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Paul C.R. Inductance.. Loop and partial (Wiley, 2009)(ISBN 0470461888)(O)(395s)_EE_

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Graduate-level electromagnetics textbook by Clayton R. Paul, filed under Phil's Transmission Lines notes as an Appendix C reference. It derives loop inductance (Faraday, vector potential, Neumann integral, energy methods) and partial inductance of wires and PCB lands. Other topics include geometric mean distance, magnetic fields of DC and time-varying currents, vias and connector pins, and a vector review appendix.

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=} INDUCTANCE 7Loop andPartial Sa ——~ siSataae >feeR.PAUL (WILEY INDUCTANCE INDUCTANCE Loop and Partial CLAYTON R. PAUL Professor of Electrical and Computer Engineering Mercer UniversityMacon, GeorgiaandEmeritus Professor of Electrical EngineeringUniversity of KentuckyLexington, Kentucky Copyright ©2010 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except aspermitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the priorwritten permission of the Publisher, or authorization through payment of the appropriate per-copy fee tothe Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400,fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permissionshould be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken,NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy orcompleteness of the contents of this book and speciﬁcally disclaim any implied warranties ofmerchantability or ﬁtness for a particular purpose. No warranty may be created or extended by salesrepresentatives or written sales materials. The advice and strategies contained herein may not be suitablefor your situation. You should consult with a professional where appropriate. Neither the publisher norauthor shall be liable for any loss of proﬁt or any other commercial damages, including but not limited tospecial, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317)572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our web site atwww.wiley.com. Library of Congress Cataloging-in-Publication Data: Paul, Clayton R. Inductance : loop and partial / Clayton R. Paul. p. cm. Includes bibliographical references and index.ISBN 978-0-470-46188-4 1. Inductance. 2. Induction coils. I. Title. QC638.P38 2010621.37’42–dc22 2009031434 Printed in the United States of America 1 0987654321 This book is dedicated to the memory of my Father and my Mother Oscar Paul and Louise Paul CONTENTS Preface xi 1 Introduction 1 1.1 Historical Background, 1 1.2 Fundamental Concepts of Lumped Circuits, 21.3 Outline of the Book, 71.4 “Loop” Inductance vs. “Partial” Inductance, 8 2 Magnetic Fields of DC Currents (Steady Flow of Charge) 13 2.1 Magnetic Field Vectors and Properties of Materials, 13 2.2 Gauss’s Law for the Magnetic Field and the Surface Integral, 15 2.3 The Biot–Savart Law, 192.4 Amp `ere’s Law and the Line Integral, 34 2.5 Vector Magnetic Potential, 47 2.5.1 Leibnitz’s Rule: Differentiate Before You Integrate, 67 2.6 Determining the Inductance of a Current Loop: A Preliminary Discussion, 71 2.7 Energy Stored in the Magnetic Field, 792.8 The Method of Images, 802.9 Steady (DC) Currents Must Form Closed Loops, 83 vii viii CONTENTS 3 Fields of Time-Varying Currents (Accelerated Charge) 87 3.1 Faraday’s Fundamental Law of Induction, 88 3.2 Amp `ere’s Law and Displacement Current, 98 3.3 Waves, Wavelength, Time Delay, and Electrical Dimensions, 102 3.4 How Can Results Derived Using Static (DC) V oltages and Currents be Used in Problems Where the V oltagesand Currents are Varying with Time?, 105 3.5 Vector Magnetic Potential for Time-Varying Currents, 1073.6 Conservation of Energy and Poynting’s Theorem, 1113.7 Inductance of a Conducting Loop, 113 4 The Concept of “Loop” Inductance 117 4.1 Self Inductance of a Current Loop from Faraday’s Law of Induction, 117 4.1.1 Rectangular Loop, 1214.1.2 Circular Loop, 1264.1.3 Coaxial Cable, 130 4.2 The Concept of Flux Linkages for Multiturn Loops, 133 4.2.1 Solenoid, 1344.2.2 Toroid, 137 4.3 Loop Inductance Using the Vector Magnetic Potential, 139 4.3.1 Rectangular Loop, 1414.3.2 Circular Loop, 144 4.4 Neumann Integral for Self and Mutual Inductances Between Current Loops, 145 4.4.1 Mutual Inductance Between Two Circular Loops, 1474.4.2 Self Inductance of the Rectangular Loop, 1504.4.3 Self Inductance of the Circular Loop, 153 4.5 Internal Inductance vs. External Inductance, 1554.6 Use of Filamentary Currents and Current Redistribution Due to the Proximity Effect, 158 4.6.1 Two-Wire Transmission Line, 1594.6.2 One Wire Above a Ground Plane, 161 4.7 Energy Storage Method for Computing Loop Inductance, 163 4.7.1 Internal Inductance of a Wire, 1644.7.2 Two-Wire Transmission Line, 1654.7.3 Coaxial Cable, 165 CONTENTS ix 4.8 Loop Inductance Matrix for Coupled Current Loops, 167 4.8.1 Dot Convention, 1694.8.2 Multiconductor Transmission Lines, 171 4.9 Loop Inductances of Printed Circuit Board Lands, 179 4.10 Summary of Methods for Computing Loop Inductance, 182 4.10.1 Mutual Inductance Between Two Rectangular Loops, 184 5 The Concept of “Partial” Inductance 195 5.1 General Meaning of Partial Inductance, 196 5.2 Physical Meaning of Partial Inductance, 2015.3 Self Partial Inductance of Wires, 2055.4 Mutual Partial Inductance Between Parallel Wires, 2095.5 Mutual Partial Inductance Between Parallel Wires that are Offset, 213 5.6 Mutual Partial Inductance Between Wires at an Angle to Each Other, 224 5.7 Numerical Values of Partial Inductances and Signiﬁcance of Internal Inductance, 239 5.8 Constructing Lumped Equivalent Circuits with Partial Inductances, 242 6 Partial Inductances of Conductors of Rectangular Cross Section 246 6.1 Formulation for the Computation of the Partial Inductances of PCB Lands, 248 6.2 Self Partial Inductance of PCB Lands, 2546.3 Mutual Partial Inductance Between PCB Lands, 2626.4 Concept of Geometric Mean Distance, 266 6.4.1 Geometrical Mean Distance Between a Shape and Itself and the Self Partial Inductance of a Shape, 273 6.4.2 Geometrical Mean Distance and Mutual Partial Inductance Between Two Shapes, 285 6.5 Computing the High-Frequency Partial Inductances of Lands and Numerical Methods, 291 7 “Loop” Inductance vs. “Partial” Inductance 307 7.1 Loop Inductance vs. Partial Inductance: Intentional Inductors vs. Nonintentional Inductors, 307 7.2 To Compute “Loop” Inductance, the “Return Path” for the Current Must be Determined, 309 x CONTENTS 7.3 Generally, There is no Unique Return Path for all Frequencies, Thereby Complicating the Calculationof a “Loop” Inductance, 311 7.4 Computing the “Ground Bounce” and “Power Rail Collapse” of a Digital Power Distribution System Using “Loop”Inductances, 312 7.5 Where Should the “Loop” Inductance of the Closed Current Path be Placed When Developing a Lumped-Circuit Model ofa Signal or Power Delivery Path?, 314 7.6 How Can a Lumped-Circuit Model of a Complicated System of a Large Number of Tightly Coupled Current Loops beConstructed Using “Loop” Inductance?, 317 7.7 Modeling Vias on PCBs, 3187.8 Modeling Pins in Connectors, 3207.9 Net Self Inductance of Wires in Parallel and in Series, 321 7.10 Computation of Loop Inductances for Various Loop Shapes, 324 7.11 Final Example: Use of Loop and Partial Inductance to Solve a Problem, 328 Appendix A: Fundamental Concepts of Vectors 335 A.1 Vectors and Coordinate Systems, 336 A.2 Line Integral, 340A.3 Surface Integral, 343A.4 Divergence, 345 A.4.1 Divergence Theorem, 347 A.5 Curl, 350 A.5.1 Stokes’s Theorem, 353 A.6 Gradient of a Scalar Field, 354A.7 Important Vector Identities, 357A.8 Cylindrical Coordinate System, 358A.9 Spherical Coordinate System, 362 Table of Identities, Derivatives, and Integrals Used in this Book 367References and Further Readings 373 Index 377 PREFACE This book has been written to provide a thorough and complete discussion of virtually all aspects of inductance: both “loop” and “partial.” There is con-siderable misunderstanding and misapplication of the important concepts ofinductance. Undergraduate electrical engineering curricula generally discuss“loop” inductance only very brieﬂy and only in one undergraduate courseat the beginning of the junior year in a four-year curriculum. However, thatcurriculum is replete with the analysis of electric circuits containing the in-ductance symbol. In all those electric circuit analysis courses, the values of the inductors are given and are not derived from physical principles. Yet inthe world of industry, the analyst must somehow obtain these values as wellas construct inductors having the chosen values of inductance used in thecircuit analysis. This book addresses that missing link: calculation of the val-ues of the various physical constructions of inductors, both intentional andunintentional, from basic electromagnetic principles and laws. In addition, today’s high-speed digital systems as well as high-frequency analog systems are using increasingly higher spectral content signals. Numer-ous “unintended” inductances such as those of the interconnection leads arebecoming increasingly important in determining whether these high-speed,high-frequency systems will function properly. This is generally classiﬁed asthe “signal integrity” of those systems and is an increasingly important aspectof digital system design as clock and data speeds increase at a dramatic rate.Some ten years ago the effects of interconnects such as printed circuit boardlands on the function of the modules that lands interconnect were not im-portant and could be ignored. Today, it is critical that circuit models of these xi xii PREFACE interconnects be included in any analysis of the overall system. The concept of “partial inductance” is the critical link in being able to model these in-terconnects. Partial inductance is not covered in any undergraduate electricalengineering course but is becoming increasingly important in digital systemdesign. A substantial portion of this book is devoted to that topic. One of the important contributions of this book is the detailed derivation of the loop and partial inductances of numerous conﬁgurations of current-carrying conductors. Although the derivations are sometimes tedious, thereis nothing we can do about it because the results are dictated by the laws ofelectromagnetics, and these can be complicated. Unlike other textbooks, allthe details regarding derivations for the inductance of inductors are given.Although these are simpliﬁed where possible, only so much simpliﬁcationcan be accepted if the reader is to have a clear and unambiguous view of howthe result is obtained. In Chapter 1 we discuss inductance and show important parallels between inductance and capacitance along with some historical details. All of thederivations of the inductance of various inductors ﬁrst require that we obtaintheir magnetic ﬁelds. Chapter 2 is devoted to this task. The fundamental lawsof Biot–Savart, Gauss, and Amp `ere are discussed, and numerous calculations of the magnetic ﬁelds are obtained from them. In addition, the vector magneticpotential method of computing the magnetic ﬁelds is also discussed, alongwith the method of images and energy stored in the magnetic ﬁeld. In Chapter 3we provide a complete explanation of how the inductance, which is computedfor dc currents, can be used to characterize the effect of time-varying currents.Maxwell’s equations for time-varying currents are discussed in detail. Aniterative solution of them is given which shows why and when the inductor,derived for dc currents, can be used to characterize the effects of time-varyingcurrents. All aspects of the derivation of the “loop” inductance of various current- carrying loops are covered in Chapter 4. The ﬂux linkage method, the vectormagnetic potential method, and the Neumann integral for determining the“loop” inductance are used, and the “loop” inductances are calculated from allthree methods. The proximity effect for closely spaced conductors is discussedalong with the loop inductance of various transmission lines. In Chapter 5 we provide details for computation of the “partial” inductances of wires. Both the self-partial inductance of wires and the mutual partialinductances between wires are derived. These generic results can then be usedto “build” a model for other current-carrying structures. Chapter 6 containsall corresponding details about the derivation of the partial inductances ofconductors of rectangular cross section, referred to as “lands.” The conceptof geometric mean distance as an aid to the calculation of partial inductancesis discussed and derived for various structures. PREFACE xiii The ﬁnal chapter of the book, Chapter 7, provides a focus on when one should use “loop” inductance and when one should use “partial” inductance fordetermining the effect of current-carrying conductors. This chapter is meantto provide a simple discussion of this in order to focus the results of previouschapters. The chapter concludes with the solution of a problem involvingcoupling between two circuit loops using the “loop” inductance method andthen using the “partial” inductance method. Both methods yield the sameanswer, as expected. This example clearly shows the advantages of using“partial” inductance to characterize “unintentional inductors” such as wiresand lands. With the present and increasing emphasis on high-speed digital systems and high-frequency analog systems, it is imperative that system designersdevelop an intimate understanding of the concepts and methods in this book.No longer can we rely on low-speed, low-frequency systems to keep us fromneeding to learn these new concepts and analysis skills. The author would like to acknowledge Dr. Albert E. Ruehli of the IBM T.J. Watson Research Center for many helpful discussions of partial inductanceover the years. Clayton R. Paul Macon, Georgia 1 INTRODUCTION The concept of inductance is simple and straightforward. However, actual computation of the inductance of various physical structures and its imple-mentation in an electric circuit model of that structure is often fraught withmisconceptions and mistakes that prevent its correct calculation and use. Thisbook is intended to ensure the correct understanding, calculation, and imple-mentation of inductance. 