R=−N1 N2ˆi11 1 +R jωL22(25) The ideal transformer-current relation is obtained if ωL22 R/greatermuch1 (26) In that case, (25) reduces to ˆi2=−N1 N2ˆi1 (27) 42 Magnetization Chapter 9 When the ideal transformer condition, (26), holds, the ﬁrst term on the left in (23) overwhelms the second. What remains if the resistance term is neglected is the statement d dt(L21i1+L22i2) =dλ2 dt= 0 (28) We conclude that for ideal transformer operation, the ﬂux linkages are negligible. This is crucial to having a transformer behave as a linear device. Whether repre- sented by the inductance matrix of (12) or by the ideal relations of (13), linear operation hinges on having a linear relation between BandHin the core, (17). By operating in the regime of (26) so that Bis small enough to avoid saturation, (17) tends to remain valid. 9.8 SUMMARY The magnetization density Mrepresents the density of magnetic dipoles. The mo- ment mof a single microscopic magnetic dipole was deﬁned in Sec. 8.2. With µom↔pwhere pis the moment of an electric dipole, the magnetic and electric dipoles play analogous roles, and so do the HandEﬁelds. In Sec. 9.1, it was there- fore natural to deﬁne the magnetization density so that it played a role analogous to the polarization density, µoM↔P. As a result, the magnetic charge density ρmwas considered to be a source of ∇ ·µoH. The relations of these sources to the magnetization density are the ﬁrst expressions summarized in Table 9.8.1. The second set of relations are diﬀerent forms of the ﬂux continuity law, including the eﬀect of magnetization. If the magnetization density is given, (9.2.2) and (9.2.3) are most useful. However, if Mis induced by H, then it is convenient to introduce the magnetic ﬂux density Bas a variable. The correspondence between the ﬁelds due to magnetization and those due to polarization is B↔D. The third set of relations pertains to linearly magnetizable materials. There is no magnetic analog to the unpaired electric charge density. In this chapter, the MQS form of Amp` ere’s law was also required to determine H. ∇ ×H=J (1) In regions where J=0,His indeed analogous to Ein the polarized EQS systems of Chap. 6. In any case, if Jis given, or if it is on perfectly conducting surfaces, its contribution to the magnetic ﬁeld intensity is determined as in Chap. 8. In Chap. 10, we introduce the additional laws required to determine Jself- consistently in materials of ﬁnite conductivity. To do this, it is necessary to give careful attention to the electric ﬁeld associated with MQS ﬁelds. In this chapter, we have generalized Faraday’s law, (9.2.11), ∇ ×E=−∂B ∂t(2) so that it can be used to determine Ein the presence of magnetizable materials. Chapter 10 brings this law to the fore as it plays a key role in determining the self-consistent J. Sec. 9.8 Summary 43 TABLE 9.8.1 SUMMARY OF MAGNETIZATION RELATIONS AND LAWS Magnetization Charge Density and Magnetization Density ρm≡ −∇ · µoM (9.2.4) σsm=−n·µo(Ma−Mb) (9.2.5) Magnetic Flux Continuity with Magnetization ∇ ·µoH=ρm (9.2.2) n·µo(Ha−Hb) =σsm (9.2.3) ∇ ·B= 0 (9.2.9) n·(Ba−Bb) = 0 (9.2.10) where B≡µo(H+M) (9.2.8) Magnetically Linear Magnetization Constitutive law M=χmH;χm≡µ µo−1 (9.4.3) B=µH (9.4.4) Magnetization source distribution ρm=−µo µH· ∇µ (9.5.21) σsm=n·µoHa¡ 1−µa µb¢ (9.5.22) R E F E R E N C E S [1] Purcell, E. M., Electricity and Magnetism , McGraw-Hill Book Co., N. Y., 2nd Ed., (1985), p. 413. 44 Magnetization Chapter 9 P R O B L E M S 9.2 Laws and Continuity Conditions with Magnetization 9.2.1 Return to Prob. 6.1.1 and replace P→M. Find ρmandσsm. 9.2.2∗A circular cylindrical rod of material is uniformly magnetized in the y/prime direction transverse to its axis, as shown in Fig. P9.2.2. Thus, for r < R,M=Mo[ixsinγ+iycosγ]. In the surrounding region, the material forces Hto be zero. (In Sec. 9.6, it will be seen that such a material is one of inﬁnite permeability.) Fig. P9.2.2 (a) Show that if H=0everywhere, both Amp` ere’s law and (9.2.2) are satisﬁed. (b) Suppose that the cylinder rotates with the angular velocity Ω so that γ= Ωt. Then, Bis time varying even though there is no H. A one- turn rectangular coil having depth din the zdirection has legs running parallel to the zaxis in the + zdirection at x=−R, y = 0 and in the−zdirection at x=R, y = 0. The other legs of the coil are perpendicular to the zaxis. Show that the voltage induced at the terminals of this coil by the time-varying magnetization density is v=−µo2RdM oΩ sin Ω t. Fig. P9.2.3 Sec. 9.3 Problems 45 Fig. P9.3.1 9.2.3 In a region between the planes y=aandy= 0, a material that moves in the xdirection with velocity Uhas the magnetization density M= Moiycosβ(x−Ut), as shown in Fig. P9.2.2. The regions above and below are constrained so that H=0there and so that the integral of H·ds between y= 0 and y=ais zero. (In Sec. 9.7, it will be clear that these materials could be the pole faces of a highly permeable magnetic circuit.) (a) Show that Amp` ere’s law and (9.2.2) are satisﬁed if H=0throughout the magnetizable layer of material. (b) A one-turn rectangular coil is located in the y= 0 plane, one leg running in the + zdirection at x=−d(from z= 0 to z=l) and another running in the −zdirection at x=d(from z=ltoz= 0). What is the voltage induced at the terminals of this coil by the motion of the layer? 9.3 Permanent Magnetization 9.3.1∗The magnet shown in Fig. P9.3.1 is much longer in the ±zdirections than either of its cross-sectional dimensions 2 aand 2 b. Show that the scalar magnetic potential is Ψ =Mo 2π½ (x−a)lnp (x−a)2+ (y−b)2 p (x−a)2+ (y+b)2 −(x+a)lnp (x+a)2+ (y−b)2 p (x+a)2+ (y+b)2 + (y−b)· tan−1¡x−a y−b¢ −tan−1¡x+a y−b¢¸ −(y+b)· tan−1¡x−a y+b¢ −tan−1¡x+a y+b¢¸¾(a) (Note Example 4.5.3.) 46 Magnetization Chapter 9 9.3.2∗In the half-space y >0,M=Mocos(βx) exp(−αy)iy, where αandβare given positive constants. The half-space y <0 is free space. Show that Ψ =Mo 28 >< >:· −2α α2−β2e−αy+e−βy α−β¸ cosβx;y >0 −eβy α+βcosβx; y <0(a) 9.3.3 In the half-space y <0,M=Mosin(βx) exp( αy)ix, where αandβare positive constants. The half-space y > 0 is free space. Find the scalar magnetic potential. Fig. P9.3.4 9.3.4 For storage of information, the cylinder shown in Fig. P9.3.4 has the mag- netization density M=Mo(r/R)p−1[ircosp(φ−γ)−iφsinp(φ−γ)] ( a) where pis a given integer. The surrounding region is free space. (a) Determine the magnetic potential Ψ. (b) A magnetic pickup is comprised of an N-turn coil located at φ= π/2. This coil has a dimension ain the φdirection that is small compared to the periodicity length 2 πR/p in that direction. Every turn is essentially at the radius d+R. Determine the output voltage voutwhen the cylinder rotates, γ= Ωt. (c) Show that if the density of information on the cylinder is to be high (pis to be high), then the spacing between the coil and the cylinder, d, must be small. 9.4 Magnetization Constitutive Laws 9.4.1∗The toroidal core of Example 9.4.1 and Demonstration 9.4.1 is ﬁlled by a material having the single-valued magnetization characteristic M=Mo tanh ( αH), where MandHare collinear. (a) Show that the B−Hcharacteristic is of the type illustrated in Fig. 9.4.4. Sec. 9.5 Problems 47 Fig. P9.5.1 (b) Show that if i=iocosωt, the output voltage is v=µoπw2N2 4d dt·N1io 2πRcosωt+MotanhµαN1io 2πRcosωt¶¸ (a) (c) Show that the characteristic is essentially linear, provided that αN1io/2πR/lessmuch1. 9.4.2 The toroidal core of Demonstration 9.4.1 is driven by a sinusoidal current i(t) and responds with the hysteresis characteristic of Fig. 9.4.6. Make qualitative sketches of the time dependence of (a)B(t) (b) the output voltage v(t). 9.5 Fields in the Presence of Magnetically Linear Insulating Materials 9.5.1∗A perfectly conducting sheet is bent into a ⊃shape to make a one-turn inductor, as shown in Fig. P9.5.1. The width wis much larger than the dimensions in the x−yplane. The region inside the inductor is ﬁlled with two linearly magnetizable materials having permeabilities µaandµb, respectively. The cross-section of the system in any x−yplane is the same. The cross-sectional areas of the magnetizable materials are AaandAb, respectively. Given that the current i(t) is uniformly distributed over the width wof the inductor, show that H= (i/w)izin both of the magnetizable materials. Show that the inductance L= (µaAa+µbAb)/w. 9.5.2 Perfectly conducting coaxial cylinders, shorted at one end, form the one- turn inductor shown in Fig. P9.5.2. The total current iﬂowing on the surface at r=bof the inner cylinder is returned through the short and the outer conductor at r=a. The annulus is ﬁlled by materials of uniform permeability with an interface at r=R, as shown. (a) Determine Hin the annulus. (A simple solution can be shown to satisfy all the laws and continuity conditions.) 48 Magnetization Chapter 9 Fig. P9.5.2 (b) Find the inductance. 9.5.3∗The piece-wise uniform material in the one-turn inductor of Fig. P9.5.1 is replaced by a smoothly inhomogeneous material having the permeability µ=−µmx/l, where µmis a given constant. Show that the inductance is L=dµml/2w. 9.5.4 The piece-wise uniform material in the one-turn inductor of Fig. P9.5.2 is replaced by one having the permeability µ=µm(r/b), where µmis a given constant. Determine the inductance. 9.5.5∗Perfectly conducting coaxial cylinders, shorted at one end, form a one-turn inductor as shown in Fig. P9.5.5. Current ﬂowing on the surface at r=b of the inner cylinder is returned on the inner surface of the outer cylinder atr=a. The annulus is ﬁlled by sectors of linearly magnetizable material, as shown. (a) Assume that in the regions (a) and (b), respectively, H=iφA/r andH=iφC/r, and show that with AandCfunctions of time, these ﬁelds satisfy Amp` ere’s law and the ﬂux continuity law in the respective regions. (b) Use the ﬂux continuity condition at the interfaces between regions to show that C= (µa/µb)A. (c) Use Amp` ere’s integral law to relate CandAto the total current iin the inner conductor. (d) Show that the inductance is L=lµaln(a/b)/[α+ (2π−α)µa/µb]. (e) Show that the surface current densities at r=badjacent to regions (a) and (b), respectively, are Kz=A/bandKz=C/b. 9.5.6 In the one-turn inductor of Fig. P9.5.1, the material of piece-wise uniform permeability is replaced by another such material. Now the region between the plates in the range 0 < z < a is ﬁlled by material having uniform permeability µa, while µ=µbin the range a < z < w . Determine the inductance. Sec. 9.6 Problems 49 Fig. P9.5.5 9.6 Fields in Piece-Wise Uniform Magnetically Linear Materials 9.6.1∗A winding in the y= 0 plane is used to produce the surface current density K=Kocosβzix. Region (a), where y >0, is free space, while region (b), where y <0, has permeability µ. (a) Show that Ψ =Kosinβz β(1 +µ/µo)½−µ µoe−βy;y >0 eβy; y <0(a) (b) Now consider the same problem, but assume at the outset that the material in region (b) has inﬁnite permeability. Show that it agrees with the limit µ→ ∞ of the ﬁrst expression of part (a). (c) In turn, use the result of part (b) as a starting point in ﬁnding an approximation to Ψ in the highly permeable material. Show that this result agrees with the limit of the second result of part (a) where µ/greatermuchµo. 9.6.2 The planar region −d < y < d is bounded from above and below by inﬁnitely permeable materials, as shown in Fig. P9.6.2. Region (a) to the right and region (b) to the left are separated by a current sheet in the plane x= 0 with the distribution K=izKosin(πy/2d). The system extends to inﬁnity in the ±xdirections and is two dimensional. (a) In terms of Ψ, what are the boundary conditions at y=±d. (b) What continuity conditions relate Ψ in regions (a) and (b) where they meet at x= 0? (c) Determine Ψ. 9.6.3∗The cross-section of a two-dimensional cylindrical system is shown in Fig. P9.6.3. A region of free space having radius Ris surrounded by material 50 Magnetization Chapter 9 Fig. P9.6.2 Fig. P9.6.3 having permeability µwhich can be considered as extending to inﬁnity. A winding at r=Ris driven by the current iand has turns density (N/2R) sinφ(turns per unit length in the φdirection). Thus, at r=R, there is a current density K= (N/2R)isinφiz. (a) Show that Ψ =(N/2)icosφ (1 +µ/µo)½R r; r > R −(µ/µo)(r/R);r < R(a) (b) An n-turn coil having a spacing between conductors of 2 ais now placed at the center. The magnetic axis of this coil is inclined at the angle αrelative to the xaxis. This coil has length lin the zdirection. Show that the mutual inductance between this coil and the one at r=RisLm=µoa lnN cosα/R[1 + ( µo/µ)]. 9.6.4 The cross-section of a motor or generator is shown in Fig. 11.7.7. The two coils comprising the stator and rotor windings and giving rise to the surface current densities of (11.7.24) and (11.7.25) have ﬂux linkages having the forms given by (11.7.26). (a) Assume that the permeabilities of the rotor and stator are inﬁnite, and determine the vector potential in the air gap. (b) Determine the self-inductances LsandLrand magnitude of the peak mutual inductance, M, in (11.7.26). Assume that the current in the +zdirection at φis returned at φ+π. Sec. 9.6 Problems 51 Fig. P9.6.5 9.6.5 A wire carrying a current iin the zdirection is suspended a height habove the surface of a magnetizable material, as shown in Fig. P9.6.5. The wire extends to “inﬁnity” in the ±zdirections. Region (a), where y >0, is free space. In region (b), where y <0, the material has uniform permeability µ. (a) Use the method of images to determine the ﬁelds in the two regions. (b) Now assume that µ/greatermuchµoand ﬁnd Hin the upper region, assuming at the outset that µ→ ∞ . (c) In turn, use this approximate result to ﬁnd the ﬁeld in the permeable material. (d) Show that the results of (b) and (c) are consistent with those from the exact analysis in the limit where µ/greatermuchµo. 9.6.6∗A conductor carries the current i(t) at a height habove the upper surface of a material, as shown in Fig. P9.6.5. The force per unit length on the conductor is f=i×µoH, where iis a vector having the direction and magnitude of the current i(t), and Hdoes not include the self-ﬁeld of the line current. (a) Show that if the material is a perfect conductor, f=µoiyi2/4πh. (b) Show that if the material is inﬁnitely permeable, f=−µoiyi2/4πh. 9.6.7∗Material having uniform permeability µis bounded from above and below by regions of inﬁnite permeability, as shown in Fig. P9.6.7. With its center at the origin and on the surface of the lower inﬁnitely permeable material is a hemispherical cavity of free space having radius athat is much less than d. A ﬁeld that has the uniform intensity Hofar from the hemispherical surface is imposed in the zdirection. (a) Assume µ/greatermuchµoand show that the approximate magnetic potential in the magnetizable material is Ψ = −Hoa[(r/a) + (a/r)2/2] cos θ. (b) In turn, show that the approximate magnetic potential inside the hemisphere is Ψ = −3Hoz/2. 9.6.8 In the magnetic tape conﬁguration of Example 9.3.2, the system is as shown in Fig. 9.3.2 except that just below the tape, in the plane y=−d/2, there is an inﬁnitely permeable material, and in the plane y=a > d/ 2 above the tape, there is a second inﬁnitely permeable material. Find the voltage vo. 52 Magnetization Chapter 9 Fig. P9.6.7 Fig. P9.6.9 9.6.9∗A cylindrical region of free space of rectangular cross-section is surrounded by inﬁnitely permeable material, as shown in Fig. P9.6.9. Surface currents are imposed by means of windings in the planes x= 0 and x=b. Show that Ψ =Koa πsinπy acoshπ a¡ x−b 2¢ cosh¡πb 2a¢ (a) 9.6.10∗A circular cylindrical hole having radius Ris cut through a material having permeability µa. A conductor passing through this hole has permeability µb and carries the uniform current density J=Joiz, as shown in Fig. P9.6.10. A ﬁeld that is uniform far from the hole, where it is given by H=Hoix, is applied by external means. Show that for r < R , and R < r , respectively, Az=(−µbJor2 4−2µbHoR (1+µb/µa)r Rsinφ −µaJoR2 2£ ln(r/R) +1 2µb µa¤ −µaHoR£r R−(µa−µb) (µa+µb)R r¤ sinφ(a) 9.6.11∗Although the introduction of a magnetizable sphere into a uniform mag- netic ﬁeld results in a distortion of that ﬁeld, nevertheless, the ﬁeld within the sphere is uniform. This fact makes it possible to determine the ﬁeld dis- tribution in and around a spherical particle even when its magnetization characteristic is nonlinear. For example, consider the ﬁelds in and around the sphere of material shown together with its B−Hcurve in Fig. P9.6.11. Sec. 9.7 Problems 53 Fig. P9.6.10 Fig. P9.6.11 (a) Assume that the magnetization density is M=Miz, where Mis a constant to be determined, and show that the magnetic ﬁeld intensity inside the sphere is uniform, zdirected, and of magnitude H=Ho− M/3, and hence that the magnetic ﬂux density, B, in the sphere is related to the magnitude of the magnetic ﬁeld intensity Hby B= 3µoHo−2µoH (a) (b) Draw this load line in the B−Hplane, showing that it is a straight line with intercepts 3 Ho/2 and 3 µoHowith the HandBaxes, respectively. (c) Show how ( B, H ) in the sphere are determined, given the applied ﬁeld intensity Ho, by graphically ﬁnding the point of intersection between theB−Hcurve of Fig. P9.6.11 and (a). (d) Show that if Ho= 4×105A/m, B= 0.75 tesla and H= 3.1×105 A/m. 9.6.12 The circular cylinder of magnetizable material shown in Fig. P9.6.12 has theB−Hcurve shown in Fig. P9.6.11. Determine BandHinside the cylinder resulting from the application of a ﬁeld intensity H=Hoixwhere Ho= 4×105A/m. 54 Magnetization Chapter 9 Fig. P9.6.12 9.6.13 The spherical coil of Example 9.6.1 is wound around a sphere of material having the B−Hcurve shown in Fig. P9.6.11. Assume that i= 800 A, N= 100 turns, and R= 10 cm, and determine BandHin the material. 9.7 Magnetic Circuits 9.7.1∗The magnetizable core shown in Fig. P9.7.1 extends a distance d into the paper that is large compared to the radius a. The driving coil, having Nturns, has an extent ∆ in the φdirection that is small compared to dimensions of interest. Assume that the core has a permeability µthat is very large compared to µo. (a) Show that the approximate Hand Ψ inside the core (with Ψ deﬁned to be zero at φ=π) are H=Ni 2πriφ; Ψ =Ni 2¡ 1−φ π¢ (a) (b) Show that the approximate magnetic potential in the central region is Ψ =∞X m=1Ni mπ(r/b)msinmφ (b) 9.7.2 For the conﬁguration of Prob. 9.7.1, determine Ψ in the region outside the core, r > a . 9.7.3∗In the magnetic circuit shown in Fig. P9.7.3, an N-turn coil is wrapped around the center leg of an inﬁnitely permeable core. The sections to right and left have uniform permeabilities µaandµb, respectively, and the gap lengths aandbare small compared to the other dimensions of these sec- tions. Show that the inductance L=N2w[(µbd/b) + (µac/a)]. 9.7.4 The magnetic circuit shown in Fig. P9.7.4 is constructed from inﬁnitely permeable material, as is the hemispherical bump of radius Rlocated on the surface of the lower pole face. A coil, having Nturns, is wound around Sec. 9.7 Problems 55 Fig. P9.7.1 Fig. P9.7.3 Fig. P9.7.4 56 Magnetization Chapter 9 Fig. P9.7.5 Fig. P9.7.6 the left leg of the magnetic circuit. A second coil is wound around the hemisphere in a distributed fashion. The turns per unit length, measured along the periphery of the hemisphere, is ( n/R) sinα, where nis the total number of turns. Given that R/lessmuchh/lessmuchw, ﬁnd the mutual inductance of the two coils. 9.7.5∗The materials comprising the magnetic circuit of Fig. P9.7.5 can be re- garded as having inﬁnite permeability. The air gaps have a length xthat is much less than aorb, and these dimensions, in turn, are much less than w. The coils to left and right, respectively, have total turns N1andN2. Show that the self- and mutual inductances of the coils are L11=N2 1Lo, L 12=L21=N1N2Lo, L22=N2 2Lo, L o≡awµ o x(1 +a/b)(a) 9.7.6 The magnetic circuit shown in Fig. P9.7.6 has rotational symmetry about thezaxis. Both the circular cylindrical plunger and the remainder of the magnetic circuit can be regarded as inﬁnitely permeable. The air gaps have Sec. 9.7 Problems 57 Fig. P9.7.7 widths xandgthat are small compared to aandd. Determine the induc- tance of the coil. 9.7.7 Two cross-sectional views of an axisymmetric magnetic circuit that could be used as an electromechanical transducer are shown in Fig. P9.7.7. Sur- rounding an inﬁnitely permeable circular cylindrical rod having a radius slightly less than ais an inﬁnitely permeable stator having a hole down its center with a radius slightly greater than a. A pair of coils, having turns N1 andN2and driven by currents i1andi2, respectively are wound around the center rod and positioned in slots in the surrounding stator. The longitudi- nal position of the rod, denoted by ξ, is limited in range so that the ends of the rod are always well inside the ends of the stator. Thus, Hin each of the air gaps is essentially uniform. Determine the inductance matrix, (9.7.12). 9.7.8 Fields in and around the magnetic circuit shown in Fig. P9.7.8 are to be considered as independent of z. The outside walls are inﬁnitely permeable, while the horizontal central leg has uniform permeability µthat is much less than that of the sides but nevertheless much greater than µo. Coils having total turns N1andN2, respectively, are wound around the center leg. These have evenly distributed turns in the planes x=l/2 and x=−l/2, respectively. The regions above and below the center leg are free space. (a) Deﬁne Ψ = 0 at the origin of the given coordinates. As far as Ψ is concerned inside the center leg, what boundary conditions must Ψ satisfy if the central leg is treated as the “inside” of an “inside- outside” problem? (b) What is Ψ in the center leg? (c) What boundary conditions must Ψ satisfy in region (a)? (d) What is Ψ, and hence H, in region (a)? (A simple exact solution is suggested by Prob. 7.5.3.) For the case where N1i1=N2i2, sketch ψ andHin regions (a) and (b). 9.7.9 The magnetic circuit shown in Fig. P9.7.9 is excited by an N-turn coil and consists of inﬁnitely permeable legs in series with ones of permeability µ, one to the right of length l2and the other to the left of length l1. This second leg has wrapped on its periphery a metal strap having thickness ∆/lessmuchw, conductivity σ, and height l1. With a terminal current i=iocosωt, determine Hwithin the left leg. 58 Magnetization Chapter 9 Fig. P9.7.8 Fig. P9.7.9 9.7.10∗The graphical approach to determining ﬁelds in magnetic circuits to be used in this and the next example is similar to that illustrated by Probs. 9.6.11–9.6.13. The magnetic circuit of a high-ﬁeld magnet is shown in Fig. P9.7.10. The two coils each have Nturns and carry a current i. (a) Show that the load line for the circuit is B=−µo d(l2+l1)H+2Niµ o d(a) (b) For N= 500 , d= 1 cm, l1= 0.8m, l 2= 0.2 m, and i= 10 amps, ﬁnd the ﬂux density Bin its air gap. 9.7.11 In the magnetic circuit of Fig. P9.7.11, the inﬁnitely permeable core has a gap with cross-sectional area Aand height a+b, where the latter is much less than the dimensions of the former. In this gap is a material having height band the M−Hrelation also shown in the ﬁgure. Within the material and in the air gap, His approximated as being uniform. Sec. 9.7 Problems 59 Fig. P9.7.10 Fig. P9.7.11 (a) Determine the load line relation between Hb, the ﬁeld intensity in the material, M, and the driving current i. (b) If Ni/a = 0.5×106amps/m and b/a= 1, what is M, and hence B? 10 MAGNETOQUASISTATIC RELAXATION AND DIFFUSION 10.0 INTRODUCTION In the MQS approximation, Amp` ere’s law relates the magnetic ﬁeld intensity Hto the current density J. ∇ ×H=J (1) Augmented by the requirement that Hhave no divergence, this law was the theme of Chap. 8. Two types of physical situations were considered. Either the current density was imposed, or it existed in perfect conductors. In both cases, we were able to determine Hwithout being concerned about the details of the electric ﬁeld distribution. In Chap. 9, the eﬀects of magnetizable materials were represented by the magnetization density M, and the magnetic ﬂux density, deﬁned as B≡µo(H+M), was found to have no divergence. ∇ ·B= 0 (2) Provided that Mis either given or instantaneously determined by H(as was the case throughout most of Chap. 9), and that Jis either given or subsumed by the boundary conditions on perfect conductors, these two magnetoquasistatic laws determine Hthroughout the volume. In this chapter, our ﬁrst objective will be to determine the distribution of E around perfect conductors. Then we shall broaden our physical domain to include ﬁnite conductors, especially in situations where currents are caused by an Ethat 1 2 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 is induced by the time rate of change of B. In both cases, we make explicit use of Faraday’s law. ∇ ×E=−∂B ∂t (3) In the EQS systems considered in Chaps. 4–7, the curl of Hgenerated by the time rate of change of the displacement ﬂux density was not of interest. Amp` ere’s law was adequately incorporated by the continuity law. However, in MQS systems, the curl of Egenerated by the magnetic induction on the right in (1) is often of primary importance. We had ﬁelds that depended on time rates of change in Chap. 7. We have already seen the consequences of Faraday’s law in Sec. 8.4, where MQS systems of perfect conductors were considered. The electric ﬁeld intensity Einside a perfect conductor must be zero, and hence Bhas to vanish inside the perfect conductor if Bvaries with time. This leads to n·B= 0 on the surface of a perfect conductor. Currents induced in the surface of perfect conductors assure the proper discontinuity of n×Hfrom a ﬁnite value outside to zero inside. Faraday’s law was in evidence in Sec. 8.4 and accounted for the voltage at terminals connected to each other by perfect conductors. Faraday’s law makes it possible to have a voltage at terminals connected to each other by a perfect “short.” A simple experiment brings out some of the subtlety of the voltage deﬁnition in MQS systems. Its description is followed by an overview of the chapter. Demonstration 10.0.1. Nonuniqueness of Voltage in an MQS System A magnetic ﬂux is created in the toroidal magnetizable core shown in Fig. 10.0.1 by driving the winding with a sinsuoidal current. Because it is highly permeable (a ferrite), the core guides a magnetic ﬂux density Bthat is much greater than that in the surrounding air. Looped in series around the core are two resistors of unequal value, R1/negationslash= R2. Thus, the terminals of these resistors are connected together to form a pair of “nodes.” One of these nodes is grounded. The other is connected to high-impedance voltmeters through two leads that follow the diﬀerent paths shown in Fig. 10.0.1. A dual-trace oscilloscope is convenient for displaying the voltages. The voltages observed with the leads connected to the same node not only diﬀer in magnitude but are 180 degrees out of phase. Faraday’s integral law explains what is observed. A cross-section of the core, showing the pair of resistors and voltmeter leads, is shown in Fig. 10.0.2. The scope resistances are very large compared to R1andR2, so the current carried by the voltmeter leads is negligible. This means that if there is a current ithrough one of the series resistors, it must be the same as that through the other. The contour Ccfollows the closed circuit formed by the series resistors. Fara- day’s integral law is now applied to this contour. The ﬂux passing through the surface Scspanning Ccis deﬁned as Φ λ. Thus,I C1E·ds=−dΦλ dt=i(R1+R2) (4) where Φλ≡Z ScB·da (5) Sec. 10.0 Introduction 3 Fig. 10.0.1 A pair of unequal resistors are connected in series around a magnetic circuit. Voltages measured between the terminals of the re- sistors by connecting the nodes to the dual-trace oscilloscope, as shown, diﬀer in magnitude and are 180 degrees out of phase. Fig. 10.0.2 Schematic of circuit for experiment of Fig. 10.0.1, showing contours used with Faraday’s law to predict the diﬀering voltages v1and v2. Given the magnetic ﬂux, (4) can be solved for the current ithat must circulate around the loop formed by the resistors. To determine the measured voltages, the same integral law is applied to con- tours C1andC2of Fig. 10.0.2. The surfaces spanning the contours link a negligible ﬂux density, so the circulation of Earound these contours must vanish. I C1E·ds=v1+iR1= 0 (6) I C2E·ds=−v2+iR2= 0 (7) 4 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 The observed voltages are found by solving (4) for i, which is then substituted into (6) and (7). v1=R1 R1+R2dΦλ dt(8) v2=−R2 R1+R2dΦλ dt(9) From this result it follows that v1 v2=−R1 R2(10) Indeed, the voltages not only diﬀer in magnitude but are of opposite signs. Suppose that one of the voltmeter leads is disconnected from the right node, looped through the core, and connected directly to the grounded terminal of the same voltmeter. The situation is even more remarkable because we now have a voltage at the terminals of a “short.” However, it is also more familiar. We recognize from Sec. 8.4 that the measured voltage is simply dλ/dt , where the ﬂux linkage is in this case Φλ. In Sec. 10.1, we begin by investigating the electric ﬁeld in the free space regions of systems of perfect conductors. Here the viewpoint taken in Sec. 8.4 has made it possible to determine the distribution of Bwithout having to determine Ein the process. The magnetic induction appearing on the right in Faraday’s law, (1), is therefore known, and hence the law prescribes the curl of E. From the introduction to Chap. 8, we know that this is not enough to uniquely prescribe the electric ﬁeld. Information about the divergence of Emust also be given, and this brings into play the electrical properties of the materials ﬁlling the regions between the perfect conductors. The analyses of Chaps. 8 and 9 determined Hin two special situations. In one case, the current distribution was prescribed; in the other case, the currents were ﬂowing in the surfaces of perfect conductors. To see the more general situation in perspective, we may think of MQS systems as analogous to networks composed of inductors and resistors, such as shown in Fig. 10.0.3. In the extreme case where the source is a rapidly varying function of time, the inductors alone determine the currents. Finding the current distribution in this “high frequency” limit is analogous to ﬁnding the H-ﬁeld, and hence the distribution of surface currents, in the systems of perfect conductors considered in Sec. 8.4. Finding the electric ﬁeld in perfectly conducting systems, the objective in Sec. 10.1 of this chapter, is analogous to determining the distribution of voltage in the circuit in the limit where the inductors dominate. In the opposite extreme, if the driving voltage is slowly varying, the induc- tors behave as shorts and the current distribution is determined by the resistive network alone. In terms of ﬁelds, the response to slowly varying sources of current is essentially the steady current distribution described in the ﬁrst half of Chap. 7. Once this distribution of Jhas been determined, the associated magnetic ﬁeld can be found using the superposition integrals of Chap. 8. In Secs. 10.2–10.4, we combine the MQS laws of Chap. 8 with those of Faraday and Ohm to describe the evolution of JandHwhen neither of these limiting cases prevails. We shall see that the ﬁeld response to a step of excitation goes from a Sec. 10.1 MQS Electric Fields 5 Fig. 10.0.3 Magnetoquasistatic systems with Ohmic conductors are gen- eralizations of inductor-resistor networks. The steady current distribution is determined by the resistors, while the high-frequency response is governed by the inductors. distribution governed by the perfect conductivity model just after the step is applied (the circuit dominated by the inductors), to one governed by the steady conduction laws for J, and Biot-Savart for Hafter a long time (the circuit dominated by the resistors with the ﬂux linkages then found from λ=Li). Under what circumstances is the perfectly conducting model appropriate? The characteristic times for this magnetic ﬁeld diﬀusion process will provide the answer. 10.1 MAGNETOQUASISTATIC ELECTRIC FIELDS IN SYSTEMS OF PERFECT CONDUCTORS The distribution of Earound the conductors in MQS systems is of engineering interest. For example, the amount of insulation required between conductors in a transformer is dependent on the electric ﬁeld. In systems composed of perfect conductors and free space, the distribution of magnetic ﬁeld intensity is determined by requiring that n·B= 0 on the perfectly conducting boundaries. Although this condition is required to make the electric ﬁeld tangential to the perfect conductor vanish, as we saw in Sec. 8.4, it is not necessary to explicitly refer to Ein ﬁnding H. Thus, in Faraday’s law of induction, (10.0.3), the right-hand side is known. The source of curlEis thus known. To determine the source of divE, further information is required. The regions outside the perfect conductors, where Eis to be found, are pre- sumably ﬁlled with relatively insulating materials. To identify the additional infor- mation necessary for the speciﬁcation of E, we must be clear about the nature of these materials. There are three possibilities: •Although the material is much less conducting than the adjacent “perfect” conductors, the charge relaxation time is far shorter than the times of interest. Thus, ∂ρ/∂t is negligible in the charge conservation equation and, as a result, the current density is solenoidal. Note that this is the situation in the MQS approximation. In the following discussion, we will then presume that if this situation prevails, the region is ﬁlled with a material of uniform conductivity, in which case Eis solenoidal within the material volume. (Of course, there 6 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 may be surface charges on the boundaries.) ∇ ·E= 0 (1) •The second situation is typical when the “perfect” conductors are surrounded by materials commonly used to insulate wires. The charge relaxation time is generally much longer than the times of interest. Thus, no unpaired charges can ﬂow into these “insulators” and they remain charge free. Provided they are of uniform permittivity, the Eﬁeld is again solenoidal within these materials. •If the charge relaxation time is on the same order as times characterizing the currents carried by the conductors, then the distribution of unpaired charge is governed by the combination of Ohm’s law, charge conservation, and Gauss’ law, as discussed in Sec. 7.7. If the material is not only of uniform conductivity but of uniform permittivity as well, this charge density is zero in the volume of the material. It follows from Gauss’ law that Eis once again solenoidal in the material volume. Of course, surface charges may exist at material interfaces. The electric ﬁeld intensity is broken into particular and homogeneous parts E=Ep+Eh (2) where, in accordance with Faraday’s law, (10.0.3), and (1), ∇ ×Ep=−∂B ∂t(3) ∇ ·Ep= 0 (4) and ∇ ×Eh= 0 (5) ∇ ·Eh= 0. (6) Our approach is reminiscent of that taken in Chap. 8, where the roles of Eand ∂B/∂tare respectively taken by Hand−J. Indeed, if all else fails, the particular solution can be generated by using an adaptation of the Biot-Savart law, (8.2.7). Ep=−1 4πZ V/prime∂B ∂t(r/prime)×ir/primer |r−r/prime|2dv/prime(7) Given a particular solution to (3) and (4), the boundary condition that there be no tangential Eon the surfaces of the perfect conductors is satisﬁed by ﬁnding a solution to (5) and (6) such that n×E= 0⇒n×Eh=−n×Ep (8) on those surfaces. Given the particular solution, the boundary value problem has been reduced to one familiar from Chap. 5. To satisfy (5), we let Eh=−∇Φ. It then follows from (6) that Φ satisﬁes Laplace’s equation. Sec. 10.1 MQS Electric Fields 7 Fig. 10.1.1 Side view of long inductor having radius aand length d. Example 10.1.1. Electric Field around a Long Coil What is the electric ﬁeld distribution in and around a typical inductor? An ap- proximate analysis for a coil of many turns brings out the reason why transformer and generator designers often speak of the “volts per turn” that must be withstood by insulation. The analysis illustrates the concept of breaking the solution into a particular rotational ﬁeld and a homogeneous conservative ﬁeld. Consider the idealized coil of Fig. 10.1.1. It is composed of a thin, perfectly conducting wire, wound in a helix of length dand radius a. The magnetic ﬁeld can be found by approximating the current by a surface current Kthat is φdirected about the zaxis of a cylindrical coodinate system having the zaxis coincident with the axis of the coil. For an N-turn coil, this surface current density is Kφ=Ni/d . If the coil is very long, d/greatermucha, the magnetic ﬁeld produced within is approximately uniform Hz=Ni d(9) while that outside is essentially zero (Example 8.2.1). Note that the surface current density is just that required to terminate Hin accordance with Amp` ere’s continuity condition. With such a simple magnetic ﬁeld, a particular solution is easily obtained. We recognize that the perfectly conducting coil is on a natural coordinate surface in the cylindrical coordinate system. Thus, we write the zcomponent of (3) in cylindrical coordinates and look for a solution to Ethat is independent of φ. The solution resulting from an integration over ris Ep=iφ½−µor 2dHz dtr < a −µoa2 2rdHz dtr > a(10) Because there is no magnetic ﬁeld outside the coil, the outside solution for Epis irrotational. If we adhere to the idealization of the wire as an inclined current sheet, the electric ﬁeld along the wire in the sheet must be zero. The particular solution does not satisfy this condition, and so we now must ﬁnd an irrotational and solenoidal Ehthat cancels the component of Eptangential to the wire. A section of the wire is shown in Fig. 10.1.2. What axial ﬁeld Ezmust be added to that given by (10) to make the net Eperpendicular to the wire? If Ezand Eφare to be components of a vector normal to the wire, then their ratio must be 8 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.1.2 With the wire from the inductor of Fig. 10.1.1 stretched into a straight line, it is evident that the slope of the wire in the inductor is essentially the total length of the coil, d, divided by the total length of the wire, 2 πaN. the same as the ratio of the total length of the wire to the length of the coil. Ez Eφ=2πaN d; r=a (11) Using (9) and (10) at r=a, we have Ez=−µoπa2N2 d2di dt(12) The homogeneous solution possesses this ﬁeld Ezon the surface of the cylinder of radius aand length d. This ﬁeld determines the potential Φ hover the surface (within an arbitrary constant). Since ∇2Φh= 0 everywhere in space and the tangential Eh ﬁeld prescribes Φ hon the cylinder, Φ his uniquely determined everywhere within an additive constant. Hence, the conservative part of the ﬁeld is determined everywhere. The voltage between the terminals is determined from the line integral of E·dsbetween the terminals. The ﬁeld of the particular solution is φ-directed and gives no contribution. The entire contribution to the line integral comes from the homogeneous solution (12) and is v=−Ezd=µoN2πa2 ddi dt(13) Note that this expression takes the form Ldi/dt , where the inductance Lis in agree- ment with that found using a contour coincident with the wire, (8.4.18). We could think of the terminal voltage as the sum of N“voltages per turn” Ezd/N. If we admit to the ﬁnite size of the wires, the electric stress between the wires is essentially this “voltage per turn” divided by the distance between wires. The next example identiﬁes the particular and homogeneous solutions in a somewhat more formal fashion. Example 10.1.2. Electric Field of a One-Turn Solenoid The cross-section of a one-turn solenoid is shown in Fig. 10.1.3. It consists of a circular cylindrical conductor having an inside radius amuch less than its length in Sec. 10.1 MQS Electric Fields 9 Fig. 10.1.3 A one-turn solenoid of inﬁnite length is driven by the distributed source of current density, K(t). Fig. 10.1.4 Tangential component of homogeneous electric ﬁeld at r=ain the conﬁguration of Fig. 10.1.3. thezdirection. It is driven by a distributed current source K(t) through the plane parallel plates to the left. This current enters through the upper sheet conductor, circulates in the φdirection around the one turn, and leaves through the lower plate. The spacing between these plates is small compared to a. As in the previous example, the ﬁeld inside the solenoid is uniform, axial, and equal to the surface current H=izK(t) (14) and a particular solution can be found by applying Faraday’s integral law to a contour having the arbitrary radius r < a , (10). Ep=Eφpiφ; Eφp≡ −µor 2dK dt(15) This ﬁeld clearly does not satisfy the boundary condition at r=a, where it has a tangential value over almost all of the surface. The homogeneous solution must have a tangential component that cancels this one. However, this ﬁeld must also be conservative, so its integral around the circumference at r=amust be zero. Thus, the plot of the φcomponent of the homogeneous solution at r=a, shown in Fig. 10.1.4, has no average value. The amplitude of the tall rectangle is adjusted so that the net area under the two functions is zero. Eφp(2π−α) =hα⇒h=Eφp¡2π α−1¢ (16) The ﬁeld between the edges of the input electrodes is approximated as being uniform right out to the contacts with the solenoid. We now ﬁnd a solution to Laplace’s equation that matches this boundary condition on the tangential component of E. Because Eφis an even function of φ,Φ is taken as an odd function. The origin is included in the region of interest, so the polar coordinate solutions (Table 5.7.1) take the form Φ =∞X n=1Anrnsinnφ (17) 10 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.1.5 Graphical representation of solution for the electric ﬁeld in the conﬁguration of Fig. 10.1.3. It follows that Eφh=−1 r∂Φ ∂φ=−∞X n=1nAnrn−1cosnφ. (18) The coeﬃcients Anare evaluated, as in Sec. 5.5, by multiplying both sides of this expression by cos( mφ) and integrating from φ=−πtoφ=π. Z−α/2 −π−Eφpcosmφdφ +Zα/2 −α/2Eφp¡2π α−1¢ cosmφdφ +Zπ α/2−Eφpcosmφdφ =−mAmam−1π(19) Thus, the coeﬃcients needed to evaluate the potential of (17) are Am=−4Eφp(r=a) m2am−1αsinmα 2(20) Finally, the desired ﬁeld intensity is the sum of the particular solution, (15), and the homogeneous solution, the gradient of (17). E=−µoa 2dK dt· r aiφ+ 4∞X n=1sinnα 2 αn¡r a¢n−1cosnφiφ + 4∞X n=1sinnα 2 nα¡r a¢n−1sinnφir¸ (21) The superposition of ﬁelds represented in this solution is shown graphically in Fig. 10.1.5. A conservative ﬁeld is added to the rotational ﬁeld. The former has Sec. 10.2 Nature of MQS Electric Fields 11 Fig. 10.2.1 Current induced in accordance with Faraday’s law circulates on contour Ca. Through Amp` ere’s law, it results in magnetic ﬁeld that follows contour Cb. a potential at r=athat is a linearly increasing function of φbetween the input electrodes, increasing from a negative value at the lower electrode at φ=−α/2, passing through zero at the midplane, and reaching an equal positive value at the upper electrode at φ=α/2. The potential decreases in a linear fashion from this high as φis increased, again passing through zero at φ= 180 degrees, and reaching the negative value upon returning to the lower input electrode. Equipotential lines therefore join points on the solenoid periphery with points at the same potential between the input electrodes. Note that the electric ﬁeld associated with this poten- tial indeed has the tangential component required to cancel that from the rotational part of the ﬁeld, the proof of this being in the last of the plots. Often the vector potential provides conveniently a particular solution. With Breplaced by ∇ ×A, ∇ ×µ E+∂A ∂t¶ = 0 (22) Suppose Ahas been determined. Then the quantity in parantheses must be equal to the gradient of a potential Φ so that E=−∂A ∂t− ∇Φ (23) In the examples treated, the ﬁrst term in this expression is the particular solution, while the second is the homogeneous solution. 10.2 NATURE OF FIELDS INDUCED IN FINITE CONDUCTORS If a conductor is situated in a time-varying magnetic ﬁeld, the induced electric ﬁeld gives rise to currents. From Sec. 8.4, we have shown that these currents prevent the penetration of the magnetic ﬁeld into a perfect conductor. How high must σbe to treat a conductor as perfect? In the next two sections, we use speciﬁc analytical models to answer this question. Here we preface these developments with a discussion of the interplay between the laws of Faraday, Amp` ere and Ohm that determines the distribution, duration, and magnitude of currents in conductors of ﬁnite conductivity. The integral form of Faraday’s law, applied to the surface Saand contour Ca of Fig. 10.2.1, isI CaE·ds=−d dtZ SaB·da (1) 12 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Ohm’s law, J=σE, introduced into (1), relates the current density circulating around a tube following Cato the enclosed magnetic ﬂux. I CaJ σ·ds=−d dtZ SaB·da (2) This statement applies to every circulating current “hose” in a conductor. Let us concentrate on one such hose. The current ﬂows parallel to the hose, and therefore J·ds=Jds. Suppose that the cross-sectional area of the hose is A(s). Then JA(s) =i, the current in the hose, and IA(s) σds=R (3) is the resistance of the hose. Therefore, iR=−dλ dt(4) Equation (2) describes how the time-varying magnetic ﬂux gives rise to a circulating current. Amp` ere’s law states how that current, in turn, produces a mag- netic ﬁeld. I CbH·ds=Z SbJ·da (5) Typically, that ﬁeld circulates around a contour such as Cbin Fig. 10.2.1, which is pierced by J. With Gauss’ law for B, Amp` ere’s law provides the relation for H produced by J. This information is summarized by the “lumped circuit” relation λ=Li (6) The combination of (4) and (6) provides a diﬀerential equation for the circuit current i(t). The equivalent circuit for the diﬀerential equation is the series inter- connection of a resistor Rwith an inductor L, as shown in Fig. 10.2.1. The solution is an exponentially decaying function of time with the time constant L/R. The combination of (2) and (5)– of the laws of Faraday, Amp` ere, and Ohm– determine JandH. The ﬁeld problem corresponds to a continuum of “circuits.” We shall ﬁnd that the time dependence of the ﬁelds is governed by time constants having the nature of L/R. This time constant will be of the form τm=µσl1l2 (7) In contrast with the charge relaxation time /epsilon1/σof EQS, this magnetic diﬀusion time depends on the product of two characteristic lengths, denoted here by l1and l2. For given time rates of change and electrical conductivity, the larger the system, the more likely it is to behave as a perfect conductor. Although we will not use the integral laws to determine the ﬁelds in the ﬁnite conductivity systems of the next sections, they are often used to make engineering Sec. 10.2 Nature of MQS Electric Fields 13 Fig. 10.2.2 When the spark gap switch is closed, the capacitor dis- charges into the coil. The contour Cbis used to estimate the average magnetic ﬁeld intensity that results. approximations. The following demonstration is quantiﬁed using rough approxima- tions in a style that typiﬁes how ﬁeld theory is often applied to practical problems. Demonstration 10.2.1. Edgerton’s Boomer The capacitor in Fig. 10.2.2, C= 25 µF, is initially charged to v= 4kV. The spark gap switch is then closed so that the capacitor can discharge into the 50-turn coil. This demonstration has been seen by many visitors to Prof. Harold Edgerton’s Strobe Laboratory at M.I.T. Given that the average radius of a coil winding a= 7 cm, and that the height of the coil is also on the order of a, roughly what magnetic ﬁeld is generated? Amp` ere’s integral law, (5), can be applied to the contour Cbof the ﬁgure to obtain an approximate relation between the average H, which we will call H1, and the coil current i1. H1≈N1i1 2πa(8) To determine i1, we need the inductance L11of the coil. To this end, the ﬂux linkage of the coil is approximated by N1times the product of the average coil area and the average ﬂux density. λ≈N1(πa2)µoH1 (9) From these last two equations, one obtains λ=L11i1, where the inductance is L11≈µoaN2 1 2(10) Evaluation gives L11= 0.1 mH. With the assumption that the combined resistance of the coil, switch, and connecting leads is small enough so that the voltage across the capacitor and the current in the inductor oscillate at the frequency ω=1√CL11(11) we can determine the peak current by recognizing that the energy1 2Cv2initially stored in the capacitor is one quarter of a cycle later stored in the inductor. 1 2L11i2 p≈1 2Cv2 p⇒ip=vpp C/L 11 (12) 14 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.2.3 Metal disk placed on top of coil shown in Fig. 10.2.2. Thus, the peak current in the coil is i1= 2,000A. We know both the capacitance and the inductance, so we can also determine the frequency with which the current oscillates. Evaluation gives ω= 20×103s−1(f= 3kHz). The Hﬁeld oscillates with this frequency and has an amplitude given by evaluating (8). We ﬁnd that the peak ﬁeld intensity is H1= 2.3×105A/m so that the peak ﬂux density is 0.3 T (3000 gauss). Now suppose that a conducting disk is placed just above the driver coil as shown in Fig. 10.2.3. What is the current induced in the disk? Choose a contour that encloses a surface Sawhich links the upward-directed magnetic ﬂux generated at the center of the driver coil. With Edeﬁned as an average azimuthally directed electric ﬁeld in the disk, Faraday’s law applied to the contour bounding the surface Sagives 2πaE φ=−d dtZ SaB·da≈ −d dt(µoH1πa2) (13) The average current density circulating in the disk is given by Ohm’s law. Jφ=σEφ=−σµoa 2dH1 dt(14) If one were to replace the disk with his hand, what current density would he feel? To determine the peak current, the derivative is replaced by ωH1. For the hand, σ≈1 S/m and (14) gives 20 mA/cm2. This is more than enough to provide a “shock.” The conductivity of an aluminum disk is much larger, namely 3 .5×107S/m. According to (14), the current density should be 35 million times larger than that in a human hand. However, we need to remind ourselves that in using Amp` ere’s law to determine the driving ﬁeld, we have ignored contributions due to the induced current in the disk. Amp` ere’s integral law can also be used to approximate the ﬁeld induced by the current in the disk. Applied to a contour that loops around the current circulating in the disk rather than in the driving coil, (2) requires that Hind≈i2 2πa≈∆aJφ 2πa(15) Here, the cross-sectional area of the disk through which the current circulates is approximated by the product of the disk thickness ∆ and the average radius a. It follows from (14) and (15) that the induced ﬁeld gets to be on the order of the imposed ﬁeld when Hind H1≈∆σµoa 4π1 |H1|¯¯dH1 dt¯¯≈τm 4πω (16) Sec. 10.2 Nature of MQS Electric Fields 15 where τm≡µoσ∆a (17) Note that τmtakes the form of (7), where l1= ∆ and l2=a. For an aluminum disk of thickness ∆ = 2 mm, a= 7 cm, τm= 6 ms, so ωτm/4π≈10, and the ﬁeld associated with the induced current is comparable to that imposed by the driving coil.1The surface of the disk is therefore one where n·B≈0. The lines of magnetic ﬂux density passing upward through the center of the driving coil are trapped between the driver coil and the disk as they turn radially outward. These lines are sketched in Fig. 10.2.4. In the terminology introduced with Example 9.7.4, the disk is the secondary of a transformer. In fact, τmis the time constant L22/Rof the secondary, where L22 andRare the inductance and resistance of a circuit representing the disk. Indeed, the condition for ideal transformer operation, (9.7.26), is equivalent to having ωτm/4π/greatermuch 1. The windings in power transformers are subject to the forces we now demonstrate. If an aluminum disk is placed on the coil and the switch closed, a number of applications emerge. First, there is a bang, correctly suggesting that the disk can be used as an acoustic transducer. Typical applications are to deep-sea acoustic sounding. The force density F(N/m3) responsible for this sound follows from the Lorentz law (Sec. 11.9) F=J×µoH (18) Note that regardless of the polarity of the driving current, and hence of the average H, this force density acts upward. It is a force of repulsion. With the current distri- bution in the disk represented by a surface current density K, and Btaken as one half its average value (the factor of 1/2 will be explained in Example 11.9.3), the total upward force on the disk is f=Z VJ×BdV≈1 2KB(πa2)iz (19) By Amp` ere’s law, the surface current Kin the disk is equal to the ﬁeld in the region between the disk and the driver, and hence essentially equal to the average H. Thus, with an additional factor of1 2to account for time averaging the sinusoidally varying drive, (19) becomes f≈fo≡1 4µoH2(πa2) (20) In evaluating this expression, the value of Hadjacent to the disk with the disk resting on the coil is required. As suggested by Fig. 10.2.4, this ﬁeld intensity is larger than that given by (8). Suppose that the ﬁeld is intensiﬁed in the gap between coil and plate by a factor of about 2 so that H/similarequal5×105A. Then, evaluation of (20) gives 103Nor more than 1000 times the force of gravity on an 80 galuminum disk. How high would the disk ﬂy? To get a rough idea, it is helpful to know that the driver current decays in several cycles. Thus, the average driving force is essentially an impulse, perhaps as pictured in Fig. 10.2.5 having the amplitude of (20) and a duration T= 1 ms. With the aerodynamic drag ignored, Newton’s law requires that MdV dt=foTuo(t) (21) 1As we shall see in the next sections, because the calculation is not self-consistent, the inequality ωτm/greatermuch1 indicates that the induced ﬁeld is comparable to and not in excess of the one imposed. 16 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.2.4 Currents induced in the metal disk tend to induce a ﬁeld that bucks out that imposed by the driving coil. These currents result in a force on the disk that tends to propel it upward. Fig. 10.2.5 Because the magnetic force on the disk is always positive and lasts for a time T shorter than the time it takes the disk to leave the vicinity of the coil, it is represented by an impulse of magnitude foT. where M= 0.08kgis the disk mass, Vis its velocity, and uo(t) is the unit impulse. Integration of this expression from t= 0−(when the velocity V= 0) to t= 0+gives MV(O+) =foT (22) For the numbers we have developed, this initial velocity is about 10 m/s or about 20 miles/hr. Perhaps of more interest is the height hto which the disk would be expected to travel. If we require that the initial kinetic energy1 2MV2be equal to the ﬁnal potential energy Mgh ( g= 9.8 m/s2), this height is1 2V2/g/similarequal5 m. The voltage and capacitance used here for illustration are modest. Even so, if the disk is thin and malleable, it is easily deformed by the ﬁeld. Metal forming and transport are natural applications of this phenomenon. 10.3 DIFFUSION OF AXIAL MAGNETIC FIELDS THROUGH THIN CONDUCTORS This and the next section are concerned with the inﬂuence of thin-sheet conductors of ﬁnite conductivity on distributions of magnetic ﬁeld. The demonstration of the previous section is typical of physical situations of interest. By virtue of Faraday’s law, an applied ﬁeld induces currents in the conducting sheet. Through Amp` ere’s law, these in turn result in an induced ﬁeld tending to buck out the imposed ﬁeld. The resulting ﬁeld has a time dependence reﬂecting not only that of the applied ﬁeld but the conductivity and dimensions of the conductor as well. This is the subject of the next two sections. A class of conﬁgurations with remarkably simple ﬁelds involves one or more sheet conductors in the shape of cylinders of inﬁnite length. As illustrated in Fig. Sec. 10.3 Axial Magnetic Fields 17 Fig. 10.3.1 A thin shell having conductivity σand thickness ∆ has the shape of a cylinder of arbitrary cross-section. The surface current density K(t) circulates in the shell in a direction perpendicular to the magnetic ﬁeld, which is parallel to the cylinder axis. 10.3.1, these are uniform in the zdirection but have an arbitrary cross-sectional geometry. In this section, the ﬁelds are zdirected and the currents circulate around thezaxis through the thin sheet. Fields and currents are pictured as independent ofz. The current density Jis divergence free. If we picture the current density as ﬂowing in planes perpendicular to the zaxis, and as essentially uniform over the thickness ∆ of the sheet, then the surface current density must be independent of the azimuthal position in the sheet. K=K(t) (1) Amp` ere’s continuity condition, (9.5.3), requires that the adjacent axial ﬁelds are related to this surface current density by −Ha z+Hb z=K (2) In a system with a single cylinder, with a given circulating surface current density K and insulating materials of uniform properties both outside (a) and inside (b), a uniform axial ﬁeld inside and no ﬁeld outside is the exact solution to Amp` ere’s law and the ﬂux continuity condition. (We saw this in Demonstration 8.2.1 and in Example 8.4.2. for a solenoid of circular cross-section.) In a system consisting of nested cylinders, each having an arbitrary cross-sectional geometry and each carrying its own surface current density, the magnetic ﬁelds between cylinders would be uniform. Then (2) would relate the uniform ﬁelds to either side of any given sheet. In general, Kis not known. To relate it to the axial ﬁeld, we must introduce the laws of Ohm and Faraday. The fact that Kis uniform makes it possible to exploit the integral form of the latter law, applied to a contour Cthat circulates through the cylinder.I CE·ds=−d dtZ SB·da (3) 18 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 To replace Ein this expression, we multiply J=σEby the thickness ∆ to relate the surface current density to E, the magnitude of Einside the sheet. K≡∆J= ∆σE⇒E=K ∆σ(4) If ∆ and σare uniform, then E(like K), is the same everywhere along the sheet. However, either the thickness or the conductivity could be functions of azimuthal position. If σand ∆ are given, the integral on the left in (3) can be taken, since K is constant. With sdenoting the distance along the contour C, (3) and (4) become KI Cds ∆(s)σ(s)=−d dtZ SB·da (5) Of most interest is the case where the thickness and conductivity are uniform and (5) becomes KP ∆σ=−d dtZ SB·da (6) with Pdenoting the peripheral length of the cylinder. The following are examples based on this model. Example 10.3.1. Diﬀusion of Axial Field into a Circular Tube The conducting sheet shown in Fig. 10.3.2 has the shape of a long pipe with a wall of uniform thickness and conductivity. There is a uniform magnetic ﬁeld H=izHo(t) in the space outside the tube, perhaps imposed by means of a coaxial solenoid. What current density circulates in the conductor and what is the axial ﬁeld intensity Hi inside? Representing Ohm’s law and Faraday’s law of induction, (6) becomes K ∆σ2πa=−d dt(µoπa2Hi) (7) Amp` ere’s law, represented by the continuity condition, (2), requires that K=−Ho+Hi (8) In these two expressions, Hois a given driving ﬁeld, so they can be combined into a single diﬀerential equation for either KorHi. Choosing the latter, we obtain dHi dt+Hi τm=Ho τm(9) where τm=1 2µoσ∆a (10) This expression pertains regardless of the driving ﬁeld. In particular, suppose that before t= 0, the ﬁelds and surface current are zero, and that when t= 0, the outside Hois suddenly turned on. The appropriate solution to (9) is the combination Sec. 10.3 Axial Magnetic Fields 19 Fig. 10.3.2 Circular cylindrical conducting shell with external axial ﬁeld intensity Ho(t) imposed. The response to a step in applied ﬁeld is a current density that initially shields the ﬁeld from the inner region. As this current decays, the ﬁeld penetrates into the interior and is ﬁnally uniform throughout. of the particular solution Hi=Hoand the homogeneous solution exp( −t/τm) that satisﬁes the initial condition. Hi=Ho(1−e−t/τm) (11) It follows from (8) that the associated surface current density is K=−Hoe−t/τm(12) At a given instant, the axial ﬁeld has the radial distribution shown in Fig. 10.3.2b. Outside, the ﬁeld is imposed to be equal to Ho, while inside it is at ﬁrst zero but then ﬁlls in with an exponential dependence on time. After a time that is long compared to τm, the ﬁeld is uniform throughout. Implied by the discontinuity in ﬁeld intensity at r=ais a surface current density that initially terminates the outside ﬁeld. When t= 0, K =−Ho, and this results in a ﬁeld that bucks out the ﬁeld imposed on the inside region. The decay of this current, expressed by (12), accounts for the penetration of the ﬁeld into the interior region. This example illustrates what one means by “perfect conductor approxima- tion.” A perfect conductor would shield out the magnetic ﬁeld forever. A physical conductor shields it out for times t/lessmuchτm. Thus, in the MQS approximation, a conductor can be treated as perfect for times that are short compared with the characteristic time τm. The electric ﬁeld Eφ≡Eis given by applying (3) to a contour having an arbitrary radius r. 2πrE =−d dt(µoHiπr2)⇒E=−µor 2dHi dtr < a (13) 2πrE =−d dt(µoHiπa2)−d dt[µoHoπ(r2−a2)]⇒ E=−µoa 2· a rdHi dt+¡r a−a r¢dHo dt¸ r > a(14) 20 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Atr=a, this particular solution matches that already found using the same integral law in the conductor. In this simple case, it is not necessary to match boundary con- ditions by superimposing a homogeneous solution taking the form of a conservative ﬁeld. We consider next an example where the electric ﬁeld is not simply the partic- ular solution. Example 10.3.2. Diﬀusion into Tube of Nonuniform Conductivity Once again, consider the circular cylindrical shell of Fig. 10.3.2 subject to an im- posed axial ﬁeld Ho(t). However, now the conductivity is a function of azimuthal position. σ=σo 1 +αcosφ(15) The integral in (5), resulting from Faraday’s law, becomes KI Cds ∆(s)σ(s)=K ∆σoZ2π 0(1 +αcosφ)adφ=2πa ∆σoK (16) and hence 2πa ∆σoK=−d dt(πa2µoHi) (17) Amp` ere’s continuity condition, (2), once again becomes K=−Ho+Hi (18) Thus, Hiis determined by the same expressions as in the previous example, except that σis replaced by σo. The surface current response to a step in imposed ﬁeld is again the exponential of (12). It is the electric ﬁeld distribution that is changed. Using (15), (4) gives E=K ∆σo(1 +αcosφ) (19) for the electric ﬁeld inside the conductor. The Eﬁeld in the adjacent free space regions is found using the familiar approach of Sec. 10.1. The particular solution is the same as for the uniformly conducting shell, (13) and (14). To this we add a homogeneous solution Eh=−∇Φ such that the sum matches the tangential ﬁeld given by (19) at r=a. The φ-independent part of (19) is already matched by the particular solution, and so the boundary condition on the homogeneous part requires that −1 a∂Φ ∂φ(r=a) =Kα ∆σocosφ⇒Φ(r=a) =−Kαa ∆σosinφ (20) Solutions to Laplace’s equation that vary as sin( φ) match this condition. Outside, the appropriate rdependence is 1 /rwhile inside it is r. With the coeﬃcients of these potentials adjusted to match the boundary condition given by (20), it follows that the electric ﬁeld outside and inside the shell is E=8 < :−µor 2dHi dtiφ− ∇Φi r < a −µoa 2· a rdHi dt+¡r a−a r¢dHo dt¸ iφ− ∇Φoa < r(21) Sec. 10.4 Transverse Magnetic Fields 21 Fig. 10.3.3 Electric ﬁeld induced in regions inside and outside shell (having conductivity that varies with azimuthal position) portrayed as the sum of a particular rotational and homogeneous conservative solu- tion. Conductivity is low on the right and high on the left, α= 0.5. where Φi=−Kα ∆σorsinφ (22) Φo=−Kαa2 ∆σosinφ r(23) These expressions can be evaluated using (11) and (12) for HiandKfor the electric ﬁeld associated with a step in applied ﬁeld. It follows that E, like the surface current and the induced H, decays exponentially with the time constant of (10). At a given instant, the distribution of Eis as illustrated in Fig. 10.3.3. The total solution is the sum of the particular rotational and homogeneous conservative parts. The degree to which the latter inﬂuences the total ﬁeld depends on α, which reﬂects the inhomogeneity in conductivity. For positive α, the conductivity is low on the right (when φ= 0) and high on the left in Fig. 10.3.3. In accordance with (7.2.8), positive unpaired charge is induced in the transition region where the current ﬂows from high to low conductivity, and negative charge is induced in the transition region from low to high conductivity. The ﬁeld of the homogeneous solution shown in the ﬁgure originates and terminates on the induced charges. We shall return to models based on conducting cylindrical shells in axial ﬁelds. Systems of conducting shells can be used to represent the nonuniform ﬂow of current in thick conductors. The model will also be found useful in determining the rate of induction heating for cylindrical objects. 22 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.4.1 Cross-section of circular cylindrical conducting shell having its axis perpendicular to the magnetic ﬁeld. 10.4 DIFFUSION OF TRANSVERSE MAGNETIC FIELDS THROUGH THIN CONDUCTORS In this section we study magnetic induction of currents in thin conducting shells by ﬁelds transverse to the shells. In Sec. 10.3, the magnetic ﬁelds were automatically tangential to the conductor surfaces, so we did not have the opportunity to explore the limitations of the boundary condition n·B= 0 used to describe a “perfect conductor.” In this section the imposed ﬁelds generally have components normal to the conducting surface. The steps we now follow can be applied to many diﬀerent geometries. We speciﬁcally consider the circular cylindrical shell shown in cross-section in Fig. 10.4.1. It has a length in the zdirection that is very large compared to its ra- diusa. Its conductivity is σ, and it has a thickness ∆ that is much less than its radius a. The regions outside and inside are speciﬁed by (a) and (b), respectively. The ﬁelds to be described are directed in planes perpendicular to the zaxis and do not depend on z. The shell currents are zdirected. A current that is directed in the + zdirection at one location on the shell is returned in the −zdirection at another. The closure for this current circulation can be imagined to be provided by perfectly conducting endplates, or by a distortion of the current paths from the z direction near the cylinder ends (end eﬀect). The shell is assumed to have essentially the same permeability as free space. It therefore has no tendency to guide the magnetic ﬂux density. Integration of the magnetic ﬂux continuity condition over an incremental volume enclosing a section of the shell shows that the normal component of Bis continuous through the shell. n·(Ba−Bb) = 0⇒Ba r=Bb r (1) Ohm’s law relates the axial current density to the axial electric ﬁeld, Jz=σEz. This density is presumed to be essentially uniformly distributed over the radial cross-section of the shell. Multiplication of both sides of this expression by the thickness ∆ of the shell gives an expression for the surface current density in the shell. Kz≡∆Jz= ∆σEz (2) Faraday’s law is a vector equation. Of the three components, the radial one is dominant in describing how the time-varying magnetic ﬁeld induces electric ﬁelds, and hence currents, tangential to the shell. In writing this component, we assume that the ﬁelds are independent of z. 1 a∂Ez ∂φ=−∂Br ∂t(3) Sec. 10.4 Transverse Magnetic Fields 23 Fig. 10.4.2 Circular cylindrical conducting shell ﬁlled by insulating material of permeability µand surrounded by free space. A magnetic ﬁeldHo(t) that is uniform at inﬁnity is imposed transverse to the cylin- der axis. Amp` ere’s continuity condition makes it possible to express the surface current density in terms of the tangential ﬁelds to either side of the shell. Kz=Ha φ−Hb φ (4) These last three expressions are now combined to obtain the desired continuity condition. 1 ∆σa∂ ∂φ(Ha φ−Hb φ) =−∂Br ∂t (5) Thus, the description of the shell is encapsulated in the two continuity conditions, (1) and (5). The thin-shell model will now be used to place in perspective the idealized boundary condition of perfect conductivity. In the following example, the conductor is subjected to a ﬁeld that is suddenly turned on. The ﬁeld evolution with time places in review the perfect conductivity mode of MQS systems in Chap. 8 and the magnetization phenomena of Chap. 9. Just after the ﬁeld is turned on, the shell acts like the perfect conductors of Chap. 8. As time goes on, the shell currents decay to zero and only the magnetization of Chap. 9 persists. Example 10.4.1. Diﬀusion of Transverse Field into Circular Cylindrical Conducting Shell with a Permeable Core A permeable circular cylindrical core having radius ais shown in Fig. 10.4.2. It is surrounded by a thin conducting shell, having thickness ∆ and conductivity σ. A uniform time-varying magnetic ﬁeld intensity Ho(t) is imposed transverse to the axis of the shell and core. The conﬁguration is long enough in the axial direction to justify representing the ﬁelds as independent of the axial coordinate z. Reﬂecting the fact that the region outside ( o) is free space while that inside (i) is the material of linear permeability are the constitutive laws Bo=µoHo;Bi=µHi(6) For the two-dimensional ﬁelds in the r−φplane, where the sheet current is in the zdirection, the scalar potential provides a convenient description of the ﬁeld. H=−∇Ψ (7) 24 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 We begin by recognizing the form taken by Ψ far from the cylinder. Ψ =−Horcosφ (8) Note that substitution of this relation into (7) indeed gives the uniform imposed ﬁeld. Given the φdependence of (8), we assume solutions of the form Ψo=−Horcosφ+Acosφ r(9) Ψi=Crcosφ where AandCare coeﬃcients to be determined by the continuity conditions. In preparation for the evaluation of these conditions, the assumed solutions are substi- tuted into (7) to give the ﬂux densities Bo=µo¡ Ho+A r2¢ cosφir−µo¡ Ho−A r2¢ sinφiφ (10) Bi=−µC(cosir−sinφiφ) Should we expect that these functions can be used to satisfy the continuity conditions at r=agiven by (1) and (5) at every azimuthal position φ? The inside and outside radial ﬁelds have the same φdependence, so we are assured of being able to adjust the two coeﬃcients to satisfy the ﬂux continuity condition. Moreover, in evaluating (5), the φderivative of Hφhas the same φdependence as Br. Thus, satisfying the continuity conditions is assured. The ﬁrst of two relations between the coeﬃcients and Hofollows from substi- tuting (10) into (1). µo¡ Ho+A a2¢ =−µC (11) The second results from a similar substitution into (5). −1 ∆σa¡ Ho−A a2¢ −C ∆σa=−µoµ dHo dt+1 a2dA dt¶ (12) With Celiminated from this latter equation by means of (11), we obtain an ordinary diﬀerential equation for A(t). dA dt+A τm=−a2dHo dt+Hoa µo∆σ¡ 1−µo µ¢ (13) The time constant τmtakes the form of (10.2.7). τm=µoσ∆a¡µ µ+µo¢ (14) In (13), the time dependence of the imposed ﬁeld is arbitrary. The form of this expression is the same as that of (7.9.28), so techniques for dealing with initial conditions and for determining the sinusoidal steady state response introduced there are directly applicable here. Sec. 10.4 Transverse Magnetic Fields 25 Response to a Step in Applied Field. Suppose there is no ﬁeld inside or outside the conducting shell before t= 0 and that Hois a step function of magnitude Hmturned on when t= 0. With Da coeﬃcient determined by the initial condition, the solution to (13) is the sum of a particular and a homogeneous solution. A=Hma2(µ−µo) (µ+µo)+De−t/τm(15) Integration of (13) from t= 0−tot= 0+shows that A(0) = −Hma2, so that Dis evaluated and (15) becomes A=Hma2· (µ−µo) (µ+µo)(1−e−t/τm)−e−t/τm¸ (16) This expression makes it possible to evaluate Cusing (11). Finally, these coeﬃcients are substituted into (9) to give the potential outside and inside the shell. Ψo=−Hma½ r a−a r· (µ−µo) (µ+µo)(1−e−t/τm)−e−t/τm¸¾ cosφ (17) Ψi=−Hmar a2µo (µ+µo)(1−e−t/τm) cosφ (18) The ﬁeld evolution represented by these expressions is shown in Fig. 10.4.3, where lines of Bare portrayed. When the transverse ﬁeld is suddenly turned on, currents circulate in the shell in such a direction as to induce a ﬁeld that bucks out the one imposed. For an applied ﬁeld that is positive, this requires that the surface current be in the −zdirection on the right and returned in the + zdirection on the left. This surface current density can be analytically expressed ﬁrst by using (10) to evaluate Amp` ere’s continuity condition Kz=Ho φ−Hi φ=· −Hoµ 1−µo µ¶ +A a2µ 1 +µo µ¶¸ sinφ (19) and then by using (15). Kz=−2Hme−t/τmsinφ (20) With the decay of Kz, the external ﬁeld goes from that for a perfect conductor (where n·B= 0) to the ﬁeld that would have been found if there were no conducting shell. The magnetizable core tends to draw this ﬁeld into the cylinder. The coeﬃcient Arepresents the amplitude of a two-dimensional dipole that has a ﬁeld equivalent to that of the shell current. Just after the ﬁeld is applied, Ais negative and hence the equivalent dipole moment is directed opposite to the imposed ﬁeld. This results in a ﬁeld that is diverted around the shell. With the passage of time, this dipole moment can switch sign. This sign reversal occurs only if µ > µ o, making it clear that it is due to the magnetization of the core. In the absence of the core, the ﬁnal ﬁeld is uniform. Under what conditions can the shell be regarded as perfectly conducting? The answer involves not only σbut also the time scale and the size, and to some extent, 26 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.4.3 When t= 0, a magnetic ﬁeld that is uniform at inﬁnity is suddenly imposed on the circular cylindrical conducting shell. The cylinder is ﬁlled by an insulating material of permeability µ= 200 µo. When t/τ= 0, an instant after the ﬁeld is applied, the surface currents completely shield the ﬁeld from the central region. As time goes on, these currents decay, until ﬁnally the ﬁeld is no longer inﬂuenced by the conducting shell. The ﬁnal ﬁeld is essentially perpendicular to the highly permeable core. In the absence of this core, the ﬁnal ﬁeld would be uniform. the permeability. For our step response, the shell shields out the ﬁeld for times that are short compared to τm, as given by (14). Demonstration 10.4.1. Currents Induced in a Conducting Shell The apparatus of Demonstration 10.2.1 can be used to make evident the shell currents predicted in the previous example. A cylinder of aluminum foil is placed on the driver coil, as shown in Fig. 10.4.4. With the discharge of the capacitor through the coil, the shell is subjected to an abruptly applied ﬁeld. By contrast with the step function assumed in the example, this ﬁeld oscillates and decays in a few cycles. However, the reversal of the ﬁeld results in a reversal in the induced shell current, so regardless of the time dependence of the driving ﬁeld, the force density J×Bis in the same direction. Sec. 10.5 Magnetic Diﬀusion Laws 27 Fig. 10.4.4 In an experiment giving evidence of the currents induced when a ﬁeld is suddenly applied transverse to a conducting cylinder, an aluminum foil cylinder, subjected to the ﬁeld produced by the experi- ment of Fig. 10.2.2, is crushed. The force associated with the induced current is inward. If the applied ﬁeld were truly uniform, the shell would then be “squashed” inward from the right and left by the ﬁeld. Because the ﬁeld is not really uniform, the cylinder of foil is observed to be compressed inward more at the bottom than at the top, as suggested by the force vectors drawn in Fig. 10.4.4. Remember that the postulated currents require paths at the ends of the cylinder through which they can circulate. In a roll of aluminum foil, these return paths are through the shell walls in those end regions that extend beyond the region of the applied ﬁeld. The derivation of the continuity conditions for a circular cylindrical shell fol- lows a format that is applicable to other geometries. Examples are a planar sheet and a spherical shell. 10.5 MAGNETIC DIFFUSION LAWS The self-consistent evolution of the magnetic ﬁeld intensity Hwith its source J induced in Ohmic materials of ﬁnite conductivity is familiar from the previous two sections. In the models so far considered, the induced currents were in thin conducting shells. Thus, in the processes of magnetic relaxation described in these sections, the currents were conﬁned to thin regions that could be represented by dynamic continuity conditions. In this and the next two sections, the conductor extends throughout at least part of a volume of interest. Like H, the current density in Amp` ere’s law ∇ ×H=J (1) is an unknown function. For an Ohmic material, it is proportional to the local electric ﬁeld intensity. J=σE (2) In turn, Eis induced in accordance with Faraday’s law ∇ ×E=−∂µH ∂t(3) The conductor is presumed to have uniform conductivity σand permeability µ. For linear magnetization, the magnetic ﬂux continuity law is ∇ ·µH= 0 (4) 28 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 In the MQS approximation the current density Jis also solenoidal, as can be seen by taking the divergence of Amp` ere’s law. ∇ ·J= 0 (5) In the previous two sections, we combined the continuity conditions implied by (1) and (4) with the other laws to obtain dynamic continuity conditions representing thin conducting sheets. The regions between sheets were insulating, and so the ﬁeld distributions in these regions were determined by solving Laplace’s equation. Here we combine the diﬀerential laws to obtain a new diﬀerential equation that takes on the role of Laplace’s equation in determining the distribution of magnetic ﬁeld intensity. If we solve Ohm’s law, (2), for Eand substitute for Ein Faraday’s law, we have in one statement the link between magnetic induction and induced current density. ∇ ×µJ σ¶ =−∂µH ∂t(6) The current density is eliminated from this expression by using Amp` ere’s law, (1). The result is an expression of Halone. ∇ ×µ∇ ×H σ¶ =−∂µH ∂t(7) This expression assumes a somewhat more familiar appearance when σandµare constants, so that they can be taken outside the operations. Further, it follows from (4) that His solenoidal so the use of a vector identity2turns (7) into 1 µσ∇2H=∂H ∂t (8) At each point in a material having uniform conductivity and permeability, the magnetic ﬁeld intensity satisﬁes this vector form of the diﬀusion equation . The distribution of current density implied by the Hfound by solving this equation with appropriate boundary conditions follows from Amp` ere’s law, (1). Physical Interpretation. With the understanding that HandJare solenoidal, the derivation of (8) identiﬁes the feedback between source and ﬁeld that underlies the magnetic diﬀusion process. The eﬀect of the (time-varying) ﬁeld on the source embodied in the combined laws of Faraday and Ohm, (6), is perhaps best appre- ciated by integrating (6) over any ﬁxed open surface Senclosed by a contour C. By Stokes’ theorem, the integration of the curl over the surface transforms into an integration around the enclosing contour. Thus, (6) implies that −I CJ σ·ds=d dtZ SµH·da (9) 2∇ × ∇ × H=∇(∇ ·H)− ∇2H Sec. 10.5 Magnetic Diﬀusion Laws 29 Fig. 10.5.1 Conﬁgurations in which cylindrically shaped conductors having axes parallel to the magnetic ﬁeld have currents transverse to the ﬁeld in x−y planes. and requires that the electromotive force around anyclosed path must be equal to the time rate of change of the enclosed magnetic ﬂux. Numerical approaches to solving magnetic diﬀusion problems may in fact approximate a system by a ﬁnite number of circuits, each representing a current tube with its own resistance and ﬂux linkage. To represent the return eﬀect of the current on H, the diﬀusion equation also incorporates Amp` ere’s law, (1). The relaxation of axial ﬁelds through thin shells, developed in Sec. 10.3, is an example where the geometry of the conductor and the symmetry make the current tubes described by (9) readily discernible. The diﬀusion of an axial magnetic ﬁeld Hzinto the volume of cylindrically shaped conductors, as shown in Fig. 10.5.1, is a generalization of the class of axial problems described in Sec. 10.3. As the only component of H, Hz(x, y) must satisfy (8). 1 µσ∇2Hz=∂Hz ∂t(10) The current density is then directed transverse to this ﬁeld and given in terms of Hzby Amp` ere’s law. J=ix∂Hz ∂y−iy∂Hz ∂x(11) Thus, the current density circulates in x−yplanes. Methods for solving the diﬀusion equation are natural extensions of those used in previous chapters for dealing with Laplace’s equation. Although we conﬁne ourselves in the next two sections to diﬀusion in one spatial dimension, the thin- shell models give an intuitive impression as to what can be expected as magnetic ﬁelds diﬀuse into solid conductors having a wide range of geometries. Consider the coaxial thin shells shown in Fig. 10.5.2 as a model for a solid cylindrical conductor. Following the approach outlined in Sec. 10.3, suppose that the exterior ﬁeld Hois an imposed function of time. Then the ﬁelds between sheets ( H1 30 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.5.2 Example of an axial ﬁeld conﬁguration composed of coaxial con- ducting shells of inﬁnite axial length. When an exterior ﬁeld Hois applied, currents circulating in the shells tend to shield out the imposed ﬁeld. andH2) and in the central region ( H3) are determined by a system of three ordinary diﬀerential equations having Ho(t) as a drive. Associated with the evolution of these ﬁelds are surface currents in the shells that tend to shield the ﬁeld from the region within. In the limit where the number of shells is inﬁnite, the ﬁeld distribution in a solid conductor could be represented by such coupled thin shells. However, the more practical approach used in the next sections is to solve the diﬀusion equation exactly. The situations considered are in cartesian rather than polar coordinates. 10.6 MAGNETIC DIFFUSION TRANSIENT RESPONSE The self-consistent distribution of current density and magnetic ﬁeld intensity in the volume of a uniformly conducting material is determined from the laws given in Sec. 10.5 and summarized by the magnetic diﬀusion equation (10.5.8). In this section, we illustrate magnetic diﬀusion phenomena by considering the transient that results when a current is abruptly turned on or oﬀ. In contrast to Laplace’s equation, the diﬀusion equation involves a time rate of change, and so it is necessary to deal with the time dependence in much the same way as the space dependence. The diﬀusion process considered in this section is in one spatial dimension, with time as the second “dimension.” Our approach builds on product solutions and the solution of boundary value problems by superposition, as introduced in Chap. 5. The class of conﬁgurations of interest is illustrated in Fig. 10.6.1. Perfectly conducting electrodes are driven along their edges at x=−bby a distributed current source. The uniformly conducting material is sandwiched between these electrodes. The current originating in the source then circulates in the xdirection through the electrode in the y= 0 plane to a point where it passes in the ydirection through the conducting material. It is then returned to the source through the other perfectly conducting plate. Note that this conﬁguration is a special case of Sec. 10.6 Magnetic Diﬀusion Transient 31 Fig. 10.6.1 A block of uniformly conducting material having length band thickness ais sandwiched between perfectly conducting electrodes that are driven along their edges at x=−bby a distributed current source. Current density and ﬁeld intensity in the block are, respectively, yandzdirected, each depending on ( x, t). that shown in Fig. 10.5.1, where the current density is transverse to a magnetic ﬁeld intensity that has only one component, Hz. If this ﬁeld and the associated current density are indeed independent of y, then it follows from (10.5.10) and (10.5.11) that Hzsatisﬁes the one-dimensional diﬀusion equation 1 µσ∂2Hz ∂x2=∂Hz ∂t(1) and the only component of the current density is related to Hzby Amp` ere’s law J=−iy∂Hz ∂x(2) Note that this one-dimensional model correctly requires that the current density, and hence the electric ﬁeld intensity, be normal to the perfectly conducting elec- trodes at y= 0 and y=a. The distributed current source, perfectly conducting sheets and conducting block form a closed path for currents that circulate in x−yplanes. These extend to inﬁnity in the + and −zdirections in the manner of an inﬁnite one-turn solenoid. The ﬁeld outside the outermost of these current paths is therefore taken as being zero. Amp` ere’s continuity condition then requires that at the surface x=−b, where the distributed current source is located, the enclosed magnetic ﬁeld intensity be equal to the imposed surface current density Ks. In the plane x= 0, the situation is similar except that there is no surface current density, and so the magnetic ﬁeld intensity must be zero. Thus, consistent with solving a diﬀerential equation that is second order in x, are the two boundary conditions Hz(−b, t) =Ks(t), H z(0, t) = 0 (3) The equation is ﬁrst order in its time dependence, suggesting that to complete the speciﬁcation of the transient solution, the initial value of Hzmust also be given. Hz(x,0) = Hi(x) (4) 32 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.6.2 Boundary and initial conditions for one-dimensional magnetic diﬀusion pictured in the x−tplane. (a) The total ﬁelds at the ends of the block are constrained to be equal to the driving surface current density and to zero, respectively, while there is one initial condition when t= 0. (b) The transient part of the solution is zero at the boundaries and satisﬁes the initial condition that makes the total solution assume the current value when t= 0. It is helpful to picture the boundary and initial conditions needed to uniquely specify solutions to (2) in the x−tplane, as shown in Fig. 10.6.2a. Here the conducting block can be pictured as extending from x= 0 to x=−b, with the ﬁeld between a function of xthat evolves in the t“direction.” Presumably, the distribution of Hzin the x−tspace is predicted by (1) with the boundary conditions of (3) at x= 0 and x=−band the initial condition of (4) when t= 0. Is the solution for Hz(t) uniquely speciﬁed by (1), the boundary conditions of (3), and the initial condition of (4)? A proof that it is can be made following a line of reasoning suggested by the EQS uniqueness arguments of Sec. 7.8. Suppose that the drive is a step function of time, so that the ﬁnal state is one of uniform steady conduction. Then, the linearity of (1) makes it possible to think of the total ﬁeld as being the superposition of this steady ﬁeld and a transient part. Hz=H∞(x) +Ht(x, t) (5) The steady solution, which presumably prevails as t→ ∞ , satisﬁes (1) with the time derivative set equal to zero, ∂2H∞ ∂x2= 0 (6) while the transient part satisﬁes the complete equation. 1 µσ∂2Ht ∂x2=∂Ht ∂t(7) Because the steady solution satisﬁes the boundary conditions for all time t >0, the boundary conditions satisﬁed by the transient part are homogeneous. Ht(−b, t) = 0; Ht(0, t) = 0 (8) However, the steady solution does not satisfy the initial condition. The transient solution is therefore adjusted so that the total solution does. Ht(x,0) = Hi(x)−H∞(x) (9) Sec. 10.6 Magnetic Diﬀusion Transient 33 The conditions satisﬁed by the transient part of the solution on the boundaries in thex−tspace are pictured in Fig. 10.6.2b. Product Solutions to the One-Dimensional Diﬀusion Equation. The ap- proach now used to ﬁnd the Htthat satisﬁes (7) and the conditions of (8) and (9) is familiar from ﬁnding Cartesian coordinate product solutions to Laplace’s equation in two dimensions in Sec. 5.4. Here the second “dimension” is tand we consider solutions that take the form Ht=X(x)T(t). Substitution into (7) and division by XTgives 1 Xd2X dx2−µσ TdT dt= 0 (10) With the ﬁrst term taken as −k2and the second as k2, it follows that 1 Xd2X dx2=−k2⇒d2X dx2+k2X= 0 (11) and −µσ TdT dt=k2⇒dT dt+k2 µσT= 0 (12) Given the boundary conditions of (8), the appropriate solution to (11) is X= sin kx; k=nπ b(13) where ncan be any integer. Associated with each of these modes is a time depen- dence given by (12) as a decaying exponential with the time constant τn=µσb2 (nπ)2(14) Thus, we are led to a transient part of the solution that is itself a superposition of modes, each satisfying the boundary conditions. Ht=∞X n=1Cnsin¡nπ bx¢ e−t/τn(15) When t= 0, the modes take the form of a Fourier series. Thus, the coeﬃcients Cn can be used to satisfy the initial condition, (9). In the following example, the coeﬃcients are evaluated for speciﬁc initial con- ditions. However, because the “short time” and “long time” ﬁeld and current dis- tributions are known at the outset, much of the dynamics can be anticipated at the outset. For times that are very short compared to the magnetic diﬀusion time µσb2, the conducting block must act as a perfect conductor. In this short time limit, we know from Chap. 8 that the current from the distributed source is conﬁned to the surface at x=−b. Thus, for early times, the distribution represented by the series of (15) tends to be an impulse function of x. After many magnetic diﬀusion 34 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 times, the current reaches a steady state and achieves a distribution that would be predicted in the ﬁrst half of Chap. 7. The following example ﬁlls in the evolution from the ﬁeld of a perfectly conducting system to that for steady conduction. Example 10.6.1. Response to a Step in Current When t= 0, suppose that there are no currents or associated ﬁelds. Then the current source suddenly becomes the constant Kp. The solution to (6) that is zero atx= 0 and is Kpatx=−bis H∞=−Kpx b(16) This is the ﬁeld associated with a constant current density Kp/bthat is uniformly distributed over the cross-section of the block. Because there is no initial magnetic ﬁeld, it follows from (9) that the initial transient part of the ﬁeld must cancel the steady part. Ht(x,0) = Kpx b(17) This must be the distribution of Htgiven by (15) when t= 0. Kpx b=∞X n=1Cnsin¡nπ bx¢ (18) Following the procedure familiar from Sec. 5.5, the coeﬃcients Cnare now evaluated by multiplying both sides of this expression by sin( mπ/b ), multiplying by dx, and integrating from x=−btox= 0. Z0 −bKpx bsin¡mπ bx¢ dx=∞X n=1CnZ0 −bsin¡nπ bx¢ sin¡mπ bx¢ dx (19) From the series on the right, only the term m=nis not zero. Carrying out the integration on the left3then gives an expression that can be solved for Cm. Replacing m→nthen gives Cn=−2Kp(−1)n nπ(20) Finally, (16) and (15) [the latter evaluated using (20)] are superimposed as required by (5) to give the desired description of how the ﬁeld evolves as a function of space and time. Hz=−Kpx b−∞X n=12Kp(−1)n nπsin¡nπx b¢ e−t/τn(21) The distribution of current density follows from this expression substituted into Amp` ere’s law, (2). Jy=Kp b+∞X n=12Kp(−1)n bcos¡nπx b¢ e−t/τn(22) 3R sin(u)udu= sin( u)−ucos(u) Sec. 10.7 Skin Eﬀect 35 Fig. 10.6.3 (a) Distribution of Hzin the conducting block of Fig. 10.6.1 in response to applying a step in current with no initial ﬁeld. In terms of time normalized to the magnetic diﬀusion time based on the length b, the ﬁeld diﬀuses into the block, ﬁnally assuming the linear distribution expected for steady conduction. (b) Distribution of Jywith normalized time as a parameter. The initial distribution is an impulse (a surface current density) at x=−b, while the ﬁnal distribution is uniform. These expressions are pictured in Fig. 10.6.3. Note that the higher the order of a term, the more rapid its exponential decay with time. As a result, the most terms in the series are needed when t= 0+. These are needed to make the initial magnetic ﬁeld intensity zero and the initial current density an impulse at x=−b. Because the lowest mode in the transient part of either HzorJyhas the longest time constant, the long-time response is dominated by the steady response and the ﬁrst term in the series. Of course, with the decay of the transient part, the ﬁeld approaches a linear xdependence while the current density assumes the uniform distribution expected for a steady current. 10.7 SKIN EFFECT If the surface current source driving the conducting block of Fig. 10.6.1 is a sinu- soidal function of time Ks(t) =ReˆKsejωt(1) the current density tends to circulate through the block in the neighborhood of the surface adjacent to the source. This tendency for the sinusoidal steady state current to return to the source through the thin zone or skin region nearest to the source gives another view of magnetic diﬀusion. To illustrate skin eﬀect in speciﬁc terms we return to the one-dimensional diﬀusion conﬁguration of Sec. 10.6, Fig. 10.6.1. Once again, the distributions of Hz 36 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 andJyare governed by the one-dimensional diﬀusion equation and Amp` ere’s law, (10.6.1) and (10.6.2). The diﬀusion equation is linear and has coeﬃcients that are independent of time. We can expect a sinusoidal steady state response having the same frequency as the drive, (1). The solution to the diﬀusion equation is therefore taken as having a product form, but with the time dependence stipulated at the outset. Hz=ReˆHz(x)ejωt(2) At a given location x, the coeﬃcient of the exponential is a complex number spec- ifying the magnitude and phase of the ﬁeld. Substitution of (2) into the diﬀusion equation, (10.6.1), shows that the com- plex amplitude has an xdependence governed by d2ˆHz dx2−γ2ˆHz= 0 (3) where γ2≡jωµσ . Solutions to (3) are simply exp( ∓γx). However, γis complex. If we note that√j= (1 + j)/√ 2, then it follows that γ=p jωµσ = (1 + j)rωµσ 2(4) In terms of the skin depth δ, deﬁned by δ≡r2 ωµσ(5) One can also write (4) as γ=(1 +j) δ(6) With C+andC−arbitrary coeﬃcients, solutions to (3) are therefore ˆHz=C+e−(1+j)x δ+C−e(1+j)x δ (7) Before considering a detailed example where these coeﬃcients are evaluated using the boundary conditions, consider the x−tdependence of the ﬁeld represented by the ﬁrst solution in (7). Substitution into (2) gives Hz=Re£ C1e−x δej(ωt−x δ)¤ (8) making it clear that the ﬁeld magnitude is an exponentially decaying function of x. Within the envelope with the decay length δshown in Fig. 10.7.1, the ﬁeld propa- gates in the xdirection. That is, points of constant phase on the ﬁeld distribution have ωt−x/δ= constant and hence move in the xdirection with the velocity ωδ. Sec. 10.7 Skin Eﬀect 37 Fig. 10.7.1 Magnetic diﬀusion wave in the sinusoidal steady state, showing envelope with decay length δand instantaneous ﬁeld at two diﬀerent times. The point of zero phase propagates with the velocity ωδ. Fig. 10.7.2 (a) One-dimensional magnetic diﬀusion in the sinusoidal steady state in the same conﬁguration as considered in Sec. 10.6. (b) Distribution of the magnitude of Hzin the conducting block of (a) as a function of the skin depth. Decreasing the skin depth is equivalent to raising the frequency. Although the phase propagation signiﬁes that at a given instant, the ﬁeld (and cur- rent density) are positive in one region while negative in another, the propagation is diﬃcult to discern because the decay is very rapid. The second solution in (7) represents a similar diﬀusion wave, but decaying and propagating in the −xrather than the + xdirection. The following illustrates how the two diﬀusion waves combine to satisfy boundary conditions. Example 10.7.1. Diﬀusion into a Conductor of Finite Thickness We consider once again the ﬁeld distribution in a conducting material sandwiched between perfectly conducting plates, as shown in either Fig. 10.7.2 or Fig. 10.6.1. The surface current density of the drive is given by (1) and it is assumed that any transient reﬂecting the initial conditions has died out. How does the frequency dependence of the ﬁeld distribution in the conducting block reﬂect the magnetic diﬀusion process? 38 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Boundary conditions on Hzare the same as in Sec. 10.6, Hz(−b, t) =Ks(t) andHz(0, t) = 0. These are satisﬁed by adjusting the complex amplitude so that ˆHz(−b) =ˆKs; ˆHz(0) = 0 (9) It follows from (7) that the second of these is satisﬁed if C+=−C−. The ﬁrst condition then serves to evaluate C+and hence C−, so that ˆHz=ˆKs¡ e−(1+j)x δ−e(1+j)x δ¢ ¡ e(1+j)b δ−e−(1+j)b δ¢ (10) This expression represents the superposition of ﬁelds propagating and decaying in the ±xdirections, respectively. Evaluated at a given location x, it is a complex number. In accordance with (2), Hzis the real part of this number multiplied by exp(jωt). The magnitude of Hzis the magnitude of (10), and is shown with the skin depth as a parameter by Fig. 10.7.2. Consider the ﬁeld distribution in two limits. First, suppose that the skin depth is very large compared to the thickness bof the conducting block. This might be the limit in which the frequency is made very low compared to the reciprocal magnetic diﬀusion time based on the conductor thickness. δ/greatermuchb⇒2 ωµσ/greatermuchb2⇒2 µσb2/greatermuchω (11) In this limit, the arguments of the exponentials in (10) are small. Using the approx- imation exp( u)≈1 +u, (10) becomes ˆHz→ˆKs£ 1−(1 +j)x δ¤ −£ 1 + (1 + j)x δ¤ £ 1 + (1 + j)b δ¤ −£ 1−(1 +j)b δ¤=−ˆKsx b(12) Substitution of this complex amplitude into (2) gives the space-time dependence. Hz→−x bReˆKsejωt(13) The ﬁeld has the linear distribution expected if the current density is uniformly distributed over the length of the conductor. In this large skin depth limit, the ﬁeld and current density spatial distributions are essentially the same as if the current source were time independent. In the opposite extreme, the skin depth is short compared to the conductor length. Perhaps this is accomplished by making the frequency very high compared to the reciprocal magnetic diﬀusion time based on the conductor length. δ/lessmuchb⇒2 µσb2/lessmuchω (14) Then, the ﬁrst term in the denominator of (10) is large compared with the second. Division of the numerator by this ﬁrst term gives ˆHz→ˆKs· e−(1+j)x+b δ−e(1+j)x−b δ¸ ≈ˆKse−(1+j)¡ x+b δ¢ (15) Sec. 10.7 Skin Eﬀect 39 Fig. 10.7.3 Skin depth as a function of frequency. In justifying the second of these expressions, remember that xis negative throughout the region of interest. Substitution of (15) into (2) shows that in this short skin depth limit Hz=Re½ ˆKse−(x+b) δ¢ ej£ ωt−(x+b) δ¤¾ (16) With the origin shifted from x= 0 to x=−b, this ﬁeld has the x−tdependence of the diﬀusion wave represented by (8). So it is that in the short skin depth limit, the distribution of the ﬁeld magnitude shown in Fig. 10.7.2 has the exponential decay typical of skin eﬀect. The skin depth, (5), is inversely proportional to the square root of ωµσ. Thus, an order of magnitude variation in frequency or in conductivity only changes δby about a factor of about 3. Even so, skin depths found under practical conditions are widely varying because these parameters have enormous ranges. In good conductors, such as copper or aluminum, Fig. 10.7.3 illustrates how δvaries from about 1 cm at 60 Hz to less than 0.1 mm at l MHz. Of interest in determining magnetically induced currents in ﬂesh is the curve for skin depth in materials having the “physiological” conductivity of about 0.2 S/m (Demonstration 7.9.1). If the frequency is high enough so that the skin depth is small compared with the dimensions of interest, then the ﬁelds external to the conductor are essentially determined using the perfect conductivity model introduced in Sec. 8.4. In Demon- stration 8.6.1, the ﬁelds around a conductor above a ground plane line were derived and the associated surface current densities deduced. If these currents are in the sinusoidal steady state, we can now picture them as actually extending into the conductors a distance that is on the order of δ. Although skin eﬀect determines the paths of current ﬂow at radio frequencies, as the following demonstrates, it can be important even at 60 Hz. Demonstration 10.7.1. Skin Eﬀect The core of magnetizable material shown in Fig. 10.7.4 passes through a slit cut from an aluminum block and through a winding that is driven at a frequency in the range of 60–240 Hz. The winding and the block of aluminum, respectively, comprise 40 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. 10.7.4 Demonstration of skin eﬀect. Currents induced in the con- ducting block tend to follow paths of minimum reactance nearest to the slot. Thus, because the aluminum block is thick compared to the skin depth, the ﬁeld intensity observed decreases exponentially with distance X. In the experiment, the block is 10 ×10×26 cm with thickness of 6 cm between the right face of the slot and the right side of the block. In aluminum at 60 Hz, δ= 1.1 cm, while at 240 Hz δis half of that. To avoid distortion of the ﬁeld, the yoke is placed at one end of the slot. the primary and secondary of a transformer. In eﬀect, the secondary is composed of one turn that is shorted on itself. The thickness bof the aluminum block is somewhat larger than a skin depth at 60 Hz. Therefore, currents circulating through the block around the leg of the magnetic circuit tend to follow the paths of least reactance closest to the slit. By making the length of the block and slit in the ydirection large compared to b, we expect to see distributions of current density and associated magnetic ﬁeld intensity at locations in the block well removed from the ends that have the xdependence found in Example 10.7.1. In the limit where δis small compared to b, the magnitude of the expected magnetic ﬂux density Bz(normalized to its value where X= 0) has the exponential decay with distance xof the inset to Fig. 10.7.4. The curves shown are for aluminum at frequencies of 60 Hz and 240 Hz. According to (5), increasing the frequency by a factor of 4 should decrease the skin depth by a factor of 2. Provision is made for measuring this ﬁeld by having a small slit milled in the block with a large enough width to permit the insertion of a magnetometer probe oriented to measure the magnetic ﬂux density in the zdirection. As we have seen in this and previous sections, currents induced in a conductor tend to exclude the magnetic ﬁeld from some region. Conductors are commonly used as shields that isolate a region from its surroundings. Typically, the conductor is made thick compared to the skin depth based on the ﬁelds to be shielded out. Sec. 10.7 Skin Eﬀect 41 Fig. 10.7.5 Perfectly conducting ⊃-shaped conductors are driven by a dis- tributed current source at the left. The magnetic ﬁeld is shielded out of the region to the right enclosed by the perfect conductors by: (a) a block of con- ductor that ﬁlls the region and has a thickness bthat is large compared to a skin depth; and (b) a sheet conductor having a thickness ∆ that is less than the skin depth. However, our studies of currents induced in thin conducting shells in Secs. 10.3 and 10.4 make it clear that this can be too strict a requirement for good shielding. The thin-sheet model can now be seen to be valid if the skin depth δislarge compared to the thickness ∆ of the sheet. Yet, we found that for a cylindrical shell of radius R, provided that ωµσ∆R/greatermuch1, a sinusoidally varying applied ﬁeld would be shielded from the interior of the shell. Apparently, under certain circumstances, even a conductor that is thin compared to a skin depth can be a good shield. To understand this seeming contradiction, consider the one-dimensional con- ﬁgurations shown in Fig. 10.7.5. In the ﬁrst of the two, plane parallel perfectly conducting electrodes again sandwich a block of conductor in a system that is very long in a direction perpendicular to the paper. However, now the plates are shorted by a perfect conductor at the right. Thus, at very low frequencies, all of the current from the source circulates through the perfectly conducting plates, bypassing the block. As a result, the ﬁeld throughout the conductor is uniform. As the frequency is raised, the electric ﬁeld generated by the time-varying magnetic ﬂux drives a current through the block much as in Example 10.7.1, with the current in the block tending to circulate through paths of least reactance near the left edge of the block. For simplicity, suppose that the skin depth δis shorter than the length of the block b, so that the decay of current density and ﬁeld into the block is essentially the exponential sketched in Fig. 10.7.5a. With the frequency high enough to make the skin depth short compared to b, the ﬁeld tends to be shielded from points within the block. In the conﬁguration of Fig. 10.7.5b, the block is replaced by a sheet having the same σandµbut a thickness ∆ that is less than a skin depth δ. Is it possible that this thin sheet could suppress the ﬁeld in the region to the right as well as the thick conductor? The answer to this question depends on the location of the observer and the extent bof the region with which he or she is associated. In the conducting block, shielding is poor in the neighborhood of the left edge but rapidly improves at 42 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 distances into the interior that are of the order of δor more. By contrast, the sheet conductor can be represented as a current divider. The surface current, Ks, of the source is tapped oﬀ by the sheet of conductivity per unit width G=σ∆/h(where h is the height of the structure) connected to the inductance (assigned to unit width) L=µbhof the single-turn inductor. The current through the single-turn inductor is Ks1/jωL¡ G+1 jωL¢=Ks 1 +jωLG(17) This current, and the associated ﬁeld, is shielded out eﬀectively when |ωLG|= ωµσb ∆/greatermuch1. With the sheet, the shielding strategy is to make equal use of all of the volume to the right for generating an electric ﬁeld in the sheet conductor. The eﬃciency of the shielding is improved by making ωµσ∆blarge: The interior ﬁeld is made small by making the shielded volume large. 10.8 SUMMARY Before tackling the concepts in this chapter, we had studied MQS ﬁelds in two limiting situations: •In the ﬁrst, currents in Ohmic conductors were essentially stationary, with distributions governed by the steady conduction laws investigated in Secs. 7.2–7.6. The associated magnetic ﬁelds were then found by using these cur- rent distributions as sources. In the absence of magnetizable material, the Biot-Savart law of Sec. 8.2 could be used for this purpose. With or without magnetizable material, the boundary value approaches of Secs. 8.5 and 9.6 were applicable. •In the second extreme, where ﬁelds were so rapidly varying that conductors were “perfect,” the eﬀect on the magnetic ﬁeld of currents induced in accor- dance with the laws of Faraday, Amp` ere, and Ohm was to nullify the magnetic ﬂux density normal to conducting surfaces. The boundary value approach used to ﬁnd self-consistent ﬁelds and surface currents in this limit was the subject of Secs. 8.4 and 8.6. In this chapter, the interplay of the laws of Faraday, Ohm, and Amp` ere has again been used to ﬁnd self-consistent MQS ﬁelds and currents. However, in this chapter, the conductivity has been ﬁnite. This has made it possible to explore the dynamics of ﬁelds with source currents that were neither distributed throughout the volumes of conductors in accordance with the laws of steady conduction nor conﬁned to the surfaces of perfect conductors. In dealing with perfect conductors in Chaps. 8 and 9, the all-important role ofEcould be placed in the background. Left for a study of this chapter was the electric ﬁeld induced by a time-varying magnetic induction. So, we began in Sec. 10.1 by picturing the electric ﬁeld in systems of perfect conductors. The approach was familiar from solving EQS (Chap. 5) and MQS (Chap. 8) boundary value problems involving Poisson’s equation. The electric ﬁeld intensity was represented by the superposition of a particular part having a curl that balanced −∂B/∂tat Sec. 10.8 Summary 43 each point in the volume, and an irrotational part that served to make the total ﬁeld tangential to the surfaces of the perfect conductors. Having developed some insight into the rotational electric ﬁelds induced by magnetic induction, we then undertook case studies aimed at forming an apprecia- tion for spatial and temporal distributions of currents and ﬁelds in ﬁnite conductors. By considering the eﬀects of ﬁnite conductivity, we could answer questions left over from the previous two chapters. •Under what conditions are distributions of current and ﬁeld quasistationary in the sense of being essentially snapshots of a sequence of static ﬁelds? •Under what conditions do they consist of surface currents and ﬁelds having negligible normal components at the surfaces of conductors? We now know that the answer comes in terms of characteristic magnetic diﬀusion (or relaxation) times τthat depend on the electrical conductivity, the permeability, and the product of lengths. τ=µσ∆b (1) The lengths in this expression make it clear that the size and topology of the conductors plays an important role. This has been illustrated by the thin-sheet models of Secs. 10.3 and 10.4 and one-dimensional magnetic diﬀusion into the bulk of conductors in Secs. 10.6 and 10.7. In each of these classes of conﬁgurations, the role played by τhas been illustrated by the step response and by the sinusoidal steady state response. For the former, the answer to the question, “When is a conductor perfect?” was literal. The conductor tended to be perfect for times that were short compared to a properly deﬁned τ. For the latter, the answer came in the form of a condition on the frequency. If ωτ/greatermuch1, the conductor tended to be perfect. In the sinusoidal state, a magnetic ﬁeld impressed at the surface of a conductor penetrates a distance δinto the conductor that is the skin depth and is given by setting ωτ=ωµσδ2= 2 and solving for δ. δ=r2 ωµσ(2) It is true that conductors will act as perfect conductors if this skin depth is much shorter than all other dimensions of interest. However, the thin sheet model of Sec. 10.4 teaches the important lesson that the skin depth may be larger than the conductor thickness and yet the conductor can still act to shield out the normal ﬂux density. Indeed, in Sec. 10.4 it was assumed that the current was uniform over the conductor cross-section and hence that the skin depth was large, not small, compared to the conductor thickness. Demonstration 8.6.1, where current passes through a cylindrical conductor at a distance labove a conducting ground plane, is an example. It would be found in that demonstration that if lis large compared to the conductor thickness, the surface current in the ground plane would distribute itself in accordance with the perfectly conducting model even if the frequency is so low that the skin depth is somewhat larger than the thickness of the ground plane. If ∆ is the ground plane thickness, we would expect the normal ﬂux density to be small so long as ωτm=ωµσ∆l/greatermuch1. Typical of such situations is that the electrical dissipation due to conduction is conﬁned to thin conductors and the magnetic 44 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 energy storage occupies relatively larger regions that are free of dissipation. Energy storage and power dissipation are subjects taken up in the next chapter. Sec. 10.2 Problems 45 Fig. P10.0.2 P R O B L E M S 10.1 Introduction 10.1.1∗In Demonstration 10.0.1, the circuit formed by the pair of resistors is re- placed by the one shown in Fig. 10.0.1, composed of four resistors of equal resistance R. The voltmeter might be the oscilloscope shown in Fig. 10.0.1. The “grounded” node at (4) is connected to the negative terminal of the voltmeter. Fig. P10.0.1 (a) Show that the voltage measured with the positive lead connected at (1), so that the voltmeter is across one of the resistors, is v= (dΦλ/dt)/4. (b) Show that if the positive voltmeter lead is connected to (2), then to (3), and ﬁnally to (4) (so that the lead is wrapped around the core once and connected to the same grounded node as the negative voltmeter lead), the voltages are, respectively, twice, three times, and four times this value. Show that this last result is as would be expected for a transformer with a one-turn secondary. 10.1.2 Plane parallel perfectly conducting plates are shorted to form the one-turn inductor shown in Fig. 10.0.2. The current source is distributed so that it supplies iamps over the width d. (a) Given that dandlare much greater than the spacing s, determine the voltage measured across the terminals of the current source by the voltmeter v2. 46 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. P10.1.2 (b) What is the voltage measured by the voltmeter v1connected as shown in the ﬁgure across these same terminals? 10.2 Magnetoquasistatic Electric Fields in Systems of Perfect Conductors 10.2.1∗In Prob. 8.4.1, the magnetic ﬁeld of a dipole surrounded by a perfectly conducting spherical shell is found. Show that E=iφµoa2 4R2dI dt£r R−(R/r)2¤ sinθ (a) in the region between the dipole and the shell. 10.2.2 The one-turn inductor of Fig. P10.1.2 is driven at the left by a current source that evenly distributes the surface current density K(t) over the width w. The dimensions are such that g/lessmucha/lessmuchw. (a) In terms of K(t), what is Hbetween the plates? (b) Determine a particular solution having the form Ep=ixExp(y, t), and ﬁnd E. 10.2.3 The one-turn solenoid shown in cross-section in Fig. P10.1.3 consists of per- fectly conducting sheets in the planes φ= 0, φ=α, and r=a. The latter is broken at the middle and driven by a current source of K(t) amps/unit length in the zdirection. The current circulates around the perfectly con- ducting path provided by the sheets, as shown in the ﬁgure. Assume that the angle α/greatermuchδand that the system is long enough in the zdirection to justify taking the ﬁelds as two dimensional. (a) In terms of K(t), what is Hin the pie-shaped region? (b) What is Ein this region? 10.2.4∗By constrast with previous examples and problems in this section, con- sider here the induction of currents in materials that have relatively low conductivity. An example would be the induction heating of silicon in the manufacture of semiconductor devices. The material in which the currents are to be induced takes the form of a long circular cylinder of radius b. Sec. 10.2 Problems 47 Fig. P10.1.3 Fig. P10.1.4 A long solenoid surrounding this material has Nturns, a length dthat is much greater than its radius, and a driving current i(t), as shown in Fig. P10.1.4. Because the material to be heated has a small conductivity, the in- duced currents are small and contribute a magnetic ﬁeld that is small com- pared to that imposed. Thus, the approach to determining the distribution of current induced in the semiconductor is 1) to ﬁrst ﬁnd H, ignoring the eﬀect of the induced current. This amounts to solving Amp` ere’s law and the ﬂux continuity law with the current density that of the excitation coil. Then, 2) with Bknown, the electric ﬁeld in the semiconductor is determined using Faraday’s law and the MQS form of the conservation of charge law, ∇·(σE) = 0. The approach to ﬁnding the ﬁelds can then be similar to that illustrated in this section. (a) Show that in the semiconductor and in the annulus, B≈(µoNi/d )iz. (b) Use the symmetry about the zaxis to show that in the semiconductor, where there is no radial component of Jand hence of Eatr=b, E=−(µoNr/2d)(di/dt )iφ. (c) To investigate the conditions under which this approximation is use- ful, suppose that the excitation is sinusoidal, with angular frequency ω. Approximate the magnetic ﬁeld intensity Hinduced associated with the induced current. Show that for the approximation to be good, Hinduced /Himposed =ωµoσb2/4/lessmuch1. 10.2.5 The conﬁguration for this problem is the same as for Prob. 10.1.4 except that the slightly conducting material is now a cylinder having a rectangular cross-section, as shown in Fig. P10.1.5. The imposed ﬁeld is therefore the same as before. 48 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. P10.1.5 (a) In terms of the coordinates shown, ﬁnd a particular solution for Ethat takes the form E=iyEyp(x, t) and satisﬁes the boundary conditions atx= 0 and x=b. (b) Determine Einside the material of rectangular cross-section. (c) Sketch the particular, homogeneous, and total electric ﬁelds, making clear how the ﬁrst two add up to satisfy the boundary conditions. (Do not take the time to evaluate your analytical formula but rather use your knowledge of the nature of the solutions and the boundary conditions that they must satisfy.) 10.3 Nature of Fields Induced in Finite Conductors 10.3.1∗The “Boomer” might be modeled as a transformer, with the disk as the one-turn secondary terminated in its own resistance. We have found here that if ωτm/greatermuch1, then the ﬂux linked by the secondary is small. In Example 9.7.4, it was shown that operation of a transformer in its “ideal” mode also implies that the ﬂux linked by the secondary be small. There it was found that to achieve this condition, the time constant L22/Rof the secondary must be long compared to times of interest. Approximate the inductance and resistance of the disk in Fig. 10.2.3 and show that L22/Ris indeed roughly the same as the time given by (10.2.17). 10.3.2 It is proposed that the healing of bone fractures can be promoted by the passage of current through the bone normal to the fracture. Using magnetic induction, a transient current can be induced without physical contact with the patient. Suppose a nonunion of the radius (a nonhealing fracture in the long bone of the forearm, as shown in Fig. P10.2.2) is to be treated. How would you arrange a driving coil so as to induce a longitudinal current along the bone axis through the fracture? 10.3.3∗Suppose that a driving coil like that shown in Fig. 10.2.2 is used to produce a magnetic ﬂux through a conductor having the shape of the circular cylin- drical shell shown in Fig. 10.3.2. The shell has a thickness ∆ and radius a. Following steps parallel to those represented by (10.2.13)–(10.2.16), show thatHind/H1isroughly ωτm, where τmis given by (10.3.10). (Assume that the applied ﬁeld is essentially uniform over the dimensions of the shell.) Sec. 10.4 Problems 49 Fig. P10.2.2 Fig. P10.3.1 10.4 Diﬀusion of Axial Magnetic Fields through Thin Conductors 10.4.1∗A metal conductor having thickness ∆ and conductivity σis formed into a cylinder having a square cross-section, as shown in Fig. P10.3.1. It is very long compared to its cross-sectional dimensions a. When t= 0, there is a surface current density Kocirculating uniformly around the shell. Show that the subsequent surface current density is K(t) =Koexp(−t/τm) where τm=µoσ∆a/4. 10.4.2 The conducting sheet of thickness ∆ shown in cross-section by Fig. P10.3.2 forms a one-turn solenoid having length lthat is large compared to the length dof two of the sides of its right-triangular cross-section. When t= 0, there is a circulating current density Jouniformly distributed in the conductor. (a) Determine the surface current density K(t) = ∆ J(t) for t >0. (b) A high-impedance voltmeter is connected as shown between the lower right and upper left corners. What v(t) is measured? (c) Now lead (1) is connected following path (2). What voltage is mea- sured? 10.4.3∗A system of two concentric shells, as shown in Fig. 10.5.2 without the center shell, is driven by the external ﬁeld Ho(t). The outer and inner shells have thicknesses ∆ and radii aandb, respectively. (a) Show that the ﬁelds H1andH2, between the shells and inside the in- 50 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 Fig. P10.3.2 ner shell, respectively, are governed by the equations ( τm=µoσ∆b/2) τmdH2 dt+H2−H1= 0 ( a) τm(b/a)dH2 dt+τm¡a b−b a¢dH1 dt+H1=Ho(t) ( b) (b) Given that Ho=Hmcosωt, show that the sinusoidal steady state ﬁelds are H1=Re{[Hm(1+jωτm)/D] expjωt}andH2=Re{[Hm/D] expjωt} where D= [1 + jωτm(a/b−b/a)](1 + jωτm) +jωτm(b/a). Fig. P10.3.4 10.4.4 The⊃-shaped perfect conductor shown in Fig. P10.3.4 is driven along its left edge by a current source having the uniformly distributed density Ko(t). At x=−athere is a thin sheet having the nonuniform conductivity σ=σo/[1 +αcos(πy/b)]. The length in the zdirection is much greater than the other dimensions. (a) Given Ko(t), ﬁnd a diﬀerential equation for K(t). (b) In terms of the solution K(t) to this equation, determine Ein the region −a < x < 0,0< y < b . Sec. 10.5 Problems 51 Fig. P10.4.1 Fig. P10.4.2 10.5 Diﬀusion of Transverse Magnetic Fields through Thin Conductors 10.5.1∗A thin planar sheet having conductivity σand thickness ∆ extends to inﬁnity in the xandzdirections, as shown in Fig. P10.4.1. Currents in the sheet are zdirected and independent of z. (a) Show that the sheet can be represented by the boundary conditions Ba y−Bb y= 0 ( a) ∂ ∂x(Ha x−Hb x) =−∆σ∂By ∂t(b) (b) Now consider the special case where the regions above and below are free space and extend to inﬁnity in the + yand−ydirections, respectively. When t= 0, there is a surface current density in the sheet K=izKosinβx, where Koandβare given constants. Show that for t >0, Kz=Koexp(−t/τ) where τ=µoσ∆/2β. 10.5.2 In the two-dimensional system shown in cross-section by Fig. P10.4.2, a planar air gap of width dis bounded from above in the surface y=dby a thin conducting sheet having conductivity σand thickness ∆. This sheet is, in turn, backed by a material of inﬁnite permeability. The region below is also inﬁnitely permeable and at the interface y= 0 there is a winding used to impose the surface current density K=K(t) cosβxiz. The system extends to inﬁnity in the ±xand±zdirections. (a) The surface current density K(t) varies so rapidly that the conducting sheet acts as a perfect conductor. What is Ψ in the air gap? (b) The current is slowly varying so that the sheet supports little induced current. What is Ψ in the air gap? (c) Determine Ψ( x, y, t ) if there is initially no magnetic ﬁeld and a step, K=Kou−1(t), is applied. Show that the early and long-time response matches that expected from parts (a) and (b). 52 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 10.5.3∗The cross-section of a spherical shell having conductivity σ, radius Rand thickness ∆ is as shown in Fig. 8.4.5. A magnetic ﬁeld that is uniform and zdirected at inﬁnity is imposed. (a) Show that boundary conditions representing the shell are Ba r−Bb r= 0 ( a) 1 Rsinθ∂ ∂θ£ sinθ(Ha θ−Hb θ)¤ =−µo∆σ∂Hr ∂t(b) (b) Given that the driving ﬁeld is Ho(t) =Re{ˆHoexp(jωt)}, show that the magnetic moment of a dipole at the origin that would have an eﬀect on the external ﬁeld equivalent to that of the shell is m=Re{−jωτ(2πR3ˆHo) 1 +jωτexp(jωt)} where τ≡µoσ∆R/3. (c) Show that in the limit where ωτ→ ∞ , the result is the same as found in Example 8.4.3. 10.5.4 A magnetic dipole, having moment i(t)a(as deﬁned in Example 8.3.2) oriented in the zdirection is at the center of a spherical shell having radius R, thickness ∆, and conductivity σ, as shown in Fig. P10.4.4. With i= Re{ˆiexp(jωt)}, the system is in the sinusoidal steady state. (a) In terms of i(t)a, what is Ψ in the neighborhood of the origin? (b) Given that the shell is perfectly conducting, ﬁnd Ψ. Make a sketch of Hfor this limit. (c) Now, with σﬁnite, determine Ψ. (d) Take the appropriate limit of the ﬁelds found in (c) to recover the result of (b). In terms of the parameters that have been speciﬁed, under what conditions does the shell behave as though it had inﬁnite conductivity? 10.5.5∗In the system shown in cross-section in Fig. P10.4.5, a thin sheet of conduc- tor, having thickness ∆ and conductivity σ, is wrapped around a circular cylinder having inﬁnite permeability and radius b. On the other side of an air gap at the radius r=ais a winding, used to impose the surface current density K=K(t) sin 2 φiz, backed by an inﬁnitely permeable material in the region a < r . (a) The current density varies so rapidly that the sheet behaves as an inﬁnite conductor. In this limit, show that Ψ in the air gap is Ψ =−aK 2£¡r b¢2+¡b r¢2¤ £¡a b¢2+¡b a¢2¤cos 2φ (a) Sec. 10.6 Problems 53 Fig. P10.4.4 Fig. P10.4.5 (b) Now suppose that the driving current is so slowly varying that the current induced in the conducting sheet is negligible. Show that Ψ =−aK 2£¡r b¢2−¡b r¢2¤ £¡a b¢2−¡b a¢2¤cos 2φ (b) (c) Show that if the ﬁelds are zero when t <0 and there is a step in current, K(t) =Kou−1(t) Ψ =aKocos 2φ ¡a b¢2−¡b a¢2½£¡r a¢2−¡a r¢2¤ ¡a b¢2+¡b a¢2e−t/τ−£¡r b¢2−¡b r¢2¤ 2¾ (c) where τ=µoσ∆b 2£¡b a¢2+¡a b¢2¤ £¡a b¢2−¡b a¢2¤ (d) Show that the early and long-time responses do indeed match the results found in parts (a) and (b). 10.5.6 The conﬁguration is as described in Prob. 10.4.5 except that the conducting shell is on the outside of the air gap at r=a, while the windings are on the inside surface of the air gap at r=b. Also, the windings are now arranged 54 Magnetoquasistatic Relaxation and Diﬀusion Chapter 10 so that the imposed surface current density is K=Ko(t) sinφ. For this conﬁguration, carry out parts (a), (b), and (c) of Prob. 10.4.5. 10.6 Magnetic Diﬀusion Laws 10.6.1∗Consider a class of problems that are analogous to those described by (10.5.10) and (10.5.11), but with Jrather than Hwritten as a solution to the diﬀusion equation. (a) Use (10.5.1)–(10.5.5) to show that ∇2(J/σ) =µ∂J ∂t(a) (b) Now consider J(rather than H) to be zdirected but independent ofz,J=Jz(x, y, t )iz, and H(rather than J) to be transverse, H= Hx(x, y, t )ix+Hy(x, y, t )iy. Show that ∇2¡Jz σ¢ =µ∂Jz ∂t(b) where Hcan be found from Jusing ∂H ∂t=−∂ ∂y¡Jz σµ¢ ix+∂ ∂x¡Jz σµ¢ iy (c) Note that these expressions are of the same form as (10.5.8), (10.5.10), and (10.5.11), respectively, but with the roles of JandHreversed. 10.7 Magnetic Diﬀusion Step Response 10.7.1∗In the conﬁguration of Fig. 10.6.1, a steady state has been established with Ks=Kp= constant. When t= 0, this driving current is suddenly turned oﬀ. Show that HandJare given by (10.6.21) and (10.6.22) with the ﬁrst term in each omitted and the sign of the summation in each reversed. 10.7.2 Consider the conﬁguration of Fig. 10.6.1 but with a perfectly conducting electrode in the plane x= 0 “shorting” the electrode at y= 0 to the one aty=a. (a) A steady driving current has been established with Ks=Kp= constant. What are the steady HandJin the conducting block? (b) When t= 0, the driving current is suddenly turned oﬀ. Determine H andJfort >0. Sec. 10.8 Problems 55 10.8 Skin Eﬀect 10.8.1∗For Example 10.7.1, the conducting block has length din the zdirection. (a) Show that the impedance seen by the current source is Z=a(1 +j) dσδ£ e(1+j)b/δ+e−(1+j)b/δ¤ £ e(1+j)b/δ−e−(1+j)b/δ¤ (a) (b) Show that in the limit where b/lessmuchδ, Z becomes the dc resistance a/dbσ . (c) Show that in the opposite extreme where b/greatermuchδ, so that the current is concentrated near the surface, the block impedance has resistive and inductive-reactive parts of equal magnitude and that the resistance is equivalent to that for a slab having thickness δin the xdirection carrying a current that is uniformly distributed with respect to x. 10.8.2 In the conﬁguration of Example 10.7.1, the perfectly conducting electrodes are terminated by a perfectly conducting electrode in the plane x= 0. (a) Determine the sinusoidal steady state response H. (b) Show that even though the current source is now “shorted” by per- fectly conducting electrodes, the high-frequency ﬁeld distribution is still given by (10.7.16), so that in this limit, the current still concen- trates at the surface. (c) Determine the impedance of a length d(in the zdirection) of the block. 11 ENERGY, POWER FLOW, AND FORCES 11.0 INTRODUCTION One way to decide whether a system is electroquasistatic or magnetoquasistatic is to consider the relative magnitudes of the electric and magnetic energy storages. The subject of this chapter therefore makes a natural transition from the quasistatic laws to the complete set of electrodynamic laws. In the order introduced in Chaps. 1 and 2, but now including polarization and magnetization,1these are Gauss’ law [(6.2.1) and (6.2.3)] ∇ ·(/epsilon1oE+P) =ρu (1) Amp` ere’s law (6.2.11), ∇ ×H=Ju+∂ ∂t(/epsilon1oE+P) (2) Faraday’s law (9.2.7), ∇ ×E=−∂ ∂tµo(H+M) (3) and the magnetic ﬂux continuity law (9.2.2). ∇ ·µo(H+M) = 0 (4) Circuit theory describes the excitation of a two-terminal element in terms of the voltage vapplied between the terminals and the current iinto and out of the respective terminals. The power supplied through the terminal pair is vi. One objective in this chapter is to extend the concept of power ﬂow in such a way that power is thought to ﬂow throughout space, and is not associated only with 1For polarized and magnetized media at rest. 1 2 Energy, Power Flow, and Forces Chapter 11 Fig. 11.0.1 If the border between two states passes between the plates of a capacitor or between the windings of a transformer, is there power ﬂow that should be overseen by the federal government? current ﬂow into and out of terminals. The basis for this extension is the laws of electrodynamics, (1)–(4). Even if a system can be represented by a circuit, the need for the generalization of the circuit-theoretical power ﬂow concept is apparent if we try to understand how electrical energy is transferred within, rather than between, circuit elements. The limitations of the circuit viewpoint would be crucial to testimony of an expert witness in litigation concerning the authority of the Federal Power Commission2to regulate power ﬂowing between states. If the view is taken that passage of current across a border is a prerequisite for power ﬂow, either of the devices shown in Fig. 11.0.1 might be installed at the border to “launder” the power. In the ﬁrst, the state line passes through the air gap between capacitor plates, while in the second, it separates the primary from the secondary in a transformer.3In each case, the current never leaves the state where it is generated. Yet in the examples shown, power generated in one state can surely be consumed in another, and a meaningful discussion of how this takes place must be based on a broadened view of power ﬂow. From the circuit-theoretical viewpoint, energy storage and rate of energy dissi- pation are assigned to circuit elements as a whole. Power ﬂowing through a terminal pair is expressed as the product of a potential diﬀerence vbetween the terminals and the current iin one terminal and out of the other. Thus, the terminal voltage vand current ido provide a meaningful description of power ﬂow into a surface S that encloses the circuit shown in Fig. 11.0.2. The surface Sdoes not pass “inside” one of the elements. Power Flow in a Circuit. For the circuit of Fig. 11.0.2, Kirchhoﬀ’s laws 2Now the Federal Energy Regulatory Commission. 3To be practical, the capacitor would be constructed with an enormous number of inter- spersed plates, so that in order to keep the state line in the air gap, a gerrymandered border would be required. Contemplation of the construction of a practical transformer, as described in Sec. 9.7, reveals that the state line would be even more diﬃcult to explain in the MQS case. Sec. 11.0 Introduction 3 Fig. 11.0.2 Circuit used to review the derivation of energy conservation statement for circuits. combine with the terminal relations for the capacitor, inductor, and resistor to give i=Cdv dt+iL+Gv (5) v=LdiL dt(6) Motivated by the objective to obtain a statement involving vi, we multiply the ﬁrst of these laws by the terminal voltage v. To eliminate the term viLon the right, we also multiply the second equation by iL. Thus, with the addition of the two relations, we obtain vi=vCdv dt+iLLdiL dt+Gv2(7) Because LandCare assumed to be constant, we can use the relation udu=d(1 2u2) to rewrite this expression as vi=dw dt+Gv2(8) where w=1 2Cv2+1 2Li2 L With its origins solely in the circuit laws, (8) can be regarded as giving no more information than inherent in the original laws. However, it gives insights into the circuit dynamics that are harbingers of what can be expected from the more general statement to be derived in Sec. 11.1. These come from considering some extremes. •If the terminals are open ( i= 0), and if the resistor is absent ( G= 0), wis constant. Thus, the energy wis conserved in this limiting case. The solution to the circuit laws must lead to the conclusion that the sum of the electric energy1 2Cv2and the magnetic energy1 2Li2 Lis constant. •Again, with G= 0, but now with a current supplied to the terminals, (8) becomes vi=dw dt(9) 4 Energy, Power Flow, and Forces Chapter 11 Because the right-hand side is a perfect time derivative, the expression can be integrated to giveZt 0vidt=w(t)−w(0) (10) Regardless of the details of how the currents and voltage vary with time, the time integral of the power viis solely a function of the initial and ﬁnal total energies w. Thus, if wwere zero to begin with and viwere positive, at some later time t, the total energy would be the positive value given by (10). To remove the total energy from the inductor and capacitor, vimust be reversed in sign until the integration has reduced wto zero. Because the process is reversible, we say that the energy wisstored in the capacitor and inductor. •If the terminals are again open ( i= 0) but the resistor is present, (8) shows that the stored energy wmust decrease with time. Because Gv2is positive, this process is not reversible and we therefore say that the energy is dissipated in the resistor. In circuit theory terms, (8) is an example of an energy conservation theorem. According to this theorem, electrical energy isnotconserved. Rather, of the electri- cal energy supplied to the circuit at the rate vi, part is stored in the capacitor and inductor and indeed conserved, and part is dissipated in the resistor. The energy supplied to the resistor is not conserved in electrical form . This energy is dissipated in heat and becomes a new kind of energy, thermal energy. Just as the circuit laws can be combined to describe the ﬂow of power between the circuit elements, so Maxwell’s equations are the basis for a ﬁeld-theoretical view of power ﬂow. The reasoning that casts the circuit laws into a power ﬂow statement parallels that used in the next section to obtain the more general ﬁeld-theoretical law, so it is worthwhile to review how the circuit laws are combined to obtain a statement describing power ﬂow. Overview. The energy conservation theorem derived in the next two sections will also not be a conservation theorem in the sense that electrical energy is con- served. Rather, in addition to accounting for the storage of energy, it will include conversion of energy into other forms as well. Indeed, one of the main reasons for our interest in power ﬂow is the insight it gives into other subsystems of the physical world [e.g. the thermodynamic, chemical, or mechanical subsystems]. This will be evident from the topics of subsequent sections. The conservation of energy statement assumes as many special forms as there are diﬀerent constitutive laws. This is one reason for pausing with Sec. 11.1 to summarize the integral and diﬀerential forms of the conservation law, regardless of the particular application. We shall reference these expressions throughout the chapter. The derivation of Poynting’s theorem, in the ﬁrst part of Sec. 11.2, is motivated by the form of the general conservation theorem. As subsequent sections evolve, we shall also make continued reference to this law in its general form. By specializing the materials to Ohmic conductors with linear polarization and magnetization constitutive laws, it is possible to make a clear identiﬁcation of the origins of electrical energy storage and dissipation in media. Such systems are considered in Sec. 11.3, where the ﬂow of power from source to “sinks” of thermal Sec. 11.1 Conservation Statements 5 Fig. 11.1.1 Integral form of energy conservation theorem applies to system within arbitrary volume Venclosed by surface S. dissipation is illustrated. Processes of energy storage and dissipation are developed in greater depth in Secs. 11.4 and 11.5. Through Sec. 11.5, the assumption is that materials are at rest. In Secs. 11.6 and 11.7, the power input is studied in the presence of motion of materials. These sections illustrate how the energy conservation law is used to determine electric and magnetic forces on macroscopic media. The discussion in these sections is conﬁned to a determination of total forces. Consistent with the ﬁeld theory point of view is the concept of a distributed force per unit volume, a force density. Rigorous derivations of macroscopic force densities are based on energy arguments paralleling those of Secs. 11.6 and 11.7. In Sec. 11.8, we shall look at microscopic models of force density distributions that provide a picture of the origin of these distributions. Finally, Sec. 11.9 is an introduction to the macroscopic force densities needed to put electromechanical coupling on a continuum basis. 11.1 INTEGRAL AND DIFFERENTIAL CONSERVATION STATEMENTS The circuit with theoretical conservation theorem (11.0.8) equates the power ﬂow- ing into the circuit to the rate of change of the energy stored and the rate of energy dissipation. In a ﬁeld, theoretical generalization, the energy must be imagined dis- tributed through space with an energy density W(joules/m3), and the power is dissipated at a local rate of dissipation per unit volume Pd(watts/m3). The power ﬂows with a density S(watts/m2), a vector, so that the power crossing a surface Sa is given byR SaS·da. With these ﬁeld-theoretical generalizations, the power ﬂowing into a volume V, enclosed by the surface Smust be given by −I SS·da=d dtZ VWdv +Z VPddv (1) where the minus sign takes care of the fact that the term on the left is the power ﬂowing intothe volume. According to the right-hand side of this equation, this input power is equal to the rate of increase of the total energy stored plus the power dissipation. The total energy is expressed as an integral over the volume of an energy density ,W. Similarly, the total power dissipation is the integral over the volume of a power dissipation density Pd. 6 Energy, Power Flow, and Forces Chapter 11 The volume is taken as being ﬁxed, so the time derivative can be taken inside the volume integration on the right in (1). With the use of Gauss’ theorem, the surface integral on the left is then converted to one over the volume and the term transferred to the right-hand side. Z V¡ ∇ ·S+∂W ∂t+Pd¢ dv= 0 (2) Because Vis arbitrary, the integrand must be zero and a diﬀerential statement of energy conservation follows. ∇ ·S+∂W ∂t+Pd= 0(3) With an appropriate deﬁnition of S, WandPd, (1) and (3) could describe the ﬂow, storage, and dissipation not only of electromagnetic energy, but of thermal, elastic, or ﬂuid mechanical energy as well. In the next section we will use Maxwell’s equations to determine these variables for an electromagnetic system. 11.2 POYNTING’S THEOREM The objective in this section is to derive a statement of energy conservation from Maxwell’s equations in the form identiﬁed in Sec. 11.1. The conservation theorem includes the eﬀects of both displacement current and of magnetic induction. The EQS and MQS limits, respectively, can be taken by neglecting those terms having their origins in the magnetic induction ∂µo(H+M)/∂ton the one hand, and in the displacement current density ∂(/epsilon1oE+P)/∂ton the other. Amp` ere’s law, including the eﬀects of polarization, is (11.0.2). ∇ ×H=Ju+∂/epsilon1oE ∂t+∂P ∂t(1) Faraday’s law, including the eﬀects of magnetization, is (11.0.3). ∇ ×E=−∂µoH ∂t−∂µoM ∂t(2) These ﬁeld-theoretical laws play a role analogous to that of the circuit equations in the introductory section. What we do next is also analogous. For the circuit case, we form expressions that are quadratic in the dependent variables. Several considerations guide the following manipulations. One aim is to derive an expression involving power dissipation or conversion densities and time rates of change of energy storages. The power per unit volume imparted to the current density of unpaired charge follows directly from the Lorentz force law (at least in free space). The force on a particle of charge qis f=q(E+v×µoH) (3) Sec. 11.2 Poynting’s Theorem 7 The rate of work on the particle is f·v=qv·E (4) If the particle density is Nand only one species of charged particles exists, then the rate of work per unit volume is Nf·v=qNv·E=Ju·E (5) Thus, one must anticipate that an energy conservation law that applies to free space must contain the term Ju·E. In order to obtain this term, one should dot multiply (1) by E. A second consideration that motivates the form of the energy conservation law is the aim to obtain a perfect divergence of density of power ﬂow. Dot multiplication of (1) by Egenerates ( ∇ ×H)·E. This term is made into a perfect divergence if one adds to it −(∇ ×E)·H, i.e., if one subtracts (2) dot multiplied by H. Indeed, (∇ ×E)·H−(∇ ×H)·E=∇ ·(E×H) (6) Thus, subtracting (2) dot multiplied by Hfrom (1) dot multiplied by Eone obtains −∇ · (E×H) =∂ ∂t¡1 2/epsilon1oE·E¢ +E·∂P ∂t +∂ ∂t¡1 2µoH·H¢ +H·∂µoM ∂t+E·Ju(7) In writing the ﬁrst and third terms on the right, we have exploited the relation u·du=d(1 2u2). These two terms now take the form of the energy storage term in the power theorem, (11.1.3). The desire to obtain expressions taking this form is a third consideration contributing to the choice of ways in which (1) and (2) were combined. We could have seen at the outset that dotting Ewith (1) and subtracting (2) after it had been dotted with Hwould result in terms on the right taking the desired form of “perfect” time derivatives. In the electroquasistatic limit, the magnetic induction terms on the right in Faraday’s law, (2), are neglected. It follows from the steps leading to (7) that in the EQS approximation, the third and fourth terms on the right of (7) are negligible. Similarly, in the magnetoquasistatic limit, the displacement current, the last two terms on the right in Amp` ere’s law, (1), is neglected. This implies that for MQS systems, the ﬁrst two terms on the right in (7) are negligible. Systems Composed of Perfect Conductors and Free Space. Quasistatic examples in this category are the EQS systems of Chaps. 4 and 5 and the MQS systems of Chap. 8, where perfect conductors are surrounded by free space. Whether quasistatic or electrodynamic, in these conﬁgurations, P= 0,M= 0; and where there is a current density Ju, the perfect conductivity insures that E= 0. Thus, 8 Energy, Power Flow, and Forces Chapter 11 the second and last two terms on the right in (7) are zero. For perfect conductors surrounded by free space, the diﬀerential form of the power theorem becomes −∇ ·S=∂W ∂t (8) with S=E×H (9) and W=1 2/epsilon1oE·E+1 2µoH·H(10) where Sis the Poynting vector andWis the sum of the electric and magnetic energy densities . The electric and magnetic ﬁelds are conﬁned to the free space regions. Thus, power ﬂow and energy storage pictured in terms of these variables occur entirely in the free space regions. Limiting cases governed by the EQS and MQS laws, respectively, are dis- tinguished by having predominantly electric and magnetic energy densities. The following simple examples illustrate the application of the power theorem to two simple quasistatic situations. Applications of the theorem to electrodynamic sys- tems will be taken up in Chap. 12. Example 11.2.1. Plane Parallel Capacitor The plane parallel capacitor of Fig. 11.2.1 is familiar from Example 3.3.1. The circular electrodes are perfectly conducting, while the region between the electrodes is free space. The system is driven by a voltage source distributed around the edges of the electrodes. Between the electrodes, the electric ﬁeld is simply the voltage divided by the plate spacing (3.3.6), E=v diz (11) while the magnetic ﬁeld that follows from the integral form of Amp` ere’s law is (3.3.10). H=r 2/epsilon1od dt¡v d¢ iφ (12) Consider the application of the integral version of (8) to the surface Senclosing the region between the electrodes in Fig. 11.2.1. First we determine the power ﬂowing into the volume through this surface by evaluating the left-hand side of (8). The density of power ﬂow follows from (11) and (12). S=E×H=−r 2/epsilon1o d2vdv dtir (13) Sec. 11.2 Poynting’s Theorem 9 Fig. 11.2.1 Plane parallel circular electrodes are driven by a dis- tributed voltage source. Poynting ﬂux through surface denoted by dashed lines accounts for rate of change of electric energy stored in the enclosed volume. The top and bottom surfaces have normals perpendicular to this vector, so the only contribution comes from the surface at r=b. Because Sis constant on that surface, the integration amounts to a multiplication. −I SE×H·da= (2πbd)¡b 2/epsilon1o d2vdv dt¢ =d dt¡1 2Cv2¢ (14) where C≡πb2/epsilon1o d Here the expression has been written as the rate of change of the energy stored in the capacitor. With Eagain given by (11), we double-check the expression for the time rate of change of energy storage. d dtZ V1 2/epsilon1oE·Edv=d dt· 1 2/epsilon1o(dπb2)¡v d¢2¸ =d dt¡1 2Cv2¢ (15) From the ﬁeld viewpoint, power ﬂows into the volume through the surface at r=b and is stored in the form of electrical energy in the volume between the plates. In the quasistatic approximation used to evaluate the electric ﬁeld, the magnetic energy storage is neglected at the outset because it is small compared to the electric energy storage. As a check on the implications of this approximation, consider the total magnetic energy storage. From (12), Z V1 2µoH·Hdv=1 2µo· 1 2/epsilon1o d¡dv dt¢¸2 dZb 0r22πrdr =µo/epsilon1ob2 16C¡dv dt¢2(16) Comparison of this expression with the electric energy storage found in (15) shows that the EQS approximation is valid provided that µo/epsilon1ob2 8¯¯dv dt¯¯2/lessmuchv2(17) For a sinusoidal excitation of frequency ω, this gives ¡bω√ 8c¢2/lessmuch1 (18) 10 Energy, Power Flow, and Forces Chapter 11 Fig. 11.2.2 One-turn solenoid surrounding volume enclosed by surface Sdenoted by dashed lines. Poynting ﬂux through this surface accounts for the rate of change of magnetic energy stored in the enclosed volume. where cis the free space velocity of light (3.1.16). The result is familiar from Example 3.3.1. The requirement that the propagation time b/cof an electromagnetic wave be short compared to a period 1/ωis equivalent to the requirement that the magnetic energy storage be negligible compared to the electric energy storage. A second example oﬀers the opportunity to apply the integral version of (8) to a simple MQS system. Example 11.2.2. Long Solenoidal Inductor The perfectly conducting one-turn solenoid of Fig. 11.2.2 is familiar from Example 10.1.2. In terms of the terminal current i=Kd, the magnetic ﬁeld intensity inside is (10.1.14), H=i diz (19) while the electric ﬁeld is the sum of the particular and conservative homogeneous parts [(10.1.15) for the particular part and Ehfor the conservative part]. E=−µo 2dHz dtriφ+Eh (20) Consider how the power ﬂow through the surface Sof the volume enclosed by the coil is accounted for by the time rate of change of the energy stored. The Poynting ﬂux implied by (19) and (20) is S=E×H=· −µoa 2d2d dt¡1 2i2¢ +i dEφh¸ ir (21) This Poynting vector has no component normal to the top and bottom surfaces of the volume. On the surface at r=a, the ﬁrst term in brackets is constant, so the integration on Samounts to a multiplication by the area. Because Ehis irrotational, the integral of Eh·ds=Eφhrdφaround a contour at r=amust be zero. For this reason, there is no net contribution of Ehto the surface integral. −I SE×H·da= 2πad¡µoa 2d2¢d dt¡1 2i2¢ =d dt¡1 2Li2¢ ; (22) Sec. 11.3 Linear Media 11 where L≡µoπa2 d Here the result shows that the power ﬂow is accounted for by the rate of change of the stored magnetic energy. Evaluation of the right hand side of (8), ignoring the electric energy storage, indeed gives the same result. d dtZ V1 2µoH·Hdv=d dt· πa2d1 2µo¡i d¢2¸ =d dt¡1 2Li2¢ (23) The validity of the quasistatic approximation is examined by comparing the mag- netic energy storage to the neglected electric energy storage. Because we are only interested in an order of magnitude comparison and we know that the homoge- neous solution is proportional to the particular solution (10.1.21), the latter can be approximated by the ﬁrst term in (20). Z V1 2/epsilon1oE·Edv/similarequal1 2/epsilon1oµ2 o 4d2¡di dt¢2£ dZa 0r22πrdr¤ =µo/epsilon1oa2 16L¡di dt¢2(24) We conclude that the MQS approximation is valid provided that the angular fre- quency ωis small compared to the time required for an electromagnetic wave to propagate the radius a of the solenoid and that this is equivalent to having an elec- tric energy storage that is negligible compared to the magnetic energy storage. µo/epsilon1oa2 8¡di dt¢2/lessmuchi2→¡ωa√ 8c¢2/lessmuch1 (25) A note of caution is in order. If the gap between the “sheet” terminals is made very small, the electric energy storage of the homogeneous part of the Eﬁeld can become large. If it becomes comparable to the magnetic energy storage, the structure approaches the condition of resonance of the circuit consisting of the gap capacitance and solenoid inductance. In this limit, the MQS approximation breaks down. In practice, the electric energy stored in the gap would be dominated by that in the connecting plates, and the resonance could be described as the coupling of MQS and EQS systems as in Example 3.4.1. In the following sections, we use (7) to study the storage and dissipation of energy in macroscopic media. 11.3 OHMIC CONDUCTORS WITH LINEAR POLARIZATION AND MAGNETIZATION Consider a stationary material described by the constitutive laws P=/epsilon1oχeE µoM=µoχmH (1) 12 Energy, Power Flow, and Forces Chapter 11 Ju=σE where the susceptibilities χeandχm, and hence the permittivity and permeability /epsilon1andµ, as well as the conductivity σ, are all independent of time. Expressed in terms of these constitutive laws for PandM, the polarization and magnetization terms in (11.2.7) become E·∂P ∂t=∂ ∂t¡1 2/epsilon1oχeE·E¢ H·∂µoM ∂t=∂ ∂t¡1 2µoχmH·H¢ (2) Because these terms now appear in (11.2.7) as perfect time derivatives, it is clear that in a material having “linear” constitutive laws, energy is stored in the polar- ization and magnetization processes. With the substitution of these terms into (11.2.7) and Ohm’s law for Ju, a conservation law is obtained in the form discussed in Sec. 11.1. For an electrically and magnetically linear material that obeys Ohm’s law, the integral and diﬀerential conservation laws are (11.1.1) and (11.1.3), respectively, with S=E×H (3a) W=1 2/epsilon1E·E+1 2µH·H(3b) Pd=σE·E (3c) The power ﬂux density Sand the energy density Wappear as in the free space con- servation theorem of Sec. 11.2. The energy storage in the polarization and magneti- zation is included by simply replacing the free space permittivity and permeability by/epsilon1andµ, respectively. The term Pdis always positive and seems to represent a rate of power loss from the electromagnetic system. That Pdindeed represents power converted to thermal form is motivated by considering the origins of the Ohmic conduction law. In terms of the bipolar conduction model introduced in Sec. 7.1, positive and negative carriers, respectively, experience the forces f+andf−. These forces are balanced by collisions with the surrounding particles, and hence the work done by the ﬁeld in forcing the migration of the particles is converted into thermal energy. If the velocity of the families of particles are, respectively, v+andv−, and the number densities N+andN−, respectively, then the rate of work performed on the carriers (per unit volume) is Pd=N+f+·v++N−f−·v− (4) Sec. 11.3 Linear Media 13 In recognition of the balance between collision forces and electrical forces, the forces of (4) are replaced by |q+|Eand−|q−|E, respectively. Pd=N+|q+|E·v+−N−|q−|E·v− (5) If, in turn, the velocities are written as the products of the respective mobilities and the macroscopic electric ﬁeld, (7.1.3), it follows that Pd= (N+|q+|µ++N−|q−|µ−)E·E=σE·E (6) where the deﬁnition of the conductivity σ(7.1.7) has been used. The power dissipation density Pd=σE·E(watts/m3) represents a rate of energy loss from the electromagnetic system to the thermal system. Example 11.3.1. The Poynting Vector of a Stationary Current Distribution In Example 7.5.2, we studied the electric ﬁelds in and around a circular cylindrical conductor fed by a battery in parallel with a disk-shaped conductor. Here we deter- mine the Poynting vector ﬁeld and explore its spatial relationship to the dissipation density. First, within the circular cylindrical conductor [region (b) in Fig. 11.3.1], the electric ﬁeld was found to be uniform, (7.5.7), Eb=v Liz (7) while in the surrounding free space region, it was [from (7.5.11)] Ea=−v L ln(a/b)£z rir+ln(r/a)iz¤ (8) and in the disk-shaped conductor [from (7.5.9)] Ec=v ln(a/b)1 rir (9) By symmetry, the magnetic ﬁeld intensity is φdirected. The φcomponent ofHis most easily evaluated from the integral form of Amp` ere’s law. The current density in the circular conductor follows from (7) as Jo=σv/L . Then, 2πrH φ=Joπr2→Hb φ=Jor 2; r < b (10) 2πrH φ=Joπb2→Ha φ=Job2 2r; b < r < a (11) The magnetic ﬁeld distribution in the disk conductor is also deduced from Amp` ere’s law. In this region, it is easiest to evaluate the rcomponent of Amp` ere’s diﬀerential law with the current density Jc=σEc, with Ecgiven by (9). Integra- tion of this partial diﬀerential equation on zthen gives a linear function of zplus 14 Energy, Power Flow, and Forces Chapter 11 Fig. 11.3.1 Distribution of Poynting ﬂux in coaxial resistors and asso- ciated free space. The conﬁguration is the same as for Example 7.5.2. A source to the left supplies current to disk-shaped and circular cylindri- cal resistive materials. The outer and right-end conductors are perfectly conducting. Note that there is a Poynting ﬂux in the free space interior region even when the currents are stationary. an “integration constant” that is a function of r. The latter is determined by the requirement that Hφbe continuous at z=−L. Hc φ=−σ ln(a/b)v r(L+z) +Job2 2r; b < r < a (12) It follows from these last four equations that the Poynting vector inside the circular cylindrical conductor, in the surrounding space, and in the disk-shaped electrode is Sb=−v LJo 2rir (13) Sa=−vb2Jo ln(a/b)2rL¡z riz−lnr air¢ (14) Sc=· −σ ln2(a/b)v2 r2(L+z) +Jov ln(a/b)b2 2r2¸ iz (15) This distribution of Sis sketched in Fig. 11.3.1. Wherever there is a dissipation density, there must be a negative divergence of S. Thus, in the conductors, the S lines terminate in the volume. In the free space region (a), Sis solenoidal. Even with the ﬁelds perfectly stationary in time, the power is seen to ﬂow through the open space to be absorbed in the volume where the dissipation takes place. The integral of the Poynting vector over the surface surrounding the inner conductor gives what we would expect either from the circuit point of view −I E×H·da= (2πbL)¡v L¢¡Job 2¢ =v(πb2Jo) =vi (16) where iis the total current through the cylinder, or from an evaluation of the right- hand side of the integral conservation law. Z VσE·Edv= (πb2L)σ¡v L¢2=v¡ πb2σv L¢ =vi (17) Sec. 11.3 Linear Media 15 An Alternative Conservation Theorem for Electroquasistatic Systems. In describing electroquasistatic systems, it is inconvenient to require that the magnetic ﬁeld intensity be evaluated. We consider now an alternative conservation theorem that is specialized to EQS systems. We will ﬁnd an alternative expression for S that does not involve H. In the process of ﬁnding an alternative distribution of S, weillustrate the danger of ascribing meaning to Sevaluated at a point, rather than integrated over a closed surface . In the EQS approximation, Eis irrotational. Thus, E=−∇Φ (18) and the power input term on the left in the integral conservation law, (11.1.1), can be expressed as −I SE×H·da=I S∇Φ×H·da (19) Next, the vector identity ∇ ×(ΦH) =∇Φ×H+ Φ∇ ×H (20) is used to write the right-hand side of (19) as −I SE×H·da=I S∇ ×(ΦH)·da−I SΦ∇ ×H·da (21) The ﬁrst integral on the right is zero because the curl of a vector is divergence free and a ﬁeld with no divergence has zero ﬂux through a closed surface. Amp` ere’s law can be used to eliminate curlHfrom the second. −I SE×H·da=−I SΦ¡ J+∂D ∂t¢ ·da (22) In this way, we have determined an alternative expression for S,valid only in the electroquasistatic approximation . S= Φ¡ J+∂D ∂t¢ (23) The density of power ﬂow, expressed by (23) as the product of a potential and total current density consisting of the sum of the conduction and displacement current densities, has a form similar to that used in circuit theory. The power ﬂux density of (23) is convenient in describing EQS systems, where the eﬀects of magnetic induction are not signiﬁcant. To be consistent with the EQS approximation, the conservation law must be used with the magnetic energy density neglected. Example 11.3.2. Alternative EQS Power Flux Density for Stationary Current Distribution 16 Energy, Power Flow, and Forces Chapter 11 Fig. 11.3.2 Distribution of electroquasistatic ﬂux density for the same sys- tem as shown in Fig. 11.3.1. Fig. 11.3.3 Arbitrary EQS system accessed through terminal pairs. To contrast the alternative EQS power ﬂow density with the Poynting ﬂux density, consider again the coaxial resistor conﬁguration of Example 11.3.1. Because the ﬁelds are stationary, the EQS power ﬂux density is S= ΦJ (24) By contrast with the Poynting ﬂux density, this vector ﬁeld is zero in the free space region. In the circular cylindrical conductor, the potential and current density are [(7.5.6) and (7.5.7)] Φb=−v Lz;Jb=σEb=σv Liz (25) and it follows that the power ﬂux density is simply S=−σv2 L2ziz (26) There is a similar, radially directed ﬂux density in the disk-shaped resistor. The alternative distribution of S, shown in Fig. 11.3.2, is clearly very diﬀerent from that shown in Fig. 11.3.1 for the Poynting ﬂux density. Poynting Power Density Related to Circuit Power Input. Suppose that the surface Sdescribed by the conservation theorem encloses a system that is accessed through terminal pairs, as shown in Fig. 11.3.3. Under what circumstances is the integral of S·daoverSequivalent to summing the voltage-current product of the terminals of the wires connected to the system? Sec. 11.4 Energy Storage 17 Two attributes of the ﬁelds on the surface Senclosing the system are required. First, the contribution of the magnetic induction to Emust be negligible on S. If this is so, then regardless of what is inside S(for example, both EQS and MQS systems), on the surface S, the electric ﬁeld can be taken as irrotational. It follows that in taking the integral over a closed surface of the Poynting power density, we can just as well use (23). −I SE×H·da=−I SΦ¡ J+∂D ∂t¢ ·da (27) By contrast with the EQS systems treated in deriving this expression, it now holds only on the surface S, not necessarily on surfaces inside the volume enclosed by S. Second, on the surface S, the contribution of the displacement current must be negligible . This is equivalent to requiring that Sis chosen parallel to the dis- placement ﬂux density. In this case, the total power into the system reduces to −I SE×H·da=−I SΦJ·da (28) The integrand has value only where the surface Sintersects a wire. If taken as perfectly conducting (but nevertheless in a region where ∂B/∂tis zero and hence E is irrotational), the wires have potentials that are uniform over their cross-sections. Thus, in (28), Φ is equal to the voltage of the terminal. In integrating the current density over the cross-section of the wire, note that dais directed out of the surface, while a positive terminal current is directed into the surface. Thus, −I SE×H·da=nX i=1viii (29) and the input power expressed by (28) is equivalent to what would be expected from circuit theory. Poynting Flux and Electromagnetic Radiation. Power cannot be supplied to or lost by a quasistatic system of ﬁnite extent through a surface at inﬁnity. Such a power supply or loss requires radiation, and electromagnetic waves are ne- glected when either the magnetic induction or the displacement current density are neglected. To prove this statement, consider an EQS system of ﬁnite net charge. Its electric ﬁeld intensity decays like 1 /r2at inﬁnity, where ris the distance to a far-oﬀ point from some origin chosen within the system. At a great distance, the currents appear equivalent to current loop sources. Hence, the magnetic ﬁeld inten- sity has the 1 /r3decay typical of a magnetic dipole. It follows that the Poynting vector decays at least as fast as 1 /r5, so that the ﬂux of E×Hintegrated over the “sphere” at inﬁnity of area 4 πr2gives zero contribution. Because it is only that part of E×Hresulting from electromagnetic radiation that contributes at inﬁnity, Poynting’s theorem is shown in Sec. 12.5 to be a powerful tool for dealing with antennae. 18 Energy, Power Flow, and Forces Chapter 11 Fig. 11.4.1 Single-valued constitutive laws showing energy density associ- ated with variables at the endpoints of the curves: (a) electric energy density; and (b) magnetic energy density. 11.4 ENERGY STORAGE In the conservation theorem, (11.2.7), we have identiﬁed the terms E·∂P/∂tand H·∂µoM/∂tas the rate of energy supplied per unit volume to the polarization and magnetization of the material. For a linear isotropic material, we found that these terms can be written as derivatives of energy density functions. In this section, we seek a more general description of energy storage. First, nonlinear materials are considered from the ﬁeld viewpoint. Then, for those systems that can be described in terms of electrical terminal pairs, energy storage is formulated in terms of terminal variables. We will ﬁnd the results of this section directly applicable to ﬁnding electric and magnetic forces in Secs. 11.6 and 11.7. Energy Densities. Consider a material in which EandD≡(/epsilon1oE+P) are collinear. With EandDrepresenting the magnitudes of these vectors, this material is presumed to be described by a constitutive law in which Eis a single- valued function of D, such as that sketched in Fig. 11.4.1a. In the case of a linear constitutive law, the curve is a straight line with a slope equal to the permittivity /epsilon1. Consider a material in which EandPare collinear (isotropic material). Then, of course, EandD≡/epsilon1oE+Pare collinear as well. One may graph the magnitude of Dversus Eand obtain a complete characterization of the material. Now the power per unit volume imparted to the polarization is E·∂P/∂t. If one adds to it the rate of energy supply to the ﬁeld per unit volume (the free space part) E·∂/epsilon1oE/∂t, one obtains for the power per unit volume E·∂ ∂t(/epsilon1oE+P) =E·∂D ∂t=E∂D ∂t (1) The power supplied to the unit volume can now be written as the time derivative of a function of D, W e(D). Indeed, if we deﬁne the area above the graph in Fig. 11.4.1 as We, then ∂We ∂t=∂We ∂D∂D ∂t=E∂D ∂t (2) Sec. 11.4 Energy Storage 19 Thus, E(∂D/∂t ) is the derivative of the function We(D). This function is the energy stored per unit volume, because the energy supplied per unit volume expressed by the integralZt −∞dtE∂D ∂t=ZD 0EδD =We(D) (3) is a function of the ﬁnal value Dof the displacement ﬂux, and we assumed that the ﬁelds EandDwere zero at t=−∞. Here, δDrepresents the diﬀerential of D, usually denoted by dD. We will use δrather than dto avoid confusion between dif- ferentials used in carrying out volume, surface and line integrals and the diﬀerential used here, which implies an integration in a “state space” having the “dimension” D. Similar arguments show that if B≡µo(H+M) and Hare collinear, and if His a single-valued function of B, then H·∂B ∂t=∂Wm ∂t (4) where Wm=Wm(B) =ZB 0HδB (5) With (1) and (4) replacing the ﬁrst four terms on the right in the energy theorem of (11.2.7), it is clear that the energy density W=We+Wm. The electric and magnetic energy densities have the geometric interpretations as areas on the graphs representing the constitutive laws in Fig. 11.4.1. Energy Storage in Terms of Terminal Variables. It was shown in Sec. 11.3 that the power input to a system could be represented by the sum of the viproducts for each of the terminal pairs, (11.3.29), provided certain conditions were met in the neighborhoods of the terminals. The description of energy storage in a loss-free system in terms of terminal variables will be found useful in determining electric and magnetic forces. With the assumption that all of the power input to a system is accounted for by a time rate of change of the energy stored, the energy conservation statement for a system becomes nX i=1viii=dw dt(6) where w=Z VWdv and the integral is carried over the volume of the system. If the system is electroqua- sistatic , conservation of charge requires that the terminal current be the time rate of change of the charge on the electrode to which the positive terminal is attached. 20 Energy, Power Flow, and Forces Chapter 11 Fig. 11.4.2 Single-valued terminal relations showing total energy stored when variables are at the endpoints of the curves: (a) electric energy storage; and (b) magnetic energy storage. ii=dqi dt (7) Further, w=we, the stored electric energy. Thus, one concludes from (6) that nX i=1vidqi dt=dwe dt⇒dwe=nX i=1vidqi (8) The second expression states that with the addition of an incremental amount of charge dqito an electrode having the voltage vigoes an incremental change in the stored energy we. Integration on the charges then gives the total energy we=nX i=1Z vidqi (9) To complete this integral, each of the terminal voltages must be a known function of the associated charges. vi=vi(q1, . . . q n) (10) Integration is then carried out along any path in the state space ( q1. . . q n) that begins at the origin and ends with the desired charges on the electrodes (and hence the desired terminal voltages). For a single terminal pair, the energy can be pictured as the area shown in Fig. 11.4.2a. If the system is magnetoquasistatic, the conservation law for a lossless system that can be described by terminal relations again takes the form of (6). However, rather than expressing the currents as derivatives of electrode charges, the voltages are derivatives of the ﬂuxes linked by the respective terminal pairs. vi=dλi dt (11) Then, (6) leads to Sec. 11.4 Energy Storage 21 Fig. 11.4.3 Capacitor partially ﬁlled by free space and by dielectric having permittivity /epsilon1. wm=nX i=1Z iidλi (12) To complete this integral, we require the terminal currents as functions of the terminal ﬂux linkages. ii=ii(λi. . . λ n) (13) For a single terminal pair system, wmis portrayed in Fig. 11.4.2b. The most general way to compute the total energy stored in a system is to integrate the energy densities given by (3) and (5) over the volumes of the respective systems. If systems can be described in terms of terminal relations and are loss free, (9) and (12) must lead to the same answers. Note that ( D, E) and ( q, v) are the ﬁeld and circuit variables in the EQS systems, while ( B, H ) and ( λ, i) have corresponding roles in MQS systems. Example 11.4.1. An Electrically Linear System A dielectric slab of permittivity /epsilon1partially ﬁlls the region between plane parallel perfectly conducting electrodes, as shown in Fig. 11.4.3. With the fringing ﬁeld ignored, we ﬁnd the total energy stored by two methods. First, the energy density is integrated over the volume. Then, the terminal relation is used to evaluate the total energy. An exact solution for the electric ﬁeld well between the electrodes is simply E=ix(v/a). Note that this ﬁeld satisﬁes the boundary conditions at the interface between the dielectric slab and the free space region above and at the electrodes. We assume that a/lessmuchband therefore neglect the fringing ﬁelds. The energy density in the linear dielectric, where D=/epsilon1E, follows from eval- uation of (3). We=ZD 0EδD =1 2D2 /epsilon1=1 2/epsilon1E2; (14) 22 Energy, Power Flow, and Forces Chapter 11 E=v a In the free space region, the same result applies with /epsilon1→/epsilon1o. Integration of these energy densities over the regions in which they apply amounts to a multiplication by the respective volumes. Thus, the total energy is we=Z VWedV=1 2/epsilon1¡v a¢2(ξca) +1 2/epsilon1o¡v a¢2[(b−ξ)ca] (15) Note that this expression takes the form we=1 2Cv2(16) where C≡c a[/epsilon1ob+ξ(/epsilon1−/epsilon1o)] In terms of the terminal variables, where q=Cv, the total energy follows from an evaluation of (9). we=Z vdq=Z q Cdq=1 2q2 C=1 2Cv2(17) Once the integration has been carried out, the last expression is written by again using the relation q=Cv. Note that the volume integration of the energy density and the integration in terms of the terminal variables give the same result. The next example considers an MQS system with two terminal pairs and thus illustrates the integration called for in evaluating the energy from the terminal relations. Also, the energy stored in coupled inductors is often of practical interest. Example 11.4.2. Coupled Coils; Transformers An example of a two terminal pair lossless MQS system is a pair of coupled coils having the terminal relations hλ1 λ2i =hL11L12 L21L22ihi1 i2i (18) In this case, (12) becomes dwm=i1dλ1+i2dλ2 (19) To evaluate this expression, we need to substitute for the currents written in terms of the ﬂux linkages. This requires the inversion of (18). For linear systems, this is easily done, but not for nonlinear systems. To avoid inversion, we rewrite the right-hand side of (19), which becomes dwm=d(i1λ1+i2λ2)−λ1di1−λ2di2 (20) and regroup terms Sec. 11.4 Energy Storage 23 Fig. 11.4.4 Integration path in state space consisting of terminal currents. dw/prime m=λ1di1+λ2di2 (21) where the coenergy is deﬁned as w/prime m= (i1λ1+i2λ2)−wm (22) Equation (21) can be integrated when the ﬂux linkages are expressed in terms of the currents, and that is the form in which the terminal relations are given by (18). Once the coenergy w/prime mhas been found, wmfollows from (22). The integration of (19) is a line integral in a state space ( i1, i2). If energy is conserved, we must be able to carry out this integration along any path that begins with the currents turned oﬀ and ends with the currents at the desired values. In the path represented by Fig. 11.4.4, the current i1is turned up ﬁrst while holding the current i2to zero. Then, with i1held ﬁxed at its ﬁnal value, the current i2is raised from zero to its ﬁnal value. For this path, the integration of (22) becomes w/prime m=Zi1 0λ1(i/prime 1,0)di/prime 1+Zi2 0λ2(i1, i/prime 2)di/prime 2 (23) Substitution for the ﬂux linkages from (18) and evaluation of the integrals then gives w/prime m=1 2L11i2 1+L12i1i2+1 2L22i2 2 (24) If the integration is carried out along a path where the roles of i1andi2are re- versed, the expression obtained is (24) with L12→L21. To make the energy stored independent of path, the mutual inductances must be equal. L12=L21 (25) This relation, which we found to hold for the transformer of Example 9.7.4, is re- quired if energy is to be conserved. The energy is now evaluated by substituting this expression and the ﬂux linkages expressed using (18) into (22) solved for wm. It follows that w/prime m=wm (26) Evaluation of the energy stored in a unity-coupled transformer, where the inductances take the form of (9.7.20), gives wm=Aµ l¡1 2N2 1i2 1+N1N2i1i2+1 2N2 2i2 2¢ (27) 24 Energy, Power Flow, and Forces Chapter 11 Operating under “ideal” conditions [in the sense that i2/i1=−N1/N2, (9.7.13)], the transformer does not store energy, wm= 0. Thus, according to the power theorem in the form of (6), under ideal operating conditions, the power input at one terminal pair instantaneously appears as a power output at the second terminal pair. Examples have so far involved linear polarization and magnetization constitu- tive laws. In the following, the EQS energy storage in a material having a nonlinear polarization constitutive law is determined. Example 11.4.3. Energy Storage in Electrically Nonlinear Material To represent the tendency of the polarization to saturate as the electric ﬁeld is raised, a constitutive law might take the form D=µ α1√ 1 +α2E2+/epsilon1o¶ E (28) Here, α1andα2are parameters descriptive of the speciﬁc material, and Dis collinear withE. This constitutive law is portrayed graphically in Fig. 11.4.5. Because Dis given as a function of Ethat is not easily solved for Eas a func- tion of D, the computation of the electric energy density using (3) is inconvenient. However, we can observe that δWe=EδD =δ(ED)−DδE (29) and then regroup terms so that the expression becomes δW/prime e=DδE (30) where W/prime e=ED−We (31) Integration now leads to the coenergy density W/prime e, but the energy density Wecan then be found using (31) and the constitutive law. Speciﬁcally, evaluation of (30) using (28) gives the coenergy density W/prime e=Z DδE =α1 α2¡p 1 +α2E2−1¢ +1 2/epsilon1oE2(32) It follows from (31) that the energy density is We=ED−W/prime e=α1 α2µ√ 1 +α2E2−1√ 1 +α2E2¶ +1 2/epsilon1oE2(33) A graphical representation of the energy and coenergy functions is given in Fig. 11.4.5. The area “under the curve” with Das the integration variable is We, (3), and the area under the curve with Eas the integration variable is W/prime e, (31). Sec. 11.5 Electromagnetic Dissipation 25 Fig. 11.4.5 Single-valued nonlinear constitutive law. Areas represent- ing energy density Wand coenergy density W/primeare not equal in this case. 11.5 ELECTROMAGNETIC DISSIPATION The heat generated by electromagnetic ﬁelds is often the controlling feature of an engineering design. Semiconductors inevitably produce heat, and the distribution and magnitude of the heat source is an important consideration whether the ap- plication is to computers or power conversion. Often, the generation of heat poses a fundamental limitation on the performance of equipment. Examples where the generation of heat is desirable include the heating coil of an electric stove and the microwave irradiation of food in a microwave oven. Ohmic conduction is the primary cause of heat generation in metals, but it also operates in semiconductors, electrolytes, and (at low frequencies) in semi-insulating liquids and solids. The mechanism responsible for this type of heating was discussed in Sec. 11.3. The dissipation density associated with Ohmic conduction is σE·E. An Ohmic current can be imposed by making electrical contact with the mate- rial, as for the heating element in a stove. If the material is a good conductor, such currents can also be induced by magnetic induction (without electrical contact). The currents induced by time-varying magnetic ﬁelds in Chap. 10 are an example. Induction heating is an MQS process and often used in processing metals. Currents induced in transformer cores by the time-varying magnetic ﬂux are an example of undesirable heating. In this context, the associated losses (which are minimized by laminating the core) are said to be due to eddy currents . Ohmic heating can also be induced by “capacitive” coupling. In the EQS examples of Sec. 7.9, dielectric heating is caused by the currents associated with the accumulation of unpaired charges. Whether due to magnetic induction or capacitive coupling, the generation of heat is described by the dissipation density Pd=σE·Eidentiﬁed in Sec. 11.3. However, the polarization and magnetization terms in the conservation theorem, (11.2.7), can also be responsible for energy dissipation. This occurs when the (elec- tric or magnetic) dipoles do not align instantaneously with the ﬁelds. The polar- ization and magnetization constitutive laws diﬀer from the laws postulated in Sec. 11.3. As an example suggesting how the polarization term in (11.2.7) can represent dissipation, picture the artiﬁcial dielectric of Demonstration 6.6.1 (the ping-pong ball dielectric) but with spheres that are highly resistive rather than perfectly con- ducting. The accumulation of charge on the poles of the spheres in response to the application of an electric ﬁeld is described by a rate, rather than a magnitude , that 26 Energy, Power Flow, and Forces Chapter 11 is proportional to the ﬁeld. Thus, we would expect ∂P/∂trather than Pto be proportional to E. With γa coeﬃcient representing the properties and geometry of the spheres, the polarization constitutive law would then take the form ∂P ∂t=γE (1) If this law is used to express the polarization term in the conservation law, the second term on the right in (11.2.7), a positive deﬁnite quantity results. E·∂P ∂t=γE·E (2) As might be expected from the physical origins of the constitutive law, the polar- ization term now represents dissipation rather than energy storage. When materials are placed in electric ﬁelds having frequencies so high that conduction eﬀects are negligible, losses due to the polarization of dipoles become the dominant heating mechanism. The artiﬁcial diamagnetic material considered in Demonstration 9.5.1 suggests how analogous losses are associated with the dynamic magnetization of a material. If the spherical particles comprising the artiﬁcial dia- magnetic material have a ﬁnite conductivity, the induced dipole moments are not in phase with an applied sinusoidal ﬁeld. What amounts to Ohmic dissipation on the particle scale is accounted for on the macroscopic scale by a modiﬁed constitutive law of magnetization. The most common losses due to magnetization are encountered in ferromag- netic materials. Hysteresis losses occur because of the coercion required to obtain alignment of ferromagnetic domains. We will end this section with the relationship between the hysteresis curve of Fig. 9.4.6 and the dissipation density. Energy Conservation for Temporally Periodic Systems. Many practical situations involve ﬁelds that vary with time in a periodic fashion. The sinusoidal steady state is the most common example. If the energy conservation law (11.0.8) is integrated over one period T, the energy storage term makes no contribution. ZT 0dw dtdt=w(T)−w(0) = 0 (3) As a result, the time average of the conservation law states that the time average of the input power goes into the time average of the dissipation. The time average of the integral form of the conservation law, (11.1.1), becomes −/angbracketleftI SSda/angbracketright=/angbracketleftZ VPddv/angbracketright (4) This expression, which assumes that the dynamics are periodic but not necessarily sinusoidal, gives us two ways to compute the total energy dissipation. Either we can use the right-hand side and integrate the power dissipation density over the Sec. 11.5 Electromagnetic Dissipation 27 volume, or we can use the left-hand side and integrate the time average of S·da over the surface enclosing the volume. Consider the sinusoidal steady state as a particular case. If PandMare related to EandHby linear diﬀerential equations, an approach can be taken that is familiar from circuit theory. The phase and amplitude of each ﬁeld at a given location are represented by a complex amplitude. For example, the electric and magnetic ﬁeld intensities are written as E=ReˆE(r)ejωt;H=ReˆH(r)ejωt(5) A complex vector ˆE(r) has three complex scalar components ˆEx(r), ˆEy(r), and ˆEz(r). The meaning of each is the same as the meaning of a com- plex voltage in circuit theory: e.g., the magnitude of ˆEx(r),|ˆEx(r)|, gives the peak amplitude of the xcomponent of the electric ﬁeld varying cosinusoidally with time, and the phase of ˆEx(r) gives the phase advance of the cosine time function. In determining the time averages of products of quantities that are in the sinusoidal steady state, it is helpful to make use of the time average theorem . With ∗designating the complex conjugate, /angbracketleftReˆAejωtReˆBejωt/angbracketright=1 2ReˆAˆB∗ (6) This can be shown by using the identity ReˆCejωt=1 2(ˆCejωt+ˆC∗e−jωt) (7) Induction Heating. In this case, the heating is represented by Ohmic con- duction and Pdgiven by (11.3.3c). The examples from Chaps. 