1.1 HISTORICAL BACKGROUND Knowledge of magnetism has a long history [3]. A type of iron ore called lodestone had been discovered in Magnesia in Asia. This material had some interesting properties of magnetic attraction at a distance of other ferromag-netic substances and was known to Plato and Socrates. In the sixteenth century,William Gilbert ﬁrst postulated that Earth was a giant spherical magnet, andA. Kirchner, in the seventeenth century, demonstrated that the two poles ofa magnet have equal strength. Pierre de Marricourt constructed a compass in1629 that allowed the determination of the direction of the North Pole of theEarth. In 1750, John Mitchell determined the universal principle that force ata distance depends on the inverse square of the distance. At the beginning ofthe nineteenth century, Alessandro V olta developed a battery (called a pile). Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 1 2 INTRODUCTION This allowed the production of a current in a conducting material such as a wire. In 1820, Hans Christian Oersted showed that a current in a wire causedthe needle of a compass to deﬂect. Around the same time, Andr ´e Amp `ere conducted a set of experiments, resulting in his famous law. At about thesame time, Jean-Baptiste Biot and Felix Savart formulated their important lawgoverning the magnetic ﬁelds produced by currents: the Biot–Savart law. Soup to this time it was known that in addition to permanent magnets, a currentwould produce a magnetic ﬁeld. In 1831, Michael Faraday discovered that atime-changing magnetic ﬁeld would also produce a current in a closed loop ofwire. This discovery formed the essential idea of the inductance of a currentloop. James Clerk Maxwell uniﬁed all this knowledge of the magnetic ﬁeldas well as the knowledge of the electric ﬁeld in 1873 in his renowned set ofequations. Extensive work on the calculation of the magnetic ﬁeld of various current distributions and the associated concept of inductance dates back to the latenineteenth and early twentieth centuries. In fact, Maxwell in his famous trea-tise discussed inductance in 1873 [23]. An enormous amount of work waspublished on the determination of inductance from 1900 to 1920. (See the ex-tensive list of references on magnetic ﬁelds in the book by Weber [11] and oninductance in the book by Grover [14].) This early work on inductance at theturn of the century was spurred by the introduction of 60-Hz ac power and itsgeneration, distribution, and use. Some books, particularly those of the earlytwentieth century, tended to give only formulas for the magnetic ﬁelds of var-ious distributions of currents and their inductance with little or no detail aboutthe derivation of formulas. In that era, computers did not exist, so that many ofthe books and papers simply gave tables of values for the magnetic ﬁeld andinductance as a function of certain parameters. Another important purpose ofthis book is to show, in considerable detail, how the results for the magneticﬁelds and the inductance are derived. All details of each derivation are shown.At the end of the book is a list of signiﬁcant references and further readings onthe subject of the computation of magnetic ﬁelds and inductance of variouscurrent-carrying structures. References to these are denoted in brackets. 1.2 FUNDAMENTAL CONCEPTS OF LUMPED CIRCUITS We construct lumped-circuit models of electrical structures using the concepts and models of resistance, capacitance, and inductance [1,2]. We then solve forthe resulting voltages and currents of that particular interconnection of circuitelements using Kirchhoff’s voltage law (KVL) (which relates the various volt-ages of the particular interconnection of circuit elements), Kirchhoff’s currentlaw (KCL) (which relates the various currents of the particular interconnec- FUNDAMENTAL CONCEPTS OF LUMPED CIRCUITS 3 tion of circuit elements), and the laws of the circuit elements (which relate the voltages of each circuit element to its currents) [1,2]. It is important to keepin mind that these lumped-circuit models are valid only if the largest physical dimension of the circuit is “electrically short” (e.g., L < λ/10), where a wave- length λis deﬁned as the ratio of the velocity of wave propagation (along the component attachment leads), v, and the frequency of the wave, f[3–6]: λ=v f(1.1) If the medium in which the circuit is immersed and through which the waves propagate along the connection leads is free space (essentially, air), the velo-city of propagation of those waves is the speed of light, which is approximatelyv 0∼=3×108m/s. For a printed circuit board (PCB), the velocity of propaga- tion of the waves traveling along the lands on that board is about 60% of thatof free space, due to the interaction of the ﬁelds with the board substrate, andthe wavelengths are consequently shorter than in free space. Hence, circuitdimensions on a PCB are electrically longer than in free space. For a sinu-soidal wave in free space at a frequency of 300 MHz, a wavelength is 1 m.At frequencies below this, the wavelength is proportionately larger than 1 m,and for frequencies above this, the wavelength is proportionately smaller. Forexample, at a frequency of 3 MHz a wavelength in free space is 100 m, andat a frequency of 3 GHz a wavelength in free space is 10 cm. Hence, forlumped-circuit concepts to be valid for a circuit having a sinusoidal source offrequency 3 MHz, the maximum physical dimension of the circuit must beless than about 10 m or about 30 ft. Similarly, for a circuit having a sinusoidalsource of frequency 3 GHz, the maximum physical dimension must be lessthan about 1 cm or about 0.4 inch for it to be modeled as a lumped circuit.Today’s digital electronics have clock and data rates on the order of 300 MHzto 3 GHz. But these digital waveforms have a spectral content consisting ofharmonics (integer multiples) of the basic repetition rate, which are generallysigniﬁcant up to at least the ﬁfth harmonic. Hence, a 300-MHz clock ratehas spectral content up to at least 1.5 GHz, and a 3-GHz clock rate has spec-tral content up to at least 15 GHz! So the lumped-circuit models (and theirconstituent components of capacitance and inductance) that were so reliablesome 10 years ago are becoming less valid today. This trend will no doubtcontinue in the future as the requirement for higher clock and data speedscontinues to increase, and the reader should keep in mind this fundamentallimitation of inductance, capacitance, and the lumped-circuit models that usethese elements. The laws governing the calculation of resistance, capacitance, and induc- tance are written in terms of the vectors of the ﬁve basic electromagnetic ﬁeld vectors, which are summarized in Table 1.1. Therefore, if we are to correctly 4 INTRODUCTION TABLE 1.1. Electromagnetic Field Vectors Symbol Vector Units J Current density A/m2 Electric ﬁeld vectors E Electric ﬁeld intensity V/meter D Electric ﬂux density C/m2 Magnetic ﬁeld vectors H Magnetic ﬁeld intensity A/meter B Magnetic ﬂux density Wb/m2=T calculate and understand the ideas of capacitance and inductance of a physical structure as well as use them correctly to construct a lumped-circuit model ofthat structure, we must understand some elementary properties of vectors andsome basic vector calculus concepts. Trying to circumvent the use of vectorcalculus ideas by relying on one’s life experiences to compute and interpretthe meaning of the capacitance and inductance of a structure properly hascaused many of the incorrect results and misunderstanding, as well as thenumerous erroneous applications that are seen throughout the literature andin conversations with engineering professionals. References [3–6] give ex-tensive details on vector algebra and vector calculus. The Appendix of thisbook contains a review of the vector algebra and vector calculus conceptsthat are required to understand and compute the inductance of all physicalstructures. The lumped-circuit elements of resistance, capacitance, and inductance are derived fundamentally for static conditions. Capacitance is derived for conductors that are supporting charges whose positions on those conductorsare ﬁxed. Resistance as well as inductance are derived for currents that arenot varying with time: that is, direct (dc) currents. For charge distributions and currents that do not vary with time, the electromagnetic ﬁeld equations(Maxwell’s equations) that govern the ﬁeld vectors simplify considerably.However, the resulting electrical elements of resistance, capacitance, and in-ductance can be used to construct lumped-circuit models of a structure whosecurrents and charge distributions vary with time. This is valid as long as thesources driving the circuit have frequency content such that the largest phys-ical dimension of the circuit is electrically small (see Section 3.4). To understand the computation of inductance (the main subject of this book), it is useful to understand the dual concept of capacitance and itscalculation. The basic idea of the capacitance of a two-conductor structureis summarized in Fig. 1.1(a). If we apply a dc voltage Vbetween two FUNDAMENTAL CONCEPTS OF LUMPED CIRCUITS 5 VV+Q -QE (a) capacitanceE E E E sI II Iψ BB B (b) inductance FIGURE 1.1. Capacitance and inductance. conductors, a charge Qis transferred to and stored on those conductors (equal magnitude on both conductors, but opposite polarity). This chargeinduces an electric ﬁeld intensity Ebetween the two conductors that is directed from the conductor containing the positive charge to the conductorcontaining the negative charge. Alternatively, we could look at this processin a different way. Place a charge on the two conductors (equal magnitudeon both conductors but opposite polarity). This charge will result in anelectric ﬁeld Ebetween the two conductors which when integrated with a line integral (see the Appendix) gives the resulting voltage between the twoconductors: V=− ⎜integraldisplay+ −E·dl (1.2) where the path for integration is from a point on the negatively charged con- ductor to a point on the positively charged conductor [3–6]. In either case,thecapacitance of the structure is the ratio of the charge stored on the two 6 INTRODUCTION conductors and the voltage between them [1–6]: C=Q V(1.3) Hence, the capacitance of a structure represents the ability of that structure tostore charge. However, the capacitance of the structure is independent of the values of the voltage Vand the charge Qand depends only on their ratio. Hence, the capacitance Cof a structure depends only on its dimensions, its shape, and the properties of the medium surrounding the conductors (e.g., freespace, Teﬂon). There is energy stored in the electric ﬁeld in the space aroundthe two conductors. That stored energy is [3–6] W E=1 2⎜integraldisplay vD·Edv=1 2ε⎜integraldisplay vE2dv (1.4) where vis the volume of the entire space surrounding the