7 and 10 involving conductors of ﬁnite conductivity oﬀer the opportunity to apply this relation to the evaluation of the right-hand side of (4). If the same total time average power is calculated using the left-hand side of this expression, it may seem that Ohm’s law is not required. However, remember that this law is also reﬂected in the ﬁeld quantities used to calculate S. Example 11.5.1. Induction Heating of the Thin Shell The thin conducting shell of Fig. 11.5.1, in a ﬁeld Ho(t) applied collinear with its axis, was described in Example 10.3.1. Here the applied ﬁeld is in the sinusoidal steady state Ho=ReˆHoejωt(8) According to (10.3.9), the complex amplitude of the response, the magnetic ﬁeld inside the shell, is ˆHi=ˆHo 1 +jωτm(9) 28 Energy, Power Flow, and Forces Chapter 11 Fig. 11.5.1 Circular cylindrical conducting shell in imposed axial magnetic ﬁeld intensity Ho(t). where τm=1 2µoσ∆a. The complex amplitude of the surface current density circulating in the shell follows from (10.3.8). ˆK=−ˆHo+ˆHo 1 +jωτm=−jωτmˆHo 1 +jωτm(10) Because the current density is uniform over the radial cross-section of the shell, the dissipation density can be written in terms of the surface current density K= ∆σE. Pd=σE·E=K2 ∆2σ(11) It follows from the application of the time average theorem, (6), that the total time average dissipation is /angbracketleftZ VPddv/angbracketright=/angbracketleftZ V1 2ReˆKˆK∗ ∆2σdv/angbracketright=2πa∆l 2∆2σReˆKˆK∗(12) where lis the shell length. To complete the derivation based on an integration of the density over the volume of the conductor, this expression can be evaluated using (10). /angbracketleftpd/angbracketright ≡ /angbracketleftZ VPddv/angbracketright=po(ωτm)2 1 + (ωτm)2; po≡πal σ∆|ˆHo|2(13) The same result is found by evaluating the time average of the Poynting ﬂux density integrated over a surface that is just outside the shell at r=a. To see this, we again use the time average theorem, (6), and recognize that the surface integral amounts to a multiplication by the surface area of the shell. −/angbracketleftI S(E×H)·da/angbracketright=I S1 2ReˆEφˆH∗ oda=2πal 2ReˆEφˆH∗ o (14) To evaluate this expression, (10) is used to determine Eφ ˆEφ=ˆK ∆σ=−jωτmˆHo ∆σ(1 +jωτm)=−jωτm(1−jωτm)ˆHo ∆σ[1 + ( ωτm)2](15) Sec. 11.5 Electromagnetic Dissipation 29 Fig. 11.5.2 Time average power dissipation density normalized to po as deﬁned with (13) as a function of the frequency normalized to the magnetic diﬀusion time deﬁned with (9). Evaluation of (14) then gives −/angbracketleftI E×H·da/angbracketright=po(ωτm)2 1 + (ωτm); po≡πal σ∆|ˆHo|2(16) which is the same result as found by integrating the dissipation density over the volume, (13). The dependence of the time average power dissipation on the normalized fre- quency is shown in Fig. 11.5.2. At very low frequencies, the induced current is not large enough to have an appreciable eﬀect on the imposed ﬁeld. Thus, the electric ﬁeld is proportional to the time rate of change of the applied ﬁeld , and because the dissipation is proportional to the square of E, the power dissipation increases as the square of ω. At high frequencies, the induced current can be no more than that required to shield the imposed ﬁeld from the region inside the shell. As a result, the dissipation reaches an asymptotic limit. Which of the two approaches is best for ﬁnding the total power dissipation? The answer depends on what ﬁeld information is available. Certainly, the notion that the total heat generated can be found by integrating over a surface that is completely outside the heated material is a fundamental consequence of Poynting’s theorem. Dielectric Heating. In the sinusoidal steady state, we can identify the power dissipation density associated with polarization by ﬁnding the time average /angbracketleftPd/angbracketright=/angbracketleftE·∂D ∂t/angbracketright (17) In view of the time average theorem, (6), this becomes /angbracketleftPd/angbracketright=1 2Rejω ˆD·ˆE∗(18) If the polarization Pdoes not follow the electric ﬁeld Einstantaneously, yet the material is still linear and isotropic, the complex vector ˆPcan be related to ˆEby 30 Energy, Power Flow, and Forces Chapter 11 Fig. 11.5.3 Deﬁnition of angle δdeﬁning the loss tangent tan( δ) in terms of the real and the negative of the imaginary parts of the complex permittivity. a complex susceptibility. Or, instead, the complex displacement ﬂux density vector ˆDis related to ˆEby a complex dielectric constant. ˆD= ˆ/epsilon1ˆE= (/epsilon1/prime−j/epsilon1/prime/prime)ˆE (19) Here ˆ /epsilon1is the complex permittivity with real and imaginary parts /epsilon1/primeand−/epsilon1/prime/prime, respec- tively. Evaluation of (18) using this constitutive law gives /angbracketleftPd/angbracketright=ω 2|ˆE|2Rejˆ/epsilon1=ω 2|ˆE|2/epsilon1/prime/prime(20) Thus, /epsilon1/prime/primerepresents the electrical dissipation associated with the polarization pro- cess. In the literature, the loss tangent tanδis often used to represent dissipation. It is the tangent of the phase angle δof the complex dielectric constant deﬁned in terms /epsilon1/primeand/epsilon1/prime/primein Fig. 11.5.3. Thus, tanδ=/epsilon1/prime/prime /epsilon1/prime; cos δ=/epsilon1/prime |ˆ/epsilon1|; sin δ=/epsilon1/prime/prime |ˆ/epsilon1|(21) From this deﬁnition, it follows from Eulers formula that /epsilon1/prime−j/epsilon1/prime/prime=|ˆ/epsilon1|(cosδ−jsinδ) =|ˆ/epsilon1|e−jδ(22) Given the complex amplitude of the electric ﬁeld, Dis D=Re{|ˆ/epsilon1|ˆEej(ωt−δ)} (23) If the electric ﬁeld is Eocos(ωt), then Dis|ˆ/epsilon1|Eocos(ωt−δ). The electric displace- ment lags the electric ﬁeld by the phase angle δ. In terms of the loss tangent deﬁned by (21), the time average electrical dissi- pation density of (20) becomes /angbracketleftPd/angbracketright=ω/epsilon1/prime 2|ˆE|2tanδ (24) Usually the loss tangent and /epsilon1/primeare measured. In the following example, we compute the complex permittivity from a model of the polarizable medium and Sec. 11.5 Electromagnetic Dissipation 31 ﬁnd the electrical dissipation on a macroscopic basis. In this special case we have the option of ﬁnding the time average loss by considering each of the dipoles on a microscopic basis. This is not generally possible, because the interactions among dipoles that are neglected in this example are usually too complicated for an analytic treatment. Example 11.5.2. An Artiﬁcial Lossy Dielectric By putting together examples considered in Chaps. 6 and 7, we can illustrate the origins of the complex permittivity. The artiﬁcial dielectric of Example 6.6.1 and Demonstration 6.6.1 had “molecules” consisting of perfectly conducting spheres. As a result, the polarization was pictured as instantaneously in step with the applied ﬁeld. We consider now the result of having spheres that have ﬁnite conductivity. The response of a single sphere having a ﬁnite conductivity σand permittivity /epsilon1surrounded by free space is a special case of Example 7.9.3. The response to a sinusoidal drive is summarized by (7.9.36), where we set σa= 0, /epsilon1a=/epsilon1o, σb=σ, and/epsilon1b=/epsilon1. All that is required from this solution for the potential is the moment of a dipole that would give rise to the same exterior ﬁeld as does the sphere. Comparison of the potential of a dipole, (4.4.10), to that given by (7.9.36a) shows that the complex amplitude of the moment is ˆp= 4π/epsilon1oR3· 1 +jωτe(/epsilon1−/epsilon1o) 2/epsilon1o+/epsilon1¸ 1 +jωτeˆE (25) where τe≡(2/epsilon1o+/epsilon1)/σ. If mutual interactions between dipoles are ignored, the polarization density Pis this moment of a single dipole multiplied by the number of dipoles per unit volume, N. For a cubic array with a distance sbetween the dipoles (the centers of the spheres), N= 1/s3. Thus, the complex amplitude of the electric displacement is ˆD=/epsilon1oˆE+ˆP=/epsilon1oˆE+ˆp s3(26) Combining this result with the moment given by (25) yields the desired constitutive law in the form ˆD= ˆ/epsilon1ˆE, where the complex permittivity is ˆ/epsilon1=/epsilon1o½ 1 + 4 π¡R s¢3· 1 +jωτe(/epsilon1−/epsilon1o) 2/epsilon1o+/epsilon1¸ 1 +jωτe¾ (27) The time average power dissipation density follows from this expression and (20). /angbracketleftPd/angbracketright=2π/epsilon1o τe¡R s¢3· 3/epsilon1o 2/epsilon1o+/epsilon1¸ (ωτe)2 1 + (ωτe)2|ˆE|2(28) The dependence of the power dissipation on frequency has the same form as for the induction heating example, Fig. 11.5.2. At low frequencies, the surface charges induced at the north and south poles of each sphere are completely determined by the external ﬁeld. Thus, the current density within the sphere that makes possible the accumulation of these surface charges is proportional to the time rate of change of the applied ﬁeld. At low frequencies, the dissipation is proportional to the square 32 Energy, Power Flow, and Forces Chapter 11 of the volume current and hence to the square of the time rate of change of the applied ﬁeld. As a result, at low frequencies, the dissipation density increases with the square of the frequency. As the frequency is raised, less surface charge is induced on the spheres. Al- though the amount of charge induced is inversely proportional to the frequency, there is a compensating eﬀect because the volume currents are responsible for the dissipation, and these are proportional to the time rate of change of the charge. Thus, the dissipation density reaches a saturation value as the frequency becomes very high. One tool used to form a picture of atomic, molecular, and domain physics is dielectric spectroscopy. Using this approach, the frequency dependence of the complex permittivity is used to gain insight into the microscopic structure. Magnetization, like polarization, can also be the source of dissipation. The time average dissipation density due to magnetization follows by taking the time averge of the third and fourth terms on the right in the basic power theorem, (11.2.7). Combined, these terms give /angbracketleftPd/angbracketright=/angbracketleftH·∂B ∂t/angbracketright (29) For small-signal applications, this source of dissipation is dealt with by in- troducing a complex permeability ˆ µsuch that ˆB= ˆµˆH. The role of the complex permeability is similar to that of the complex permittivity. The artiﬁcial diamag- netic material of Example 9.5.2 and Demonstration 9.5.1 can be used to exemplify the concept. Instead of perfectly conducting spheres that give rise to a magnetic moment instantaneously induced antiparallel to the applied ﬁeld, spherical shells of ﬁnite conductivity would be used. The dipole moment induced in the individual spherical shells would be deduced following the same approach as in Sec. 10.4. The resulting dipole moment would not be in phase with an applied sinusoidally varying magnetic ﬁeld. The derivation of an equivalent complex permeability would follow from the same line of reasoning as used in the previous example. Hysteresis Losses. Under periodic conditions in magnetizable solids, Band Hare related by the hysteresis curve described in Sec. 9.4 and illustrated again in Fig. 11.5.4. What time average power dissipation is implied by the hysteresis? As before, BandHare collinear. However, neither is now a single-valued function of the other. Evaluation of (29) is accomplished by breaking the cycle into two parts, each involving a single-valued relationship between BandH. The ﬁrst is the upswing “trajectory” from A→Cin Fig. 11.5.4. Over this half-cycle, which takes Bfrom BAtoBC, the trajectory is H+(B). With Btaken as BAwhen t= 0, it follows from (11.4.4) and (11.4.5) that ZT/2 0H·∂B ∂tdt=ZBC BAdWm dtdt=ZBC BAH+δB (30) This is the area under the curve of Hversus Bbetween AandCin Fig. 11.5.4, traversed on the “upswing.” A similar evaluation for the “downswing,” where the Sec. 11.6 Macroscopic Electrical Forces 33 Fig. 11.5.4 With the application of a sinusoidal magnetic ﬁeld intensity, a steady state is reached in which the hysteresis loop shown in the B−Hplane is traced out in the direction shown. The dashed area represents the energy density associated with upward traversal from AtoC. The dotted area inside the loop represents the energy density dissipated per traversal of the loop. trajectory is H−(B), gives ZT T/2H·∂B ∂tdt=ZBA BCH−δB=−ZBC BAH−δB (31) The time average power dissipation, (29), then is the sum of these two contributions divided by T. /angbracketleftPd/angbracketright=1 T£ZBC BAH+δB−ZBC BAH−δB¤ (32) Thus, the area within the hysteresis loop is the energy dissipated in one cycle. 11.6 ELECTRICAL FORCES ON MACROSCOPIC MEDIA Electrical forces on macroscopic materials have their origins in the forces exerted on the microscopic particles of which the materials are composed. Macroscopic ﬁelds have been used to describe conduction, polarization, and magnetization. In Chaps. 6, 7 and 9, polarization, current, and magnetization densities, respectively, were related to the macroscopic ﬁeld variables through constitutive laws. Typically, the parameters in these laws are determined from measurements. Thus, the experimen- tally determined relations make it unnecessary to take detailed account of how the microscopic ﬁelds are averaged. Because the deﬁnition of the average is already implicit in our macroscopic formulation of Maxwell’s equations, we must now take care that our use of macro- scopic ﬁeld quantities for representing electromagnetic forces is self-consistent. The 34 Energy, Power Flow, and Forces Chapter 11 Fig. 11.6.1 (a) Electroquasistatic system having one electrical terminal pair and one mechanical degree of freedom. (b) Schematic representation of EQS subsystem with coupling to external mechanical system represented by a me- chanical terminal pair. force on a macroscopic volume element ∆ Vof a material is the sum of the forces on the charged particles and magnetic dipoles constituting the material. Consider the simple case in which no magnetic dipoles are present. Then f=X iqi{E(ri) +vi×µoH(ri)} (1) where the summation is over all the charges within ∆ Vat their respective positions. Now, the ﬁelds E(ri) and H(ri) are the microscopic ﬁelds that vary greatly from point rito point rjin the material. The macroscopic ﬁelds E(r) and H(r) are averaged (smoothed) versions of these ﬁelds, whose sources are the averaged charge densities ρ≡X iqi ∆V(2) and J≡X iqivi ∆V(3) where the velocity viof the microscopic particles should be distinguished from that of the macroscopic material in which they are embedded or through which they move. The average of a product is not equal to the product of the averages. Thus, one could not ﬁnd the force density F=f/∆Vfrom the expression ρE+J×µoH, as the product of the averaged charge density and averaged electric ﬁeld plus averaged current density times averaged magnetic ﬂux density. Other methods have to be used to determine the force. One of the most useful is the energy method. Given the constitutive law for the material, which represents the interrelationship between macroscopic ﬁeld variables, conservation of energy provides a way of deducing the self-consistent force acting on the material. In this and the next section, we illustrate how total forces can be determined using conservation of energy as a premise . In this section, the EQS systems consid- ered have only one mechanical degree of freedom and only one electrical terminal pair. In the next section, MQS systems are considered and the approach is broad- ened to a somewhat more general class of systems. A parallel approach determines the force density rather than the total force. After expanding on microscopic forces in Sec. 11.8, we shall review macroscopic force densities in Sec. 11.9. Typical of the electroquasistatic problems considered in this section is the pair of metallic electrodes shown in Fig. 11.6.1. With the application of a voltage, Sec. 11.6 Macroscopic Electrical Forces 35 unpaired charges of opposite polarity are induced on the electrode surfaces. The electrical state of the system is speciﬁed by giving the geometry and the potential diﬀerence vbetween the electrodes. Here we picture one electrode as movable, with its position denoted by ξ. The two terminal pair system of Fig. 11.6.1b is useful to include mechanical eﬀects via an additional terminal pair. If we think of the net unpaired charge qon the electrode as an electrical terminal variable complementing v, then the force of electrical origin fcomplements the mechanical displacement ξ. Given the electrical terminal relation v=v(q, ξ), we now use an energy con- servation principle to determine the force f=f(q, ξ) that acts to increase the displacement ξ. The electrical terminal relation can either be regarded as a mea- sured function or be predicted using the macroscopic ﬁeld laws and constitutive laws for the materials within the “box.” It is now assumed that there is no conversion of electrical energy to thermal form within the box of Fig. 11.6.1b. Mechanisms for conversion of energy to heat are modeled by elements outside the box. For example, the ﬁnite conductivity of any dielectric is taken into account by a resistance external to the system. Thus, the electrical power input to what is deﬁned as the “box,” the electroquasistatic subsystem, must either result in a change in the electrical energy stored or mechan- ical power expended as the force facts on the mechanical system. The integral form of the power conservation theorem, (11.1.1), is generalized to include the rate of work by the force f −I SS·da=d dtZ VWedv+fdξ dt(4) In Sec. 11.4, we represented the quasistatic net electrical power input on the left in this expression in circuit theory terms. With the total energy wedeﬁned as the integral of the energy density over the entire volume of the system, (4) becomes vdq dt=dwe dt+fdξ dt(5) where the electrical power input is the product vi=vdq/dt . Multiplication of (4) bydtconverts a statement of power ﬂow to one of energy conservation. vdq=dwe+fdξ (6) If an increment of charge dqis placed on an electrode at potential v, an increment of energy vdqis added to the system that produces a change in the total stored energy dwe, an increment of work fdξdone on an external mechanical system, or some combination of both. Here, f(q, ξ) is the as yet unknown force. Solved for dwe, this energy conservation statement is dwe=vdq−fdξ (7) This expression describes what might be termed a quasistatic electrical and mechan- icalsubsystem . The state of this subsystem is speciﬁed by prescribing the geometry (ξ) and the charge on the electrode, for then the voltage of the electrode follows 36 Energy, Power Flow, and Forces Chapter 11 Fig. 11.6.2 Path of line integration in state space ( q, ξ) used to ﬁnd energy at location C. from the terminal relation v(q, ξ). The state of the subsystem is fully determined by the variables ( q, ξ), which are therefore regarded as independent variables . In terms of the two terminal pairs shown in Fig. 11.6.1b, one of each pair of terminal variables has been chosen as an independent variable. The incremental change in we(q, ξ) associated with incremental changes of dq anddξin the independent variables is dwe=∂we ∂qdq+∂we ∂ξdξ (8) Because qandξcan be independently speciﬁed, (7) and (8) must hold for any combination of dqand d ξ. For example, they must hold if the position of the electrode is held ﬁxed so that dξ= 0 and the charge is changed by the incremental amount dq. They must also describe the change in energy resulting from making an incremental displacement dξof the electrode under open circuit conditions, where dq= 0. Indeed, (7) and (8) hold if qandξare changed by arbitrary incremental amounts, and so it follows that the coeﬃcients of dqin (7) and (8) must be equal to each other, as must the coeﬃcients of d ξ. v=∂ ∂qwe(q, ξ); f=−∂ ∂ξwe(q, ξ) (9) Given the total energy, written in terms of the independent variables ( q, ξ), the second of these relations provides the desired force. Integration of the energy density over the volume of the system is one way to determine we. Another is to integrate (7) along a line in the state space ( q, ξ) designed so that the integral can be carried out without having to know f. we(q, ξ) =Z (vdq−fdξ) (10) Such a path4is shown in Fig. 11.6.2, where it is assumed that the force of electrical origin fis zero if the charge qis zero. Thus, in integrating along the contour q= 0 from A→B, dq = 0 and f= 0, so there is no contribution. The remainder of the integral, from B→C, is carried out with ξﬁxed, so dξ= 0, and (10) reduces to 4Note the analogy with the line integralR (Exdx+Eydy) of a two-dimensional conservation ﬁeld that results in the potential φ(x, y). Sec. 11.6 Macroscopic Electrical Forces 37 we=−Zξ 0f(0, ξ/prime)dξ/prime+Zq 0v(q/prime, ξ)dq/prime=Zq 0v(q/prime, ξ)dq/prime (11) We have accounted for the energy required to place the subsystem in the state ( q, ξ). In physical terms, the mathematical steps represent ﬁrst assembling the subsystem mechanically with no electrical excitation. Because there is no force acting on the electrode as it is put in place, no work is involved. Then, with its location ﬁxed, the electrode is charged by means of an electrical source. Suppose that the subsystem is electrically linear, so that either as a result of mathematical modeling or of measurements on the actual system, the electrical terminal relation takes the form v=q C(ξ)(12) Then, with this relation used to evaluate (11), it follows that the energy is we=Zq 0q/prime Cdq/prime=1 2q2 C(13) Finally, the desired force of electrical origin follows from substituting this expression into (9b). f=−1 2q2dC−1 dξ=1 2¡q C¢2dC dξ(14) Note that with a similar substitution into (9a), the terminal relation of (12) is obtained. Once the partial derivative with respect to ξhas been taken while holding the proper independent variable ( q) ﬁxed, the force can be written in terms of variables other than the independent ones. Thus, with the use of the terminal relation, (12), the force is written in terms of the terminal voltage vas f=1 2v2dC dξ (15) The following example gives the opportunity to apply this result to a speciﬁc conﬁguration. Example 11.6.1. Force on a Capacitor Plate The region between the plane parallel electrodes shown in Fig. 11.6.3 is ﬁlled by a layer of dielectric having permittivity /epsilon1and thickness band an air gap ξ. The total distance between electrodes, b+ξ, is small compared to the linear dimensions of the plates, so fringing ﬁelds will be ignored. Thus, the electric ﬁelds EaandEbin the air gap and in the dielectric, respectively, are uniform. What force on the upper electrode results from applying the voltage vbetween the electrodes? 38 Energy, Power Flow, and Forces Chapter 11 Fig. 11.6.3 Speciﬁc example of EQS systems having one electrical and one mechanical terminal pair. First we determine the charge qon the upper electrode. To this end, the integral of Efrom the upper electrode to the lower one must be equal to the applied voltage, so v=ξEa+bEb (16) Further, there is presumably no unpaired surface charge at the interface between the dielectric layer and the air gap. Thus, Gauss’ continuity condition requires that /epsilon1oEa=/epsilon1Eb (17) Elimination of Ebbetween these equations gives Ea=v ξ+b/epsilon1o /epsilon1(18) In terms of this electric ﬁeld at the surface of the upper electrode, Gauss’ continuity condition shows that the total charge on the upper electrode is q=DaA=/epsilon1oEaA (19) and so it follows from (18) that the electrical terminal relation can be written in terms of a capacitance C. q=Cv; C≡/epsilon1oA ξ+b/epsilon1o /epsilon1(20) Because the dielectric is described by a linear constitutive law, we have obtained an electrical terminal relation where vis a linear function of q. The force acting on the upper electrode follows from a substitution of (20) into (15). f=−1 2v2/epsilon1oA¡ ξ+b/epsilon1o /epsilon1¢2(21) By deﬁnition, if fis positive, it acts in the direction of ξ. Here we ﬁnd that regardless of the polarity of the applied voltage, fis negative. This is to be expected, because charges of one polarity on the upper electrode are attracted toward those of opposite polarity on the lower electrode. In describing energy conversion, a minus sign can be extremely important. For example, vdqis the incremental energy intothe electroquasistatic subsystem, while fdξis the energy leaving that subsystem as the force of electrical origin acts onthe external mechanical system. Thus, if fis positive, it acts on the mechanical system in such a direction as to increase the associated displacement. Sec. 11.6 Macroscopic Electrical Forces 39 Fig. 11.6.4 Apparatus used to demonstrate ampliﬁcation of voltage as the upper electrode is raised. (The electrodes are initially charged and then the voltage source is removed so q= constant.) The electrodes, consisting of foil mounted on insulating sheets, are about 1 m ×1 m, with the upper one insulated from the frame, which is used to control its position. The voltage is measured by the electrostatic voltmeter, which “loads” the system with a capacitance that is small compared to that of the electrodes and (at least on a dry day) a negligible resistance. Rotating motors and generators are examples where the conversion of energy between electrical and mechanical form is a cyclic process. In these cases, the sub- system returns to its original state once each cycle. The energy converted per cycle is determined by integrating the energy conservation law, (6), around the closed path in the state space representing this process. I vdq=I dwe+I fdξ (22) Because the energy stored in the system must return to its original value, there is no net contribution of the energy storage term in (22). For a cyclic process, the net electrical energy input per cycle must be equal to the net mechanical power output per cycle.I vdq=I fdξ (23) The following demonstration is primarily intended to give further insight into the implications of the conservation of energy principle for a cyclic process. Demonstration 11.6.1. An Energy Conversion Cycle The experiment shown in Fig. 11.6.4 is based on the plane parallel capacitor con- ﬁguration analyzed in Example 11.6.1. The lower electrode, aluminum foil mounted on a table top, is covered by a thin sheet of plastic. The upper one is also foil, but taped to an insulating sheet which is attached to a frame. This electrode can then be manually raised and lowered to eﬀectively control the displacement ξ. With the letters Athrough Dused to designate states of the system, we consider the following energy conversion cycle. •A→B. With v= 0, the upper electrode rests on the plastic sheet. A voltage Vois applied. •B→C. With the voltage source removed so that the upper electrode is electrically isolated, it is raised to the position ξ=L. •C→D. The upper electrode is shorted, so that its voltage returns to zero. •D→A. The upper electrode is returned to its original position at ξ= 0. 40 Energy, Power Flow, and Forces Chapter 11 Fig. 11.6.5 Closed paths followed in cyclic conversion of energy from mechanical to electrical form: (a) in ( q, v) plane; and (b) in ( f, ξ) plane. Is electrical energy converted to mechanical form, or vice versa? The process in carrying out the closed integrals on the left-hand and right- hand sides of (23) as the cycle is carried out can be pictured in the ( q, v) and ( f, ξ) planes, respectively, as shown in Fig. 11.6.5. From A→B, q =C(0)v, where C(ξ) is given by (20). Thus, the trajectory in the ( q, v) plane is a straight line ending at the voltage v=Voof the source. Because the upper electrode has remained at its original position, the trajectory in the ( f, ξ) plane is along the ξ= 0 axis. The force on the electrode caused by raising its voltage to Vofollows from (21). f=−1 2v2C2 /epsilon1oA⇒fB=−1 2[VoC(0)]2 /epsilon1oA(24) Now, from B→C, the voltage source is removed so that as the upper electrode is raised to ξ=L, its charge is conserved. This means that the trajectory in the (q, v) plane is one where q= constant = VoC(0). The voltage reached by the upper electrode can be found by requiring that qbe conserved. VoC(0) = V C(L)⇒V=VoC(0) C(L)(25) In the experiment, the thickness bof the dielectric sheet is a fraction of a millimeter, while the ﬁnal elevation ξ=Lmight be 20 cm. (If the displacement is larger than this, the fringing ﬁeld comes into play and the expression for the capacitance is no longer valid.) Thus, as the sheet is raised, an original voltage of 500 V is easily ampliﬁed to 10 - 20 kV. This is readily observed by means of an electrostatic voltmeter attached to the upper electrode, as shown in Fig. 11.6.4. To determine the trajectory B→Cin the ( f, ξ) plane, observe from (14) or (24) that as a function of q, the force is independent of ξ. f=−1 2q2 /epsilon1oA(26) The trajectory B→Cin the ( f, ξ) plane is therefore one of constant fBgiven by (24). In general, f(q, ξ) is not independent of ξ, but in plane parallel geometry, it is. The system is now returned to its original state in two steps. First, from C→D, the upper electrode remains at ξ=Land is shorted to ground. In the ( q, v) plane, the state returns to the origin along the straight line given by q=C(L)v, (20). In the ( f, ξ) plane, the force drops to zero with ξ=L. Second, from D→A, Sec. 11.6 Macroscopic Electrical Forces 41 the upper electrode is returned to its original position. The values of ( q, v) remain zero, while the trajectory in the ( f, ξ) plane is f= 0. The experiment is simple enough so that we can use physical reasoning to decide the direction of energy conversion. Although the force of gravity is likely to exceed the electrical force of attraction between the electrodes, as far as the electrical subsystem is concerned, the upper electrode is raised against a downward electrical force. Because the charge is removed before it is lowered, there is no electrical force on the electrode as it is lowered. Thus, net work is done on the EQS subsystem. The right-hand side of (23), the net work done by the subsystem on the external mechanical system, is thus negative. Evaluation of one or the other of the two sides of the energy conversion law, (23), provides two other ways to determine the direction of energy conversion. Con- sider ﬁrst the electrical input energy. The integral has contributions from A→B (where the source is used to charge the upper electrode) and from C→D(where the electrode is discharged). The areas under the respective triangles representing the integral of vdqare I vdq=ZB Avdq+ZD Cvdq=1 2C(0)V2 o−1 2C(L)V2(27) In view of the expression for C(ξ), (20), this expression can be written as I vdq=1 2C(0)V2 o· 1−C(0) C(L)¸ =−1 2C(0)V2 oL b/epsilon1 /epsilon1o(28) This expression is clearly negative, indicating that the net electrical energy ﬂow is out of the electrical terminal pair. This is consistent with having a net mechanical energy input to the system. The net mechanical output energy per cycle expressed by the right-hand side of (23) should be equal to (28). To see that this is so, we recognize that the integral consists of two possible contributions, from B→CandD→A. During the latter, f= 0, so the magnitude of the integral is simply the area of the rectangle in Fig. 11.6.5b. I fdξ=−1 2V2 oC2(0) /epsilon1oAL=−1 2C(0)V2 oL b/epsilon1 /epsilon1o(29) As required by the conservation law, this mechanical energy output is negative and is equal to the net electrical input energy given by (28). With the sequence of electrical and mechanical terminal constraints described above, there is a net conversion of energy from mechanical to electrical form. The system acts as an electrical generator with energy provided by whoever raises and lowers the upper electrode. With the voltage applied when the electrode is at its largest displacement, and the electrode grounded before it is raised, the energy ﬂow is from the electrical voltage source to the mechanical system. In this case, the system acts as a motor. We will have the opportunity to exemplify a practical motor in the next section. Most motors are MQS rather than EQS. However, a practical EQS device that is designed to convert energy from mechanical to electrical form is the capacitor microphone, a version of which was described in Example 6.3.3. Electrical forces have their origins in forces on unpaired charges and on dipoles. The force on the upper capacitor plate of Example 11.6.1 is due to the unpaired charges. The equal and opposite force on the combination of electrode and dielectric 42 Energy, Power Flow, and Forces Chapter 11 Fig. 11.6.6 Slab of dielectric partially extending between capacitor plates. The spacing, a, is much less than either bor the depth cof the system into the paper. Further, the upper surface at ξis many spacings aaway from the upper and lower edges of the capacitor plates, as is the lower surface as well. is in part due to unpaired charges and in part to dipoles induced in the dielectric. The following exempliﬁes a total force that is entirely due to polarization. Example 11.6.2. Force on a Dielectric Material We return to the conﬁguration of Example 11.4.1, where a dielectric slab extends a distance ξinto the region between plane parallel electrodes, as shown in Fig. 11.6.6. The capacitance was found in Example 11.4.1 to be (11.4.16). C=c a[/epsilon1ob+ξ(/epsilon1−/epsilon1o)] (30) It follows from (15) that there is a force tending to draw the dielectric into the region between the electrodes. f=1 2v2c a(/epsilon1−/epsilon1o) (31) This force results because dipoles induced by the fringing ﬁeld experience forces in theξdirection that are passed on to the material in which they are embedded. Even though the force is due to the fringing ﬁelds, the net force does not depend on the details of that ﬁeld. This is evident from our energy arguments because, at least as long as the upper edge of the slab is well within the region between the plates and the lower edge never reaches the vicinity of the electrodes, the energy storage in the fringing ﬁelds does not change when the slab is moved . Further discussion of the force density responsible for the force on the dielectric will be given in Sec. 11.9. Its physical reality is demonstrated next. Demonstration 11.6.2. Force on a Liquid Dielectric In the experiment shown in Fig. 11.6.7, capacitor plates are dipped into a dish full of dielectric liquid. Thus, with the application of the voltage, it rises against gravity. To demonstrate the relationship between the voltage and the force, the spacing between the electrodes is a slowly varying function of radial position. With rdenoting the radial distance from an axis where an extension of the electrodes would join, and α Sec. 11.6 Macroscopic Electrical Forces 43 Fig. 11.6.7 In a demonstration of the polarization force, a pair of conducting transparent electrodes are dipped into a liquid (corn oil dyed with food coloring). They are closer together at the upper right than at the lower left, so when a voltage is applied, the electric ﬁeld intensity decreases with increasing distance, r, from the apex. As a result, the liquid is seen to rise to a height that varies as 1 /r2. The electrodes are about 10 cm ×10 cm, with an electric ﬁeld exceeding the nominal breakdown strength of air at atmospheric pressure, 3 ×106V/m. The experiment is therefore carried out under pressurized nitrogen. the angle between the electrodes, the spacing at a distance risαr. Thus, in (31), the spacing between electrodes a→αrand the force per unit radial distance tending to push the liquid upward is a function of the radial position. f c=1 2v2 αr(/epsilon1−/epsilon1o) (32) This force must raise a column of liquid having a height ξand width αr. With the mass density of the liquid deﬁned as ρ, the total mass per radial distance raised by this force is therefore αrξρ. Force equilibrium is therefore represented by setting the force per unit radial length equal to this mass multiplied by the gravitational acceleration g. 1 2v2 αr(/epsilon1−/epsilon1o) =αrξρg (33) This expression can be solved for ξ(r). ξ=1 2v2(/epsilon1−/epsilon1o) α2ρg1 r2(34) In the experiment,5the electrodes are constructed from tin-oxide coated glass. They are then both conducting and transparent. As a result, the height to which the liquid rises can be seen to obey (34), both as to its magnitude and its radial dependence on r. 5See ﬁlm Electric Fields and Moving Media from series by National Committee for Electrical Engineering Films, Education Development Center, 39 Chapel St., Newton, Mass. 02160. 44 Energy, Power Flow, and Forces Chapter 11 Fig. 11.7.1 (a) Magnetoquasistatic system with two electrical terminal pairs and one mechanical degree of freedom. (b) MQS subsystem representing (a). To obtain an appreciable rise of the liquid without exceeding the ﬁeld strength for electrical breakdown between the electrodes, the atmosphere over the liquid must be pressurized. Also, to avoid eﬀects of unpaired charges injected at high ﬁeld strengths by the electrodes, the applied voltage is alternating and, because the force is proportional to the square of the applied ﬁeld, the height of rise is proportional to the rms value of the voltage. 11.7 MACROSCOPIC MAGNETIC FORCES In this section, an energy principle is applied to the determination of net forces in MQS systems. With Sec. 11.6 as background, it is appropriate to include the case of multiple terminal pairs. As in Sec. 11.4, the coenergy is again found to be a convenient alternative to the energy. The MQS system is shown schematically in Fig. 11.7.1. It has two electrical terminal pairs and one mechanical degree of freedom. The magnetoquasistatic sub- system now described by an energy principle excludes electrical dissipation and all aspects of the mechanical system, mechanical energy storage and dissipation. The energy principle then states that the input of electrical power through the electrical terminal pairs either goes into a rate of change of the stored magnetic energy or into a rate of change of the work done on the external mechanical world. v1i1+v2i2=dwm dt+fdξ dt(1) As in Sec. 11.6, our starting point in ﬁnding the force is a postulated principle of energy conservation. Because the system is presumably MQS, in accordance with (11.3.29), the left-hand side represents the net ﬂux of power into the system. With the addition of the last term and the inherent assumption that there is no electrical dissipation in the subsystem being described, (1) is more than the recasting of Poynting’s theorem. In an MQS system, the voltages are the time rates of change of the ﬂux linkages. With these derivatives substituted into (1) and the expression multiplied bydt, it becomes Sec. 11.7 Macroscopic Magnetic Forces 45 i1dλ1+i2dλ2=dwm+fdξ (2) This energy principle states that the increments of electrical energy put into the MQS subsystem (as increments of ﬂux dλ1anddλ2through the terminals multiplied by their currents i1andi2, respectively) either go into the total energy, which is increased by the amount dwm, or into work on the external mechanical system, subject to the force fand experiencing a displacement dξ. With the energy principle written as in (2), the ﬂux linkages are the indepen- dent variables. We saw in Example 11.4.2 that it is inconvenient to specify the ﬂux linkages as functions of the currents. With the objective of casting the currents as the independent variables, we now recognize that i1dλ1=d(i1λ1)−λ1di1; i2dλ2=d(i2λ2)−λ2di2 (3) and substitute into (2) to obtain dw/prime m=λ1di1+λ2di2+fdξ (4) where the coenergy function, seen before in Sec. 11.4, is deﬁned as w/prime m=i1λ1+i2λ2−wm (5) We picture the MQS subsystem as having ﬂux linkages λ1andλ2, a force fand a total energy wmthat are speciﬁed once the currents i1andi2and the displacement ξare stipulated. According to (4), the coenergy is a function of the independent variables i1, i2, and ξ,w/prime m=w/prime m(i1, i2, ξ), and the change in w/prime mcan also be written as dw/prime m=∂w/prime m ∂i1di1+∂w/prime m ∂i2di2+∂w/prime m ∂ξdξ (6) Because the currents and displacement are independent variables, (4) and (6) can hold only if the coeﬃcients of like terms on the right are equal. Thus, λ1=∂w/prime m ∂i1; λ2=∂w/prime m ∂i2; f=∂w/prime m ∂ξ (7) The last of these three expressions is the key to ﬁnding the force f. Reciprocity Condition. Before we ﬁnd the coenergy and hence f, consider the implication of the ﬁrst two expressions in (7) for the electrical terminal relations. Taking the derivative of λ1with respect to i2, and of λ2with respect to i1, shows that ∂λ1 ∂i2=∂2w/prime m ∂i1∂i2=∂2w/prime m ∂i2∂i1=∂λ2 ∂i1 (8) 46 Energy, Power Flow, and Forces Chapter 11 Although this reciprocity condition must reﬂect conservation of energy for any loss- less system, magnetically linear or not, consider its implications for a system de- scribed by the linear terminal relations. · λ1 λ2¸ =· L11L12 L21L22¸· i1 i2¸ (9) Application of (8) shows that energy conservation requires the equality of the mu- tual inductances. L12=L21 (10) This relation has been derived in Example 11.4.2 from a related but diﬀerent point of view. Finding the Coenergy. To ﬁnd w/prime m, we integrate (4) along a path in the state space ( i1, i2, ξ) arranged so that the integral can be carried out without having to know f. w/prime m=Z (λ1di1+λ2di2+fdξ) (11) Thus, the ﬁrst leg of the line integral is carried out on ξwith the currents equal to zero. Provided that f= 0 in the absence of these currents, this means that the integral of fdξmakes no contribution. The payoﬀ from our formulation in terms of the coenergy rather than the energy comes in being able to carry out the remaining integration using terminal relations in which the ﬂux linkages are expressed in terms of the currents. For the linear terminal relations of (9), this line integration was illustrated in Example 11.4.2, where it was found that w/prime m=1 2L11i2 1+L12i1i2+1 2L22i2 2 (12) Evaluation of the Force. In general, the inductances in this expression are functions of ξ. Thus, the force ffollows from substituting this expression into (7c). f=1 2i2 1dL11 dξ+i1i2dL12 dξ+1 2i2 2dL22 dξ (13) Of course, this expression applies to systems having a single electrical terminal pair as a special case where i2= 0. This generalization of the energy method to multiple electrical terminal pair systems suggests how systems with two or more mechanical degrees of freedom are treated. Sec. 11.7 Macroscopic Magnetic Forces 47 Fig. 11.7.2 Cross-section of axially symmetric transducer similar to the ones used to drive dot matrix printers. Example 11.7.1. Driver for a Matrix Printer A transducer that is similar to those used to drive an impact printer is shown in Fig. 11.7.2. The device, which is symmetric about the axis, might be one of seven used to drive wires in a high-speed matrix printer. The objective is to transduce a current ithat drives the N-turn coil into a longitudinal displacement ξof the permeable disk at the top. This disk is attached to one end of a wire, the other end of which is used to impact the ribbon against paper, imprinting a dot. The objective here is to determine the force facting on the plunger at the top. For simplicity, we make a highly idealized model in which the magnetizable material surrounding the coil and ﬁlling its core, as well as that of the movable disk, is regarded as perfectly permeable. Moreover, the air gap spacing ξis small compared to the radial dimension a, so the magnetic ﬁeld intensity is approximated as uniform in the air gap. The wire is so ﬁne that the magnetizable material removed to provide clearance for the wire can be disregarded. With HaandHcdeﬁned as shown by the inset to Fig. 11.7.2, Amp` ere’s integral law is applied to a contour passing upward through the center of the core, across the air gap at the center, radially outward in the disk, and then downward across the air gap and through the outer part of the stator to encircle the winding in the inﬁnitely permeable material. Haξ+Hcξ=Ni (14) A second relation between HaandHcfollows from requiring that the net ﬂux out of the disk must be zero. µoHaπ(a2−b2) =µoHcπc2(15) Using this last expression to replace Hcin (14) results in Hc=Ni ξ¡ 1 +c2 a2−b2¢ (16) 48 Energy, Power Flow, and Forces Chapter 11 Fig. 11.7.3 Cross-section of perfectly conducting current-carrying wire over a perfectly conducting ground plane. The magnetic ﬂux linking every turn in the coil is µoHcπc2. Thus, the total ﬂux linked by the coil is λ=Li; L=µoN2πc2 ξ¡ 1 +c2 a2−b2¢ (17) Finally, the force follows from an evaluation of (13) (specialized to the single terminal pair system of this example). f=1 2i2dL dξ=−1 2i2µoN2πc2 ξ2¡ 1 +c2 a2−b2¢ (18) As might have been expected by one who has observed magnetizable materials pulled into a magnetic ﬁeld, the force is negative. Given the deﬁnition of ξin Fig. 11.7.2, the application of a current will tend to close the air gap. To write a dot, the current is applied. To provide for a return of the plunger to its original position when the current is removed, a spring is inserted in the air gap. In the magnetic transducer of the previous example, the force on the driver disk is due to magnetization. The next example illustrates the force associated with the current density. Example 11.7.2. Force on a Wire over a Perfectly Conducting Plane The cross-section of a perfectly conducting wire with its center a distance ξabove a perfectly conducting ground plane is shown in Fig. 11.7.3. The conﬁguration is familiar from Demonstration 8.6.1. The current carried by the wire is returned in the ground plane. The distribution of this current on the surfaces of the wire and ground plane is consistent with the requirement that there be no ﬂux density normal to the perfectly conducting surfaces. What is the force per unit length facting on the wire? The inductance per unit length is half of that for a pair of conductors having the center-to-center spacing 2 ξ. Thus, it is half of that given by (8.6.12). L=µo 2πln· ξ R+r¡ξ R¢2−1¸ (19) Sec. 11.7 Macroscopic Magnetic Forces 49 Fig. 11.7.4 The force tending to levitate the wire of Fig. 11.7.3 as a function of the distance to the ground plane normalized to the radius R of wire. The force per unit length in the ξdirection then follows from an evaluation of (13) (again adapted to the single terminal pair situation). f=1 2i2dL dξ=fo1p (ξ/R)2−1; fo=µoi2 4πR(20) The dependence of this force on the elevation above the ground plane is shown in Fig. 11.7.4. In the limit where the elevation is large compared to the radius of the conductor, (20) becomes f→1 4µoi2 πξ(21) In Sec. 11.8, we will identify the force density acting on materials carrying a current density Jas being J×µoH. Note that the upward force predicted by (20) is indeed consistent with the direction of this force density. The force on a thin wire, (21), can be derived from this force density by recognizing that the contribution of the self-ﬁeld of the wire to the total force per unit length is zero. Thus, the force per unit length can be computed using for B the ﬂux density caused by the image current a distance 2 ξaway. The ﬂux density due to this image current has a magnitude that follows from Amp` ere’s integral law asµoi/2π(2ξ). This ﬁeld is essentially uniform over the cross-section of the wire, so the integral of the force density J×Bover the cross-section of the wire amounts to an integration of the current density Jover the cross-section. The latter is the total current i, and so we are led to a force per unit length of magnitude µoi2/2π(2ξ), which is in agreement with (21). The following is a demonstration of the force on current-carrying conductors exempliﬁed previously. It also provides a dramatic demonstration of the existence of induced currents. Demonstration 11.7.1. Steady State Magnetic Levitation In the experiment shown in Fig. 11.7.5, the current-carrying wire of the previous example has been wound into a pancake shaped coil that is driven by about 20 amps of 60 Hz current. The conductor beneath is an aluminum sheet of 1.3 cm thickness. Even at 60 Hz, this conductor tends to act as a perfect conductor. This follows from 50 Energy, Power Flow, and Forces Chapter 11 Fig. 11.7.5 When the pancake coil is driven by an ac current, it ﬂoats above the aluminum plate. In this experiment, the coil consists of 250 turns of No. 10 copper wire with an outer radius of 16 cm and an inner one of 2.5 cm. The aluminum sheet has a thickness of 1.3 cm. With a 60 Hz current iof about 20 amp rms, the height above the plate is 2 cm. Fig. 11.7.6 (a) The force f, acting through a lever-arm of length r, produces a torque τ=rf. (b) Mechanical terminal pair representing a rotational degree of freedom. an evaluation of the product of the angular frequency ωand the time constant τm estimated in Sec. 10.2. From (10.2.17), ωτm=ωµoσ∆a≈20, where the average radius is a= 9 cm, ∆ is the sheet thickness and the sheet conductivity is given by Table 7.1.1. The time average force, of the type described in Example 11.7.2, is suﬃciently large to levitate the coil. As the current is increased, its height above the aluminum sheet increases, as would be expected from the dependence of the force on the height for a single wire, Fig. 11.7.4. The Torque of Electrical Origin. In some of the most important transducers, the mechanical response takes the form of a rotation rather than a translation. The shaft shown in Fig. 11.7.6a might be attached to the rotor of a motor or generator. A force facting through a lever arm of length rthat rotates the shaft through an incremental angle dθcauses a displacement dξ=rdθ. Thus, the incremental work done on the mechanical system fdξbecomes fdξ→τdθ (22) where τis deﬁned as the torque . If the two terminal pair MQS system of Fig. 11.7.1 had a rotational rather than a displacement degree of freedom, the representation would be the same as has been outlined, except that f→τandξ→θ. The mechanical terminal pair is now represented as in Fig. 11.7.6b. Sec. 11.7 Macroscopic Magnetic Forces 51 Fig. 11.7.7 Cross-section of rotating machine. The torque follows from (13) as τ=1 2i2 1dL11 dθ+i1i2dL12 dθ+1 2i2 2dL22 dθ(23) Among the types of magnetic rotating motors and generators that could be used to exemplify the torque of (23), we now choose a synchronous machine. Al- though other types of motors are more common, it is a near certainty that if these words are being read with the aid of electrical illumination, the electricity used is being generated by means of a synchronous generator. Example 11.7.3. A Synchronous Machine The cross-section of a stator and rotor modeling a rotating machine is shown in Fig. 11.7.7. The rotor consists of a highly permeable circular cylindrical material mounted on a shaft so that it can undergo a rotation measured by the angle θ. Surrounding this rotor is a stator, composed of a highly permeable material. In slots, on the inner surface of the stator and on the outer surface of the rotor, respectively, are windings with sinusoidally varying turn densities. These windings, driven by the currents i1 andi2, respectively, give rise to current distributions that might be modeled by surface current densities Kz=i1Nssinφatr=a (24) Kz=i2Nrsin(φ−θ) at r=b (25) where NsandNrare constants descriptive of the windings. Thus, the current dis- tribution shown on the stator in Fig. 11.7.7 is ﬁxed and gives rise to a magnetic ﬁeld having the ﬁxed vertical axis shown in the ﬁgure. The rotor coil gives rise to a similar ﬁeld except that its axis is at the angle θof the rotor. The rotor magnetic axis, also shown in Fig. 11.7.7, therefore rotates with the rotor. Electrical Terminal Relations. With the rotor and stator materials taken as inﬁnitely permeable, the air gap ﬁelds are determined by using (24) and (25) to 52 Energy, Power Flow, and Forces Chapter 11 write boundary conditions on the tangential Hand then solving Laplace’s equation for the air gap magnetic ﬁelds (Secs. 9.6 and 9.7). The ﬂux linked by the respective coils is then of the form hλ1 λ2i =hL11 L12(θ) L12(θ)L22ihi1 i2i =hLs Mcosθ Mcosθ L rihi1 i2i (26) where the self-inductances LsandLrand peak mutual inductance Mare constants. The dependence of the inductance matrix on the angle of the rotor, θ, can be reasoned physically. The rotor is modeled as a smooth circular cylinder, so in the absence of a rotor current i2, there can be no eﬀect of the rotor angle θon the ﬂux linked by the stator winding. Hence, the stator self-inductance is independent of θ. Similar reasoning shows that the rotor self-inductance must be independent of rotor angle θ. The θdependence of the mutual inductance is plausible because the ﬂux λ1 linked by the stator, due to the current in the rotor, must peak when the magnetic axes of the coils are aligned ( θ= 0) and must be zero when they are perpendicular (θ= 90 degrees). Torque Evaluation. The magnetic torque on the rotor follows directly from using (26) to evaluate (23). τ=i1i2d dθMcosθ=−i1i2Msinθ (27) This torque depends on the currents and θin such a way that the magnetic axis of the rotor tends to align with that of the stator. With θ= 0, the axes are aligned and there is no torque. If θis slightly positive and the currents are both positive, the torque is negative. This is as would be expected with the magnetic axis of the stator vertical and that of the rotor in the ﬁrst quadrant (as in Fig. 11.7.7). Synchronous Operation. In the synchronous mode of operation, the stator current is constrained to be sinusoidal while that on the rotor is a constant. To avoid having to describe the mechanical system, we will assume that the shaft is attached to a mechanical load that makes the angular velocity Ω constant. Thus, the electrical and mechanical terminals are constrained so that i1=I1cosωt i2=I2 (28) θ= Ωt−γ Under what circumstances can we derive a time average torque on the shaft, and hence a net conversion of energy with each rotation? With the constraints of (28), the torque follows from (27) as τ=−I1I2Mcosωtsin(Ω t−γ) (29) Sec. 11.7 Macroscopic Magnetic Forces 53 Fig. 11.7.8 (a) Time average torque as a function of angle γ. (b) The rotor magnetic axis lags the clockwise rotating component of the stator magnetic ﬁeld axis by the angle γ. A trigonometric identity6makes the implications of this result more apparent. τ=−I1I2M 2© sin[(ω+ Ω)t−γ] + sin[ −(ω−Ω)t−γ]ª (30) There is no time average value of either of these sinusoidal functions of time unless one or the other of the frequencies, ( ω+ Ω) and ( ω−Ω), is zero. For example, with the rotation frequency equal to that of the excitation, ω= Ω (31) the time average torque is /angbracketleftτ/angbracketright=I1I2M 2sinγ (32) The dependence of the time average torque on the phase angle γis shown in Fig. 11.7.8a. With γbetween 0 and 180 degrees, there is a positive time average torque acting on the external mechanical system in the direction of rotation Ω. In this range, the machine acts as a motor to convert energy from electrical to mechanical form. In the range of γfrom 180 degrees to 360 degrees, energy is converted from mechanical to electrical form and operation is as a generator. Stator Field Analyzed into Traveling Waves. From (30), it is clear that a time-average torque results from either a forward (Ω = ω) or a backward (Ω = −ω) rotation. This suggests that the ﬁeld produced by the stator winding is the superposition of ﬁelds having magnetic axes rotating in the clockwise and counterclockwise directions. Formally, this can be seen by rewriting the stator surface current density, (24), using the electrical and mechanical constraints of (28). With the use once again of the double-angle trigonometric identity, the distribution of surface current density is separated into two parts. Kz=NsI1cosωtsinφ=NsI1 2[sin(φ+ωt) + sin( φ−ωt)] (33) 6sin(x) cos( y) =1 2(sin(x+y) + sin( x−y)) 54 Energy, Power Flow, and Forces Chapter 11 Fig. 11.7.9 With the sinusoidally distributed stator current excited by a current that varies sinusoidally with time, the surface current is a standing wave which can be analyzed into the sum of oppositely prop- agating traveling waves. The sinusoidal excitation produces a standing-wave surface current with nodes at θ= 0 and 180 degrees. This is the ﬁrst distribution in Fig. 11.7.9. Analyzed as it is on the right in (33), and pictured in Fig. 11.7.9, it is the sum of two countertraveling waves. The magnetic axis of the wave traveling to the right is at φ=ωt. We now have the following picture of the synchronous operation found to give rise to the time average torque. The ﬁeld of the stator is composed of rotating parts, one with a magnetic axis that rotates in a clockwise direction at angular velocity ω, and the other rotating in the opposite direction. The frequency condition of (31) therefore represents a synchronous condition in which the “forward” component of the stator ﬁeld and the magnetic axis of the rotor rotate at the same angular velocity. In view of the deﬁnition of γgiven in (28), if γis positive, the rotor magnetic axis lags the stator axis by the angle γ, as shown in Fig. 11.7.8. When the machine operates as a motor, the forward component of the stator magnetic ﬁeld “pulls” the rotor along. When the device operates as a generator, γis negative and the rotor magnetic axis leads that of the forward component of the stator ﬁeld. For generator operation, the rotor magnetic axis “pulls” the forward component of the stator ﬁeld. 11.8 FORCES ON MICROSCOPIC ELECTRIC AND MAGNETIC DIPOLES The energy principle was used in the preceding sections to derive the macroscopic forces on polarizable and magnetizable materials. The same principle can also be applied to derive the force distributions, the force densities. For this purpose, one needs more than a purely electromagnetic description of the system. In order to develop the simple model for the force density distribution, we need the expression for the force on an electric dipole for polarizable media, and on a magnetic dipole for magnetizable media. The force on an electric dipole will be derived simply from the Lorentz force law. We have not stated a corresponding force law for magnetic charges. Even though these are not found in nature as isolated charges but only Sec. 11.8 Microscopic Forces 55 Fig. 11.8.1 An electric dipole experiences a net electric force if the positive charge qis subject to an electric ﬁeld E(r+d) that diﬀers from E(r) acting on the negative charge q. as dipoles, it is nevertheless convenient to state such a law. This will be done by showing how the electric force law follows from the energy principle. By analogy a corresponding law on magnetic charges will be derived from which the force on a magnetic dipole will follow. Force on an Electric Dipole. The force on a stationary electric charge is given by the Lorentz law with v= 0. f=qE (1) A dipole is the limit of two charges of equal magnitude and opposite sign spaced a distance dapart, in the limit limq→∞ |d|→0(qd) =p withpbeing ﬁnite. Charges qof opposite polarity, separated by the vector distance d, are shown in Fig. 11.8.1. The total force on the dipole is the sum of the forces on the individual charges. f=q[E(r+d)−E(r)] (2) Unless the electric ﬁeld at the location r+dof the positive charge diﬀers from that at the location rof the negative charge, the separate contributions cancel. In order to develop an expression for the force on the dipole in the limit where the spacing dof the charges is small compared to distances over which the ﬁeld varies appreciably, (2) is written in Cartesian coordinates and the ﬁeld at the positive charge expanded about the position of the negative charge. Thus, the x component is fx=q[Ex(x+dx, y+dy, z+dz)−Ex(x, y, z )] =q· Ex(x, y, z ) +dx∂Ex ∂x+dy∂Ex ∂y+dz∂Ex ∂z +. . .−Ex(x, y, z )¸(3) 56 Energy, Power Flow, and Forces Chapter 11 Fig. 11.8.2 Dipole having ydirection and positioned on the xaxis in ﬁeld of (6) experiences force in the xdirection. The ﬁrst and last terms cancel. In more compact notation, this expression is there- fore fx=p· ∇Ex (4) where we have identiﬁed the dipole moment p≡qd. The other force components follow in a similar fashion, with yandzplaying the role of x. The three components are then summarized in the vector expression f=p· ∇E (5) The derivation provides an explanation of how p·∇Eis evaluated in Cartesian coordinates. The i-th component of (5) is obtained by dotting pwith the gradient of the i-th component of E. Illustration. Force on a Dipole Suppose that a dipole ﬁnds itself in the ﬁeld Φ =−Voxy a2;E=−∇Φ =Vo a2(yix+xiy) (6) which is familiar from Example 4.1.1. It follows from (5) that the force is f=Vo a2(pyix+pxiy) (7) According to this expression, the y-directed dipole on the xaxis in Fig. 11.8.2 experiences a force in the xdirection. The y-directed force is zero because Eyis the same at the respective locations of the charges. The x-directed force exists because Exgoes from being positive just above the xaxis to negative just below. Thus, the x-directed contributions to the force of each of the charges is in the same direction. Sec. 11.8 Microscopic Forces 57 Fig. 11.8.3 (a) Electric charge brought into ﬁeld created by permanent po- larization. (b) Analogous magnetic charge brought into ﬁeld of permanent magnet. Force on Electric Charge Derived from Energy Principle. The force on an electric charge is stated in the Lorentz law. This law is also an ingredient in Poynting’s theorem, and in the identiﬁcation of energy and power ﬂow. Indeed, E·Juwas recognized from the Lorentz law as the power density imparted to the current density of unpaired charge. The energy principle can be used to derive the force law on a microscopic charge “in reverse”. This seems to be the hard way to obtain the Lorentz law of force on a stationary charge. Yet we go through the derivation for three reasons. (a) The derivation of force from the EQS energy principle is shown to be consistent with the Lorentz force on a stationary charge. (b) The derivation shows that the ﬁeld can be produced by permanently polarized material objects, and yet the energy principle can be employed in a straight- forward manner. (c) The same principle can be applied to derive the microscopic MQS force on a magnetic charge. Let us consider an EQS ﬁeld produced by charge distributions and permanent polarizations Ppin free space as sketched in Fig. 11.8.3a. By analogy, we will then have found the force on a magnetic charge in the ﬁeld of a permanent magnet, Fig. 11.8.3b. The Poynting theorem identiﬁes the rate of energy imparted to the polarization per unit volume as rate of change of energy volume=E·∂ ∂t/epsilon1oE+E·∂Pp ∂t(8) Because Ppis a permanent polarization, ∂Pp/∂t= 0, and the permanent polarization does not contribute to the change in energy associated with introducing a point charge. Hence, as charge is brought into the vicinity of the permanent polarization, the change of energy density is δWe=E·δ(/epsilon1oE) (9) where δstands for the diﬀerential change of /epsilon1oE. The change of energy is δwe=Z VdvE·δ(/epsilon1oE) (10) where Vincludes all of space. The electroquasistatic Eﬁeld is the negative gradient of the potential E=−∇Φ (11) 58 Energy, Power Flow, and Forces Chapter 11 Introducing this into (10), one has δwe=−Z Vdv∇Φ·δ(/epsilon1oE) =−Z Vdv∇ ·[Φδ(/epsilon1oE)] +Z VdvΦ∇ ·(δ/epsilon1oE)(12) where we have “integrated by parts,” using an identity.7The ﬁrst integral can be written as an integral over the surface enclosing the volume V. Since Vis all of space, the surface is at inﬁnity. Because EΦ vanishes at inﬁnity at least as fast as 1 /r3 (1/r2forE,1/rfor Φ, where ris the distance from the origin of a coordinate system mounted within the electroquasistatic structure), the surface integral vanishes. Now ∇ ·δ(/epsilon1oE) =δρu (13) from Gauss’ law, where δρuis the change of unpaired charge. Thus, from (12), δwe=Z VΦδρudv (14) so the change of energy is equal to the charge increment δρudvintroduced at r times the potential Φ at r, summed over all the charges. Suppose that one introduces only a small test charge q, so that ρudv=qat point r. Then δwe(r) =qΦ(r) (15) The change of energy is the potential Φ at the point at which the charge is in- troduced times the charge. This form of the energy interprets the potential of an EQS ﬁeld as the work to be done in bringing a charge from inﬁnity to the point of interest. If the charge is introduced at r+∆r, then the change in total energy associated with introducing that charge is δwe(r+ ∆r) =q[Φ(r) + ∆r· ∇Φ(r)] (16) Introduction of a charge qatr, subsequent removal of the charge, and introduction of the charge at r+ ∆ris equivalent to the displacement of the charge from rto r+ ∆r. If there is a net energy decrease, then work must have been done by the forcefexerted by the ﬁeld on the charge. The work done by the ﬁeld on the charge is −[δwe(r+ ∆r)−δwe(r)] =−q∆r· ∇Φ(r) = ∆ r·f (17) and therefore f=−q∇Φ =qE (18) Thus, the Lorentz law for a stationary charge is implied by the EQS laws. Before we attack the problem of force on a magnetic charge, we explore some features of the electroquasistatic case. In (17), qis a small test charge. Electric test 7(∇ψ)·A=∇ ·(Aψ)−(∇ ·A)ψ Sec. 11.8 Microscopic Forces 59 charges are available as electrons. But suppose that in analogy with the magnetic case, no free electric charge was available. Then one could still produce a test charge by the following artiﬁce. One could polarize a very long-thin rod of cross-section a, with a uniform polarization density Palong the axis of the rod (Sec. 6.1). At one end of the rod, there would be a polarization charge q=Pa, at the other end there would be a charge of equal magnitude and opposite sign. If the rod were of very long length, while the end with positive charge could be used as the “test charge,” the end of opposite charge would be outside the ﬁeld and experience no force. Here the charge representing the polarization of the rod has been treated as unpaired. We are now ready to derive the force on a magnetic charge. Force on a Magnetic Charge and Magnetic Dipole. The attraction of a magnetizable particle to a magnet is the result of the force exerted by a magnetic ﬁeld on a magnetic dipole. Even in this case, because the particle is macroscopic, the force is actually the sum of forces acting on the microscopic atomic constituents of the material. As pointed out in Secs. 9.0 and 9.4, the magnetization characteristics of macroscopic media such as the iron particle relate back to the magnetic moment of molecules, atoms, and even individual electrons. Given that a particle has a magnetic moment mas deﬁned in Sec. 8.2, what is the force on the particle in a magnetic ﬁeld intensity H? The particle can be comprised of a macroscopic material such as a piece of iron. However, to distinguish between forces on macroscopic media and microscopic particles, we should consider here that the force is on an elementary particle, such as an atom or electron. We have shown how one derives the force on an electric charge in an elec- tric ﬁeld from the energy principle. The electric ﬁeld could have been produced by permanently polarized dielectric bodies. In analogy, one could produce a magnetic ﬁeld by permanently magnetized magnetic bodies. In the EQS case, the test charge could have been produced by a long, uniformly and permanently polarized cylin- drical rod. In the magnetic case, an “isolated” magnetic charge could be produced by a long, uniformly and permanently magnetized rod of cross-sectional area a. If the magnetization density is M, then the analogy is ∇ ·P↔ ∇ · µoM, Φ↔Ψ, q E↔qmH (19) where, for the uniformly magnetized rod, and the magnetic charge qm=µoMa (20) is located at one end of the rod, the charge −qmat the other end of the rod (Example 9.3.1). The force on a magnetic charge is thus, in analogy with (18), f=qmH (21) which is the extension of the Lorentz force law for a stationary electric charge to the magnetic case. Of course, the force on a dipole is, in analogy to (5) (see Fig. 11.8.4), 60 Energy, Power Flow, and Forces Chapter 11 Fig. 11.8.4 Magnetic dipole consisting of positive and negative magnetic charges qm. f=qmd·H=µom· ∇H (22) where mis the magnetic dipole moment. We have seen in Example 8.3.2 that a magnetic dipole of moment mcan be made up of a circulating current loop with magnitude m=ia, where iis the current andathe area of the loop. Thus, the force on a current loop could also be evaluated from the Lorentz law for electric currents as f=iZ ds×µoH (23) with ithe total current in the loop. Use of vector identities indeed yields (22) in the MQS case. Thus, this could be an alternate way of deriving the force on a magnetic dipole. We prefer to derive the law independently via a Lorentz force law for stationary magnetic charges, because an important dispute on the validity of the magnetic dipole model rested on the correct interpretation of the force law[1−3]. While the details of the dispute are beyond the scope of this textbook, some of the issues raised are fundamental and may be of interest to the reader who wants to explore how macroscopic formulations of the electrodynamics of moving media based on magnetization represented by magnetic charge (Chap. 9) or by circulating currents are reconciled. The analogy between the polarization and magnetization was emphasized by Prof. L. J. Chu[2], who taught the introductory electrical engineering course in electromagnetism at MIT in the ﬁfties. He derived the force law for moving magnetic charges, of which (21) is the special case for a stationary charge. His approach was soon criticized by Tellegen[3], who pointed out that the accepted model of magnetization is that of current loops being the cause of magnetization. While this in itself would not render the magnetic charge model invalid, Tellegen pointed out that the force computed from (23) in a dynamic ﬁeld does not lead to (22), but to f=µom· ∇H−µo/epsilon1om×∂E ∂t(24) Because the force is diﬀerent depending upon whether one uses the magnetic charge model or the circulating current model for the magnetic dipole, so his reasoning went, and because the circulating current model is the physically correct one, the Sec. 11.8 Microscopic Forces 61 magnetic charge model is incorrect. The issue was ﬁnally settled[4]when it was shown that the force (24) as computed by Tellegen was incorrect. Equation (23) assumes that icould be described as constant around the current loop and pulled out from under the integral. However, in a time-varying electric ﬁeld, the charges induced in the loop cause a current whose contribution precisely cancels the second term in (24). Thus, both models lead to the same force on a magnetic dipole and it is legitimate to use either model. The magnetic model has the advantage that a stationary dipole contains no “moving parts,” while the current model does con- tain moving charges. Hence the circulating current formalism is by necessity more complicated and more likely to lead to error. Comparison of Coulomb’s Force on an Electron to the Force on its Mag- netic Dipole. Why is it possible to accurately describe the motions of an electron in vacuum by the Lorentz force law without including the magnetic force associ- ated with its dipole moment? The answer is that the magnetic dipole eﬀect on the electron is relatively small. To obtain an estimate of the magnitude of the mag- netic dipole eﬀect, we compare the forces produced by a typical (but large) electric ﬁeld achievable without electrical breakdown in air on the charge eof the electron, and by a typical (but large) magnetic ﬁeld gradient acting on the magnetic dipole moment of the electron. Taking for Ethe value 106V/m, with e≈1.6×10−19 coulomb, fe=eE≈1.6×10−13N (25) ABof 1 tesla (10,000 gauss) is a typical large ﬂux density produced by an iron core electromagnet. Let us assume that a ﬂux density variation of this order can be produced over a distance of 1 cm, which is, in practice, a rather high gradient. Yet taking this value and a moment of one Bohr magneton (9.0.1), we obtain from (22) for the force on the electron fm= 1.8×10−25N (26) Note that the electric force associated with the net charge is much greater than the magnetic one due to the magnetic dipole moment. Because of the large ratio fe/fm for ﬁelds of realistic magnitudes, experiments designed to detect magnetic dipole eﬀects on fundamental particles did not utilize particles having a net charge, but rather used neutral atoms (most notably, the Stern-Gerlach experiment8). Indeed, a stray electric ﬁeld on the order of 10−6V/m would deﬂect an electron as strongly as a magnetic ﬁeld gradient of the very large magnitude assumed in calculating (26). The small magnetic dipole moment of the electron can become very important in solid matter because macroscopic solids are largely neutral. Hence, the forces exerted upon the positive and negative charges within matter by an applied electric ﬁeld more or less cancel. In such a case, the forces on the electronic magnetic dipoles in an applied magnetic ﬁeld can dominate and give rise to the signiﬁcant macroscopic force observed when an iron ﬁling is picked up by a magnet. 8W. Gerlach and O. Stern, “Uber die Richtungsquantelung im Magnetfeld,” Ann. d. Physik , 4th series, Vol. 74, (1924), pp. 673-699. 62 Energy, Power Flow, and Forces Chapter 11 Fig. 11.8.5 By dint of its ﬁeld gradient, a magnet can be used to pick up a spherical magnetizable particle. Example 11.8.1. Magnetization Force on a Macroscopic Particle Suppose that we wanted to know the force exerted on an iron particle by a magnet. Could the microscopic force, (22), be used? The energy method derivation shows that, provided the particle is surrounded by free space, the answer is yes. The parti- cle is taken as being spherical, with radius R, as shown in Fig. 11.8.5. It is assumed to have such a large magnetizability that its permeability can be taken as inﬁnite. Further, the radius Ris much smaller than other dimensions of interest, especially those characterizing variations in the applied ﬁeld in the neighborhood of the par- ticle. Because the particle is small compared to dimensions over which the ﬁeld varies signiﬁcantly, we can compute its moment by approximating the local ﬁeld as uniform. Thus, the magnetic potential is determined by solving Laplace’s equation in the region around the particle subject to the conditions that Hbe the uniform ﬁeld Hoat “inﬁnity” and Ψ be constant on the surface of the particle. The calculation is fully analogous to that for the electric potential surrounding a perfectly conducting sphere in a uniform electric ﬁeld. In the electric analog, the dipole moment was found to be (6.6.5), p= 4π/epsilon1oR3E. Therefore, it follows from the analogy provided by (19) that the magnetic dipole moment at the particle location is µom= 4πµoR3H⇒m= 4πR3H (27) Directly below the magnet, Hhas only a zcomponent. Thus, the dipole mo- ment follows from (27) as m= 4πR3Hziz (28) Evaluation of (22) therefore gives fz= 4πR3Hz∂Hz ∂z(29) where Hzand its derivative are evaluated at the location of the particle. A typical axial distribution of Hzis shown in Fig. 11.8.6 together with two pictures aimed at gaining insight into the origins of the magnetic dipole force. In Sec. 11.9 Macroscopic Force Densities 63 Fig. 11.8.6 In an increasing axial ﬁeld, the force on a dipole is upward whether the dipole is modeled as a pair of magnetic charges or as a circulating current. the ﬁrst, the dipole is again depicted as a pair of magnetic monopoles, induced to form a moment collinear with the H. Because the ﬁeld is more intense at the north pole of the particle than at the south pole, there is then a net force. Alternatively, suppose that the dipole is actually a circulating current, so that the force is given by (23). Even though the energy argument makes it clear that the force is again given by (22), the physical picture is diﬀerent. Because His solenoidal, an intensity that increases with zimplies that the ﬁeld just oﬀ axis has a component that is directed radially inward. It is this radial component of the ﬂux density crossed with the current density that results in an upward force on each segment of the loop. R E F E R E N C E S [1] P. Penﬁeld, Jr., and H. A. Haus, Electrodynamics of Moving Media , MIT Press, Cambridge, Mass. (1967). [2] R. M. Fano, L. J. Chu, and R. B. Adler, Electromagnetic Fields, Energy, and Forces , John Wiley and Sons, New York (1960). [3] D. B. H. Tellegen, “Magnetic-dipole models,” Am. J. Phys. Vol. 30 (Sept. 1962), pp. 650-652. [4] H. A. Haus and P. Penﬁeld, Jr., “Force on a current loop,” Phys. Lett. , Vol. 26A (March 1968), pp. 412-413. 11.9 MACROSCOPIC FORCE DENSITIES A macroscopic force density F(r) is the force per unit volume acting on a medium in the neighborhood of r. Fundamentally, the electromagnetic force density is the result of forces acting on those microscopic particles embedded in the material that are charged, or that have electric or magnetic dipole moments. The forces acting on these individual particles are passed along through interparticle forces to the 64 Energy, Power Flow, and Forces Chapter 11 macroscopic material as a whole. In the limit where that volume becomes small, the force density can then be regarded as the sum of the microscopic forces over a volume element ∆ V. F= lim ∆V→0X ∆Vf (1) Of course, the linear dimensions of ∆ Vare large compared to the microscopic scale. Strictly, the forces in this sum should be evaluated using the microscopic ﬁelds. However, we can gain insight concerning the form taken by the force den- sity by using the macroscopic ﬁelds in this evaluation. This is the basis for the following discussions of the force densities associated with unpaired charges and with conduction currents (the Lorentz force density) and with the polarization and magnetization of media (the Kelvin force density). To be certain that the usage of macroscopic ﬁelds in describing the force den- sities is consistent with that implicit in the constitutive laws already introduced to describe conduction, polarization, and magnetization, the electromagnetic force densities should be derived using energy arguments. These derivations are exten- sions of those of Secs. 11.6 and 11.7 for forces. We end this section with a discussion of the results of such derivations and of circumstances under which they will predict the same total forces or even material deformations as those derived here. The Lorentz Force Density. Without restricting the generality of the result- ing force density, suppose that the electrical force on a material is due to two species of charged particles. One has N+particles per unit volume, each with a charge q+, while the other has density N−and a charge equal to −q−. With vdenoting the velocity of the macroscopic material and v±representing the respective velocities of the carriers relative to that material, the Lorentz force law gives the force on the individual particles. f+=q+[E+ (v++v)×µoH] (2) f−=−q−[E+ (v−+v)×µoH] (3) Note that q−is a positive number. In typical solids and ﬂuids, the charged particles are either bonded to the material or migrate relative to the material, suﬀering many collisions with the neutral material during times of interest. In either case, the inertia of the particles is inconsequential, so that on the average, the forces on the individual particles are passed along to the macroscopic material. In either situation, the force density on the material is the sum of (2) and (3), respectively, multiplied by the charged particle densities. F=N+f++N−f− (4) Substitution of (2) and (3) into this expression gives the Lorentz force density F=ρuE+J×µoH (5) where ρuis the unpaired charge density (7.1.6) and Jis the current density. ρu=N+q+−N−q−;J=N+q+(v+v+)−N−q−(v+v−) (6) Sec. 11.9 Macroscopic Force Densities 65 Fig. 11.9.1 The electric Lorentz force density ρuis proportional to the net charge density because the charges individually pass their force to the material in which they are embedded. Fig. 11.9.2 The magnetic Lorentz force density J×µoH. Because the material is in motion, with velocity v, the current density Jhas not only the contribution familiar from Sec. 7.1 (7.1.4) due to the migration of the carriers relative to the material, but one due to the net charge carried by the moving material as well. In EQS systems, the ﬁrst term in (5) usually outweighs the second, while in MQS systems (where the unpaired charge density is negligible), the second term tends to dominate. The derivation and Fig. 11.9.1 suggest why the electric term is proportional to the netcharge density. In a given region, the force density resulting from the positively charged particles tends to be canceled by that due to the negatively charged particles, and the net force density is therefore proportional to the diﬀerence in absolute magnitudes of the charge densities. We exploited this fact in Chap. 7 to let electrically induced material motions evidence the distribution of the unpaired charge density. For example, in Demonstration 7.5.1, the unpaired charge density was restricted to an interface, and as a result, the motion of the ﬂuid was suppressed by constraining the interface. A more recent example is the force on the upper electrode in the capacitor transducer of Example 11.6.1. Here again the force density is conﬁned to a thin region on the surface of the conducting electrode. The magnetic term in (6), pictured in Fig. 11.9.2 as acting on a current- carrying wire, is also familiar. This force density was responsible for throwing the metal disk into the air in the experiment described in Sec. 10.2. The force responsible for the levitation of the pancake coil in Demonstration 11.7.1 was also the net eﬀect of the Lorentz force density, acting either over the volume of the coil conductors or over that of the conducting sheet below. In MQS systems, where the contribution of the “convection” current ρuvis negligible, the current density is typically due 66 Energy, Power Flow, and Forces Chapter 11 Fig. 11.9.3 The electric Kelvin force density results because the force on the individual dipoles is passed on to the neutral medium. to conduction. Note that this means that the velocity of the charge carriers is determined by the electric ﬁeld they experience in the conductor, and not simply by the motion of the conductor. The current density Jin a moving conductor is generally not in the direction of motion.9 The Kelvin Polarization Force Density. If microscopic particles carrying a net charge were the only contributors to a macroscopic force density, it would not be possible to explain the forces on polarized materials that are free of unpaired charge. Example 11.6.2 and Demonstration 11.6.2 highlighted the polarization force. The experiment was carried out in such a way that the dielectric material did not support unpaired charge, so the force is not explained by the Lorentz force density. In EQS cases where ρu= 0, the macroscopic force density is the result of forces on the microscopic particles with dipole moments. The resulting force density is fundamentally diﬀerent from that due to unpaired charges; the forces p· ∇Eon the individual microscopic particles are passed along by interparticle forces to the medium as a whole. A comparison of Fig. 11.9.3 with Fig. 11.9.1 emphasizes this point. For a single species of particle, the force density is the force on a single dipole multiplied by the number of dipoles per unit volume Np. By deﬁnition, the polarization density P=Npp, so it follows that the force density due to polarization is F=P· ∇E (7) This is often called the Kelvin polarization force density . Example 11.9.1. Force on a Dielectric Material In Fig. 11.9.4, the cross-section of a pair of electrodes that are dipped into a liquid dielectric is shown. The picture might be of a cross-section from the experiment 9Indeed, it is fortunate that the carriers do not have the same velocity as the material, for if they did, it would not be possible to use the magnetic Lorentz force density for electromechanical energy conversion. If we recognize that the rate at which a force fdoes work on a particle that moves at the velocity visv·f, then it follows from the Lorentz force law, (1.1.1), that the rate of doing work on individual particles through the agent of the magnetic ﬁeld is v·(v×µoH). The cross-product is perpendicular to v, so this rate of doing work must be zero. Sec. 11.9 Macroscopic Force Densities 67 Fig. 11.9.4 In terms of the Kelvin force density, the dielectric liquid is pushed into the ﬁeld region between capacitor plates because of the forces on individual dipoles in the fringing ﬁeld. of Demonstration 11.6.2. With the application of a potential diﬀerence to the elec- trodes, the dielectric rises between the electrodes. According to (7), what is the distribution of force density causing this rise? For the liquid dielectric, the polarization constitutive law is taken as linear [(6.4.2) and (6.4.4)] P= (/epsilon1−/epsilon1o)E (8) so that with the understanding that /epsilon1is a function of position (uniform in the liquid, /epsilon1oin the gas, and taking a step at the interface), the force density of (7) becomes F= (/epsilon1−/epsilon1o)E· ∇E (9) By using a vector identity10and invoking the EQS approximation where ∇×E= 0, this expression is written as F=1 2(/epsilon1−/epsilon1o)∇(E·E) (10) A second identity11converts this expression into one that will now prove useful in picturing the distribution of force density. F=1 2∇[(/epsilon1−/epsilon1o)E·E]−1 2E·E∇(/epsilon1−/epsilon1o) (11) Provided that the interface is well removed from the fringing ﬁelds at the top and bottom edges of the electrodes, the electric ﬁeld is uniform not only in the dielectric and gas above and below the interface between the electrodes, but through the interface as well. Thus, throughout the region between the electrodes, there is no gradient of E, and hence, according to (7), no Kelvin force density. The Kelvin force density is therefore conﬁned to the fringing ﬁeld region where the ﬂuid surrounds the lower edges of the electrodes. In this region, /epsilon1is uniform, so the force density reduces to the ﬁrst term in (11). Expressed by this term, the direction and 10A· ∇A= (∇ ×A)×A+1 2∇(A·A) 11∇(ψφ) =ψ∇φ+φ∇ψ 68 Energy, Power Flow, and Forces Chapter 11 magnitude of the force density is determined by the gradient of the scalar E·E. Thus, where Eis varying in the fringing ﬁeld, it is directed generally upward and into the region of greater ﬁeld intensity, as suggested by Fig. 11.9.4. The force on the dipole shown by the inset lends further credence to the dipolar origins of the force density. Although there is no physical basis for doing so, it might seem reasonable to take the force density caused by polarization as being ρpE. After all, it is the polarization charge density ρpthat was used in Chap. 6 to represent the eﬀect of the media on the macroscopic electric ﬁeld intensity E. The experiment of Demon- stration 11.6.2, pictured in Fig. 11.6.7, makes it clear that this force density is not correct. With the interface well removed from the fringing ﬁelds, there is no polariza- tion charge density anywhere in the liquid, either at the interface or in the fringing ﬁeld. If ρpEwere the correct force density, it would be zero throughout the ﬂuid volume except at the interfaces with the conducting electrodes. There, the forces are perpendicular to the surface of the electrodes. Such a force distribution could not cause the ﬂuid to rise. The Kelvin Magnetization Force Density. Forces caused by magnetization are probably the most commonly experienced electromagnetic forces. They account for the attraction between a magnet and a piece of iron. In Example 11.7.1, this force density acts on the disk of magnetizable material. Given that the magnetizable material is made up of microscopic dipoles, each experiencing a force of the nature of (11.8.22), and that the magnetization density Mis the number of these per unit volume multiplied by m, it follows from the arguments of the preceding section that the force density due to magnetization is F=µoM· ∇H (12) This is sometimes called the Kelvin magnetization force density . Example 11.9.2. Force Density in a Magnetized Fluid With the dielectric liquid replaced by a ferroﬂuid having a uniform permeability µ, and the electrodes replaced by the pole faces of an electromagnet, the physical conﬁguration shown in Fig. 11.9.4 becomes the one of Fig. 11.9.5, illustrating the magnetization force density. In such ﬂuids[1], the magnetization results from an essentially permanent suspension of magnetized particles. Each particle comprises a magnetic dipole and passes its force on to the liquid medium in which it is suspended. Provided that the magnetization obeys a linear law, the discussion of the distribution of force density given in Example 11.9.1 applies equally well here. Alternative Force Densities. We now return to comments made at the beginning of this section. The ﬁelds used to express the Lorentz and Kelvin force densities are macroscopic. To assure consistency between the averages implied by these force densities and those already inherent in the constitutive laws, an energy principle can be used. The approach is a continuum version of that exempliﬁed Sec. 11.9 Macroscopic Force Densities 69 Fig. 11.9.5 In an experiment that is the magnetic analog of that shown in Fig. 11.9.4, a magnetizable liquid is pushed upward into the ﬁeld region between the pole faces by the forces on magnetic dipoles in the fringing region at the bottom. for lumped parameter systems in Secs. 11.7 and 11.8. In the lumped parameter systems, electrical terminal relations were used to determine a total energy, and energy conservation was used to determine the force. In the continuum system[2], the electrical constitutive law is used to ﬁnd an energy density, and energy conservation used, in turn, to ﬁnd a force density. This energy method, like the one exempliﬁed in Secs. 11.7 and 11.8 for lumped parameter systems, describes systems that are loss free. In making practical use of the result, it is assumed that it will be applicable even if there are losses. A more general method, which invokes a principle of virtual power[3], allows for dissipation but requires more empirical information than the polarization or magnetization constitutive law as a starting point. Force densities derived from more rigorous arguments than given here can have very diﬀerent distributions from the superposition of the Lorentz and Kelvin force densities. We would expect that the arguments break down when the microscopic particles become so densely packed that the ﬁeld experienced by one is signiﬁcantly altered by its nearest neighbor. But surely the diﬀerence between the magnetic force density of Lorentz and Kelvin (LK) FLK=J×µoH+µoM· ∇H (13) we have derived here and the Korteweg-Helmholtz force density (KH) for incom- pressible media FKH=J×B−1 2H·H∇µ (14) cited in the literature[2]is not due to interactions between microscopic particles. This latter force density is often obtained for an incompressible material from en- ergy arguments. [Note that with −∇µandH·H, respectively, playing the roles of dL/dξ andi2, the magnetization term in (14) takes a form found for the force on a magnetizable material in Sec. 11.7.] 70 Energy, Power Flow, and Forces Chapter 11 In Example 11.9.2 (where J=0), we found the force density of (13) to be conﬁned to the fringing ﬁeld. By contrast, (14) gives no force density in the fringing region (where µis uniform), but rather puts it all at the interface. According to this latter equation, through the agent of a surface force density (a force density that is a spatial impulse at the interface), the ﬁeld pulls upward on the interface. The question may then be asked whether, and how, the two force density expressions can be reconciled. The answer is that if µoM= (µ−µo)H, they predict the same motion for any volume-conserving material deformations such as those of an incompressible ﬂuid. We shall demonstrate this for the case of a liquid, such as shown in Fig. 11.9.5, but allowing for the action of a current Jas well. As the ﬁrst step in the derivation, we shall show that (13) and (14) diﬀer by the gradient of a scalar, π(r). To see this, use a vector identity12to write (13) as FLK=J×µoH+ (µ−µo)[(∇ ×H)×H+1 2∇(H·H)] (15) The MQS form of Amp` ere’s law makes it possible to substitute ∇×HforJin this expression, which then becomes FLK=J×B+1 2(µ−µo)∇(H·H) (16) The second term in this expression is then expanded using a second vector identity13 FLK=J×B−1 2H·H∇µ+∇£1 2(µ−µo)H·H] (17) This expression diﬀers from (14) by the last term, which indeed takes the form ∇π where π=1 2(µ−µo)H·H (18) Now consider Newton’s force law for an elemental volume of material. Using the Korteweg-Helmholtz force density, (14), it takes the form Finertial =Fm− ∇p+FKH (19) where pis the internal ﬂuid pressure and Fmis the sum of all other mechanical contributions to the force density. Alternatively, using (13) written as (17) as the force density, this same law is represented by Finertial =Fm− ∇p+FKH+∇π=Fm− ∇p/prime+FKH (20) For an incompressible material, none of the other laws needed to describe the con- tinuum (such as mass conservation) involve the pressure.14Thus, if (19) is used, p 12A· ∇A= (∇ ×A)×A+1 2∇(A·A) 13∇(ψφ) =ψ∇φ+φ∇ψ 14For example, for a compressible ﬂuid, the pressure depends on mass density and tempera- ture, so the pressure does appear in the physical laws. Indeed, in the constitutive law relating these values, the pressure has a well-deﬁned value. However, in an incompressible ﬂuid, the constitutive law relating the pressure to mass density and temperature is not relevant to the prediction of material motion. Sec. 11.9 Macroscopic Force Densities 71 Fig. 11.9.6 The block having uniform permeability and conductivity carries a uniform current density in the ydirection which produces a z-directed magnetic ﬁeld intensity. Although the force densities of (13) and (14) have very diﬀerent distributions in the block, they predict the same net force. appears only in that equation and if (20) is used, p/prime≡p−πappears only in that expression. This means that pandp/primeplay identical roles in predicting the defor- mation. In an incompressible material, it is the role of the pressure to adjust itself so that only volume conserving deformations are allowed.15The two formulations would diﬀer in what one would call the pressure, but would result in the same ma- terial deformation and velocity. An example is the height of rise of the ﬂuid between the parallel plates in Fig. 11.9.5. Included in the class of incompressible deformations are rigid body motions. If used self-consistently, force densities that diﬀer by the gradient of a “ π” will predict the same motions of rigid bodies. Thus, the net force on a body surrounded by free space will be the same whether found using the Lorentz-Kelvin or the Korteweg- Helmholtz force density. The following example illustrates this concept. Example 11.9.3. Magnetic Force on a Magnetizable Current-Carrying Material A block of conducting material having permeability µis shown in Fig. 11.9.6 sand- wiched between perfectly conducting plates. A current source, distributed over the left edges of these electrodes, drives a constant surface current density Kin the + x direction along the left edge of the lower electrode. This current passes through the block in the ydirection as a current density J=K biy (21) and is returned to the source in the −xdirection at the left edge of the upper electrode. The thickness aof the block is small compared to its other two dimensions, so the magnetic ﬁeld between the electrodes is zdirected and dependent only on x. 15Like the “perfectly permeable material” of magnetic circuits, in which Bremains ﬁnite as Hgoes to zero, the “perfectly incompressible” material is one in which the pressure remains ﬁnite even as the material becomes inﬁnitely “stiﬀ” to all but those deformations that conserve volume. 72 Energy, Power Flow, and Forces Chapter 11 From Amp` ere’s law it follows that ∂Hz ∂x=−Jy⇒H=−izK bx (22) in the conducting block. The alternative force densities, (13) and (14), have very diﬀerent distributions in the block. Yet we must ﬁnd that the net force on the block, found by integrating each over its volume, is the same. To see that this is so, consider ﬁrst the sum of the Lorentz and Kelvin force densities, (13). There is no xcomponent of the magnetic ﬁeld intensity, so for this particular conﬁguration, the magnetization term makes no contribution to (13). Evaluation of the ﬁrst term using (19) and (20) then gives (Fx)LK=−K2 b2µox (23) Integration of this force density over the volume amounts to a multiplication by the cross-sectional area ad, and integration on x. Thus, the net force predicted by using the force density of Lorentz and Kelvin is (fx)LK=adZ0 −b−K2µox b2dx=1 2adK2µo (24) Now, the Korteweg-Helmholtz force density given by (14) is evaluated. The permeability µis uniform throughout the interior of the block, so the magnetization term is again zero there. However, µis a step function at the ends of the block, where x=−bandx= 0. Thus, ∇µis an impulse there and we must take care to include the contributions from the surface regions in our integration. Evaluation of thexcomponent of (14) using (21) and (22) gives (Fx)KH=−µK2 b2x−1 2H2 z∂µ ∂x(25) Integration of (25) over the volume of the block therefore gives (fx)KH=ad·Z0 −b−K2µ b2xdx−Z−b+ −b−1 2H2 z∂µ ∂xdx¸ (26) Note that Hzis constant through the interface at x=−b. Thus, the integration of the last term can be carried out. Simpliﬁcation of this expression gives the same total force as found before, (24). The distributions of the force densities given by (13) and (14) are generally diﬀerent, even very diﬀerent. It is therefore natural to ask which of the two is the “right” one. In general, until the “other” force densities acting on the medium in question are speciﬁed, this question cannot be answered. Here, where a discussion of continuum mechanics is beyond our purview, we have identiﬁed a class of mechan- ical deformations (namely, those that are volume conserving or “incompressible”), where these force densities are equally valid. In fact, any other force density diﬀer- ing from these by a term having the form ∇πwould also be valid. The combined Sec. 11.10 Summary 73 Lorentz and Kelvin force densities have the advantage of a satisfying physical in- terpretation. However, the derivation has the weakness of making an ad hoc use of the macroscopic ﬁelds. Force densities resulting from an energy argument have the advantage of dealing rigorously with the macroscopic ﬁelds. R E F E R E N C E S [1] R. E. Rosensweig, “Magnetic Fluids,” Scientiﬁc American , (Oct. 1982), pp. 136-145. [2] J. R. Melcher, Continuum Electromechanics , MIT Press, Cambridge, Mass. (1981), chap. 3. [3] P. Penﬁeld and H. A. Haus, Electrodynamics of Moving Media , MIT Press, Cambridge, Mass. (1967). 11.10 SUMMARY Far reaching as they are, the laws summarized by Maxwell’s equations are directly applicable to the description of only one of many physical subsystems of scien- tiﬁc and engineering interest. Like those before it, this chapter has been concerned with the electromagnetic subsystem. However, by casting the electromagnetic laws into statements of power ﬂow, we have come to recognize how the electromagnetic subsystem couples to the thermodynamic subsystem through the power dissipa- tion density and to the mechanical subsystem through forces and force densities of electromagnetic origin. The basis for a self-consistent macroscopic description of any continuum sub- system is a power ﬂow statement having the forms identiﬁed in Sec. 11.1. Describing the energy and power ﬂow in and into a volume Venclosed by a surface S, the in- tegral conservation of energy statement takes the form (11.1.1). −I SS·da=d dtZ VWdv +Z VPddv (1) The diﬀerential form of the conservation of energy statement is implied by the above. −∇ ·S=∂W ∂t+Pd (2) Poynting’s theorem, the subject of Sec. 11.2, is obtained starting from the laws of Faraday and Amp` ere to obtain an expression of the form of (2). For materials that are Ohmic ( J=σE) and that are linearly polarizable and magnetizable ( D=/epsilon1E andB=µH), the power ﬂux density S(orPoynting’s vector ),energy density W, andpower dissipation density Pdwere shown in Sec. 11.3 to be S=E×H (3) 74 Energy, Power Flow, and Forces Chapter 11 W=1 2/epsilon1E·E+1 2µH·H (4) Pd=σE·E (5) Of course, taking the free space limit where /epsilon1andµassume their free space values andσ= 0 gives the free space conservation statement discussed in Sec. 11.2. In Sec. 11.3, we found that in EQS systems, an alternative to Poynting’s vector is (11.3.24). S= Φ¡ J+∂D ∂t¢ (6) This expression is of practical importance, because it can be evaluated without determining H, which is generally not of interest in EQS systems. An important application of the integral form of the energy conservation state- ment is to lumped parameter systems. In these cases, the surface Sof (1) encloses a system that is connected to the outside world through terminals. It is then conve- nient to describe the power ﬂow in terms of the terminal variables. It was shown in Sec. 11.3 (11.3.29), that the net power into the system represented by the left-hand side of (1) becomes −I SE×H·da=nX i=1viii (7) provided that the magnetic induction and the electric displacement current through the surface Sare negligible. This set the stage for the application of the integral form of the energy con- servation theorem to lumped parameter systems. In Sec. 11.4, attention focused on the energy storage term, the ﬁrst terms on the right in (1) and (2). The energy density concept was broadened to include materials having constitutive laws relating the ﬂux densities to the ﬁeld intensities that were single valued and collinear. With E, D, H , and Brepresenting the ﬁeld magnitudes, the energy density was found to be the sum of electric and magnetic energy densities. W=We+Wm; We=ZD 0E(D/prime)δD/prime; Wm=ZB 0H(B/prime)δB/prime(8) Integrated over the volume Vof a system, this function leads to the total energy w. For quasistatic lumped parameter systems, the total electric or magnetic en- ergy is often conveniently found following a diﬀerent route. First the terminal rela- tions are determined and then the total energy is found by adding up the increments of energy put into the system as it is energized. In the case of an nterminal pair EQS system, where the relation between terminal voltage viand associated charge qiisvi(q1, q2, . . . q n), the increment of energy is vidqi, and the total electric energy is (11.4.9). we=nX i=1Z vidqi (9) Sec. 11.10 Summary 75 The line integration in an n-dimensional space representing the nindependent qi’s was illustrated by Example 11.4.2. Similarly, for an nterminal pair MQS system where the current iiis related to the ﬂux linkage λibyii=ii(λ1, λ2, . . . λ n), the total energy is (11.4.12). wm=nX i=1Z iidλi (10) Note the analogy between these expressions for the total energy of EQS and MQS lumped parameter systems and the electric and magnetic energy densities, respectively, of (8). The transition from the ﬁeld picture aﬀorded by the energy densities to the lumped parameter characterization is made by E→v, D→qand byH→i, B→λ. Especially in using the energy to evaluate forces of electrical origin, we found it convenient to deﬁne coenergy density functions. W/prime e=DE−We; W/prime m=BH−Wm (11) It followed that these functions were natural when it was desirable to use EandH as the independent variables rather than DandB. W/prime e=ZE 0D(E/prime)δE/prime; W/prime m=ZH 0B(H/prime)δH/prime(12) The total coenergy functions for lumped parameter EQS and MQS systems could be found either by integrating these densities over the volume or by again viewing the system in terms of its terminal variables. With the total coenergy functions deﬁned by w/prime e=nX i=1qivi−we; w/prime m=nX i=1λiii−wm (13) it followed that the coenergy functions could be determined from the terminal relations by again carrying out line integrations, but this time with the voltages and currents as the independent variables. For EQS systems, w/prime e=nX i=1Z qidvi (14) while for MQS systems, w/prime m=nX i=1Z λidii (15) Again, note the analogy to the respective terms in (12). The remaining sections of the chapter developed some of the possible implica- tions of the “dissipation” term in the energy conservation statement, the last terms in (1) and (2). In Sec. 11.5, coupling to a thermal subsystem was discussed. In 76 Energy, Power Flow, and Forces Chapter 11 this section, the disparity between the power input and the rate of increase of the energy stored was accounted for by heating. In addition to Ohmic heating, caused by collisions between the migrating carriers and the neutral media, we considered losses associated with the dynamic polarization and magnetization of materials. In Secs. 11.6–11.9, we considered coupling to a mechanical subsystem as a second mechanism by which energy could be extracted from (or put into) the elec- tromagnetic subsystem. With the displacement of an object denoted by ξ, we used an energy conservation postulate to infer the total electric or magnetic force acting on the object from the energy functions [(11.6.9), and its magnetic analog] fe=−∂we(q1. . . q n, ξ) ∂ξ; fm=−∂wm(λ1. . . λ n, ξ) ∂ξ(16) or from the coenergy functions [(11.7.7) and the analogous expression for electric systems]. fe=∂w/prime e(v1. . . v n, ξ) ∂ξ; fm=∂w/prime m(i1. . . i n, ξ) ∂ξ(17) In Sec. 11.8, where the Lorentz force on a particle was generalized to account for electric and magnetic dipole moments, one objective was a microscopic picture that would lend physical insight into the forces on polarized and magnetized mate- rials. The Lorentz force was generalized to include the force on stationary electric and magnetic dipoles, respectively. f=p· ∇E;f=µom· ∇H (18) The total macroscopic forces resulting from microscopic forces had already been encountered in the previous two sections. The force density describes the interac- tion between a volume element of the electromagnetic subsystem and a mechanical continuum. The force density inferred by averaging over the forces identiﬁed in Sec. 11.8 as acting on microscopic particles was F=ρuE+J×µoH+P· ∇E+µoM· ∇H (19) A more rigorous approach to ﬁnding the force density could be based on a gener- alization of the energy method introduced in Secs. 11.6 and 11.7. As background for further pursuit of this subject, we have illustrated the importance of including the mechanical continuum with which the force density acts. Before there can be a meaningful answer to the question, “Which force density is correct?” the other force densities acting on the material must be speciﬁed. As an illustration, we found that very diﬀerent electric or magnetic force densities would result in the same de- formations of an incompressible material and in the same net force on an object surrounded by free space[1,2]. R E F E R E N C E S [1] P. Penﬁeld, Jr., and H. A. Haus, Electrodynamics of Moving Media , MIT Press, Cambridge, Mass. (1967). [2] J. R. Melcher, Continuum Electromechanics , MIT Press, Cambridge, Mass. (1981), chap. 3. Sec. 11.3 Problems 77 P R O B L E M S 11.1 Introduction 11.1.1∗A capacitor C, an inductor L, and a resistor Rare in series, driven by the voltage v(t) and carrying the current i(t). With vcdeﬁned as the voltage across the capacitor, show that vi=dw/dt +i2Rwhere w=1 2Cv2 c+1 2Li2. Argue that wis the energy stored in the inductor and capacitor, while i2R is the power dissipated in the resistor. 11.2 Integral and Diﬀerential Conservation Statements 11.2.1∗Consider a system in which the ﬁelds are yand/or zdirected and indepen- dent of yandz. Then S=Sx(x, t)ix, W=W(x, t), and Pd=Pd(x, t). (a) Show that for a volume having area Ain any y−zplane and located between x=x1andx=x2, (1) becomes −[ASx(x1)−ASx(x2)] =d dtAZx1 x2Wdx +AZx1 x2Pddx (a) (b) Take the limit where x1−x2= ∆x→0 and show that the one- dimensional form of (3) results. (c) Based on (a), argue that Sxis the power ﬂux density in the xdirection. 11.3 Poynting’s Theorem 11.3.1∗The perfectly conducting plane parallel electrodes of Fig. 13.1.1 are driven at the left by a voltage source Vd(t) and are “open circuit” at the right, as shown in Fig. 13.1.4. The system is EQS. (a) Show that the power ﬂux density is S=iy(−/epsilon1oy/a2)VddVd/dt. (b) Using S, show that the power input is d(1 2CV2 d)/dt, where C= /epsilon1obw/a . (c) Evaluate the right-hand side of (11.1.1) to show that if the magnetic energy storage is neglected, the same result is obtained. (d) Show that the magnetic energy storage is indeed negligible if b/cis much shorter than times of interest. 11.3.2 The perfectly conducting plane parallel electrodes of Fig. 13.1.1 are driven at the left by a current source Id(t), as shown in Fig. 13.1.3. The system is MQS. 78 Energy, Power Flow, and Forces Chapter 11 Fig. P11.3.2 (a) Determine S. (b) From S, ﬁnd the input power. (c) Evaluate the right-hand side of (11.1.1) for a volume enclosing the region between the electrodes, and show that if the electric energy storage is neglected, it is indeed equal to the left-hand side. (d) Under what conditions is the electric energy storage negligible? 11.4 Ohmic Conductors with Linear Polarization and Magnetization 11.4.1∗In Example 7.3.2, a three-dimensional dipole current source drives circu- lating currents through a uniformly conducting material. This source is so slowly varying with time that time rates of change have a negligible eﬀect. Consider ﬁrst the power ﬂow as pictured in terms of the Poynting ﬂux density, (3). (a) Show that E×H=µipd 4π¶21 σ¯¯¯¯¡−2 cosθsinθ r5¢ iθ+sin2θ r5ir¯¯¯¯(a) (b) Show that Pd=µipd 4π¶21 σ(1 + 3 cos2θ) r6(b) (c) Using these results, show that (11.1.3) is indeed satisﬁed. (d) Now, using the alternative EQS power theorem, evaluate Sas given by (23) and again show that (11.1.3) is satisﬁed. (e) Observe that the latter evaluation is much simpler to carry out and that the latter power ﬂux density is easier to picture. 11.4.2 Coaxial perfectly conducting circular cylindrical electrodes make contact with a uniformly conducting material of conductivity σin the annulus b < r < a , as shown in Fig. P11.3.2. The length lis large compared to a. A voltage source vdrives the system at the left, while the electrodes are “open” at the right. Assume that v(t) is so slowly varying that the voltage can be regarded as independent of z. Sec. 11.4 Problems 79 Fig. P11.3.3 (a) Determine E,Φ, and Hin the annulus. (b) Evaluate the Poynting power ﬂux density S[as given by (3)] in the annulus. (c) Use Sto evaluate the total power dissipation by integration over the surface enclosing the annulus. (d) Show that the same result is obtained by integrating Pdover the volume. (e) Evaluate Sas given by (23), and use that distribution of the power ﬂux density to determine the total power dissipation. (f) Make sketches of the alternative distributions of S. (g) Show that the input power is vi, where iis the total current from the voltage source. 11.4.3∗A pair of perfectly conducting circular plates having a spacing dform par- allel electrodes in a system having cylindrical symmetry about the zaxis and the cross-section shown by Fig. P11.3.3. The central region between the plates is ﬁlled out to the radius bby a uniformly conducting material having conductivity σand uniform permittivity /epsilon1, while the surrounding region, where b < r < a , is free space. A distributed voltage source v(t) con- strains the potential diﬀerence between the outer edges of the electrodes. Assume that the system is EQS. (a) Show that the Poynting power ﬂux density is S=−ir( r 2¡σv d+/epsilon1 ddv dt¢v d; r < b 1 2r£1 d(/epsilon1b2+/epsilon1o(r2−b2))dv dt+σb2 dv¤v d;b < r < a(a) (b) Integrate this ﬂux density over a surface enclosing the region between the plates, and show that it is equal to the sum of the rate of change of electric energy storage and the power dissipation. (c) Now show that the alternative power ﬂux density given by (23) is S=−v d(z−d)iz½σv d+/epsilon1o ddv dt;r < b /epsilon1o ddv dt; b < r < a(b) (d) Carry out part (b) using this distribution of S, and show that the result is the same. (e) Show that the power input is equal to vi, where iis the total current from the voltage source. 80 Energy, Power Flow, and Forces Chapter 11 11.4.4 In Example 7.5.1, the steady current distribution in and around a con- ducting circular cylindrical rod immersed in a conducting material was determined. Assume that Eois so slowly varying that it can be regarded as static. (a) Determine the distribution of Poynting power ﬂux density S, as given by (3). (b) Determine the alternative Sgiven by (23). (c) Find the power dissipation density Pdin and around the rod. (d) Show that the diﬀerential energy conservation law [(11.1.3) with ∂W/∂t = 0] is satisﬁed at each point in and around the rod using either of these distributions of S. 11.5 Energy Storage 11.5.1∗In Example 8.5.1, the inductance Lof a spherically shaped coil was found by “adding up” the ﬂux linkages of the individual windings. Taking an alternative approach to ﬁnding L, use the ﬁelds found in that example to determine the total energy storage, wm. Then use the fact that wm=1 2Li2 to show that Lis as given by (8.5.20). 11.5.2 In Prob. 9.6.3, a coil has turns at the interface between a magnetizable material and a circular cylindrical core of free space, as shown in Fig. P9.6.3. Assume that the system has a length lin the zdirection and determine the total energy, wm. (Assume that the rotatable coil carries no current.) Use the fact that wm=1 2Li2to ﬁnd L. 11.5.3∗In Example 8.6.4, the ﬁelds of a coil distributed throughout a volume were found. Using these ﬁelds to evaluate the total energy storage, show that the inductance is as given by (8.6.35). 11.5.4 The magnetic circuit described in Prob. 9.7.5 and shown in Fig. P9.7.5 has two electrical excitations. Determine the total magnetic coenergy, w/prime m(i1, i2, x). 11.5.5 The cross-section of a motor or generator is shown in Fig. 11.7.7. (a) Determine the magnetic coenergy density W/prime m, and hence the total coenergy w/prime m. (b) By writing w/prime min the form of (11.4.24), determine L11, L12, and L22. 11.5.6∗The material in the system of Fig. 11.4.3 has the constitutive law of (28). Show that the total coenergy is w/prime e=·α1 α2µr 1 +α2v2 a2−1¶ +1 2/epsilon1ov2 a2¸ ξca+1 2/epsilon1ov2 a(b−ξ)c (a) Sec. 11.6 Problems 81 11.5.7 Consider the system shown in Fig. P9.5.1 but with µa=µoand the region where B=µbHnow ﬁlled with a material having the constitutive law B=¡ µo+α1/p 1 +α2H2)H (a) (a) Determine BandHin each region. (b) Find the coenergy density in each region and hence the total coenergy w/prime mas a function of the driving current i. 11.6 Electromagnetic Dissipation 11.6.1∗In Example 7.9.2, the Maxwell capacitor has an area A(perpendicular tox), and the terminals are driven by a source v=Re[ˆvexp(jωt)]. The sinusoidal steady state has been established. Show that the time average power dissipation in the lossy dielectrics is /angbracketleftPd/angbracketright=A 2[aσa(σ2 b+ω2/epsilon12 b) +bσb(σ2 a+ω2/epsilon12 a)] (bσa+aσb)2+ω2(b/epsilon1a+a/epsilon1b)2|ˆv|2(a) 11.6.2 In Example 7.9.3, the potential is found in the EQS approximation in and around a lossy dielectric sphere embedded in a lossy dielectric and stressed by a uniform ﬁeld having a sinusoidal dependence on time (7.9.36). (a) Find the time average power dissipation density in each region. (b) What is the total time average power dissipated in the sphere? 11.6.3∗Plane parallel perfectly conducting plates having the spacing dare shorted by a perfectly conducting sheet in the plane x= 0, as shown in Fig. P11.5.3. A sheet having thickness ∆ and conductivity σis in the plane x=−band makes contact with the perfectly conducting plates above and below. At their left edges, in the plane x=−(a+b), a source of surface current density, K(t), is connected to the plates. The regions to left and right of the resistive sheet are free space, and wis large compared to a, b, and d. 82 Energy, Power Flow, and Forces Chapter 11 Fig. P11.5.3 (a) Show that the total power dissipation and magnetic energy stored as deﬁned on the right in (11.1.1), are Z VPddv= ∆σwdµ2 ob2¡dHb dt¢2;Z VWdv =1 2µodw(bH2 b+aK2) ( a) (b) Show that the integral on the left in (11.1.1) over the surface indicated by the dashed line in the ﬁgure gives the same result as found in part (a). 11.6.4 In Example 10.4.1, the applied ﬁeld is Ho(t) =Hmcos(ωt) and sinsuoidal steady state conditions prevail. Determine the time average power dissipa- tion in the conducting sheet. Fig. P11.5.5 11.6.5∗The cross-section of an N-turn circular solenoid having radius ais shown in Fig. P11.5.5. It surrounds a thin cylindrical shell of square cross-section, with length bon a side. This shell has thickness ∆ and conductivity σ, and is ﬁlled by a material having permeability µ. Both the shell and the solenoid have a length dperpendicular to the paper that is large compared toa. (a) Given that the terminals of the solenoid are driven by the current i1=iocosωtand the sinusoidal steady state has been established, integrate the time average power dissipation density over the volume of the shell to show that the total time-average power dissipation is pd=2b σ∆dN2i2 o·(ωτm)2 1 + (ωτm)2¸ (a) (b) In the sinusoidal steady state, the time average Poynting ﬂux through a surface enclosing the shell goes into the time average dissipation. Use this fact to obtain (a). 11.6.6 In describing the response of macroscopic media to ﬁelds in the sinusoidal steady state, it is convenient to use complex constitutive laws. The complex permittivity is introduced by (19). Here we introduce and illustrate the complex permeability . Suppose that ﬁeld quantities take the form E=ReˆE(x, y, z )ejωt; ˆH=ReˆH(x, y, z )ejωt(a) Sec. 11.6 Problems 83 Fig. P11.5.6 (a) Show that in a region where there is no macroscopic current density, the MQS laws require that ∇ × ˆE=−jωˆB (b) ∇ × ˆH= 0 ( c) ∇ ·ˆB= 0 ( d) (c) Given that the spherical shell of Prob. 10.4.3 comprises each element in the cubic array of Fig. P11.5.6, each sphere with spacing ssuch that s/greatermuchR, what is the complex permeability µdeﬁned such that ˆB= ˆµˆH? (d) A macroscopic material composed of this array of spheres is placed in the one-turn solenoid of rectangular cross-section shown in Fig. P11.5.6. This conﬁguration is long enough in the zdirection so that fringing ﬁelds can be ignored. At their left edges, the perfectly con- ducting plates composing the top and bottom of the solenoid are driven by a distributed current source, K(t). With the fringing ﬁelds in the neighborhood of the left end ignored, the resulting ﬁelds take the form H=Hz(x, t)izandE=Ey(x, t)iy. Use an evaluation of the Poynting ﬂux to determine the total time average power dissipated in the length l, width d, and height aof the material. 11.6.7∗In the limit where the skin depth δis small compared to the length b, the magnetic ﬁeld distribution in the conductor of Fig. 10.7.2 is given by (10.7.15). Show that (per unit y−zarea) the time average power dissipa- tion associated with the current ﬂowing in the “skin” region is |Ks|2/2σδ watts/m2. 11.6.8 The conducting block shown in Fig. 10.7.2 has a length din the zdirection. (a) Determine the total time average power dissipation. (b) Show that in the case δ/lessmuchbthis expression reduces to that obtained in Prob. 11.5.7, while in the limit δ/greatermuchb, the result is i2Rwhere Ris the dc resistance of the slab and iis the total current. 11.6.9∗The toroid of Fig. 9.4.1 is ﬁlled with an insulating material having the magnetization constitutive law of Prob. 9.4.3. Show that from the terminals 84 Energy, Power Flow, and Forces Chapter 11 of the N1-turn coil, the circuit is equivalent to one having an inductance L=µoN2 1w2/8Rin series with a resistance Rm=µoγN2 1w2/8R. 11.6.10 The toroid of Fig. 9.4.1 is ﬁlled by a material having the magnetization characteristic shown in Fig. P11.5.10. A sinusoidal current is supplied with a particular amplitude, i= (2Hc2πR/N 1) cos( ωt). Fig. P11.5.10 (a) Draw a dimensioned plot of B(t). (b) Find the terminal voltage v(t) and also make a dimensioned plot. (c) Compute the time average power input, deﬁned as /angbracketleftvi/angbracketright=1 TZt+T tvidt (a) where T= 2π/ω. (d) Show that the result of part (c) can also be found by recognizing that, during one cycle, there is an energy/unit volume dissipated which is equal to the area enclosed by the B−Hcharacteristic. 11.7 Electrical Forces on Macroscopic Media 11.7.1∗A pair of perfectly conducting plates, the upper one ﬁxed and the lower one free to move with the horizontal displacement ξ, have a ﬁxed spacing aas shown in Fig. P11.6.1. Show that the force of electrical origin acting on the lower electrode in the ξdirection is f=−/epsilon1ov2d/2a. Fig. P11.6.1 11.7.2 In Example 4.6.3, the capacitance per unit length of the pair of parallel circular cylindrical conductors shown in Fig. 4.6.6 was found. Determine the force per unit length acting on the right cylinder in the xdirection. Sec. 11.8 Problems 85 Fig. P11.6.4 11.7.3∗The electric transducer shown in cross-section by Fig. P11.6.3 has cylindri- cal symmetry about the center line. A coaxial pair of perfectly conducting electrodes having length lare excited at the left end by a voltage source v(t). A perfectly insulating dielectric material having permittivity /epsilon1is free to slide in and out of the annular region between electrodes. Fig. P11.6.3 (a) Show that the force of electric origin acting on the dielectric material in the axial direction is f=v2π(/epsilon1−/epsilon1o)/ln(a/b). (b) Show that if the electrical terminals are constrained by the circuit shown, Ris very small and the plunger suﬀers the displacement ξ(t) the output voltage is vo=−2πRV (/epsilon1−/epsilon1o)(dξ/dt )/ln(a/b). 11.7.4 The electrometer movement shown in Fig. P11.6.4 consists of concentric, perfectly conducting tubes, the inner one free to move in the axial direction. (a) Ignore the fringing ﬁeld and determine the force of electrical origin acting in the direction of ξ. (b) For the energy conversion cycle of Demonstration 11.6.1, but for this transducer, make dimensioned plots of the cycle in the ( q, v) and ( f, ξ) planes (analogous to those of Fig. 11.6.5). (c) By calculating both, show that the electrical energy input in one cycle is equal to the work done on the external mechanical system. 11.7.5∗Show that the vertical force on the nonlinear dielectric material of Prob. 11.4.6 is f=·α1 α2µr 1 +α2v2 a2−1¶ +1 2/epsilon1ov2 a2¸ ca−/epsilon1ov2c 2a(a) 86 Energy, Power Flow, and Forces Chapter 11 Fig. P11.7.3 Fig. P11.7.4 11.8 Macroscopic Magnetic Fields 11.8.1∗Show that the force acting in the xdirection on the movable element of Prob. 9.7.5 (Note Prob. 11.4.4.) is f=−µoaw 2x2(1 +a/b)(N2 1i2 1+ 2N1N2i1i2+N2 2i2 2) ( a) 11.8.2 Determine the force f(i, ξ) acting in the xdirection on the plunger of the magnetic circuit shown in Fig. P9.7.6. 11.8.3∗The magnetic transducer shown in Fig. P11.7.3 consists of a magnetic cir- cuit in which the lower element is free to move in the xandydirections. From the energy principle, ignoring fringing ﬁelds, show that the force on this element is f=µon2di2 2a·a−2x yix−x(a−x) y2iy¸ (a) 11.8.4 The magnetic circuit shown in cross-section by Fig. P11.7.4 has cylindrical symmetry. A plunger of permeability µhaving outer and inner radii aand bcan suﬀer a displacement ξinto the annular gap of a magnetic circuit otherwise made of inﬁnitely permeable material. The coil has Nturns. Assume that the left end of the plunger is well within the magnetic circuit, so that fringing ﬁelds can be ignored, and determine the force f(i, ξ) acting to displace the plunger in the ξdirection. 11.8.5∗The “variable reluctance” motor shown in cross-section in Fig. P11.7.5 consists of an inﬁnitely permeable yoke and an inﬁnitely permeable rotor Sec. 11.9 Problems 87 Fig. P11.7.5 element forming a magnetic circuit with two air gaps of length ∆ /lessmuchR. The system has depth d/greatermuch∆ into the paper. Assume that 0 < θ < α , as shown, and show that the torque caused by passing a current ithrough the twoN-turn coils is τ=−µoRdN2i2/∆. 11.8.6 A “two-phase” synchronous machine is constructed having a cross-section like that shown in Fig. 11.7.7, except that there is an additional winding on the stator. This is identical to the one shown except that it is rotated 90 degrees in the clockwise direction. The current in the stator winding shown in Fig. 11.7.7 is denoted by ia, while that in the additional winding is ib. Thus, the magnetic axes of iaandib, respectively, are upward and to the right. With Ls, Lr, and Mgiven constants, the inductance matrix is "λa λb λr# ="Ls 0 Mcosθ 0 Ls Msinθ Mcosθ M sinθ L r#"ia ib ir# (a) (a) Determine the coenergy w/prime m(ia, ib, θ). (b) Find the torque on the rotor, τ(ia, ib, θ). (c) With ia=Icos(ωt) and ib=Isin(ωt), where Iandωare given constants, argue that the magnetic axis produced by the stator rotates with the angular velocity ω. (d) Using these current constraints together with ir=Irandθ= Ωt− γ, where Ir, γand Ω are constants, show that under synchronous conditions (where ω= Ω), the torque is τ=MII rsin(γ). 11.9 Forces on Microscopic Electric and Magnetic Dipoles 11.9.1∗In a uniform electric ﬁeld E, a perfectly conducting particle having radius R has a dipole moment p= 4π/epsilon1oR3E. Provided that Ris short compared to 88 Energy, Power Flow, and Forces Chapter 11 distances over which the ﬁeld varies, this gives a good approximation to p, even where the ﬁeld is not uniform. Such a particle is shown at the location x=X, y =Yin Fig. P11.8.1, where it is subject to the ﬁeld produced by a periodic potential Φ = Vocos(βx) imposed in the plane y= 0. (a) Show that the potential imposed in the region 0 < yisVocos(βx) exp (−βy). (b) Show that, provided that the particle has no net charge, the force on the particle is f=−4π/epsilon1oR3(Voβ)2βiye−2βy(a) Fig. P11.8.1 11.9.2 The perfectly conducting particle described in Prob. 11.8.1, carrying no net charge but polarized by the imposed electric ﬁeld, is subjected to the ﬁeld of a charge Qlocated at the origin of a spherical coordinate system. In terms of its location Rrelative to the charged particle at the origin, determine the force on the particle. Fig. P11.8.3 11.9.3∗In Fig. P11.8.3, permanent magnets in the lower half-space are represented by the magnetization density M=Mocos(βx)iy, where Moandβare given positive constants. (a) Show that the resulting magnetic potential in the upper half-space is Ψ = ( Mo/2β) cos( βx) exp(−βy) (b) A small inﬁnitely permeable particle having the radius Ris located atx=X, y =Y. Show that the magnetization force on the particle is as given by (a) of Prob. 11.8.1, with Vo→(Mo/2β) and /epsilon1o→µo. 11.9.4 A small “inﬁnitely permeable” particle of radius Ris a distance Zabove an inﬁnitely permeable plane, as shown in Fig. P11.8.4. A uniform ﬁeld Sec. 11.10 Problems 89 Fig. P11.8.4 H=Hoizis imposed. Assume that R/lessmuchZ, and use (27) to approximate the dipole moment induced in the particle. The eﬀect of the inﬁnitely permeable plane on the ﬁeld induced by this dipole is equivalent to that of a second image dipole located at z=−Z. Thus, there is a force of attraction between the magnetized particle and the inﬁnite plane that is equivalent to that attracting the dipole to its image. Determine the force in the zdirection on the particle. 11.10 Macroscopic Force Densities 11.10.1 In Prob. 11.7.2, the total force on a magnetizable plunger is found (Fig. P9.7.6). Find this same force by integrating the force density, (14), over the volume of the plunger. 11.10.2∗In Example 10.3.1, the transient current induced by applying a magnetic ﬁeld intensity Hoto a conducting shell is determined. (a) Show that there is a radial magnetic force per unit area acting on the shell Tr=µoK(Ho+Hi)/2. (Note that the thin-shell model implies thatHvaries in an essentially linear fashion with Rinside the shell.) (b) Speciﬁcally, show that Tr=−µoH2 o 2¡ 2−e−t/τm¢ e−t/τm(a) 11.10.3 In Example 10.4.1, the transient current induced in a conducting shell by the application of a transverse magnetic ﬁeld is found. Suppose that the magnetizable core is absent. (a) Show that the radial force per unit area acting on the shell is Tr= µoK(Ho φ+Hi φ)/2. (Note that according to the thin-shell model, H has an essentially linear dependence on rwithin the shell.) (b) Determine Tr(φ, t) and relate the result to Demonstration 10.4.1. 12 ELECTRODYNAMIC FIELDS: THE SUPERPOSITION INTEGRAL POINT OF VIEW 12.0 INTRODUCTION This chapter and the remaining chapters are concerned with the combined eﬀects of the magnetic induction ∂B/∂tin Faraday’s law and the electric displacement current ∂D/∂tin Amp` ere’s law. Thus, the full Maxwell’s equations without the quasistatic approximations form our point of departure. In the order introduced in Chaps. 1 and 2, but now including polarization and magnetization, these are, as generalized in Chaps. 6 and 9, ∇ ·(/epsilon1oE) =ρu− ∇ · P (1) ∇ ×H=Ju+∂ ∂t(/epsilon1oE+P) (2) ∇ ×E=−∂ ∂tµo(H+M) (3) ∇ ·(µoH) =−∇ · (µoM) (4) One may question whether a generalization carried out within the formalism of electroquasistatics and magnetoquasistatics is adequate to be included in the full dynamic Maxwell’s equations, and some remarks are in order. Gauss’ law for the electric ﬁeld was modiﬁed to include charge that accumulates in the polarization process. The accounting for the charge leaving a designated volume was done under no restrictions of quasistatics, and thus (1) can be adopted in the fully dynamic case. Subsequently, Amp` ere’s law was modiﬁed to preserve the divergence-free char- acter of the right-hand side. But there was more involved in that step. The term ∂P/∂tcan be identiﬁed unequivocally as the current density associated with a time dependent polarization process, provided that the medium as a whole is at rest. Thus, (2) is the correct generalization of Amp` ere’s law for polarizable media 1 2 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 at rest. If the medium moves with the velocity v, a term ∇ ×(P×v) has to be added to the right-hand side[1,2]. The generalization of Gauss’ law and Faraday’s law for magnetic ﬁelds is by analogy. If the material is moving and magnetized, a term−µo∇ ×(M×v) must be added to the right-hand side of (3). We shall not consider such moving polarized or magnetized media in the sequel. Throughout this chapter, we are generally interested in electromagnetic ﬁelds in free space. If the region of interest is ﬁlled by a material having an appreciable polarization and or magnetization, the constitutive laws are presumed to represent a linear and isotropic material D≡/epsilon1oE+P=/epsilon1E (5) B≡µo(H+M) =µH (6) and/epsilon1andµare assumed uniform throughout the region of interest.1Maxwell’s equations in linear and isotropic media may be rewritten more simply ∇ ·/epsilon1E=ρu (7) ∇ ×H=Ju+∂ ∂t/epsilon1E (8) ∇ ×E=−∂ ∂tµH (9) ∇ ·µH= 0 (10) Our approach in this chapter is a continuation of the one used before. By ex- pressing the ﬁelds in terms of superposition integrals, we emphasize the relationship between electrodynamic ﬁelds and their sources. Next we take into account the ef- fect of conducting bodies upon the electromagnetic ﬁeld, introducing the boundary value approach. We began Chaps. 4 and 8 by expressing an irrotational Ein terms of a scalar potential Φ and a solenoidal Bin terms of a vector potential A. We start this chapter in Sec. 12.1 with the generalization of these potentials to represent the electric and magnetic ﬁelds under electrodynamic conditions. Poisson’s equation related Φ to its source in Chap. 4 and Ato the current density Jin Chap. 8. What equation relates these potentials to their sources when quasistatic approximations do not apply? In Sec. 12.1, we develop the inhomogeneous wave equation, which assumes the role played by Poisson’s equation in the quasistatic cases. It follows from this equation that for linearly polarizable and magnetizable materials, the superposition principle applies to electrodynamics. The ﬁelds associated with source singularities are the next topic, in analogy either with Chaps. 4 or 8. In Sec. 12.2, we start with the ﬁeld of an elemental charge and build up the ﬁeld of a dynamic electric dipole. Here we exemplify the launching of an electromagnetic wave and see how the quasistatic electric dipole ﬁelds relate to the more general electrodynamic ﬁelds. The section concludes by deriving the electrodynamic ﬁelds associated with a magnetic dipole from the ﬁelds 1To make any relation in this chapter apply to free space, let /epsilon1=/epsilon1oandµ=µo. Sec. 12.1 Electrodynamic Potentials 3 for an electric dipole by exploiting the symmetry of Maxwell’s equations in source- free regions. The superposition integrals developed in Sec. 12.3 provide particular solutions to the inhomogeneous wave equations, just as those of Chaps. 4 and 8, respectively, gave solutions to the scalar and vector Poisson’s equations. In describing the op- eration of antennae, the ﬁelds that radiate away from the source are of primary interest. The superposition integrals for these radiation ﬁelds are used to ﬁnd an- tenna radiation patterns in Sec. 12.4. The discussion of antennae is continued in Sec. 12.5, which has as a theme the complex form of Poynting’s theorem. This theorem makes it possible to model the impedance of antennae as “seen” by their driving sources. In Sec. 12.6, the ﬁeld sources take the form of surface currents and surface charges. It is generally not convenient to ﬁnd the associated ﬁelds by making direct use of the superposition integrals. Nevertheless, the sources are a “given,” and any method that results in the associated ﬁelds amounts to solving the superposition integrals. This section provides a ﬁrst view of the solutions to the wave equation in Cartesian coordinates that will be derived from the boundary value point of view in Chap. 13. In preparation for the boundary value approach of the next chapter, boundary conditions are satisﬁed by appropriate choices of sources. Thus, the parallel plate waveguide considered from the boundary value point of view in Chap. 13 is seen here from the point of view of waves initiated by given sources. The method of images, taken up in Sec. 12.7, provides further examples of this approach to satisfying boundary conditions. When boundaries are introduced in this chapter, they are presumed to be perfectly conducting. In Chap. 13, the boundaries can also be interfaces between perfectly insulating dielectrics. In both of these chapters, the theme is dynamical phenomena related to the propagation and reﬂection of electromagnetic waves. The dynamics are characterized by one or more electromagnetic transit times, τem. Dy- namical phenomena associated with charge relaxation or magnetic diﬀusion, char- acterized by τeandτm, are excluded. We will look at these again in Chaps. 14 and 15. 12.1 ELECTRODYNAMIC FIELDS AND POTENTIALS In this section, we extend the use of the scalar and vector potentials to the de- scription of electrodynamic ﬁelds. In regions of interest, the current density Jof unpaired charge and the charge density ρuare prescribed functions of space and time. If there is any material present, it is of uniform permittivity /epsilon1and permeabil- ityµ,D=/epsilon1EandB=µH. For quasistatic ﬁelds in such regions, the potentials Φ andAare governed by Poisson’s equation. In this section, we see the role of Pois- son’s equation for quasistatic ﬁelds taken over by the inhomogeneous wave equation for electrodynamic ﬁelds. In both Chaps. 4 and 8, potentials were introduced so as to satisfy automati- cally the one of the two laws that was source free. In Chap. 4, we made E=−∇Φ so that Ewas automatically irrotational, ∇×E= 0. In Chap. 8 we let B=∇×A so that Bwas automatically solenoidal, ∇ ·B= 0. Of the four laws compris- ing Maxwell’s equations, (12.0.7)–(12.0.10), those of Gauss and Amp` ere involve 4 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 sources, while the last two, Faraday’s law and the magnetic ﬂux continuity law, do not. Following the approach used before, potentials should be introduced that automatically satisfy Faraday’s law and the magnetic ﬂux continuity law, (12.0.9) and (12.0.10). This is the objective of the following steps. Given that the magnetic ﬂux density remains solenoidal, the vector potential Acan be deﬁned just as it was in Chap. 8. B=µH=∇ ×A (1) With µHrepresented in this way, (12.0.10) is again automatically satisﬁed and Faraday’s law, (12.0.9), becomes ∇ ×¡ E+∂A ∂t¢ = 0 (2) This expression is also automatically satisﬁed if we make the quantity in brackets equal to −∇Φ. E=−∇Φ−∂A ∂t (3) With HandEdeﬁned in terms of Φ and Aas given by (1) and (3), the last two of the four Maxwell’s equations, (12.0.9–12.0.10), are automatically satisﬁed. Note, however, that the potentials that represent given ﬁelds HandEare not fully speciﬁed by (1) and (3). We can add to Athe gradient of any scalar function, thus changing both Aand Φ without aﬀecting HorE. A further speciﬁcation of the potentials will therefore be given shortly. We now turn to ﬁnding the equations that Aand Φ must obey if the laws of Gauss and Amp` ere, the ﬁrst two of (12.0.9-12.0.10), are to be satisﬁed. Substitution of (1) and (3) into Amp` ere’s law, (12.0.8), gives ∇ ×(∇ ×A) =µ/epsilon1∂ ∂t¡ − ∇Φ−∂A ∂t¢ +µJu (4) A vector identity makes it possible to rewrite the left-hand side so that this equation is ∇(∇ ·A)− ∇2A=µ/epsilon1∂ ∂t¡ − ∇Φ−∂A ∂t¢ +µJu (5) With the gradient and time derivative operators interchanged, this expression is ∇¡ ∇ ·A+µ/epsilon1∂Φ ∂t¢ − ∇2A=−µ/epsilon1∂2A ∂t2+µJu (6) To uniquely specify A, we must not only stipulate its curl, but give its di- vergence as well. This point was made in Sec. 8.0. In Sec. 8.1, where we were concerned with MQS ﬁelds, we found it convenient to make Asolenoidal. Here, Sec. 12.1 Electrodynamic Potentials 5 where we have kept the displacement current, we set the divergence of Aso that the term in brackets on the left is zero. ∇ ·A=−µ/epsilon1∂Φ ∂t (7) This choice of ∇ ·Ais called the choice of the Lorentz gauge . In this gauge, the expression representing Amp` ere’s law, (6), reduces to one involving Aalone, to the exclusion of Φ. ∇2A−µ/epsilon1∂2A ∂t2=−µJu(8) The last of Maxwell’s equations, Gauss’ law, is satisﬁed by making Φ obey the diﬀerential equation that results from the substitution of (3) into (12.0.7). ∇ ·/epsilon1¡ − ∇Φ−∂A ∂t¢ =ρu⇒ ∇2Φ +∂ ∂t(∇ ·A) =−ρu /epsilon1(9) We can substitute for ∇ ·Ausing (7), thus eliminating Afrom this expression. ∇2Φ−µ/epsilon1∂2Φ ∂t2=−ρu /epsilon1 (10) In summary, with HandEdeﬁned in terms of the vector potential Aand scalar potential Φ by (1) and (3), the distributions of these potentials are governed by the vector and scalar inhomogeneous wave equations (8) and (10), respectively. Theunpaired charge density and the unpaired current density are the “sources” in these equations. In representing the ﬁelds in terms of the potentials, it is understood that the “gauge” of Ahas been set so that Aand Φ are related by (7). The time derivatives in (8) and (10) are the result of retaining both the displacement current and the magnetic induction. Thus, in the quasistatic limits, these terms are neglected and we return to vector and scalar potentials governed by Poisson’s equation. Superposition Principle. The inhomogeneous wave equations satisﬁed by A and Φ [(8) and (10)] as well as the gauge condition, (7), are linear when the sources on the right are prescribed. That is, if solutions Aaand Φ aare associated with sources Jaandρa, (Ja, ρa)⇒(Aa,Φa) (11) and similarly, Jbandρbproduce the potentials Ab,Φb, (Jb, ρb)⇒(Ab,Φb) (12) 6 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 then the potentials resulting from the sum of the sources is the sum of the potentials. [(Ja+Jb),(ρa+ρb)]⇒[(Aa+Ab),(Φa+ Φb)] (13) The formal proof of this superposition principle follows from the same reasoning used for Poisson’s equation in Sec. 4.3. In prescribing the charge and current density on the right in (8) and (10), it should be remembered that these sources are related by the law of charge conser- vation. Thus, although Φ and Aappear in (8) and (10) to be independent, they are actually coupled. This interdependence of the sources is reﬂected in the link between the scalar and vector potentials established by the gauge condition of (7). Once Ahas been found, it is often convenient to use this relation to determine Φ. Continuity Conditions. Each of Maxwell’s equations, (12.0.7)–(12.0.10), as well as the charge conservation law obtained by combining the divergence of Amp` ere’s laws with Gauss’ law, implies a continuity condition. In the absence of polarization and magnetization, these conditions were derived from the integral laws in Chap. 1. Generalized to include polarization and magnetization in Chaps. 6 and 9, the continuity conditions for (12.0.7)–(12.0.10) are, respectively, n·(/epsilon1aEa−/epsilon1bEb) =σsu (14) n×(Ha−Hb) =Ku (15) n×(Ea−Eb) = 0 (16) n·(µaHa−µbHb) = 0 (17) The derivation of these conditions is the same as given at the end of the sections introducing the respective integral laws in Chap. 1, except that µoHis replaced by µHin Faraday’s law and /epsilon1oEby/epsilon1Ein Amp` ere’s law. In Secs. 12.6 and 12.7, and in the following chapters, these conditions are used to relate electrodynamic ﬁelds to surface currents and surface charges. At the outset, we recognize that two of these continuity conditions are, like Faraday’s law and the law of magnetic ﬂux continuity, not independent of each other. Further, just as the laws of Amp` ere and Gauss imply the charge conservation relation between Juandρu, the continuity conditions associated with these laws imply the charge conservation continuity condition obeyed by the surface currents and surface charge densities. To see the ﬁrst interdependence, Faraday’s law is integrated over a surface S enclosed by a contour Clying in the plane of the interface , as shown in Fig. 12.1.1a. Stokes’ theorem is then used to write I CE·ds=−d dtZ SµH·da (18) Sec. 12.1 Electrodynamic Potentials 7 Fig. 12.1.1 (a) Surface Sjust above or just below the interface. (b) Volume Vof incremental thickness henclosing a section of the interface. Whether taken on side (a) or side (b) of the interface, the line integral on the left is the same. This follows from Faraday’s continuity law (16). Thus, if we take the diﬀerence between (18) evaluated on side (a) and on side (b), we obtain d dt(µaHa−µbHb)·n= 0 (19) By making the tangential electric ﬁeld continuous, we have assured the conti- nuity of the time derivative of the normal magnetic ﬂux density. For a sinusoidally time-dependent process, matching the tangential electric ﬁeld automatically assures the matching of the normal magnetic ﬂux densities. In particular, consider a surface of a conductor that is “perfect” in the MQS sense. The electric ﬁeld inside such a conductor is zero. From (16), the tangential component of Ejust outside the conductor must also be zero. In view of (19), we conclude that the normal ﬂux density at a perfectly conducting surface must be time independent. This boundary condition is familiar from the last half of Chap. 8.2 Given that the divergence of Amp` ere’s law combines with Gauss’ law to give conservation of charge, ∇ ·Ju+∂ρu ∂t= 0 (20) we should expect that there is a second relationship among the conditions of (14)– (17), this time between the surface charge density and surface current density that appear in the ﬁrst two. Integration of (20) over the volume of the “pillbox” shown in Fig. 12.1.1b gives lim h→0 ∆A→0·I SJu·da+d dtZ VρudV¸ = 0 (21) In the limit where ﬁrst the thickness hand then the area ∆ Ago to zero, these integrals reduce to ∆ Atimes n·(Ja u−Jb u) +∇Σ·Ku+∂σu ∂t= 0 (22) 2Note that the absence of a time-varying normal ﬂux density does not imply that there is no tangential E. The surface of a material that is an inﬁnite conductor in one direction but an insulator in the other might have no normal µHand yet support a tangential Ein the direction of zero conductivity. 8 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 The ﬁrst term is the contribution to the ﬁrst integral in (21) from the surfaces on the (a) and (b) sides of the interface, respectively, having normals + nand−n. The second term, which is written in terms of the “surface divergence” deﬁned in terms of a vector Fby ∇Σ·F≡lim ∆A→0I CF·indl (23) results because the surface current density makes a ﬁnite contribution to the ﬁrst integral in (21) even though the thickness hof the volume goes to zero. [In (23), in is the unit normal to the volume V, as shown in the ﬁgure.] Such a surface current density can be used to represent currents imposed over a region having a thickness that is small compared to other dimensions of interest. It can also represent the current on the surface of a perfect conductor. (In using the conservation of charge continuity condition in Secs. 7.6 and 7.7, this term was not present because the surfaces described by this continuity condition were not carrying surface currents.) In terms of coordinates local to the point of interest, the surface divergence can be thought of as a two-dimensional divergence. The last term in (22) results from the integration of the charge density over the volume. Because there is a surface charge density, there is net charge inside the volume even in the limit where h→0. When we specify Kuandσuin (14) and (15), it is with the understanding that they obey the charge conservation continuity condition, (22). But, we also conclude that the charge conservation law is implied by the laws of Amp` ere and Gauss, and so we know that if (14) and (15) are satisﬁed, then so too is (22). When perfectly conducting boundaries are described in Chaps. 13 and 14, the surface current and charge found on a perfectly conducting boundary using the continuity conditions from the laws of Amp` ere and Gauss will automatically satisfy the charge conservation condition. Further, a zero tangential electric ﬁeld on a perfect conductor automatically implies that the normal magnetic ﬂux density vanishes. With the inhomogeneous wave equation playing the role of Poisson’s equation, the stage is now set for a scenario paralleling that for electroquasistatics in Chap. 4 and for magnetoquasistatics in Chap. 8. The next section identiﬁes the ﬁelds associated with source singularities. Section 12.3 develops superposition integrals for the response to given distributions of the sources. Henceforth, in this and the next chapter, we shall drop the subscript ufrom the source quantities. 12.2 ELECTRODYNAMIC FIELDS OF SOURCE SINGULARITIES Given the response to an elemental source, the ﬁelds associated with an arbi- trary distribution of sources can be found by superposition. This approach will be formalized in the next section and can be utilized for determining the radiation pat- terns of many antenna arrays. The ﬁelds resulting from this superposition principle form a particular solution that can be combined with solutions to the homogeneous wave equation to satisfy the boundary conditions imposed by perfectly conducting boundaries. We begin by identifying the potential Φ associated with a time varying point charge q(t). In a closed system, where the net charge is invariant, an increase in Sec. 12.2 Fields of Source Singularities 9 Fig. 12.2.1 A point charge located at the origin of a spherical coordinate system. charge at one point must be compensated by a decrease in charge elsewhere. Thus, as we shall see in identifying the ﬁelds of an electric dipole, physically meaningful ﬁelds are the superposition of those produced by at least two point charges of opposite sign. Conservation of charge further requires that this shift in the distribution of net charge from one region to another be accounted for by a current. This current is the source term in the inhomogeneous wave equation for the vector potential. Potential of a Point Charge. Consider the potential Φ predicted by the in- homogeneous wave equation, (12.1.10), for a time varying point charge q(t) located at the origin of the spherical coordinate system shown in Fig. 12.2.1. By deﬁnition, ρis zero everywhere except at the origin, where it is singular.3In the immediate neighborhood of the origin, we should expect that the potential varies so rapidly with rthat the Laplacian would dominate the second time derivative in the inhomogeneous wave equation, (12.1.10). Then, in the vicinity of the origin, we should expect the potential for a point charge to be the same as for Poisson’s equation, namely q(t)/(4π/epsilon1r) (4.4.1). From Sec. 3.1, we have a hint as to how the combined eﬀects of the magnetic induction and electric displacement current represented by the second time derivative in the inhomogeneous wave-equation, (12.1.10), should aﬀect this potential. We can expect that the response at a radial position rwill be delayed by the time required for an electromagnetic wave to reach that position from the origin. For a wave propagating at the velocity c, this time isr/c. Thus, we make the educated guess that the solution to (12.1.10) for a point charge at the origin is Φ =q¡ t−r c¢ 4π/epsilon1r(1) where c= 1/√µ/epsilon1. According to (1), given that the time dependence of the point charge is q(t), the potential at radius ris given by the familiar potential for a point charge, provided that t→(t−r/c). Veriﬁcation that Φ of (1) is a solution to the inhomogeneous wave equation (12.1.10) takes two steps. First, the expression is substituted into the homogeneous wave equation [(12.1.10) with no source] to see that it is satisﬁed everywhere except at the origin. In carrying out this step, note that Φ is a function of the spherical 3Of course, charge conservation requires that there be a current supplying this time-varying charge and that through action of this current, if charge accumulates at the origin, there must be a reduction of charge somewhere else. The simplest example of a source obeying charge conservation is the dipole. 10 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 radial coordinate ralone. Thus, ∇2Φ is simply r−2∂(r2∂Φ/∂r)/∂r. This operation gives the same result as the operation r−1∂2(rΦ)/∂r2. Thus, evaluated using the potential of (1), the terms on the left in the inhomogeneous wave equation, (12.1.10), become ∇2Φ−1 c2∂2Φ ∂t2=1 4π/epsilon1·1 r∂2 ∂r2q¡ t−r c¢ −1 r1 c2∂2 ∂t2q¡ t−r c¢¸ = 0 (2) forr/negationslash= 0. In carrying out this evaluation, note that ∂q/∂r =−q/prime/cand∂q/∂t =q/prime where the prime indicates a derivative with respect to the argument. Thus, the homogeneous wave equation is satisﬁed everywhere except at the origin. In the second step, we conﬁrm that (1) is the dynamic potential of a point charge. We integrate the inhomogeneous wave equation in the neighborhood of r= 0, (12.1.10), over a small spherical volume of radius rcentered on the origin. Z Vµ − ∇ · ∇ Φ +µ/epsilon1∂2Φ ∂t2¶ dv=Z Vρ /epsilon1dv (3) The Laplacian has been written in terms of its deﬁnition in anticipation of using Gauss’ theorem to convert the ﬁrst integral to one over the surface at r. In the limit where ris small, the integration of the second time derivative term gives no contribution. Z Vµ/epsilon1∂2Φ ∂t2dv=µ/epsilon1∂2 ∂t2lim r→0Z VΦdv =µ/epsilon1∂2 ∂t2lim r→0Zr 0q4πr2 4π/epsilon1rdr= 0(4) Integration of the ﬁrst term on the left in (3) is familiar from Chap. 4, because Gauss’ theorem converts the volume integration to one over the enclosing surface and we therefore have −Z ∇ · ∇ Φdv=−I S∇Φ·da=−4πr2∂Φ ∂r =−4πr2¡ −q 4π/epsilon1r2¢ =q /epsilon1(5) In the limit where r→0, the integral on the right in (3) gives q//epsilon1. Thus, it reduces to the same expression obtained using (1) to evaluate the left-hand side of (3). We conclude that (1) is indeed the solution to the inhomogeneous wave equation for a point charge at the origin. Electric Dipole Field. An electric dipole consists of a pair of charges ±q(t) separated by the distance d, as shown in Fig. 12.2.2. As one charge increases in magnitude at the expense of the other, there is an elemental current i(t) directed between the two along the zaxis. Charge conservation requires that Sec. 12.2 Fields of Source Singularities 11 Fig. 12.2.2 A dynamic dipole in which the time-variation of the charge is accounted for by the elemental current i(t). i=dq dt (6) This current can be pictured as a singularity in the distribution of the current density Jz. In fact, the role played by ρ//epsilon1as the source of Φ on the right in (12.1.10) is played by µJzin determining Azin (12.1.8). Just as qcan be regarded as the integral of the charge density ρover the elemental volume occupied by that charge density, µidisµJzﬁrst integrated over the cross-sectional area in the x−yplane of the current tube joining the charges (to give µi) and then integrated over the length dof the tube. Thus, we exploit the analogy between the zcomponent of the vector inhomogeneous wave equation for Azand that for Φ, (12.1.8) and (12.1.10), to write the vector potential associated with an incremental current element at the origin. The solution to (12.1.8) is the same as that to (12.1.10) with q//epsilon1→µid. Az=µdi¡ t−r c¢ 4πr (7) Remember that ris a spherical coordinate, so it is best to convert this ex- pression into spherical coordinates. Figure 12.2.3 shows that Ar=Azcosθ; Aθ=−Azsinθ (8) Thus, in spherical coordinates, (7) becomes the vector potential for an electric dipole. A=µd 4π·i¡ t−r c¢ rcosθir−i¡ t−r c¢ rsinθiθ¸ (9) The dipole scalar potential is the superposition of the potentials due to the individual charges, (5). The positive charge is located on the zaxis at z=d, while the negative one is at the origin, so superposition gives Φ =1 4π/epsilon1½q£ t−¡r c−d ccosθ¢¤ r−dcosθ−q£ t−r c¤ r¾ (10) 12 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.2.3 The z-directed potential is analyzed into its components in spherical coordinates. where, in a way familiar from Sec. 4.4, the distance from the point of observation to the charge at z=dis approximated by r−dcosθ. With q/primeindicating a derivative with respect to the argument, expansion in a Taylor’s series based on dcosθ/lessmuchr gives Φ/similarequal1 4π/epsilon1½µ1 r+dcosθ r2¶ q¡ t−r c¢ +d ccosθ rq/prime¡ t−r c¢ −q¡ t−r c¢ r¾ (11) and keeping terms that are linear in dresults in the desired scalar potential for the electric dipole. Φ =d 4π/epsilon1·q¡ t−r c¢ r2+q/prime¡ t−r c¢ cr¸ cosθ (12) The vector potential (9) and scalar potential (12) obey (12.1.7), as can be con- ﬁrmed by diﬀerentiation and use of the conservation law (6). We can now evaluate the magnetic and electric ﬁelds associated with these scalar and vector potentials. The magnetic ﬁeld intensity follows by evaluating (12.1.1) using (9). [Remember that conservation of charge requires that q/prime=i, in accordance with (6).] H=d 4π·i/prime¡ t−r c¢ cr+i¡ t−r c¢ r2¸ sinθiφ (13) To ﬁnd E, (12.1.3) is evaluated using (9) and (12). E=d 4π/epsilon1½ 2·q¡ t−r c¢ r3+q/prime¡ t−r c¢ cr2¸ cosθir +·q¡ t−r c¢ r3+q/prime¡ t−r c¢ cr2+q/prime/prime¡ t−r c¢ c2r¸ sinθiθ¾ (14) As can be seen by comparing (14) to (4.4.10), in the limit where c→ ∞ , this electric ﬁeld becomes the electric ﬁeld found from the electroquasistatic dipole potential. Note that the quasistatic ﬁeld is proportional to q(rather than its ﬁrst or second temporal derivative) and decays as 1 /r3. The ﬁrst and second time deriva- tives of qare of order q/τandq/τ2respectively, where τis the typical time interval Sec. 12.2 Fields of Source Singularities 13 Fig. 12.2.4 Far ﬁelds constituting a plane wave propagating in the radial direction. within which qexperiences an appreciable change. Thus, these time derivative terms are small compared to the quasistatic terms if r/c/lessmuchτ. What we have found gives substance to the arguments given for the EQS approximation in Sec. 3.3. That is, we have found that the quasistatic approximation is justiﬁed if the condition of (3.3.5) prevails. The combination of electric displacement current and magnetic induction lead- ing to the inhomogeneous wave equation has three dramatic eﬀects on the dipole ﬁelds. First, the response at a location ris delayed4by the transit time r/c. Second, the electric ﬁeld is not only proportional to q(t−r/c), but also to q/prime(t−r/c) and q/prime/prime(t−r/c). Third, the part of the electric ﬁeld that is proportional to q/prime/primedecreases with radius in proportion to 1 /r. Associated with this “far ﬁeld” is a magnetic ﬁeld, the ﬁrst term in (13), that similarly decreases as 1 /r. Together, these ﬁelds comprise an electromagnetic wave propagating radially outward from the dipole antenna. lim r→∞H→d 4πi/prime¡ t−r c¢ sinθiφ cr lim r→∞E→d 4π/epsilon1q/prime/prime¡ t−r c¢ c2rsinθiθ (15) Note that these ﬁeld components are orthogonal to each other and transverse to the radial direction of propagation, as shown in Fig. 12.2.4. To appreciate the signiﬁcance of the 1 /rdependence of the ﬁelds in (15), consider the Poynting ﬂux, (11.2.9), associated with these ﬁelds. lim r→∞[E×H] =¡d 4π¢2p µ//epsilon1£ q/prime/prime¡ t−r c¢¤2 c2r2sin2θ (16) The power ﬂow out through a spherical surface at the radius rfollows from this expression as P=I E×H·da=Zπ 0¡d 4π¢2p µ//epsilon1(q/prime/prime)2 c2r2sin2θ2πr2sinθdθ =d2 6πp µ//epsilon1£ q/prime/prime¡ t−r c¢¤2 c2(17) 4In addition to the retarded response highlighted here, an “advanced” response, where t− r/c→t+r/c, is also a solution to the inhomogeneous wave equation. Because it does not ﬁt with our idea of causality, it is not used here. 14 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.2.5 (a) Time dependence of the dipole charge q(t) as well as its ﬁrst and second derivatives. (b) The radial dependence of the functions needed to evaluate the dipole ﬁelds resulting from the turn-on transient of (a) when t > T . Because the far ﬁelds of the dipole vary as 1 /r, and hence the power ﬂux density is proportional to 1 /r2, and because the area of the surface at rincreases asr2, we conclude that there is a net power ﬂowing outward from the dipole at inﬁnity. These far ﬁeld components are called the radiation ﬁeld . Example 12.2.1. Turn-on Fields of an Electric Dipole To help establish the physical signiﬁcance of the electric dipole expressions for EandH, (13) and (14), consider the ﬁelds associated with charging an electric dipole through the transient shown in Fig. 12.2.5a. Over a period T, the charge increases from zero to Qwith a continuous ﬁrst derivative but a second derivative that suﬀers a ﬁnite discontinuity, as shown in the ﬁgure. Multiplied by appropriate factors of 1 /r,1/r2, and 1 /r3, the ﬁeld distributions are made up of these three functions, with treplaced by t−r/c. Thus, at a given instant in time, the factors q(t−r/c), q/prime(t−r/c), and q/prime/prime(t−r/c) have the radial distributions shown in Fig. 12.2.5b. The electric and magnetic ﬁelds are shown at three successive instants in time in Fig. 12.2.6. The transient part of the ﬁeld is conﬁned to an annular region with its outside radius at r=ct(the wave front) and inner radius at r=c(t−T). Inside this latter radius, the ﬁelds are static and composed only of those terms varying as 1/r3. Thus, when t=T(Fig. 12.2.6a), all of the ﬁeld is transient, because the source has just reached a constant state. At the subsequent times t= 2Tandt= 3T, the ﬁelds left behind by the outward propagating rear of the wave transient, the Eﬁeld of a static electric dipole and H=0, are as shown in Figs. 12.2.6b and 6c. The ﬂow of charges to the poles of the dipole produces an electromagnetic wave which reveals its identity once the annular region of the transient ﬁelds propagates out of the range of the near ﬁeld. Note that the electric and magnetic ﬁelds shown in the outward propagating wave of Fig. 12.2.6c are mutually sustaining. In accordance with Faradays’ law, the curl of E, which is φdirected and tends to be largest midway Sec. 12.2 Fields of Source Singularities 15 Fig. 12.2.6 Electric ﬁelds (solid lines) and magnetic ﬁelds resulting from turning on an electric dipole in accordance with the temporal de- pendence indicated in Fig. 12.2.5. The ﬁelds are zero outside the wave front indicated by the outermost broken line. (a) For t < T , the entire ﬁeld is in a transient state. (b) By the time t > T , the ﬁelds due to the transient are seen to be propagating outward between the expanding spherical surfaces at r=ctandr=c(t−T). Inside the latter surface, which is also indicated by a broken line, the ﬁelds are static. (c) At still later times, the propagating wave divorces itself from the dipole as the electric ﬁeld generated by the magnetic induction, and the magnetic ﬁeld generated by the displacement current, become self-sustaining. between the front and back of the wave, is balanced by a time rate of change of B which also has its largest value in the same region.5Similarly, to satisfy Amp` ere’s law, the θ-directed curl of H, which also peaks midway between the front and back of the wave, is balanced by a time rate of change of Dthat peaks in the same region. It is instructive to review the discussion given in Sec. 3.3 of EQS and MQS approximations and their relation to electromagnetic waves. The electric dipole considered here in detail is the prototype system sketched in Fig. 3.3.1a. We have indeed found that if the condition of (3.3.5) is met, the EQS ﬁelds dominate. We should expect that if the current carried by the elemental loop of the prototype MQS system of Fig. 3.3.1b is a rapidly varying function of time, then the magnetic dipole (considered in the MQS limit in Sec. 8.3) also gives rise to a radiation ﬁeld 5In discerning a time rate of change implied by the ﬁgure, remember that the ﬁelds in the region of the spherical shell indicated by the two broken-line circles in Figs. 12.2.6b and 12.2.6c are propagating outward. 16 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 much like that discussed here. These ﬁelds are considered at the conclusion of this section. Electric Dipole in the Sinusoidal Steady State. In the sections that follow, the ﬁelds of the electric dipole will be superimposed to obtain ﬁeld patterns from antennae used at radio and microwave frequencies. In most of these practical sit- uations, the ﬁeld sources, qandi, are essentially in the sinusoidal steady state. In particular, i= Re ˆiejωt(18) where ˆiis a complex number representing both the phase and amplitude of the current. Then, the general expression for the vector potential of the electric dipole, (7), becomes Az= Reµdˆi 4πrejω(t−r c)(19) Separation of the time dependence from the space dependence in solving the inho- mogeneous wave equation is accomplished by the use of complex vector functions of space multiplied by exp jωt. With the understanding that the time dependence is recovered by multiplying by exp( jωt) and taking the real part, we will now deal with the complex amplitudes of the ﬁelds and drop the factor exp jωt. Thus, (19) becomes Az= Re ˆAzejωt; ˆAz=µdˆi 4πe−jkr r(20) where the wave number k≡ω/c. In terms of complex amplitudes, the magnetic and electric ﬁeld intensities of the electric dipole follow from (13) and (14) as [by substituting q→Re (ˆi/jω) exp (−jkr) exp( jωt)] ˆH=jkdˆi 4π¡1 jkr+ 1¢ sinθe−jkr riφ (21) ˆE=jkdˆi 4πp µ//epsilon1½ 2·1 (jkr)2+1 jkr¸ cosθir +·1 (jkr)2+1 jkr+ 1¸ sinθiθ¾e−jkr r(22) The far ﬁelds are given by terms with the 1 /rdependence. ˆHφ=jkdˆi 4πsinθe−jkr r (23) ˆEθ=p µ//epsilon1ˆHφ (24) Sec. 12.2 Fields of Source Singularities 17 Fig. 12.2.7 Radiation pattern of short electric dipole, shown in the range π/2< φ < 3π/2. These ﬁelds, which are a special case of those pictured in Fig. 12.2.4, propagate radially outward. The far ﬁeld pattern is a radial progression of the ﬁelds shown between the broken lines in Fig. 12.2.6c. (The response shown is the result of one half of a cycle.) It follows from (23) and (24) that for a short dipole in the sinusoidal steady state, the power radiated per unit solid angle is6 4πr2/angbracketleftSr/angbracketright 4π=r21 2ReˆE×ˆH∗·ir=1 2p µ//epsilon1(kd)2 (4π)2|ˆi|2sin2θ (25) Equation (25) expresses the dependence of the radiated power on the direction (θ, φ), and can be called the radiation pattern. Often, only the functional depen- dence, Ψ( θ, φ) is identiﬁed with the “radiation pattern.” In the case of the short electric dipole, Ψ(θ, φ) = sin2θ (26) and the radiation pattern is as shown in Fig. 12.2.7. The Far-Field and Uniformly Polarized Plane Waves. For an observer far from the dipole, the variation of the ﬁeld with respect to radius is more noticeable than that with respect to the angle θ. Further, if kris large, the radial variation represented by exp( −jkr) dominates over the much weaker dependence due to the factor 1 /r. This term makes the ﬁelds tend to repeat themselves every wavelength λ= 2π/k. At frequencies of the order used for VHF television, the wavelength is on the order of a meter, while the station antenna is typically kilometers away. Thus, over the dimensions of a receiving antenna, the variations due to the factor 1/rand the θvariation in (23) and (24) are insigniﬁcant. By contrast, the receiving antenna has dimensions on the order of λ, and so the radial variation represented by exp(−jkr) is all-important. Far from the dipole, where spatial variations transverse to the radial direction of propagation are unimportant, and where the slow decay due to the 1 /rterm is negligible, the ﬁelds take the form of uniform plane waves . With the local spherical coordinates replaced by Cartesian coordinates, as shown in Fig. 12.2.8, the ﬁelds then take the form E=Ez(y, t)iz;H=Hx(y, t)ix (27) 6Here we use the time average theorem of (11.5.6). 18 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.2.8 (a) Radiation ﬁeld of electric dipole. (b) Cartesian representation in neighborhood of remote point. That is, the ﬁelds depend only on y, which plays the role of r, and are directed transverse to y. Instead of the far ﬁelds given by (15), we have traveling-wave ﬁelds that, by virtue of their independence of the transverse coordinates, are called plane waves . To emphasize that the dipole has indeed launched a plane wave, in (15) we replace d 4π/epsilon1q/prime/prime¡ t−r c¢ c2rsinθ→E+¡ t−y c¢ (28) and recognize that i/prime=q/prime/prime, (6), so that E=E+¡ t−y c¢ iz;H=p /epsilon1/µE +¡ t−y c¢ ix (29) The dynamics of such plane waves are described in Chap. 14. Note that the ratio of the magnitudes of EandHis the intrinsic impedance ζ≡p µ//epsilon1. In free space, ζ=ζo≡p µo//epsilon1o≈377Ω. Magnetic Dipole Field. Given the magnetic and electric ﬁelds of an elec- tric dipole, (13) and (14), what are the electrodynamic ﬁelds of a magnetic dipole? We answer this question by exploiting a far-reaching property of Maxwell’s equa- tions, (12.0.7)–(12.0.10), as they apply where Ju= 0 and ρu= 0. In such regions, Maxwell’s equations are replicated by replacing Hby−E,EbyH, /epsilon1byµ, and µ by/epsilon1. It follows that because (13) and (14) are solutions to Maxwell’s equations, then so are the ﬁelds. E=−d 4π¡q/prime/prime m cr+q/prime m r2¢ sinθiφ (30) H=d 4πµ· 2¡qm r3+q/prime m cr2¢ cosθir+¡qm r3+q/prime m cr2+q/prime/prime m c2r¢ sinθiθ¸ (31) Of course, qmmust now be interpreted as a source of divergence of H, i.e., a magnetic charge. Substitution shows that these ﬁelds do indeed satisfy Maxwell’s equations with J= 0 and ρ= 0, except at the origin. To discover the source singularity at the origin giving rise to these ﬁelds, they are examined in the limit Sec. 12.2 Fields of Source Singularities 19 Fig. 12.2.9 Magnetic dipole giving rise to the ﬁelds of (33) and (34). where r→0. Observe that in the neighborhood of the origin, terms proportional to 1/r3dominate Has given by (31). Close to the source, Htakes the form of a magnetic dipole. This can be seen by a comparison of this near ﬁeld to that given by (8.3.20) for a magnetic dipole. dqm¡ t−r c¢ =µm¡ t−r c¢ (32) With this identiﬁcation of the source, (30) and (31) become E=−µ 4π·m/prime/prime¡ t−r c¢ cr+m/prime¡ t−r c¢ r2¸ sinθiφ (33) H=1 4π½ 2·m¡ t−r c¢ r3+m/prime¡ t−r c¢ cr2¸ cosθir +·m¡ t−r c¢ r3+m/prime¡ t−r c¢ cr2+m/prime/prime¡ t−r c¢ c2r¸ sinθiθ¾ (34) The small current loop of Fig. 12.2.9, which has a magnetic moment m=πR2i, could be the source of the ﬁelds given by (33) and (34). If the current driving this loop were turned on in a manner analogous to that considered in Example 12.2.1, the ﬁeld left behind the outward propagating pulse would be the magnetic dipole ﬁeld derived in Example 8.3.2. The complex amplitudes of the far ﬁelds for the magnetic dipole are the counterpart of the ﬁelds given by (23) and (24) for an electric dipole. They follow from the ﬁrst term of (33) and the last term of (34) as ˆHθ=−k2 4πˆmsinθe−jkr r (35) ˆEφ=−p µ//epsilon1ˆHθ (36) 20 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 In Sec. 12.4, it will be seen that the radiation ﬁelds of the electric dipole can be superimposed to describe the radiation patterns of current distributions and of antenna arrays. A similar application of (35) and (36) to describing the radiation patterns of antennae composed of arrays of magnetic dipoles is illustrated by the problems. 12.3 SUPERPOSITION INTEGRAL FOR ELECTRODYNAMIC FIELDS With the identiﬁcation in Sec. 12.2 of the ﬁelds associated with point charge and current sources, we are ready to construct ﬁelds produced by an arbitrary distribution of sources. Just as the superposition integral of Sec. 4.5 was based on the linearity of Poisson’s equation, the superposition principle for the dynamic ﬁelds hinges on the linear nature of the inhomogeneous wave equations of Sec. 12.1. Transient Response. The scalar potential for a point charge qat the origin, given by (12.2.1), can be generalized to describe a point charge at an arbitrary source position r/primeby replacing the distance rby|r−r/prime|(see Fig. 4.5.1). Then, the point charge is replaced by the charge density ρevaluated at the source position multiplied by the incremental volume element dv/prime. With these substitutions in the scalar potential of a point charge, (12.2.1), the potential at an observer location r is the integrand of the expression Φ(r, t) =Z V/primeρ¡ r/prime, t−|r−r/prime| c¢ 4π/epsilon1|r−r/prime|dv/prime (1) The integration over the source coordinates r/primethen superimposes the ﬁelds at rdue to all of the sources. Given the charge density everywhere, this integral comprises the solution to the inhomogeneous wave equation for the scalar potential, (12.1.10). In Cartesian coordinates, any one of the components of the vector inhomoge- neous wave-equation, (12.1.8), obeys a scalar equation. Thus, with ρ//epsilon1→µJi, (1) becomes the solution for Ai, whether ibex, y orz. A(r, t) =µZ V/primeJ¡ r/prime, t−|r−r/prime| c¢ 4π|r−r/prime|dv/prime (2) We should keep in mind that conservation of charge implies a relationship between the current and charge densities of (1) and (2). Given the current density, the charge density is determined to within a time-independent distribution. An alternative, and often less involved, approach to ﬁnding Eavoids the computation of the charge density. Given J,Ais found from (2). Then, the gauge condition, (12.1.7), is used to ﬁnd Φ. Finally, Eis found from (12.1.3). Sec. 12.4 Antennae Radiation Fields 21 Sinusoidal Steady State Response. In many practical situations involving radio, microwave, and optical frequency systems, the sources are essentially in the sinusoidal steady state. ρ= Re ˆ ρ(r)ejωt⇒Φ = Re ˆΦ(r)ejωt(3) Equation (1) is evaluated by using the charge density given by (3), with r→r/primeand t→t− |r−r/prime|/c Φ = ReZ V/primeˆρ(r/prime)ejω¡ t−|r−r/prime| c¢ 4π/epsilon1|r−r/prime|dv/prime = Re·Z V/primeˆρ(r/prime)e−jk|r−r/prime| 4π/epsilon1|r−r/prime|dv/prime¸ ejωt(4) where k≡ω/c. Thus, the quantity in brackets in the second expression is the complex amplitude of Φ at the location r. With the understanding that the time dependence will be recovered by multiplying this complex amplitude by exp( jωt) and taking the real part, the superposition integral for the complex amplitude of the potential is ˆΦ =Z V/primeˆρ(r/prime)e−jk|r−r/prime| 4π/epsilon1|r−r/prime|dv/prime (5) From (2), the same reasoning gives the superposition integral for the complex am- plitude of the vector potential. ˆA=µ 4πZ V/primeˆJ(r/prime)e−jk|r−r/prime| |r−r/prime|dv/prime (6) The superposition integrals are often used to ﬁnd the radiation patterns of driven antenna arrays. In these cases, the distribution of current, and hence charge, is independently prescribed everywhere. Section 12.4 illustrates this application of the superposition integral. If ﬁelds are to be found in conﬁned regions of space, with part of the source distribution on boundaries, the ﬁelds given by the superposition integrals represent particular solutions to the inhomogeneous wave equations. Following the same ap- proach as used in Sec. 5.1 for solving boundary value problems involving Poisson’s equation, the boundary conditions can then be satisﬁed by superimposing on the solution to the inhomogeneous wave equation solutions satisfying the homogeneous wave equation. 