conductors, εis thepermittivity of the surrounding medium, and we have used the relation D=εE. In terms of capacitance this stored energy is [1,2] WE=1 2CV2(1.5) The dual concept is that of inductance, illustrated in Fig. 1.1(b). If we pass a steady (dc) current Iaround a conducting loop of wire, the current will produce a magnetic ﬂux density Bthat circulates about the wire with its direction about the wire determined by the right-hand rule: If we place the thumb of the right hand in the direction of the current, the ﬁngers will showthe direction of the resulting magnetic ﬁeld that is circumferential about thecurrent. This causes a magnetic ﬁeld Bto penetrate the surface that is enclosed by the loop of current. The total magnetic ﬂux penetrating the surface enclosedby the current loop is obtained with a surface integral (see the Appendix) as[3–6], ψ= ⎜integraldisplay sB·ds (1.6) where sis the surface of the loop that is surrounded by the current. The inductance of the loop is the ratio of the total magnetic ﬂux penetrating theloop and the current that produced it [1–6]: L=ψ I(1.7) If the surrounding medium is not ferromagnetic (iron is an example of a ferro- magnetic material), that is, is not magnetizeable, the inductance is independentof the values of the ﬂux and the current and depends only on the dimensionof the loop, its shape, and the properties of the medium surrounding the con- OUTLINE OF THE BOOK 7 ductor (e.g., free space). There is energy stored in the magnetic ﬁeld in the space around the conductor loop. That stored energy is [3–6] WM=1 2⎜integraldisplay vB·Hdv=1 2μ⎜integraldisplay vH2dv (1.8) where vis the volume of the entire space surrounding the conductors, μis thepermeability of the surrounding medium, and we have used the relation B=μH. In terms of inductance, the stored energy is [1,2] WM=1 2LI2(1.9) The duality between the concept of capacitance and the corresponding concept of inductance is striking. However, the methods and techniques forcomputing them are generally different in both concept and method. Visual-izing how to go about calculating the capacitance of a particular structure isusually much easier to understand than is the visualization of how to go aboutcalculating inductance. 1.3 OUTLINE OF THE BOOK In Chapter 2 we summarize the fundamental electromagnetic ﬁeld laws gov- erning the magnetic ﬁeld, those of Gauss, Amp `ere, and Biot–Savart, on which the inductance calculation is based. The magnetic ﬁelds Bof various conﬁg- urations carrying a dc current are derived from these laws. This is a necessaryﬁrst step in computing the inductance of a structure since the magnetic ﬂux ψ penetrating the surface that comprises the inductance must be computed fromBvia (1.6). The inductance of the structure is then obtained as the ratio of the ﬂux and the current producing it via (1.7). The derivation of the Bﬁeld for a particular structure that carries a dc current generally involves the settingup and evaluation of somewhat complicated integrals. An extensive table ofintegrals is given by Dwight [7]. Furthermore, the next step in calculation ofthe inductance of a structure requires a further integration of Bas in (1.6). An alternative way of computing the Bﬁeld of a current-carrying structure is obtained using the vector magnetic potential A. In some cases it is easier to compute Adirectly and from this obtain Bby differentiation. The method of images for simplifying problems involving currents over large “groundplanes” is also discussed. The commonly assumed fact that all dc currentsmust “return to their source” and therefore must comprise closed loops isproven. The important ideas that arise when the currents are, instead of dc, varying with time are discussed in Chapter 3. The fundamental law that provides an 8 INTRODUCTION understanding of how an inductance produces a voltage between its two termi- nals is Faraday’s law of induction, which is examined in detail. The importantnotion of displacement current in Amp `ere’s law, which affords an understand- ing of how a capacitance can conduct a time-varying current through it, is alsodiscussed. The important concepts of waves, wavelength, time delay, and elec-trical dimensions that allow these static ideas of capacitance and inductanceto be incorporated into lumped circuits which have time-varying sources driv-ing them are examined. The important ability of being able to use a quantitythat is derived for static (dc) currents (e.g., inductance and capacitance) ina circuit where the currents vary with time is shown in terms of an iterativeexpansion of the electromagnetic ﬁelds. Finally, conservation of energy in theelectromagnetic ﬁeld and Poynting’s theorem are reviewed. With this requisite background, we are able to understand how to calculate and interpret the meaning of the “loop inductance” of a closed loop of current,which is given in Chapter 4. The “loop inductances” for various structures arealso derived in Chapter 4 using several methods. The remaining chapters are devoted to the concept of “partial inductance,” which is rapidly becoming important in today’s high-speed digital electronics.The general concept of “partial inductance” is examined in Chapter 5, and theself and mutual partial inductances of straight wire segments are determined.The self and mutual partial inductances of conductors of rectangular crosssection, which the “lands” on printed circuit boards (PCBs) represent, aredetermined in Chapter 6. Chapter 7 is devoted to a critical examination ofthe relative merits of using loop inductances to characterize current loopsversus the use of partial inductances. A fairly complex structure is analyzedby ﬁrst characterizing it with loop inductances and then characterizing it withpartial inductances. This example is quite useful in bringing together all theconcepts of the previous chapters and in comparing their relative merits anddeﬁciencies. 1.4 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE It is critically important that the reader understand the following two dis- tinctions with regard to inductance. In undergraduate electrical engineeringcourses, only the concept of the inductance of a complete loop of current isstudied. (It is shown in Section 2.9 that dc currents must form closed loops.)This “loop inductance” is given in (1.7) and requires that we be able to com-pute the magnetic ﬂux ψthat passes through the enclosed surface of a closed current loop, as illustrated in Fig. 1.1(b). Therefore, computation of the loopinductance of a structure requires that we be able to identify the complete cur- rent loop. For “intentional” inductors this current loop is rather obvious. For “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE 9 example, if we wind several turns of wire around a ferromagnetic toroid core, the loop area of the current that the magnetic ﬂux passes through is evident.Hence, the concept of loop inductance of intentional inductors is useful in thatit allows us to characterize those as lumped-circuit elements. On the other hand, if we want to assign an “inductance” to segments of a conductor on a printed circuit board (referred to as lands), there are severalproblems in trying to use the concept of loop inductance to do so. The ﬁrstproblem is that we must be able to determine the complete current loop path in order to calculate loop inductance of that current loop. In other words, wemust be able to identify not only the “going down” path from the source to theload (which is relatively easy to do) but also the “return current” path of thecurrent back to the source in order to determine the complete current loop.I n today’s densely packed integrated circuits and printed circuit boards carryingcurrents having ever-increasing spectral content, this has become virtuallyimpossible to do! Furthermore, the complete path for the current depends onthe frequency of the current. For one frequency, the return current will take aparticular path, but for a higher frequency the path of the return current maybe entirely different! So the ﬁrst problem with using loop inductance to model the conductors of a loop is that at different frequencies, the return path of the loop current maybe different. This is best illustrated by the situation of a coaxial cable abovea ground plane shown in Fig. 1.2. (See [5] for an analysis of this problem.)At dc and low frequencies, the current Itakes its return path, I G, through the massive ground plane. However, at higher frequencies, the current Itakes its return path up through the shield, IS. Therefore, the return paths and hence the complete current loops are different for different frequencies. So if wewere to compute the loop inductance it would appear that we would have twodifferent values, depending on the frequency of the current. The ﬁnal and most important problem in trying to use loop inductance to allocate inductances to the individual lands on a PCB is that the total loop inductance of a current loop cannot be placed in any unique position inthat loop. For example, the current loop in Fig. 1.1(b) is said to present aninductance at its input terminals. But that is a loop inductance that cannot, I IS IS I IGR VSI FIGURE 1.2. “Return currents” of different frequencies may take different paths. 10 INTRODUCTION VPR V 5+ VGBVGBLGBpower supply groundL-HI L-HIH−LI LV LGBLPR VL tV 5HIGH LOW FIGURE 1.3. The problem in using “loop inductance” to characterize the inductance of PCB lands. a priori, be divided into portions that are associated with segments of that loop! Figure 1.3 illustrates this problem of trying to use loop inductance to model the inductance of portions of the PCB lands. We have shown a CMOS inverterthat is attached to a capacitive load (perhaps representing the input to anotherCMOS inverter). The +5-V output of the power supply is attached to the +5-V power pin of the CMOS module via a land on the PCB. Similarly, theground terminal of the power supply is attached to the ground pin of the CMOSmodule with another land on the PCB. As the inverter switches from the low tohigh state, the current, I L-H, is drawn from the power supply through the +5-V land and through the inverter to charge the capacitor to put the loadvoltage, V L,i nt h e high state and returns to the power supply through the ground land. When the inverter switches, the load capacitor then dischargesvia current I H-Lthrough the inverter via a different loop: from the capacitor, through the inverter, and back to the capacitor. We have shown the conductorsas each having associated individual inductances. The land connecting the+5-V output of the power supply to the +5-V pin of the inverter is shown as having an inductance L PR. The land connecting the ground of the power supply to the ground pin of the inverter and the land connecting the bottom “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE 11 of the capacitor to the ground pin of the inverter are also shown as having inductances LGB. (Although these two inductances have the same symbol, they obviously have different values, due to the different lengths of thesereturn paths.) The current I L-Hclearly forms a loop: from the power supply through the inverter, through the capacitor, and back to the power supply.When the inverter is switching from low tohigh, the current I L-Hthrough the+5-V land increases in value in order to charge the capacitor. Hence, a voltage is developed across LPRof VPR=LPRdIL-H dt This is referred to as power rail collapse, since the voltage of the power pin of the inverter is 5 −VPR, and hence the voltage of the +5-V pin of the inverter module drops in value from +5-V . Similarly, the voltage at the ground pin of the inverter goes from zero to VGB: VGB=LGBdIL-H dt This is referred to as ground bounce. On the other hand, when the load voltage is transitioning from high tolow state, a voltage is developed across the VGB of the other ground land as the current IH-Lfrom the capacitor discharges through the inverter and returns to the capacitor. Although at ﬁrst glance this seems to be a straightforward characterization of the individual lands with an inductance, it is not. What do we mean bythe inductances of the two lands, L PRandLGB? These certainly are not loop inductances because the total inductance of a loop cannot be placed uniquelyin any segment of the loop. In fact, these are “partial inductances.” But the typeof diagram shown in Fig. 1.3 is seen throughout the literature. The problemhere is that few people know how to compute L PRandLGB. Even worse, they often mistakenly compute LPRandLGBusing a formula for a loop