12.4 ANTENNA RADIATION FIELDS IN THE SINUSOIDAL STEADY STATE Antennae are designed to transmit and receive electromagnetic waves. As we know from Sec. 12.2, the superposition integrals for the scalar and vector potentials result 22 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.4.1 Incremental current element at r/primeis source for radiation ﬁeld at (r, θ, φ ). in both the radiation and near ﬁelds. If we conﬁne our interest to the ﬁelds far from the antenna, extensive simpliﬁcations are achieved. Many types of antennae are composed of driven conducting elements that are extremely thin. This often makes it possible to use simple arguments to approxi- mate the distribution of current over the length of the conductor. With the current distribution speciﬁed at the outset, the superposition integrals of Sec. 12.3 can then be used to determine the associated ﬁelds. An element idz/primeof the current distribution of an antenna is pictured in Fig. 12.4.1 at the source location r/prime. If this element were at the origin of the spheri- cal coordinate system shown, the associated radiation ﬁelds would be as given by (12.2.23) and (12.2.24). With the distance to the current element r/primemuch less than r, how do we adapt these expressions so that they represent the ﬁelds when the incremental source is located at r/primerather than at the origin? The current elements comprising the antenna are typically within a few wave- lengths of the origin. By contrast, the distance r(say, from a TV transmitting antenna, where the wavelength is on the order of 1 meter, to a receiver 10 kilome- ters away) is far larger. For an observer in the neighborhood of a point ( r, θ, φ ), there is little change in sin θ/r, and hence in the magnitude of the ﬁeld, caused by a displacement of the current element from the origin to r/prime. However, the phase of the electromagnetic wave launched by the current element is strongly inﬂuenced by changes in the distance from the element to the observer that are of the order of a wavelength. This is seen by writing the argument of the exponential term in terms of the wavelength λ, jkr =j2πr/λ . With the help of Fig. 12.4.1, we see that the distance from the source to the observer is r−r/prime·ir. Thus, for the current element located at r/primein the neighborhood of the origin, the radiation ﬁelds given by (12.2.23) and (12.2.24) are ˆHφ/similarequaljk 4πsinθe−jk(r−r/prime·ir) ri(r/prime)dz/prime (1) Sec. 12.4 Antennae Radiation Fields 23 Fig. 12.4.2 Line current distribution as source of radiation ﬁeld. ˆEθ/similarequalrµ /epsilon1ˆHφ(2) Because EandHare vector ﬁelds, yet another approximation is implicit in writing these expressions. In shifting the current element, there is a slight shift in the coordinate directions at the observer location. Again, because ris much larger than |r/prime|, this slight change in the direction of the ﬁeld can be ignored. Thus, radiation ﬁelds due to a superposition of current elements can be found by simply superimposing the ﬁelds as though they were parallel vectors. Distributed Current Distribution. A wire antenna, driven by a given current distribution Re [ ˆi(z) exp( jωt)], is shown in Fig. 12.4.2. At the terminals, the complex amplitude of this current is ˆi=Ioexp(jωt+αo). It follows from (1) and the superposition principle that the magnetic radiation ﬁeld for this antenna is ˆHφ/similarequaljk 4πsinθe−jkr rZ ˆi(z/prime)ejkr/prime·irdz/prime(3) Note that the role played by idfor the incremental dipole is now played by i(z/prime)dz/prime. For convenience, we deﬁne a ﬁeld pattern function ψo(θ) that gives the θdependence of the EandHﬁelds ψo(θ)≡sinθ lZˆi(z/prime) Ioej(kr/prime·ir−αo)dz/prime (4) where lis the length of the antenna and ψo(θ) is dimensionless. With the aid of ψo(θ), one may write (3) in the form ˆHφ/similarequaljkl 4πe−jkr rIoejαoψo(θ)(5) 24 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.4.3 Center-fed wire antenna with standing-wave distribution of current. By deﬁnition, ˆi(ˆz/prime) =Ioexp(jαo) ifz/primeis evaluated at the terminals of the antenna. Thus, ψois neither a function of Ionor of αo. In order to evaluate (5), one needs to know the current dependence on z/prime,ˆi(z/prime). One can show that the current dis- tribution on a (open-ended) thin wire is made up of a “standing wave” with the dependence sin(2 πs/λ ) upon the coordinate smeasured along the wire, from the end of the wire. The proof of this statement will be presented in Chapter 14, when we shall discuss the current distribution in a coaxial cable. Example 12.4.1. Radiation Pattern of Center-Fed Wire Antenna A wire antenna, fed at its midpoint and on the zaxis, is shown in Fig. 12.4.3. The current distribution is “given” according to the above remarks. ˆi=−Iosink(|z| −l/2) sin(kl/2)ejαo(6) In setting up the radiation ﬁeld superposition integral, (5), observe that r/prime·ir= z/primecosθ. ψo=sinθ lZl/2 −l/2−sink(|z/prime| −l/2) sin(kl/2)ejkz/primecosθdz/prime(7) Evaluation of the integral7then gives ψo=2 klsin(kl/2)·cos(kl/2)−cos¡kl 2cosθ¢ sinθ¸ (8) The radiation pattern of the wire antenna is proportional to the absolute value squared of the θ-dependent factor of ψo Ψ(θ) =·cos(kl/2)−cos¡kl 2cosθ¢ sinθ¸2 (9) 7To carry out the integration, ﬁrst express the integration over the positive and negative segments of z/primeas separate integrals. With the sine functions represented by the sum of complex exponentials, the integration is reduced to a sum of integrations of complex exponentials. Sec. 12.4 Antennae Radiation Fields 25 Fig. 12.4.4 Radiation patterns for center-fed wire antennas. In viewing the plots of this radiation pattern shown in Fig. 12.4.4, remember that it is the same in any plane of constant φ. Thus, a three-dimensional picture of the function Ψ( θ, φ) is generated by rotating one of these patterns about the zaxis. The radiation pattern for a half-wave antenna diﬀers little from that for the short dipole, shown in Fig. 12.2.7. Because of the interference between waves gen- erated by segments having diﬀerent phases and amplitudes, the pattern for longer wires is more complex. As the length of the antenna is increased to many wave- lengths, the number of lobes increases. Arrays. Desired radiation patterns are often obtained by combining driven elements into arrays. To illustrate, consider an array of 1 + nelements, the ﬁrst at the origin and designated by “0”. The others are designated by i= 1. . . n and respectively located at ai. We can ﬁnd the radiation pattern for the array by summing over the contributions of the separate elements. Each of these takes the form of (5), with r→r−ai·ir, Io→Ii, αo→αi, and ψo(θ)→ψ(θ). ˆHφ/similarequaljkl 4πrnX i=0e−jk(r−ai·ir)Iiejαiψi(θ) (10) In the special case where the magnitude (but not the phase) of each element is the same and the elements are identical, so that Ii=Ioandψi=ψo, this expression can be written as ˆHθ/similarequaljkl 4πIoejαoe−jkr rψo(θ)ψa(θ, φ) (11) where the array factor is ψa(θ, φ)≡nX i=0ejkai·irej(αi−αo)(12) 26 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.4.5 Array consisting of two elements with spacing, a. Note that the radiation pattern of the array is represented by the square of the product of ψo, representing the pattern for a single element, and the array factor ψa. If the n+ 1 element array is considered one element in a second array, these same arguments could be repeated to show that the radiation pattern of the array of arrays is represented by the square of the product of ψo, ψaand the square of the array factor of the second array. Example 12.4.2. Two-Element Arrays The elements of an array have a spacing a, as shown in Fig. 12.4.5. The array factor follows from evaluation of (12), where ao= 0 and a1=aix. The projection of ir intoixgives (see Fig. 12.4.5) a1·ir=asinθcosφ (13) It follows that ψa= 1 + ej(kasinθcosφ+α1−αo)(14) It is convenient to write this expression as a product of a part that determines the phase and a part that determines the amplitude. ψa= 2ej(kasinθcosφ+α1−αo)/2cos£ka 2sinθcosφ+1 2(α1−αo)¤ (15) Dipoles in Broadside Array. With the elements short compared to a wavelength, the individual patterns are those of a dipole. It follows from (4) that ψo= sin θ (16) With the dipoles having a half-wavelength spacing and driven in phase, a=λ 2⇒ka=π, α 1−αo= 0 (17) The magnitude of the array factor follows from (15). |ψa|= 2¯¯cos¡π 2sinθcosφ¢¯¯ (18) Sec. 12.4 Antennae Radiation Fields 27 Fig. 12.4.6 Radiation pattern of dipoles in phase, half-wave spaced, is product of pattern for individual elements multiplied by the array factor. The radiation pattern for the array follows from (16) and (18). Ψ =|ψo|2|ψa|2= 4 sin2θcos2¡π 2sinθcosφ¢ (19) Figure 12.4.6 geometrically portrays how the single-element pattern and ar- ray pattern multiply to provide the radiation pattern. With the elements a half- wavelength apart and driven in phase, electromagnetic waves arrive in phase at points along the yaxis and reinforce. There is no radiation in the ±xdirections, because a wave initiated by one element arrives out of phase with the wave being initiated by that second element. As a result, the waves reinforce along the yaxis, the “broadside” direction, while they cancel along the xaxis. Dipoles in End-Fire Array. With quarter-wave spacing and driven 90 degrees out of phase, a=λ 4, α 1−αo=π 2(20) 28 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.4.7 Radiation pattern for dipoles quarter-wave spaced, 90 de- grees out of phase. the magnitude of the array factor follows from (15) as |ψa|= 2¯¯cos£π 4sinθcosφ+π 4¤¯¯ (21) The radiation pattern follows from (16) and (21). Ψ =|ψo|2|ψa|2= 4 sin2θcos2£π 4sinθcosφ+π 4¤ (22) Shown graphically in Fig. 12.4.7, the pattern is now in the −xdirection. Waves initiated in the −xdirection by the element at x=aarrive in phase with those originating from the second element. Thus, the wave being initiated by that second element in the −xdirection is reinforced. By contrast, the wave initiated in the + x direction by the element at x= 0 arrives 180 degrees out of phase with the wave being initiated in the + xdirection by the other element. Thus, radiation in the + x direction cancels, and the array is unidirectional. Finite Dipoles in End-Fire Array. Finally, consider a pair of ﬁnite length elements, each having a length l, as in Fig. 12.4.3. The pattern for the individual elements is given by (8). With the elements spaced as in Fig. 12.4.5, with a=λ/4 and driven 90 degrees out of phase, the magnitude of the array factor is given by (21). Thus, the amplitude of the radiation pattern is Ψ = 4·cos¡kl 2¢ −cos¡kl 2cosθ¢ sinθ¸2µ cos2£π 4sinθcosφ+π 4¤¶2 (23) For elements of length l= 3λ/2 (kl= 3π), this pattern is pictured in Fig. 12.4.8. Gain. The time average power ﬂux density, /angbracketleftSr(θ, φ)/angbracketright, normalized to the power ﬂux density averaged over the surface of a sphere, is called the gain of an antenna. G=/angbracketleftSr(θ, φ)/angbracketright 1 4πr2Rπ 0R2π 0/angbracketleftSr/angbracketrightrsinθdφrdθ(24) Sec. 12.5 Complex Poynting’s Theorem 29 Fig. 12.4.8 Radiation pattern for two center-fed wire antennas, quarter-wave spaced, 90 degrees out-of-phase, each having length 3 λ/2. If the direction is not speciﬁed, it is implied that Gis the gain in the direction of maximum gain. The radial power ﬂux density is the Poynting ﬂux, deﬁned by (11.2.9). Using the time average theorem, (11.5.6), and the fact that the ratio of EtoHfor the radiation ﬁeld isp µ//epsilon1, (2), gives /angbracketleftSr/angbracketright=1 2ReˆE×ˆH∗=1 2ReˆEθˆH∗ φ=1 2p /epsilon1/µ|ˆEθ|2(25) Because the radiation pattern expresses the ( θ, φ) dependence of |Eθ|2with a multiplicative factor that is in common to the numerator and denominator of (24), Gcan be evaluated using the radiation pattern Ψ for /angbracketleftSr/angbracketright. Example 12.4.3. Gain of an Electric Dipole For the electric dipole, it follows from (1) and (2) that the radiation pattern is proportional to sin2(θ). The gain in the θdirection is then G=sin2θ 1 2Rπ 0sin3θdθ=3 2sin2θ (26) and the “gain” is 3 /2. 12.5 COMPLEX POYNTING’S THEOREM AND RADIATION RESISTANCE To the generator supplying its terminal current, a radiating antenna appears as a load with an impedance having a resistive part. This is true even if the antenna is made from perfectly conducting material and therefore incapable of converting 30 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 electrical power to heat. The power radiated away from the antenna must be sup- plied through its terminals, much as if it were dissipated in a resistor. Indeed, if there is no electrical dissipation in the antenna, the power supplied at the terminals is that radiated away. This statement of power conservation makes it possible to determine the equivalent resistance of the antenna simply by using the far ﬁelds that were the theme of Sec. 12.4. Complex Poynting’s Theorem. For systems in the sinusoidal steady state, a useful alternative to the form of Poynting’s theorem introduced in Secs. 11.1 and 11.2 results from writing Maxwell’s equations in terms of complex amplitudes before they are combined to provide the desired theorem. That is, we assume at the outset that ﬁelds and sources take the form E=ReˆE(x, y, z )ejωt(1) Suppose that the region of interest is composed either of free space or of perfect conductors. Then, substitution of complex amplitudes into the laws of Amp` ere and Faraday, (12.0.8) and (12.0.9), gives ∇ × ˆH=ˆJ+jω/epsilon1ˆE (2) ∇ × ˆE=−jωµˆH (3) The manipulations that are now used to obtain the desired “complex Poynting’s theorem” parallel those used to derive the real, time-dependent form of Poynting’s theorem in Sec. 11.2. We dotEwith the complex conjugate of (2) and subtract the dot product of the complex conjugate of Hwith (3). It follows that8 −∇ · (ˆE×ˆH∗) =ˆE·ˆJ∗+jω(µˆH·ˆH∗−/epsilon1ˆE·ˆE∗) (4) The object of this manipulation was to obtain the “perfect” divergence on the left, because this expression can then be integrated over a volume Vand Gauss’ theorem used to convert the volume integral on the left to an integral over the enclosing surface S. −I S1 2(ˆE×ˆH∗)·da=jω2Z V1 4(µˆH·ˆH∗−/epsilon1ˆE·ˆE∗)dv +Z V1 2ˆE·ˆJ∗ udv(5) This expression has been multiplied by1 2, so that its real part represents the time average ﬂow of power, familiar from Sec. 11.5. Note that the real part of the ﬁrst term on the right is zero. The real part of (5) equates the time average of the Poynting vector ﬂux into the volume with the time average of the power imparted to the current density of unpaired charge, Ju, by the electric ﬁeld. This information 8∇ ·(A×B) =B· ∇ × A−A· ∇ × B Sec. 12.5 Complex Poynting’s Theorem 31 Fig. 12.5.1 Surface Sencloses the antenna but excludes the source. Spherical part of Sis at “inﬁnity.” is equivalent to the time average of the (real form of) Poynting’s theorem. The imaginary part of (5) relates the diﬀerence between the time average magnetic and electric energies in the volume Vto the imaginary part of the complex Poynting ﬂux into the volume. The imaginary part of the complex Poynting theorem conveys additional information. Radiation Resistance. Consider the perfectly conducting antenna system surrounded by the spherical surface, S, shown in Fig. 12.5.1. To exclude sources from the enclosed volume, this surface is composed of an outer surface, Sa, that is far enough from the antenna so that only the radiation ﬁeld makes a contribution, a surface Sbthat surrounds the source(s), and a surface Scthat can be envisioned as the wall of a system of thin tubes connecting SatoSbin such a way that Sa+Sb+Scis indeed the surface enclosing V. By making the connecting tubes very thin, contributions to the integral on the left in (5) from the surface Scare negligible. We now write, and then explain, the terms in (5) as they describe this radiation system. −Z Sa1 2p /epsilon1/µ|ˆEθ|2da+nX i=11 2ˆviˆi∗ i=j2ωZ V1 2¡1 2µ|ˆH|2−1 2/epsilon1|ˆE|2¢ dv (6) The ﬁrst term is the contribution from integrating the radiation Poynting ﬂux over Sa, where (12.4.2) serves to eliminate H. The second term comes from the surface integral in (5) of the Poynting ﬂux over the surface, Sb, enclosing the sources (generators). Think of the generators as enclosed by perfectly conducting boxes powered by terminal pairs (coaxial cables) to which the antennae are attached. We have shown in Sec. 11.3 (11.3.29) that the integral of the Poynting ﬂux over Sbis 32 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.5.2 End-loaded dipole and equivalent circuit. equivalent to the sum of voltage-current products expressing power ﬂow from the circuit point of view. The ﬁrst term on the right is the same as the ﬁrst on the right in (5). Finally, the last term in (5) makes no contribution, because the only regions where Jexists within Vare those modeled here as perfectly conducting and hence where E= 0. Consider a single antenna with one input terminal pair. The antenna is a linear system, so the complex voltage must be proportional to the complex terminal current. ˆv=Zantˆi (7) Here, Zantis the impedance of the antenna. In terms of this impedance, the time average power can be written as 1 2Re(ˆvˆi∗) =1 2Re(Zant)|ˆi|2=1 2Re(Zant)|ˆIo|2(8) It follows from the real part of (6) that the radiation resistance ,Rrad, is Re(Zant)≡Rrad=p /epsilon1/µZ Sa|Eθ|2da |ˆIo|2(9) The imaginary part of Zoutdescribes the reactive power supplied to the antenna. Im(Zant) =ωR (µ|ˆH|2−/epsilon1|ˆE|2)dv |Io|2(10) The radiation ﬁeld contributions to this integral cancel out. If the antenna elements are short compared to a wavelength, contributions to (10) are dominated by the quasistatic ﬁelds. Thus, the electric dipole contributions are dominated by the elec- tric ﬁeld (and the reactance is capacitive), while those for the magnetic dipole are inductive. By making the antenna on the order of a wavelength, the magnetic and electric contributions to (10) are often made to essentially cancel. An example is a half-wavelength version of the wire antenna in Example 12.4.1. The equivalent circuit for such resonant antennae is then solely the radiation resistance. Example 12.5.1. Equivalent Circuit of an Electric Dipole An “end-loaded” electric dipole is composed of a pair of perfectly conducting metal spheres, each of radius R, as shown in Fig. 12.5.2. These spheres have a spacing, d, that is short compared to a wavelength but large compared to the radius, R, of the spheres. Sec. 12.5 Complex Poynting’s Theorem 33 The equivalent circuit is also shown in Fig. 12.5.2. The statement that the sum of the voltage drops around the circuit is zero requires that ˆv=ˆi jωC+Rradˆi (11) A statement of power ﬂow is obtained by multiplying this expression by the complex conjugate of the complex amplitude of the current. 1 2ˆvˆi∗=−jω 2Cˆqˆq∗+1 2Rradˆiˆi∗; ˆi≡jωˆq (12) Here the dipole charge, q, is deﬁned such that i=dq/dt . The real part of this expression takes the same form as the statement of complex power ﬂow for the antenna, (6). Thus, with ˆEθprovided by (12.2.23) and (12.2.24), we can solve for the radiation resistance: Rrad=p µ//epsilon1(kd)2 (4π)2Zπ 0sin2θ(2πrsinθ)rdθ r2=(kd)2 6πp µ//epsilon1 (13) Note that because k≡ω/c, this radiation resistance is proportional to the square of the frequency. The imaginary part of the impedance is given by the right-hand side of (10). The radiation ﬁeld contributions to this integral cancel out. In integrating over the near ﬁeld, the electric energy storage dominates and becomes essentially that associated with the quasistatic capacitance of the pair of spheres. We assume that the spheres are connected by wires that are extremely thin, so that their eﬀect can be ignored. Then, the capacitance is the series capacitance of two isolated spheres, each having a capacitance of 4 π/epsilon1R. C= 2π/epsilon1R (14) Radiation ﬁelds are solutions to the full Maxwell equations. In contrast, EQS ﬁelds were analyzed ignoring the magnetic ﬂux linkage in Faraday’s law. The ap- proximation is justiﬁed if the size of the system is small compared with a wavelength. The following example treats the scattering of particles that are small compared with the wavelength. The ﬁelds around the particles are EQS, and the currents induced in the particles are deduced from the EQS approximation. These currents drive radiation ﬁelds, resulting in Rayleigh scattering. The theory of Rayleigh scat- tering explains why the sky is blue in color, as the following example shows. Example 12.5.2. Rayleigh Scattering Consider a spatial distribution of particles in the ﬁeld of an inﬁnite parallel plane wave. The particles are assumed to be small as compared to the wavelength of the plane wave. They get polarized in the presence of an electric ﬁeld Ea, acquiring a dipole moment p=/epsilon1oαEa (15) where αis the polarizability. These particles could be atoms or molecules, such as the molecules of nitrogen and oxygen of air exposed to visible light. They could also 34 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 be conducting spheres of radius R. In the latter case, the dipole moment produced by an applied electric ﬁeld Eais given by (6.6.5) and the polarizability is α= 4πR3(16) If the frequency of the polarizing wave is ωand its propagation constant k=ω/c, the far ﬁeld radiated by the particle, expressed in a spherical coordinate system with itsθ= 0 axis aligned with the electric ﬁeld of the wave, is, from (12.2.22), Eθ=−r µo /epsilon1okωˆp 4πsinθe−jkr r(17) where ˆid=jωˆpis used in the above expression. The power radiated by each dipole, i.e., the power scattered by a dipole, is PScatt=Z Sa1 2r /epsilon1o µo|ˆEθ|2=1 2r µo /epsilon1o¯¯¯¯ωkˆp 4π¯¯¯¯21 r2Zπ 0Z2π 0dφr2sin3θdθ =1 2r µo /epsilon1o¯¯¯¯ωkˆp 4π¯¯¯¯28π 3=1 12πr µo /epsilon1oω4 c2/epsilon12 oα2ˆE2(18) The scattered power increases with the fourth power of frequency when αis not a function of frequency. The polarizability of N2andO2is roughly frequency indepen- dent. Of the visible radiation, the blue (high) frequencies scatter much more than the red (low) frequencies. This is the reason for the blue color of the sky. The same phenomenon accounts for the polarization of the scattered radiation. Along a line L at a large angle from the line from the observer Oto the sun S, only the electric ﬁeld perpendicular to the plane LOS produces radiation visible at the observer position (note the sin2θdependence of the radiation). Thus, the scattered radiation observed atOhas an electric ﬁeld perpendicular to LOS. The present analysis has made two approximations. First, of course, we as- sumed that the particle is small compared with a wavelength. Second, we computed the induced polarization from the unperturbed ﬁeld Eθof the incident plane wave. This assumes that the particle perturbs the wave negligibly, that the scattered power is very small compared to the power in the wave. Of course, the incident wave de- creases in intensity as it proceeds through the distribution of scatterers, but this macroscopic change can be treated as a simple attenuation proportional to the den- sity of scatterers. 12.6 PERIODIC SHEET-SOURCE FIELDS: UNIFORM AND NONUNIFORM PLANE WAVES This section introduces the electrodynamic ﬁelds associated with surface sources. The physical systems analyzed are generalizations, on the one hand, of such EQS situations as Example 5.6.2, where sinusoidal surface charge densities produced a Laplacian ﬁeld decaying away from the surface charge source. On the other hand, the MQS sinusoidal surface current sources producing magnetic ﬁelds that decay Sec. 12.6 Periodic Sheet-Source 35 away from their source (for example, Prob. 8.6.9) are generalized to the fully dy- namic case. In both cases, one expects that the quasistatic approximation will be contained in the limit where the spatial period of the source is much smaller than the wavelength λ= 2π/ω√µ/epsilon1. When the spatial period of the source approaches, or exceeds, the wavelength, new phenomena ought to be revealed. Speciﬁcally, dis- tributions of surface current density Kand surface charge density σsare given in thex−zplane. K=Kx(x, t)ix+Kz(x, t)iz (1) σs=σs(x, t) (2) These are independent of zand are typically periodic in space and time, extending to inﬁnity in the xandzdirections. Charge conservation links Kandσs. A two-dimensional version of the charge conservation law, (12.1.22), requires that there must be a time rate of decrease of surface charge density σswherever there is a two-dimensional divergence of K. ∂Kx ∂x+∂Kz ∂z+∂σs ∂t= 0⇒∂Kx ∂x+∂σs ∂t= 0 (3) The second expression results because Kzis independent of z. Under the assumption that the only sources are those in the x−zplane, it follows that the ﬁelds can be pictured as the superposition of those due to ( Kx, σs) given to satisfy (3) and due to Kz. It is therefore convenient to break the ﬁelds produced by these two kinds of sources into two categories. Transverse Magnetic (TM) Fields. The source distribution ( Kx, σs) does not produce a zcomponent of the vector potential A, Az= 0. This follows because there is no zcomponent of the current in the superposition integral for A, (12.3.2). However, there are both current and charge sources, so that the superposition integral for Φ requires that in addition to an Athat lies in x−yplanes, there is an electric potential as well. A=Ax(x, y, t )ix+Ay(x, y, t )iy; Φ = Φ( x, y, t ) (4) Because the source distribution is independent of z, we have taken these potentials to be also two dimensional. It follows that His transverse to the x−ycoordinates upon which the ﬁelds depend , while Elies in the x−yplane. H=1 µ∇ ×A=Hz(x, y, t )iz E=−∇Φ−∂A ∂t=Ex(x, y, t )ix+Ey(x, y, t )iy (5) Sources and ﬁelds for these transverse magnetic (TM) ﬁelds have the relative ori- entations shown in Fig. 12.6.1. We will be concerned here with sources that are in the sinusoidal steady state. Although Aand Φ could be used to derive the ﬁelds, in what follows it is more 36 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.6.1 Transverse magnetic and electric sources and ﬁelds. convenient to deal directly with the ﬁelds themselves. The complex amplitude of Hz, the only component of H, is conveniently used to represent Ein the free space regions to either side of the sheet. This can be seen by using the xandycomponents of Amp` ere’s law to write Ex=1 jω/epsilon1∂ˆHz ∂y (6) ˆEy=−1 jω/epsilon1∂ˆHz ∂x (7) for the only two components of E. The relationship between Hand its source is obtained by taking the curl of the vector wave equation for A, (12.1.8). The curl operator commutes with the Laplacian and time derivative, so that the result is the inhomogeneous wave equation for H. ∇2H−µ/epsilon1∂2H ∂t2=−∇ × J(8) For the sheet source, the driving term on the right is zero everywhere except in the x−zplane. Thus, in the free space regions, the zcomponent of this equation gives a diﬀerential equation for the complex amplitude of Hz. ¡∂2 ∂x2+∂2 ∂y2¢ˆHz+ω2µ/epsilon1ˆHz= 0 (9) This expression is a two-dimensional example of the Helmholtz equation . Given sinusoidal steady state source distributions of ( Kx, σs) consistent with charge con- servation, (3), the continuity conditions can be used to relate these sources to the ﬁelds described by (6), (7), and (9). Sec. 12.6 Periodic Sheet-Source 37 Product Solutions to the Helmholtz Equation. One theme of this section is the solution to the Helmholtz equation, (9). Note that this equation resulted from the time-dependent wave equation by separation of variables, by assuming solu- tions of the form Hz(x, y)T(t), where T(t) = exp( jωt). We now look for solutions expressing the x−ydependence that take the product form X(x)Y(y). The process is familiar from Sec. 5.4, but the resulting family of solutions is of wider variety, and it is worthwhile to focus on their nature before applying them to particular examples. With the substitution of the product solution Hz=X(x)Y(y), (9) becomes 1 Xd2X dx2+1 Yd2Y dy2+ω2µ/epsilon1= 0 (10) This expression is satisﬁed if the ﬁrst and second terms are constants −k2 x−k2 y+ω2µ/epsilon1= 0 (11) and it follows that parts of the total solution are governed by the ordinary diﬀer- ential equations d2X dx2+k2 xX= 0;d2Y dy2+k2 yY= 0. (12) Although kxandkyare constants, as long as they satisfy (11) they can be real or imaginary. In this chapter, we are interested in solutions that are periodic in the x direction, so we can think of kxas being real and k2 x>0. Furthermore, the value ofkxis ﬁxed by the assumed functional form of the surface currents and charges. Equation (11) then determines kyfrom given values of kxandω. In solving (11) forky, we must take the square root of a quantity that can be positive or negative. By way of distinguishing the two roots of (11) solved for ky, we deﬁne β≡½ |p ω2µ/epsilon1−k2x|; ω2µ/epsilon1 > k2 x −j|p k2x−ω2µ/epsilon1|;ω2µ/epsilon1 < k2 x (13) and write the two solutions to (11) as ky≡ ∓β (14) Thus, βis deﬁned as either positive real or negative imaginary, and what we have found for the product solution X(x)Y(y) are combinations of products ˆHzα½ coskxx sinkxx¾½ e−jβy ejβy¾ (15) Note that if ω2µ/epsilon1 < k2 x, kyis deﬁned by (13) and (14) such that the ﬁeld, which is periodic in the xdirection, decays in the + ydirection for the upper solution but decays in the −ydirection for the lower solution. These ﬁelds resemble solutions to 38 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.6.2 Standing wave of surface charge density. Laplace’s equation. Indeed, in the limit where ω2µ/epsilon1/lessmuchk2 x, the Helmholtz equation becomes Laplace’s equation. As the frequency is raised, the rate of decay in the ±ydirections decreases until kybecomes real, at which point the solutions take a form that is in sharp contrast to those for Laplace’s equation. With ω2µ/epsilon1 > k2 x, the solutions that we assumed to be periodic in the xdirection are also periodic in the ydirection . The wave propagation in the ydirection that renders the solutions periodic inyis more evident if the product solutions of (15) are written with the time dependence included. Hzα½ coskxx sinkxx¾ ej(ωt∓βy)(16) Forω2µ/epsilon1 > k2 x, the upper and lower signs in (16) [and hence in (14)] correspond to waves propagating in the positive and negative ydirections, respectively. By taking a linear combination of the trigonometric functions in (16), we can also form the complex exponential exp( jkxx). Thus, another expression of the solutions given by (16) is as Hzα ej(ωt∓βy∓kxx)(17) Instead of having standing waves in the xdirection, as represented by (16), we now have solutions that are traveling in the ±xdirections. Examples 12.6.1 and 12.6.2, respectively, illustrate how standing-wave and traveling-wave ﬁelds are excited. Example 12.6.1. Standing-Wave TM Fields Consider the ﬁeld response to a surface charge density that is in the sinusoidal steady state and represented by σs=Re£ ˆσosinkxxejωt¤ (18) The complex coeﬃcient ˆ σo, which determines the temporal phase and magnitude of the charge density at any given location x, is given. Figure 12.6.2 shows this function represented in space and time. The charge density is always zero at the locations kxx=nπ, where nis any integer, and oscillates between positive and negative peak amplitudes at locations in between. When it is positive in one half-period between nulls, it is negative in the adjacent half-periods. It has the xdependence of a standing wave. Sec. 12.6 Periodic Sheet-Source 39 The current density that is consistent with the surface charge density of (18) follows from (3). Kx=Re· jωˆσo kxcoskxxejωt¸ (19) With the surface current density in the zdirection zero, the ﬁelds excited by these surface sources above and below the sheet are TM. The continuity conditions, (12.1.14)–(12.1.17), relate the ﬁelds to the given surface source distributions. We start with Amp` ere’s continuity condition, the xcomponent of (12.1.15) Ha z−Hb z=Kxaty= 0 (20) because it determines the xdependence of Hzas cos kxx. Of the possible combina- tions of solutions given by (15), we let ˆHz=½ ˆAcoskxxe−jβy ˆBcoskxxejβy (21) The upper solution pertains to the upper region. Note that we select a ydependence that represents either a wave propagating in the + ydirection (for ω2µ/epsilon1 > 0) or a ﬁeld that decays in that direction (for ω2µ/epsilon1 < 0). The lower solution, which applies in the lower region, either propagates in the −ydirection or decays in that direction. One of two conditions on the coeﬃcients in (21) it follows from substitution of these equations into Amp` ere’s continuity condition, (20). ˆA−ˆB=jωˆσo kx(22) A second condition follows from Faraday’s continuity condition, (12.1.16), which requires that the tangential electric ﬁeld be continuous. ˆEa x−ˆEb x= 0 at y= 0 (23) Substitution of the solutions, (21), into (6) gives ˆEa xandˆEb x, from which follows ˆA=−ˆB (24) Combining (2) and (3) we ﬁnd ˆA=−ˆB=jωˆσo 2kx(25) The remaining continuity conditions are now automatically satisﬁed. There is no normal ﬂux density, so the ﬂux continuity condition of (12.1.17) is automatically satisﬁed. But even if there were a ycomponent of H, continuity of tangential Eas expressed by (23) would guarantee that this condition is satisﬁed. In summary, the coeﬃcients given by (25) can be used in (21), and those expressions introduced into (6) and (7), to determine the ﬁelds as Hz=Re±jωˆσo 2kxcoskxxej(ωt∓βy)(26) 40 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Ex=Re−jˆσoβ 2/epsilon1kxcoskxxej(ωt∓βy)(27) Ey=Re±ˆσo 2/epsilon1sinkxxej(ωt∓βy)(28) These ﬁelds are pictured in Fig. 12.6.3 for the case where σois real. In the ﬁrst ﬁeld distribution, the frequency is low enough so that ω2µ/epsilon1 < k2 x. Thus, βas given by (13) has a negative imaginary value. The electric ﬁeld pattern is shown when t= 0. At this instant, H= 0. When ωt=π/2,His as shown while E= 0. The Eand Hare 90 degrees out of temporal phase. The ﬁelds decay in the ydirection, much as they would for a spatially periodic surface charge distribution in the EQS limit. Because they decay in the ±ydirections for reasons that do not involve dissipation, these ﬁelds are sometimes called evanescent waves . The decay has its origins in the nature of quasistatic ﬁelds, shaped as they are by Laplace’s equation. Indeed, with ω2µ/epsilon1/lessmuchk2 x,E=−∇Φ, and we are dealing with scalar solutions Φ to Laplace’s equation. The ﬁeld pattern corresponds to that of Example 5.6.2. As the frequency is raised, the rate of decay in the ydirection decreases. The rate of decay, |β|, reaches zero as the frequency reaches ω=kx/√µ/epsilon1=kxc. The physical signiﬁcance of this condition is seen by recognizing that kx= 2π/λx, where λxis the wavelength in thexdirection of the imposed surface charge density, and that ω= 2π/T where Tis the temporal period of the excitation. Thus, as the frequency is raised to the point where the ﬁelds no longer decay in the ±ydirections, the period Thas become T=λx/c, and so has become as short as the time required for an electromagnetic wave to propagate the wavelength λx. The second distribution of Fig. 12.6.3 illustrates what happens to the ﬁelds as the frequency is raised beyond the cutoﬀ frequency, when ω2µ/epsilon1 > k2 x. In this case, bothEandHare shown in Fig. 12.6.3 when t= 0. Fields above and below the sheet propagate in the ±ydirections, respectively. As time progresses, the evolution of the ﬁelds in the respective regions can be pictured as a translation of these distributions in the ±ydirections with the phase velocities ω/ky. Transverse Electric (TE) Fields. Consider the case of a z-directed surface current density K=kziz. Then, the surface charge density σsis zero. It follows from the superposition integral for Φ, (12.3.1), that Φ = 0 and from the superposition integral for A, (12.3.2), that A=Aziz. Equation (12.1.3) then shows that Eis in the zdirection, E=Eziz.The electric ﬁeld is transverse to the xandyaxes, while the magnetic ﬁeld lines are in x−yplanes. These are the ﬁeld directions summarized in the second part of Fig. 12.6.1. In the sinusoidal steady state, it is convenient to use Ezas the function from which all other quantities can be derived, for it follows from Faraday’s law that the two components of Hcan be written in terms of Ez. ˆHx=−1 jωµ∂ˆEz ∂y (29) ˆHy=1 jωµ∂ˆEz ∂x (30) Sec. 12.6 Periodic Sheet-Source 41 Fig. 12.6.3 TM waves due to standing wave of sources in y= 0 plane. In the free space regions to either side of the sheet, each of the Cartesian components of EandHsatisﬁes the wave equation. We have already seen this for H. To obtain an expression playing a similar role for E, we could again return to the wave equations for Aand Φ. A more direct derivation begins by taking the curl of Faraday’s law. ∇ × ∇ × E=−∇ ×∂µH ∂t⇒ ∇(∇ ·E)− ∇2E=−µ∂ ∂t(∇ ×H) (31) On the left, a vector identity has been used, while on the right the order of taking the time derivative and the curl has been reversed. Now, if we substitute for the divergence on the left using Gauss’ law, and for the curl on the right using Amp` ere’s law, it follows that ∇2E−µ/epsilon1∂2E ∂t2=∇¡ρ /epsilon1¢ +µ∂J ∂t (32) In the free space regions, the driving terms on the right are absent. In the case of transverse electric ﬁelds, ˆEzis the only ﬁeld component. From (32), an assumed 42 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 time dependence of the form E=Re[ˆEzizexp(jωt)] leads to the Helmholtz equation forˆEz µ∂2 ∂x2+∂2 ∂y2¶ ˆEz+ω2µ/epsilon1ˆEz= 0 (33) In retrospect, we see that the TE ﬁeld relations are obtained from those for the TM ﬁelds by replacing H→ −E,E→H, /epsilon1→µandµ→/epsilon1. This could have been expected, because in the free space regions to either side of the source sheet, Maxwell’s equations are replicated by such an exchange of variables. The discussion of product solutions to the Helmholtz equation, given following (9), is equally applicable here. Example 12.6.2. Traveling-Wave TE Fields This example has two objectives. One is to illustrate the TE ﬁelds, while the other is to provide further insights into the nature of electrodynamic ﬁelds that are periodic in time and in one space dimension. In Example 12.6.1, these ﬁelds were induced by a standing wave of surface sources. Here the source takes the form of a wave traveling in the xdirection . Kz=ReˆKoej(ωt−kxx)(34) Again, the frequency of the source current, ω, and its spatial dependence, exp(−jkxx), are prescribed. The traveling-wave x, tdependence of the source sug- gests that solutions take the form of (17). ˆEz=½ˆAe−jβye−jkxx;y >0 ˆBejβye−jkxx;y <0(35) Faraday’s continuity condition, (12.1.16), requires that ˆEa z=ˆEb z at y= 0 (36) and this provides the ﬁrst of two conditions on the coeﬃcients in (35). ˆA=ˆB (37) Amp` ere’s continuity condition, (12.1.15), further requires that −(ˆHa x−ˆHb x) =ˆKz at y= 0 (38) With Hxfound by substituting (35) into (29), this condition shows that ˆA=ˆB=−ωµˆKo 2β(39) Sec. 12.6 Periodic Sheet-Source 43 Fig. 12.6.4 TE ﬁelds induced by traveling-wave source in the y= 0 plane. With the substitution of these coeﬃcients into (35), we have Ez=Re−ωµ 2βˆKoej(ωt−kxx)ne−jβy;y >0 ejβy;y <0(40) Provided that βis as deﬁned by (13), these relations are valid regardless of the frequency. However, to emphasize the eﬀect on the ﬁeld when the frequency is such thatω2µ/epsilon1 < k2 x, these expressions are written for that case as Ez=Re−jωµ 2|β|ˆKoej(ωt−kxx)½ e−|β|y;y >0 e|β|y;y <0(41) The space time dependence of Ez, and Has found by using (40) to evaluate (29) and (30), is illustrated in Fig. 12.6.4. For ω2µ/epsilon1 > k2 x, the response to the traveling wave of surface current is waves with lines of constant amplitude given by kxx±kyy= constant + ωt (42) Thus, points of constant phase are lines of slope ∓kx/ky. The velocity of these lines in the xdirection, ω/kx, is called the phase velocity of the wave in the x 44 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 direction. The respective waves also have phase velocities in the ±ydirections, in this case ±ω/ky. The response to the traveling-current sheet in this high-frequency regime is a pair of uniform plane waves. Their direction of propagation is along the gradient of (42), and it is sometimes convenient to describe such plane waves by a vector wave number khaving the direction of propagation of the planes of constant phase. The waves in the half-plane Y >0 possess the kvector k=kxix+kyiy (43) At frequencies low enough so that ω2µ/epsilon1 < k2 x, points of constant phase lie on lines perpendicular to the xaxis. At a given location along the xaxis, the ﬁelds vary in synchronism but decay in the ±ydirections. In the limit when ω2µ/epsilon1/lessmuchk2 x (orf/lessmuchc/λx, where ω≡2πf), the Hﬁelds given by (29) and (30) become the MQS ﬁelds of a spatially periodic current sheet that happens to be traveling in thexdirection. These “waves” are similar to those predicted by Laplace’s equation except that for a given wavelength 2 π/kxin the xdirection, they reach out further in theydirection. (A standing-wave version of this MQS ﬁeld is exempliﬁed by Prob. 8.6.9.) In recognition of the decay in the ydirection, they are sometimes called nonuniform plane waves orevanescent waves . Note that the frequency demarcating propagation in the ±ydirections from evanescence or decay in the ±ydirections is f=c/λxor the frequency at which the spatial period of the imposed current sheet is equal to one wavelength for a plane wave propagating in free space. 12.7 ELECTRODYNAMIC FIELDS IN THE PRESENCE OF PERFECT CONDUCTORS The superposition integral approach is directly applicable to the determination of electrodynamic ﬁelds from sources speciﬁed throughout all space. In the presence of materials, sources are induced as well as imposed. These sources cannot be speciﬁed in advance. For example, if a perfect conductor is introduced, surface currents and charges are induced on its surface in just such a way as to insure that there is neither a tangential electric ﬁeld at its surface nor a magnetic ﬂux density normal to its surface. We have already seen how the superposition integral approach can be used to ﬁnd the ﬁelds in the vicinity of perfect conductors, for EQS systems in Chap. 4 and for MQS systems in Chap. 8. Fictitious sources are located in regions outside that of interest so that they add to those from the actual sources in such a way as to satisfy the boundary conditions. The approach is usually used to provide simple analytical descriptions of ﬁelds, in which case its application is a bit of an art– but it can also be the basis for practical numerical analyses involving complex systems. We begin with a reminder of the boundary conditions that represent the inﬂuence of the sources induced on the surface of a perfect conductor. Such a conductor is deﬁned as one in which E→0 because σ→ ∞ . Because the tangential electric ﬁeld must be continuous across the boundary, it follows from Faraday’s continuity condition that just outside the surface of the perfect conductor (having the unit normal n) n×E= 0 (1) Sec. 12.7 Perfect Conductors 45 In Sec. 8.4, and again in Sec. 12.1, it was argued that (1) implies that the normal magnetic ﬂux density just outside a perfectly conducting surface must be constant. ∂ ∂t(n·µH) = 0 (2) The physical origins and limitations of this boundary condition were one of the subjects of Chap. 10. Method of Images. The symmetry considerations used to satisfy boundary conditions in Secs. 4.7 and 8.6 on certain planes of symmetry are equally applicable here, even though the ﬁelds now suﬀer time delays under transient conditions and phase delays in the sinusoidal steady state. We shall illustrate the method of images for an incremental dipole. It follows by superposition that the same method can be used with arbitrary source distributions. Suppose that we wished to determine the ﬁelds associated with an electric dipole over a perfectly conducting ground plane. This dipole is the upper one of the two shown in Fig. 12.7.1. The associated electric and magnetic ﬁelds were determined in Sec. 12.2, and will be called EpandHp, respectively. To satisfy the condition that there be no tangential electric ﬁeld on the perfectly conducting plane, that plane is made one of symmetry in an equivalent conﬁguration in which a second “image” dipole is mounted, having a direction and intensity such that at any instant, its charges are the negatives of those of the ﬁrst dipole. That is, the + charge of the upper dipole is imaged by a negative charge of equal magnitude with the plane of symmetry perpendicular to and bisecting a line joining the two. The second dipole has been arranged so that at each instant in time, it produces a tangential E=Ehthat just cancels that of the ﬁrst at each location on the symmetry plane. With E=Eh+Ep (3) we have made Esatisfy (1) and hence (2) on the ground plane. There are two ways of conceptualizing the “method of images.” The one given here is consistent with the superposition integral point of view that is the theme of this chapter. The second takes the boundary value point of view of the next chapter. These alternative points of view are familiar from Chaps. 4 and 5 for EQS systems and from the ﬁrst and second halves of Chap. 8 for MQS systems. From the boundary value point of view, in the upper half-space, EpandHpare particular solutions, satisfying the inhomogeneous wave equation everywhere in the volume of interest. In this region, the ﬁelds EhandHhdue to the image dipole are then solutions to the homogeneous wave equation. Physically, they represent ﬁelds induced by sources on the perfectly conducting boundary. To emphasize that the symmetry arguments apply regardless of the temporal details of the excitations, the ﬁelds shown in Fig. 12.7.1 are those of the electric dipole during the turn-on transient discussed in Example 12.2.1. At an arbitrary point on the ground plane, the “real” dipole produces ﬁelds that are not necessarily in the plane of the paper or perpendicular to it. Yet symmetry requires that the tangential Edue to the sum of the ﬁelds is zero on the ground plane, and Faraday’s law requires that the normal His zero as well. 46 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.7.1 Dipoles over a ground plane together with their images: (a) electric dipole; and (b) magnetic dipole. In the case of the magnetic dipole over a ground plane shown in Fig. 12.7.1b, ﬁnding the image dipole is easiest by nulling the magnetic ﬂux density normal to the ground plane, rather than the electric ﬁeld tangential to the ground plane. The ﬁelds shown are the dual [(12.2.33)–(12.2.34)] of those for the electric dipole turn-on transient of Example 12.2.1. If we visualize the dipole as due to magnetic charge, the image charge is now of the same sign, rather than opposite sign, as the source. Image methods are commonly used in extending the superposition integral techniques to antenna ﬁeld patterns in order to treat the eﬀects of a ground plane and of reﬂectors. Example 12.7.1. Ground Planes and Reﬂectors Quarter-Wave Antenna above a Ground Plane. The center-fed wire antenna of Example 12.4.1, shown in Fig. 12.7.2a, has a plane of symmetry, θ= π/2, on which there is no tangential electric ﬁeld. Thus, provided the terminal current remains the same, the ﬁeld in the upper half-space remains unaltered if a perfectly conducting ground plane is placed in this plane. The radiation electric ﬁeld is therefore given by (12.4.2), (12.4.5), and (12.4.8). Note that the lower half of the wire antenna serves as an image for the top half. Whether used for AM broadcasting or as a microwave mobile antenna (on the roof of an automobile), the height is usually a quarter-wavelength. In this case, kl=π, and these relations give |ˆEθ|=1 4p µ//epsilon1Io r|ψo(θ)| (4) Sec. 12.7 Perfect Conductors 47 Fig. 12.7.2 Equivalent image systems for three physical systems. where the radiation intensity pattern is ψo=2 πcos¡π 2cosθ¢ sinθ(5) Although the radiation pattern for the quarter-wave ground plane is the same as that for the half-wave center-fed wire antenna, the radiation resistance is half as 48 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 much. This follows from the fact that the surface of integration in (12.5.9) is now a hemisphere rather than a sphere. Rrad=1 2πp µ//epsilon1Zπ/2 0cos2¡π 2cosθ¢ sinθdθ=p µ//epsilon10.61 2π(6) The integral can be converted to a sine integral, which is tabulated.9In free space, this radiation resistance is 37Ω. Two-Element Array over Ground Plane. The radiation pattern from an array of elements vertical to a ground plane can be deduced using the same image arguments. The pair of center-fed half-wave elements shown in Fig. 12.7.2c have lower elements that serve as images for the quarter-wave vertical elements over a ground plane shown in Fig. 12.7.2d. If we consider elements with a half-wave spacing that are driven 180 degrees out of phase, the array factor is given by (12.4.15) with ka=πandα1−αo=π. Thus, with ψofrom (5), the electric radiation ﬁeld is |ˆEθ|=1 4p µ//epsilon1Io r|ψo(θ)ψa(θ, φ)| (7) where ψoψa=4 πcos£π 2(sinθcosφ+ 1)¤cos¡π 2cosθ¢ sinθ(8) The radiation pattern is proportional to the square of this function and is sketched in Fig. 12.7.2d. The ﬁeld initiated by one element arrives in the far ﬁeld at φ= 0 andφ=πwith a phase that reinforces that from the second element. The ﬁelds produced from the elements arrive out of phase in the “broadside” directions, and so the pattern nulls in those directions ( φ=±π/2). Phased arrays of two or more verticals are often used by AM stations to provide directed broadcasting, with the ground plane preferably wet land, often with buried “radial” conductors to make the ground plane more nearly like a perfect conductor. Ground-Plane with Reﬂector. The radiation pattern for the pair of vertical elements has no electric ﬁeld tangential to a vertical plane located midway between the elements. Thus, the eﬀect of one of the elements is equivalent to that of a reﬂector having a distance of a quarter-wavelength from the vertical element. This is the conﬁguration shown in Fig. 12.7.2f. The radiation resistance of the vertical quarter wave element with a reﬂector follows from (12.5.9), evaluated using (7). Now the integration is over the quarter- sphere which, together with the ground plane and the reﬂector plane, encloses the element at a radius of many wavelengths. Rrad=p µ//epsilon1 π2Zπ/2 0Zπ/2 −π/2cos2£π 2(sinθcosφ+ 1)¤ cos2¡π 2cosθ¢ sinθdφdθ (9) 9It is perhaps easiest to carry out the integral numerically, as can be done with a pro- grammable calculator. Note that the integrand is zero at θ= 0. Sec. 12.7 Perfect Conductors 49 Demonstration 12.7.1. Ground-Planes, Phased Arrays, and Reﬂectors The experiment shown in Fig. 12.7.3 demonstrates the eﬀect of the phase shift on the radiation pattern of the array considered in Example 12.7.1. The spacing and length of the vertical elements are 7.9 cm and 3.9 cm, respectively, which corresponds toλ/2 and λ/4 respectively at a frequency of 1.9 GHz. The ground plane consists of an aluminum sheet, with the array mounted on a section of the sheet that can be rotated. Thus, the radiation pattern in the plane θ=π/2 can be measured by rotating the array, keeping the receiving antenna, which is many wavelengths away, ﬁxed. An audible tone can be used to indicate the amplitude of the received signal. To this end, the 1.9 GHz source is modulated at the desired audio frequency and detected at the receiver, ampliﬁed, and made audible through a loud speaker. The 180 degree phase shift between the drives for the two driven elements is obtained by inserting a “line stretcher” in series with the coaxial line feeding one of the elements. By eﬀectively lengthening the transmission line, the delay in the transmission line wave results in the desired phase delay. (Chapter 14 is devoted to the dynamics of signals propagating on such transmission lines.) The desired 180 degree phase shift is produced by rotating the array to a broadside position (the elements equidistant from the receiving antenna) and tuning the line stretcher so that the signals are nulled. With a further 90 degree rotation so that the elements are in the end-ﬁre array position (in line with the receiving antenna), the detected signal should peak. One vertical element can be regarded as the image for the other in a physical situation in which one element is backed at a quarter-wavelength by a reﬂector. This quarter-wave ground plane with a reﬂector is demonstrated by introducing a sheet of aluminum halfway between the original elements, as shown in Fig. 12.7.3. With the introduction of the sheet, the “image” element is shielded from the receiving antenna. Nevertheless, the detected signal should be essentially unaltered. The experiment suggests many other interesting and practical conﬁgurations. For example, if the line stretcher is used to null the signal with the elements in end- ﬁre array position, the elements are presumably driven in phase. Then, the signal should peak if the array is rotated 90 degrees so that it is broadside to the receiver. Boundaries at the Nodes of Standing Waves. The TM ﬁelds found in Example 12.6.1 were those produced by a surface charge density taking the form of a standing wave in the y= 0 plane. Examination of the analytical expressions for E, (12.6.27)–(12.6.28), and of their graphical portrayal, Fig. 12.6.3, shows that at every instant in time, Ewas normal to the planes where kxx=nπ(nany integer), whether the waves were evanescent or propagating in the ±ydirections. That is, the ﬁelds have nodal planes (of no tangential E) parallel to the y−zplane. These ﬁelds would therefore remain unaltered by the introduction of thin, perfectly conducting sheets in these planes. Example 12.7.2. TM Fields between Parallel Perfect Conductors To be speciﬁc, suppose that the ﬁelds found in Example 12.6.1 are to “ﬁt” within a region bounded by perfectly conducting surfaces in the planes x= 0 and x=a. The conﬁguration is shown in Fig. 12.7.4. We adjust kxso that kxa=nπ⇒kx=nπ a(10) 50 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. 12.7.3 Demonstration of phase shift on radiation pattern. Fig. 12.7.4 Then= 1 TM ﬁelds between parallel plates (a) evanes- cent in ydirection and (b) propagating in ydirection. where nindicates the number of half-wavelengths in the xdirection of the ﬁelds shown in Fig. 12.6.3 that have been made to ﬁt between the perfect conductors. To make the ﬁelds satisfy the wave equation, kymust be given by (12.6.14) and (12.6.13). Thus, from this expression and (10), we see that for the n-th mode of the TM ﬁelds between the plates, the wave number in the ydirection is related to the frequency by ky=β=½p ω2µ/epsilon1−(nπ/a )2; ω2µ/epsilon1 > (nπ/a )2 −jp (nπ/a )2−ω2µ/epsilon1;ω2µ/epsilon1 < (nπ/a )2(11) Sec. 12.8 Summary 51 We shall encounter these modes and this dispersion equation again in Chap. 13, where waves propagating between parallel plates will be considered from the boundary value point of view. There we shall superimpose these modes and, if need be, comparable TM ﬁeld modes, to satisfy arbitrary source conditions in the plane y= 0. The sources in the plane y= 0 will then represent an antenna driving a parallel plate waveguide. The standing-wave ﬁelds of Example 12.6.1 are the superposition of two trav- eling waves that exactly cancel at the nodal planes to form the standing wave in the xdirection. To see this, observe that a standing wave, such as that for the surface charge distribution given by (12.6.18), can be written as the sum of two traveling waves.10 σs= Re£ ˆσosinkxxejωt¤ = Re·jˆσo 2ej(ωt−kxx)−jˆσo 2ej(ωt+kxx)¸ (12) By superposition, the ﬁeld responses therefore must take this same form. For ex- ample, Eyas given by (12.6.28) can be written as Ey= Re∓ˆσo 4/epsilon1j£ ej(ωt∓βy−kxx)−ej(ωt∓βy+kxx)¤ (13) where the upper and lower signs again refer to the regions above and below the sheet of charge density. The ﬁrst term represents the response to the component of the surface current density that travels to the right while the second is the response from the component traveling to the left. The planes of constant phase for the component waves traveling to the right, as well as their respective directions of propagation, are as for the TE ﬁelds of Fig. 12.6.4. Because the traveling wave components of the standing wave have phases that advance in the ydirection with the same velocity, have the same wavelength in the xdirection and the same frequency, their electric ﬁelds in the y-direction exactly cancel in the planes x= 0 and x=aat each instant in time. With this recognition, we may construct TE modes of the parallel plate conductor structure of Fig. 12.7.4 by superposition of two countertraveling waves, one of which was studied in Example 12.6.2. 12.8 SUMMARY This chapter has been concerned with the determination of the electrodynamic ﬁelds associated with given distributions of current density J(r, t) and charge density ρ(r, t). We began by extending the vector potential Aand scalar potential Φ to situations where both the displacement current density and the magnetic induction are important. The resulting ﬁeld-potential relations, the ﬁrst two equations in Table 12.8.1, are familiar from quasistatics, except that −∂A/∂tis added to −∇Φ. As deﬁned here, with Aand Φ related by the gauge condition of (12.1.7) in the table, the current density Jis the source of A, while the charge density ρis the source of 10sinu= (exp( ju)−exp(−ju))/2j 52 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 TABLE 12.8.1 ELECTRODYNAMIC SOURCE-POTENTIAL RELATIONS B=µH=∇ ×A (12.1.1) E=−∇Φ−∂A ∂t(12.1.3) ∇2A−µ/epsilon1∂2A ∂t2=−µJ (12.1.8) A=µZ V/primeJ¡ r/prime, t−|r−r/prime| c¢ 4π|r−r/prime|dv/prime(12.3.2) ∇2Φ−µ/epsilon1∂2Φ ∂t2=−ρ /epsilon1(12.1.10) Φ =Z V/primeρ¡ r/prime, t−|r−r/prime| c¢ 4π/epsilon1|r−r/prime|dv/prime(12.3.1) ∇ ·A+µ/epsilon1∂Φ ∂t= 0 (12.1.7) ∇ ·J+∂ρ ∂t= 0 (12.1.20) Φ. This is evident from the ﬁnding that, written in terms of Aand Φ, Maxwell’s equations imply the inhomogeneous wave equations summarized by (12.1.8) and (12.1.10) in Table 12.8.1. Given the sources everywhere, solutions to the inhomogeneous wave equations are given by the respective superposition integrals of Table 12.8.1. As a reminder that the sources in these integrals are related, the charge conservation law, (12.1.20), is included. The relation between Jandρin the superposition integrals implied by charge conservation underlies the gauge relation between Aandρ, (12.1.7). The derivation of the superposition integrals began in Sec. 12.2 with the iden- tiﬁcation of the potentials, and hence ﬁelds, associated with dipoles. Here, in re- markably simple terms, it was seen that the eﬀect on the ﬁeld at rof the source at r/primeis delayed by the time required for a wave to propagate through the intervening distance at the velocity of light, c. In the quasistatic limit, where times of interest are long compared to this delay time, the electric and magnetic dipoles considered in Sec. 12.2 are those familiar from electroquasistatics (Sec. 4.4) and magnetoqua- sistatics (Sec. 8.3), respectively. With the complete description of electromagnetic radiation from these dipoles, we could place the introduction to quasistatics of Sec. 3.3 on ﬁrmer ground. For the purpose of determining the radiation pattern and radiation resistance of antennae, the radiation ﬁelds are of primary interest. For the sinusoidal steady state, Section 12.4 illustrated how the radiation ﬁelds could be superimposed to describe the radiation from given distributions of current elements representing an antenna, and how the ﬁelds from these elements could be combined to represent the radiation from an array. The elementary solutions from which these ﬁelds were constructed are those of an electric dipole, as summarized in Table 12.8.2. A similar Sec. 12.8 Summary 53 TABLE 12.8.2 DIPOLE RADIATION FIELDS ˆHφ=jkd 4πˆisinθe−jkr r(12.2.23) ˆEφ=−p µ//epsilon1ˆHθ (12.2.36) k≡ω/c ˆEθ=p µ//epsilon1ˆHφ (12.2.24) ˆHθ=−k2 4πˆmsinθe−jkr r(12.2.35) k≡ω/c use can be made of the magnetic dipole radiation ﬁelds, which are also summarized for reference in the table. The ﬁelds associated with planar sheet sources, the subject of Sec. 12.6, will be encountered again in the next chapter. In Sec. 12.6, the surface sources were taken as given. We found that sources having distributions that were dependent on (x, t) (independent of z) could be classiﬁed in accordance with the ﬁelds they produced, as summarized by the ﬁgures in Table 12.8.3. The TM and TE sources and ﬁelds, respectively, are described in terms of HzandEzby the relations given in the table. In the limit ω2µ/epsilon1/lessmuchk2 x, these source and ﬁeld cases are EQS and MQS, respectively. This condition on the frequency means that the period 2 π/ωis much longer than the time λx/cfor an electromagnetic wave to propagate a distance equal to a wavelength λx= 2π/kxin the xdirection. In the form of uniform and nonuniform plane waves, the Cartesian coordinate solutions to the homogeneous wave equation for these two-dimensional ﬁelds are summarized by the last equations in Table 12.8.3. In this chapter, we have thought ofkxas being imposed by the given source distribution. As the frequency is raised, 54 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 TABLE 12.8.3 TWO-DIMENSIONAL ELECTRODYNAMIC FIELDS ¡∂2 ∂x2+∂2 ∂y2¢ˆHz+ω2µ/epsilon1ˆHz= 0 (12.6.9)¡∂2 ∂x2+∂2 ∂y2¢ˆEz+ω2µ/epsilon1ˆEz= 0 (12.6.33) ˆEx=1 jω/epsilon1∂ˆHz ∂y(12.6.6) ˆHx=−1 jωµ∂ˆEz ∂y(12.6.29) ˆEy=−1 jω/epsilon1∂ˆHz ∂x(12.6.7) ˆHy=1 jωµ∂ˆEz ∂x(12.6.30) · ˆHzˆEz¸ ∝Reej(ωt∓βy−kxx); β≡½ |p ω2µ/epsilon1−k2x|, ω2µ/epsilon1 > k2 x −j|p k2x−ω2µ/epsilon1|, ω2µ/epsilon1 < k2 x(12.6.13) with the wavelength along the xdirection λx= 2π/kxﬁxed, the ﬁelds at ﬁrst decay in the ±ydirections (are evanescent in those directions) and are in temporal synchronism with the sources. These are the EQS and MQS limits. As the frequency is raised, the ﬁelds extend further and further in the ±ydirections. At the frequency f=c/λx, the ﬁeld decay in the ±ydirections gives way to propagation. In the next chapter, these ﬁeld solutions will be found fundamental to the description of ﬁelds in the presence of perfect conductors and dielectrics. R E F E R E N C E S Sec. 12.8 Summary 55 [1] H. A. Haus and P. Penﬁeld, Jr., Electrodynamics of Moving Media , MIT Press, Cambridge, Mass. (1967). [2] J. R. Melcher, Continuum Electromechanics , Secs. 2.8 and 2.9, MIT Press, Cambridge, Mass. (1982). 56 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 P R O B L E M S 12.1 Electrodynamic Fields and Potentials 12.1.1∗In Sec. 10.1, the electric ﬁeld in an MQS system was divided into a partic- ular part Epsatisfying Faraday’s law, and an irrotational part Eh. The lat- ter was adjusted to make the sum satisfy appropriate boundary conditions. Show that in terms of Φ and A, as deﬁned in this section, Ep=−∂A/∂t andEh=−∇Φ, where these potentials satisfy (12.1.8) and (12.1.10) with the time derivatives neglected. 12.1.2 In Sec. 3.3, dimensional arguments were used to show that the quasistatic limits were valid in a system having a typical length Land time τifL/c/lessmuch τ. Use similar arguments to show that the second term on the right in either (12.1.8) or (12.1.10) is negligible when this condition prevails. Note that the resulting equations are those for MQS (8.1.5) and EQS (4.2.2) systems. 12.2 Electrodynamic Fields of Source Singularities 12.2.1 An electric dipole has q(t) = 0 for t <0 and t > T . When 0 < t < T, q (t) = Q[1−cos(2 πt/T )]/2. Use sketches similar to those of Figs. 12.2.5 and 12.2.6 to show the ﬁeld distributions when t < T andT < t . 12.2.2∗Use the “interchange of variables” property of Maxwell’s equations to show that the sinusoidal steady state far ﬁelds of a magnetic dipole, (12.2.35) and (12.2.36), follow directly from (12.2.23), (12.2.24), and (12.2.32). 12.2.3∗A magnetic dipole has a moment m(t) having the time dependence shown in Fig. 12.2.5a where dq(t)→µm(t). Show that the ﬁelds are then much as shown in Fig. 12.2.6 with E→H,H→ −E, and /epsilon1↔µ. 12.4 Antenna Radiation Fields in the Sinusoidal Steady State 12.4.1 An “end-fed” antenna consists of a wire stretching between z= 0, where it is driven by the current Iocos(ωt−αo), and z=l. At z=l, it is terminated in such a resistance that the current distribution over its length is a wave traveling with the velocity of light in the zdirection; i(z, t) = Re [Ioexp[j(ωt−kz+αo)]] where k≡ω/c. (a) Determine the radiation pattern, Ψ( θ). Sec. 12.4 Problems 57 (b) For a one-wavelength antenna ( kl= 2π), use a plot of |Ψ(θ)|to show that the lobes of the radiation pattern tend to be in the direction of the traveling wave. 12.4.2∗An antenna is modeled by a distribution of incremental magnetic dipoles, as shown in Fig. P12.4.2. Deﬁne M(z) as a dipole moment per unit length so that for an incremental dipole located at z/prime,ˆm→ M (z/prime)dz/prime. Given M, show that ˆEφ=k2l 4πp µ//epsilon1e−jkr rMoejαoψo(θ) ( a) where ψo(θ)≡sinθ lZM(z/prime) Moej(kr/prime·ir−αo)dz/prime(b) Fig. P12.4.2 12.4.3 A linear distribution of magnetic dipoles, described in general in Prob. 12.4.2, is excited so that M(z/prime) =−M oexp(jαo) sinβ(z−l)/sinβl,0≤ z≤lwhere βis a given parameter (not necessarily ω/c). Determine ψo(θ). 12.4.4∗For the three-element array shown in Fig. P12.4.4, the spacing is λ/4. (a) Show that the array factor is ψa(θ, φ) =£ 1 +ej¡ π 2cosφsinθ+α1−αo)+ej(πcosφsinθ+α2−αo)¤ (a) (b) Show that for an array of in-phase short dipoles, the “broadside” radiation intensity pattern is |ψo|2|ψa|2=£ 1 + 2 cos¡π 2cosφsinθ¢¤2sin2θ (b) 58 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 (c) Show that for an array of short dipoles diﬀering progressively by 90 degrees so that α1−αo=π/2 and α2−αo=π, the end-ﬁre radiation pattern is |ψo|2|ψa|2=© 1 + 2 cos£π 2(cosφsinθ+ 1)¤ª2sin2θ (c) Fig. P12.4.4 12.4.5 Collinear elements have the half-wave spacing and conﬁguration shown in Fig. P12.4.5. (a) Determine the array factor ψa(θ). (b) What is the radiation pattern if the elements are “short” dipoles driven in phase? (c) What is the gain G(θ) for the array of part (b)? 12.5 Complex Poynting’s Theorem and Radiation Resistance 12.5.1∗A center-fed wire antenna has a length of 3 λ/2. Show that its radiation resistance in free space is 104Ω. (The deﬁnite integral can be evaluated numerically.) 12.5.2 The spherical coil of Example 8.5.1 is used as a magnetic dipole antenna. Its diameter is much less than a wavelength, and its equivalent circuit is an inductance Lin parallel with a radiation resistance Rrad. In terms of the radius R, number of turns N, and frequency ω, what are LandRrad? 12.6 Periodic Sheet Source Fields: Uniform and Nonuniform Plane Waves Sec. 12.6 Problems 59 Fig. P12.4.5 12.6.1∗In the plane y= 0, Kz= 0 and the surface charge density is given as the traveling wave σs= Re σoexp[j(ωt−kxx)] = Re [ σoexp(−jkxx) exp( jωt)], where σo, ω, and kxare given real numbers. (a) Show that the current density is Kx= Reωσo kxej(ωt−kxx)= Reµωσoe−jkxx kx¶ ejωt(a) (b) Show that the ﬁelds are H=izRe£ ±ωσo 2kxe∓jβyej(ωt−kxx)¤ (b) E= Re· ixµ−βσo 2/epsilon1kxe∓jβy¶ +iy¡ ±σo 2/epsilon1¢ e∓jβy¸ ej(ωt−kxx)(c) where upper and lower signs, respectively, refer to the regions where 0< yand 0 > y. (c) Sketch the ﬁeld distributions at a given instant in time for βimaginary andβreal. 12.6.2 In the plane y= 0, the surface current density is a standing wave, K= Re[izKo sin(kxx) exp( jωt)], and there is no surface charge density. (a) Determine EandH. (b) Sketch these ﬁelds at a given instant in time for βreal and βimagi- nary. (c) Show that these ﬁelds can be decomposed into waves traveling in the ±xdirections with the phase velocities ±ω/kx. 60 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 12.6.3∗In the planes y=±d/2, shown in Fig. P12.6.3, there are surface current densities Kz= Re ˆKexp[j(ωt−kxx)], where ˆK=ˆKaaty=d/2 and ˆK=ˆKbaty=−d/2. The surface charge density is zero in each plane. (a) Show that Ez= Re−ωµ 2β( ˆKa2 4exp£ −jβ¡ y−d 2¢¤ exp£ jβ¡ y−d 2¢¤ exp£ jβ¡ y−d 2¢¤3 5+ ˆKb2 4exp£ −jβ¡ y+d 2¢¤ exp£ −jβ¡ y+d 2¢¤ exp£ jβ¡ y+d 2¢¤3 5) ej(ωt−kxx);d 2< y ;−d 2< y <d 2 ; y <−d 2(a) (b) Show that if ˆKb=−ˆKaexp(−jβd), the ﬁelds cancel in region (b) where ( y <−d/2), so that the combined radiation is unidirectional. (c) Show that under this condition, the ﬁeld in the region y > d/ 2 is Ez= Re−ωµj βˆKae−jβd(sinβd)ej£ ωt−β(y−d 2)−kxx¤ ;d 2< y (b) (d) With the structure used to impose the surface currents such that kx is ﬁxed, show that to maximize the wave radiated in region (a), the frequency should be ω=1√µ/epsilon1r k2x+£(2n+ 1)π 2d¤2(c) and that under this condition, the direction of the radiated wave is k=kxix+ [(2n+ 1)π/2d]iywhere n= 0,1,2, . . .. 12.6.4 Surface charges in the planes y=±d/2 shown in Fig. P12.6.3 have the densities σs= Re ˆ σexp[j(ωt−kxx)] where ˆ σ= ˆσaaty=d/2 and ˆ σ= ˆσb aty=−d/2. (a) How should ˆ σaand ˆσbbe related to produce ﬁeld cancellation in region (b)? (b) Under this condition, what is Hzin region (a)? (c) What frequencies give a maximum Hzin region (a), and what is the direction of propagation under this condition? 12.7 Electrodynamic Fields in the Presence of Perfect Conductors 12.7.1∗An antenna consists of a ground plane with a 3 λ/4 vertical element in which a “quarter-wave stub” is used to make the current in the top half- wavelength in phase with that in the bottom quarter-wavelength. In each Sec. 12.7 Problems 61 Fig. P12.6.3 section, the current has the sinusoidal distribution shown in Fig. P12.7.1. Show that the radiation intensity factor is |ψo|2|ψa|2= (2/π)2|cos[(π/2) cos θ]|2|1 + 2 cos( πcosθ)|2/|sinθ|2 Fig. P12.7.1 Fig. P12.7.2 12.7.2 A vertical half-wave antenna with a horizontal perfectly conducting ground plane is shown in Fig. P12.7.2. What is its radiation resistance? 62 Electrodynamic Fields: The Superposition Integral Point of View Chapter 12 Fig. P12.7.3 12.7.3 Plane parallel perfectly conducting plates in the planes x=±a/2 form the walls of a waveguide, as shown in Fig. 12.7.3. Waves in the free- space region between are excited by a sheet of surface charge density σs= Re σocos(πx/a ) exp( jωt) and Kz= 0. (a) Find the ﬁelds in regions 0 < y and 0 > y. (Guess solutions that meet both the continuity conditions at the sheet and the boundary conditions on the perfectly conducting plates.) Are they TE or TM? (b) What are the distributions of σsandKon the perfectly conducting plates? (c) What is the “dispersion equation” relating ωtoβ? (d) Sketch EandHforβimaginary and real. 12.7.4 Consider the conﬁguration of Prob. 12.7.3, but with σs= 0 and Kz= Re ˆKo cos(πx/a ) exp( jωt) in the plane y= 0. Complete parts (a)-(d) of Prob. 12.7.3. 13 ELECTRODYNAMIC FIELDS: THE BOUNDARY VALUE POINT OF VIEW 13.0 INTRODUCTION In the treatment of EQS and MQS systems, we started in Chaps. 4 and 8, re- spectively, by analyzing the ﬁelds produced by speciﬁed (known) sources. Then we recognized that in the presence of materials, at least some of these sources were induced by the ﬁelds themselves. Induced surface charge and surface current den- sities were determined by making the ﬁelds satisfy boundary conditions. In the volume of a given region, ﬁelds were composed of particular solutions to the gov- erning quasistatic equations (the scalar and vector Poisson equations for EQS and MQS systems, respectively) and those solutions to the homogeneous equations (the scalar and vector Laplace equation, respectively) that made the total ﬁelds satisfy appropriate boundary conditions. We now embark on a similar approach in the analysis of electrodynamic ﬁelds. Chapter 12 presented a study of the ﬁelds produced by speciﬁed sources (dipoles, line sources, and surface sources) and obeying the inhomogeneous wave equation. Just as in the case of EQS and MQS systems in Chap. 5 and the last half of Chap. 8, we shall now concentrate on solutions to the homogeneous source-free equations. These solutions then serve to obtain the ﬁelds produced by sources lying outside (maybe on the boundary) of the region within which the ﬁelds are to be found. In the region of interest, the ﬁelds generally satisfy the inhomogeneous wave equation. However in this chapter, where there are no sources in the volume of interest, they satisfy the homogeneous wave equation. It should come as no surprise that, following this systematic approach, we shall reencounter some of the previously obtained solutions. In this chapter, ﬁelds will be determined in some limited region such as the volume Vof Fig. 13.0.1. The boundaries might be in part perfectly conducting in the sense that on their surfaces, Eis perpendicular and the time-varying His tangential. The surface current and charge densities implied by these conditions 1 2 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.0.1 Fields in a limited region are in part due to sources induced on boundaries by the ﬁelds themselves. are not known until after the ﬁelds have been found. If there is material within the region of interest, it is perfectly insulating and of piece-wise uniform permittivity /epsilon1 and permeability µ.1Sources Jandρare speciﬁed throughout the volume and ap- pear as driving terms in the inhomogeneous wave equations, (12.6.8) and (12.6.32). Thus, the HandEﬁelds obey the inhomogeneous wave-equations. ∇2H−µ/epsilon1∂2H ∂t2=−∇ × J (1) ∇2E−µ/epsilon1∂2E ∂t2=∇¡ρ /epsilon1¢ +µ∂J ∂t(2) As in earlier chapters, we might think of the solution to these equations as the sum of a part satisfying the inhomogeneous equations throughout V(partic- ular solution), and a part satisfying the homogeneous wave equation throughout that region. In principle, the particular solution could be obtained using the su- perposition integral approach taken in Chap. 12. For example, if an electric dipole were introduced into a region containing a uniform medium, the particular solution would be that given in Sec. 12.2 for an electric dipole. The boundary conditions are generally not met by these ﬁelds. They are then satisﬁed by adding an appropriate solution of the homogeneous wave equation.2 In this chapter, the source terms on the right in (1) and (2) will be set equal to zero, and so we shall be concentrating on solutions to the homogeneous wave equation. By combining the solutions of the homogeneous wave equation that satisfy boundary conditions with the source-driven ﬁelds of the preceding chapter, one can describe situations with given sources and given boundaries. In this chapter, we shall consider the propagation of waves in some axial direction along a structure that is uniform in that direction. Such waves are used to transport energy along pairs of conductors (transmission lines), and through 1If the region is one of free space, /epsilon1→/epsilon1oandµ→µo. 2As pointed out in Sec. 12.7, this is essentially what is being done in satisfying boundary conditions by the method of images. Sec. 13.1 TEM Waves 3 waveguides (metal tubes at microwave frequencies and dielectric ﬁbers at optical frequencies). We conﬁne ourselves to the sinusoidal steady state. Sections 13.1-13.3 study two-dimensional modes between plane parallel con- ductors. This example introduces the mode expansion of electrodynamic ﬁelds that is analogous to the expansion of the EQS ﬁeld of the capacitive attenuator (in Sec. 5.5) in terms of the solutions to Laplace’s equation. The principal and higher order modes form a complete set for the representation of arbitrary boundary conditions. The example is a model for a strip transmission line and hence serves as an intro- duction to the subject of Chap. 14. The higher-order modes manifest properties much like those found in Sec. 13.4 for hollow pipe guides. The dielectric waveguides considered in Sec. 13.5 explain the guiding prop- erties of optical ﬁbers that are of great practical interest. Waves are guided by a dielectric core having permittivity larger than that of the surrounding medium but possess ﬁelds extending outside this core. Such electromagnetic waves are guided because the dielectric core slows the eﬀective velocity of the wave in the guide to the point where it can match the velocity of a wave in the surrounding region that propagates along the guide but decays in a direction perpendicular to the guide. The ﬁelds considered in Secs. 13.1–13.3 oﬀer the opportunity to reinforce the notions of quasistatics. Connections between the EQS and MQS ﬁelds studied in Chaps. 5 and 8, respectively, and their corresponding electrodynamic ﬁelds are made throughout Secs. 13.1–13.4. 13.1 INTRODUCTION TO TEM WAVES TheEandHﬁelds of transverse electromagnetic waves are directed transverse to the direction of propagation. It will be shown in Sec. 14.2 that such TEM waves propagate along structures composed of pairs of perfect conductors of arbitrary cross-section. The parallel plates shown in Fig. 13.1.1 are a special case of such a pair of conductors. The direction of propagation is along the yaxis. With a source driving the conductors at the left, the conductors can be used to deliver electrical energy to a load connected between the right edges of the plates. They then function as aparallel plate transmission line . We assume that the plates are wide in the zdirection compared to the spacing, a, and that conditions imposed in the planes y= 0 and y=−bare independent of z, so that the ﬁelds are also zindependent. In this section, discussion is limited to either “open” electrodes at y= 0 or “shorted” electrodes. Techniques for dealing with arbitrarily terminated transmission lines will be introduced in Chap. 14. The “open” or “shorted” terminals result in standing waves that serve to illustrate the relationship between simple electrodynamic ﬁelds and the EQS and MQS limits. These ﬁelds will be generalized in the next two sections, where we ﬁnd that the TEM wave is but one of an inﬁnite number of modes of propagation along the y axis between the plates. If the plates are open circuited at the right, as shown in Fig. 13.1.1, a voltage is applied at the left at y=−b, and the ﬁelds are EQS, the Ethat results is x directed. (The plates form a parallel plate capacitor.) If they are “shorted” at the right and the ﬁelds are MQS, the Hthat results from applying a current source at the left is zdirected. (The plates form a one-turn inductor.) We are now looking 4 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.1.1 Plane parallel plate transmission line. for solutions to Maxwell’s equations (12.0.7)–(12.0.10) that are similarly transverse to the yaxis. E=Exix;H=Hziz (1) Fields of this form automatically satisfy the boundary conditions of zero tan- gential Eand normal H(normal B) on the surfaces of the perfect conductors. These ﬁelds have no divergence, so the divergence laws for EandH[(12.0.7) and (12.0.10)] are automatically satisﬁed. Thus, the remaining laws, Amp` ere’s law (12.0.8) and Faraday’s law (12.0.9) fully describe these TEM ﬁelds. We pick out the only com- ponents of these laws that are not automatically satisﬁed by observing that ∂Ex/∂t drives the xcomponent of Amp` ere’s law and ∂Hz/∂tis the source term of the z component of Faraday’s law. ∂Hz ∂y=/epsilon1∂Ex ∂t (2) ∂Ex ∂y=µ∂Hz ∂t (3) The other components of these laws are automatically satisﬁed if it is assumed that the ﬁelds are independent of the transverse coordinates and thus depend only on y. The eﬀect of the plates is to terminate the ﬁeld lines so that there are no ﬁelds in the regions outside. With Gauss’ continuity condition applied to the respective plates, Exterminates on surface charge densities of opposite sign on the respective electrodes. σs(x= 0) = /epsilon1Ex; σs(x=a) =−/epsilon1Ex (4) These relationships are illustrated in Fig. 13.1.2a. The magnetic ﬁeld is terminated on the plates by surface current densities. With Amp` ere’s continuity condition applied to each of the plates, Ky(x= 0) = −Hz; Ky(x=a) =Hz (5) Sec. 13.1 TEM Waves 5 Fig. 13.1.2 (a) Surface charge densities terminating Eof TEM ﬁeld between electrodes of Fig. 13.1.1. (b) Surface current densities terminating H. these relationships are represented in Fig. 13.1.2b. We shall be interested primarily in the sinusoidal steady state. Between the plates, the ﬁelds are governed by diﬀerential equations having constant coeﬃcients. We therefore assume that the ﬁeld response takes the form Hz= Re ˆHz(y)ejωt; Ex= Re ˆEx(y)ejωt(6) where ωcan be regarded as determined by the source that drives the system at one of the boundaries. Substitution of these solutions into (2) and (3) results in a pair of ordinary constant coeﬃcient diﬀerential equations describing the ydependence ofExandHz. Without bothering to write these equations out, we know that they too will be satisﬁed by exponential functions of y. Thus, we proceed to look for solutions where the functions of yin (6) take the form exp( −jkyy). Hz= Re ˆhzej(ωt−kyy); Ex= Re ˆ exej(ωt−kyy)(7) Once again, we have assumed a solution taking a product form. Substitution into (2) then shows that ˆex=−ky ω/epsilon1ˆhz (8) and substitution of this expression into (3) gives the dispersion equation ky=±β; β≡ω√µ/epsilon1=ω c(9) For a given frequency, there are two values of ky. A linear combination of the solutions in the form of (7) is therefore Hz= Re [ A+e−jβy+A−ejβy]ejωt(10) The associated electric ﬁeld follows from (8) evaluated for the ±waves, respectively, using ky=±β. Ex=−Rep µ//epsilon1[A+e−jβy−A−ejβy]ejωt(11) The amplitudes of the waves, A+andA−, are determined by the boundary conditions imposed in planes perpendicular to the yaxis. The following example 6 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.1.3 (a) Shorted transmission line driven by a distributed cur- rent source. (b) Standing wave ﬁelds with EandHshown at times diﬀering by 90 degrees. (c) MQS ﬁelds in limit where wavelength is long compared to length of system. illustrates how the imposition of these longitudinal boundary conditions determines the ﬁelds. It also is the ﬁrst of several opportunities we now use to place the EQS and MQS approximations in perspective. Example 13.1.1. Standing Waves on a Shorted Parallel Plate Transmission Line In Fig. 13.1.3a, the parallel plates are terminated at y= 0 by a perfectly conducting plate. They are driven at y=−bby a current source Iddistributed over the width w. Thus, there is a surface current density Ky=Id/w≡Koimposed on the lower plate at y=−b. Further, in this example we will assume that a distribution of sources is used in the plane y=−bto make this driving surface current density uniform over that plane. In summary, the longitudinal boundary conditions are Ex(0, t) = 0 (12) Hz(−b, t) =−ReˆKoejωt(13) To make Exas given by (11) satisfy the ﬁrst of these boundary conditions, we must have the amplitudes of the two traveling waves equal. A+=A−(14) With this relation used to eliminate A+in (10), it follows from (13) that A+=−ˆKo 2 cosβb(15) We have found that the ﬁelds between the plates take the form of standing waves. Hz=−ReˆKocosβy cosβbejωt(16) Sec. 13.1 TEM Waves 7 Ex=−RejˆKop µ//epsilon1sinβy cosβbejωt(17) Note that EandHare 90◦out of temporal phase.3When one is at its peak, the other is zero. The distributions of EandHshown in Fig. 13.1.3b are therefore at diﬀerent instants in time. Every half-wavelength π/β from the short, Eis again zero, as sketched in Fig. 13.1.3b. Beginning at a distance of a quarter-wavelength from the short, the magnetic ﬁeld also exhibits nulls at half-wavelength intervals. Adjacent peaks in a given ﬁeld are 180 degrees out of temporal phase. The MQS Limit. If the driving frequency is so low that a wavelength is much longer than the length b, we have 2πb λ=βb/lessmuch1 (18) In this limit, the ﬁelds are those of a one-turn inductor. That is, with sin( βy)≈βy and cos( βy)≈1, (16) and (17) become Hz→ −ReˆKoejωt(19) Ex→ −ReˆKojωµyejωt(20) The magnetic ﬁeld intensity is uniform throughout and the surface current density circulates uniformly around the one-turn loop. The electric ﬁeld increases in a linear fashion from zero at the short to a maximum at the source, where the source voltage is v(t) =Za 0Ex(−b, t)dx= Re ˆKojωµbaejωt=dλ dt(21) To make it clear that these are the ﬁelds of a one-turn solenoid (Example 8.4.4), the ﬂux linkage λhas been identiﬁed as λ=Ldi dt; i= Re ˆKowejωt; L=abµ w(22) where Lis the inductance. The MQS Approximation. In Chap. 8, we would have been led to these same limiting ﬁelds by assuming at the outset that the displacement current, the term on the right in (2), is negligible. Then, this one-dimensional form of Amp` ere’s law and (1) requires that ∇ ×H≈0⇒∂Hz ∂y≈0⇒Hz=Hz(t) =−ReˆKoejωt(23) If we now use this ﬁnding in Faraday’s law, (3), integration on yand use of the boundary condition of (12) gives the same result for Eas found taking the low- frequency limit, (20). 3In making this and the following deductions, it is helpful to take ˆKoas being real. 8 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.1.4 (a) Open circuit transmission line driven by voltage source. (b) EandHat times that diﬀer by 90 degrees. (c) EQS ﬁelds in limit where wavelength is long compared to b. In the previous example, the longitudinal boundary conditions (conditions imposed at planes of constant y) could be satisﬁed exactly using the TEM mode alone. The short at the right and the distributed current source at the left each imposed a condition that was, like the TEM ﬁelds, independent of the transverse coordinates. In almost all practical situations, longitudinal boundary conditions which are independent of the transverse coordinates (used to describe transmission lines) are approximate. The open circuit termination at y= 0, shown in Fig. 13.1.4, is a case in point, as is the source which in this case is not distributed in the x direction. If a longitudinal boundary condition is independent of z, the ﬁelds are, in principle, still two dimensional. Between the plates, we can therefore think of sat- isfying the longitudinal boundary conditions using a superposition of the modes to be developed in the next section. These consist of not only the TEM mode con- sidered here, but of modes having an xdependence. A detailed evaluation of the coeﬃcients specifying the amplitudes of the higher-order modes brought in by the transverse dependence of a longitudinal boundary condition is illustrated in Sec. 13.3. There we shall ﬁnd that at low frequencies, where these higher-order modes are governed by Laplace’s equation, they contribute to the ﬁelds only in the vicinity of the longitudinal boundaries. As the frequency is raised beyond their respective cutoﬀ frequencies, the higher-order modes begin to propagate along the yaxis and so have an inﬂuence far from the longitudinal boundaries. Here, where we wish to restrict ourselves to situations that are well described by the TEM modes, we restrict the frequency range of interest to well below the lowest cutoﬀ frequency of the lowest of the higher-order modes. Given this condition, “end eﬀects” are restricted to the neighborhood of a longitudinal boundary. Approximate boundary conditions then determine the dis- tribution of the TEM ﬁelds, which dominate over most of the length. In the open Sec. 13.1 TEM Waves 9 Fig. 13.1.5 The surface current density, and hence, Hzgo to zero in the vicinity of the open end. circuit example of Fig. 13.1.4a, application of the integral charge conservation law to a volume enclosing the end of one of the plates, as illustrated in Fig. 13.1.5, shows that Kymust be essentially zero at y= 0. For the TEM ﬁelds, this implies the boundary condition4 Hz(0, t) = 0 (24) At the left end, the vertical segments of perfect conductor joining the voltage source to the parallel plates require that Exbe zero over these segments. We shall show later that the higher-order modes do not contribute to the line integral of E between the plates. Thus, in so far as the TEM ﬁelds are concerned, the requirement is that Vd(t) =Za 0Ex(−b, t)dx⇒Ex(−b, t) =Vd a(25) Example 13.1.2. Standing Waves on an Open-Circuit Parallel Plate Transmission Line Consider the parallel plates “open” at y= 0 and driven by a voltage source at y=−b. Boundary conditions are then Hz(0, t) = 0; Ex(−b, t) = Re ˆVdejωt/a (26) Evaluation of the coeﬃcients in (10) and (11) so that the boundary conditions in (26) are satisﬁed gives A+=−A−=−ˆVd 2acosβbp /epsilon1/µ (27) It follows that the TEM ﬁelds between the plates, (10) and (11), are Hz= Re jˆVd ap /epsilon1/µsinβy cosβbejωt(28) Ex= ReˆVd acosβy cosβbejωt(29) These distributions of HandEare shown in Fig. 13.1.4 at times that diﬀer by 90 degrees. The standing wave is similar to that described in the previous example, except that it is now Erather than Hthat peaks at the open end. 4In the region outside, the ﬁelds are not conﬁned by the plates. As a result, there is actually some radiation from the open end of the line, and this too is not represented by (24). This eﬀect is small if the plate spacing is small compared to a wavelength. 10 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 The EQS Limit. In the low frequency limit, where the wavelength is much longer than the length of the plates so that βb/lessmuch1, the ﬁelds given by (28) and (29) become Hz→RejˆVd aω/epsilon1yejωt(30) Ex→ReˆVd aejωt(31) At low frequencies, the ﬁelds are those of a capacitor. The electric ﬁeld is uniform and simply equal to the applied voltage divided by the spacing. The magnetic ﬁeld varies in a linear fashion from zero at the open end to its peak value at the voltage source. Evaluation of −Hzatz=−bgives the surface current density, and hence the current i, provided by the voltage source. i= Re jω/epsilon1bw aˆVdejωt(32) Note that this expression implies that i=dq dt; q=CVd; C=/epsilon1bw a(33) so that the limiting behavior is indeed that of a plane parallel capacitor. EQS Approximation. How would the quasistatic ﬁelds be predicted in terms of the TEM ﬁelds? If quasistatic, we expect the system to be EQS. Thus, the magnetic induction is negligible, so that the right-hand side of (3) is approximated as being equal to zero. ∇ ×E≈0⇒∂Ex ∂y≈0 (34) It follows from integration of this expression and using the boundary condition of (26b) that the quasistatic Eis Ex=Vd a(35) In turn, this result provides the displacement current density in Amp` ere’s law, the right-hand side of (2). ∂Hz ∂y/similarequal/epsilon1d dt¡Vd a¢ (36) The right-hand side of this expression is independent of y. Thus, integration with respect to y, with the “constant” of integration evaluated using the boundary condition of (26a), gives Hz/similarequal/epsilon1d dtVdy a(37) For the sinusoidal voltage drive assumed at the outset in the description of the TEM waves, this expression is consistent with that found in taking the quasistatic limit, (30). Sec. 13.2 Parallel Plate Modes 11 Demonstration 13.1.1. Visualization of Standing Waves A demonstration of the ﬁelds described by the two previous examples is shown in Fig. 13.1.6. A pair of sheet metal electrodes are driven at the left by an oscillator. A ﬂuorescent lamp placed between the electrodes is used to show the distribution of the rms electric ﬁeld intensity. The gas in the tube is ionized by the oscillating electric ﬁeld. Through the ﬁeld-induced acceleration of electrons in this gas, a suﬃcient velocity is reached so that collisions result in ionization and an associated optical radiation. What is seen is a time average response to an electric ﬁeld that is oscillating far more rapidly than can be followed by the eye. Because the light is proportional to the magnitude of the electric ﬁeld, the observed 0.75 m distance between nulls is a half-wavelength. It can be inferred that the generator frequency is f=c/λ= 3×108/1.5 = 200 MHz. Thus, the frequency is typical of the lower VHF television channels. With the right end of the line shorted, the section of the lamp near that end gives evidence that the electric ﬁeld there is indeed as would be expected from Fig. 13.1.3b, where it is zero at the short. Similarly, with the right end open, there is a peak in the light indicating that the electric ﬁeld near that end is maximum. This is consistent with the picture given in Fig. 13.1.4b. In going from an open to a shorted condition, the positions of peak light intensity, and hence of peak electric ﬁeld intensity, are shifted by λ/4. 