inductance they ﬁnd in a handbook that does not apply to these inductances, therebygiving erroneous results for the magnitudes of V PRandVGB! So loop inductance is not useful in modeling an “unintended inductance” to obtain the voltage developed between its two ends, due to a rate of changeof current through it. However, using the concept of loop inductance to model“intended” physical inductors such as a toroid or a solenoid is a useful appli-cation of that concept. On the other hand, the concept of partial inductanceallows us to represent the lands on a PCB as well as other types of conductorswith inductances and to compute the values of those inductances uniquely todetermine the correct voltage drop between two ends of the conductor. Unlikeloop inductances, we can compute partial inductances without the necessity of having to be able to identify the return paths for the currents ! We simply model all conductors with their partial inductances (self and mutual betweenthis and other conductors in the circuit), build a lumped-circuit model using 12 INTRODUCTION these partial inductances, and “turn the crank” (analyze the resulting lumped- circuit model) to ﬁnd the return paths for the currents rather than trying toguess their paths a priori. There is a dual concept to partial inductance that isreferred to by the author as generalized capacitance (see references [5,8] for a discussion). Prior to a decade ago, when the digital clock and data speeds and their spectral content were below about 100 MHz and the density of electroniccircuits was not what it is today, the concept of partial inductance was not asimportant. Today, it is a virtual necessity if we are to cope with the rapidlyescalating densities of electronic circuits whose conductors carry currentshaving increasingly higher spectral content. The units of the quantities are named for the great scientists who made major contributions to the discovery of these phenomena. Throughout this book, weuse the abbreviations A, Wb, H, and T, respectively, for the units amperes,webers, henrys, and tesla. The standard is to use lowercase for the ﬁrst letterof each of the names of these units and capital letters in their abbreviations. 2 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) As discussed in Chapter 1, inductance is intimately related to a closed loop of dc current which produces magnetic ﬂux through the surface surrounded bythe current loop. So our ﬁrst priority is to understand the computation of themagnetic ﬁelds of steady (dc) currents that do not vary with time for variousconﬁgurations of those currents. 2.1 MAGNETIC FIELD VECTORS AND PROPERTIES OF MATERIALS The fundamental magnetic ﬁeld vectors are the magnetic ﬁeld intensity H, whose units are A/m, and magnetic ﬂux density B, whose units are Wb /m 2= T. In a simple (but very common) linear, homogeneous, and isotropic medium,BandHare related as [3–6] B=μH=μ0μrH (2.1) where μis the permeability of the medium. The permeability can be written as the product of the relative permeability μrand the permeability of free Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 13 14 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) space (essentially, air), μ0=4π×10−7H/m: μ=μrμ0 (2.2) The units of permeability are named for Joseph Henry of Albany, New York, who essentially discovered Faraday’s law at about the time Faraday did butdid not publish his results until much later. Hence, for linear, homogeneous,and isotropic media, BandHcan be freely interchanged according to (2.1). Dielectrics and metals that are not magnetizeable, such as copper, aluminum,and brass, are linear and isotropic with regard to magnetic ﬁelds and have μ r= 1. However, materials that are magnetizeable have μr>1 and are generally nonlinear. There are common materials such as iron and steel that are magnetizable. These are said to be nonlinear with respect to magnetic ﬁelds. Ferromagneticmaterials such as iron and steel have BandHrelated by the common “hystere- sis curve” shown in Fig. 2.1. Suppose that we wind N turns of wire around atoroid of nonlinear magnetic material such as iron and pass a current Ithrough the turns of wire as illustrated in Fig. 2.2. The turns of wire produce a mag-netic ﬁeld intensity of approximately H=NI. Starting from an unmagnetized toroid, B=H=0, we start at the origin in Fig. 2.1. Increasing the current Iwe move up and to the right, reaching a point where further increases in I (and H) cause little or no change in B. At this point the material is said to be in saturation. Upon reducing Iwe proceed to a point where I=0 (and H=0) but the magnetic ﬂux density Bhas not decreased to zero. Further reductions ofIfor negative values reduces Bto zero where His negative. This process continues as we cycle around the hysteresis curve. B H FIGURE 2.1. “Hysteresis curve” for nonlinear magnetic media. GAUSS’S LAW FOR THE MAGNETIC FIELD AND THE SURFACE INTEGRAL 15 I N turnsB FIGURE 2.2. Toroid. The instanteous slope at a point on the hysteresis curve is the incremental permeability of the material: /Delta1μ=B H(2.3) The slope of the hysteresis curve at B=H=0 when the material is unmag- netized is called the initial permeability and is typically stated in the brochures of manufacturers of the material. Clearly, this material is nonlinear and therecan be no numerical value for a “permeability” stated for it. Because of thisdifﬁculty we deal only with materials such as air and copper, for which μ r=1, or nonlinear magnetic materials where the applied currents and consequentlythe levels of Hare sufﬁciently small that we can consider the material to be lin- ear, having its initial permeability. We could also deal with situations where,for example, the sinusoidal variations of the current and consequentially of H are sufﬁciently small that we can use an incremental permeability to charac-terize this nonlinear magnetic material. Common ferromagnetic materials aresteel (μ r=2000), iron (μ r=1000), and nickel ( μr=600), as well as cer- tain powdered ferrites such as nickel–zinc ( μr∼=600) and manganese–zinc (μr∼=1200). Certain exotic materials such as Mu-metal ( μr=30,000) have very large relative permeabilities (at low frequencies, e.g., 1 kHz, and lowvalues of H). 2.2 GAUSS’S LA W FOR THE MAGNETIC FIELD AND THE SURFACE INTEGRAL In the case of ﬁxed distributions of charge, electric ﬁeld lines that begin on a positive charge must end on a negative charge, as illustrated in Fig. 2.3. So 16 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) surface sE EQ1 Q2 Q3Q4 Qnet=Q1-Q2+Q3 FIGURE 2.3. Static electric ﬁeld of ﬁxed distributions of charge. we can view charges as a source of the electric ﬁeld intensity vector E, whose units are V /m. Gauss’s law for the electric ﬁeld is stated as [3–6] ⎜contintegraldisplay sD·ds=Qenclosed (2.4) where Dis the electric ﬂux density whose units are C/m2andD=εE, where εis the permittivity of the surrounding medium. The surface integral in (2.4) gives the net ﬂux of the Dﬁeld out of the closed surface s(see the Appendix for a discussion of the surface integral). Hence, Gauss’s law in (2.4) simplyprovides that if we take the products of the differential surface elements dsand the components of Dthat are perpendicular to the surface s and add them over the closed surface s, we will obtain the netpositive charge enclosed by the closed surface s. This is a sensible result because there are two components of DandEat a point on the surface s: One component is parallel to the surface and the other is perpendicular to the surface. Only the component perpendicular to the surface contributes to the net ﬂux of the electric ﬁeld entering or leavingthe surface s. However, in the case of magnetic ﬁelds, there are no known sources or sinks for the magnetic ﬁeld, so that the magnetic ﬁeld lines must form closed loops. If we cut a permanent magnet into two pieces, we do not create isolatedsources of the magnetic ﬁeld, as illustrated in Fig. 2.4. Gauss’s law for themagnetic ﬁeld states this important fact in terms of a surface integral [3–6]: ψ=⎜contintegraldisplay sB·ds=0 (2.5) GAUSS’S LAW FOR THE MAGNETIC FIELD AND THE SURFACE INTEGRAL 17 N SB BBB N S N SB B FIGURE 2.4. Permanent magnets. This law provides that if we take the surface integral of the magnetic ﬂux density Bover a closed surface s as illustrated in Fig. 2.5 giving the net magnetic ﬂux,ψ, out of the closed surface, we will obtain a result of zero for any closed surface: There is no netmagnetic ﬂux entering or leaving a closed surface. The units of Bare Wb/m2=T. Hence, the units of the magnetic ﬂux ψleaving the closed surface s are webers. The surface integral in (2.5) simply provides that if we take the products of the differential surface elements ds and the components of Bthat are perpendicular to the surface s and add them over the closed surface, we will obtain a result of zero. This is a sensible N SB B surface s FIGURE 2.5. Gauss’s law for the magnetic ﬁeld. 18 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) result if the magnetic ﬁeld lines must close on themselves since there are two components of Bat a point on the surface: One component is parallel to the surface and the other is perpendicular to the surface. Only the componentperpendicular to the surface contributes to the net ﬂux of the magnetic ﬁeld out of the surface s. The laws of Gauss in (2.4) and (2.5) are said to be in integral form; that is, they apply to broad regions of space. The point forms of these laws apply to speciﬁc points in space and are [3–6] ∇·D=ρ (2.6) for the electric ﬁeld, where ρis the volume charge density at the point whose units are C/m3and ∇·B=0 (2.7) for the magnetic ﬁeld. The notation ∇·Fdenotes the divergence of the vector ﬁeld F(see the Appendix). These point forms can be derived from the integral forms using the divergence theorem (see the Appendix): ⎜contintegraldisplay sD·ds=⎜integraldisplay v(∇·D)dv =Qenclosed =⎜integraldisplay vρdv and ⎜contintegraldisplay sB·ds=⎜integraldisplay v(∇·B)dv =0 where the closed surface sencloses the volume v. Comparing both sides gives the point forms in (2.6) and (2.7). Gauss’s law for the electric ﬁeld in(2.6) provides that the divergence or net outﬂow of the electric ﬁeld lines from a point equals the net positive volume charge density at the point (seethe Appendix for a discussion of divergence). Gauss’s law for the magneticﬁeld in (2.7) simply provides that there is no divergence of the magnetic ﬁeld lines: There are no isolated sources or sinks for the magnetic ﬁeld, andthe magnetic ﬁeld lines must therefore form closed loops. In a rectangularcoordinate system consisting of mutually orthogonal axes x,y, and z,w em a y write the “del operator” as (see the Appendix) ∇= a x∂ ∂x+ay∂ ∂y+az∂ ∂z(2.8) THE BIOT–SA V ART LAW 19 and Gauss’s laws become ∇·D=∂Dx ∂x+∂Dy ∂y+∂Dz ∂z=ρ (2.9) ∇·B=∂Bx ∂x+∂By ∂y+∂Bz ∂z=0 (2.10) Note that if the vector components of Bareindependent of x, y, and z, re- spectively [i.e., Bx(y, z),B y(x, z), andBz(x, y)], the divergence of Bwill automatically be zero. But there are obviously many cases where the vectorcomponents of Bare functions of some or all of the axis variables x,y, and z, yet the divergence of Bis still zero. 2.3 THE BIOT–SA V ART LA W Perhaps the most fundamental law that allows computation of the magnetic ﬁeld due to a dc current is the Biot–Savart law [3–6,9–11]: B=μ0 4π⎜integraldisplay vJ×aR R2dv (2.11) The dc current density vector is denoted as J, whose units are A /m2, and vis the volume containing this current. A differential segment or “chunk” of this current density vector contains Jdvampere-meters, and the distance from this chunk of current (the source of the Bﬁeld) to the point at which we are computing the magnetic ﬁeld Bis denoted as R. The unit vector aRis directed from this differential chunk of current tothe point at which we are computing B. The resulting Bﬁeld is perpendicular to the plane containing JandaRaccording to the right-hand rule (see the Appendix). Note that the Biot–Savart law is an inverse-square law like Coulomb’s law and the law ofgravity since it depends on the inverse of the square of the distance betweenBand the differential segment of the current density vector (the source of the ﬁeld). Throughout this book we generally concentrate on line currents, denoted as I, whose units are amperes. Considering a differential length of these currentsas a small cylinder of length dland cross-sectional area dswith current density distributed uniformly over the cross section (as will be the case fordc currents [3]), Jd s=I, so that Jdv=Jd sd l =Id l. In this