12 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.2.1 (a) Plane parallel perfectly conducting plates. (b) Coaxial ge- ometry in which z-independent ﬁelds of (a) might be approximately obtained without edge eﬀects. 13.2 TWO-DIMENSIONAL MODES BETWEEN PARALLEL PLATES This section treats the boundary value approach to ﬁnding the ﬁelds between the perfectly conducting parallel plates shown in Fig. 13.2.1a. Most of the mathematical ideas and physical insights that come from a study of modes on perfectly conducting structures that are uniform in one direction (for example, parallel wire and coaxial transmission lines and waveguides in the form of hollow perfectly conducting tubes) are illustrated by this example. In the previous section, we have already seen that the plates can be used as a transmission line supporting TEM waves. In this and the next section, we shall see that they are capable of supporting other electromagnetic waves. Because the structure is uniform in the zdirection, it can be excited in such a way that ﬁelds are independent of z. One way to make the structure approximately uniform in the zdirection is illustrated in Fig. 13.2.1b, where the region between the plates becomes the annulus of coaxial conductors having very nearly the same radii. Thus, the diﬀerence of these radii becomes essentially the spacing aand the zcoordinate maps into the φcoordinate. Another way is to make the plates very wide (in the zdirection) compared to their spacing, a. Then, the fringing ﬁelds from the edges of the plates are negligible. In either case, the understanding is that the ﬁeld excitation is uniformly distributed in the zdirection. The ﬁelds are now assumed to be independent of z. Because the ﬁelds are two dimensional, the classiﬁcations and relations given in Sec. 12.6 and summarized in Table 12.8.3 serve as our starting point. Cartesian coordinates are appropriate because the plates lie in coordinate planes. Fields either haveHtransverse to the x−yplane and Ein the x−yplane (TM) or have E transverse and Hin the x−yplane (TE). In these cases, HzandEzare taken as the functions from which all other ﬁeld components can be derived. We consider sinusoidal steady state solutions, so these ﬁelds take the form Hz= Re ˆHz(x, y)ejωt(1) Ez= Re ˆEz(x, y)ejωt(2) Sec. 13.2 Parallel Plate Modes 13 These ﬁeld components, respectively, satisfy the Helmholtz equation, (12.6.9) and (12.6.33) in Table 12.8.3, and the associated ﬁelds are given in terms of these components by the remaining relations in that table. Once again, we ﬁnd product solutions to the Helmholtz equation, where Hz andEzare assumed to take the form X(x)Y(y). This formalism for reducing a partial diﬀerential equation to ordinary diﬀerential equations was illustrated for Helmholtz’s equation in Sec. 12.6. This time, we take a more mature approach, based on the observation that the coeﬃcients of the governing equation are inde- pendent of y(are constants). As a result, Y(y) will turn out to be governed by a constant coeﬃcient diﬀerential equation. This equation will have exponential solu- tions. Thus, with the understanding that kyis a yet to be determined constant (that will turn out to have two values), we assume that the solutions take the speciﬁc product forms ˆHz=ˆhz(x)e−jkyy(3) ˆEz= ˆez(x)e−jkyy(4) Then, the ﬁeld relations of Table 12.8.3 become TM Fields: d2ˆhz dx2+p2ˆhz= 0 (5) where p2≡ω2µ/epsilon1−k2 y ˆex=−ky ω/epsilon1ˆhz (6) ˆey=−1 jω/epsilon1dˆhz dx(7) TE Fields: d2ˆez dx2+q2ˆez= 0 (8) where q2≡ω2µ/epsilon1−k2 y ˆhx=ky ωµˆez (9) ˆhy=1 jωµdˆez dx(10) The boundary value problem now takes a classic form familiar from Sec. 5.5. What values of pandqwill make the electric ﬁeld tangential to the plates zero? For the TM ﬁelds, ˆ ey= 0 on the plates, and it follows from (7) that it is the derivative ofHzthat must be zero on the plates. For the TE ﬁelds, Ezmust itself be zero at the plates. Thus, the boundary conditions are TM Fields: dˆhz dx(0) = 0;dˆhz dx(a) = 0 (11) 14 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.2.2 Dependence of fundamental ﬁelds on x. TE Fields: ˆez(0) = 0; ˆ ez(a) = 0 (12) To check that all of the conditions are indeed met at the boundaries, note that if (11) is satisﬁed, there is neither a tangential Enor a normal Hat the boundaries for the TM ﬁelds. (There is no normal Hwhether the boundary condition is satisﬁed or not.) For the TE ﬁeld, Ezis the only electric ﬁeld, and making Ez=0 on the boundaries indeed guarantees that Hx= 0 there, as can be seen from (9). Representing the TM modes, the solution to (5) is a linear combination of sin(px) and cos( px). To satisfy the boundary condition, (11), at x= 0, we must select cos( px). Then, to satisfy the condition at x=a, it follows that p=pn= nπ/a, n = 0,1,2, . . . ˆhz∝cospnx (13) pn=nπ a, n = 0,1,2, . . . (14) These functions and the associated values of pare called eigenfunctions andeigen- values , respectively. The solutions that have been found have the xdependence shown in Fig. 13.2.2a. From the deﬁnition of pgiven in (5), it follows that for a given frequency ω (presumably imposed by an excitation), the wave number kyassociated with the n-th mode is ky≡ ±βn; βn≡½p ω2µ/epsilon1−(nπ/a )2; ω2µ/epsilon1 > (nπ/a )2 −jp (nπ/a )2−ω2µ/epsilon1;ω2µ/epsilon1 < (nπ/a )2(15) Similar reasoning identiﬁes the modes for the TE ﬁelds. Of the two solutions to (8), the one that satisﬁes the boundary condition at x= 0 is sin( qx). The second boundary condition then requires that qtake on certain eigenvalues, qn. ˆez∝sinqnx (16) qn=nπ a(17) Sec. 13.2 Parallel Plate Modes 15 Thexdependence of Ezis then as shown in Fig. 13.2.2b. Note that the case n= 0 is excluded because it implies a solution of zero amplitude. For the TE ﬁelds, it follows from (17) and the deﬁnition of qgiven with (8) that5 ky≡ ±βn; βn≡½p ω2µ/epsilon1−(nπ/a )2; ω2µ/epsilon1 > (nπ/a )2 −jp (nπ/a )2−ω2µ/epsilon1;ω2µ/epsilon1 < (nπ/a )2(18) In general, the ﬁelds between the plates are a linear combination of all of the modes. In superimposing these modes, we recognize that ky=±βn. Thus, with coeﬃcients that will be determined by boundary conditions in planes of constant y, we have the solutions TM Modes: Hz=Re£ A+ oe−jβoy+A− oejβoy +∞X n=1¡ A+ ne−jβny+A− nejβny¢ cosnπ ax¤ ejωt(19) TE Modes: Ez= Re∞X n=1¡ C+ ne−jβny+C− nejβny¢ sinnπ ax ejωt(20) We shall refer to the n-th mode represented by these ﬁelds as the TM nor TE n mode, respectively. We now make an observation about the TM 0mode that is of far-reaching signiﬁcance. Its distribution of Hzhas no dependence on x[(13) with pn= 0]. As a result, Ey= 0 according to (7). Thus, for the TM 0mode, bothEandHare transverse to the axial direction y. This special mode, represented by the n= 0 terms in (19), is therefore the transverse electromagnetic (TEM) mode featured in the previous section. One of its most signiﬁcant features is that the relation between frequency ωand wave number in the ydirection, ky, [(15) with n= 0] isky=±ω√µ/epsilon1=±ω/c, the same as for a uniform electromagnetic plane wave. Indeed, as we saw in Sec. 13.1, it isa uniform plane wave. The frequency dependence of kyfor the TEM mode and for the higher-order TMnmodes given by (15) are represented graphically by the ω−kyplot of Fig. 13.2.3. For a given frequency, ω, there are two values of kywhich we have called ±βn. The dashed curves represent imaginary values of ky. Imaginary values correspond to exponentially decaying and “growing” solutions. An exponentially “growing” solution is in fact a solution that decays in the −ydirection. Note that the switch from exponentially decaying to propagating ﬁelds for the higher-order modes occurs at the cutoﬀ frequency ωcn=1√µ/epsilon1¡nπ a¢ (21) 5For the particular geometry considered here, it has turned out that the eigenvalues pnand qnare the same (with the exception of n= 0). This coincidence does not occur with boundaries having other geometries. 16 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.2.3 Dispersion relation for TM modes. Fig. 13.2.4 Dispersion relation for TE modes. The velocity of propagation of points of constant phase (for example, a point at which a ﬁeld component is zero) is ω/ky. Figure 13.2.3 emphasizes that for all but the TEM mode, the phase velocity is a function of frequency. The equation relating ωtokyrepresented by this ﬁgure, (15), is often called the dispersion equation . The dispersion equation for the TE modes is shown in Fig. 13.2.4. Although the ﬁeld distributions implied by each branch are very diﬀerent, in the case of the plane parallel electrodes considered here, the curves are the same as those for the TMn/negationslash=0modes. The next section will provide greater insight into the higher-order TM and TE modes. Sec. 13.3 TE and TM Standing Waves 17 13.3 TE AND TM STANDING WAVES BETWEEN PARALLEL PLATES In this section, we delve into the relationship between the two-dimensional higher- order modes derived in Sec. 13.2 and their sources. The examples are chosen to relate directly to case studies treated in quasistatic terms in Chaps. 5 and 8. The matching of a longitudinal boundary condition by a superposition of modes may at ﬁrst seem to be a purely mathematical process. However, even quali- tatively it is helpful to think of the inﬂuence of an excitation in terms of the resulting modes. For quasistatic systems, this has already been our experience. For the pur- pose of estimating the dependence of the output signal on the spacing bbetween excitation and detection electrodes, the EQS response of the capacitive attenuator of Sec. 5.5 could be pictured in terms of the lowest-order mode. In the electrody- namic situations of interest here, it is even more common that one mode dominates. Above its cutoﬀ frequency, a given mode can propagate through a waveguide to re- gions far removed from the excitation. Modes obey orthogonality relations that are mathematically useful for the evaluation of the mode amplitudes. Formally, the mode orthogonality is implied by the diﬀerential equations governing the transverse dependence of the fundamental ﬁeld components and the associated boundary conditions. For the TM modes, these are (13.2.5) and (13.2.11). TM Modes: d2ˆhzn dx2+p2 nˆhzn= 0 (1) where dˆhzn dx(a) = 0;dˆhzn dx(0) = 0 and for the TE modes, these are (13.2.8) and (13.2.12). TE Modes: d2ˆezn dx2+q2 nˆezn= 0 (2) where ˆezn(a) = 0; ˆ ezn(0) = 0 The word “orthogonal” is used here to mean that Za 0ˆhznˆhzmdx= 0; n/negationslash=m (3) Za 0ˆeznˆezmdx= 0; n/negationslash=m (4) These properties of the modes can be seen simply by carrying out the integrals, using the modes as given by (13.2.13) and (13.2.16). More fundamentally, they can be deduced from the diﬀerential equations and boundary conditions themselves, (1) and (2). This was illustrated in Sec. 5.5 using arguments that are directly applicable here [(5.5.20)–(5.5.26)]. 18 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.3.1 Conﬁguration for excitation of TM waves. The following two examples illustrate how TE and TM modes can be excited in waveguides. In the quasistatic limit, the conﬁgurations respectively become identical to EQS and MQS situations treated in Chaps. 5 and 8. Example 13.3.1. Excitation of TM Modes and the EQS Limit In the conﬁguration shown in Fig. 13.3.1, the parallel plates lying in the planes x= 0 andx=aare shorted at y= 0 by a perfectly conducting plate. The excitation is provided by distributed voltage sources driving a perfectly conducting plate in the plane y=b. These sources constrain the integral of Eacross narrow insulating gaps of length ∆ between the respective edges of the upper plate and the adjacent plates. All the conductors are modeled as perfect. The distributed voltage sources maintain the two-dimensional character of the ﬁelds even as the width in the zdirection becomes long compared to a wavelength. Note that the conﬁguration is identical to that treated in Sec. 5.5. Therefore, we already know the ﬁeld behavior in the quasistatic (low frequency) limit. In general, the two-dimensional ﬁelds are the sum of the TM and TE ﬁelds. However, here the boundary conditions can be met by the TM ﬁelds alone. Thus, we begin with Hz, (13.2.19), expressed as a single sum. Hz= Re£∞X n=0(A+ ne−jβny+A− nejβny) cosnπ ax¤ ejωt(5) This ﬁeld and the associated Esatisfy the boundary conditions on the parallel plates atx= 0 and x=a. Boundary conditions are imposed on the tangential Eat the longitudinal boundaries, where y= 0 Ex(x,0, t) = 0 (6) Sec. 13.3 TE and TM Standing Waves 19 and at the driving electrode, where y=b. We assume here that the gap lengths ∆ are small compared to other dimensions of interest. Then, the electric ﬁeld within each gap is conservative and the line integral of Exacross the gaps is equal to the gap voltages ±v. Over the region between x= ∆ and x=a−∆, the perfectly conducting electrode makes Ex= 0. Za a−∆Ex(x, b, t )dx=v;Z∆ 0Ex(x, b, t )dx=−v (7) Because the longitudinal boundary conditions are on Ex, we substitute Hz as given by (5) into the xcomponent of Faraday’s law [(12.6.6) of Table 12.8.3] to obtain Ex= Re£∞X n=0−βn ω/epsilon1(A+ ne−jβny−A− nejβny¢ cosnπ ax¤ ejωt(8) To satisfy the condition at the short, (6), A+ n=A− nand (8) becomes Ex= Re£∞X n=02jβn ω/epsilon1A+ nsinβnycosnπ ax¤ ejωt(9) This set of solutions satisﬁes the boundary conditions on three of the four boundaries. What we now do to satisfy the “last” boundary condition diﬀers little from what was done in Sec. 5.5. The A+ n’s are adjusted so that the summation of product solutions in (9) matches the boundary condition at y=bsummarized by (7). Thus, we write (9) with y=bon the right and with the function representing (7) on the left. This expression is multiplied by the m’th eigenfunction, cos( mπx/a ), and integrated from x= 0 to x=a. Za 0ˆEx(x, b) cosmπx adx=Za 0∞X n=02jβnA+ n ω/epsilon1sinβnb· cosnπ axcosmπ axdx(10) Because the intervals where ˆEx(x, b) is ﬁnite are so small, the cosine function can be approximated by a constant, namely ±1 as appropriate. On the right-hand side of (10), we exploit the orthogonality condition so as to pick out only one term in the inﬁnite series. ˆv[−1 + cos mπ] =2jβm ω/epsilon1sinβmb¡a 2¢ A+ m (11) Of the inﬁnite number of terms in the integral on the right in (10), only the term where n=mhas contributed. The coeﬃcients follow from solving (11) and replacing m→n. A+ n=(0; neven −2ω/epsilon1ˆv jβnasinβnb;nodd(12) 20 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 With the coeﬃcients A+ n=A− nnow determined, we can evaluate all of the ﬁelds. Substitution into (5), and (8) and into the result using (12.6.7) from Table 12.8.3 gives Hz= Re·∞X n=1 odd4jω/epsilon1ˆv βnacosβny sinβnbcosnπ ax¸ ejωt(13) Ex= Re·∞X n=1 odd−4ˆv asinβny sinβnbcosnπ ax¸ ejωt(14) Ey= Re·∞X n=1 odd4nπ aˆv (βna)cosβny sinβnbsinnπ ax¸ ejωt(15) Note the following aspects of these ﬁelds (which we can expect to see in Demon- stration 13.3.1). First, the magnetic ﬁeld is directed perpendicular to the x−yplane. Second, by making the excitation symmetric, we have eliminated the TEM mode. As a result, the only modes are of order n= 1 and higher. Third, at frequencies below the cutoﬀ for the TM 1mode, βyis imaginary and the ﬁelds decay in the y direction.6Indeed, in the quasistatic limit where ω2µ/epsilon1/lessmuch(π/a)2, the electric ﬁeld is the same as that given by taking the gradient of (5.5.9). In this same quasistatic limit, the magnetic ﬁeld would be obtained by using this quasistatic Eto evaluate the displacement current and then solving for the resulting magnetic ﬁeld subject to the boundary condition that there be no normal ﬂux density on the surfaces of the perfect conductors. Fourth, above the cutoﬀ frequency for the n= 1 mode but below the cutoﬀ for the n= 2 mode, we should ﬁnd standing waves having a wavelength 2π/β1. Finally, note that each of the expressions for the ﬁeld components has sin( βnb) in its denominator. With the frequency adjusted such that βn=nπ/b, this function goes to zero and the ﬁelds become inﬁnite. This resonance condition results in an inﬁnite response, because we have pictured all of the conductors as perfect. It occurs when the frequency is adjusted so that a wave reﬂected from one boundary arrives at the other with just the right phase to reinforce, upon a second reﬂection, the wave currently being initiated by the drive. The following experiment gives the opportunity to probe the ﬁelds that have been found in the previous example. In practical terms, the structure considered might be a parallel plate waveguide. Demonstration 13.3.1. Evanescent and Standing TM Waves The experiment shown in Fig. 13.3.2 is designed so that the ﬁeld distributions can be probed as the excitation is varied from below to above the cutoﬀ frequency of the TM 1mode. The excitation structures are designed to give ﬁelds approximating those found in Example 13.3.1. For convenience, a= 4.8 cm so that the excitation frequency ranges above and below a cut-oﬀ frequency of 3.1 GHz. The generator is modulated at an audible frequency so that the amplitude of the detected signal is converted to “loudness” of the tone from the loudspeaker. In this TM case, the driving electrode is broken into segments, each insulated from the parallel plates forming the waveguide and each attached at its center to a 6sin(ju) =jsinh(u) and cos( ju) = cosh( u) Sec. 13.3 TE and TM Standing Waves 21 Fig. 13.3.2 Demonstration of TM evanescent and standing waves. coaxial line from the generator. The segments insure that the ﬁelds applied to each part of the electrode are essentially in phase. (The cables feeding each segment are of the same length so that signals arrive at each segment in phase.) The width of the structure in the zdirection is of the order of a wavelength or more to make the ﬁelds two dimensional. (Remember, in the vicinity of the lowest cutoﬀ frequency, ais about one-half wavelength.) Thus, if the feeder were attached to a contiguous electrode at one point, there would be a tendency for standing waves to appear on the excitation electrode, much as they did on the wire antennae in Sec. 12.4. In the experiment, the segments are about a quarter-wavelength in the zdirection but, of course, about a half-wavelength in the xdirection. In the experiment, His detected by means of a one-turn coil. The voltage induced at the terminals of this loop is proportional to the magnetic ﬂux perpendic- ular to the loop. Thus, for the TM ﬁelds, the loop detects its greatest signal when it is placed in an x−yplane. To avoid interference with E, the coaxial line connected to the probe as well as the loop itself are kept adjacent to the conducting walls (where Hzpeaks anyway). The spatial features of the ﬁeld, implied by the normalized ωversus kyplot of Fig. 13.3.2, can be seen by moving the probe about. With the frequency below cutoﬀ, the ﬁeld decays in the −ydirection. This exponential decay or evanescence decreases to a linear dependence at cutoﬀ and is replaced above cutoﬀ by standing waves. The value of kyat a given frequency can be deduced from the experiment by measuring the quarter-wave distance from the short to the ﬁrst null in the magnetic ﬁeld. Note that if there are asymmetries in the excitation that result in excitation of the TEM mode, the standing waves produced by this mode will tend to obscure 22 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 the TM 1mode when it is evanescent. The TEM waves do not have a cutoﬀ! As we have seen once again, the TM ﬁelds are the electrodynamic generaliza- tion of two-dimensional EQS ﬁelds. That is, in the quasistatic limit, the previous example becomes the capacitive attenuator of Sec. 5.5.7 We have more than one reason to expect that the two-dimensional TE ﬁelds are the generalization of MQS systems. First, this was seen to be the case in Sec. 12.6, where the TE ﬁelds associated with a given surface current density were found to approach the MQS limit as ω2µ/epsilon1/lessmuchk2 y. Second, from Sec. 8.6 we know that for every two-dimensional EQS conﬁguration involving perfectly conducting boundaries, there is an MQS one as well.8In particular, the MQS analog of the capacitor attenuator is the conﬁguration shown in Fig. 13.3.3. The MQS Hﬁeld was found in Example 8.6.3. In treating MQS ﬁelds in the presence of perfect conductors, we recognized that the condition of zero tangential Eimplied that there be no time-varying normal B. This made it possible to determine Hwithout regard for E. We could then delay taking detailed account of Euntil Sec. 10.1. Thus, in the MQS limit, a system involving essentially a two-dimensional distribution of Hcan (and usually does) have an Ethat depends on the third dimension. For example, in the conﬁguration of Fig. 13.3.3, a voltage source might be used to drive the current in the zdirection through the upper electrode. This current is returned in the perfectly conducting /unionsq- shaped walls. The electric ﬁelds in the vicinities of the gaps must therefore increase in the zdirection from zero at the shorts to values consistent with the voltage sources at the near end. Over most of the length of the system, Eis across the gap and therefore in planes perpendicular to the zaxis. This MQS conﬁguration does not excite pure TE ﬁelds. In order to produce (approximately) two-dimensional TE ﬁelds, provision must be made to make Eas well as Htwo dimensional. The following example and demonstration give the opportunity to further develop an appreciation for TE ﬁelds. Example 13.3.2. Excitation of TE Modes and the MQS Limit An idealized conﬁguration for exciting standing TE modes is shown in Fig. 13.3.4. As in Example 13.3.1, the perfectly conducting plates are shorted in the plane y= 0. In the plane y=bis a perfectly conducting plate that is segmented in the zdirection. Each segment is driven by a voltage source that is itself distributed in the xdirection. In the limit where there are many of these voltage sources and perfectly conducting segments, the driving electrode becomes one that both imposes a z-directed Eand has no zcomponent of B. That is, just below the surface of this electrode, wEzis equal to the sum of the source voltages. One way of approximately realizing this idealization is used in the next demonstration. Let Λ be deﬁned as the ﬂux per unit length (length taken along the zdirection) into and out of the enclosed region through the gaps of width ∆ between the driving electrode and the adjacent edges of the plane parallel electrodes. The magnetic ﬁeld 7The example which was the theme of Sec. 5.5 might equally well have been called the “microwave attenuator,” for a section of waveguide operated below cutoﬀ is used in microwave circuits to attenuate signals. 8TheHsatisfying the condition that n·B= 0 on the perfectly conducting boundaries was obtained by replacing Φ →Azin the solution to the analogous EQS problem. Sec. 13.3 TE and TM Standing Waves 23 Fig. 13.3.3 Two-dimensional MQS conﬁguration that does not have TE ﬁelds. Fig. 13.3.4 Idealized conﬁguration for excitation of TE standing waves. normal to the driving electrode between the gaps is zero. Thus, at the upper surface, Hyhas the distribution shown in Fig. 13.3.5a. Faraday’s integral law applied to the contour Cof Fig. 13.3.4 and to a similar contour around the other gap shows that Ez(x, b, t ) =−dΛ dt⇒ˆEz=−jωˆΛ (16) 24 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.3.5 Equivalent boundary conditions on normal Hand tan- gential Eaty=b. Thus, either the normal Bor the tangential Eon the surface at y=bis speciﬁed. The two must be consistent with each other, i.e., they must obey Faraday’s law. It is perhaps easiest in this case to deal directly with Ezin ﬁnding the coeﬃcients ap- pearing in (13.2.20). Once they have been determined (much as in Example 13.3.1), Hfollows from Faraday’s law, (12.6.29) and (12.6.30) of Table 12.8.3. Ez= Re∞X m=1 odd−4jˆΛω mπsinβmy sinβmbsinmπx aejωt(17) Hx= Re∞X m=1 odd4βmˆΛ µmπcosβmy sinβmbsinmπx aejωt(18) Hy= Re∞X m=1 odd−4ˆΛ µasinβmy sinβmbcosmπx aejωt(19) In the quasistatic limit, ω2µ/epsilon1/lessmuch(mπ/a )2, this magnetic ﬁeld reduces to that found in Example 8.6.3. A few observations may help one to gain some insights from these expressions. First, if the magnetic ﬁeld is sensed, then the detection loop must have its axis in thex−yplane. For these TE modes, there should be no signal sensed with the axis of the detection loop in the zdirection. This probe can also be used to verify that Hnormal to the perfectly conducting surfaces is indeed zero, while its tangential value peaks at the short. Second, the same decay of the ﬁelds below cutoﬀ and appearance of standing waves above cutoﬀ is predicted here, as in the TM case. Third, because Eis perpendicular to planes of constant z, the boundary conditions onE, and hence H, are met, even if perfectly conducting plates are placed over the open ends of the guide, say in the planes z= 0 and z=w. In this case, the guide becomes a closed pipe of rectangular cross-section. What we have found are then a subset of the three-dimensional modes of propagation in a rectangular waveguide. Demonstration 13.3.2. Evanescent and Standing TE Waves The apparatus of Demonstration 13.3.1 is altered to give TE rather than TM waves by using an array of “one-turn inductors” rather than the array of “capacitor plates.” These are shown in Fig. 13.3.6. Sec. 13.4 Rectangular Waveguide Modes 25 Fig. 13.3.6 Demonstration of evanescent and standing TE waves. Each member of the array consists of an electrode of width a−2∆, driven at one edge by a common source and shorted to the perfectly conducting backing at its other edge. Thus, the magnetic ﬂux through the closed loop passes into and out of the guide through the gaps of width ∆ between the ends of the one-turn coil and the parallel plate (vertical) walls of the guide. Eﬀectively, the integral of Ezcreated by the voltage sources in the idealized model of Fig. 13.3.4 is produced by the integral ofEzbetween the left edge of one current loop and the right edge of the next. The current loop can be held in the x−zplane to sense Hyor in the y−z plane to sense Hxto verify the ﬁeld distributions derived in the previous example. It can also be observed that placing conducting sheets against the open ends of the parallel plate guide, making it a rectangular pipe guide, leaves the characteristics of these two-dimensional TE modes unchanged. 13.4 RECTANGULAR WAVEGUIDE MODES Metal pipe waveguides are often used to guide electromagnetic waves. The most common waveguides have rectangular cross-sections and so are well suited for the exploration of electrodynamic ﬁelds that depend on three dimensions. Although we conﬁne ourselves to a rectangular cross-section and hence Cartesian coordinates, the classiﬁcation of waveguide modes and the general approach used here are equally applicable to other geometries, for example to waveguides of circular cross-section. The parallel plate system considered in the previous three sections illustrates 26 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.4.1 Rectangular waveguide. much of what can be expected in pipe waveguides. However, unlike the parallel plates, which can support TEM modes as well as higher-order TE modes and TM modes, the pipe cannot transmit a TEM mode. From the parallel plate system, we expect that a waveguide will support propagating modes only if the frequency is high enough to make the greater interior cross-sectional dimension of the pipe greater than a free space half-wavelength. Thus, we will ﬁnd that a guide having a larger dimension greater than 5 cm would typically be used to guide energy having a frequency of 3 GHz. We found it convenient to classify two-dimensional ﬁelds as transverse mag- netic (TM) or transverse electric (TE) according to whether EorHwas trans- verse to the direction of propagation (or decay). Here, where we deal with three- dimensional ﬁelds, it will be convenient to classify ﬁelds according to whether they haveEorHtransverse to the axial direction of the guide . This classiﬁcation is used regardless of the cross-sectional geometry of the pipe. We choose again the y coordinate as the axis of the guide, as shown in Fig. 13.4.1. If we focus on solutions to Maxwell’s equations taking the form Hy= Re ˆhy(x, z)ej(ωt−kyy)(1) Ey= Re ˆ ey(x, z)ej(ωt−kyy)(2) then all of the other complex amplitude ﬁeld components can be written in terms of the complex amplitudes of these axial ﬁelds, HyandEy. This can be seen from substituting ﬁelds having the form of (1) and (2) into the transverse components of Amp` ere’s law, (12.0.8), −jkyˆhz−∂ˆhy ∂z=jω/epsilon1ˆex (3) Sec. 13.4 Rectangular Waveguide Modes 27 ∂ˆhy ∂x+jkyˆhx=jω/epsilon1ˆez (4) and into the transverse components of Faraday’s law, (12.0.9), −jkyˆez−∂ˆey ∂z=−jωµˆhx (5) ∂ˆey ∂x+jkyˆex=−jωµˆhz (6) If we take ˆhyand ˆeyas speciﬁed, (3) and (6) constitute two algebraic equations in the unknowns ˆ exandˆhz. Thus, they can be solved for these components. Similarly, ˆhxand ˆezfollow from (4) and (5). ˆhx=µ −jky∂ˆhy ∂x−jω/epsilon1∂ˆey ∂z¶ /(ω2µ/epsilon1−k2 y) (7) ˆhz=µ −jky∂ˆhy ∂z+jω/epsilon1∂ˆey ∂x¶ /(ω2µ/epsilon1−k2 y) (8) ˆex=µ jωµ∂ˆhy ∂z−jky∂ˆey ∂x¶ /(ω2µ/epsilon1−k2 y) (9) ˆez=µ −jωµ∂ˆhy ∂x−jky∂ˆey ∂z¶ /(ω2µ/epsilon1−k2 y) (10) We have found that the three-dimensional ﬁelds are a superposition of those associated with Ey(so that the magnetic ﬁeld is transverse to the guide axis ), the TM ﬁelds, and those due to Hy, the TE modes. The axial ﬁeld components now play the role of “potentials” from which the other ﬁeld components can be derived. We can use the ycomponents of the laws of Amp` ere and Faraday together with Gauss’ law and the divergence law for Hto show that the axial complex amplitudes ˆ eyandˆhysatisfy the two-dimensional Helmholtz equations. TM Modes (Hy= 0): ∂2ˆey ∂x2+∂2ˆey ∂z2+p2ˆey= 0 (11) where p2=ω2µ/epsilon1−k2 y and TE Modes (Ey= 0): ∂2ˆhy ∂x2+∂2ˆhy ∂z2+q2ˆhy= 0 (12) 28 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 where q2=ω2µ/epsilon1−k2 y These relations also follow from substitution of (1) and (2) into the ycomponents of (13.0.2) and (13.0.1). The solutions to (11) and (12) must satisfy boundary conditions on the per- fectly conducting walls. Because Eyis parallel to the perfectly conducting walls, it must be zero there. TM Modes: ˆey(0, z) = 0; ˆ ey(a, z) = 0; ˆ ey(x,0) = 0; ˆ ey(x, w) = 0 (13) The boundary condition on Hyfollows from (9) and (10), which express ˆ ex and ˆezin terms of ˆhy. On the walls at x= 0 and x=a,ˆez= 0. On the walls at z= 0, z=w,ˆex= 0. Therefore, from (9) and (10) we obtain TE Modes: ∂hy ∂x(0, z) = 0;∂hy ∂x(a, z) = 0;∂hy ∂z(x,0) = 0;∂hy ∂z(x, w) = 0 (14) The derivative of ˆhywith respect to a coordinate perpendicular to the boundary must be zero. The solution to the Helmholtz equation, (11) or (12), follows a pattern that is familiar from that used for Laplace’s equation in Sec. 5.4. Either of the complex amplitudes representing the axial ﬁelds is represented by a product solution. ·ˆey ˆhy¸ ∝X(x)Z(z) (15) Substitution into (11) or (12) and separation of variables then gives d2X dx2+γ2X= 0 (16) d2Z dz2+δ2Z= 0 where −γ2−δ2+µ p2 q2¶ = 0 (17) Solutions that satisfy the TM boundary conditions, (13), are then TM Modes: X∝sinγmx; γm=mπ a, m = 1,2, . . . (18) Sec. 13.4 Rectangular Waveguide Modes 29 Z∝sinδnz; δn=nπ w, n = 1,2, . . . so that p2 mn=¡mπ a¢2+¡nπ w¢2; m= 1,2, . . . , n = 1,2, . . . (19) When either mornis zero, the ﬁeld is zero, and thus mandnmust be equal to an integer equal to or greater than one. For a given frequency ωand mode number (m, n), the wave number kyis found by using (19) in the deﬁnition of passociated with (11) ky=±βmn with βmn≡8 < :q ω2µ/epsilon1−¡mπ a¢2−¡nπ w¢2; ω2µ/epsilon1 >¡mπ a¢2+¡nπ w¢2 −jq¡mπ a¢2+¡nπ w¢2−ω2µ/epsilon1;ω2µ/epsilon1 <¡mπ a¢2+¡nπ w¢2(20) Thus, the TM solutions are Ey= Re∞X m=1∞X n=1(A+ mne−jβmny+A− mnejβmny) sinmπ axsinnπ wz ejωt(21) For the TE modes, (14) provides the boundary conditions, and we are led to the solutions TE Modes: X∝cosγmx; γm=mπ a;m= 0,1,2, . . . (22) Z∝cosδnz; δn=nπ a;n= 0,1,2, . . . Substitution of γmandδninto (17) therefore gives q2 mn=¡mπ a¢2+¡nπ w¢2; m= 0,1,2, . . . , n = 0,1,2, . . . , (23) (m, n)/negationslash= (0,0) The wave number kyis obtained using this eigenvalue in the deﬁnition of qasso- ciated with (12). With the understanding that either morncan now be zero, the expression is the same as that for the TM modes, (20). However, both mandn cannot be zero. If they were, it follows from (22) that the axial Hwould be uniform over any given cross-section of the guide. The integral of Faraday’s law over the cross-section of the guide, with the enclosing contour Cadjacent to the perfectly conducting boundaries as shown in Fig. 13.4.2, requires that I E·ds=−µAdHy dt(24) 30 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.4.2 Cross-section of guide with contour adjacent to perfectly con- ducting walls. where Ais the cross-sectional area of the guide. Because the contour on the left is adjacent to the perfectly conducting boundaries, the line integral of Emust be zero. It follows that for the m= 0, n= 0 mode, Hy= 0. If there were such a mode, it would have both EandHtransverse to the guide axis. We will show in Sec. 14.2, where TEM modes are considered in general, that TEM modes cannot exist within a perfectly conducting pipe. Even though the dispersion equations for the TM and TE modes only diﬀer in the allowed lowest values of ( m, n), the ﬁeld distributions of these modes are very diﬀerent.9The superposition of TE modes gives Hy= Re∞X m=0∞X n=0(C+ mne−jβmny+C− mnejβmny)·cosmπ axcosnπ wz ejωt(25) where m·n/negationslash= 0. The frequency at which a given mode switches from evanescence to propagation is an important parameter. This cutoﬀ frequency follows from (20) as ωc=1√µ/epsilon1r¡mπ a¢2+¡nπ w¢2(26) TM Modes: m/negationslash= 0, n/negationslash= 0 TE Modes: mand nnot both zero Rearranging this expression gives the normalized cutoﬀ frequency as functions of the aspect ratio a/wof the guide. ωc≡ωcw cπ=p (w/a)2m2+n2 (27) These normalized cutoﬀ frequencies are shown as functions of w/ain Fig. 13.4.3. The numbering of the modes is standardized. The dimension wis chosen as w≤a, and the ﬁrst index mgives the variation of the ﬁeld along a. The TE 10 9In other geometries, such as a circular waveguide, this coincidence of pmnandqmnis not found. Sec. 13.4 Rectangular Waveguide Modes 31 Fig. 13.4.3 Normalized cutoﬀ frequencies for lowest rectangular waveguide modes as a function of aspect ratio. mode then has the lowest cutoﬀ frequency and is called the dominant mode. All other modes have higher cutoﬀ frequencies (except, of course, in the case of the square cross-section for which TE 01has the same cutoﬀ frequency). Guides are usually designed so that at the frequency of operation only the dominant mode is propagating, while all higher-order modes are “cutoﬀ.” In general, an excitation of the guide at a cross-section y= constant excites all waveguide modes. The modes with cutoﬀ frequencies higher than the frequency of excitation decay away from the source. Only the dominant mode has a sinusoidal dependence upon yand thus possesses ﬁelds that are periodic in yand “dominate” the ﬁeld pattern far away from the source, at distances larger than the transverse dimensions of the waveguide. Example 13.4.1. TE10Standing Wave Fields The section of rectangular guide shown in Fig. 13.4.4 is excited somewhere to the right of y= 0 and shorted by a conducting plate in the plane y= 0. We presume that the frequency is above the cutoﬀ frequency for the TE 10mode and that a > w as shown. The frequency of excitation is chosen to be below the cutoﬀ frequency for all higher order modes and the source is far away from y= 0 (i.e., at y/greatermucha). The ﬁeld in the guide is then that of the TE 10mode. Thus, Hyis given by (25) with m= 1 and n= 0. What is the space-time dependence of the standing waves that result from having shorted the guide? Because of the short, Ez(x, y= 0, z) = 0. In order to relate the coeﬃcients C+ 10andC− 10, we must determine ˆ ezfrom ˆhyas given by (25) using (10) Ez= Re jωµa π(C+ 10e−jβ10y+C− 10ejβ10y) sinπx aejωt(28) and because ˆ ez= 0 at the short, it follows that C+ 10=−C− 10 (29) 32 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.4.4 Fields and surface sources for TE 10mode. so that Ez= Re· 2ωµa πC+ 10sinβ10ysinπ axejωt¸ (30) and this is the only component of the electric ﬁeld in this mode. We can now use (29) to evaluate (25). Hy=−Re· 2jC+ 10sinβ10ycosπ axejωt¸ (31) In using (7) to evaluate the other component of H, remember that in the C+ mnterm of (25), ky=βmn, while in the C− mnterm, ky=−βmn. Hx= Re· 2jβ10a πC+ 10cosβ10ysinπ axejωt¸ (32) To sketch these ﬁelds in the neighborhood of the short and deduce the associ- ated surface charge and current densities, consider C+ 10to be real. The jin (31) and (32) shows that HxandHyare 90 degrees out of phase with the electric ﬁeld. Thus, in the ﬁeld sketches of Fig. 13.4.4, EandHare shown at diﬀerent instants of time, sayEwhen ωt=πandHwhen ωt=π/2. The surface charge density is where Ez terminates and originates on the upper and lower walls. The surface current density can be inferred from Amp` ere’s continuity condition. The temporal oscillations of these ﬁelds should be pictured with Hequal to zero when Epeaks, and with E equal to zero when Hpeaks. At planes spaced by multiples of a half-wavelength along the yaxis,Eis always zero. Sec. 13.5 Optical Fibers 33 Fig. 13.4.5 Slotted line for measuring axial distribution of TE 10ﬁelds. The following demonstration illustrates how a movable probe designed to cou- ple to the electric ﬁeld is introduced into a waveguide with minimal disturbance of the wall currents. Demonstration 13.4.1. Probing the TE 10Mode. A waveguide slotted line is shown in Fig. 13.4.5. Here the line is shorted at y= 0 and excited at the right. The probe used to excite the guide is of the capacitive type, positioned so that charges induced on its tip couple to the lines of electric ﬁeld shown in Fig. 13.4.4. This electrical coupling is an alternative to the magnetic coupling used for the TE mode in Demonstration 13.3.2. Theydependence of the ﬁeld pattern is detected in the apparatus shown in Fig. 13.4.5 by means of a second capacitive electrode introduced through a slot so that it can be moved in the ydirection and not perturb the ﬁeld, i.e., the wall is cut along the lines of the surface current K. From the sketch of Kgiven in Fig. 13.4.4, it can be seen that Kis in the ydirection along the center line of the guide. The probe can be used to measure the wavelength 2 π/kyof the standing waves by measuring the distance between nulls in the output signal (between nulls in Ez). With the frequency somewhat below the cutoﬀ of the TE 10mode, the spatial decay away from the source of the evanescent wave also can be detected. 13.5 DIELECTRIC WAVEGUIDES: OPTICAL FIBERS Waves can be guided by dielectric rods or slabs and the ﬁelds of these waves occupy the space within and around these dielectric structures. Especially at optical wavelengths, dielectric ﬁbers are commonly used to guide waves. In this section, we develop the properties of waves guided by a planar sheet of dielectric material. The waves that we ﬁnd are typical of those found in integrated optical systems and in the more commonly used optical ﬁbers of circular cross-section. A planar version of a dielectric waveguide is pictured in Fig. 13.5.1. A dielectric of thickness 2 dand permittivity /epsilon1iis surrounded by a dielectric of permittivity /epsilon1 < /epsilon1 i. The latter might be free space with /epsilon1=/epsilon1o. We are interested in how this 34 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.5.1 Dielectric slab waveguide. structure might be used to guide waves in the ydirection and will conﬁne ourselves to ﬁelds that are independent of z. With a source somewhere to the left (for example an antenna imbedded in the dielectric), there is reason to expect that there are ﬁelds outside as well as inside the dielectric. We shall look for ﬁeld solutions that propagate in the ydirection and possess ﬁelds solely inside and near the layer. The ﬁelds external to the layer decay to zero in the ±xdirections. Like the waves propagating along waveguides, those guided by this structure have transverse components that take the form Ez= Re ˆ ez(x)ej(ωt−kyy)(1) both inside and outside the dielectric. That is, the ﬁelds inside and outside the dielectric have the same frequency ω, the same phase velocity ω/ky, and hence the same wavelength 2 π/kyin the ydirection. Of course, whether such ﬁelds can actually exist will be determined by the following analysis. The classiﬁcation of two-dimensional ﬁelds introduced in Sec. 12.6 is applica- ble here. The TM and TE ﬁelds can be made to independently satisfy the boundary conditions so that the resulting modes can be classiﬁed as TM or TE.10Here we will conﬁne ourselves to the transverse electric modes. In the exterior and interior regions, where the permittivities are uniform but diﬀerent, it follows from substi- tution of (1) into (12.6.33) (Table 12.8.3) that d2ˆez dx2−α2 xˆez= 0; αx=q k2y−ω2µ/epsilon1;d < x andx <−d (2) d2ˆez dx2+k2 xˆez= 0; kx=q ω2µ/epsilon1i−k2y;−d < x < d (3) A guided wave is one that is composed of a nonuniform plane wave in the exterior regions, decaying in the ±xdirections and propagating with the phase velocity ω/kyin the ydirection. In anticipation of this, we have written (2) in 10Circular dielectric rods do not support simple TE or TM waves; in that case, this classiﬁ- cation of modes is not possible. Sec. 13.5 Optical Fibers 35 terms of the parameter αx, which must then be real and positive. Through the continuity conditions, the exterior wave must match up to the interior wave at the dielectric surfaces. The solutions to (3) are sines and cosines if kxis real. In order to match the interior ﬁelds onto the nonuniform plane waves on both sides of the guide, it is necessary that kxbe real. We now set out to ﬁnd the wave numbers kythat not only satisfy the wave equations in each of the regions, represented by (2) and (3), but the continuity conditions at the interfaces as well. The conﬁguration is symmetric about the x= 0 plane so we can further divide the modes into those that have even and odd functions Ez(x). Thus, with Aan arbitrary factor, appropriate even solutions to (2) and (3) are ˆez=8 >>< >>:Ae−αx(x−d);d < x Acoskxx coskxd;−d < x < d Aeαx(x+d);x <−d(4) To simplify the algebra, we have displaced the origin in the exterior solutions so that just the coeﬃcient, A, is obtained when ˆ ezis evaluated at the respective interfaces. With a similar objective, the interior solution has been divided by the constant cos( kxd) so that at the boundaries, ˆ ezalso becomes A. In this way, we have adjusted the interior coeﬃcient so that ˆ ezis continuous at the boundaries. Because this transverse ﬁeld is the only component of E, all of the continuity conditions on Eare now satisﬁed. The permeabilities of all regions are presumed to be the same, so both tangential and normal components of Hmust be continuous at the boundaries. From (12.6.29), the continuity of normal µHis guaranteed by the continuity of Ezin any case. The tangential ﬁeld is obtained using (12.6.30). ˆhy=1 jωµdˆez dx(5) Substitution of (4) into (5) gives ˆhy=1 jωµ8 < :−αxAe−αx(x−d);d < x −kxAsinkxx coskxd; −d < x < d αxAeαx(x+d); x <−d(6) The assumption that Ezis even in xhas as a consequence the fact that the continu- ity condition on tangential His satisﬁed by the same relation at both boundaries. −αxA=−kxAtankxd⇒αx kx= tan kxd (7) Our goal is to determine the propagation constant kyfor a given ω. If we were to substitute the deﬁnitions of αxandkxinto this expression, we would have this dispersion equation, D( ω,ky), implicitly relating kytoω. It is more convenient to solve for αxandkxﬁrst, and then for ky. Elimination of kybetween the expressions for αxandkxgiven with (2) and (3) gives a second expression for αx/kx. αx kx=s ω2µ/epsilon1id2 (kxd)2¡ 1−/epsilon1 /epsilon1i¢ −1 (8) 36 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.5.2 Graphical solution to (7) and (8). The solutions for the values of the normalized transverse wave numbers ( kxd) can be pictured as shown in Fig. 13.5.2. Plotted as functions of kxdare the right-hand sides of (7) and (8). The points of intersection, kxd=γm, are the desired solutions. For the frequency used to make Fig. 13.5.2, there are two solutions. These are designated by even integers because the odd modes (Prob. 13.5.1) have roots that interleave these even modes. As the frequency is raised, an additional even TE-guided mode is found each time the curve representing (8) reaches a new branch of (7). This happens at fre- quencies ωcsuch that αx/kx= 0 and kxd=mπ/2, where m= 0,2,4, . . .From (8), ωc=mπ 2d1p µ(/epsilon1i−/epsilon1)(9) Them= 0 mode has no cutoﬀ frequency. To ﬁnally determine kyfrom these eigenvalues, the deﬁnition of kxgiven with (3) is used to write kyd=p ω2µ/epsilon1id2−(kxd)2 (10) and the dispersion equation takes the graphical form of Fig. 13.5.3. To make Fig. 13.5.2, we had to specify the ratio of permittivities, so that ratio is also implicit in Fig. 13.5.3. Features of the dispersion diagram, Fig. 13.5.3, can be gathered rather simply. Where a mode is just cutoﬀ because ω=ωc, αx= 0, as can be seen from Fig. 13.5.2. From (2), we gather that ky=ωc√µ/epsilon1. Thus, at cutoﬀ, a mode must have a propagation constant kythat lies on the straight broken line to the left, shown in Fig. 13.5.3. At cutoﬀ, each mode has a phase velocity equal to that of a plane wave in the medium exterior to the layer. In the high-frequency limit, where ωgoes to inﬁnity, we see from Fig. 13.5.2 thatkxd approaches the constant kx→(m+ 1)π/2d. That is, in (3), kxbecomes a constant even as ωgoes to inﬁnity and it follows that in this high frequency limit ky→ω√µ/epsilon1i. Sec. 13.5 Optical Fibers 37 Fig. 13.5.3 Dispersion equation for even TE modes with /epsilon1i//epsilon1= 6.6. Fig. 13.5.4 Distribution of transverse Efor TE 0mode on dielectric waveg- uide of Fig. 13.5.1. The physical reasons for this behavior follow from the nature of the mode pattern as a function of frequency. When αx→0, as the frequency approaches cutoﬀ, it follows from (4) that the ﬁelds extend far into the regions outside of the layer. The wave approaches an inﬁnite parallel plane wave having a propagation constant that is hardly aﬀected by the layer. In the opposite extreme, where ωgoes to inﬁnity, the decay of the external ﬁeld is rapid, and a given mode is well conﬁned inside the layer. Again, the wave assumes the character of an inﬁnite parallel plane wave, but in this limit, one that propagates with the phase velocity of a plane wave in a medium with the dielectric constant of the layer. The distribution of Ezof the m= 0 mode at one frequency is shown in Fig. 13.5.4. As the frequency is raised, each mode becomes more conﬁned to the layer. 38 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. 13.5.5 Dielectric waveguide demonstration. Demonstration 13.5.1. Microwave Dielectric Guided Waves In the experiment shown in Fig. 13.5.5, a dielectric slab is demonstrated to guide microwaves. To assure the excitation of only an m= 0 TE-guided wave, but one as well conﬁned to the dielectric as possible, the frequency is made just under the cutoﬀ frequency ωc2. (For a 2 cm thick slab having /epsilon1i//epsilon1o= 6.6, this is a frequency just under 6 GHz.) The m= 0 wave is excited in the dielectric slab by means of a vertical element at its left edge. This assures excitation of Ezwhile having the symmetry necessary to avoid excitation of the odd modes. The antenna is mounted at the center of a metal ground plane. Thus, without the slab, the signal at the receiving antenna (which is oriented to be sensitive to Ez) is essentially the same in all directions perpendicular to the zaxis. With the slab, a sharply increased signal in the vicinity of the right edge of the slab gives qualitative evidence of the wave guidance. The receiving antenna can also be used to probe the ﬁeld decay in the xdirection and to see that this decay increases with frequency.11 13.6 SUMMARY There are two perspectives from which this chapter can be reviewed. First, it can be viewed as a sequence of speciﬁc examples that are useful for dealing with radio frequency, microwave, and optical systems. Secs. 13.1–13.3 are concerned with the propagation of energy along parallel plates, ﬁrst acting as a transmission line and then as a waveguide. Practical systems to which the derived properties of the TEM and higher-order modes are directly applicable are strip lines used at frequencies 11To make the excitation independent of z, a collinear array of in-phase dipoles could be used for the excitation. This is not necessary to demonstrate the qualitative features of the guide. Sec. 13.6 Summary 39 that extend from dc to the microwave range. The rectangular waveguide of Sec. 13.4 might well be a section of “plumbing” from a microwave communication system, and the dielectric waveguide of Sec. 13.5 has many of the properties of an optical ﬁber. Second, the mathematical analysis of waves exempliﬁed in this chapter is generally applicable to other more complex systems that are uniform in one direction. When the structures described in this chapter are used to transport energy from one location to another, they are generally not terminated in “shorts” and “opens” and hence, generally, do not simply support standing waves. The object is usually to carry energy from an antenna to a receiver or from a generator to a load whether that be an antenna or a light bulb. Such energy transport is accomplished by the traveling waves featured in the next chapter. 40 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 P R O B L E M S 13.1 Introduction to TEM Waves 13.1.1∗With a short at y= 0, it is possible to ﬁnd the ﬁelds for Example 13.1.1 by recognizing at the outset that standing wave solutions meeting the homoge- neous boundary condition of (12) are of the form Ex= Re Asin(βy) exp( jωt). (a) Use (13.1.2) and (13.1.3) to determine the associated Hzand the dispersion equation (relation between βandω). (b) Now use the boundary condition at y=−bto show that the ﬁelds are as given by (13.1.16) and (13.1.17). 13.1.2∗Take the approach outlined in Prob. 13.1.1 for ﬁnding the ﬁelds [(13.1.28) and (13.1.29)] in Example 13.1.2. 13.1.3 Assume that ˆKois real and express the standing wave of (13.1.17) so as to make it evident that it is the sum of equal-amplitude waves traveling in the±ydirections, each with a magnitude of phase velocity ω/β=cand wavelength 2 π/β. 13.1.4∗Coaxial perfectly conducting circular cylinders having outer and inner radii aandb, respectively, form the transmission line shown in Fig. P13.1.4. (a) If the conductors were “open circuit” at z= 0 and driven by a voltage source Vatz=−l, show that the EQS electric ﬁeld is radial and given by V/[r ln(a/b)]. (b) If the conductors were “shorted” at z= 0 and driven by a current source Iatz=−l, show that the MQS magnetic ﬁeld intensity is φ directed and given by I/2πr. (c) With the motivation provided by these limiting solutions, show that solutions to all of Maxwell’s equations (in the region between the conductors) that satisfy the boundary conditions on the surfaces of the coaxial conductors are E=irV(z, t) ln¡a b¢ r;H=iφI(z, t) 2πr(a) provided that VandIare now functions not only of tbut of zas well that satisfy equations taking the same form as (13.1.2) and (13.1.3). ∂I ∂z=−C∂V ∂t; C≡2π/epsilon1 ln¡a b¢ (b) ∂V ∂z=−L∂I ∂t; L≡ln¡a b¢ µ 2π(c) Sec. 13.2 Problems 41 Fig. P13.1.4 13.1.5 For the coaxial conﬁguration of Prob. 13.1.4, there is a perfectly conducting “short” at z= 0, and the conductors are driven by a current source I= Re[Ioejωt] atz=−l. (a) Find I(z, t) and V(z, t) and hence EandH. (b) Take the low frequency limit where ω√µ/epsilon1l/lessmuch1 and show that Eand Hare the same as for a coaxial inductor. (c) Find EandHdirectly from the MQS laws and show that they agree with the results of part (b). 13.1.6 For the coaxial conﬁguration of Prob. 13.1.4, the conductors are “open circuited” at z= 0 and driven by a voltage source V= Re [ Voexp(jωt] at x=−l. (a) Find I(z, t) and V(z, t) and hence EandH. (b) Take the low-frequency limit where ω√µ/epsilon1l/lessmuch1 and show that Eand Hare the same as for a coaxial capacitor. (c) Find EandHdirectly from the EQS laws and show that they agree with the results of (b). 13.2 Two-Dimensional Modes Between Parallel Plates 13.2.1∗Show that each of the higher-order modes propagating in the + ydirec- tion, represented by A+ nandC+ nin (13.2.19) and (13.2.20), respectively, can be regarded as the sum of plane waves propagating in the directions represented by the vector wave number k=±nπ aix+βniy (a) and interfering in the planes x= 0 and x=aso as to satisfy the boundary conditions. 13.2.2 The TM and TE modes can themselves be classiﬁed into odd or even modes that, respectively, have ˆhzor ˆezodd or even functions of x. With this in mind, the origin of the coordinate system is moved so that it is midway between the perfectly conducting plates, as shown in Fig. P13.2.2. 42 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. P13.2.2 (a) Find the odd TM and TE solutions. Note that when the boundary condition is met at x=d≡a/2 for these functions, it is automatically met at x=−d. (b) Find the even TM and TE solutions, again noting that if the condi- tions are met at x=d, then they are at x=−das well. 13.3 TE and TM Standing Waves between Parallel Plates 13.3.1∗Starting with (13.3.1) (for TM modes) and (13.3.2) (for TE modes) use steps similar to those illustrated by (5.5.20)–(5.5.26) to obtain the orthog- onality conditions of (13.3.3) and (13.3.4), respectively. 13.3.2 In the system of Example 13.3.1, the wall at y= 0 is replaced by that shown in Fig. P13.3.2. A strip electrode is embedded in, but insulated from, the wall at y= 0. The resistance Ris low enough so that Etangential to the boundary at y= 0, even at the insulating gaps between the strip electrode and the surrounding wall, is negligible. (a) Determine the output voltage voin terms of v. (b) For b/a= 2, describe the dependence of |vo|on frequency over the range ω√µ/epsilon1 a = 0→πp 5/4, specifying the low-frequency range where the response has a linear dependence on frequency and the resonance frequencies. (c) What is the distribution of Hz(x, y) at the resonance frequencies? 13.3.3∗In the two-dimensional system of Fig. P13.3.3, each driven electrode has the same nature as the one in Fig. 13.3.1. The origin of the yaxis has been chosen to be in the plane of symmetry. (a) Use the symmetry to argue that Hz(y= 0) = 0. Sec. 13.3 Problems 43 Fig. P13.3.2 Fig. P13.3.3 (b) Show that in the interior region, Hz= Re∞X n=1 odd−4jω/epsilon1ˆv βnasinβny cosβnbcosnπx aejωt(a) 13.3.4 The one-turn loop of Fig. P13.3.4 has dimensions that are small compared toa, b, or wavelengths of interest and has area Ain the x−yplane. (a) It is used to detect the TM Hﬁeld at the middle of the bottom electrode in Fig. 13.3.1. Assume that the resistance is large enough so that the current induced in this loop gives rise to a magnetic ﬁeld that is negligible compared to that already found. In terms of Hz, what is vo? (b) At what locations x=Xof the loop is |vo|a maximum? (c) If the same loop were in the plate at y= 0 in the conﬁguration of Fig. 13.1.3 and used to detect Hzaty= 0 for the TEM ﬁelds of Example 44 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. P13.3.4 13.1.1, what would be the dependence of |vo|on the location x=X of the loop? (d) If the loop were located in the plate at y= 0 in the TE conﬁguration of Fig. 13.3.4, how should the loop be oriented to detect H? 13.3.5 In the system shown in Fig. P13.3.5, ∆ /lessmuchdand the driving sources v= Re[ˆvexp(jωt)] are uniformly distributed in the zdirection so that the ﬁelds are two dimensional. Thus, the driving electrode is like that of Fig. 13.3.1 except that it spans the width drather than the full width a. Find Hand Ein terms of v. 13.3.6 In the system shown in Fig. P13.3.6, the excitation electrode is like that for Fig. 13.3.4 except that it has a width drather than a. Find HandE in terms of ˆΛ. 13.4 Rectangular Waveguide Modes 13.4.1∗Show that an alternative method of exciting and detecting the TE 10mode in Demonstration 13.4.1 is to introduce one-turn loops as shown in Fig. P13.4.1. The excitation loop is inserted through a hole in the conducting wall while the detection loop passes through a slot, so that it can be moved in the ydirection. The loops are each in the y−zplane. To minimize disturbance of the ﬁeld, the detection loop is terminated in a high enough impedance so that the ﬁeld from the current in the loop is negligible. Com- pare the ydependence of the detected signal to that measured using the electric probe. 13.4.2 A rectangular waveguide has w/a= 0.75. Presuming that all TE and TM modes are excited in the guide, in what order do the lowest six modes begin to propagate in the ydirection as the frequency is raised? 13.4.3∗The rectangular waveguide shown in Fig. P13.4.3 is terminated in a per- fectly conducting plate at y= 0 that makes contact with the guide walls. An electrode at y=bhas a gap of width ∆ /lessmuchaand ∆ /lessmuchwaround its Sec. 13.4 Problems 45 Fig. P13.3.5 Fig. P13.3.6 Fig. P13.4.1 edges. Distributed around this gap are sources that constrain the ﬁeld from the edges of the plate to the guide walls to v(t)/∆ = Re(ˆ v/∆) exp( jωt). (a) Argue that the ﬁelds should be TM and use the boundary condition 46 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 Fig. P13.4.3 aty= 0 to show that ˆEy=∞X m=1∞X n=12A+ mncosβmnysinmπ axsinnπ wz (a) [Hint: If (13.4.9) and (13.4.10) are used, remember that ky= +βmn for the A+ mnmode but ky=−βmnfor the A− mnmode. ] (b) Show that, for mandnboth odd, A+ mn= 8ˆv(ω2µ/epsilon1−β2 mn)/nmπ2βmnsin(kmnb), while for either morneven, A+ mn= 0. (c) Show that for these modes the resonance frequencies (normalized to 1/√µ/epsilon1a) are ω√µ/epsilon1a=πr m2+¡a wn¢2+¡a bp¢2(b) where m, n, and pare integers, mandnodd. (d) Show that under quasistatic conditions, the ﬁeld which has been found is consistent with that implied by the EQS potential given by (5.10.10) and (5.10.15). 13.4.4 The rectangular waveguide shown in Fig. P13.4.3 is terminated in a per- fectly conducting plate at y= 0 that makes contact with the guide walls. However, instead of the excitation electrode shown, at y=bthere is the perfectly conducting plate with a square hole cut in its center, shown in Fig. P13.4.4. In this hole, the pole faces of a magnetic circuit are ﬂush with the plate and are used to excite ﬁelds within the guide. Approximate the normal ﬁelds over the surface of the pole faces as Hy=½ˆHofora 2< x <a+∆ 2andw−∆ 2< z <w+∆ 2 −ˆHofora−∆ 2< x <a 2andw−∆ 2< z <w+∆ 2(a) Sec. 13.5 Problems 47 Fig. P13.4.4 where ˆHois a complex constant. (Note that, if the magnetic circuit is driven by a one turn coil, the terminal voltage v=jω(∆2/2)µˆHo.) Determine Hy, and hence EandH, inside the guide. 13.5 Dielectric Waveguides: Optical Fibers 13.5.1∗For the dielectric slab waveguide of Fig. 13.5.1, consider the TE modes that have Ezan odd function of z. (a) Show that the dispersion relation between ωandkyis again found from (13.5.10), but with ( kxd) found by simultaneously solving (13.5.8) andαx kx=−cotkxd (a) (b) Sketch the graphical solution for kxd≡γm(modd) and show that the cutoﬀ frequency is again given by (13.5.9), but with modd rather than even. (c) Show that these odd modes also have the asymptote of unity slope shown in Fig. 13.5.3. (d) Sketch the odd mode dispersion relation on that for the even modes (Fig. 13.5.3). 13.5.2 For the dielectric slab waveguide shown in Fig. 13.5.1, /epsilon1i//epsilon1= 2.5, µ=µo, andd= 1 cm. In Hz, what is the highest frequency that can be used to guide only one TE mode. (Note the result of Prob. 13.5.1.) 13.5.3∗The dielectric slab waveguide of Fig. 13.5.1 is the same as that considered in this problem except that it now has a permeability µithat diﬀers from that outside, where it is µ. (a) Show that (13.5.7) and (13.5.8), respectively, are replaced by αx kx=µ µi· tankxd −cotkxd¸ ; even ; odd(a) 48 Electrodynamic Fields: The Boundary Value Point of View Chapter 13 αx kx=s ω2µi/epsilon1id2 (kxd)2¡ 1−µ/epsilon1 µi/epsilon1i¢ −1 ( b) (b) Show that making µi> µlowers the cutoﬀ frequency. (c) For a given frequency, does making µi/µ > 1 increase or decrease the wavelength λ≡2π/ky? 13.5.4 The dielectric slab of Fig. 13.5.1 has permittivity /epsilon1iand permeability µi, while in the surrounding regions these are /epsilon1andµ, respectively. Consider the TM modes. (a) Determine expressions analogous to (13.5.7), (13.5.8), and (13.5.10) that can be used to determine the dispersion relation ω=ω(ky) for modes that have Hzeven and odd functions of x. (b) What are the cutoﬀ frequencies? (c) For µi=µand/epsilon1i=/epsilon1= 2.5, draw the dispersion plot for the lowest three modes that is analogous to that of Fig. 13.5.3. 14 ONE-DIMENSIONAL WAVE DYNAMICS 14.0 INTRODUCTION Examples of conductor pairs range from parallel conductor transmission lines car- rying gigawatts of power to coaxial lines carrying microwatt signals between com- puters. When these lines become very long, times of interest become very short, or frequencies become very high, electromagnetic wave dynamics play an essential role. The transmission line model developed in this chapter is therefore widely used. Equally well described by the transmission line model are plane waves, which are often used as representations of radiation ﬁelds at radio, microwave, and optical frequencies. For both qualitative and quantitative purposes, there is again a need to develop convenient ways of analyzing the dynamics of such systems. Thus, there are practical reasons for extending the analysis of TEM waves and one-dimensional plane waves given in Chap. 13. The wave equation is ubiquitous. Although this equation represents most ac- curately electromagnetic waves, it is also applicable to acoustic waves, whether they be in gases, liquids or solids. The dynamic interaction between excitation ampli- tudes ( EandHﬁelds in the electromagnetic case, pressure and velocity ﬁelds in the acoustic case) is displayed very clearly by the solutions to the wave equation. The developments of this chapter are therefore an investment in understanding other more complex dynamic phenomena. We begin in Sec. 14.1 with the distributed parameter ideal transmission line. This provides an exact representation of plane (one-dimensional) waves. In Sec. 14.2, it is shown that for a wide class of two-conductor systems, uniform in an axial direction, the transmission line equations provide an exact description of the TEM ﬁelds. Although such ﬁelds are in general three dimensional, their propagation in the axial direction is exactly represented by the one-dimensional wave equation to the extent that the conductors and insulators are perfect. The distributed parameter 1 2 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.1.1 Incremental length of distributed parameter transmission line. model is also commonly used in an approximate way to describe systems that do not support ﬁelds that are exactly TEM. Sections 14.3–14.6 deal with the space-time evolution of transmission line volt- age and current. Sections 14.3–14.4, which concentrate on the transient response, are especially applicable to the propagation of digital signals. Sections 14.5-14.6 concentrate on the sinusoidal steady state that prevails in power transmission and communication systems. The eﬀects of electrical losses on electromagnetic waves, propagating through lossy media or on lossy structures, are considered in Secs. 14.7–14.9. The distributed parameter model is generalized to include the electrical losses in Sec. 14.7. A limiting form of this model provides an “exact” representation of TEM waves in lossy media, either propagating in free space or along pairs of perfect conductors embedded in uniform lossy media. This limit is developed in Sec. 14.8. Once the conductors are taken as being “perfect,” the model is exact and the model is equivalent to the physical system. However, a second limit of the lossy transmission line model, which is exempliﬁed in Sec. 14.9, is not “exact.” In this case, conductor losses give rise to an electric ﬁeld in the direction of propagation. Thus, the ﬁelds are not TEM and this section gives a more realistic view of how quasi-one-dimensional models are often used. 14.1 DISTRIBUTED PARAMETER EQUIVALENTS AND MODELS The theme of this section is the distributed parameter transmission line shown in Fig. 14.1.1. Over any ﬁnite axial length of interest, there is an inﬁnite set of the basic units shown in the inset, an inﬁnite number of capacitors and inductors. The parameters LandCare deﬁned per unit length. Thus, for the segment shown between z+ ∆zandz, L∆zis the series inductance (in Henrys) of a section of the distributed line having length ∆ z, while C∆zis the shunt capacitance (in Farads). In the limit where the incremental length ∆ z→0, this distributed parameter transmission line serves as a model for the propagation of three types of electro- magnetic ﬁelds.1 1To facilitate comparison with quasistatic ﬁelds, the direction of wave propagation for TEM waves in Chap. 13 was taken as y. It is more customary to make it z. Sec. 14.1 Distributed Parameter Model 3 •First, it gives an exact representation of uniformly polarized electromagnetic plane waves. Whether these are waves in free space, perhaps as launched by the dipole considered in Sec. 12.2, or TEM waves between plane parallel perfectly conducting electrodes, Sec. 13.1, these ﬁelds depend only on one spatial coordinate and time. •Second, we will see in the next section that the distributed parameter trans- mission line represents exactly the ( z, t) dependence of TEM waves propagat- ing on pairs of axially uniform perfect conductors forming transmission lines of arbitrary cross-section. Such systems are a generalization of the parallel plate transmission line. By contrast with that special case, however, the ﬁelds generally depend on the transverse coordinates. These ﬁelds are therefore, in general, three dimensional. •Third, it represents in an approximate way, the ( z, t) dependence for sys- tems of large aspect ratio, having lengths over which the ﬁelds evolve in the zdirection (e.g., wavelengths) that are long compared to the transverse di- mensions. To reﬂect the approximate nature of the model and the two- or three-dimensional nature of the system it represents, it is sometimes said to bequasi-one-dimensional . We can obtain a pair of partial diﬀerential equations governing the transmis- sion line current I(z, t) and voltage V(z, t) by ﬁrst requiring that the currents into the node of the elemental section sum to zero I(z)−I(z+ ∆z) =C∆z∂V ∂t(1) and then requiring that the series voltage drops around the circuit also sum to zero. V(z)−V(z+ ∆z) =L∆z∂I ∂t(2) Then, division by ∆ zand recognition that lim ∆z→0f(z+ ∆z)−f(z) ∆z=∂f ∂z(3) results in the transmission line equations . ∂I ∂z=−C∂V ∂t (4) ∂V ∂z=−L∂I ∂t (5) The remainder of this section is an introduction to some of the physical situations represented by these laws. 4 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.1.2 Possible polarization and direction of propagation of plane wave described by the transmission line equations. Plane-Waves. In the following sections, we will develop techniques for de- scribing the space-time evolution of ﬁelds on transmission lines. These are equally applicable to the description of electromagnetic plane waves. For example, suppose the ﬁelds take the form shown in Fig. 14.1.2. E=Ex(z, t)ix;H=Hy(z, t)iy (6) Then, the xandycomponents of the laws of Amp` ere and Faraday reduce to2 −∂Hy ∂z=/epsilon1∂Ex ∂t(7) ∂Ex ∂z=−µ∂Hy ∂t(8) These laws are identical to the transmission line equations, (4) and (5), with Hy↔I, E x↔V, /epsilon1 ↔C, µ ↔L (9) With this identiﬁcation of variables and parameters, the discussion is equally appli- cable to plane waves, whether we are considering wave transients or the sinusoidal steady state in the following sections. Ideal Transmission Line. The TEM ﬁelds that can exist between the parallel plates of Fig. 14.1.3 can either be regarded as plane waves that happen to meet the boundary conditions imposed by the electrodes or as a special case of transmission line ﬁelds. The following example illustrates the transition to the second viewpoint. Example 14.1.1. Plane Parallel Plate Transmission Line In this case, the ﬁelds ExandHypictured in Fig. 14.1.2 and described by (7) and (8) can exist unaltered between the plates of Fig. 14.1.3. If the voltage and current are deﬁned as V=Exa; I=Hyw (10) 2Compare with (13.1.2) and (13.1.3) for ﬁelds in x−zplane and propagating in the y direction. Sec. 14.1 Distributed Parameter Model 5 Fig. 14.1.3 Example of transmission line where conductors are parallel plates. Equations (7) and (8) become identical to the transmission line equations, (4) and (5), with the capacitance and inductance per unit length deﬁned as C=w/epsilon1 a; L=aµ w(11) Note that these are indeed the CandLthat would be found in Chaps. 5 and 8 for the pair of perfectly conducting plates shown in Fig. 14.1.3 if they had unit length in the zdirection and were, respectively, “open circuited” and “short circuited” at the right end. As an alternative to a ﬁeld description, the distributed L−Ctransmission line model gives circuit theory interpretation to the physical processes at work in the actual system. As expressed by (1) and hence (4), the current Ican be a function ofzbecause some of it can be diverted into charging the “capacitance” of the line. This is an alternative way of representing the eﬀect of the displacement current density on the right in Amp` ere’s law, (7). The voltage Vis a function of zbecause the inductance of the line causes a voltage drop, even though the conductors are pictured as having no resistance. This follows from (2) and (5) and embodies the same information as did Faraday’s diﬀerential law (8). The integral of Efrom one conductor to the other at some location zcan diﬀer from that at another location because of the ﬂux linked by a contour consisting of these integration paths and closing by contours along the perfect conductors. In the next section, we will generalize our picture of TEM waves and see that (4) and (5) exactly describe transverse waves on pairs of perfect conductors of arbitrary cross-section. Of course, LandCare the inductance per unit length and capacitance per unit length of the particular conductor pair under consideration. The ﬁelds depend not only on the independent variables ( z, t) appearing explicitly in the transmission line equations, but upon the transverse coordinates as well. Thus, the parallel plate transmission line and the generalization of that line considered in the next section are examples for which the distributed parameter model is exact. In these cases, TEM waves are exact solutions to the boundary value problem at all frequencies, including frequencies so high that the wavelength of the TEM wave is comparable to, or smaller than, the transverse dimensions of the line. As one would expect from the analysis of Secs. 13.1–13.3, higher-order modes propa- gating in the zdirection are also valid solutions. These are not described by the transmission line equations (4) and (5). 6 One-Dimensional Wave Dynamics Chapter 14 Quasi-One-Dimensional Models. The distributed parameter model is also often used to represent ﬁelds that are not quite TEM. As an example where an approximate model consists of the distributed L−Cnetwork, suppose that the region between the plane parallel plate conductors is ﬁlled to the level x=d < a by a dielectric of one permittivity with the remainder ﬁlled by a material having a diﬀerent permittivity. The region between the conductors is then one of nonuniform permittivity. We would ﬁnd that it is not possible to exactly satisfy the boundary conditions on both the tangential and normal electric ﬁelds at the interface between dielectrics with an electric ﬁeld that only had components transverse to z.3Even so, if the wavelength is very long compared to the transverse dimensions, the distributed parameter model provides a useful approximate description. The capacitance per unit length used in this model reﬂects the eﬀect of the nonuniform dielectric in an approximate way. 14.2 TRANSVERSE ELECTROMAGNETIC WAVES The parallel plates of Sec. 13.1 are a special case of the general conﬁguration shown in Fig. 14.2.1. The conductors have the same cross-section in any plane z= constant, but their cross-sectional geometry is arbitrary.4The region between the pair of perfect conductors is ﬁlled by a material having uniform permittivity /epsilon1and permeability µ. In this section, we show that such a structure can support ﬁelds that are transverse to the axial coordinate z, and that the z−tdependence of these ﬁelds is described by the ideal transmission line model. Two common transmission line conﬁgurations are illustrated in Fig. 14.2.2. The TEM ﬁelds are conveniently pictured in terms of the vector and scalar potentials, Aand Φ, generalized to describe electrodynamic ﬁelds in Sec. 12.1. This is because such ﬁelds have only an axial component of A. A=Az(x, y, z, t )iz (1) Indeed, evaluation in Cartesian coordinates, shows that even though Azis in general not only a function of the transverse coordinates but of the axial coordinate zas well, there is no longitudinal component of H. To insure that the electric ﬁeld is also transverse to the zaxis, the zcomponent of the expression relating EtoAand Φ (12.1.3) must be zero. Ez=−∂Φ ∂z−∂Az ∂t= 0 (2) A second relation between Φ and Azis the gauge condition, (12.1.7), which in view of (1) becomes ∂Az ∂z=−µ/epsilon1∂Φ ∂t(3) 3We can see that a uniform plane wave cannot describe such a situation because the propa- gational velocities of plane waves in dielectrics of diﬀerent permittivities diﬀer. 4The direction of propagation is now zrather than y. Sec. 14.2 Transverse Waves 7 Fig. 14.2.1 Conﬁguration of two parallel perfect conductors supporting TEM ﬁelds. Fig. 14.2.2 Two examples of transmission lines that support TEM waves: (a) parallel wire conductors; and (b) coaxial conductors. These last two equations combine to show that both Φ and Azmust satisfy 8 One-Dimensional Wave Dynamics Chapter 14 the one-dimensional wave equation. For example, elimination of ∂2Az/∂z∂t between thezderivative of (2) and the time derivative of (3) gives ∂2Φ ∂z2=µ/epsilon1∂2Φ ∂t2(4) A similar manipulation, with the roles of zandtreversed, shows that Azalso satisﬁes the one-dimensional wave equation. ∂2Az ∂z2=µ/epsilon1∂2Az ∂t2(5) Even though the potentials satisfy the one-dimensional wave equations, in general they depend on the transverse coordinates. In fact, the diﬀerential equa- tion governing the dependence on the transverse coordinates is the two-dimensional Laplace’s equation. To see this, observe that the three-dimensional Laplacian con- sists of a part involving derivatives with respect to the transverse coordinates and a second derivative with respect to z. ∇2=∇2 T+∂2 ∂z2(6) In general, Φ and Asatisfy the three-dimensional wave equation, the homogeneous forms of (12.1.8) and (12.1.10). But, in view of (4) and (5), these expressions reduce to ∇2 TΦ = 0 (7) ∇2 TAz= 0 (8) where the Laplacian ∇2 Tis the two-dimensional Laplacian, written in terms of the transverse coordinates. Even though the ﬁelds actually depend on z,the transverse dependence is as though the ﬁelds were quasistatic and two dimensional. The boundary conditions on the surfaces of the conductors require that there be no tangential Eand no normal B. The latter condition prevails if Azis constant on the surfaces of the conductors. This condition is familiar from Sec. 8.6. With Az deﬁned as zero on the surface S1of one of the conductors, as shown in Fig. 14.2.1, it is equal to the ﬂux per unit length passing between the conductors when evaluated anywhere on the second conductor. Thus, the boundary conditions imposed on Az are Az= 0 on S1; Az= Λ(z, t) onS2 (9) As described in Sec. 8.6, where two-dimensional magnetic ﬁelds were represented in terms of Az,Λ is the ﬂux per unit length passing between the conductors. Because Eis transverse to zandAhas only a zcomponent, Eis found from Φ by taking the transverse gradient just as if the ﬁelds were two dimensional. The boundary condition on E, met by making Φ constant on the surfaces of the conductors, is therefore familiar from Chaps. 4 and 5. Φ = 0 on S1; Φ = V(z, t) onS2 (10) Sec. 14.2 Transverse Waves 9 By deﬁnition, Λ is equal to the inductance per unit length Ltimes the total current Icarried by the conductor having the surface S2. Λ =LI (11) The ﬁrst of the transmission line equations is now obtained simply by evalu- ating (2) on the boundary S2of the second conductor and using the deﬁnition of Λ from (11). ∂V ∂z+L∂I ∂t= 0(12) The second equation follows from a similar evaluation of (3). This time we introduce the capacitance per unit length by exploiting the relation LC=µ/epsilon1, (8.6.14). ∂I ∂z+C∂V ∂t= 0(13) The integral of Ebetween the conductors within a given plane of constant zisV, and can be interpreted as the voltage between the two conductors. The total current carried in the + zdirection through a plane of constant zby one of the conductors and returned in the −zdirection by the other is I.Because eﬀects of magnetic induction are important, Vis a function of z. Similarly, because the displacement current is important, the current Iis also a function of z. Example 14.2.1. Parallel Plate Transmission Line Between the perfectly conducting parallel plates of Fig. 14.1.3, solutions to (7) and (8) that meet the boundary conditions of (9) and (10) are Az= Λ(z, t)¡ 1−x a¢ =aµ w¡ 1−x a¢ I(z, t) (14) Φ =¡ 1−x a¢ V(z, t) (15) In the EQS context of Chap. 5, the latter is the potential associated with a uniform electric ﬁeld between plane parallel electrodes, while in the MQS context of Example 8.4.4, (14) is the vector potential associated with the uniform magnetic ﬁeld inside a one-turn solenoid. The inductance per unit length follows from (11) and the eval- uation of (14) on the surface S2, and one way to evaluate the capacitance per unit length is to use the relation LC=µ/epsilon1. L=µa w; C=µ/epsilon1 L=/epsilon1w a(16) Every two-dimensional example from Chap. 4 with perfectly conducting bound- aries is a candidate for supporting TEM ﬁelds that propagate in a direction per- pendicular to the two dimensions. For every solution to (7) meeting the boundary 10 One-Dimensional Wave Dynamics Chapter 14 conditions of (10), there is one to (8) satisfying the conditions of (9). This follows from the antiduality exploited in Chap. 8 to describe the magnetic ﬁelds with per- fectly conducting boundaries (Example 8.6.3). The next example illustrates how we can draw upon results from these earlier chapters. Example 14.2.2. Parallel Wire Transmission Line For the parallel wire conﬁguration of Fig. 14.2.2a, the capacitance per unit length was derived in Example 4.6.3, (4.6.27). C=π/epsilon1 ln· l R+q¡l R¢2−1¸ (17) The inductance per unit length was derived in Example 8.6.1, (8.6.12). L=µ πln· l R+r¡l R¢2−1¸ (18) Of course, the product of these is µ/epsilon1. At any given instant, the electric and magnetic ﬁelds have a cross-sectional distribution depicted by Figs. 4.6.5 and 8.6.6, respectively. The evolution of the ﬁelds with zandtare predicted by the one-dimensional wave equation, (4) or (5), or a similar equation resulting from combining the transmission line equations. Propagation is in the zdirection. With the understanding that the ﬁelds have transverse distributions that are identical to the EQS and MQS patterns, the next sections focus on the evolution of the ﬁelds with zandt. No TEM Fields in Hollow Pipes. From the general description of TEM ﬁelds given in this section, we can see that TEM modes will not exist inside a hollow perfectly conducting pipe. This follows from the fact that both Azand Φ must be constant on the walls of such a pipe, and solutions to (7) and (8) that meet these conditions are that Azand Φ, respectively, are equal to these constants throughout. From Sec. 5.2, we know that these solutions to Laplace’s equation are unique. The EandHthey represent are zero, so there can be no TEM ﬁelds. This is consistent with the ﬁnding for rectangular waveguides in Sec. 13.4. The parallel plate conﬁguration considered in Secs. 13.1–13.3 could support TEM modes because it was assumed that in any given cross-section (perpendicular to the axial position), the electrodes were insulated from each other. Power-ﬂow and Energy Storage. The transmission line model expresses the ﬁelds in terms of VandI. For the TEM ﬁelds, this is not an approximation but rather an elegant way of dealing with a class of three-dimensional time-dependent ﬁelds. To emphasize this point, we now show the equivalence of power ﬂow and energy storage as derived from the transmission line model and from Poynting’s theorem. Sec. 14.2 Transverse Waves 11 Fig. 14.2.3 Incremental length of transmission line and its cross-section. An incremental length, ∆ z, of a two-conductor system and its cross-section are pictured in Fig. 14.2.3. A one-dimensional version of the energy conservation law introduced in Sec. 11.1 can be derived from the transmission line equations using manipulations analogous to those used to derive Poynting’s theorem in Sec. 11.2. We multiply (14.1.4) by Vand (14.1.5) by Iand add. The result is a one- dimensional statement of energy conservation. −∂ ∂z(V I) =∂ ∂t¡1 2CV2+1 2LI2¢ (19) This equation has intuitive “appeal.” The power ﬂowing in the zdirection isV I, and the energy per unit length stored in the electric and magnetic ﬁelds is 1 2CV2and1 2LI2, respectively. Multiplied by ∆ z, (19) states that the amount by which the power ﬂow at zexceeds that at z+ ∆zis equal to the rate at which energy is stored in the length ∆ zof the line. We can obtain the same result from the three-dimensional Poynting’s integral theorem, (11.1.1), evaluated using (11.3.3), and applied to a volume element of incremental length ∆ zbut one having the cross-sectional area Aof the system (if need be, one extending to inﬁnity). −hZ AE×H·izda¯¯ z+∆z−Z AE×H·izda¯¯ zi =∂ ∂tZ A¡1 2/epsilon1E·E+1 2µH·H¢ da∆z(20) Here, the integral of Poynting’s ﬂux density, E×H, over a closed surface Shas been converted to one over the cross-sectional areas Ain the planes zandz+ ∆z. The closed surface is in this case a cylinder having length ∆ zin the zdirection 12 One-Dimensional Wave Dynamics Chapter 14 and a lateral surface described by the contour Cin Fig. 14.2.3b. The integrals of Poynting’s ﬂux density over the various parts of this lateral surface (having circum- ference Cand length ∆ z) either are zero or cancel. For example, on the surfaces of the conductors denoted by C1andC2, the contributions are zero because Eis perpendicular. Thus, the contributions to the integral over Scome only from inte- grations over Ain the planes z+∆zandz. Note that in writing these contributions on the left in (20), the normal to Son these surfaces is izand−iz, respectively. To see that the integrals of the Poynting ﬂux over the cross-section of the system are indeed simply V I,Eis written in terms of the potentials (12.1.3).Z AE×H·izda=Z A¡ − ∇Φ−∂A ∂t¢ ×H·izda (21) The surface of integration has its normal in the zdirection. Because Ais also in the zdirection, the cross-product of ∂A/∂twithHmust be perpendicular to z, and therefore makes no contribution to the integral. A vector identity then converts the integral to Z AE×H·izda=Z A−∇Φ×H·izda =−Z A∇ ×(ΦH)·izda +Z AΦ∇ ×H·izda(22) In Fig. 14.2.3, the area A, enclosed by the contour C, is insulating. Thus, because J= 0 in this region and the electric ﬁeld, and hence the displacement current, are perpendicular to the surface of integration, Amp` ere’s law tells us that the integrand in the second integral is zero. The ﬁrst integral can be converted, by Stokes’ theorem, to a line integral. Z AE×H·izda=−I CΦH·ds (23) On the contour, Φ = 0 on C1and at inﬁnity. The contributions along the segments connecting C1andC2to inﬁnity cancel, and so the only contribution comes from C2. On that contour, Φ = V, so Φ is a constant. Finally, again because the displacement current is perpendicular to ds, Amp` ere’s integral law requires that the line integral ofHon the contour C2enclosing the conductor having potential Vbe equal to −I. Thus, (23) becomesZ AE×H·izda=−VI C2H·ds=V I (24) The axial power ﬂux pictured by Poynting’s theorem as passing through the insu- lating region between the conductors can just as well be represented by the current and voltage of one of the conductors. To formalize the equivalence of these points of view, (24) is used to evaluate the left-hand side of Poynting’s theorem, (20), and that expression divided by ∆ z. −[V(z+ ∆z)I(z+ ∆z)−V(z)I(z)] ∆z =∂ ∂tZ A¡1 2/epsilon1E·E+1 2µH·H¢ da(25) Sec. 14.3 Transients on Inﬁnite 13 In the limit ∆ z→0, this statement is equivalent to that implied by the transmission line equations, (19), because the electric and magnetic energy storages per unit length are 1 2CV2=Z A1 2/epsilon1E·Eda;1 2LI2=Z A1 2µH·Hda (26) In summary, for TEM ﬁelds, we are justiﬁed in thinking of a transmission line as storing energies per unit length given by (26) and as carrying a power V Iin the zdirection. 