case the 20 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) R ld IθdB into the page Ra FIGURE 2.6. Biot–Savart law. differential contribution of this current to the magnetic ﬂux density vector at a point is dB=μ0I 4πR2dl×aR (2.12a) as illustrated in Fig. 2.6. The direction of the vector differential length dlof this ﬁlamentary current segment is in the direction of the current. Note that the magnetic ﬁeld depends on the cross product dl×aR, where the unit vector from the current element to the point aRis directed from the current tothe point (see the Appendix for a review of the cross product). Hence, the magnetic ﬁeld is directed into the page (perpendicular to the planecontaining dland the unit vector a Raccording to the right-hand rule). Hence, in terms of the angle θbetween these two vectors, we can write the Biot–Savart law as dB=μ0Idl 4πR2sinθan (2.12b) where anis a unit vector perpendicular to the plane containing dlandaR according to the right-hand rule in the order dlan=dl×aR(i.e., pointing into the page). So the magnetic ﬁeld is a maximum along a line perpendicularto the current element and is zero off the ends of the current element. If we placethe current element along the zaxis of a cylindrical coordinate system (see the Appendix), the magnetic ﬁeld will be directed circumferentially aroundthe current in the φdirection at allpoints around the current. EXAMPLE As an example, we use the Biot–Savart law to determine the magnetic ﬁeld about a current of ﬁnite length L. Since dc currents must form closed loops THE BIOT–SA V ART LAW 21 xyz 2Lz= 2Lz− =dz rR IIr B (a ( ) b)Bθ αx yRa FIGURE 2.7. Current of length Land the magnetic ﬁeld about it. (see Section 2.9), we use the magnetic ﬁelds of ﬁnite lengths of current to construct the ﬁelds of closed current loops by the superposition of theﬁelds of the current segments of the closed current loop. Hence, deter-mining the magnetic ﬁelds of ﬁnite lengths of current is useful from thatstandpoint. First, set up a rectangular coordinate system and orient the current along thezaxis and centered on the origin with the current directed in the positive z direction, as shown in Fig. 2.7(a). We will determine the magnetic ﬂux densityat a point that is a distance r= ⎜radicalbig x2+y2from the midpoint of the current and along a line that is perpendicular to the current. The contribution to the magnetic ﬁeld at a distance rfrom the origin of the coordinate system that is due to a differential length of the current dzwhich is at a distance Rfrom the point is dB=μ0Idz 4πR2sinθ The direction of this Bﬁeld is, according to the Biot–Savart law, perpendicular to the plane containing the positive zaxis and the unit vector from the current element Id zto the point according to the right-hand rule: az×aR. Hence, it is directed circumferentially about the current. This is in the φdirection in a cylindrical coordinate system, aφ(see the Appendix for a discussion of the cylindrical coordinate system). The sine of the angle involved in the cross 22 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) product is sinθ=sin(α+90◦)=cosα =r R and the distance Ris R=⎜radicalbig z2+r2 Hence, the total magnetic ﬁeld at the point is B=μ0I 4π⎜integraldisplayL/2 z=−L/2r R3dz =μ0I 4π⎜integraldisplayL/2 z=−L/2r ⎜parenleftbigr2+z2⎜parenrightbig3/2dz =μ0Ir 4π⎜bracketleftbiggz r2√ r2+z2⎜bracketrightbiggL/2 z=−L/2 =μ0Ir 4π⎡ ⎣L/2 r2⎜radicalBig r2+(L/2)2−−L/2 r2⎜radicalBig r2+(−L/2 )2⎤ ⎦ =μ0Ir 4πL r2⎜radicalBig r2+(L/2)2 =μ0I 4πrL⎜radicalbig r2+L2/4 =μ0I 2πrL√ 4r2+L2 We have used integral 200.03 from the table of integrals by Dwight [7]: ⎜integraldisplay1 ⎜parenleftbiga2+x2⎜parenrightbig3/2dx=x a2√ a2+x2(D200.03) (Note: Throughout this book we evaluate the somewhat complicated integrals we encounter using the extensive table of integrals by Dwight [7]. Theseintegrals will be denoted as (Dxxx.xx) according to the integral number inDwight.) The magnetic ﬂux density vector is directed in the circumferen-tial direction about the wire which corresponds to the φcoordinate of the THE BIOT–SA V ART LAW 23 cylindrical coordinate system. Hence, the result can be written as a vector: B=μ0I 4πrL⎜radicalBig r2+(L/2)2aφ =μ0I 2πrL√ 4r2+L2aφ (2.13) where aφ=az×ar. For an inﬁnite length of current, L→∞ , (2.13) reduces to a very funda- mental result that we will use on numerous occasions: B=μ0I 2πraφL→∞ (2.14) Determination of the direction of the magnetic ﬁeld of a current is obtained with the famous right-hand rule. [The ofﬁcial symbol of the Institute of Electrical and Electronics Engineers (IEEE) memorializes this very funda-mental rule.] According to the Biot–Savart law, if we place the thumb of ourright hand in the direction of the current I, the ﬁngers of that hand will give the resulting direction of the Bﬁeld, which is perpendicular to the plane contain- ing (1) the current Iand (2) the vector pointing from the current to the point at which we desire to determine the Bﬁeld; a φ=az×ar. Hence, the magnetic ﬁeld about an inﬁnitely long current is directed circumferentially about thewire at all points along it according to the right-hand rule, decays inverselywith distance from the wire, and is constant in magnitude at distances r fromthe wire as illustrated in Fig. 2.7(b). An important difference between the magnetic ﬁelds of a wire of inﬁnite length and one of ﬁnite length is that thelatter has fringing ﬁelds at its endpoints. EXAMPLE In the preceding example we centered the current on the origin of the coordi- nate system and determined the Bﬁeld at a radial distance rfrom that center and on a line perpendicular to the midpoint of the current . We next generalize this result to obtain the magnetic ﬁeld of a current that is of ﬁnite length but atany point about the current which is not necessarily on a line perpendicular to its midpoint, as shown in Fig. 2.8. We again orient the current along the zaxis and center the current on the origin of that coordinate system, but the Bﬁeld is determined at a general point that is at a horizontal distance r(the cylindrical coordinate system variable) 24 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) xyz 2Lz= 2Lz− =dzR IBr 1θ 2θZ Raθ FIGURE 2.8 from the zaxis and is located at an arbitrary value of z=Z. Using the Biot– Savart law we see that the Bﬁeld is again circumferential about the current and, for example, is perpendicular to the yzplane. The Bﬁeld is again in the aφ=az×aRdirection. The Biot–Savart law again gives dB=μ0I 4πR2sinθd z The distance Rfrom the current element to the point is R=⎜radicalBig (Z−z)2+r2 and sinθ=r R Hence, the integral to be evaluated is B=μ0Ir 4π⎜integraldisplayL/2 z=−L/21 ⎜bracketleftBig (Z−z)2+r2⎜bracketrightBig3/2dz THE BIOT–SA V ART LAW 25 Using a change of variables, Z−z=λ,dλ=−dzgives B=μ0Ir 4π⎜integraldisplayZ+L/2 λ=Z−L/21 ⎜parenleftbigλ2+r2⎜parenrightbig3/2dz Using Dwight [7] (D200.03) again gives B=μ0I 4πr⎡ ⎣Z+L/2⎜radicalBig (Z+L/2)2+r2−Z−L/2⎜radicalBig (Z−L/2)2+r2⎤ ⎦aφ(2.15) which, of course, reduces to (2.13) for Z=0. For a current of inﬁnite length, L→∞ , (2.15) reduces to (2.14). In terms of the angles θ1andθ2between the zaxis and lines drawn from the ends of the current to the point, this becomes B=μ0I 4πr(cosθ2−cosθ1)aφ (2.16) In the case of a current of inﬁnite length, L→∞ ,θ1→π, andθ2→0 and (2.16) reduces to (2.14). EXAMPLE The principle of superimposing the contributions of several currents to give the total ﬁeld at a point is a powerful technique for linear media. We showin Section 2.9 that steady (dc) currents must form closed loops . Hence, we use this principle of superimposing the contributions of the segments of thecurrent of a closed current loop to obtain the total magnetic ﬁeld of closedloops of current. In this example we determine the total magnetic ﬁeld at adistance dfrom the center of a rectangular loop of current having sides of length wandlandalong a line that is perpendicular to the loop at a distance d from its center , as shown in Fig. 2.9. We restrict this solution to a point along a line from the center of the loop because the equation for the Bﬁeld at any other point about the loop is very difﬁcult to derive and the resultis extraordinarily complicated (see [9], p. 286). Treat this as four currentswhose Bﬁelds are given by (2.13) and superimpose the ﬁelds. The Bﬁeld due to each side is perpendicular to a line drawn from the center of eachcurrent to the point. Considering two pairs of opposite sides, we see fromFig. 2.9 that the horizontal contributions (in the xyplane) cancel and we are left with the total in the zdirection. (Use the right-hand rule to determine the direction of the magnetic ﬁeld that is due to each current.) Hence, the 26 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) w l xyz IIR IIB z=dBz 2l 2wRB α ααα FIGURE 2.9. Magnetic ﬁeld at a distance dalong a line perpendicular to the center of a rectangular loop. magnetic ﬁeld at the point due to two of the opposite sides each of length wis, using (2.13), B=2μ0I 2πw R√ 4R2+w2cosαaz where R=⎜radicalBigg⎜parenleftbiggl 2⎜parenrightbigg2 +d2 =1 2⎜radicalbig l2+4d2 and cosα=l/2 R Hence, the total from two of the opposite sides is B=2μ0I πwl ⎜parenleftbigl2+4d2⎜parenrightbig√ l2+w2+4d2az THE BIOT–SA V ART LAW 27 Adding the contributions from the other two opposite sides gives the total as B=2μ0I π⎜bracketleftBigg wl ⎜parenleftbigl2+4d2⎜parenrightbig√ l2+w2+4d2 +lw ⎜parenleftbigw2+4d2⎜parenrightbig√ w2+l2+4d2⎜bracketrightBigg az (2.17) At the center of the loop, d=0, this reduces to B=2μ0I π⎜parenleftbiggw l√ l2+w2+l w√ w2+l2⎜parenrightbigg azd=0 =2μ0I π√ l2+w2 wlazd=0 (2.18) For a square loop, w=l, (2.17) reduces to B=2√ 2μ0I πl2 ⎜parenleftbigl2+4d2⎜parenrightbig√ l2+2d2azl=w (2.19) At the center of a square loop, w=landd=0, (2.18) becomes B=2√ 2μ0I πlaz w=l, d=0 (2.20) EXAMPLE Consider a sheet of current lying in the yzplane as shown in Fig. 2.10(a). The sheet carries a surface current Kwhose units are A/m that is parallel to the yzplane and directed in the zdirection. The sheet extends to inﬁnity in all directions. Viewing this as currents of inﬁnite length directed in the z direction whose values are I=Kd y , we can superimpose their Bﬁelds using the results for an inﬁnite current obtained in (2.14). The Bﬁeld due to one of the currents at a point along the +xaxis at x=d(perpendicular to the plane containing the surface current) as shown in Fig. 2.10(b) is dB=μ0K 2πRdy where R=⎜radicalBig d2+y2 28 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) xz yto∞ to∞to∞to∞ (a) (b)y K KBdR Ry yBnet B B α αRa Raα αmAK KmA FIGURE 2.10. Inﬁnite current sheet. The direction of the Bﬁeld from each current is, according to the Biot–Savart law, perpendicular to the plane containing the current and a unit vector directedfrom the current to the point: a z×aR. The xcomponents of the ﬁelds of two symmetrically disposed currents cancel as shown in Fig. 2.10(b), giving thenet ﬁeld in the ydirection as B net=2μ0K 2π⎜integraldisplay∞ y=01 Rcosαd y ay THE BIOT–SA V ART LAW 29 where cosα=d R Substituting gives Bnet=2μ0K 2π⎜integraldisplay∞ y=01 Rcosαd y ay =μ0Kd π⎜integraldisplay∞ y=01 d2+y2dyay =⎧ ⎪⎪⎨ ⎪⎪⎩μ0K 2ay forx>0 −μ0K 2ay forx<0(2.21) and we have used integral 120.1 from Dwight [7]: ⎜integraldisplay1 a2+x2dx=1 atan−1x a(D120.1) Hence, the magnetic ﬂux density at any distance from a current sheet is di- rected parallel to the sheet and is independent of distance from the sheet. Thisresult applies also to the ﬁeld on the other side of the sheet, but the directionof the ﬁeld is in the –y direction on that side. EXAMPLE We can generalize the result for an inﬁnite current sheet obtained in the pre- ceding example to one that has a ﬁnite width Wand ﬁnite length L. We will determine the magnetic ﬂux density vector Bat a point that is a distance x=d from the center of the sheet , as illustrated in Fig. 2.11. Again viewing this re- sult as a superposition of the ﬁelds due to two symmetrically disposed butﬁnite-length currents I=Kd y of length Lthat are parallel to the zaxis, we can use the result obtained in (2.13) for the ﬁeld at a point a distance rfrom the midpoint of a ﬁnite-length current and write the net Bﬁeld as B net=2μ0K 4π⎜integraldisplayW/2 y=0L R⎜radicalbig R2+(L/2)2cosαd y ay Substituting R=⎜radicalbig d2+y2and cos α=d/R gives Bnet=μ0KLd 2π⎜integraldisplayW/2 y=01⎜parenleftbigy2+d2⎜parenrightbig⎜radicalbig y2+d2+(L/2)2dyay 30 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) z y (a) (b)y xK KBdααR Ry yBnet B BL W WaR RaBnet mAK KmA FIGURE 2.11. Current sheet of ﬁnite length and width. Using integral 387 from Dwight [7], ⎜integraldisplaydx⎜parenleftbigax2+b⎜parenrightbig⎜radicalbig fx2+g=1√ b√ag−bftan−1x√ag−bf√ b⎜radicalbig fx2+g (D387) gives Bnet=μ0K πtan−1 (W/2)(L/2) d⎜radicalbig d2+(W/2)2+(L/2)2ayx>0 (2.22) THE BIOT–SA V ART LAW 31 On the back side of the plate, x<0, the result in (2.22) must be negated according to the right-hand rule. Taking the limit as L→∞ gives the result for an inﬁnitely long current strip of width Wat a distance dfrom its center and perpendicular to the strip surface as Bnet=μ0K πtan−1W 2dayL→∞ (2.23) This result in (2.23) can be derived directly by using the result for an inﬁnite current in (2.14), B=(μ0I/2πr)aφ: B=2μ0K 2π⎜integraldisplayW/2 y=01 Rcosαd y ayL→∞ =μ0K π⎜integraldisplayW/2 y=0d d2+y2dyay =μ0K πtan−1W 2day where we again used integral 120.1 from Dwight [7]. Taking the limit of this asW→∞ gives an inﬁnite current sheet and the result derived directly in (2.21). EXAMPLE A loop of current of radius ais centered on the origin of a rectangular coor- dinate system and lies in the xyplane as shown in Fig. 2.12. Determine the B ﬁeld at a point z=don the zaxis. A segment of the current loop is of length ad φ, where φis the cylindrical coordinate system variable. The distance R from the differential segment to the point is R=√ a2+d2, and the cosine of the angle between the differential contribution dBand the zaxis is cosα=a R Asφvaries from φ=0t oφ=2πthe horizontal components (in the xyplane) ofdBcancel, leaving the Bﬁeld along the zaxis in the positive zdirection as B=μ0I 4π⎜integraldisplay2π φ=01 R2cosαad φ az =μ0I 4π⎜integraldisplay2π φ=0a2 ⎜parenleftbiga2+d2⎜parenrightbig3/2dφaz 32 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) z xyI a a dφφRdB B α d z= Ra FIGURE 2.12. Current loop. But this is a simple integral since the integrand does not depend on φ: B=μ0I 2a2 ⎜parenleftbiga2+d2⎜parenrightbig3/2azz≥0 (2.24) At the center of the loop, the ﬁeld is B=μ0I 2aazd=0 (2.25) At very large distances from the loop compared to the loop radius, d/greatermucha, (2.24) simpliﬁes to B=μ0Ia2 2d3az =μ0m 2πd3azd/greatermucha (2.26) Themagnetic dipole moment m is deﬁned as the product of the current and the area of the current loop: m=πa2I (2.27) THE BIOT–SA V ART LAW 33 z xyd z= a Ia I zBz FIGURE 2.13. Helmholtz coil. Observe that at relatively large distances from the loop, d/greatermucha, the magnetic ﬁeld decays with distance as inverse-distance cubed. A Helmholtz coil, shown in Fig. 2.13, is a pair of current loops that are used to provide a fairly uniform magnetic ﬁeld. Superimposing the results for themagnetic ﬁeld on the axis of each coil obtained in (2.24) gives the magneticﬁeld along the zaxis as B z=μ0Ia2 2⎡ ⎢⎣1 ⎜parenleftbiga2+z2⎜parenrightbig3/2+1 ⎜parenleftBig a2+(z−d)2⎜parenrightBig3/2⎤ ⎥⎦ (2.28) To examine the change in the ﬁeld along the zaxis between these two coils, we differentiate (2.28) with respect to zto give ∂Bz ∂z=3μ0Ia2 2⎡ ⎢⎣−z ⎜parenleftbiga2+z2⎜parenrightbig5/2−z−d ⎜parenleftBig a2+(z−d)2⎜parenrightBig5/2⎤ ⎥⎦ (2.29) This derivative is precisely zero midway between the two coils at z=d/2, meaning that the rate of change of the ﬁeld along the zaxis midway between the two coils is zero. If we take the second derivative, it can also be made zeroatz=d/2 if we choose the separation between the two coils equal to their 34 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) radii,d=a, thereby giving a further uniform nature of the Bﬁeld between the two coils. 2.4 AMP `ERE’S LA W AND THE LINE INTEGRAL In Section 2.3 we showed how to calculate the magnetic ﬁelds of currents using the Biot–Savart law. The calculations required the evaluation of certainintegrals. Amp `ere’s law allows the direct solution of many of those problems without the evaluation of any integrals , but the problem must exhibit a certain symmetry to be able to do so. Amp `ere’s law for dc currents is stated as [3,6] ⎜contintegraldisplay cH·dl=Ienclosed (2.30) where His the magnetic ﬁeld intensity vector. Recall that for a linear, homoge- neous, and isotropic surrounding medium, BandHcan be freely interchanged using B=μH, and μ=μrμ0is the permeability of the surrounding medium. Amp `ere’s law essentially provides that if we sum the product of the differ- ential segments of the path, dl, and the components of Hthat are tangent to a closed path c, we obtain the netcurrent that penetrates the surface sthat is enclosed by the closed path illustrated in Fig. 2.14. The direction of the closedcontour cand the direction of the enclosed current are related by the right- hand rule: Place the ﬁngers of the right hand in the direction of cand the thumb will point in the direction of I enclosed . The integral on the left-hand side c sI c dlH-I I I Inet= 3I-I = 2I FIGURE 2.14. Amp `ere’s law. AMP `ERE’S LAW AND THE LINE INTEGRAL 35 of Amp `ere’s law is said to be a line integral (see the Appendix for a review of the line integral). The line integral adds the products of the differential pathlengths dland the components of Hthat are tangent to the contour path c. There are two components of H: One is parallel to the path and the other is perpendicular to the path. It is sensible that only the components of Hthat are parallel to the path should contribute to the line integral. Amp `ere’s law is similar to Gauss’s law for the electric ﬁeld given in (2.4), which provides that the sum of the products of the components of Dthat are perpendicular to aclosed surface sand the differential surface areas dswill give the net positive charge enclosed by the closed surface. Amp `ere’s law can be used to compute the Hﬁeld for current distributions by using symmetry. To use Amp `ere’s law to determine H, we must be able to choose a closed contour cencircling the current so that the Hﬁeld along that contour has two properties. The ﬁrst property is that Hmust be tangent to the closed contour c at every point on it. This allows us to remove the dot productand write Amp `ere’s law solely in terms of the magnitudes of Handdlas ⎜contintegraldisplay cHd l=Ienclosed (2.31a) The second property is that Hmust be constant at all points along the contour c. This will allow us to remove Hfrom the integral in (2.31a) and write Amp `ere’s law as H⎜contintegraldisplay cdl ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright total length of contour c=Ienclosed (2.31b) Hence, if contour ccan be chosen such that it has these two properties, His simply the total current enclosed divided by the total length of contour c. EXAMPLE Determine the magnetic ﬁeld intensity about a current that is inﬁnite in length. This was solved in Section 2.3 using the Biot–Savart law, which requiredsetting up and evaluating an integral. To solve this problem using Amp `ere’s law, we again orient the current along the zaxis with the current directed in the+zdirection as shown in Fig. 2.15. We observe that because of the Biot– Savart law, the Hﬁeld will be circumferentially directed about the current at 36 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) (b)rIz xyto∞ to∞Hc r dφφ (a) Ic Hrx y r dφ FIGURE 2.15. Using Amp `ere’s law to determine the Hﬁeld about an inﬁnitely long current. all points along it. Because of the assumption that the current is inﬁnite in length, we may choose a closed contour cthat is a circle of radius rcentered on the current and place it at any point along the wire. The Hﬁeld will be tangent to all points on this contour. This allows us to remove the dot productfrom Amp `ere’s law: ⎜contintegraldisplay cHd l=⎜integraldisplay2π φ=0Hrdφ =I AMP `ERE’S LAW AND THE LINE INTEGRAL 37 In addition, the Biot–Savart law shows that the Hﬁeld magnitude will be constant in value at all points on cthat are a distance rfrom the current, so that we may remove Hfrom the integral and obtain H⎜contintegraldisplay cdl=H⎜integraldisplay2π φ=0rd φ =2πrH =I giving the Hﬁeld as H=I 2πraφ (2.32) Substituting B=μ0H gives the result in (2.14) that was derived with the Biot–Savart law but required the evaluation of an integral. EXAMPLE Currents ﬂow through wires of circular, cylindrical cross section whose radii rw, although small, are nonzero. If the wire is isolated from (or far from) other currents, the current inside it will be distributed symmetrically aboutthe wire axis. (We investigate the inﬂuence of nearby currents on the currentredistribution, the proximity effect, in Section 4.6). In the case of dc currents,the total current Icarried by the wire will also be uniformly distributed over the wire cross section with a current density over the cross section of J=I πr2wA/m2 For the purpose of determining the magnetic ﬁeld external to this isolated wire, we can replace the wire and its current with a ﬁlament of current locatedon the axis of the wire that contains the total current I. If the wire is further assumed to be inﬁnite in length (or very long), we can then use the basic resultfor a ﬁlamentary current of inﬁnite length in (2.14) to compute the magneticﬁeld external to the wire, and the actual radius of the wire does not enter intothis. In this example we demonstrate the validity of this important principle. Consider an isolated wire of radius r wcarrying a total current Ithrough its cross section. Certainly because of symmetry, the current is symmetric aboutthe wire axis. But as the frequency fof the current increases from zero (dc), 38 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) it will become concentrated near the wire surface in an annulus at the wire surface having a thickness of a few skin depths [3,6], where the skin depthparameter is δ=1 √πfμ 0σm The conductivity of the wire material is denoted as σ(copper has σ=5.8× 107S/m), and the wire material is assumed to be nonmagnetic, μr=1. At dc,f=0, the skin depth is inﬁnite, showing that a dc current is distributed uniformly over the wire cross section. We ﬁrst determine the magnetic ﬁeld of an isolated wire of radius rwand inﬁnite length that is carrying a total dc current Iin its cross section by using Amp `ere’s law. To determine the ﬁeld external to the wire using Amp `ere’s law, we surround the wire with a circular contour of radius ras shown in Fig. 2.16(a). Since the current is distributed uniformly over the wire cross rrw Bφ (a) r>rw (b) r<rw rrw φB2 2 wmA rIJ π= 2 2 wmA rIJ π= FIGURE 2.16. Using Amp `ere’s law to determine the magnetic ﬁeld of an isolated wire. AMP `ERE’S LAW AND THE LINE INTEGRAL 39 section and therefore symmetrically about the wire axis, the magnetic ﬁeld intensity vector is in the circumferential or φdirection (in a cylindrical coor- dinate system) about the axis of the wire and is constant around that contour.Hence, from Amp `ere’s law we obtain ⎜contintegraldisplay cH·dl=Hφ2πr=I and we again obtain the basic result in (2.14): Bφ=μ0Hφ =μ0I 2πrr>r w (2.33a) The magnetic ﬁeld external to the wire is the same as if we concentrate the entire current in a ﬁlament on the wire axis and is independent of the wireradius. Next, we determine the magnetic ﬁeld internal to the wire. Again surround an interior portion of the wire with a circular contour of radius r(r<r w) centered on the wire axis as shown in Fig. 2.16(b). The current density for atotal current of Ithat is uniformly distributed over the wire cross section is J=I πr2wA/m2 Hence, this contour encloses a total current of Ienclosed =I πr2wπr2 =Ir2 r2wA Again, by symmetry about the wire axis, the magnetic ﬁeld is directed cir- cumferentially around this contour and is constant at points on it. Hence, byAmp `ere’s law we obtain B φ=μ0Hφ =μ0 2πrIr2 r2w =μ0Ir 2πr2wr<r w (2.33b) The magnetic ﬁeld is plotted versus the radius of the contour in Fig. 2.17. Finally, we determine these results directly by integration as shown in Fig. 2.18. The current density over the wire cross section is again J=I πr2wA/m2 40 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) rwrφB rI πμ 202 w0 2rr I πμ FIGURE 2.17. Plot of the magnetic ﬁeld for an isolated wire. At a radius r/primeand angle φ, a differential area is r/primedφ dr/prime, which contains a differential current of I πr2wr/primedφ dr/primeA Treat this as an inﬁnite-length ﬁlament of current and use the result in (2.14) to determine the differential contribution to the magnetic ﬁeld at a radius r rR Rr′ r′φ φα α() ( ) r d d r′ ′ = φ Area wrA2 wr d d r rI′ ′φ π 2 2 wmA rIJ π=α dB dB φdBα FIGURE 2.18. Determining the magnetic ﬁeld of a wire with the Biot–Savart law. AMP `ERE’S LAW AND THE LINE INTEGRAL 41 from the wire axis as shown in Fig. 2.18: dB=μ0 2πRI πr2wr/primedφ dr/prime where the distance from this differential current to the point where we desire to compute the ﬁeld is (using the