14.3 TRANSIENTS ON INFINITE TRANSMISSION LINES The transient response of transmission lines or plane waves is of interest for time- domain reﬂectometry and for radar. In these applications, it is the delay and shape of the response to pulse-like signals that provides the desired information. Even more common is the use of pulses to represent digitally encoded information car- ried by various types of cables and optical ﬁbers. Again, pulse delays and reﬂections are often crucial, and an understanding of how these are endemic to common com- munications systems is one of the points in this and the next section. The next four sections develop insights into dynamic phenomena described by the one-dimensional wave equation. This and the next section are concerned with transients and focus on initial as well as boundary conditions to create an awareness of the key role played by causality. Then, with the understanding that eﬀects of the turn-on transient have died away, the sinusoidal steady state response is considered in Secs. 14.5–14.6, The evolution of the transmission line voltage V(z, t), and hence the associated TEM ﬁelds, is governed by the one-dimensional wave equation. This follows by combining the transmission line equations, (14.1.4)-(5), to obtain one expression forV. ∂2V ∂z2=1 c2∂2V ∂t2; c≡1√ LC=1√µ/epsilon1(1) This equation has a remarkably general pair of solutions V=V+(α) +V−(β) (2) where V+andV−arearbitrary functions of variables αandβthat are deﬁned as particular combinations of the independent variables zandt. α=z−ct (3) β=z+ct (4) To see that this general solution in fact satisﬁes the wave equation, it is only nec- essary to perform the derivatives and substitute them into the equation. To that end, observe that ∂V± ∂z=V/prime ±;∂V± ∂t=∓cV/prime ± (5) 14 One-Dimensional Wave Dynamics Chapter 14 where primes indicate the derivative with respect to the argument of the function. Carrying out the same process once more gives the second derivatives required to evaluate the wave equation. ∂2V± ∂z2=V/prime/prime ±;∂2V± ∂t2=c2V/prime/prime ± (6) Substitution of these expression for the derivatives in (1) shows that (1) is satisﬁed. Functions having the form of (2) are indeed solutions to the wave equation. According to (2), Vis a superposition of ﬁelds that propagate, without chang- ing their shape, in the positive and negative zdirections. With αmaintained con- stant, the component V+is constant. With αa constant, the position zincreases with time according to the law z=α+ct (7) The shape of the second component of (2) remains invariant when βis held constant, as it is if the zcoordinate decreases at the rate c. The functions V+(z−ct) and V−(z+ct) represent forward and backward waves proceeding without change of shape at the speed cin the + zand−zdirections respectively. We conclude that the voltage can be represented as a superposition of forward and backward waves, V+andV−, which, if the space surrounding the conductors is free space (where /epsilon1=/epsilon1oandµ=µo), propagate with the velocity c/similarequal3×108m/s of light. Because I(z, t) also satisﬁes the one-dimensional wave equation, it also can be written as the sum of traveling waves. I=I+(α) +I−(β) (8) The relationships between these components of Iand those of Vare found by substi- tution of (2) and (8) into either of the transmission line equations, (14.1.4)–(14.1.5), which give the same result if it is remembered that c= 1/√ LC. In summary, as fundamental solutions to the equations representing the ideal transmission line, we have V=V+(α) +V−(β) (9) I=1 Zo[V+(α)−V−(β)] (10) where α=z−ct; β=z+ct (11) Here, Zois deﬁned as the characteristic impedance of the line. Zo≡p L/C (12) Sec. 14.3 Transients on Inﬁnite 15 Fig. 14.3.1 Waves initiated at z=αandz=βpropagate along the lines of constant αandβto combine at P. Typically, Zois the intrinsic impedancep µ//epsilon1multiplied by a function of the ratio of dimensions describing the cross-sectional geometry of the line. Illustration. Characteristic Impedance of Parallel Wires For example, the parallel wire transmission line of Example 14.2.2 has the charac- teristic impedance p L/C =1 πln· l R+r¡l R¢2−1¸p µ//epsilon1 (13) where for free space,p µ//epsilon1≈377Ω. Response to Initial Conditions. The speciﬁcation of the distribution of V andIat an initial time, t= 0, leads to two traveling waves. It is helpful to picture the ﬁeld evolution in the z−tplane shown in Fig. 14.3.1. In this plane, the α= constant and β= constant characteristic lines are straight and have slopes ±c, respectively. When t= 0, we are given that along the zaxis, V(z,0) = Vi(z) (14) I(z,0) = Ii(z) (15) What are these ﬁelds at some later time, such as at Pin Fig. 14.3.1? We answer this question in two steps. First, we use the initial conditions to establish the separate components V+andV−at each position when t= 0. To this end, the initial conditions of (14) and (15) are substituted for the quantities on the left in (9) and (10) to obtain two equations for these unknowns. V++V−=Vi (16) 1 Zo(V+−V−) =Ii (17) 16 One-Dimensional Wave Dynamics Chapter 14 These expressions can then be solved for the components in terms of the initial conditions. V+=1 2³ IiZo+Vi´ (18) V−=1 2³ −IiZo+Vi´ (19) The second step combines these components to determine the ﬁeld at Pin Fig. 14.3.1. Here we use the invariance of V+along the line α= constant and the invariance of V−along the line β= constant. The way in which these components combine at Pto give VandIis summarized by (9) and (10). The total voltage at Pis the sum of the components, while the current is the characteristic admittance Z−1 omultiplied by the diﬀerence of the components. The following examples illustrate how the initial conditions determine the invariants (the waves V±propagating in the ±zdirections) and how these invariants in turn determine the ﬁelds at a subsequent time and diﬀerent position. They show how the response at Pin Fig. 14.3.1 is determined by the initial conditions at just two locations, indicated in the ﬁgure by the points z=αandz=β. Implicit in our understanding of the dynamics is causality. The response at the location Pat some later time is the result of conditions at ( z=α, t= 0) that propagate with the velocity cin the + zdirection and conditions at ( z=β, t= 0) that propagate in the −zdirection with velocity c. Example 14.3.1. Initiation of a Pure Traveling Wave In Example 3.1.1, we were introduced to a uniform plane wave composed of a single component traveling in the + zdirection. The particular initial conditions for Ex andHy[(3.1.9) and (3.1.10)] were selected so that the response would be composed of just the wave propagating in the + zdirection. Given that the initial distribution ofExis Ex(z,0) = Ei(z) =Eoe−z2/2a2(20) can we now show how to select a distribution of Hysuch that there is no part of the response propagating in the −zdirection? In applying the transmission line to plane waves, we make the identiﬁcation (14.1.9) V↔Ex, I↔Hy, C↔/epsilon1o, L↔µo⇒Zo↔r µo /epsilon1o(21) We are assured that E−= 0 by making the right-hand side of (19) vanish. Thus, we make Hi=r /epsilon1o µoEi=r /epsilon1o µoEoe−z2/2a2(22) Sec. 14.3 Transients on Inﬁnite 17 It follows from (18) and (19) that along the characteristic lines passing through (z,0), E+=Ei; E−= 0 (23) and from (9) and (10) that the subsequent ﬁelds are Ex=E+=Eoe−(z−ct)2/2a2(24) Hy=r /epsilon1o µoE+=r /epsilon1o µoEoe−(z−ct)2/2a2(25) These are the traveling electromagnetic waves found “the hard way” in Example 3.1.1. The following example gives further substance to the two-step process used to deduce the ﬁelds at Pin Fig. 14.3.1 from those at ( z=α, t= 0) and ( z=β, t= 0). First, the components V+andV−, respectively, are deduced at ( z=α, t= 0) and (z=β, t= 0) from the initial conditions. Because V+is invariant along the line α= constant while V−is invariant along the line β= constant, we can then combine these components to determine the ﬁelds at P. Example 14.3.2. Initiation of a Wave Transient Suppose that when t= 0 there is a uniform voltage Vpbetween the positions z=−d andz=d, but that outside this range, V= 0. Further, suppose that initially, I= 0 over the entire length of the line. Vi=nVp;−d < z < d 0; z <−dandd < z(26) What are the subsequent distributions of VandI? Once we have found these re- sponses, we will see how such initial conditions might be realized physically. The initial conditions are given a pictorial representation in Fig. 14.3.2, where V(z,0) = ViandI(z,0) = Iiare shown as the solid and broken distributions when t= 0. It follows from (18) and (19) that V+=n0; α <−d, d < α 1 2Vp;−d < α < d, V −=n0; β <−d, d < β 1 2Vp;−d < β < d(27) Now that the initial conditions have been used to identify the wave components V±, we can use (9) and (10) to establish the subsequent VandI. These are also shown in Fig. 14.3.2 using the axis perpendicular to the z−tplane to represent either V(z, t) (the solid lines) or I(z, t) (the dashed lines). Shown in this ﬁgure are the initial and two subsequent ﬁeld distributions. At point P1, both V+andV−are zero, so that both VandIare also zero. At points like P2, where the wave propagating from z=dhas arrived but that from z=−dhas not, V+isVp/2 while V−remains zero. At points like P3, neither the wave propagating in the −zdirection from z=dor that propagating in the + zdirection from z=−dhas yet arrived, V+andV−are given by (27), and the ﬁelds remain the same as they were initially. By the time t=d/c, the wave transient has resolved itself into two pulses propagating in the + zand−zdirections with the velocity c. These pulses consist 18 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.3.2 Wave transient pictured in the z−tplane. When t= 0, I= 0 and Vassumes a uniform value over the range −d < z < d and is zero outside this range. of a voltage and a current that are in a constant ratio equal to the characteristic impedance, Zo. With the help of the step function u−1(z), deﬁned by u−1(z)≡n0;z <0 1; 0 < z(28) we can carry out these same steps in analytical terms. The initial conditions are I(z,0) = 0 V(z,0) = Vp[u−1(z+d)−u−1(z−d)] (29) The wave components follow from (18) and (19) and are expressed in terms of the variables αandβbecause they are invariant along lines where these parameters, respectively, are constant. V+=1 2Vp[u−1(α+d)−u−1(α−d)] V−=1 2Vp[u−1(β+d)−u−1(β−d)](30) Sec. 14.4 Transients on Bounded Lines 19 Fig. 14.3.3 Thunderstorm over power line modeled by initial conditions of Fig. 14.3.2. The voltage and current at the point Pin Fig. 14.3.1 follow from substitution of these expresions into (9) and (10). With αandβexpressed in terms of ( z, t) using (11), it follows that V=1 2Vp[u−1(z−ct+d)−u−1(z−ct−d)] +1 2Vp[u−1(z+ct+d)−u−1(z+ct−d)] I=1 2Vp Zo[u−1(z−ct+d)−u−1(z−ct−d)] −1 2Vp Zo[u−1(z+ct+d)−u−1(z+ct−d)](31) These are analytical expressions for the the functions depicted by Fig. 14.3.2. When our lights blink during a thunderstorm, it is possibly due to circuit interruption resulting from a power line transient initiated by a lightning stroke. Even if the discharge does not strike the power line, there can be transients resulting from an accumulation of charge on the line imaging the charge in the cloud above, as shown in Fig. 14.3.3. When the cloud is discharged to ground by the lightning stroke, initial conditions are established that might be modeled by those considered in this example. Just after the lightning discharge, the images for the charge accumulated on the line are on the ground below. 14.4 TRANSIENTS ON BOUNDED TRANSMISSION LINES Transmission lines are generally connected to a source and to a load, as shown in Fig. 14.4.1a. More complex systems composed of interconnected transmission lines can usually be decomposed into subsystems having this basic conﬁguration. A generator at z= 0 is connected to a load at z=lby a transmission line having the length l. In this section, we build upon the traveling wave picture introduced in Sec. 14.3 to describe transients at a boundary initiated by a source. 20 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.4.1 (a)Transmission line with terminations. (b) Initial and boundary conditions in z−tplane. In picturing the evolution with time of the voltage V(z, t) and current I(z, t) on a terminated line, it is again helpful to use the z−tplane shown in Fig. 14.4.1b. The load and generator impose boundary conditions at z=landz= 0. In addition to satisfying these conditions, the distributions of VandImust also satisfy the respective initial values V=Vi(z) and I=Ii(z) when t= 0, introduced in Sec. 14.3. Thus, our goal is to ﬁnd VandIin the ⊂-shaped region of z−tspace shown in Fig. 14.4.1b. In Sec. 14.3, we found that the transmission line equations, (14.1.4) and (14.1.5), have solutions V=V+(α) +V−(β) (1) I=1 Zo[V+(α)−V−(β)] (2) where α=z−ct; β=z+ct (3) andc= 1/√ LCandZo=p L/C. A mathematical way of saying that V+andV−, respectively, represent waves traveling in the + zand−zdirections is to say that these quantities are invariants on the characteristic lines α= constant and β= constant in the z−tplane. There are two steps in ﬁnding VandI. •First, the initial conditions, and now the boundary conditions as well, are used to determine V+andV−along the two families of characteristic lines in the region of the z−tplane of interest. This is done with the understanding that causality prevails in the sense that the dynamics evolve in the “direction” of increasing time. Thus it is where a characteristic line enters the ⊂-shaped region of Fig. 14.4.1b and goes to the right that the invariant for that line is set. •Second, the solution at a given point of intersection for the lines α= con- stant and β= constant are found in accordance with (1) and (2). This second step can be pictured as in Fig. 14.4.1b. In physical terms, the total voltage or Sec. 14.4 Transients on Bounded Lines 21 Fig. 14.4.2 Characteristic lines originating on initial conditions. Fig. 14.4.3 Characteristic line originating on load. current is the superposition of traveling waves propagating along the charac- teristic lines that intersect at the point of interest. To complete the ﬁrst step, note that a characteristic line passing through a given point Phas three possible origins. First, it can originate on the t= 0 axis, in which case the invariants, V±, are determined by the initial conditions. This was the only possibility on the inﬁnite transmission line considered in Sec. 14.3. The initial voltage and current where the characteristic line originates when t= 0 in Fig. 14.4.2 is used to evaluate (1) and (2), and the simultaneous solution of these expressions then gives the desired invariants. V+=1 2(Vi+ZoIi) (4) V−=1 2(Vi−ZoIi) (5) The second origin of a characteristic line is the boundary at z=l, as shown in Fig. 14.4.3. In particular, we consider the load resistance RLas the termination that imposes the boundary condition V(l, t) =RLI(l, t) (6) 22 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.4.4 Characteristic lines originating on generator end of line. The problems will illustrate how the same approach illustrated here can also be used to describe terminations composed of arbitrary circuits. Certainly, the case where the load is a pure resistance is the most important type of termination, for reasons that will be clear shortly. Again, because phenomena proceed in the + t“direction,” the incident wave V+and the boundary condition at z=lconspire to determine the reﬂected wave V−on the characteristic line β= constant originating on the boundary at z=l (Fig. 14.4.3). To say this mathematically, we substitute (1) and (2) into (6) V++V−=RL Zo(V+−V−) (7) and solve for V−. V−=V+ΓL; Γ L≡¡RL Zo−1¢ ¡RL Zo+ 1¢ (8) Here, V+andV−are evaluated at z=l, and hence with α=l−ctandβ=l+ct. Given the incident wave V+, we multiply it by the reﬂection coeﬃcient ΓLand determine V−. The third possible origin of a characteristic line passing through the given point Pis on the boundary at z= 0, as shown in Fig. 14.4.4. Here the line has been terminated in a source modeled as an ideal voltage source, Vg(t), in series with a resistance Rg. In this case, it is the wave traveling in the −zdirection (represented byV−and incident on the boundary from the left in Fig. 14.4.4) that combines with the boundary condition there to determine the reﬂected wave V+. The boundary condition is the constraint of the circuit on the voltage and current at the terminals. V(0, t) =Vg−RgI(0, t) (9) Substitution of (1) and (2) then gives an expression that can be solved for V+, given V−andVg(t). Sec. 14.4 Transients on Bounded Lines 23 V+=Vg Rg Zo+ 1+V−Γg; Γ g≡¡Rg Zo−1¢ ¡Rg Zo+ 1¢ (10) The following examples illustrate the two steps necessary to determine the transient response. First, V±are found over the range of time of interest using the initial conditions [(4) and (5) and Fig. 14.4.2] and boundary conditions [(8) and Fig. 14.4.3 and (10) and Fig. 14.4.4]. Then, the wave-components are superimposed to ﬁndVandI((1) and (2) and Fig. 14.4.1.) To appreciate the space-time signiﬁcance of the equations used in this process, it is helpful to have in mind the associated z−tsketches. Matching. The reﬂection of waves from the terminations of a line results in responses that can persist long after a signal has propagated the length of the transmission line. As a practical matter, it is therefore often desirable to eliminate reﬂections by matching the line. From (8), it follows that wave reﬂection is eliminated at the load by making the load resistance equal to the characteristic impedance of the line, RL=Zo. Similarly, from (10), there will be no reﬂection of the wave V−at the source if the resistance Rgis made equal to Zo. Consider ﬁrst an example in which the response is made simple because the line is matched to its load. Example 14.4.1. Matching In the conﬁguration shown in Fig. 14.4.5a, the load has a resistance RLwhile the generator is an ideal voltage source Vg(t) in series with the resistor Rg. The load is matched to the line, RL=Zo. As a result, according to (8), there are no V−waves on characteristics originating at the load. RL=Zo⇒V−= 0 (11) Suppose that the driving voltage consists of a pulse of amplitude Vpand duration T, as shown in Fig. 14.4.5b. Further, suppose that when t= 0 the line voltage and current are both zero, Vi= 0, and Ii= 0. Then, it follows from (4) and (5) that V+andV−are both zero on the respective characteristic lines originating on the t= 0 axis, as shown in Fig. 14.4.5b. By design, (8) gives V−= 0 for the β= constant characteristics originating at the load. Finally, because V−= 0 for all characteristic lines incident on the source (whether they originate on the initial conditions or on the load), it follows from (10) that on characteristic lines originating atz= 0, V+is as shown in Fig. 14.4.5b. We now know V+andV−everywhere. It follows from (1) and (2) that VandIare as shown in Fig. 14.4.5. Because V−= 0, the voltage and current both take the form of a pulse of temporal duration Tand spatial length cT, propagating from source to load with the velocity c. To express analytically what has been found, we know that at z= 0,V−= 0 and in turn from (10) that at z= 0, V+=Vg(t)¡Rg Zo+ 1¢ (12) 24 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.4.5 (a) Matched line. (b) Wave components in z−tplane. (c) Response in z−tplane. This is the value of V+along any line of constant αoriginating on the z= 0 axis. For example, along the line α=−ct/primepassing through the z= 0 axis when t=t/prime, V+=Vg(t/prime)¡Rg Zo+ 1¢ (13) We can express this result in terms of z−tby introducing α=−ct/primeinto (3), solving that expression for t/prime, and introducing that expression for t/primein (13). The result is just what we have already pictured in Fig. 14.4.5. V(z, t) =V+(α) =Vg¡ t−z c¢ ¡Rg Zo+ 1¢ (14) Regardless of the shape of the voltage pulse, it appears undistorted at some location zbut delayed by z/c. Note that at any location on the matched line, including the terminals of the generator, V/I =Zo. The matched line appears to the generator as a resistance equal to the characteristic impedance of the line. We have assumed in this example that the initial voltage and current are zero over the length of the line. If there were ﬁnite initial conditions, their response with the generator voltage set equal to zero would add to that obtained here because the wave equation is linear and superposition holds. Initial conditions give rise to waves V+andV−propagating in the + zand−zdirections, respectively. However, because there are no reﬂected waves at the load, the eﬀect of the initial conditions could not last longer at the generator than the time l/crequired for V−to reach z= 0 from Sec. 14.4 Transients on Bounded Lines 25 Fig. 14.4.6 (a) Open line. (b) Wave components in z−tplane. (c) Response in z−tplane. z=l. They would not last longer at the load than the time 2 l/c, when any resulting wave reﬂected from the generator would return to the load. Open circuit and short circuit terminations result in complete reﬂection. For the open circuit, I= 0 at the termination, and it follows from (2) that V+=V−. For a short, V= 0, and (1) requires that V+=−V−. Note that these limiting relations follow from (8) by making RLinﬁnite and zero in the respective cases. In the following example, we see that an open circuit termination can result in a voltage that is momentarily as much as twice that of the generator. Example 14.4.2. Open Circuit Termination The transmission line of Fig. 14.4.6 is terminated in an inﬁnite load resistance and driven by a generator modeled as a voltage source in series with a resistance Rg equal to the characteristic impedance Zo. As in the previous example, the driving voltage is a pulse of time duration T, as shown in Fig. 14.4.6b. When t= 0, V andIare zero. In this example, we illustrate the eﬀect of matching the generator resistance to the line and of having complete reﬂection at the load. The boundary at z=l, (8), requires that V−=V+ (15) while that at the generator, (10), is simply V+=Vg 2(16) Because the generator is matched, this latter condition establishes V+on character- istic lines originating on the z= 0 axis without regard for V−. These are summarized along the taxis in Fig. 14.4.6b. 26 One-Dimensional Wave Dynamics Chapter 14 To establish the values of V−on characteristic lines originating at the load, we must know values of the incident V+. Because IandVare both initially zero, the incident V+at the load is zero until t= l/c. From (15), V−is also zero. From t=l/cuntil t=l/c+T, the incident V+=Vp/2 and V−on the characteristic lines originating at the open circuit during this time interval follows from (15) as Vp/2. Finally, for all greater times, the incident wave is zero at the load and so also is the reﬂected wave. The values of V±for characteristic lines originating on the associated segments of the boundaries and on the t= 0 axis are summarized in Fig. 14.4.6c. With the values of V±determined, we now use (1) and (2) to make the picture also shown in Fig. 14.4.6c of the distributions of VandIat progressive instants in time. Because of the matched condition at the generator, the transient is over by the time the pulse has made one round trip. To make the current at the open circuit termination zero, the voltage doubles during that period when both incident and reﬂected waves exist at the termination. The conﬁguration of Fig. 14.4.6 was regarded in the previous example as an “open circuit transmission line” driven by a voltage source in series with a resistor. If we had been given the same conﬁguration in Chap. 7, we would have taken it to be a “capacitor” in series with the resistor and the voltage source. The next example puts the EQS approximation in perspective by showing how it represents the dynamics when the resistance Rgis large compared to Zo. A clue as to what happens when this ratio is large comes from writing it in the form Rg Zo=Rgp C/L =RgCl l√ CL=RgCl (l/c)(17) Here, RgClis the charging time of the capacitor and l/cis the electromagnetic wave transit time. When this ratio is large, the time for the transient to complete itself is many wave transit times. Thus, as will now be seen, the exponential charging of the capacitor is made up of many small steps associated with the electromagnetic wave passing “to and fro” over the length of the line. Example 14.4.3. Quasistatic Transient as the Limit of an Electrodynamic Transient The transmission line shown to the left in Fig. 14.4.7 is open at z=land driven atz= 0 by a step in voltage, Vg=Vpu−1(t). We are especially interested in the response with the series resistance, Rg, very large compared to Zo. For simplicity, we assume that the initial voltage and current are zero. The boundary condition imposed at the open termination, where z=l, is I= 0. From (2), V+=V− (18) while at the source, (10) pertains with Vg=Vpa constant5 V+=Vg+V−Γg; Vg≡Vp¡Rg Zo+ 1¢ (19) 5Be careful to distinguish the constant Vgas deﬁned in this example from the source voltage Vg(t) =VpU−1(t). Sec. 14.4 Transients on Bounded Lines 27 Fig. 14.4.7 Wave components of open line to a step in voltage in series with a high resistance. with the reﬂection coeﬃcient of the generator deﬁned as Γg≡Rg Zo−1 Rg Zo+ 1(20) Starting with characteristic lines originating at t= 0, where the initial conditions determine that V+andV−are zero, we can now use these boundary conditions to determine V−on lines originating at the load and V+on lines originating at z= 0. These values are shown in Fig. 14.4.7. Thus, V±are now known everywhere in the ⊂-shaped region. The voltage and current now follow from (1) and (2). In particular, consider the response at the generator terminals, where z= 0. In Fig. 14.4.7, the taxis has been divided into intervals of duration 2 l/c, the ﬁrst denoted by N= 1, the second byN= 2, etc. We have found that the wave components incident on and reﬂected from the z= 0 boundary in the N-th interval are V−=VgN−2X n=0Γn g (21) V+=VgN−1X n=0Γn g (22) It follows from (2) that the current at z= 0 during this time interval is I(0, t) =Vp Rg1 1 +Zo RgΓN−1 g; 2( N−1)l c< t < 2Nl c(23) 28 One-Dimensional Wave Dynamics Chapter 14 In turn, this current can be used to evaluate the terminal voltage. V(0, t) =Vpµ 1−ΓN−1 g 1 +Zo Rg¶ (24) With Rg/Zovery large, it follows from (20) that Γg→¡ 1−2Zo Rg¢ (25) In this same limit, the term 1+ Zo/Rgin (24) is essentially unity. Thus, (24) becomes approximately V(0, t)→Vp· 1−¡ 1−2Zo Rg¢N−1¸ ; 2( N−1)l c< t <2Nl c(26) We suspect that in the limit where the round-trip transit time 2 l/cis short compared to the charging time τ=RgCl, this voltage becomes the step response of the series capacitor and resistor. V(0, t)→Vp(1−e−t/τ] (27) To see that this is indeed the case, we exploit the fact that lim x→0(1−x)1/x=e−1(28) by writing (26) in the form V(0, t)→Vp½ 1−·¡ 1−2Zo Rg¢1/(2Zo/Rg)¸(N−1)(2Zo/Rg)¾ (29) It follows that in the limit where Zo/Rgis small, V(0, t)→Vp£ 1−e−(N−1)(2Zo/Rg)¤ ; 2( N−1)l c< t < 2Nl c(30) Remember that Nrepresents the interval of time during which the expression is valid. If we take the time as being that when the interval begins, then 2(N−1)l c∼t⇒2(N−1) =t (l/c)(31) Substitution of this expression for 2( N−1) into (30) and use of (17) then shows that in this high-resistance limit, the voltage does indeed take the exponential form for a charging capacitor, (27), with a charging time τ=RgCl. In the example of V(0, t) shown in Fig. 14.4.8, there are 10 round-trip transit times in one charging time, RgCl= 20l/c. Sec. 14.4 Transients on Bounded Lines 29 Fig. 14.4.8 Response of open circuit transmission line to step in voltage in series with a high resistance. The smooth curve is predicted by the EQS model. Fig. 14.4.9 Oscilloscope displays voltage at terminals of line under conditions of Examples 14.4.1-3. The following demonstration is typical of a variety of demonstrations that are easily carried out using a good oscilloscope and a stretch of transmission line. Demonstration 14.4.1. Transmission Line Matching, Reﬂection, and Qua- sistatic Charging The apparatus shown in Fig. 14.4.9 is all that is required to demonstrate the phenomena described in the examples. In a typical experiment, a 10 m length of cable is used, in which case the wave transit time is about 0.05 µs. Thus, to resolve the transient, the oscilloscope must have a frequency response that extends to 100 MHz. To achieve matching of the generator, as called for in Example 14.4.2, Rg=Zo. Typically, for a coaxial cable, this is 50 Ω. 30 One-Dimensional Wave Dynamics Chapter 14 To see the charging transient of Example 14.4.3 with 10 round trip transit times in the capacitive charging time, it follows from (17) that we should make Rg/Zo= 20. Thus, for a coaxial cable having Zo= 50Ω , Rg= 1kΩ. 14.5 TRANSMISSION LINES IN THE SINUSOIDAL STEADY STATE The method used in Sec. 14.4 is equally applicable to ﬁnding the response to a sinusoidal excitation of an ideal transmission line. Rather than exciting the line by a voltage step or a voltage pulse, as in the examples of Sec. 14.4, the source may produce a sinusoidal excitation. In that case, there is a part of the response that is in the sinusoidal steady state and a part that accounts for the initial conditions and the transient associated with turning on the source. Provided that the boundary conditions are (like the transmission line equations) linear, we can express the response as a superposition of these two parts. V(z, t) =Vs(z, t) +Vt(z, t) (1) Here, Vsis the sinusoidal steady state response, determined without regard for the initial conditions but satisfying the boundary conditions. Added to this to make the total solution satisfy the initial conditions is Vt. This transient solution is deﬁned to satisfy the boundary conditions with the drive equal to zero and to make the total solution satisfy the initial conditions. If we were interested in it, this transient solution could be found using the methods of the previous section. In an actual physical situation, this part of the solution is usually dissipated in the resistances of the terminations and the line itself. Then the sinusoidal steady state prevails. In this and the next section, we focus on this part of the solution. With the understanding that the boundary conditions, like those describing the transmission line, are linear diﬀerential equations with constant coeﬃcients, the response will be sinusoidal and at the same frequency, ω, as the drive. Thus, we assume at the outset that V= Re ˆV(z)ejωt; I= Re ˆI(z)ejωt(2) Substitution of these expressions into the transmission line equations, (14.1.4)– (14.1.5), shows that the zdependence is governed by the ordinary diﬀerential equa- tions dˆI dz=−jωCˆV(3) dˆV dz=−jωLˆI(4) Sec. 14.5 Sinusoidal Steady State 31 Fig. 14.5.1 Termination at z= 0 in load impedance. Again because of the constant coeﬃcients, these linear equations have two solutions, each having the form exp( −jkz). Substitution shows that ˆV=ˆV+e−jβz+ˆV−ejβz(5) where β≡ω√ LC. In terms of the same two arbitrary complex coeﬃcients, it also follows from substitution of this expression into (14.1.5) that ˆI=1 Zo¡ˆV+e−jβz−ˆV−ejβz¢ (6) where Zo=p L/C. What we have found are solutions having the same traveling wave forms as identiﬁed in Sec. 14.3, (14.3.9)–(14.3.10). This can be seen by using (2) to recover the time dependence and writing these two expressions as V= Re· ˆV+e−jβ¡ z−ω βt¢ +ˆV−ejβ¡ z+ω βt¢¸ (7) I= Re1 Zo· ˆV+e−jβ¡ z−ω βt¢ −ˆV−ejβ¡ z+ω βt¢¸ (8) The velocity of the waves is ±ω/β= 1/√ LC. Because the coeﬃcients ˆV±are com- plex, they represent both the amplitude and phase of these traveling waves. Thus, the solutions could be sinusoids, cosinusoids, or any combination of these having the given arguments. In working with standing waves in Sec. 13.2, we demonstrated how the coeﬃcients could be adjusted to satisfy simple boundary conditions. Here we introduce a point of view that is convenient in dealing with complicated termi- nations. Transmission Line Impedance. The transmission line shown in Fig. 14.5.1 is terminated in a load impedance ZL. By deﬁnition, ZLis the complex number ˆV(0) ˆI(0)=ZL (9) 32 One-Dimensional Wave Dynamics Chapter 14 In general, it could represent any linear system composed of resistors, in- ductors, and capacitors. The complex amplitudes ˆV±are determined by this and another boundary condition. This second condition represents the termination of the line somewhere to the left in Fig. 14.5.1. At any location on the line, the impedance is found by taking the ratio of (5) and (6). Z(z)≡ˆV(z) ˆI(z)=Zo1 + Γ Le2jβz 1−ΓLe2jβz(10) Here, Γ Lis the reﬂection coeﬃcient of the load. ΓL≡ˆV− ˆV+ (11) Thus, Γ Lis simply the ratio of the complex amplitudes of the traveling wave com- ponents. At the location z= 0, where the line is connected to the load and (9) applies, this expression becomes ZL Zo=1 + Γ L 1−ΓL (12) The boundary condition, expressed by (12), is suﬃcient to determine the reﬂection coeﬃcient. That is, from (12) it follows that ΓL=(ZL/Zo−1) (ZL/Zo+ 1)(13) Given the load impedance, Γ Lfollows from this expression. The line impedance at a location zto the left then follows from the use of this expression to evaluate (10). The following examples lead to important implications of (11) while indicating the usefulness of the impedance point of view. Example 14.5.1. Impedance Matching Given an incident wave V+, how can we eliminate the reﬂected wave represented byV−? By deﬁnition, there is no reﬂected wave if the reﬂection coeﬃcient, (11), is zero. It follows from (13) that ΓL= 0⇒ZL=Zo (14) Note that Zois real, which means that the matched load is equivalent to a resistance, RL=Zo. Thus, our ﬁnding is consistent with that of Sec. 14.4, where we found that Sec. 14.5 Sinusoidal Steady State 33 such a termination would eliminate the reﬂected wave, sinusoidal steady state or not. It follows from (10) that the line has the same impedance, Zo, at any location z, when terminated in its characteristic impedance. Because V−=0, it follows from (7) that the voltage takes the form V= Re ˆV+ej(ωt−βz)(15) The voltage has the distribution in space and time of a sinusoid traveling in the z direction with the velocity 1 /√ LC. At any given location, the voltage is sinusoidal in time at the (angular) frequency ω. The amplitude is the same, regardless of z.6 The previous example illustrated that at any location, a transmission line terminated in a resistance equal to its characteristic impedance has an impedance which is also resistive and equal to Zo. The next example illustrates what happens in the opposite extreme, where the termination dissipates no energy and the response is a pure standing wave rather than the pure traveling wave of the matched line. Example 14.5.2. Short Circuit Impedance and Standing Waves With a short circuit at z= 0, (5) makes it clear that V−=−V+. Thus, the reﬂection coeﬃcient deﬁned by (11) is Γ L=−1. We come to the same conclusion from the evaluation of (13). ZL= 0⇒ΓL=−1 (16) The impedance at some location z then follows from (10) as Z(−l) Zo≡jX Zo=jtanβl (17) In view of the deﬁnition of β, βl=ωl c= 2πl λ(18) and so we can think of βlas being proportional either to the frequency or to the length of the line measured in wavelengths λ. The impedance of the line is a reactance Xhaving the dependence on either of these quantities shown in Fig. 14.5.2. At low frequencies (or for a length that is short compared to a quarter- wavelength), Xis positive and proportional to ω. As should be expected from either Chap. 8 or Example 13.1.1, the reactance is that of an inductor. βl/lessmuch1⇒X→(βl)p L/C =ωLl (19) As the frequency is raised to the point where the line is a quarter-wavelength long, the impedance is inﬁnite. A shorted quarter-wavelength line has the impedance of an open circuit! As the frequency is raised still further, the reactance becomes ca- pacitive, decreasing with increasing frequency until the half-wavelength line exhibits 6By contrast with Demonstration 13.1.1, where the light emitted by the ﬂuorescent tube indicated that the electric ﬁeld peaked at some locations and nulled at others, the distribution of light for a matched line would be “ﬂat.” 34 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.5.2 Reactance as a function of normalized frequency for a shorted line. Fig. 14.5.3 A quarter-wave matching section. the impedance of the termination, a short. That the impedance repeats itself as the line is increased in length by a half-wavelength is evident from Fig. 14.5.2. We consider next an example that illustrates one of many methods for match- ing a load resistance RLto a line having a characteristic impedance not equal to RL. Example 14.5.3. Quarter-Wave Matching Section A quarter-wavelength line, as shown in Fig. 14.5.3, has the useful property of converting a normalized load impedance ZL/Zoto a normalized impedance that is the reciprocal of that impedance, Zo/ZL. To see this, we evaluate the impedance, (10), a quarter-wavelength from the load, where βz=−π/2, and then use (12). Z¡ βz=−π 2¢ =Z2 o ZL(20) Thus, if we wanted to match a line having the characteristic impedance Za oto a load resistance ZL=RL, we could interpose a quarter-wavelength section of line having as its characteristic impedance a Zothat is the geometric mean of the load resistance and the characteristic impedance of the line to be matched. Zo=p ZaoRL (21) The idea of using quarter-wavelength sections to achieve matching will be continued in the next example. The transmission line model is equally well applicable to electromagnetic plane waves. The equivalence was pointed out in Sec. 14.1. When these waves are opti- cal, the permeability of common materials remains µo, and the polarizability is Sec. 14.5 Sinusoidal Steady State 35 Fig. 14.5.4 (a) Cascaded quarter-wave transmission line sections. (b) Optical coating represented by (a). described by the index of refraction ,n, deﬁned such that D=n2/epsilon1oE (22) Thus, n2/epsilon1otakes the place of the dielectric constant, /epsilon1. The appropriate value of n2/epsilon1ois likely to be very diﬀerent from the value of /epsilon1used for the same material at low frequencies.7 The following example illustrates the application of the transmission line view- point to an optical problem. Example 14.5.4. Quarter-Wave Cascades for Reduction of Reﬂection When onequarter-wavelength line is used to transform from one speciﬁed impedance to another, it is necessary to specify the characteristic impedance of the quarter- wave section. In optics, where it is desirable to minimize reﬂections that result from the passage of light from one transparent medium to another, it is necessary to specify the index of refraction of the quarter wave section. Given other constraints on the materials, this often is not possible. In this example, we see how the use of multiple layers gives some ﬂexibility in the choice of materials. The matching section of Fig. 14.5.4a consists of mpairs of quarter-wave sec- tions of transmission line, respectively, having characteristic impedances Za oandZb o. This represents equally well the cascaded pairs of quarter wave layers of dielectric shown in Fig. 14.5.4b, interposed between materials of dielectric constants /epsilon1and/epsilon1i. Alternatively, these layers are represented by their indices of refraction, naandnb, interposed between materials having indices niandn. First, we picture the matching problem in terms of the transmission line. The load resistance RLrepresents the material to the right of the cascade. This region is pictured as an inﬁnite transmission line having characteristic impedance Zo. Thus, it presents a load to the cascade of resistance RL=Zo. To determine the impedance at the other side of the cascade, we make repeated use of the impedance transformation for a quarter-wave section, (20). To begin with, the impedance at the terminals of the ﬁrst quarter-wave section is Z=(Za o)2 Zo(23) 7With ﬁelds described in the frequency domain, /epsilon1, and hence n2, are in general complex functions of frequency, as in Sec. 11.5. 36 One-Dimensional Wave Dynamics Chapter 14 With this taken as the load resistance in (20), the impedance at the terminals of the second section is Z=³Zb o Zao´2 Zo (24) This can now be regarded as the impedance transformation for the pair of quarter- wave sections. If we now make repeated use of (24) to represent the impedance trans- formation for the quarter-wave sections taken in pairs, we ﬁnd that the impedance at the terminals of mpairs is Z=³Zb o Zao´2m Zo (25) Now, to apply this result to the optics conﬁguration, we identify (14.1.9) RL→p µo//epsilon1≡ζL=ζo n; Za o=p µo//epsilon1a≡ζa=ζo na; (26) Zb o=p µo//epsilon1b≡ζb=ζo nb and have from (25) for the intrinsic impedance of the cascade ζ=ζL¡ζb ζa¢2m(27) In terms of the indices of refraction, n ni=³na nb´2m (28) If this condition on the optical properties and number of the layer pairs is fulﬁlled, the wave can propagate through the interface between regions of indices niandn without reﬂection. Given materials having na/nbless than n/ni, it is possible to pick the number of layer pairs, m, to satisfy the condition (at least approximately). Coatings are commonly used on lenses to prevent reﬂection. In such appli- cations, the waves processed by the lens generally have a spectrum of frequencies. Thus, optimization of the matching coatings is more complex than pictured here, where it has been assumed that the light is at a single frequency (is monochromatic). It has been assumed here that the electromagnetic wave has normal incidence at the dielectric interface. Waves arriving at the interface at an angle can also be pictured in terms of the transmission line. In practical applications, the design of lens coatings to prevent reﬂection over a range of angles of incidence is a further complication.8 8H. A. Haus, Waves and Fields in Optoelectronics , Prentice-Hall, Inc., Englewood Cliﬀs, N.J. (1984), pp. 43-46. Sec. 14.6 Reﬂection Coeﬃcient 37 Fig. 14.6.1 (a)Transmission line conventions. (b) Reﬂection coeﬃcient de- pendence on zin the complex Γ plane. 14.6 REFLECTION COEFFICIENT REPRESENTATION OF TRANSMISSION LINES In Sec. 14.5, we found that a quarter-wavelength of transmission line turned a short circuit into an open circuit. Indeed, with an appropriate length (or driven at an appropriate frequency), the shorted line could have an inductive or a capacitive reactance. In general, the impedance observed at the terminals of a transmission line has a more complicated dependence on the termination. Typical microwave measurements are made with a length of transmission line between the observation point and the terminals of the device under study, whether that be an antenna or a transistor. In this section, the objective is a way of visu- alizing the relation between the impedance at the “generator” terminals and the impedance of the “load.” We will ﬁnd that a representation of the variables in the reﬂection coeﬃcient plane is valuable both conceptually and practically. At a location z, the impedance of the transmission line shown in Fig. 14.6.1a is (14.5.10) Z(z) Zo=1 + Γ( z) 1−Γ(z) (1) where the reﬂection coeﬃcient at the location zis deﬁned as the complex function Γ(z) =ˆV− ˆV+ej2βz (2) At the load position, where z= 0, the reﬂection coeﬃcient is equal to Γ Las deﬁned by (14.5.11). Like the impedance, the reﬂection coeﬃcient is a function of z. Unlike the impedance, Γ has an easily pictured zdependence. Regardless of z, the magnitude of Γ is the same. Thus, as pictured in the complex Γ plane of Fig. 14.6.1b, it is a complex vector of magnitude |ˆV−/ˆV+|and angle θ+ 2βz, where θis the angle at 38 One-Dimensional Wave Dynamics Chapter 14 the position z= 0. With zdeﬁned as increasing from the generator to the load, the dependence of the reﬂection coeﬃcient on zis as summarized in the ﬁgure. As we move from the generator toward the load, zincreases and hence Γ rotates in the counterclockwise direction. In summary, once the complex number Γ is established at one location z, its variation as we move toward the load or toward the generator can be pictured as a rotation at constant magnitude in the counterclockwise or clockwise direc- tions, respectively. Typically, Γ is established at the location of the load, where the impedance, ZL, is known. Then Γ at any location zfollows from (1) solved for Γ. Γ =¡Z Zo−1¢ ¡Z Zo+ 1¢ (3) With the magnitude and phase of Γ established at the load, the reﬂection coeﬃcient can be found at another location by a simple rotation through an angle 4π(z/λ), as shown in Fig. 14.6.1b. The impedance at this second location would then follow from evaluation of (1). Smith Chart. We save ourselves the trouble of evaluating (1) or (3), either to establish Γ at the load or to infer the impedance implied by Γ at some other location, by mapping Z/Z oin the Γ plane of Fig. 14.6.1b. To this end, we deﬁne the normalized impedance as having a resistive part rand a reactive part x Z Zo=r+jx (4) and plot the contours of constant rand of constant xin the Γ plane. This makes it possible to see directly what Zis implied by each value of Γ. Eﬀectively, such a mapping provides a graphical solution of (1). The next few steps summarize how this mapping of the contours of constant randxin the Γ r−Γiplane can be made with ruler and compass. First, (1) is written using (4) on the left and Γ = Γ r+jΓion the right. The real and imaginary parts of this equation must be equal, so it follows that r=(1−Γ2 r−Γ2 i) (1−Γr)2+ Γ2 i(5) x=2Γi (1−Γr)2+ Γ2 i(6) These expressions are quadratic in Γ rand Γ i. By completing the squares, they can be written as³ Γr−r r+ 1´2 + Γ2 i=³1 1 +r´2 (7) (Γr−1)2+¡ Γi−1 x¢2=¡1 x¢2(8) Sec. 14.6 Reﬂection Coeﬃcient 39 Fig. 14.6.2 (a) Circle of constant normalized resistance, r, in Γ plane. (b) Circle of constant normalized reactance, x, in Γ plane. Fig. 14.6.3 Smith chart. Thus, the contours of constant normalized resistance, r, and of constant normalized reactance, x, are the circles shown in Figs. 14.6.2a–14.6.2b. Putting these contours together gives the lines of constant randxin the complex Γ plane shown in Fig. 14.6.3. This is called a Smith chart . Illustration. Impedance with Simple Terminations How do we interpret the examples of Sec. 14.5 in terms of the Smith chart? •Quarter-wave Section. In Example 14.5.3 we found that a normalized re- sistive load rLwas transformed into its reciprocal by a quarter-wave line. Suppose that rL= 2 (the load resistance is 2 Zo) and x= 0. Then, the load is 40 One-Dimensional Wave Dynamics Chapter 14 atAin Fig. 14.6.3. A quarter-wavelength toward the generator is a rotation of 180 degrees in a clockwise direction, with Γ following the trajectory from A→Bin Fig. 14.6.3. Note that the impedance at Bis indeed the reciprocal of that at A, r= 0.5, x= 0. •Impedance of Short Circuit Line. Consider next the shorted line of Ex- ample 14.5.2. The load resistance rLis 0, and reactance xLis 0 as well, so we begin at the point Cin Fig. 14.6.3. Now, we can trace out the impedance as we move away from the short toward the generator by rotating along the trajectory of unit radius in the clockwise direction. Note that all along this trajectory, r= 0. The normalized reactance then traces out the values given in Fig. 14.5.2, ﬁrst taking on positive (inductive) values until it becomes inﬁnite atλ/4 (rotation of 180 degrees), and then negative (capacitive) values until it returns to C, when the line has a length of λ/2. •Matched Line. For the matched load of Example 14.5.1, we start out with rL= 1 and xL= 0. This is point Dat the origin in Fig. 14.6.3. Thus, the trajectory of Γ is a circle of zero radius, and the impedance remains rL= 1 over the length of the line. While taking measurements on a transmission line terminated in a particular device, the Smith chart is often used to have an immediate picture of the impedance at the terminals. Even though the chart could be replaced by a programmable calculator, the overview provided by the Smith chart is important. Not only does it provide insight concerning the impedance, it can be used to picture the spatial evolution of the voltage and current, as we now see. Standing Wave Ratio. Once the reﬂection coeﬃcient has been established, the voltage and current distributions are determined (to within a factor determined by the source). That is, in terms of Γ, (14.5.5) becomes ˆV=ˆV+e−jβz[1 + Γ( z)] (9) The exponential factor has an amplitude that is independent of z. Thus, [1 + Γ( z)] represents the zdependence of the voltage amplitude. This complex quantity can be pictured in the Γ plane as shown in Fig. 14.6.4a. Remember, as we move from load to generator, Γ rotates in the clockwise direction. As it does so, 1+Γ varies between a maximum value of 1 + |Γ|and a minimum value of 1 − |Γ|. According to (9), we can now picture the spatial distribution of the voltage amplitude. Convenient for describing this distribution is the voltage standing wave ratio (VSWR), deﬁned as the ratio of the maximum voltage amplitude to the minimum voltage amplitude. From Fig. 14.6.4a, we can see that this ratio is VSWR =(1 + Γ) max (1 + Γ) min=1 +|Γ| 1− |Γ| (10) The distribution of voltage amplitude is shown for several VSWR’s in Fig. 14.6.4b. We have already seen such distributions in two extremes. With the short Sec. 14.6 Reﬂection Coeﬃcient 41 Fig. 14.6.4 (a) Normalized line voltage 1 + Γ. (b) Distribution of voltage amplitude for three VSWR’s. circuit or open circuit terminations considered in Sec. 13.1, the reﬂection coeﬃcient was on the unit circle and the VSWR was inﬁnite. Indeed, the inﬁnite VSWR envelope of Fig. 14.6.4b is that of a standing wave, with nulls every half-wavelength. The opposite extreme is also familiar. Here, the line is matched and the reﬂection coeﬃcient is on a circle of zero radius. Thus, the VSWR is unity and the distribution of voltage amplitude is uniform. Measurement of the VSWR and the location of a voltage null provides the information needed to determine a line termination. This follows by ﬁrst using (10) to evaluate the magnitude of the reﬂection coeﬃcient from the measured VSWR. |Γ|=VSWR −1 VSWR + 1(11) Thus, the radius of the circle representing the voltage distribution on the line has been determined. Second, a determination of the position of a null is tantamount to locating (to within a half-wavelength) the position on the line where Γ passes through the negative real axis. The distance from this point to the load, in wave- lengths, then determines where the load is located on this circle. The corresponding impedance is that of the load. Demonstration 14.6.1. VSWR and Load Impedance In the slotted line shown in Fig. 14.6.5, a movable probe with its attached detector provides a measure of the line voltage as a function of z. The distance between the load and the voltage probe can be measured directly. By using a frequency of 3 GHz and an air-insulated cable (having a permittivity that is essentially that of free space, so that the wave velocity is 3 ×108m/s), the wavelength is conveniently 10 cm. The characteristic impedance of the coaxial cable is 50 Ω, so with terminations of 50 Ω, 100 Ω, and a short, the observed distribution of voltage is as shown in Fig. 14.6.4b for VSWR’s of 1, 2, and ∞. (To plot data points on these curves, the 42 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.6.5 Demonstration of distribution of voltage magnitude as function of VSWR. measured values should be normalized to match the peak voltage of the appropriate distribution.) Figure 14.6.5 illustrates how a measurement of the VSWR and position of a null can be used to infer the termination. Addition of a half-wavelength to lmeans an additional revolution in the Γ plane, so which null is used to deﬁne the distance lmakes no diﬀerence. The trajectory drawn in the illustration is for the 100 Ω termination. Admittance in the Reﬂection Coeﬃcient Plane. Commonly, transmission lines are interconnected in parallel. It is then convenient to work with the ad- mittance rather than the impedance. The Smith chart describes equally well the evolution of the admittance with z. With Yo= 1/Zodeﬁned as the characteristic admittance, it follows from (1) thatY Yo=1−Γ 1 + Γ(12) If Γ→ − Γ, this expression becomes identical to that relating the normalized impedance to Γ, (1). Thus, the contours of constant normalized conductance, g, and normalized susceptance, y, Y Yo≡g+jy (13) are those of the normalized impedance, randx, rotated by 180 degrees. Rotate by 180 degrees the impedance form of the Smith chart and the admittance form is obtained! The contours of randx, respectively, become those of gandy.9 9Usually, Γ is not explicitly evaluated. Rather, the admittance is given at one point on the Γ circle (and hence on the chart) and determined (by a rotation through the appropriate angle on the chart) at another point. Thus, for most applications, the chart need not even be rotated. However, if Γ is to be evaluated directly from the admittance, it should be remembered that the coordinates are actually −Γrand−Γi. Sec. 14.6 Reﬂection Coeﬃcient 43 Fig. 14.6.6 (a) Single stub matching. (b) Admittance Smith chart. The admittance form of the Smith chart is used in the following example. Example 14.6.1. Single Stub Matching In Fig. 14.6.6a, the load admittance YLis to be matched to a transmission line having characteristic admittance Yoby means of a “stub” consisting of a shorted section of line having the same characteristic admittance Yo. Variables that can be used to accomplish the matching are the distance lfrom the load to the stub and the length lsof the stub. Matching is accomplished in two steps. First, the length lis adjusted so that the real part of the admittance at the position where the stub is attached is equal to Yo. Then the length of the shorted stub is adjusted so that it’s susceptance cancels that of the line. Here, we see the reason for using the admittance form of the Smith chart, shown in Fig. 14.6.6b. The stub and the line are connected in parallel so that their admittances add. The two steps are pictured in Fig. 14.6.6b for the case where the normalized load admittance is g+jy= 0.5, at Aon the chart. The real part of the admittance becomes equal to the characteristic admittance on the circle g= 1; we adjust the length lso that the stub is connected at B, where the |Γ|constant curve intersects theg= 1 circle. In the particular example shown, this length is l= 0.152λ. From the chart, one reads oﬀ a positive susceptibility at this point of about y= 0.7. We can determine the stub length lsthat gives the negative of this susceptance by again using the chart. The desired admittance of the stub is at C, where g= 0 and y= −0.7. In the case of the stub, the “load” is the short, where the admittance is inﬁnite, atDon the chart. Following the |Γ|= 1 circle in the clockwise direction (from the “load” toward the “generator”) from the short at Dto the desired admittance at C then gives the length of the stub. For the example, ls= 0.153λ. To the left of the point where the stub is attached, the line should have a unity VSWR. The following demonstrates this concept. Demonstration 14.6.2. Single Stub Matching 44 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.6.7 Single stub matching demonstration. Fig. 14.7.1 Incremental section of lossy distributed line. In Fig. 14.6.7, the previous demonstration has been terminated with an adjustable length of line (a line stretcher) and a stub. The slotted line makes it possible to see the eﬀect on the VSWR of matching the line. With a load of Z= 100Ω and the stub and line stretcher adjusted to the values found in the previous example, the voltage amplitude is found to be independent of the position of the probe in the slotted line. Of course, we can add a half-wavelength to either lorlsand obtain the same condition. 14.7 DISTRIBUTED PARAMETER EQUIVALENTS AND MODELS WITH DISSIPATION The distributed parameter transmission line of Sec. 14.1 is now generalized to include certain types of dissipation by using the incremental circuit shown in Fig. 14.7.1 . The capacitance per unit length Cis shunted by a conductance per unit length Gand in series with the inductance per unit length Lis the resistance per unit length R. If the line really were made up of so many lumped parameter elements that it could be described by continuum equations, Gwould be the conductance per unit length of the lossy capacitors (Sec. 7.9) and Rwould be the resistance per unit length of the inductors. More often, GandR(like LandC) are either equivalent to or a model of a physical system. Examples are discussed in the next two sections. Sec. 14.7 Equivalents and Models 45 The steps leading to the generalized transmission line equations are suggested by Eqs. 14.1.1 through 14.1.5. In requiring that the currents at the terminal on the right sum to zero, there is now an additional current through the shunt conductance. In the limit where ∆ z→0, ∂I ∂z=−C∂V ∂t−GV(1) Similarly, in summing the voltages around the loop, there is now a voltage drop across the series resistance. Again, in the limit ∆ z→0, ∂V ∂z=−L∂I ∂t−RI(2) As should be expected, with the introduction of dissipation represented by G andR, V (z, t) and I(z, t) no longer take the form of waves propagating without distortion. That is, substitution shows that solutions no longer take the form of (14.4.1)–(14.4.3). As a result, we would have to work considerably harder than in Secs. 14.3–14.4 to describe transients on lossy transmission lines. However, although somewhat more involved then before, the sinusoidal steady state response follows from the approach illustrated in Secs. 14.5–14.6. With the objective of describing the sinusoidal steady state, complex ampli- tude representations of VandI(14.5.2) are substituted into (1) and (2) to give dˆI dz=−(jωC+G)ˆV (3) dˆV dz=−(jωL+R)ˆI (4) To obtain an expression for the voltage alone, (3) is substituted into the derivative of (4). d2ˆV dz2−(jωL+R)(jωC+G)ˆV= 0 (5) With the voltage found from this equation, the current follows from (4). ˆI=−1 R+jωLdˆV dz(6) Albeit complex, the coeﬃcient in (5) is constant, so it is again appropriate to look for exponential solutions. Using the convention established in Sec. 14.5, we look for solutions exp( −jkz). Substitution into (5) then shows that k2=−(jωL+R)(jωC+G) (7) So as to be clear in distinguishing the two roots of this dispersion equation, we deﬁne βas having a positive real part and write the roots as 46 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.7.2 Roots of (7) as functions of normalized ω. The real and imag- inary parts, respectively, of the complex wave number βrelate to the phase velocity and rate of decay of the wave, as shown. k=±β; β≡p (ω2LC−RG)−jω(RC+LG); Re β >0 (8) These roots, k=kr+jki, are pictured as a function of frequency in Fig. 14.7.2. From (7), k2is in the lower half-plane. It follows that the value of khaving a positive real part (deﬁned as β) has a negative imaginary part. These deﬁnitions take on physical signiﬁcance when the solutions to (5) are written as ˆV=ˆV+e−jβz+ˆV−ejβz(9) because it is then clear that we have deﬁned βso that V+represents a wave with points of constant phase propagating in the + zdirection. Note that because βiis negative, this wave decays in the + zdirection, as shown in Fig. 14.7.2. Similarly, ˆV−is the complex amplitude at z= 0 of a wave that decays in the −zdirection and has phases propagating in the −zdirection.10 In terms of the two coeﬃcients ˆV±, the current expression follows from sub- stituting (9) into (6) ˆI=1 Zo(ˆV+e−jβz−ˆV−ejβz) (10) where the complex characteristic impedance is now deﬁned as 10It is important not to generalize from this ﬁnding. The direction of propagation of points of constant phase (the phase propagation direction, PPG, is not, in general, indicative of the directions of propagation of the wave; i.e., a source positioned on an inﬁnite line at z= 0 does not necessarily cause waves with positive PPG to go away from the source in the + zdirection and with negative PPG to go away in the −zdirection. The direction of propagation is usually determined by the direction of the group velocity (GV). In the presence of loss, waves with positive GV decay in the + z-direction, with negative GV in the −z-direction. Sec. 14.7 Equivalents and Models 47 Fig. 14.7.3 Open circuit lossy line. Zo≡(R+jωL) jβ (11) Comparison of (9) and (10) with (14.5.5) and (14.5.6) shows that the impedance and reﬂection coeﬃcient descriptions are applicable, provided we generalize βand Zoto be the complex numbers given by (8) and (11). That these quantities are now complex is an inconvenience11and a warning that some ideas established for the ideal line need to be reexamined. For example, reasoning as in Sec. 14.5 shows that a pure resistance can no longer be used to match the line. Further, it is not possible to match the line at all frequencies with any ﬁnite number of lumped elements. Example 14.7.1. Signal Attenuation on an Open Circuit Line The lossy transmission line shown in Fig. 14.7.3 is open at the right and driven by a voltage source of complex amplitude Vgat the left. What is the voltage measured at the open circuit? The open circuit at z= 0 requires that I(0) = 0, and hence [from (10)] that V+=V−. Thus, the voltage, as given by (9), is ˆV=ˆV+(e−jβz+ejβz) =ˆVg(e−jβz+ejβz) (ejβl+e−jβl)(12) Here we have adjusted the coeﬃcient ˆV+so that the voltage is ˆVgat the left end, where the voltage source is connected. As a function of time, the voltage distribution is therefore V(z, t) = Re ˆVg(e−jβz+ejβz) (ejβl+e−jβl)ejωt(13) Much of the complicated phenomenon represented by this simple expression is en- capsulated in the complex wave number. To illustrate, consider the voltage measured atz= 0, which from (13) has the complex amplitude ˆV(0) =ˆVg cosβl(14) If a calculator or computer is not available for evaluating the cosine of a complex number, then we can use the double-angle identity to write cosβl= cos( βrl+jβil) = cos βrlcoshβil−jsinβrlsinhβil (15) 11Circumvented by having a calculator programmed to carry out operations on complex variables. 48 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.7.4 Frequency response for open circuit lossy line. For the case where all of the loss is due to the shunt conductance ( R= 0), the frequency response is illustrated in Fig. 14.7.4. The four responses shown are for increasing amounts of loss, perhaps introduced by increasing G. At very low frequencies, the output voltage simply follows the driving voltage. This is because we have considered the case where R= 0 and with the frequency so low that the inductor has no eﬀect, the output terminals are connected to the source through a negligible impedance. As the frequency is raised, consider ﬁrst the response with no loss ( l/l∗= 0). As the frequency approaches that required to make the line a quarter-wavelength long, the impedance of the line at the generator approaches zero and the current approaches inﬁnity. This resonance condition results in the inﬁnite response at z= 0. (As the frequency is raised further, these resonance conditions occur each time the frequency is such that the line has a length equal to a quarter-wavelength plus a multiple of a half-wavelength.) With the addition of a slight amount of loss ( l/l∗= 0.1), the response is ﬁnite even under the resonance conditions. Further increasing the loss ( l/l∗= 1) results in a response with a dull peak at the resonance point. Still larger losses ( l/l∗= 10) bring in skin eﬀect and monotonic attenuation of the voltage over the length of the line. Phenomena underlying this response are discussed in the next section. 14.8 UNIFORM AND TEM WAVES IN OHMIC CONDUCTORS (R= 0) Transverse electromagnetic (TEM) waves that propagate in the zdirection and are polarized in the xdirection have electric and magnetic ﬁelds E=Ex(z, t)ix;H=Hy(z, t)iy (1) We consider here how waves having this form propagate through a material with not only uniform permittivity /epsilon1and permeability µ(as in Secs. 14.1–14.6) but now Sec. 14.8 Waves in Ohmic Conductors 49 Fig. 14.8.1 (a) TEM ﬁelds in uniform lossy material. (b) Perfectly conduct- ing electrodes with material having uniform σas well as /epsilon1andµbetween. With fringing ﬁeld ignored, ﬁelds in the material have the transverse character of (a). a uniform conductivity σas well. As suggested by Fig. 14.8.1a, these ﬁelds might constitute plane waves in “inﬁnite” media. They might also be the ﬁelds between the perfectly conducting planar electrodes of Fig. 14.8.1b. Our ﬁrst objective in this section is to see that both the TEM ﬁelds in “inﬁnite media” and on the “strip line” are described exactly by the distributed parameter model of Sec. 14.7 with R= 0. The ohmic loss introduces a conductance per unit length G. With the introduction of ﬁelds having the form of (1) into the laws of Faraday and Amp` ere, only two of the six equations are not automatically satisﬁed. These are thexcomponent of Amp` ere’s law with Ohm’s law used to express the conduction current −∂Hy ∂z=σEx+∂/epsilon1Ex ∂t(2) and the ycomponent of Faraday’s law ∂Ex ∂z=−∂µH y ∂t(3) Except for the conduction current, the ﬁrst term on the right in (2), these are the same equations as featured in Secs. 14.1-14.6. They become the plane parallel transmission-line equations if (2) is multiplied by the plate width wand (3) is multiplied by −a, where ais the plate spacing ∂I ∂z=−GV−C∂V ∂t(4) ∂V ∂z=−L∂I ∂t(5) where the current Iand voltage Vare I≡Kzw=−Hyw; V≡ −Exa (6) and G=σw a; C=/epsilon1w a; L=µa w(7) 50 One-Dimensional Wave Dynamics Chapter 14 These are the same equations found for the distributed parameter line of Sec. 14.7 with R= 0. Thus, whether representing a plane wave in a uniform lossy material or the transmission-line ﬁlled with such a uniform medium, the distributed parameter model is “exact.” From the point of view taken in Sec. 14.2, the geometry of the parallel con- ductors in the “strip line” is a special case. In the problems, the derivation of the transmission line equations given in Sec. 14.2 is generalized to include the eﬀects of a uniformly conducting material in the space between the perfect conductors. Regardless of their cross-sectional geometry, so long as the pair of conductors are perfectly conducting and the material between them is uniform, the ﬁelds are ex- actly TEM and exactly represented by the distributed parameter model. Not only is√ LC=√µ/epsilon1(8.6.14), but so also is C/G =/epsilon1/σ(7.6.4), regardless of the cross- sectional geometry. We now suppose that sinusoidal steady state conditions have been established. Then the voltage and current are represented by (14.7.9) and (14.7.10) with the wave number and characteristic impedance given by (14.7.8) and (14.7.11) with R= 0 and L, C, and Ggiven by (7). β=p ω2µ/epsilon1−jωµσ (8) Zo=ωµa wβ(9) Example 14.8.1. TEM Fields in a Lossy Material between Plane Parallel Plates Terminated in an Open Circuit With an “open circuit” at z= 0 and driven by a voltage source at z=−l, the parallel plate conﬁguration of Fig. 14.7.1b is equivalent to the open circuit transmission line of Example 14.7.1. With the understanding that βis now given by (8), it follows from (14.7.13) that V= Re ˆVg[e−jβl(z/l)+ejβl(z/l)] ejβl+e−jβlejωt(10) Using the values of ˆV+=ˆV−implied by (14.7.12) in (14.7.10) results in the current distribution I= ReˆVg Zo[e−jβl(z/l)−ejβl(z/l)] ejβl+e−jβlejωt(11) where Zois evaluated using (9). The voltage and current have been speciﬁed as a superposition of forward and backward waves, which, respectively, decay in the directions in which their phases propagate. The distribution of the square of the magnitude of V(ofEx) predicted by (10) (and hence of the time average dissipation density or time average electric energy density) is illustrated by Fig. 14.8.2. In this case, the electrical dissipation is small enough so that a standing wave pattern is evident. The wave propagating and decaying to the right interferes with the wave propagating and decaying to the left in such a way that the boundary condition at z= 0 is satisﬁed. In the neighborhood ofz= 0, where the wave traveling to the left has not yet decayed appreciably, the two waves interfere to form the familiar standing wave pattern. However, at the left, the wave traveling to the left has largely decayed and so interferes with the wave Sec. 14.8 Waves in Ohmic Conductors 51 Fig. 14.8.2 Square of the magnitude of ˆVas a function of position zfor a slightly damped electromagnetic wave. ω= 12×1010rad/s, l= 0.1 m, /epsilon1=/epsilon1o, µ=µo, and σ= 0.1 S/m, so ω/epsilon1/σ =ωτe= 10.6 and l∗= 0.0265 m. traveling to the right to produce no more than a ripple in the total ﬁeld magnitude. Further insights concerning these ﬁeld distributions will come from considering some limits of the dispersion equation discussed next. The ﬁrst and second terms under the radical in the dispersion equation, (8), represent the displacement and conduction current densities, respectively. The rela- tive importance of these current densities is determined by the relationship between the frequency and the reciprocal charge relaxation time. This is evident if (8) is written as β=ω√µ/epsilon1r 1−j ωτe(12) where τe≡/epsilon1/σis the charge relaxation time. Displacement Current Much Greater Than Conduction Current:. ωτe/greatermuch1. In this limit, the waves are essentially electromagnetic, with some damping due to the ﬁnite conductivity. The second term under the radical in (12) is small compared to the ﬁrst. Thus, the expression can be given a convenient approximation by using the ﬁrst two terms in a binomial expansion.12 β≈ω√µ/epsilon1−j 2l∗(13) The natural distance over which the electromagnetic wave decays by 1 /eis 2l∗, where the characteristic length l∗is deﬁned in (13) as l∗≡1 σp /epsilon1/µ(14) 12With x≡ −j/ωτ e,(1 +x)1/2≈(1 +1 2x) 52 One-Dimensional Wave Dynamics Chapter 14 Note that this length is the reciprocal of the intrinsic impedance conductivity prod- uct. With τe=C/G, the dependence of βonωgiven by (13) approximates the high-frequency range of Fig. 14.7.2. In the case of Fig. 14.8.2, ω/epsilon1/σ = 10 .6, so that the conditions for this low loss limit are met. Because the attenuation length for the electric ﬁeld is 2 l∗, the attenuation length for the square of the magnitude of Exisl∗, as shown in the ﬁgure, and the length of the system lis several times larger than the characteristic length l∗. Conduction Current Much Greater Than Displacement Current: ωτe/lessmuch1.In this limit, the eﬀects of displacement current are ignored altogether so that the ﬁrst term under the radical is neglected compared to the second, and the dispersion equation is approximated by β≈p −jωµσ =(1−j) δ(15) Here, δ≡p 2/ωµσ is the skin depth, familiar from Sec. 10.7.13Thus, in this regime, the decay length is δwhile the wavelength is 2 πδ. The dependence of βonωgiven by (15) approximates βin the low-frequency range of Fig. 14.7.2. Example 14.8.2. Overview of TEM Fields in Open Circuit Transmission Line Filled with Lossy Material Given the properties and dimensions of a simple system and a characteristic time for the dynamics, what are its dominant electromagnetic features? In this example, with the voltage source driving the system of Fig. 14.8.1b (so that the characteristic time is 1 /ω), the system could be essentially a: (1) resistor, in which case Exwould be uniform (Sec. 7.2) (2) lossy capacitor, also with an essentially uniform Ex(Sec. 7.9) (3) distribution of inductors and resistors with the distribution of Exgoverned by magnetic diﬀusion (Sec. 10.7) (4) lossy transmission line supporting slightly damped electromagnetic waves With the objective of having a summary way of picturing these possibilities, we recognize that the ﬁeld distributions [(10) and (11)] are exponential functions ofβl(z/l). Thus, βlencapsulates the ﬁeld distribution. In terms of dimensionless parameters, ωτe(representing the frequency) and l/l∗(representing the length), we write (12) as βl=p ωτe(ωτe−j)l l∗(16) and conclude that the ﬁeld distributions are governed by two parameters, the length of the system relative to the characteristic length ( l/l∗) and the frequency relative to 13As deﬁned by (10.7.2), the product of ωand the magnetic diﬀusion time based on this length, µσδ2, is equal to 2. Sec. 14.8 Waves in Ohmic Conductors 53 Fig. 14.8.3 In the length-frequency plane, regimes for TEM ﬁelds in material of uniform /epsilon1, µ, and σbetween perfect conductors having length l(Fig. 14.8.1a). The length is normalized to l∗= (σp µ//epsilon1)−1 and the angular frequency to τe=/epsilon1/σ. the reciprocal charge relaxation time ( ωτe). The logs of these variables are the coordi- nates in Fig. 14.8.3. The origin of the plot is at the length equal to the characteristic length, l∗, and the angular frequency equal to the reciprocal charge relaxation time, (/epsilon1/σ)−1. These coordinates provide for a systematic overview of the electromagnetic regimes. The approximate expressions for the wave number given by (13) and (15), respectively, apply to the right and left of the vertical axis, as indicated at the top of Fig. 14.8.3. It is tempting to jump to the conclusion that there are simply two regimes, the one to the right where the ﬁelds are composed of slightly damped electromagnetic waves, and the one to the left involving “skin-eﬀect.” However, this is not the whole story, because it does not take into account the length of the system. It is really βl, and not βalone, that determines the ﬁeld distribution between the plates. Whether representing a slightly damped electromagnetic wave or magnetic diﬀusion (skin eﬀect), a small value of βlmeans that there is little variation of the voltage over the length of the system. In the cases where |βl| /lessmuch1, the exponentials expressing the zdependence can be approximated by the ﬁrst terms in a Taylor’s series. Thus, in this regime, the voltage ( Ex) follows from (10) as being essentially uniform V≈ReˆVgejωt(17) and the current ( Hy) given by (11) takes on an esssentially linear distribution. I≈ˆVg Zo(−jβl)z lejωt= Re−ˆVgwσ a(1 +jωτe)zejωt(18) 54 One-Dimensional Wave Dynamics Chapter 14 The regime in Fig. 14.8.3 where this limit pertains, follows from the dispersion equation, (16). |βl|=l l∗(ωτe)1/2[(ωτe)2+ 1]1/4/lessmuch1 (19) The line along which βl= 1 can be conveniently pictured by making this expression an equality, solving for l/l∗and taking the log. log¡l l∗¢ =−1 2log(ωτe)−1 4log[(ωτe)2+ 1] (20) This makes it clear that for large values of ωτe, the line of demarcation has a slope of −1, while for small values, its slope is −1/2. This line is shown in Fig. 14.8.3. In the region well to the southwest of this line, the electric ﬁeld distribution is essentially uniform. The circuits drawn on the respective regions in Fig. 14.8.3 picture the four limiting cases. In terms of this ﬁgure, picture what happens as the frequency is raised for systems that are larger than the matching length, l/greatermuchl∗. In this case, raising the frequency follows a trajectory in the upper half-plane from the left to the right. With the frequency very low, the voltage and current distributions are approximated by (17) and (18). Because ωτe/lessmuch1, it follows from the latter equation that the system is essentially a resistor. As the line βl= 1 is approached, ωτeis still small, so that eﬀects of the displacement current are negligible. That the variation of the ﬁelds that comes into play is due to magnetic diﬀusion is clear from the appropriate limiting expression for β, (15). Indeed, the line βl= 1 in this quadrant approaches the line along which the angular frequency is equal to the reciprocal magnetic diﬀusion time based on the length l, ωτm=ωµσl2= 1 (21) as can be seen by rewriting this expression as log¡l l∗¢ =−1 2log(ωτe) (22) In the neighborhood of this line, in the second quadrant where the magnetic diﬀusion line is shown (the distributed transmission line of Sec. 14.7 with R= 0 and C= 0), the system is magnetoquasistatic (MQS). As the frequency is raised still further, the displacement current begins to come into eﬀect. The wave number makes a transition from representing the heavily damped waves of magnetic diﬀusion to the slightly damped electromagnetic waves of the ﬁrst quadrant. Consider the contrasting nature of the system with its length much less than the characteristic length, l/lessmuchl∗, as the frequency is raised. As before, to the far left of the ﬁgure, the electric ﬁeld is uniform and the current is that characteristic of a resistor. This regime, like that just above in the second quadrant, is one of quasi-steady conduction. In this regime, the ﬁelds are described by the steady conduction approximation which was the subject of the ﬁrst half of Chap. 7. By contrast with the situation in the upper half-plane, the ﬁelds now remain uniform until the angular frequency passes well beyond the reciprocal charge re- laxation time, the vertical axis. Note that in this range, the voltage and current are those for a distribution of conductances shunting perfectly conducting plates, as shown in Fig. 14.8.3. Thus, all of the conductances and capacitances can be lumped together. Up to this frequency range, the system is electroquasistatic (EQS). Sec. 14.8 Waves in Ohmic Conductors 55 Fig. 14.8.4 (a) Strip line, used for microwave transmission on circuit boards and chips. (b) “Twin lead” commonly used for TV antennae. As the frequency is raised still further, eﬀects of magnetic induction come into play and conspire with the displacement current to create ﬁeld distributions typical of electromagnetic waves. This happens in the range where |βl|= 1, because in this quadrant, the angular frequency is equal to the electromagnetic delay time based on the length of the system, ωτem≡ω√µ/epsilon1l= 1 (23) as can be seen by writing this expression as log¡l l∗¢ =−log(ωτe) (24) In this frequency range and further to the right, the ﬁeld distributions are those for slightly damped electromagnetic waves. In summary, far to the left in Fig. 14.8.3, quasi-steady conduction prevails. The dynamic process that ﬁrst comes into play as the frequency is raised is determined by the length of the system relative to the characteristic length. Systems large enough to be in the upper half-plane are in the MQS regime. Those small enough to be in the lower half-plane are in the EQS regime. The distributed parameter transmission line is often used to represent the evolution of ﬁelds on pairs of conductors surrounded by inhomogeneous dielectrics. Practical examples are shown in Fig. 14.8.4, where the dielectric is piece-wise uni- form. In these cases, even if the conductors can be represented as perfectly con- ducting, so that R= 0, the ﬁelds between the conductors are notexactly TEM. Completely transverse waves would propagate with diﬀerent velocities in the two dielectric regions, and it would not be possible to match boundary conditions at the interfaces. Nevertheless, with CandG, respectively, taken as being the EQS capacitance and conductance per unit length of the open circuit conductors, and L the MQS inductance per unit length of the short circuit conductors, the distributed parameter line of Sec. 14.7 can provide an excellent model. In cases where the material between the conductors is inhomogeneous, the distributed parameter model provides a good approximation of the principal mode of propagation, provided that the frequency is low enough to insure that the wave- length in the zdirection is long compared to the cross-sectional dimensions. For 56 One-Dimensional Wave Dynamics Chapter 14 example, it is shown in the problems that there is a z-directed Ein the strip line of Fig. 14.8.4a, so the ﬁelds are not TEM. However, if one neglects fringing ﬁelds one can show that Ezis small compared to the transverse ﬁelds if b(βa) a+b¯¯¯¯1−/epsilon1a /epsilon1b¯¯¯¯/lessmuch1 (25) Thus, the distributed parameter model is exact if the dielectric is uniform ( /epsilon1a=/epsilon1b), and approximately correct if βa= 2πa/λ is small enough to fulﬁll the inequality. 14.9 QUASI-ONE-DIMENSIONAL MODELS (G= 0) The transmission line model of Sec. 14.7 can also represent the losses in the parallel conductors. With the conductors of ﬁnite conductivity, currents in the zdirection cause a component of Ein that direction. Because the tangential Eis continuous at the surfaces of the conductors, this axial electric ﬁeld extends into the insulating region between the conductors as well. We conclude that the ﬁelds are no longer exactly TEM when the conductor losses are ﬁnite. Under what circumstances can the series distributed resistance Rbe used to represent the conductor losses? We will ﬁnd that the conductivity must be suﬃ- ciently low so that the skin depth is large compared to the conductor thickness. One might expect that this model applies only to the case of large R. Interestingly, we ﬁnd that this “constant resistance” model can remain valid even under circum- stances where line losses are small, in the sense that the decay of a wave within a distance of the order of a wavelength is small. This occurs when |ωL| /greatermuchR, i.e., the eﬀect of the distributed inductance is much larger than that of the series re- sistance. In the opposite extreme, where the eﬀect of the series resistance is large compared to that of the inductance, the model represents EQS charge diﬀusion. A demonstration is used to exemplify physical situations modeled by this distributed R-C line. These include solid state electronic devices and physiological systems. We conclude this section with a model that is appropriate if the skin depth is much less than the conductor thickness. By restricting the model to the sinusoidal steady state, the series distributed resistance Rcan be replaced by a “frequency dependent” resistance. This approximate model is typical of those used for repre- senting losses in metallic conductors at radio frequencies and above. We assume conductors in which the conduction current dominates the dis- placement current. In the sinusoidal steady state, this is true if ω/epsilon1 σ≡ωτe/lessmuch1 (1) Thus, as the frequency is raised, the distribution of current density in the conduc- tors is at ﬁrst determined by quasi-stationary conduction (ﬁrst half of Chap. 7) and then by the magnetic diﬀusion processes discussed in Secs. 10.3-10.7. That is, with the frequency low enough so that magnetic diﬀusion is essentially instantaneous, the current density is uniformly distributed over the conductor cross-sections. Intu- itively, we should expect that the constant resistance Ronly represents conductor Sec. 14.9 Models 57 Fig. 14.9.1 (a) Faraday’s integral law and, (b) Amp` ere’s integral law applied to an incremental length ∆ zof line. losses at frequencies suﬃciently low so that the distribution of current density in the conductors does not depend on rates of change. The equations used to describe the incremental circuit in Sec. 14.7 express the integral laws of Faraday and Amp` ere for incremental lengths of the transmission line. The “current loop equation” for loop C1in the circuit of Fig. 14.9.1a can be derived by applying Faraday’s law to the surface S1enclosed by the contour C1, also shown in that ﬁgure. Zb aE·ds+Zd cE·ds+Ez¯¯ (1)∆z−Ez¯¯ (2)∆z=−∂ ∂tZ S1µH·da (2) With the line integrals between conductors deﬁned as the voltages and the ﬂux through the surface as ∆ zLI, this expression becomes V(z+ ∆z)−V(z) +Ez¯¯ (1)∆z−Ez¯¯ (2)∆z=−∂ ∂tZ S1µH·da (3) and in the limit where ∆ z→0, we obtain ∂V ∂z=−L∂I ∂t−Ez¯¯ (1)+Ez¯¯ (2)(4) 58 One-Dimensional Wave Dynamics Chapter 14 The ﬁeld equivalent of charge conservation for the circuit node enclosed by the surface S2in Fig. 14.9.1b is Amp` ere’s integral law applied to the surface S2 enclosed by the contour C2, also shown in that ﬁgure. Note that C2almost encircles one of the conductors with oppositely directed adjacent segments completing the z-directed parts of the contour. For a surface S2of incremental length ∆ z, Amp` ere’s integral law requires that Zb aH·ds+Zd cH·ds=∂ ∂tZ S2/epsilon1E·da (5) where the contributions from the oppositely directed legs in the zdirection cancel. Amp` ere’s integral law requires that the integral of H·dson the contours essentially surrounding the conductor be the enclosed current I. Gauss’ integral law requires that the surface integral of /epsilon1E·dabe equal to ∆ zCV. Thus, (5) becomes −I(z+ ∆z) +I(z) =C∆z∂V ∂t(6) and in the limit, the second transmission line equation. ∂I ∂z=−C∂V ∂t(7) If the current density is uniformly distributed over the cross-sectional areas A1andA2of the respective conductors, it follows that the current densities are related to the total current by I=A1Jz1=−A2Jz2 (8) In each conductor, Jz=σEz, so the axial electric ﬁelds required to complete (4) are related to Iby Ez1=Jz1 σ1=I σ1A1; Ez2=Jz2 σ1=−I σ2A2(9) and indeed, the voltage equation is the same as for the distributed line, ∂V ∂z=−L∂I ∂t−RI (10) where the resistance per unit length has been found to be R≡1 σ1A1+1 σ2A2(11) Example 14.9.1. Low-Frequency Losses on Parallel Plate Line In the parallel plate transmission line shown in Fig. 14.9.2, the conductor thickness isband the cross-sectional areas are A1=A2=bw. It follows from (11) that the resistance is R=2 bwσ(12) Sec. 14.9 Models 59 Fig. 14.9.2 Parallel plate transmission line with conductor thickness bthat is small compared to skin depth. Under the assumption that the conductor thickness, b, is much less than the plate spacing,14a, the inductance per unit length is the same as found in Example 14.1.1, as is also the capacitance per unit length. L=aµ w; C=w/epsilon1 a(13) As the frequency is raised, the current distribution over the cross-sections of the conductors becomes nonuniform when the skin depth δ(10.7.5) gets to be on the order of the plate thickness. Thus, for the model to be valid using the resistance given by (12), δ≡r 2 ωµσ/greatermuchb⇒ωµσb/lessmuch2 b(14) With this inequality we require that the eﬀects of magnetic induction in determining the distribution of current in the conductors be negligible. Under what conditions are we justiﬁed in ignoring this eﬀect of magnetic induction but nevertheless keeping that represented by the distributed inductance? Put another way, we ask if the inductive reactance jωLcan be large compared to the resistance Rand still satisfy the condition of (14). ωL/greatermuchR⇒2 a/lessmuchωµσb (15) Combined, these last two conditions require that b a/lessmuch1 (16) We conclude that as long as the conductor thicknesses are small compared to their spacing, Rrepresents the loss over the full frequency range from dc to the frequency at which the current in the conductors ceases to be uniformly distributed. This is true because the time constant τm≡L/R =µσab that determines the frequency at which the resistance is equal to the inductive reactance15is much larger than the magnetic diﬀusion time µσb2based on the thickness of the conductors. 14So that the magnetic energy stored in the plates themselves is negligible compared to that between the plates. 15Familiar from Sec. 10.3. 60 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.9.3 Charge diﬀusion or R-C transmission line. Fig. 14.9.4 Charge diﬀusion line. Charge Diﬀusion Transmission Line. If the resistance is large enough so that the inductance has little eﬀect, the lossy transmission line becomes an EQS model. The line is simply composed of the series resistance shunted by the distributed capacitance of Fig. 14.9.3. To see that the voltage (and hence charge) and current on this line are governed by the diﬀusion equation, (10) is solved for I, with Lset equal to zero, I=−1 R∂V ∂z(17) and that expression substituted into the zderivative of (7). ∂2V ∂z2=RC∂V ∂t(18) By contrast with the charge relaxation process undergone by charge in a uniform conductor, the charge in this heterogeneous system diﬀuses. The distributed R-C line is used to model EQS processes that range from those found in neural conduction to relaxation in semiconductors. We can either view the solution of (17) and (18) as a special case from Sec. 14.7 or exploit the complete analogy to the magnetic diﬀusion processes described in Secs. 10.6 and 10.7. Demonstration 14.9.1. Charge Diﬀusion Line A simple demonstration of the charge diﬀusion line is shown in Fig. 14.9.4. A thin insulating sheet is sandwiched between a resistive sheet on top (the same Teledeltos paper used in Demonstration 7.6.2) and a metal plate on the bottom. Sec. 14.9 Models 61 With sinusoidal steady state conditions established by means of a voltage source at z=−land a short circuit at the right, the voltage distribution is the analog of that described for magnetic diﬀusion in Example 10.7.1. The “skin depth” for the charge diﬀusion process is given by (10.7.5) with µσ→RC. δ=r 2 ωRC(19) With this the new deﬁnition of δ, the magnitude of the voltage measured by means of the high-impedance voltmeter can be compared to the theory, plotted in Fig. 10.7.2. Typical values are /epsilon1= 3.5/epsilon1o, bσ= 4.5×10−4S (where bσis the surface conductivity of the conducting sheet), and a= 25 µm, in which case RC=/epsilon1/(bσ)a= 2.7×10−3 sec/m2andδ/similarequal0.5 m at a frequency of 500 Hz. In the previous example, we found that the transmission line model is appli- cable provided that the conductor thicknesses were small compared to their spac- ing and to the skin-depth. That the model could be self-consistent from dc up to frequencies at which the inductive impedance dominates resistance is in part attributable to the plane parallel geometry. To see this, consider a transmission line composed of a circular cylindrical conductor and a thin sheet, as shown in Fig. 8.6.7. In Demonstration 8.6.1, it was found that the condition n·B= 0 on the conductor surfaces is met at frequencies for which the skin depth is far greater than the thickness of the thin sheet conduc- tor. The examples of Sec. 10.4 show why this is possible. The eﬀective magnetic diﬀusion time that determines the frequency at which currents in the conducting sheet make a transition from a quasi-stationary distribution to one consistent with n·B= 0 is µσ∆l, where ∆ is the thickness of the conductor and lis the distance between conductors. This is also the L/R time constant governing the transition from resistance to inductance domination in the distributed electrodynamic model. We conclude that, even though the current may be essentially uniform over the conductor cross section, as the frequency changes from dc to the inductance domi- nated range, the current can shift its distribution over the conductor surface. Thus, in non-planar geometries, the constant Rmodel can be inadequate even over a frequency range where the skin depth is large compared to the conductor thickness. Skin Depth Small Compared to All Dimensions of Interest. In transmission lines used at radio frequencies and higher, it is usual for the skin depth to be much less than the conductor thickness, δ/lessmuchb. In the case of Fig. 14.9.2, 2 /lessmuchωµσb2. Provided that a > b , it follows from (15) that the inductive reactance dominates resistance. Although the line is then very nearly ideal, it is often long enough so that losses cannot be neglected. We therefore conclude this section by developing a model, restricted to the sinusoidal steady state, that accounts for losses when the skin depth is small compared to all dimensions of interest. In this case, the axial conduction currents are conﬁned to within a few skin depths of the conductor surfaces. Within a few skin depths, the tangential magnetic ﬁeld decays from its value at the conductor surface to zero. Because the magnetic ﬁeld decays so rapidly along a coordinate perpendicular to a given point on the conductor surface, the eﬀects on the magnetic diﬀusion of spatial variations along 62 One-Dimensional Wave Dynamics Chapter 14 Fig. 14.9.5 Parallel plate transmission line with conductors that are thick compared to the skin depth. the conductor surface are negligible. For this reason, ﬁelds in the conductors can be approximated by the one-dimensional magnetic diﬀusion process described in Sec. 10.7. The following example illustrates this concept. Example 14.9.2. High-Frequency Losses on Parallel Plate Line The parallel plate transmission line is shown again in Fig. 14.9.5, this time with the axial current distribution in the conductors in thin regions on the inner surfaces of the conductors rather than uniform. In the conductors, the displacement current is negligible, so that the magnetic ﬁeld is governed by the magnetic diﬀusion equation, (10.5.8). In the sinusoidal steady state, the ycomponent of this equation requires that 1 µσ· ∂2ˆHy ∂x2+∂2ˆHy ∂z2¸ =−jωˆHy (20) The ﬁrst term on the left is of the order of Hy/(δ)2, while the second is of the order of Hyk2=Hy(2π/λ)2[where λis the wavelength in the axial ( z) direction]. Thus, the derivative with respect to zcan be ignored compared to that with respect tox, provided that 1 δ2/greatermuchk2⇒δ/lessmuchλ 2π(21) In this case, (20) becomes the one-dimensional magnetic diﬀusion equation studied in Sec. 10.7. In the lower conductor, the magnetic ﬁeld diﬀuses in the −xdirection, so the appropriate solution to (20) is ˆHy=ˆHoe(1+j)x/δ(22) where Hois the magnetic ﬁeld intensity at the surface of the lower conductor [see (10.7.8)]. Amp` ere’s law gives the current density associated with this ﬁeld distribu- tion ˆJz=∂ˆHy ∂x=ˆHo(1 +j) δe(1+j)x/δ(23) It follows from either integrating this expression over the cross-section of the lower conductor or appealing to Amp` ere’s integral law that the the total current in the lower conductor is ˆI=wZ0 −∞ˆJzdx=wˆHo (24) Sec. 14.10 Summary 63 The axial electric ﬁeld intensity at the surface of the lower conductor can now be written in terms of this total current by ﬁrst using Ohm’s law and the current density of (23) evaluated at the surface and then using (24) to express this ﬁeld in terms of the total current. ˆEz(0) =ˆJz(0) σ=ˆI w(1 +j) σδ(25) A similar derivation gives an axial electric ﬁeld at the surface of the upper conductor that is the negative of this result. Thus, we can complete the sinusoidal steady state version of the voltage transmission-line equation, (4). dˆV dz=−· jωL+2(1 + j) wσδ¸ ˆI (26) Because the magnetic energy stored within the conductor is usually negligible com- pared to that in the region between conductors, ωL/greatermuch2 wσδ(27) and (26) becomes the ﬁrst of the two sinusoidal steady state transmission line equa- tions. dˆV dz=−¡ jωL+2 wσδ¢ˆI (28) The second follows directly from (7). dˆI dz=−jωCˆV (29) Comparison of these expressions with those describing the line operating with the conductor thickness much less than the skin depth, (10) and (7), shows that here there is an equivalent distributed resistance. Req=2 wσδ=1 wr 2ωµ σ(30) (Here, µis the permeability of the conductor, not of the region between conduc- tors.) Note that this is the series dc resistance of conductors having width wand thickness δ. Because δis inversely proportional to the square root of the frequency, this equivalent resistance increases with the square root of the frequency. 14.10 SUMMARY The theme in this chapter has been the transmission line. It has been used to represent the evolution of electromagnetic ﬁelds through structures generally comprised of a pair of “conductors” embedded in a less conducting, and often highly insulating, medium. We have conﬁned ourselves to systems that are uniform in the direction of evolution, the zdirection. 64 One-Dimensional Wave Dynamics Chapter 14 If the conductors can be regarded as perfectly conducting, and the medium in which they are embedded as having uniform permeability, permittivity, and conduc- tivity, the ﬁelds are exactly TEM, regardless of the cross-sectional geometry. The relevant laws and distributed parameter model are summarized in Table 14.10.1. Identiﬁcation of variables as illustrated in the table make the transmission line ex- actly equivalent to a plane wave. Whether L, C, orGrepresent ﬁelds propagating along the conductors or a plane wave, LC=µ/epsilon1andC/G =/epsilon1/σ. Much of this chapter is devoted to describing the limit where the conduc- tors are not only perfect, but the medium has negligible conductivity ( G= 0). Transients on this “ideal” transmission line are described by using the relations summarized in Table 14.10.2. The voltage and current at any position and time (z, t) are superpositions of wave components that propagate with the velocity c. These forward and backward wave components are, respectively, invariant on lines in the z−tplane of constant αandβ. Along those lines originating on initial condi- tions, the wave components are as summarized in the second row of the table. The last two rows summarize how the reﬂected wave component is determined from the incident component at two common terminations. A summary of the relations used to describe the ideal line in the sinusoidal steady state is given in Table 14.10.3. Because it has a magnitude that is constant and a phase that increases linearly with z, the evolution of voltage and current and of their ratio, the impedance Z(z), is conveniently pictured in terms of the complex reﬂection coeﬃcient, Γ( z). Relations and the complex Γ plane are illustrated in the ﬁrst row. The mappings of the impedance and of the admittance onto this plane, respectively, are summarized by the second and third rows. Because the magnitude of Γ is constant over a uniform length of line, the trajectory of Z(z) orY(z) is on a circle of constant radius in the directions of the generator or the load, as indicated. These Smith charts give a convenient overview of how the impedance and admittance vary with position. In Sec. 14.7, a shunt conductance per unit length, G, (to represent losses in the material between the transmission line conductors) and a series resistance per unit length, R, (for losses in the conductors themselves) was added to the distributed parameter transmission line representation. For the limiting case where the con- ductors were inﬁnitely conducting, R= 0, and the material between of uniform properties, the ﬁelds represented by the line were exactly TEM. In the case where the material properties did vary over the cross section, the distributed parameter picture provided a useful model for the line provided that the wavelength was long compared to the cross-sectional dimensions. In speciﬁc terms, this model gave the opportunity to consider the dynamical processes considered in Chaps. 7, 10, and 12 (charge relaxation, magnetic diﬀusion and electromagnetic wave propagation, respectively) in one self consistent situation. What was learned will be generalized in the review of the processes given in Secs. 15.3–15.4. In Sec. 14.9, where G= 0 but Rwas ﬁnite, the speciﬁc objective was to understand how the transmission line concept could be used to approximate con- ductor losses. A broader objective was to again illustrate the use of the distributed parameter line as a model, representing the ﬁelds at frequencies suﬃciently low so that the wavelength is long compared with the transverse dimensions. Sec. 14.0 Problems 65 TABLE 14.10.1 TRANSMISSION LINE EQUIVALENTS ∂I ∂z=−C∂V ∂t−GV(14.8.4) ∂V ∂z=−L∂I ∂t(14.8.5) I→Hy V→Ex C→/epsilon1= n2/epsilon1o L→µ C G→/epsilon1/σ 66 One-Dimensional Wave Dynamics Chapter 14 TABLE 14.10.2 WAVE TRANSIENTS V=V+(α) +V−(β) (14.3.9) I=1 Zo(V+(α)−V−(β)) (14.3.10) α=z−ct;β=z+ct (14.3.11) Zo=p L/C (14.3.12) c= 1/√ LC (14.3.1) V+=1 2(Vi+ZoIi) (14.3.18) V−=1 2(Vi−ZoIi) (14.3.19) V−=V+¡RL Zo−1¢ ¡RL Zo+ 1¢(14.4.8) V+=Vg ¡Rg Zo+ 1¢+V−¡Rg Zo−1¢ ¡Rg Zo+ 1¢(14.4.10) Sec. 14.0 Problems 67 TABLE 14.10.3 SINUSOIDAL-STEADY-STATE (R= 0, G= 0) ˆV=ˆV+e−jβz[1 + Γ( z)] (14.5.5) ˆI=ˆV+e−jβz Zo[1−Γ(z)](14.5.6) Γ≡ˆV− ˆV+ej2βz (14.6.2) Zo=p L/C (14.3.12) β=ω√ LC (14.5.5) Z(z) Zo=1 + Γ( z) 1−Γ(z)≡r+jx (14.6.1) Γ =Z Zo−1 Z Zo+ 1(14.6.3) Y(z) Yo=1−Γ(z) 1 + Γ( z)=g+jy (14.6.12) Γ =1−Y Yo 1 +Y Yo Yo=p C/L 68 One-Dimensional Wave Dynamics Chapter 14 P R O B L E M S 14.1 Distributed Parameter Equivalents and Models 14.1.1 The “strip line” shown in Fig. P14.1.1 is an example where the ﬁelds are notexactly TEM. Nevertheless, wavelengths long compared to aandb, the distributed parameter model is applicable. The lower perfectly conducting plate is covered by a planar perfectly insulating layer having properties (/epsilon1b, µb=µo). Between this layer and the upper electrode is a second per- fectly insulating material having properties ( /epsilon1a, µa=µo). The width wis much greater than a+b, so fringing ﬁelds can be ignored. Determine Land Cand hence the transmission line equations. Show that LC/negationslash=µ/epsilon1unless /epsilon1a=/epsilon1b. Fig. P14.1.1 Fig. P14.1.2 14.1.2 An incremental section of a “backward wave” transmission line is as shown in Fig. P14.1.2. The incremental section of length ∆ zshown has a reciprocal capacitance per unit length ∆ zC−1and reciprocal inductance per unit length ∆ zL−1. Show that, by contrast with (4) and (5), in this case the transmission line equations are L∂2I ∂t∂z=−V; C∂2V ∂t∂z=−I (a) Sec. 14.4 Problems 69 14.2 Transverse Electromagnetic Waves 14.2.1∗For the coaxial conﬁguration of Fig. 14.2.2b, (a) Show that, deﬁned as zero on the outer conductor, Azand Φ are Az=−µIln(r/a)/2π; Φ = −λlln(r/a)/2π/epsilon1 (a) where λlis the charge per unit length on the inner conductor. (b) Using these expressions, show that the LandCneeded to complete the transmission line equations are L=µ 2πln¡a b¢ ; C= 2π/epsilon1/ln¡a b¢ (b) and hence that LC=µ/epsilon1. 14.2.2 A transmission line consists of a conductor having the cross-section shown in Fig. P4.7.5 adjacent to an L-shaped return conductor comprised of “ground planes” in the planes x= 0 and y= 0, intersecting at the ori- gin. Assuming that the region between these conductors is free space, what are the transmission line parameters LandC? 14.3 Transients on Inﬁnite Transmission Lines 14.3.1 Show that the characteristic impedance of a coaxial cable (Prob. 14.2.1) is Zo=p µ//epsilon1ln (a/b)/2π (a) For a dielectric having /epsilon1= 2.5/epsilon1oandµ=µo, evaluate Zofor values of a/b= 2,10,100, and 1000. Would it be reasonable to design such a cable to have Zo= 1KΩ? 14.3.2 For the parallel conductor line of Fig. 14.2.2 in free space, what value of l/Rshould be used to make Zo= 300 ohms? 14.3.3 The initial conditions on an inﬁnite line are V= 0 and I=Ipfor−d < z < d andI= 0 for z <−dandd < z . Determine V(z, t) and I(z, t) for 0< t, presenting the solution graphically, as in Fig. 14.3.2. 14.3.4 On an inﬁnite line, when t= 0, V=Voexp(−z2/2a2), and I= 0, deter- mine analytical expressions for V(z, t) and I(z, t). 14.3.5∗In the energy conservation theorem for a transmission line, (14.2.19), V I is the power ﬂow. Show that at any location, z, and time, t, it is correct to 70 One-Dimensional Wave Dynamics Chapter 14 think of power ﬂow as the superposition of power carried by the + wave in the + zdirection and −wave in the −zdirection. V I=1 Zo[V2 +−V2 −] ( a) 14.3.6 Show that the traveling wave solutions of (2) are not solutions of the equa- tions for the “backward wave” transmission line of Prob. 14.1.2. 14.4 Transients on Bounded Transmission Lines 14.4.1 A transmission line, terminated at z=lin an “open circuit,” is driven at z= 0 by a voltage source Vgin series with a resistor, Rg, that is matched to the characteristic impedance of the line, Rg=Zo. For t <0, Vg=Vo= constant. For 0 < t, V g= 0. Determine the distribution of voltage and current on the line for 0 < t. 14.4.2 The transient is to be determined as in Prob. 14.4.1, except the line is now terminated at z=lin a “short circuit.” 14.4.3 The transmission line of Fig. 14.4.1 is terminated in a resistance RL=Zo. Show that, provided that the voltage and current over the length of the line are initially zero, the line has the same eﬀect on the circuit connected atz= 0 as would a resistance Zo. 14.4.4 A transmission line having characteristic impedance Zais terminated at z=l+Lin a resistance Ra=Za. At the other end, where z=l, it is connected to a second transmission line having the characteristic impedance Zb. This line is driven at z= 0 by a voltage source Vg(t) in series with a resistance Rb=Zb. With Vg= 0 for t <0, the driving voltage makes a step change to Vg=Vo, a constant voltage. Determine the voltage V(0, t). 14.4.5 A pair of transmission lines is connected as in Prob. 14.4.4. However, rather than being turned on when t= 0, the voltage source has been on for a long time and when t= 0 is suddenly turned oﬀ. Thus, Vg=Vofort <0 andVg= 0 for 0 < t. The lines have the same wave velocity c. Determine V(0, t). (Note that, by contrast with the situation in Prob. 14.4.4, the line having characteristic impedance Zanow has initial values of voltage and current.) 14.4.6 A transmission line is terminated at z=lin a “short” and driven at z= 0 by a current source Ig(t) in parallel with a resistance Rg. For 0 < t < T, I g=Io= constant, while for t <0 and T < t, I g= 0. For Rg=Zo, determine V(0, t). Sec. 14.5 Problems 71 14.4.7 With Rgnot necessarily equal to Zo, the line of Prob. 14.4.6 is driven by a step in current; for t <0, Ig= 0, while for 0 < t, I g=Io= constant. (a) Using an approach suggested by Example 14.4.3, determine the cur- rentI(0, t). (b) If the transmission line is MQS, the system can be represented by a parallel inductor and resistor. Find I(0, t) assuming such a model. (c) Show that in the limit where the round-trip transit time 2 l/cis short compared to the time τ=lL/R g, the current I(0, t) found in (a) approaches that predicted by the MQS model. 14.4.8 The transmission line shown in Fig. P14.4.8 is terminated in a series load resistance, RL, and capacitance CL. (a) Show that the algebraic relation between the incident and reﬂected wave at z=l, given by (8) for the load resistance alone, is replaced by the diﬀerential equation at z=l ZoCLµRL Zo+ 1¶dV− dt+V−=ZoCLµRL Zo−1¶dV+ dt+V+ (a) which can be solved for the reﬂected wave V−(l, t) given the incident wave V+(l, t). (b) Show that if the capacitor voltage is Vcwhen t= 0, then V−(l,0) =Vc¡RL Zo+ 1¢+V+(l,0)¡RL Zo−1¢ ¡RL Zo+ 1¢ (b) (c) Given that Vg(t) = 0 for t <0, Vg(t) =Vo= constant for 0 < t, and thatRg=Zo, determine V(0, t). Fig. P14.4.8 14.5 Transmission Lines in the Sinusoidal Steady State 14.5.1 Determine the impedance of a quarter-wave section of line that is termi- nated, ﬁrst, in a load capacitance CL, and second, in a load inductance LL. 14.5.2 A line having length lis terminated in an open circuit. 72 One-Dimensional Wave Dynamics Chapter 14 (a) Determine the line admittance Y(−l) and sketch it as a function of ωl/c. (b) Show that the low-frequency admittance is that of a capacitor lC. 14.5.3∗A line is matched at z= 0 and driven at z=−lby a voltage source Vg(t) = Vosin(ωt) in series with a resistance equal to the characteristic impedance of the line. Thus, the line is as shown in Fig. 14.4.5 with Rg=Zo. Show that in the sinusoidal steady state, V= Re1 2ˆVge−jβ(z+l)ejωt; I= Re1 2ZoˆVge−jβ(z+l)ejωt where ˆVg≡ −jVo. 14.5.4 In Prob. 14.5.3, the drive is zero for t <0 and suddenly turned on when t= 0. Thus, for 0 < t, V g(t) is as in Prob. 14.5.3. With the solution written in the form of (1), where Vs(z, t) is the sinusoidal steady state solution found in Prob. 14.5.3, what are the initial and boundary conditions on the transient part of the solution? Determine V(z, t) and I(z, t). 14.6 Reﬂection Coeﬃcient Representation of Transmission Lines 14.6.1∗The normalized load impedance is ZL/Zo= 2 + j2. Use the Smith chart to show that the impedance of a quarter-wave line with this termination is Z/Z o= (1−j)/4. Check this result using (20). 14.6.2 For a normalized load impedance ZL/Zo= 2 + j2, use (3) to evaluate the reﬂection coeﬃcient, |Γ|, and hence the VSWR, (10). Use the Smith chart to check these results. 14.6.3 For the system shown in Fig. 14.6.6a, the load admittance is YL= 2Yo. Determine the position, l, and length, ls, of a shorted stub, also having the characteristic admittance Yo, that matches the load to the line. 14.6.4 In practice, it may not be possible or convenient to control the position l of the stub, as required for single stub matching of a load admittance YL to a line having characteristic admittance Yo. In that case, a “double stub” matching approach can be used, where two stubs at arbitrary locations but with adjustable lengths are used. At the price of restricting the range of loads that can be matched, suppose that the ﬁrst stub is attached in parallel with the load and shorted at length l1, and that the second stub is shorted at length l2and connected in parallel with the line at a given distance lfrom the load. The stubs have the same characteristic admittance as the line. Describe how, given the load admittance and the distance lto the second stub, the lengths l1andl2would be designed to match the load to the line. (Hint: The ﬁrst stub can be adjusted in length to locate Sec. 14.7 Problems 73 the eﬀective load anywhere on the circle on the Smith chart having the normalized conductance gLof the load.) Demonstrate for the case where YL= 2Yoandl= 0.042λ. 14.6.5 Use the Smith chart to obtain the VSWR on the line to the left in Fig. 14.5.3 if the load resistance is RL/Zo= 2 and Za o= 2Z0. (Hint: Remember that the impedance of the Smith chart is normalized to the characteristic impedance at the position in question. In this situation, the lines have diﬀerent characteristic impedances.) 14.7 Distributed Parameter Equivalents and Models with Dissipation 14.7.1 Following the steps exempliﬁed in Section 14.1, derive (1) and (2). 14.7.2 For Example 14.7.1, (a) Determine I(z, t). (b) Find the impedance at z=−l. (c) In the long wave limit, |βl| /lessmuch1, what is this impedance and what equivalent circuit does it imply? 14.7.3 The conﬁguration is as in Example 14.7.1 except that the line is shorted at z= 0. Determine V(z, t) and I(z, t), and hence the impedance at z=−l. In the long wave limit, |βl| /lessmuch1, what is this impedance and what equivalent circuit does it imply? 14.7.4∗Following steps suggested by the derivation of (14.2.19), (a) Use (1) and (2) to derive the power theorem −∂ ∂z(V I) =∂ ∂t¡1 2CV2+1 2LI2¢ +I2R+V2G (a) (b) The product of two sinusoidally varying quantities is a constant (time average) part plus a part that varies sinusoidally at twice the fre- quency. In complex notation, ReˆAejωtReˆBejωt=1 2ReˆAˆB∗+1 2ReˆAˆBe2jωt(b) Use (11.5.7) to prove this identity. (c) Show that, in describing the sinusoidal steady state, the time average of the power theorem becomes −d dz¡1 2ReˆVˆI∗¢ =1 2Re(ˆIˆI∗R+ˆVˆV∗G) ( c) 74 One-Dimensional Wave Dynamics Chapter 14 Show that for Example 14.7.1, it follows that the time average power input is equal to the integral over the length of the time average power dissipation per unit length. 1 2ReˆVˆI∗¯¯ z=−l=Z0 −l1 2Re(ˆIˆI∗R+ˆVˆV∗G)dz (d) (d) Evaluate the time average input power on the left in this relation and the integral of the time average dissipation per unit length on the right and show that they are indeed equal. 14.8 Uniform and TEM Waves in Ohmic Conductors 14.8.1 In the general TEM conﬁguration of Fig. 14.2.1, the material between the conductors has uniform conductivity, σ, as well as uniform permittivity, /epsilon1. Following steps like those leading to 14.2.12 and 14.2.13, show that (4) and (5) describe the waves, regardless of cross-sectional geometry. Note the relationship between GandCsummarized by (7.6.4). 14.8.2 Although associated with the planar conﬁguration of Fig. 14.8.1 in this section, the transmission line equations, (4) and (5), represent exact ﬁeld solutions that are, in general, functions of the transverse coordinates as well as z. Thus, the transmission line represents a large family of exact solutions to Maxwell’s equations. This follows from Prob. 14.8.1, where it is shown that the transmission line equations apply even if the regions between conductors are coaxial, as shown in Fig. 14.2.2b, with a material of uniform permittivity, permeability, and conductivity between z=−land z= 0. At z= 0, the transmission line conductors are “open circuit.” At z=−l, the applied voltage is Re ˆVgexp(jωt). Determine the electric and magnetic ﬁelds in the region between transmission line conductors. Include the dependence of the ﬁelds on the transverse coordinates. Note that the axial dependence of these ﬁelds is exactly as described in Examples 14.8.1 and 14.8.2. 14.8.3 The terminations and material between the conductors of a transmission line are as described in Prob. 14.8.2. However, rather than being coaxial, the perfectly conducting transmission line conductors are in the parallel wire conﬁguration of Fig. 14.2.2a. In terms of Φ( x, y, z, t ) and Az(x, y, z, t ), determine the electric and magnetic ﬁelds over the length of the line, in- cluding their dependencies on the transverse coordinates. What are L, C, andGand hence βandZo? 14.8.4∗The transmission line model for the strip line of Fig. 14.8.4a is derived in Prob. 14.1.1. Because the permittivity is not uniform over the cross-section of the line, the waves represented by the model are not exactly TEM. The approximation is valid as long as the wavelength is long enough so that (25) Sec. 14.9 Problems 75 is satisﬁed. In the approximation, Exis taken as being uniform with xin each of the dielectrics, EaandEb, respectively. To estimate the longitudinal ﬁeldEzand compare it to Ea, (a) Use the integral form of the law of induction applied to an incremental surface between z+ ∆zandzand between the perfect conductors to derive Faraday’s transmission line equation written in terms of Ea. ¡ a+/epsilon1a /epsilon1bb¢∂Ea ∂z=−µo(a+b)∂Hy ∂t(a) (b) Then carry out this same procedure using a surface that again has edges at z+ ∆zandzon the upper perfect conductor, but which has its lower edge at the interface between dielectrics. With the axial electric ﬁeld at the interface deﬁned as Ez, show that Ez=−aµo∂Hy ∂t−a∂Ea ∂z=−abµo¡/epsilon1a /epsilon1b−1¢ ¡ a+/epsilon1a /epsilon1bb¢∂Hy ∂t(b) (c) Now show that in order for this ﬁeld to be small compared to Ea, (25) must hold. 14.9 Quasi-One-Dimensional Models (G= 0) 14.9.1 The transmission line of Fig. 14.2.2a is comprised of wires having a ﬁnite conductivity σ, with the dielectric between of negligible conductivity. With the distribution of VandIdescribed by (7) and (10), what are C, L, and R, and over what frequency range is this model valid? (Note Examples 4.6.3 and 8.6.1.) Give a condition on the dimensions R→aandlthat must be satisﬁed to have the model be self-consistent over frequencies ranging from where the resistance dominates to where the inductive reactance dominates. 14.9.2 In the coaxial transmission line of Fig. 14.2.2b, the outer conductor has a thickness ∆. Each conductor has the conductivity σ. What are C, L, and R, and over what frequency range are (7) and (10) valid? Give a condition on the transverse dimensions that insures the model being valid into the frequency range where the inductive reactance dominates the resistance. 14.9.3 Find V(z, t) on the charge diﬀusion line of Fig. 14.9.4 in the case where the applied voltage has been zero for t <0 and suddenly becomes Vp= constant for 0 < tand the line is shorted at z= 0. (Note Example 10.6.1.) 14.9.4 Find V(z, t) under the conditions of Prob. 14.9.3 but with the line “open circuited” at z= 0. 15 OVERVIEW OF ELECTROMAGNETIC FIELDS 15.0 INTRODUCTION In developing the study of electromagnetic ﬁelds, we have followed the course sum- marized in Fig. 1.0.1. Our quest has been to make the laws of electricity and magnetism, summarized by Maxwell’s equations, a basis for understanding and innovation. These laws are both general and simple. But, as a consequence, they are mastered only after experience has been gained through many speciﬁc exam- ples. The case studies developed in this text have been aimed at providing this experience. This chapter reviews the examples and intends to foster a synthesis of concepts and applications. At each stage, simple conﬁgurations have been used to illustrate how ﬁelds relate to their sources, whether the latter are imposed or induced in materials. Some of these conﬁgurations are identiﬁed in Section 15.1, where they are used to outline a comparative study of electroquasistatic, magnetoquasistatic, and electrodynamic ﬁelds. A review of much of the outline (Fig. 1.0.1) can be made by selecting a particular class of conﬁgurations, such as cylinders and spheres, and using it to exemplify the material in a sequence of case studies. The relationship between ﬁelds and their sources is the theme in Section 15.2. Again, following the outline in Fig. 1.0.1, electric ﬁeld sources are unpaired charges and polarization charges, while magnetic ﬁeld sources are current and (paired) mag- netic charges. Beginning with electroquasistatics, followed by magnetoquasistatics and ﬁnally by electrodynamics, our outline ﬁrst focused on physical situations where the sources were constrained and then were induced by the presence of media. In this text, magnetization has been represented by magnetic charge. An alternative commonly used formulation, in which magnetization is represented by “Amp` erian” currents, is discussed in Sec. 15.2. As a starting point in the discussions of EQS, MQS, and electrodynamic ﬁelds, we have used idealized models for media. The limits in which materials behave as 1 2 Overview of Electromagnetic Fields Chapter 15 “perfect conductors” and “perfect insulators” and in which they can be said to have “inﬁnite permittivity or permeability” provide yet another way to form an overview of the material. Such an approach is taken at the end of Sec. 15.2. Useful as these idealizations are, their physical signiﬁcance can be appreciated only by considering the relativity of perfection. Although we have introduced the eﬀects of materials by making them ideal, we have then looked more closely and seen that “perfection” is a relative concept. If the ﬁelds associated with idealized models are said to be “zero order,” the second part of Sec. 15.2 raises the level of maturity reﬂected in the review by considering the “ﬁrst order” ﬁelds. What is meant by a “perfect conductor” in EQS and MQS systems is a part of Sec. 15.2 that naturally leads to a review in Sec. 15.3 of how characteristic times can be used to understand electromagnetic ﬁeld interactions with media. Now that we can see EQS and MQS systems from the perspective of electrodynamics, Sec. 15.3 is aimed at an overview of how the spatial scale, time scale (frequency), and material properties determine the dominant processes. The objective in this section is not only to integrate material, but to add insight into the often iterative process by which a model is made to both encapsulate the essential physics and serve as a basis of engineering innovation. Energy storage and dissipation, together with the associated forces on macro- scopic media, provide yet another overview of electromagnetic systems. This is the theme of Sec. 15.4, which summarizes the reasons why macroscopic forces can usu- ally be classiﬁed as being either EQS or MQS. 15.1 SOURCE AND MATERIAL CONFIGURATIONS We can use any one of a number of conﬁgurations to review physical phenomena outlined in Fig. 1.0.1. The sections, examples, and problems associated with a given physical situation are referenced in the tables used to trace the evolution of a given conﬁguration. Incremental Dipoles. In homogeneous media, dipole ﬁelds are simple solu- tions to Laplace’s equation or the wave equation in two or three dimensions and have been used to represent the range of situations summarized in Table 15.1.1. As introduced in Chap. 4, the dipole represented closely spaced equal and opposite electric charges. Perhaps these charges were produced on a pair of closely spaced conducting objects, as shown in Fig. 3.3.1a. In Chap. 6, the electric dipole was used to represent polarization, and a distinction was made between unpaired and paired (polarization) charges. In representing conduction phenomena in Chap. 7, the dipole represented a closely spaced pair of current sources. Rather than being a source in Gauss’ law, the dipole was a source in the law of charge conservation. In magnetoquasistatics, there were two types of dipoles. First was the small current loop, where the dipole moment was the product of the area, a, and the circulating current, i. The dipole ﬁelds were those from a current loop, far from the loop, such as shown in Fig. 3.3.1b. As we will discuss in Sec. 15.2, we could have used current loop dipoles to represent magnetization. However, in Chap. 9, Sec. 15.1 Conﬁgurations 3 TABLE 15.1.1 SUMMARY OF INCREMENTAL DIPOLES Electroquasistatic charge : Point; Sec. 4.4, Line; Prob. 4.4.1, Sec. 5.7 Electroquasistatic polarization : Sec. 6.1 Stationary conduction current : Point; Example 7.3.2 Line; Prob. 7.3.3 Magnetoquasistatic current : Point; Example 8.3.2 Line; Example 8.1.2 Magnetoquasistatic magnetization : Sec. 9.1 Electric Electrodynamic : Point; Sec. 12.2 Magnetic Electrodynamic : Point; Sec. 12.2 4 Overview of Electromagnetic Fields Chapter 15 magnetization was represented by magnetic dipoles, a pair of equal and opposite magnetic charges. Thus, the developments of polarization in Chap. 6 were directly applicable to magnetization. To create the time-varying positive and negative charges of the electric dipole, a current is required. In Fig. 3.3.1a, this current is supplied by the voltage source. In the EQS limit, the magnetic ﬁeld associated with this current is negligible, as are the eﬀects of the associated magnetic ﬁeld. In Chap. 12, where the laws of Faraday and Amp` ere were made self-consistent, the coupling between these laws was found to result in electromagnetic radiation. Electric dipole radiation existed because the charging currents created some magnetic ﬁeld and that, in turn, induced a rotational electric ﬁeld. In the case of the magnetic dipole shown last in Table 15.1.1, electromagnetic waves resulted from a displacement current induced by the time-varying magnetic ﬁeld that, in turn, produced a more rotational magnetic ﬁeld. Planar Periodic Conﬁgurations. Solutions to Laplace’s equation in Cartesian coordinates are all that is required to study the quasistatic and “steady” situations outlined in Table 15.1.2. The ﬁelds used to study these physical situations, which are periodic in a plane that “extends to inﬁnity,” are by nature decaying in the direction perpendicular to that plane. The electrodynamic ﬁelds studied in Sec. 12.6 have this same decay in a direction perpendicular to the direction of periodicity as the frequency becomes low. From the point of view of electromagnetic waves, these low frequency, essentially Laplacian, ﬁelds are represented by nonuniform plane waves. As the frequency is raised, the nonuniform plane waves become waves that propagate in the direction in which they formerly decayed. Solutions to the wave equation can be spatially periodic in both directions. The TE and TM electrodynamic ﬁeld conﬁgurations that conclude Table 15.1.2 help put into perspective those aspects of the EQS and MQS conﬁgurations that do not involve losses. Cylindrical and Spherical. A few simple solutions to Laplace’s equation are suﬃcient to illustrate the nature of ﬁelds in and around cylindrical and spherical material objects. Table 15.1.3 shows how a sequence of case studies begins with EQS and MQS ﬁelds, respectively, in systems of “perfect” insulators and “perfect” conductors and culminates in the very diﬀerent inﬂuences of ﬁnite conductivity on EQS and MQS ﬁelds. Fields Between Plane Parallel Plates. Uniform and piece-wise uniform qua- sistatic ﬁelds are suﬃcient to illustrate phenomena ranging from EQS, the “capac- itor,” to MQS “magnetic diﬀusion through thin conductors,” Table 15.1.4. Closely related TEM ﬁelds describe the remaining situations. Axisymmetric (Coaxial) Fields. The case studies summarized in Table 15.1.4 under this category parallel those for ﬁelds between plane parallel conductors. Sec. 15.1 Conﬁgurations 5 TABLE 15.1.2 PLANAR PERIODIC CONFIGURATIONS Field Solutions Laplace’s equation: Sec. 5.4 Wave equation: Sec. 12.6 Electroquasistatic (EQS) Constrained Potentials and Surface Charge: Examp. 5.6.2 Constrained Potentials and Volume Charge: Examp. 5.6.1 Probs. 5.6.1-4 Constrained Potentials and Polarization: Probs. 6.3.1-4 Charge Relaxation: Probs. 7.9.7-8 Steady Conductor (MQS or EQS) Constrained Potential and Insulating Boundary: Prob. 7.4.3 Magnetoquasistatic (MQS) Magnetization: Examp. 9.3.2 Magnetic diﬀusion through Thin Conductors: Probs. 10.4.1-2 Electrodynamic Imposed Surface Sources: Examps. 12.6.1-2 Probs. 12.6.1-4 Imposed Sources with Perfectly Examp. 12.7.2 Conducting Boundaries: Probs. 12.7.3-4 Probs. 13.2.1 Perfectly Insulating Boundaries: Sec. 13.5 Probs. 13.2.3-4 Probs. 13.5.1-4 6 Overview of Electromagnetic Fields Chapter 15 TABLE 15.1.3 CYLINDRICAL AND SPHERICAL CONFIGURATIONS Field Solutions to Laplace’s Equation : Cylindrical; Sec. 5.7 Spherical; Sec. 5.9 Electroquasistatic Equipotentials: Examp. 5.8.1 Examp. 5.9.2 Polarization: Permanent: Prob. 6.3.6 Examp. 6.3.1 Prob. 6.3.5 Induced: Examp. 6.6.2 Probs. 6.6.1-2 Charge Relaxation: Probs. 7.9.4-5 Examp. 7.9.3 Prob. 7.9.6 Steady Conduction (MQS or EQS) Imposed Current: Examp. 7.5.1 Probs. 7.5.1-2 Magnetoquasistatic Imposed Current: Probs. 8.5.1-2 Examp. 8.5.1 Perfect Conductor: Probs. 8.4.2-3 Examp. 8.4.3 Prob. 8.4.1 Magnetization: Probs. 9.6.3-4,10,12 Probs. 9.6.11,13 Magnetic Diﬀusion: Examp. 10.4.1 Probs. 10.4.3-4 Probs. 10.4.5-6 TM and TE Fields with Longitudinal Boundary Conditions. The case stud- ies under this heading in Table 15.1.4 oﬀer the opportunity to see the relationship Sec. 15.1 Conﬁgurations 7 TABLE 15.1.4. SPECIAL CONFIGURATIONS Fields Between Plane Parallel Plates Capacitor: Examps. 3.3.1, 6.3.3 Probs. 6.5.1-4, 6.6.8, 11.2.1 11.3.3, 11.6.1 Resistor: Examps. 7.2.1, 7.5.2 Inductor: Examp. 8.4.4, Probs. 9.5.1,3,6 Charge Relaxation: Examp. 7.9.2 Magnetic Diﬀusion though: Thin Conductors: Prob. 10.3.4 Thick Conductors (TEM): Examps. 10.6.1, 10.7.1 Probs. 10.3.4, 10.6.1-2, 10.7.1-2 Principle (TEM) Waveguide Modes Examps. 13.1.1-2 Transmission Line: Examps. 14.1.1, 14.8.2 Axisymmetric (Coaxial) Fields Capacitor:Probs 6.5.5-6 Resistor: Examps. 7.5.2 Probs. 7.2.1,4,8 Inductor: Examp. 3.4.1 Probs. 9.5.2,4-5 Charge Relaxation: Prob. 7.9.1 TEM Transmission Line Prob. 13.1.4 TM and TE Fields with Longitudinal Boundary Conditions Capacitive Attenuator: Sec. 5.5 TM Waveguide Fields: Examp. 13.3.1 Inductive Attenuator: Examp. 8.6.3 TE Waveguide Fields: Examp. 13.3.2 Cylindrical Conductor-Pair and Conductor-Plane EQS Perfect Conductors: Examp. 4.6.3 MQS Perfect Conductors: Examp. 8.6.1 TEM Transmission Line: Examp. 14.2.2 8 Overview of Electromagnetic Fields Chapter 15 between ﬁelds and their sources, in the quasistatic limits and as electromagnetic waves. The EQS and MQS limits, illustrated by Demonstrations 5.5.1 and 8.6.2, respectively, become the shorted TM and TE waveguide ﬁelds of Demonstrations 13.3.1 and 13.3.2. Cylindrical Conductor Pair and Conductor Plane. The ﬁelds used in these conﬁgurations are ﬁrst EQS, then MQS, and ﬁnally TEM. The relationship between the EQS and MQS ﬁelds and the physical world is illustrated by Demonstrations 4.7.1 and 8.6.1. Regardless of cross-sectional geometry, TEM waves on pairs of perfect conductors are much of the same nature regardless of geometry, as illustrated by Demonstration 13.1.1. 15.2 MACROSCOPIC MEDIA Source Representation of Macroscopic Media. The primary sources of the EQS electric ﬁeld intensity were the unpaired and paired charge densities, respectively, describing the inﬂuence of macroscopic media on the ﬁelds through conduction and polarization (Chap. 6). Although in Chap. 8 the primary source of the MQS magnetic ﬁeld due to conduction was the unpaired current density, in Chap. 9, magnetization was modeled as the result of orientation of permanent magnetic dipoles made up of a pair of magnetic charges, positive and negative. This is not the conventional way of introducing magnetization. However, the magnetic charge model made possible an analogy between polarization and magnetization that enabled us to introduce magnetization into the ﬁeld equations by analogy to polarization. More conventional is the approach that treats magnetization as the result of circulating Amp` erian currents. The two approaches lead to the same ﬁ- nal result, only the model is diﬀerent. To illustrate this, let us rewrite Maxwell’s equations (12.0.1)–(12.0.4) in terms of B, rather than H ∇ ×E=−∂ ∂tB (1) ∇ ×B µo=∇ ×M+Ju+∂ ∂t/epsilon1oE+∂ ∂tP (2) ∇ ·/epsilon1oE=−∇ ·P+ρu (3) ∇ ·B= 0 (4) Thus, if Bis considered to be the fundamental ﬁeld variable, rather than H, then the presence of magnetization manifests itself by the appearance of the term ∇×Mnext toJuin Amp` ere’s law. Like Ju, the Amp` erian current density, ∇×M, is the source responsible for driving B/µo. Because Bis solenoidal, no sources of divergence appear in Maxwell’s equations reformulated in terms of B. The fundamental source representing magnetization is now a current ﬂowing around a small loop (magnetic Sec. 15.3 Characteristic Times 9 dipole). Equations (1)–(4) are, of course, identical in content to (12.0.1)–(12.0.4) because they resulted from the latter by a simple substitution of B/µo−MforH. Yet the model of magnetization was changed by this substitution. As mentioned in Sec. 11.8, both models lead to the same result even when relativistic eﬀects are included, but the Amp` erian model calls for greater care and sophistication, because it contains moving parts (currents) in the rest frame. This is the other reason we chose the magnetic charge model extensively developed by L. J. Chu. Material Idealizations. Much of our analysis of electromagnetic ﬁelds has been based on source idealizations. In the case of sources produced by or induced in media, idealizations were made of the media and of the boundary conditions implied by the induced sources. These are summarized by the ﬁrst and second parts of Table 15.2.1. The case studies listed in Tables 15.1.2–15.1.4 can be used as themes to ex- emplify these idealizations. The Relativity of Perfection. We began modeling EQS and MQS ﬁelds in the presence of media by postulating “perfect” conductors. When we studied materials in more detail, we learned that “perfection” is a relative concept. Useful as are the idealizations summarized in Table 15.2.1, they must be used with proper regard for the approximations made. Those idealizations that involve conductivity depend not only on relative material properties for their validity but on size and time-rates of change as well. These are reviewed in the next section. In each of the three “inﬁnite parameter” idealizations listed in the table, the parameter in one region is large compared to that in another region. The appropriate boundary condition depends on the region of ﬁeld excitation. The idealization makes it possible to approximate the ﬁeld in an “inside” region without regard for what is “outside.” One of the continuity conditions on the surface of the “inside” region is approximated as being homogeneous. Then the ﬁelds in the “outside” region are found by starting with the other continuity condition. Our ﬁrst introduction to this “inside-outside” approach came in Sec. 7.5. With appropriate regard for replacing a source of curl with a source of divergence, the general discussion given in Sec. 9.6 for magnetizable materials is applicable to the other situations as well. 15.3 CHARACTERISTIC TIMES, PHYSICAL PROCESSES, AND APPROXIMATIONS Self-Consistency of Approximate Laws. By dealing with EQS and MQS systems, we concentrated on phenomena that result from approximate forms of Maxwell’s equations. Terms in the “exact” equations were ignored, and ﬁeld con- ﬁgurations were derived from these truncated forms of the equations. This way of solving problems is not unique to electromagnetic ﬁeld theory. Very often it is 10 Overview of Electromagnetic Fields Chapter 15 TABLE 15.2.1 IDEALIZATIONS Idealization Source Constraint Section EQS Perfect Insulator Charges Constrained 4.3-5 Perfectly Polarized PConstrained 6.3 MQS Perfect “Insulator” Currents Constrained 8.1-3 Perfectly Magnetized MConstrained 9.3 Resonant/Traveling-Wave Electrodynamic SystemsSelf-Consistent Charge and Current12.2-4, 12.6 Idealization Boundary Condition Section EQS Perfect Conductor Perfectly Conducting Surfaces Equipotentials4.6-7, 5.1-10 Steady Conduction “Inﬁnite Conductivity”n×E≈0 orn·J≈0 on surface7.2, 9.6 “Inﬁnite” Permittivity n×E≈0 orn·D≈0 on surface9.6 “Inﬁnite” Permeability n×H≈Korn·B≈0 on surface9.6 MQS Perfect Conductor ∂n·B/∂t≈0 on perfectly conducting surfaces8.4, 8.6 10.1, 12.7 13.1-4 necessary to ignore terms that appear in a “more exact” formulation of a physical problem. When this is done, it is necessary to be fully cognizant of the consequences of such approximations. Thus, the energy conservation relations used in the EQS and MQS approximations are special limiting cases of the Poynting theorem obeyed by the full Maxwell equations. The neglect of the displacement current or magnetic induction is equivalent to the neglect of the electric or magnetic energy storage. Next, one needs to ascertain whether the problem has been suﬃciently speci- ﬁed by the approximate form of the equations and which boundary conditions have to be retained, which discarded. The development of the EQS and MQS approxi- mations, with the proof of the uniqueness theorem, provided examples of the devel- opment of a self-consistent formalism within the framework of a set of approximate equations. In systems composed of “perfectly conducting” and “perfectly insulat- ing” media, it is relatively easy to decide whether or not there are subsystems that are EQS or MQS. Sec. 15.3 Characteristic Times 11 A system of perfect conductors surrounded by perfect insulators is likely to be EQS, if it is “open circuit” at zero frequency (a system of capacitors), and MQS, if it is “short circuit” at zero frequency (a system of inductors). However, we are generally not confronted with physical situations in which the materials are labeled as “perfect conductors” or “perfect insulators.” Indeed, with the last half of Chap. 7 and Chap. 10 as background, there comes an awareness that in EQS and MQS systems the term “perfect” usually has very diﬀerent meanings. Presented with a physical object connected to an electrical source, how do we sort the dominant from the inconsequential electromagnetic phenomena? Generally, this is an iterative process with the ﬁrst “guess” based on experience and intuition. With the understanding that the combinations of materials and geometries that are of practical interest are far too diverse to make a few simple rules universally applicable, this section is nevertheless aimed at organizing what we have learned so as to promote the insight required to identify dominant physical processes. From the examination of how ﬁnite conductivity inﬂuences the distribution of the charge density in the EQS systems of Chap. 7 and the current density in the MQS systems of Chap. 10, and from the discussion of the electrodynamics of lossy materials, we have a good idea of what questions must be asked to determine the electromagnetic nature of simple subsystems. A speciﬁc example, familiar from Sec. 14.8, is the conducting block sandwiched between perfectly conducting plane parallel electrodes, shown in Fig. 14.8.1. •First, what are the electrical properties of the materials? Here this question has been reduced to, What are σ, /epsilon1, and µ? The most widely ranging of these parameters is the conductivity σ, which can vary from 10−14S/m in com- mon hydrocarbon liquids to almost 108S/m in copper. Indeed, vacuum and superconducting materials extend this range from absolute zero to inﬁnity. •Second, what is the size scale l? In common engineering systems, lengths of interest range from the submicrometer scales of semiconductor junctions to lengths for power transmission systems in excess of 1000 kilometers. Of course, even this range is small compared to the subnuclear to supergalactic range provided by nature. •Third, what time scale τis of interest? Perhaps the system is driven by a sinusoidally varying source. Then, the time scale would most likely be the reciprocal of the angular frequency 1 /ω. In common engineering practice, frequencies range from 10−2Hz used to characterize insulation to optical fre- quencies in the range of 1015Hz. Again, nature provides frequencies that range even more widely, including the reciprocal of millions of years for terrestrial magnetic ﬁelds in one extreme and the frequencies of gamma rays in the other. Similitude and Maxwell’s Equations. Consider an arbitrary system, shown in Fig. 15.3.1, having the typical length land properties /epsilon1/epsilon1(r), σσ (r), µµ (r) (1) where /epsilon1, σ, and µare typical magnitudes of dielectric constant, conductivity and permeability, and /epsilon1(r), σ(r), and µ(r) are the spatial distributions, normalized so that their peak values are of the order of unity. 12 Overview of Electromagnetic Fields Chapter 15 Fig. 15.3.1 Arbitrary system having typical length l, permittivity /epsilon1, con- ductivity σ, and permeability µ. TABLE 15.3.1 SECTIONS EXEMPLIFYING CHARACTERISTIC TIMES Electroquasistatic charge relaxation time: Sec. 7.7, 7.9 Magnetoquasistatic magnetic (current) diﬀusion time:Sec. 10.2-7 Electromagnetic wave transit time: Sec. 12.2-7, 14.3-4 From our studies of ohmic conductors in EQS and MQS systems, we know that ﬁeld distributions are governed by the charge relaxation time τeand the magnetic diﬀusion time τm, respectively. Moreover, from our study of electromagnetic waves, we know that the transit time for an electromagnetic wave, τem, comes into play with electrodynamic eﬀects. Sections in which these three times were exempliﬁed are listed in Table 15.3.1. Thus, we expect to ﬁnd that in systems having one typical size scale, there are no more than three times that determine the nature of the ﬁelds. τe≡/epsilon1 σ;τm≡µσl2;τem≡l c=l√µ/epsilon1 (2) Actually, the electromagnetic transit time is the geometric mean of the other two times, so that only two of these times are independent. τem=√τeτm (3) With an excitation having the angular frequency ω, the relative distribution of sources and ﬁelds in a system is determined by the product of ωand any pair of these times. This can be seen by writing Maxwell’s equations in normalized form. To that end, we use underbars to denote normalized (dimensionless) variables and normalize the spatial coordinates to the typical length l. The time is normalized to the reciprocal of the angular frequency. (x, y, z ) = (xl, yl, zl), t=t/ω (4) Sec. 15.3 Characteristic Times 13 The ﬁelds and charge density are normalized to a typical electric ﬁeld intensity E. E=EE,H=Er/epsilon1 µH, ρ u=/epsilon1E lρu(5) Then, Maxwell’s equations (12.0.7)–(12.0.10), with the constitutive laws of (1), become ∇·/epsilon1E=ρu(6) ∇×H=ωτem¡1 ωτeE+∂/epsilon1E ∂t¢ (7a) =1 ωτemωτmE+ωτem∂/epsilon1E ∂t(7b) ∇×E=−ωτem∂H ∂t(8) ∇·µH= 0 (9) In writing the alternative forms of Amp` ere’s law, (3) has been used. In a system having the constitutive laws of (1), two parameters specify the ﬁelds predicted by Maxwell’s equations, (6)–(9). These are any pair of the three ratios of the characteristic times of (2) to the typical time of interest. For the sinu- soidal steady state, the time of interest is 1 /ω. Thus, using the version of Amp` ere’s law given by (7a), the dimensionless parameters ( ωτem, ωτe) specify the ﬁelds. Using (7b), the parameters are ( ωτem, ωτm). Characteristic Times and Lengths. Evidently, the three dimensionless pa- rameters formed by multiplying the characteristic times of (2) by the frequency, ω, (or the reciprocal of some other time typifying the dynamics), are the key to sorting out physical processes. ωτe=ω/epsilon1 σ; ωτm=ωµσl2; ωτem=ωl√µ/epsilon1 (10) Given two of these parameters and hence the third, we have some clues as to what physical processes are dominant. However, even in a subsystem typiﬁed by one permittivity, one conductivity, and one permeability, other parameters may be needed to specify the geometry. Every ratio of dimensions is another dimensionless parameter! To begin with, suppose that we are dealing with a system where all of the dimensions are on the order of the typical length l. The characteristic times make evident why quasistatic systems are either EQS or MQS. They also determine how the eﬀects of ﬁnite conductivity come into play either through charge relaxation or magnetic diﬀusion as the frequency is raised. Since the electromagnetic transit time is the geometric mean of the charge relaxation and magnetic diﬀusion times, (3), τemmust lie between the other two times. Thus, the three times are in one of two orders. Either τm< τe, in which case 14 Overview of Electromagnetic Fields Chapter 15 Fig. 15.3.2 Ordering of reciprocal of characteristic times on the frequency axis. the order of reciprocal times is as shown in Fig. 15.3.2a, or the reverse is true, and the order is as in Fig. 15.3.2b. Moreover, if τeis well removed from τem, then we are assured that τmis also very diﬀerent from τem. As the frequency is raised, we ﬁrst encounter either the charge relaxation phenomena typical of EQS subsystems (Fig. 15.3.2a) or the magnetic diﬀusion phenomena of MQS subsystems (Fig. 15.3.2b). The respective quasistatic laws for EQS and MQS systems apply for frequencies ranging above the ﬁrst reciprocal time but below the reciprocal electromagnetic transit time. In both cases, the frequency is well below the reciprocal of the electromagnetic delay time. The EQS laws follow from (6)–(9) using the ﬁrst form of (7). A physical situation is characterized by the EQS laws, when the term on the right hand side of Faraday’s law, (8), is negligible. From Amp` ere’s law we gather that His of the order of ωτemEwhen ωτe>1, and of order τem/τewhen ωτe<1. In the former case, in which the displacement current density dominates over the conduction current density, one ﬁnds for the right hand side in Faraday’s law: ( ωτem)2E. In the latter case, in which the conduction current density is larger than the displacement current density, the right hand side of (8) is ωτ2 em/τeE. Thus the source of curl in Faraday’s law can be neglected when ( ωτem)2/lessmuch1 orωτem/τe/lessmuch1 whichever is a more stringent limit on ω. The laws of EQS prevail. An analogous, but simpler, argument arrives at the laws of MQS. The argument is simpler, because there is no analog to unpaired electric charge. In cases where the ordering of characteristic times is as in Fig. 15.3.2b, the MQS laws apply for frequencies beyond the reciprocal magnetic diﬀusion time but again falling short of the electromagnetic transit time. This can be seen from the normalized Maxwell’s equations, this time using (7b). Because ωτem/lessmuch1, the last term in (7b) (the displacement current) is negligible. Thus, we are led to the primary MQS laws, Amp` ere’s law with the displacement current neglected and the continuity law for the magnetic ﬂux density (9). This time, it follows from Amp` ere’s law [(7b) with the last term neglected] that H≈(ωτm/ωτem)E, so that the right-hand side of Faraday’s law, (8), is of the order of ωτm. Thus, the MQS laws are (10.0.1)–(10.0.3). As the frequency is raised, so that we move from left to right along the fre- quency axes of Fig. 15.3.2, we expect dynamical phenomena associated with charge relaxation, electromagnetic waves, and magnetic diﬀusion to come into play as the frequency comes into the range of the respective reciprocal characteristic times. Actually, because the dynamics can establish their own length scales (for example, the skin depth), matters are sometimes not so simple. However, insight is gained by observing that the length scale lorders these critical frequencies. With the ob- jective of picturing the electromagnetic phenomena in a plane, in which one axis reﬂects the eﬀect of the frequency while the other axis represents the length scale, Sec. 15.3 Characteristic Times 15 Fig. 15.3.3 In plane where the vertical axis denotes the log of the length scale normalized to the characteristic length deﬁned by (14), and the horizontal axis is the angular frequency multiplied by the charge relaxation time τe, the three lines denote possible boundaries between regimes. we normalize the frequency to the one characteristic time, τe, that does not de- pend on the length. Thus, the frequency conditions for eﬀects of charge relaxation, magnetic diﬀusion, and electromagnetic waves to be important are, respectively, ωτe= 1 (11) ωτm= 1⇒ωτe= (l/l∗)−2(12) ωτem= 1⇒ωτe= (l/l∗)−1(13) where the characteristic length l∗is l∗≡1 σp /epsilon1/µ (14) In a plane in which the coordinates are essentially the length scale and the frequency, the lines along which the frequency is equal to the respective reciprocal characteristic times are shown in Fig. 15.3.3. The vertical axis denotes the log of the length scale normalized to the characteristic length, while the horizontal axis is the log of the frequency multiplied by the charge relaxation time. Thus, the origin is where the length is equal to l∗and the frequency is equal to 1 /τe. Note that for systems having a typical length lless than the reciprocal of the characteristic impedance conductivity product, l∗, the ordering of times is as in Fig. 15.2.1a. If the length is greater than this characteristic length, then the ordering is as in Fig. 15.2.1b. At least for systems having one length scale land one characteristic time 1 /ω,the system can be MQS only if lis larger than l∗and can be EQS only if lis smaller than l∗. The MQS and EQS regimes of Fig. 15.3.3 both reduce to quasistationary conduction (QSC) at frequencies such that ωτm/lessmuch1 and ωτe/lessmuch1, respectively. Since σis such a widely varying parameter, the values of l∗also have a wide range. Table 15.3.2 illustrates this fact. In water having physiological conductivity 16 Overview of Electromagnetic Fields Chapter 15 (in ﬂesh), the characteristic times would coincide if the length scale were about 12 cm at a characteristic frequency ( ωτe= 1) f = 45 MHz. For lengths less than about 12 cm, the ordering would be as in Fig. 15.3.2a and for longer lengths, as in Fig. 15.3.2b. However, in copper it would require that the characteristic length be less than an atomic distance to make τeexceed τm. On such a short length scale, the conductivity model is not valid.1In the opposite extreme, a layer of corn oil about 60,000 miles thick would be required to make τmexceed τe! Example 15.3.1. Overview of TEM Fields in Open Circuit Transmission Line Filled with Lossy Material (continued) In Sec. 14.8, we considered the nature of the electromagnetic ﬁelds in a conductor sandwiched between “perfectly conducting” plates. Example 14.8.2 was devoted to an overview of electromagnetic regimes pictured in the length-time plane, Fig. 14.8.3, redrawn as Fig. 15.3.3. As the frequency was raised in that example with l/greatermuchl∗, the lineωτm= 1 indicated that quasi-stationary conduction had given way to magnetic diﬀusion (the resistor had become a system of distributed resistors and inductors). In that speciﬁc example, this was the line at which the long wave approximation broke down, βl≈1. With l/lessmuchl∗, we have seen that as the frequency was raised, the crossing of the line ωτe= 1 denoted that a resistor had changed into a system of distributed resistors in parallel with distributed capacitors. This example has a misleading simplicity that can be traced to the fact that it actually possesses more than one length scale and conductivity. To impose the TEM ﬁelds by means of the source, it was necessary to envision the slab of conductor as making perfect electrical contact with perfectly conducting plates. In reality, the boundary condition used to represent these plates implies conditions on still other parameters, notably the electrical properties and thickness of the plates. As the frequency is raised for a system in the upper half-plane ( llarger than the matching length), why do we not see a transition to electromagnetic waves at ωτem= 1 rather than ωτe= 1? The perfectly conducting plates force the displacement current to compete with the conduction current on its “own” length scale (either the skin depth or the electromagnetic wavelength). Thus, in this example, we do not make a transition from magnetic diﬀusion (with a penetration length determined by the skin depth δ) to a damped electromagnetic wave (with a decay length of twice l∗) until the electromagnetic wavelength λ= 2π/√µ/epsilon1ωhas become as short as the skin depth. Both are decreasing with increasing frequency. However, the skin depth (which decreases as 1 /√ω) is equal to the wavelength (which decreases as 1 /ω) only as the frequency reaches ωτe= 2π2(for present purposes, “ ωτe= 1”). In the lower half-plane, where systems are smaller than the characteristic length, why was the transition at ωτe= 1 evident in the surface current density in the plates but not in the spatial distribution of the ﬁelds? The electric ﬁeld was found to remain uniform until the frequency had been raised to ωτem= 1. Here again, the “perfectly conducting” plates obscure the general situation. The conducting block has uniform conductivity. As a result, it can support no volume charge density, regardless of the frequency. In the EQS limit, it is the charge density that shapes the electric ﬁeld distribution. Here the only charges are at the interfaces between the block and the perfectly conducting plates. Until magnetic induction comes into play at ωτem= 1, these surface charges assume whatever distribution they must 1Put another way, on a time scale as short as the charge relaxation time in a metal, the inertia of the electrons responsible for the conduction would come into play. (S. Gruber, “On Charge Relaxation in Good Conductors,” Proc. IEEE , Vol. 61 (1973), pp. 237-238. The inertial force is not included in the conductivity model. Sec. 15.4 Energy, Power, and Force 17 to be consistent with an irrotational electric ﬁeld. As a result, the plates make the EQS ﬁelds essentially uniform, and the appropriate model simpliﬁes to one lumped parameter Cin parallel with one lumped parameter R. 15.4 ENERGY, POWER, AND FORCE Maxwell’s equations attribute an excitation ( EandH) to every point in space. Consistent with this view, energy density and power ﬂow density must be associ- ated with every point in space as well. Poynting’s theorem, Sec. 11.2, does that. Poynting’s theorem identiﬁes energy storage and dissipation associated with the polarization and magnetization processes. Each self-consistent macroscopic set of equations must possess an energy con- servation principle, maybe including terms describing transformation of energy into other forms, like heat, if dissipation is present. An example was given in Sec. 11.3 of a conservation principle for the approximate description of EQS ﬁelds with a density of power ﬂow vector that was diﬀerent from E×H. This alternate form of an energy conservation principle was better suited to the EQS description, because it did not contain the Hﬁeld which is not usually evaluated in the EQS approxi- mation. Instead, the charge conservation law (derived from Amp` ere’s law) was used to ﬁnd the currents ﬂowing in the system. An important application of the concept of energy was the derivation of the force on macroscopic material. The force on a dielectric or magnetic object com- puted from energy change can include correctly the contributions to the net force from fringing ﬁelds even though the ﬁeld expressions neglect them, if the energy associated with the fringing ﬁeld does not change in a small displacement of the object. Energy and Quasistatics. Because magnetic and electric energy storages, respectively, are negligible in EQS and MQS systems, a comparison of energy den- sities can also be used to establish the validity of a quasistatic approximation. Speciﬁcally, we will see that in systems characterized by one length scale, the ratio of magnetic to electric energy storage takes the form wm we=K¡l l∗¢2(1) where l∗is the characteristic length l∗≡1 σp /epsilon1/µ (2) familiar from Secs. 14.82and 15.3 and Kis of the order of unity. 2In Sec. 14.8, twice this length was found to be the decay length for an electromagnetic wave. 18 Overview of Electromagnetic Fields Chapter 15 Fig. 15.4.1 Low-frequency equivalent circuits and associated ordering to reciprocal times. Energy arguments can also be the basis for simple models that modestly extend the frequency range of quasi-stationary conduction. A second object in this section is the illustration of how these models are deduced. As the frequency is raised, one of two processes leads to a modiﬁcation in the ﬁeld sources, and hence of the ﬁelds. If lis less than l∗, so that 1 /τeis the ﬁrst reciprocal characteristic time encountered as ωis raised, then the current density is progressively altered to supply unpaired charge to regions of nonuniform σand/epsilon1. Alternatively, if lis larger than l∗, so that 1 /τmis the shortest reciprocal characteristic time, magnetic induction alters the current density notonly in its magnitude and time dependence but in its spatial distribution as well. Fully dynamic ﬁelds, in which all three (or more) characteristic times are of the same order of magnitude are diﬃcult to analyze because the distribution of sources is not known until the ﬁelds have been solved selfconsistently, often a diﬃcult task. However, if the frequency is lower than the lowest reciprocal time, the ﬁeld distributions still approximate those for stationary conduction. This makes it possible to approximate the energy storages, and hence to identify both the conditions for the system to be EQS or MQS and to develop models that are appropriate for frequencies approaching the lowest reciprocal characteristic time. The ﬁrst step in this process is to determine the quasi-stationary ﬁelds. The second is to use these ﬁelds to evaluate the total electric and magnetic energy storages as well as the total energy dissipation. we=Z V1 2/epsilon1E·Edv;wm=Z V1 2µH·Hdv;pd=Z VσE·Edv (3) If it is found that the ratio of magnetic to electric energy storage takes the form of (1), and that if lis either very small or very large compared to the characteristic length, then we can presumably model the system by either the R-C or the L-R circuit of Fig. 15.4.1. As the third step, parameters in these circuits are determined by compar- ingwe, wm, and pd, as found from the QSC ﬁelds using (3), to these quantities determined in terms of the circuit variables. we=1 2Cv2; wm=1 2Li2; pd=Ri2(4) In general, the circuit models are valid only up to frequencies approaching, but not equal to, the lowest reciprocal time for the system. In the following example, we Sec. 15.4 Energy, Power, and Force 19 will ﬁnd that the R-C circuit is an exact model for the EQS system, so that the model is valid even for frequencies beyond 1 /τe. However, because the ﬁelds can be strongly altered by rate processes if the frequency is equal to the lowest reciprocal time, it is generally not appropriate to use the equivalent circuits except to take into account energy storage eﬀects coming into play as the frequency approaches 1/RC orR/L. Example 15.4.1. Energy Method for Deriving an Equivalent Circuit The block of uniformly conducting material sandwiched between plane parallel perfectly conducting plates, as shown in Fig. 14.8.1, was the theme of Sec. 14.8. This gives the opportunity to see how the low-frequency model developed here ﬁts into the general picture provided by that section. In the conducting block, the quasi-stationary conduction (QSC) ﬁelds have the distributions E=v aix;H=σv aziy (5) The total electric and magnetic energies and total dissipation follow from an integration of the respective densities over the volume of the system in accordance with (3) we=wal1 2/epsilon1v2 a2;wm=waµ 6¡σv a¢2l3;pd=a wlσi2(6) where vandiare the terminal voltage and current. Comparison of (4) and (6) shows that C=lw/epsilon1 a; L=aµl 3w; R=a lwσ(7) Because the entire volume of the system considered here has uniform prop- erties, there are no sources of the electric ﬁeld (charge densities) in the volume of the system. As a result, the capacitance Cfound here is no diﬀerent than if the vol- ume were ﬁlled with a perfectly insulating material. By contrast, if the slab were of nonuniform conductivity, as in Example 7.2.1, the capacitance, and hence equivalent circuit, found by this energy method would not be so “obvious.” The inductance of the equivalent circuit does reﬂect a distribution of the source of the magnetic ﬁeld, for the current density is distributed throughout the volume of the slab. By using the energy argument, we have acknowledged that there is a distribution of current paths, each having a diﬀerent ﬂux linkage. Strictly, when the ﬂux linked by any current path is the same, inductance is only deﬁned for perfectly conducting current paths. Which equivalent circuit is appropriate? Here we decide by comparing the stored energies. wm we=1 3¡l l∗¢2(8) Thus, as we anticipated with (1), the system can be EQS if l/lessmuchl∗and MQS ifl/greatermuchl∗. The appropriate equivalent circuit in Fig. 15.4.1 is the R−Ccircuit if l/lessmuchl∗and is the L−Rcircuit if l/greatermuchl∗. The simple circuits of Fig. 15.4.1 are not generally valid if the frequency reaches the reciprocal of the longest characteristic time, since the ﬁeld distributions 20 Overview of Electromagnetic Fields Chapter 15 have changed by then. In terms of the circuit elements, this means that in order for the circuits to be equivalent to the physical system, the time rates of change must remain slow enough so that ωRC < 1 orωL/R < 1. Sec. 15.3 Problems 21 P R O B L E M S 15.1 Source and Material Conﬁgurations 15.1.1 A theme from Chap. 5 on has been the use of orthogonal modes to represent ﬁeld solutions and satisfy boundary conditions. Make a table identifying examples and problems illustrating this theme. 15.2 Macroscopic Media 15.2.1 Field lines in the vicinity of a spherical interface between materials (a) and (b) are shown in Fig. P15.2.1. In each case, describe four idealized physical situations for which the ﬁeld lines would be appropriate. Fig. P15.2.1 Fig. P15.2.2 15.2.2 Dipoles at the center of a spherical region and associated ﬁelds are shown in Fig. P15.2.2. In each case, describe four appropriate idealized physical situations. 15.3 Characteristic Times, Physical Processes, and Approximations 22 Overview of Electromagnetic Fields Chapter 15 15.3.1 In Fig. 15.3.3, a typical length and time are considered the independent parameters. Suppose that we wish to see the eﬀect of varying the conduc- tivity with the size held ﬁxed. For example, with not only the size but the frequency ﬁxed, the material might be cooling from a very high tem- perature where it is molten and an ionic conductor to a low temperature where it is a good insulator. Using the conductivity rather than the length for the vertical axis, select a normalization time for the horizontal axis that is independent of conductivity, and construct a diagram analogous to Fig. 15.3.3. Identify a “characteristic” conductivity, σ∗, for normalizing the conductivity. 15.3.2 Figure 7.5.3 shows a circular conductor carrying a current that is returned through a coaxial “perfectly” conducting “can.” For suﬃciently low fre- quencies, the electric ﬁeld and surface charge densities are as shown in Fig. 7.5.4. The magnetic ﬁeld is described in Example 11.3.1 where the eﬀect of the washer-shaped conductor is neglected. (a) Sketch EandH, as well as the distribution of ρuandJu. (b) Suppose that the length Lis on the order of the radius (a), and (b) is not much smaller than (a). As the frequency is raised, argue that either charge relaxation will ﬁrst dominate in revising the ﬁeld distribution as in Fig. P15.3.2a, or magnetic diﬀusion will dominate as in Fig. P15.3.2b. In the latter case, describe the current distribution in the conductor by associating it with an example and a demonstration in this text. (c) With Lallowed to be large compared to (a), under what circum- stances will the system behave as the lossy transmission line of Fig. 14.7.1 with G= 0? Discuss the EQS and MQS limits where this model applies. Fig. P15.3.2 15.4 Energy, Power, and Force 15.4.1 For the system considered in Prob. 15.3.2, use the energy approach to Sec. 15.4 Problems 23 identify the parameters in the low frequency equivalent circuits of Fig. 15.4.1, and write the ratio of energies in the form of (1). Ignore the eﬀect of the washer-shaped conductor. 1 APPENDIX 1.1 VECTOR OPERATIONS A vector is a quantity which possesses magnitude and direction. In order to describe a vector mathematically, a coordinate system having orthogonal axes is usually cho- sen. In this text, use is made of the Cartesian, circular cylindrical, and spherical coordinate systems. In these three-dimensional systems, any vector is completely described by three scalar quantities. For example, in Cartesian coordinates, a vec- tor is described with reference to mutually orthogonal coordinate axes. Then the magnitude and orientation of the vector are described by specifying the three pro- jections of the vector onto the three coordinate axes. In representing a vector1Amathematically, its direction along the three or- thogonal coordinate axes must be given. The direction of each axis is represented by a unit vector i, that is, a vector of unit magnitude directed along the axis. In Cartesian coordinates, the three unit vectors are denoted ix,iy,iz. In cylindrical coordinates, they are ir,iφ,iz, and in spherical coordinates, ir,iθ,iφ.A, then, has three vector components, each component corresponding to the projection of Aonto the three axes. Expressed in Cartesian coordinates, a vector is deﬁned in terms of its components by A=Axix+Ayiy+Aziz (1) These components are shown in Fig. A.1.1. 1Vectors are usually indicated either with boldface characters, such as A, or by drawing a line (or an arrow) above a character to indicate its vector nature, as in ¯Aor/vectorA. 1 2 Appendix Chapter 1 Fig. A.1.1 Vector Arepresented by its components in Cartesian coordinates and unit vectors i. Fig. A.1.2 (a) Graphical representation of vector addition in terms of spe- ciﬁc coordinates. (b) Representation of vector addition independent of speciﬁc coordinates. Vector Addition. The sum of two vectors A=Axix+Ayiy+AzizandB= Bxix+Byiy+Bzizis eﬀected by adding the coeﬃcients of each of the components, as shown in two dimensions in Fig. A.1.2a. A+B= (Ax+Bx)ix+ (Ay+By)iy+ (Az+Bz)iz (2) From (2), then, it should be clear that vector addition is both commutative, A+B= B+A, and associative, ( A+B) +C=A+ (B+C). Graphically, vector summation can be performed without regard to the coor- dinate system, as shown in Fig. A.1.2b, by noticing that the sum A+Bis a vector directed along the diagonal of a parallelogram formed by AandB. It should be noted that the representation of a vector in terms of its com- ponents is dependent on the coordinate system in which it is carried out. That is, changes of coordinate system will require an appropriate vector transformation. Fur- ther, the variables used must also be transformed. The transformation of variables and vectors from one coordinate system to another is illustrated by considering a transformation from Cartesian to spherical coordinates. Example 1.1.1. Transformation of Variables and Vectors We are given variables in terms of x, y, and zand vectors such as A=Axix+ Ayiy+Aziz. We wish to obtain variables in terms of r, θ, and φand vectors ex- pressed as A=Arir+Aθiθ+Aφiφ. In Fig. A.1.3a, we see that the point Phas two Sec. 1.1 Appendix 3 Fig. A.1.3 Speciﬁcation of a point Pin Cartesian and spherical co- ordinates. (b) Transformation from Cartesian coordinate xto spherical coordinates. (c) Transformation of unit vector in xdirection into spher- ical coordinate coordinates. representations, one involving the variables x, yandzand the other, r, θandφ. In particular, from Fig. A.1.3b, xis related to the spherical coordinates by x=rsinθcosφ (3) In a similar way, the variables yandzevaluated in spherical coordinates can be shown to be y=rsinθsinφ (4) z=rcosθ (5) The vector Ais transformed by resolving each of the unit vectors ix,iy,iz in terms of the unit vectors in spherical coordinates. For example, ixcan ﬁrst be 4 Appendix Chapter 1 Fig. A.1.4 Illustration for deﬁnition of dot product. resolved into components in the orthogonal coordinates ( x/prime, y/prime, z) shown in Fig. A.1.3c. By deﬁnition, y/primeis along the intersection of the φ= constant and the x−y planes. Also in the x−yplane is x/prime, which is perpendicular to the y/prime−zplane. Thus, sinφ,cosφ, and 0 are the components of ixalong the x/prime, y/prime, and zaxes respectively. These components are in turn resolved into components along the spherical coordi- nate directions by recognizing that the component sin φalong the x/primeaxis is in the −iφdirection while the component of cos φalong the y/primeaxis resolves into components cosφcosθin the direction of iθ, and cos φsinθin the irdirection. Thus, ix= sin θcosφir+ cos θcosφiθ−sinφiφ (6) Similarly, iy= sin θsinφir+ cos θsinφiθ+ cos φiφ (7) iz= cos θir−sinθiθ (8) It must be emphasized that the concept of a vector is independent of the coordinate system. (In the same sense, in Chaps. 2 and 4, vector operations are deﬁned independently of the coordinate system in which they are expressed.) A vector can be visualized as having the direction and magnitude of an arrow-tipped line element. This picture makes it possible to deal with vectors in a geometrical language that is independent of the choice of a particular coordinate system, one that will now be used to deﬁne the most important vector operations. For analytical or numerical purposes, the operations are usually carried out in coordinate notation. Then, as illustrated, either in the text that follows or in the problems, each operation will be evaluated in a Cartesian coordinate system. Deﬁnition of Scalar Product. Given vectors AandBas illustrated in Fig. A.1.4, the scalar, or dot product, between the two vectors is deﬁned as A·B=|A||B|cosθ (9) where θis the angle between the two vectors. It follows directly from its deﬁnition that the scalar product is commutative. A·B=B·A (10) The scalar product is also distributive. (A+B)·C=A·C+B·C (11) Sec. 1.1 Appendix 5 Fig. A.1.5 Illustration for deﬁnition of vector-product. To see this, note that A·Cis the projection of AontoCtimes the magnitude of C,|C|, and B·Cis the projection of BontoCtimes |C|. Because projections are additive, (11) follows. These two properties can be used to deﬁne the scalar product in terms of the vector components in Cartesian coordinates. According to the deﬁnition of the unit vectors, ix·ix=iy·iy=iz·iz= 1 ix·iy=ix·iz=iy·iz= 0 (12) With AandBexpressed in terms of these components, it follows from the dis- tributive and commutative properties that A·B=AxBx+AyBy+AzBz (13) Thus, in agreement with (9), the square of the magnitude of a vector is A·A=|A|2=A2 x+A2 y+A2 z (14) Deﬁnition of Vector Product. The cross-product of vectors AandBis a vector Chaving a magnitude |C|=|A||B|sinθ (15) and having a direction perpendicular to both AandB. Geometrically, the mag- nitude of Cis the area of the parallelogram formed by the vectors AandB. The vector Chas the direction of advance of a right-hand screw, as though driven by rotating AintoB. Put another way, a right-handed coordinate system is formed byA−B−C, as is shown in Fig. A.1.5. The commonly accepted notation for the cross-product is C=A×B (16) It is useful to note that if the vector Ais resolved into two mutually per- pendicular vectors, A=A⊥+A/bardbl, where A⊥lies in the plane of AandBand is perpendicular to BandA/bardblis parallel to B, then A×B=A⊥×B (17) 6 Appendix Chapter 1 Fig. A.1.6 Graphical representation showing that the vector-product is dis- tributive. This equality follows from the fact that both cross-products have equal magnitude (since |A⊥×B|=|A⊥||B|and|A|⊥|=|A|sinθ) and direction (perpendicular to bothAandB). The distributive property for the cross-product, (A+B)×D=A×D+B×D (18) can be shown using (17) and the geometrical construction in Fig. A.1.6 as follows. First, note that ( A+B)⊥= (A⊥+B⊥), where ⊥denotes a component in the planes of AandDorBandD, respectively, and perpendicular to D. Thus, (A+B)×D= (A+B)⊥×D= (A⊥+B⊥)×D (19) Now, we need only show that (A⊥+B⊥)×D=A⊥×D+B⊥×D (20) This equation is given graphical expression in Fig. A.1.6 by the vectors A⊥,B⊥, and their sum. To within a factor of |D|, the three vectors A⊥×D,B⊥×D, and their sum, are, respectively, the vectors A⊥,B⊥, and their sum, rotated by 90 degrees. Thus, the vector addition property already shown for A⊥+B⊥also applies to A⊥×D+B⊥×D. Because interchanging the order of two vectors calls for a reassignment of the direction of the product vector (the direction of Cin Fig. A.1.5), the commutative property does not hold. Rather, A×B=−B×A (21) Using the distributive law, the vector product of two vectors can be con- structed in terms of their Cartesian coordinates by using the following properties of the vector products of the unit vectors. ix×ix= 0 ix×iy=iz iy×iy= 0 iy×iz=−iz×iy=ix iz×iz= 0 ix×iz=−iz×ix=−iy (22) Sec. 1.1 Appendix 7 Fig. A.1.7 Graphical representation of scalar triple product. Thus, A×B=ix(AyBz−AzBy) +iy(AzBx−AxBz) +iz(AxBy−AyBx)(23) A useful mnemonic for ﬁnding the cross-product in Cartesian coordinates is realized by noting that the right-hand side of (23) is the determinant of a matrix: A×B=¯¯¯¯¯ixiyiz AxAyAz BxByBz¯¯¯¯¯(24) The Scalar Triple Product. The deﬁnition of the scalar triple product of vectors A,B, and Cfollows from Fig. A.1.7, and the deﬁnition of the scalar and vector products. A·(B×C) = [|A|cos(A,B×C)][|B||C|sin(B,C)] (25) The scalar triple product is equal to the volume of the parallelepiped having the three vectors for its three bases. That is, in (25) the second term in square brackets is the area of the base parallelogram in Fig. A.1.7 while the ﬁrst is the height of the parallelopiped. The scalar triple product is positive if the three vectors form a right-handed coordinate system in the order in which they are written; otherwise it is negative. Hence, a cyclic rearrangement in the order of the vectors leaves the value of the product unchanged. A·(B×C) =B·(C×A) =C·(A×B) (26) It follows that the placing of the cross and the dot in a scalar triple product is arbitrary. The cross and dot can be interchanged without aﬀecting the product. Using the rules for evaluating the dot product and the cross-product in Carte- sian coordinates, we have A·(B×C) =Ax(ByCz−BzCy) +Ay(BzCx−BxCz) +Az(BxCy−ByCx) (27) The Double Cross-Product. Consider the vector product A×(B×C). Is there another, sometimes more useful, way of expressing this double cross-product? 8 Appendix Chapter 1 Fig. A.1.8 Graphical representation of double cross-product. Since the product B×Cis perpendicular to the plane deﬁned by BandC, then the ﬁnal product A×(B×C) must lie in the plane of BandC. Hence, the vector product must be expressible as a linear combination of the vectors BandC. One way to ﬁnd the coeﬃcients of this linear combination is to evaluate the product in Cartesian coordinates. Here we prefer to use a geometric derivation. Because the vector B×Cis perpendicular to the plane deﬁned by the vectors BandC, it follows from Fig. A.1.7 that A×(B×C) =A/prime×(B×C) (28) where A/primeis the projection of Aonto the plane deﬁned by BandC. Next, we separate the vector Cinto a component parallel to B,C/bardbl, and a component per- pendicular to B,C⊥, as shown by Fig. A.1.8, so that A×(B×C) =A/prime×(B×C⊥) (29) Then, according to the properties of the cross-product, the magnitude of the vector product is given by |A×(B×C)|=|A/prime||B||C⊥| (30) and the direction of the vector product is orthogonal to A/primeand lies in the plane deﬁned by the vectors BandC, as shown in Fig. A.1.8 A rule for constructing a vector perpendicular to a given vector, A/prime, in an x−yplane is as follows. First, the two components of A/primewith respect to any two orthogonal axes ( x, y) are determined. Here these are the directions of C⊥andB with components A/prime·C⊥, andA/prime·B, respectively. Then, a new vector is constructed by interchanging the xandycomponents and changing the sign of one of them. According to this rule, Fig. A.1.8 shows that the vector A×(B×C) is given by A×(B×C) = (A/prime·C⊥)B−(A/prime·B)C⊥ (31) Now, because C/bardblhas the same direction as B, (A/prime·B)C/bardbl= (A/prime·C/bardbl)B, (32) and addition of (31) gives A×(B×C) =A/prime·(C⊥+C/bardbl)B−(A/prime·B)(C⊥+C/bardbl) (33) Sec. 1.1 Appendix 9 Now observe that A/prime·C=A·CandA/prime·B=A·B(which follow from the deﬁnition ofA/primeas the projection of Ainto the B−Cplane), and the double cross-product becomes A×(B×C) = (A·C)B−(A·B)C (34) This result is particularly convenient because it does not contain any special nota- tion or projections. The vector identities found in this Appendix are summarized in Table III at the end of the text. 2 APPENDIX 2.1 LINE AND SURFACE INTEGRALS Consider a path connecting points (a) and (b) as shown in Fig. A.2.1. Assume that a vector ﬁeld A(r) exists in the space in which the path is situated. Then the line integral of A(r) is deﬁned by Z(b) (a)A·ds (1) To interpret (1), think of the path between (a) and (b) as subdivided into diﬀerential vector segments ds. At every vector segment, the vector A(r) is evaluated and the dot product is formed. The line integral is then deﬁned as the sum of these dot products in the limit as dsapproaches zero. A line integral over a path that closes on itself is denoted by the symbolH A·ds. Fig. A.2.1 Conﬁguration for integration of vector ﬁeld Aalong line having diﬀerential length dsbetween points (a) and (b). 1 2 Appendix Chapter 2 Fig. A.2.2 Integration line having shape of quarter segment of a circle with radius Rand diﬀerential element ds. To perform a line integration, the integral must ﬁrst be reduced to a form that can be evaluated using the rules of integral calculus. This is done with the aid of a coordinate system. The following example illustrates this process. Example 2.1.1. Line Integral Given the two-dimensional vector ﬁeld A=xix+axyiy (2) ﬁnd the line integral along a quarter circle of radius Ras shown in Fig. A.2.2. Using a Cartesian coordinate system, the diﬀerential line segment dshas the components dxanddy. ds=ixdx+iydy (3) Nowxandyare not independent but are constrained by the fact that the integration path follows a circle deﬁned by the equation x2+y2=R2(4) Diﬀerentiation of (4) gives 2xdx+ 2ydy= 0 (5) and therefore dy=−x ydx (6) Thus, the dot product A·dscan be written as a function of the variable xalone. A·ds=xdx+a xydy = (x−ax2)dx (7) When the path is described in the sense shown in Fig. A.2.4, xdecreases from Rto zero. Therefore, Z A·ds=Z0 R(x−ax2)dx=¡x2 2−ax3 3¢¯¯¯¯0 R=aR3 3−R2 2(8) If the path is not expressible in terms of an analytic function, the evaluation of the line integral becomes diﬃcult. If everything else fails, numerical methods can be employed. Sec. 2.2 Appendix 3 Surface Integrals. Given a vector ﬁeld A(r) in a region of space containing a speciﬁed (open or closed) surface S, an important form of the surface integral of AoverSis Z SA·da (9) The vector dahas a magnitude that represents the diﬀerential area of a surface element and a direction that is normal to that area. To interpret (9), think of the surface Sas subdivided into these diﬀerential area elements da. At each area element, the diﬀerential scalar A·dais evaluated and the surface integral is deﬁned as the sum of these dot products over Sin the limit as daapproaches zero. The surface integralR SA·dais also called the “ﬂux” of the vector Athrough the surface S. To evaluate a surface integral, a coordinate system is introduced in which the integration can be performed according to the methods of integral calculus. Then the surface integral is transformed into a double integral in two independent variables. This is best illustrated with the aid of a speciﬁc example. Example 2.1.2. Surface Integral Given the vector ﬁeld A=ixx (10) ﬁnd the surface integralR SA·da, where Sis one eighth of a spherical surface of radius Rin the ﬁrst octant of a sphere (0 ≤φ≤π/2,0≤θ≤π/2). Because the surface lies on a sphere, it is best to carry out the integration in spherical coordinates. To transform coordinates from Cartesian to spherical, recall from (A.1.3) that the xcoordinate is related to r, θ, and φby x=rsinθcosφ (11) and from (A.1.6), the unit vector ixis ix= sin θcosφir+ cos θcosφiθ−sinφiφ (12) Therefore, because the area element dais da=irR2sinθdθdφ (13) the surface integral becomes Z SA·da=Zπ/2 0dθZπ/2 0dφR3sin3θcos2φ =πR3 4Zπ/2 0dθsin3θ=πR3 6(14) A surface integral of a vector Aover a closed surface is indicated by I SA·da (15) 4 Appendix Chapter 2 Note also that we use a single integral sign for a surface integral, even though, in fact, two integrations are involved when the integral is actually evaluated in terms of a coordinate system. 2.2 PROOF THAT THE CURL OPERATION RESULTS IN A VECTOR The deﬁnition [curlA]n= lim a→01 aI A·ds (1) assigns a scalar, [ curlA]n, to each direction nat the point Punder consideration. The limit must be independent of the shape of the contour C(as long as all its points approach the point Pin the limit as the area aof the contour goes to zero). The identiﬁcation of curlAas a vector also implies a proper dependence of this limit upon the orientation of the normal nofa. The purpose of this appendix is to show that these two requirements are indeed satisﬁed by (1). We shall prove the following facts: 1. At a particular point ( x, y, z ) lying in the plane speciﬁed by its normal vector n, the quantity on the right in (1) is independent of the shape of the con- tour. (The notation [ curlA]n, is introduced at this stage only as a convenient abbreviation for the expression on the right.) 2. If [ curlA]nis indeed the component of a vector [ curlA] in the ndirection andnis a unit normal in the ndirection, then [curlA]n= [curl A]·n (2) where [ curlA] is a vector deﬁned at the point ( x, y, z ). The proof of (1) follows from the fact that any closed contour integral can be built up from a superposition of contour integrals around a large number of rectangular contours Ci, as shown in Fig. A.2.3. All rectangles have sides ∆ ξ,∆η. If the entire contour containing the rectangles is small ( a→0), then the contour integral around each rectangle diﬀers from that for the contour Coat the origin only by a term on the order of the linear dimension of the contour, a1/2, times the area ∆ ξ∆η. This is true provided that the distance from the origin to any point on the contour does not exceed a1/2by an order of magnitude and that Ais once diﬀerentiable in the neighborhood of the origin. We have 1 ∆ξ∆ηI CiA·ds=1 ∆ξ∆ηI CoA·ds+O(a1/2) (3) Therefore, 1 aI CA·ds=X i1 aI CiA·ds=∆ξ∆η aX i1 ∆ξ∆ηI CiA·ds =N∆ξ∆η a·1 ∆ξ∆ηI CoA·ds+O(a1/2)¸ (4) Sec. 2.2 Appendix 5 Fig. A.2.3 Separation of closed contour integral into large number of inte- grals over rectangular contours. Fig. A.2.4 Arbitrary incremental contour integral having normal nanalyzed into integration contours enclosing surface, having normals in the directions of the Cartesian coordinates. where Nis the number of rectangles into which the contour Chas been subdivided. However, N=a/(∆ξ∆η), and therefore we ﬁnd lim a→0X i1 aI CiA·ds= lim a→01 ∆ξ∆ηI CoA·ds (5) The expression on the left refers to the original contour, while the expression on the right refers to the rectangular contour at the origin. Since a contour of arbitrary shape can be constructed by a proper arrangement of rectangular contours, we have proven that the expression lim a→0R A·ds/ais independent of the shape of the contour as long as (3) holds. Turning to the proof that (1) deﬁnes the component of a vector, we recognize that the shape of the contour is arbitrary when evaluatingR A·ds/a. We displace the plane in which the contour lies by a diﬀerential amount away from the point P(x, y, z ), as shown in Fig. A.2.4 which does not aﬀect the value of [ curlA]nas deﬁned in (1). The intersection of the plane with the three coordinate planes through Pis a triangle. We pick the triangle for the contour Cin (1). It follows from Fig. A.2.4 that the contour integral around the triangular contour in the plane perpendicular to ncan also be written as the sum of three integrals around the three triangular contours in the respective coordinate planes. Indeed, each of the added sections of line are traversed in one contour integration in the opposite direction, so that the integrals over the added sections of the line cancel upon summation and we have I nA·ds=I xA·ds+I yA·ds+I zA·ds (6) 6 Appendix Chapter 2 where each contour integral is denoted by the subscript taken from the unit vector normal to the plane of the contour. We further note that the areas ax, ay, azof the three triangles in the respec- tive coordinate planes are the projections of the area aonto the corresponding coordinate plane. ax=aix·n (7) ay=aiy·n (8) az=aiz·n (9) Thus, by dividing (6) by aand making use of (7), (8), and (9), we have: 1 aI nA·ds=1 axI xA·dsix·n+1 ayI yA·dsiy·n +1 azI zA·dsiz·n(10) Now, since the contours are already taken around diﬀerential area elements, the limit a→0 is already implied in (10). Thus, we have the quantities [curlA]x= lim ax→0I xA·ds/ax. . . (11) But (10) is the deﬁnition of the component in the ndirection of a vector: curlA= [curlA]xix+ [curlA]yiy+ [curlA]ziz (12) It is therefore legitimate to deﬁne at every point x, y, z in space a vector quantity, curlA, whose x-,y-, and z-components are evaluated as the limiting expressions of (1).