law of cosines) R=⎜radicalBig r2+r/prime2−2rr/primecosφ The horizontal components of dBfrom symmetrically disposed elements can- cel, leaving the net ﬁeld in the φdirection as dBφ=2μ0 2πRI πr2wr/primedφ dr/primecosα =μ0I π⎜parenleftbigπr2w⎜parenrightbigr/prime⎜parenleftbigr−r/primecosφ⎜parenrightbigdφ dr/prime r2+r/prime2−2rr/primecosφ and cosα=r−r/primecosφ R Integrating this from r/prime=0t or/prime=rwand from φ=0t oφ=πgives the total magnetic ﬁeld of the wire: Bφ=μ0I π⎜parenleftbigπr2w⎜parenrightbig⎜integraldisplayrw r/prime=0r/prime⎜bracketleftbigg⎜integraldisplayπ φ=0r−r/primecosφ r2+r/prime2−2rr/primecosφdφ⎜bracketrightbigg dr/prime The interior integral can be evaluated using the remarkable integral 859.124 of Dwight [7]: ⎜integraldisplayπ 0(a−bcosx)dx a2+b2−2abcosx=⎧ ⎨ ⎩π aa>b>0 0 b>a>0(D859.124) Forr>r wthe result is Bφ=μ0I π⎜parenleftbigπr2w⎜parenrightbig⎜integraldisplayrw r/prime=0r/prime⎜bracketleftbigg⎜integraldisplayπ φ=0r−r/primecosφ r2+r/prime2−2rr/primecosφdφ⎜bracketrightbigg dr/prime =μ0I π⎜parenleftbigπr2w⎜parenrightbig⎜integraldisplayrw r/prime=0r/primeπ rdr/prime =μ0I π⎜parenleftbigπr2w⎜parenrightbigπ rr2 w 2 =μ0I 2πrr>r w (2.34a) 42 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) again giving the fundamental result in (2.14). For r<r wwe break the integral into two pieces with respect to r/primein order to use (D859.124): Bφ=μ0I π⎜parenleftbigπr2w⎜parenrightbig⎜integraldisplayrw r/prime=0r/prime⎜bracketleftbigg⎜integraldisplayπ φ=0r−r/primecosφ r2+r/prime2−2rr/primecosφdφ⎜bracketrightbigg dr/prime =μ0I π⎜parenleftbigπr2w⎜parenrightbig⎜bracketleftbigg⎜integraldisplayr r/prime=0π rr/primedr/prime+⎜integraldisplayrw r/prime=r(0)r/primedr/prime⎜bracketrightbigg =μ0I π⎜parenleftbigπr2w⎜parenrightbigπ rr2 2 =μ0Ir 2πr2wr<r w (2.34b) as was derived using Amp `ere’s law. EXAMPLE Next we derive the magnetic ﬁeld of a coaxial cable. The coaxial cable has an inﬁnite (or very long) length and consists of an inner wire of radius rw contained within an overall shield of inner radius rsand thickness t, as shown in Fig. 2.19(a). A current Iis passed down the inner wire and returns in the shield. To determine the magnetic ﬁeld, we surround the inner wire with a circular contour of radius ras shown in Fig. 2.19(b). The dc current Iis uniformly distributed over the cross section of the wire and over the cross section of theshield. Hence, the magnetic ﬁeld is in the circumferential or φdirection tangent to the contour and is constant around that contour. In the region between thewire and the shield, r w<r<r s, the total current enclosed by the contour is I. Hence, Amp `ere’s law gives for rw<r<r s, ⎜contintegraldisplay cH·dl=Hφ2πr =Ir w<r<r s and the magnetic ﬂux density in the region between the wire and the shield is Bφ=μ0I 2πrrw<r<r s (2.35a) AMP `ERE’S LAW AND THE LINE INTEGRAL 43 wrI I (a) (b)rwr φB I It tsr srto∞ to∞ FIGURE 2.19. Coaxial cable. The magnetic ﬂux density inside the wire is, as in the preceding example, Bφ=μ0Ir 2πr2wr<r w (2.35b) The dc current −Iin the shield is also uniformly distributed over the shield cross section and has a current density of Jshield=−I π⎜bracketleftbig(rs+t)2−r2s⎜bracketrightbig A/m2rs<r<r s+t Hence, expanding the contour to within the shield, rs<r<r s+t, encloses a total current of Ienclosed =I−Iπr2−πr2 s π⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig =I(rs+t)2−r2 (rs+t)2−r2sA rs<r<r s+t 44 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) By Amp `ere’s law, the magnetic ﬁeld in the shield is circumferentially directed and becomes Bφ=μ0I 2πr(rs+t)2−r2 (rs+t)2−r2srs<r<r s+t (2.35c) Expanding the contour to enclose the entire coaxial cable, r>r s+t, shows, by Amp `ere’s law, that the magnetic ﬁeld is Bφ=0 r>r s+t (2.35d) since the total current enclosed is zero because of the equal but oppositely directed currents. These results, easily obtained using Amp `ere’s law, can also be obtained using direct integration in the same fashion as in the preceding example. Theresults for the ﬁelds in a coaxial cable obtained by using Amp `ere’s law in this example and given in (2.35b) for r<r w, and in (2.35a) for rw<r<r s, were obtained by direct integration in the preceding example. The ﬁnal resultin (2.35c) for r s<r<r s+tcan also be obtained by direct integration by reference to Fig. 2.20. Again we set up the integration as in the preceding example. The differential currents in the shield are −I π⎜bracketleftbig(rs+t)2−r2s⎜bracketrightbig ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright JshieldA/m2r/primedφ dr/primeA rR Rr′r′ φ φα α αα() ( ) dr rd′ ′ = φ Area dB dB φ dBt I –Isr[]A ) (2 2r d d r r t rI s s′ ′ − +− φ π FIGURE 2.20. Determining the magnetic ﬁeld within the shield by direct integration. AMP `ERE’S LAW AND THE LINE INTEGRAL 45 Hence, the magnetic ﬁeld in the shield for rs<r<r s+tdue to the currents in the shield is Bφ=2μ0 2π−I π⎜bracketleftbig(rs+t)2−r2s⎜bracketrightbig ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright Jshield A/m2⎜integraldisplayrs+t r/prime=rsr/prime⎜bracketleftbigg⎜integraldisplayπ φ=0r−r/primecosφ r2+r/prime2−2rr/primecosφdφ⎜bracketrightbigg dr/prime =−μ0I π1 π⎜bracketleftbig(rs+t)2−r2s⎜bracketrightbig⎜bracketleftbigg⎜integraldisplayr r/prime=rsπ rr/primedr/prime+⎜integraldisplayrs+t r/prime=r(0)r/primedr/prime⎜bracketrightbigg =−μ0I πr⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig⎜bracketleftBigg r/prime2 2⎜bracketrightBiggr r/prime=rs =−μ0I 2πr⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig⎜parenleftBig r2−r2 s⎜parenrightBig rs<r<r s+t where we have again separated the integration from r/prime=rstor/prime=rs+tinto two parts in order to use integral 859.124: ⎜integraldisplayπ 0(a−bcosx)dx a2+b2−2abcosx=⎧ ⎨ ⎩π aa>b>0 0 b>a>0(D859.124) To this we add the contribution to the ﬁeld within the shield due to the current of the interior wire: Bφ=μ0I 2πrrs<r<r s+t Combining these two contributions yields (2.35c) Bφ=μ0I 2πr−μ0I 2πr⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig⎜parenleftBig r2−r2 s⎜parenrightBig =μ0I 2πr(rs+t)2−r2 (rs+t)2−r2srs<r<r s+t (2.35c) For the ﬁelds external to the cable, r>r s+t, the integral above is, according to (D859.124), Bφ=2μ0 2π−I π⎜bracketleftbig(rs+t)2−r2s⎜bracketrightbig ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright Jshield A/m2⎜integraldisplayrs+t r/prime=rsr/prime⎜bracketleftbigg⎜integraldisplayπ φ=0r−r/primecosφ r2+r/prime2−2rr/primecosφdφ⎜bracketrightbigg dr/prime =−μ0I π1 π⎜bracketleftbig(rs+t)2−r2s⎜bracketrightbig⎜bracketleftbigg⎜integraldisplayrs+t r/prime=rsπ rr/primedr/prime⎜bracketrightbigg 46 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) =−μ0I πr⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig⎜bracketleftBigg r/prime2 2⎜bracketrightBigg(rs+t) r/prime=rs =−μ0I 2πr⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig⎜bracketleftBig (rs+t)2−r2 s⎜bracketrightBig =−μ0I 2πrrs+t<r which, combined with the ﬁeld due to the interior wire, gives a result of zero, which is (2.35d). EXAMPLE Use Amp `ere’s law to determine the Hﬁeld of the inﬁnite current sheet shown in Fig. 2.10. View the sheet from the top in the xyplane and construct a rectangular closed contour cas shown in Fig. 2.21. By symmetry we see that the Hﬁeld must be directed parallel to the sheet. Hence, the Hﬁeld is y xK Kc cc cH Hto∞ to∞HHl FIGURE 2.21. Inﬁnite current sheet and Amp `ere’s law. VECTOR MAGNETIC POTENTIAL 47 parallel to the sides and perpendicular to the tops (and contributes nothing to Amp `ere’s law along the tops of contour c). If the contour has tops of width w and sides of length l, the total current enclosed by the contour is Ienclosed =lK m×A/m=A Hence, Amp `ere’s law around the entire contour is ⎜contintegraldisplay cH·dl=2⎜integraldisplay wH·dl ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0+2⎜integraldisplay lH·dl =2⎜integraldisplay lHd l =2lH =Ienclosed =lK Hence, the Hﬁeld is H=⎧ ⎪⎪⎨ ⎪⎪⎩K 2ay forx>0 −K 2ay forx<0(2.36) Substituting B=μ0H gives the result in (2.21) that was derived by the Biot–Savart law but required evaluation of an integral. 2.5 VECTOR MAGNETIC POTENTIAL Since the magnetic ﬁeld has no sources or sinks, it must form closed loops everywhere. Hence, the divergence of the magnetic ﬁeld, according to Gauss’s law for the magnetic ﬁeld, is zero: ∇·B=0 (2.37) In the Appendix it is shown that the divergence of the curl of any vector ﬁeld is zero: ∇·∇×A=0 (2.38) The divergence ∇·Frepresents the net ﬂux or outﬂow of the vector ﬁeld F from a point, whereas the curl ∇×Frepresents the circulation of the vector 48 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) ﬁeld Fabout a point . Although the identity in (2.38) is proven directly in the Appendix, it is a sensible identity. If the vector ﬁeld Ahas nonzero circulation at a point, ∇×A/=0, its curl ∇×Ashould have no outﬂow from the point and the identity is satisﬁed. On the other hand, suppose that the vector ﬁeldhas no circulation at a point, ∇×A=0. Then the divergence of this is clearly zero. The identity in (2.38), combined with Gauss’s law for the magnetic ﬁeld in (2.37), ∇·B=0, allows us to deﬁne another, auxiliary ﬁeld as B=∇ ×A (2.39) This new vector ﬁeld Ais called the vector magnetic potential . This is very similar to deﬁning the scalar electric potential or voltage φfor a static (dc) electric ﬁeld Efrom∇×E=0 and using the identity from the Appendix of∇×∇φ=0 to deﬁne the electric ﬁeld in terms of the scalar potential function φasE=− ∇ φ[3,6]. It turns out (see [3]) that to deﬁne a vector ﬁeld completely, we must deﬁne the curl of that ﬁeld as well as its divergence.Equation (2.39) has deﬁned the curl of A. It also turns out that we can, without any contradiction in doing so, deﬁne the divergence of Aas zero: ∇·A=0 (2.40) thereby completely deﬁning this new magnetic potential vector A. To determine an equation relating the vector magnetic potential to the cur- rents that produce it, we employ Amp `ere’s law given in (2.30): ⎜contintegraldisplay cH·dl=Ienclosed (2.30) Using Stokes’s theorem (see the Appendix), we can write Amp `ere’s law as ⎜contintegraldisplay cH·dl=⎜integraldisplay s(∇×H)·ds =Ienclosed =⎜integraldisplay sJ·ds (2.41) where sis the open surface surrounded by the closed contour cas illustrated in Fig. 2.22. The current density vector throught the surface sis denoted as J, whose units are A /m2. The direction of the normal to the surface sas well as the direction of the contour cthat encloses sare related by the right-hand rule. Place the ﬁngers of the right hand in the direction of cand the thumb will point in the direction of ds. Comparing both sides of (2.41) gives Amp `ere’s law in point form as ∇×H=J (2.42) VECTOR MAGNETIC POTENTIAL 49 cH dl sJ J FIGURE 2.22. Amp `ere’s law and Stokes’s theorem. Substituting the relation between BandHusing the permeability of the surrounding medium (assumed not to be ferromagnetic), B=μ0H,g i v e s ∇×B=μ0J (2.43) Substituting the deﬁnition of the vector magnetic potential given in (2.39) gives ∇×(∇×A)=μ0J (2.44) The curl of the curl of a vector ﬁeld can be written as [3,6] ∇×(∇×A)=∇ (∇·A)−∇2A (2.45) We deﬁned the divergence of Aas zero in (2.40), ∇·A=0, to complete the deﬁnition of the vector magnetic potential A. Hence, (2.44) becomes ∇2A=−μ0J (2.46) The solution to (2.46) is [3,6] A=μ0 4π⎜integraldisplay vJdv R(2.47a) where vis the volume enclosing the current density J(which is the source ofA). The distance between the point where we are determining Aand a differential volume of the current that contains Jdvampere-meters is denoted 50 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) asR. If the current is conﬁned to a surface, this reduces to A=μ0 4π⎜integraldisplay sKds R(2.47b) where sis the surface containing the surface current density Kwhose units are A/m. The distance between the point where we are determining Aand a differential surface of the current that contains Kdsampere-meters is denoted asR. For line currents we consider the current Ito be contained in a differential cylinder of length dland cross-sectional area ds. If the current density is uniformly distributed over the cylinder cross section (as will be the case fordc currents [3]), the total current is Jd s=I, so that Jdv=Jd sd l =Id l, and the result becomes A=μ0 4π⎜integraldisplay lI Rdl (2.47c) where a vector differential length of the line in the direction of the current is denoted as dland contains Idlampere-meters. Again Ris the distance between the point where we are determining Aand the differential current segment. The units of the vector magnetic potential are Wb/m, magnetic ﬂuxper length. We will learn the reason for these units in Chapter 4. It should be noted that we obtained the basic result for the computation of A in (2.47a) from the solution of (2.46), ∇ 2A=−μ0J. But we obtained (2.46) by deﬁning the divergence of Ain (2.45) as zero (i.e., ∇·A=0). However, the basic result for computing Ain (2.47a) does not require that ∇·A=0. The Helmholtz theorem [3,6] establishes the fact that to deﬁne a vector ﬁeldsuch as Acompletely requires that its curl andits divergence be deﬁned. But the choices for these are arbitrary and are not related. We can show this bydemonstrating that taking the curl of (2.47a) gives B=∇ ×A, where the resulting Bis the Biot–Savart law given in (2.11). To show this we take the curl of (2.47a): B=∇ ×A =μ 0 4π∇×⎜integraldisplay vJ Rdv =μ0 4π⎜integraldisplay v∇×J Rdv (2.48) We can interchange the order of differentiation and integration using Leibnitz’s rule [12] since the ∇operator takes derivatives with respect to the coordinates of the location of BandA, whereas the volume integral is with respect to the coordinates of the current J. Using a vector identity [3], VECTOR MAGNETIC POTENTIAL 51 we can write the curl of the integrand as ∇×J R=∇⎜parenleftbigg1 R⎜parenrightbigg ×J+1 R(∇×J)⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0 =∇⎜parenleftbigg1 R⎜parenrightbigg ×J =−J×∇⎜parenleftbigg1 R⎜parenrightbigg (2.49) The del operator takes the derivatives with respect to the coordinates of the location of BandA: the ﬁeld point. Hence, the curl of Jis zero here since J involves only the coordinates of the location of the source current. You canverify in spherical coordinates (see the Appendix) that ∇⎜parenleftbigg1 R⎜parenrightbigg =−1 R2aR (2.50) where aRis a unit vector pointing from the current to the ﬁeld point. The vector identity in (2.49) is, of course, sensible since it is the vector counterpart to thescalar result using the chain rule. Therefore, we obtain B=∇ ×A =μ 0 4π∇×⎜integraldisplay vJ Rdv =μ0 4π⎜integraldisplay v∇×J Rdv =μ0 4π⎜integraldisplay vJ×aR R2dv (2.51) which is the Biot–Savart law for determining Bgiven in (2.11). If the current forms a closed loop (as all dc currents must), ∇·A=0, but it is not necessary to deﬁne the divergence of Ato be zero in order to obtain (2.47). The solutions in (2.47) apply to any coordinate system. If we specialize them to a rectangular coordinate system, we obtain Ax=μ0 4π⎜integraldisplay vJx Rdv Ay=μ0 4π⎜integraldisplay vJy Rdv Az=μ0 4π⎜integraldisplay vJz Rdv (2.52a) This is a signiﬁcant result because it says that (1) each component of Jpro- duces the corresponding component of A, and (2) the direction of the resulting 52 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) Ax,Ay,Azis the same as the direction of the corresponding Jx,Jy,Jzthat pro- duced it! In other words, a current that is directed solely in the zdirection will produce a vector magnetic potential that is solely in the zdirection parallel to theJzatall points in the space around the current! For a current distributed over a surface s, these results become Ax=μ0 4π⎜integraldisplay sKx Rds Ay=μ0 4π⎜integraldisplay sKy Rds Az=μ0 4π⎜integraldisplay sKz Rds (2.52b) If the current is a line current whose contour is lsuch as in a wire, these become Ax=μ0 4π⎜integraldisplay lIx Rdl=μ0 4π⎜integraldisplay lI Rdlx Ay=μ0 4π⎜integraldisplay lIy Rdl=μ0 4π⎜integraldisplay lI Rdly Az=μ0 4π⎜integraldisplay lIz Rdl=μ0 4π⎜integraldisplay lI Rdlz (2.52c) Note that unlike the Biot–Savart law, which is an inverse-square law where Bdepends on the square of the inverse distance between the current and the B ﬁeld, the magnetic vector potential simply depends on the inverse of the dis-tance Rbetween the current and the component of Athat it produces. In some problems it is simpler to determine the components of Afrom (2.47), which for rectangular coordinates are given in (2.52), and then simply determineBby computing its curl mechanically from B=∇ ×Ain the appropriate coordinate system, than it is to compute Bdirectly using the Biot–Savart law. One of the main advantages of ﬁrst computing the three components of the vector magnetic potential Avia (2.47) or, in rectangular coordinates, from (2.52) and then determining Bby differentiation via B=∇ ×Ais that we do not need to integrate vector quantities to obtain A! The expansion of (2.47) for rectangular coordinates given in (2.52) shows this. Determining Bdirectly by integration by applying the Biot–Savart law may require that we integratevector quantities. To avoid integrating vector quantities, we utilize symmetryand resolve Binto components, thereby restricting the solution only to points about the current where we can exploit symmetry (see, e.g., Figures 2.10and 2.11). VECTOR MAGNETIC POTENTIAL 53 Observe that the vector magnetic potential in (2.47) and (2.52) requires the integral of an inverse distance, 1/R. Hence, the resulting vector magneticpotential typically involves a natural logarithm (ln) function as the result.Thus, we expect to see these natural log functions in the following resultsfor various conﬁgurations. The Bﬁeld obtained from (2.39) then requires the derivatives of these natural log functions. In some of the earlier problems we solved for the magnetic ﬁeld of a very idealized case: a current of inﬁnite length that is directed in the z direction. The magnetic ﬂux density of a current of inﬁnite length is ﬁnite andgiven by B=μ 0I 2πraφ From the relation B=∇ ×Ain cylindrical coordinates, Ais related to Bas B=⎜parenleftbigg∂Ar ∂z−∂Az ∂r⎜parenrightbigg aφ =−∂Az ∂raφ This is because the Bﬁeld has only a φcomponent, and the current and resulting vector magnetic potential is directed in the zdirection. Integrating this, we obtain Az=−μ0I 2πlnr+C current of inﬁnite length where Cis a constant of integration. However, when we are using B=∇ ×A to determine the Bﬁeld for an inﬁnite-length current by ﬁrst determining A, we can ignore the integration constant Cbecause we differentiate Azin order to determine Bφ. Hence, for a current of inﬁnite length we may assume a form ofAzto be Az=−μ0I 2πlnr current of inﬁnite length (2.53) EXAMPLE Determine the vector magnetic potential at a distance rfrom the center of a current of length Land on a line perpendicular to its midpoint as shown in Fig. 2.23. Then determine the magnetic ﬁeld Bfrom that result. Although steady (dc) currents must form closed loops , we again use the solutions for currents of ﬁnite length to construct, by superposition, the magnetic ﬁelds of 54 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) xyz 2Lz= 2Lz− =dzR I rA FIGURE 2.23. Vector magnetic potential of a current. closed current loops (see the discussion in Section 2.9). Hence, determining the vector magnetic potential for currents of ﬁnite length is useful for thatpurpose. The problem again ﬁts a cylindrical coordinate system (see the Appendix for a discussion of the cylindrical coordinate system). From (2.52c) andFig. 2.23 we see that since the current is directed solely in the zdirection, the vector magnetic potential will be parallel to it at all points and directed inthezdirection also. From (2.52c), A z=μ0 4π⎜integraldisplayL/2 z=−L/2I Rdz =μ0I 4π⎜integraldisplayL/2 −L/21√ r2+z2dz =μ0I 4π⎜bracketleftBig ln⎜parenleftBig z+⎜radicalbig z2+r2⎜parenrightBig⎜bracketrightBigL/2 −L/2 =μ0I 4πlnL/2+⎜radicalbig (L/2)2+r2 −(L/2) +⎜radicalbig (L/2)2+r2(2.54a) and we have used integral 200.01 from Dwight [7]: ⎜integraldisplay1√ x2+a2dx=ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig (D200.01) VECTOR MAGNETIC POTENTIAL 55 This result can be simpliﬁed to Az=μ0I 4πln(L/2r)+⎜radicalbig (L/2r)2+1 −(L/2 r)+⎜radicalbig (L/2r)2+1 =μ0I 2πln⎡ ⎣L 2r+⎜radicalBigg⎜parenleftbiggL 2r⎜parenrightbigg2 +1⎤ ⎦ (2.54b) and we have used log( A/B) =logA−logBand an important natural logarithm identity: ln⎜parenleftBig x+⎜radicalbig x2+1⎜parenrightBig =− ln⎜parenleftBig −x+⎜radicalbig x2+1⎜parenrightBig which can be proven directly (log AB=logA+logB): ln⎜parenleftBig x+⎜radicalbig x2+1⎜parenrightBig +ln⎜parenleftBig −x+⎜radicalbig x2+1⎜parenrightBig =ln(1)=0 The result in (2.54b) can be written in an alternative form using the inverse hyperbolic sine function: sinh−1x≡ln⎜parenleftBig x+⎜radicalbig x2+1⎜parenrightBig =− sinh−1(−x) (D700.1) Hence, we could write (2.54b) as Az=μ0I 2πsinh−1L 2r(2.54c) The simpliﬁed result in (2.54b) could also have been obtained directly by utilizing symmetry and integrating from z=0t oz=L/2 and doubling the result: Az=2μ0 4π⎜integraldisplayL/2 z=0I Rdz =μ0I 2π⎜bracketleftBig ln⎜parenleftBig z+⎜radicalbig z2+r2⎜parenrightBig⎜bracketrightBigL/2 0 =μ0I 2π⎡ ⎣ln⎛⎝ L 2+⎜radicalBigg⎜parenleftbiggL 2⎜parenrightbigg2 +r2⎞ ⎠−lnr⎤ ⎦ =μ0I 2πln⎡ ⎣L 2r+⎜radicalBigg⎜parenleftbiggL 2r⎜parenrightbigg2 +1⎤ ⎦ =μ0I 2πsinh−1L 2r 56 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) Taking the curl of Ato give Bvia (2.39) gives (see the Appendix for the curl in cylindrical coordinates) B=∇ ×A(r, φ, z ) =⎜parenleftbigg1 r∂Az ∂φ−∂Aφ ∂z⎜parenrightbigg ar+⎜parenleftbigg∂Ar ∂z−∂Az ∂r⎜parenrightbigg aφ+⎜bracketleftBigg 1 r∂⎜parenleftbigrAφ⎜parenrightbig ∂r−1 r∂Ar ∂φ⎜bracketrightBigg az =−∂Az ∂raφ Substituting Azfrom (2.54b) and performing the differentiation gives B=−∂Az ∂raφ =μ0I 2πrL/2⎜radicalbig (L/2)2+r2aφ =μ0I 2πrL√ 4r2+L2aφ (2.55) which is the same as (2.13), which was obtained with the Biot–Savart law. We have used Azfrom (2.54b) and the derivative d drln⎡ ⎣a r+⎜radicalBigg⎜parenleftbigga r⎜parenrightbigg2 +1⎤ ⎦=−a r√ a2+r2 Alternatively, d dxsinh−1a x=d dxcsch−1x a =−a |x|√ x2+a2(D728.8) For an inﬁnitely long current, L→∞ , (2.55) evaluates to B=μ0I 2πraφL→∞ (2.56) which is (2.14) again. EXAMPLE Determine the vector magnetic potential of a current of length Lat some general point that is at a distance rfrom it (the cylindrical coordinate system variable) and at z=Z, as shown in Fig. 2.24, and then determine the Bﬁeld. VECTOR MAGNETIC POTENTIAL 57 xyz 2Lz= 2Lz− =dzR IrZ A FIGURE 2.24. Vector magnetic potential of a current of ﬁnite length. From Fig. 2.24 and (2.52c), we again see that since the current is directed solely in the zdirection, the vector magnetic potential will be parallel to it at all points and directed in the zdirection also. From (2.52c), Az=μ0I 4π⎜integraldisplayL/2 z=−L/21 Rdz where R=⎜radicalBig (Z−z)2+r2. Making a change of variables as λ=Z−zand dλ=−dzgives Az=μ0I 4π⎜integraldisplayZ+L/2 λ=Z−L/21√ λ2+r2dλ =μ0I 4πln(Z+L/2)+⎜radicalBig (Z+L/2)2+r2 (Z−L/2)+⎜radicalBig (Z−L/2)2+r2 =μ0I 4π⎜parenleftbigg sinh−1Z+L/2 r−sinh−1Z−L/2 r⎜parenrightbigg =μ0I 4π⎜parenleftbigg sinh−1Z+L/2 r+sinh−1L/2−Z r⎜parenrightbigg (2.57) 58 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) and we have again used integral 200.01 of Dwight [7]: ⎜integraldisplaydx√ x2+a2=ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig (D200.01) Also, we have again used the alternative relation sinh−1x≡ln⎜parenleftBig x+⎜radicalbig x2+1⎜parenrightBig =− sinh−1(−x) (D700.1) Obtain Bby taking the curl of Ain cylindrical coordinates, giving B=−∂Az ∂raφ =μ0I 4πr⎡ ⎣Z+L/2⎜radicalBig (Z+L/2)2+r2−Z−L/2⎜radicalBig (Z−L/2)2+r2⎤ ⎦aφ(2.58) which is the same as (2.15) obtained with the Biot–Savart law. We have used the derivative d drln⎜parenleftBig a+⎜radicalbig a2+r2⎜parenrightBig =−a r√ a2+r2+1 r and log(A/B) =logA−logB. Alternatively, we could use (D728.8). EXAMPLE Determine the vector magnetic potential Adue to a current loop of radius a that is centered on the origin of a rectangular coordinate system and lying inthexyplane as shown in Fig. 2.25. From that, determine B. For the following computation, we determine Aat a point located in the xz plane at x=ρ,y=0, and z. Because of the symmetry of the current loop, the ﬁeld will then be determined at any general point located at (ρ,φ,z )in a cylindrical coordinate system or, equivalently, at any general point (r, θ, φ )in a spherical coordinate system and will be independent of φin either case. If we pair off current segments at ±φmeasured with respect to the xaxis, from symmetry we see that Ais in the ydirection or, equivalently, in the φdirection at this point. At the point of interest we obtain, from (2.47c), A φ=μ0I 4π⎜contintegraldisplaydlφ R VECTOR MAGNETIC POTENTIAL 59 z xIa a dφθ rR Rφ x=ρφ –φA FIGURE 2.25. Current loop. The component of a differential length of the loop, dl=ad φ, in the direction ofAat this point is dlφ=ad φ cosφ Using the law of cosines, we obtain R2=a2+r2−2arsinθcosφ Hence, at the ﬁeld point Aφ=2μ0I 4π⎜integraldisplayπ φ=0acosφ⎜radicalbig a2+r2−2arsinθcosφdφ =μ0I 2π⎜integraldisplayπ φ=0acosφ⎜radicalbig a2+ρ2+z2−2aρcosφdφ (2.59) sincer2=z2+ρ2andρ=rsinθ. The integral in (2.59) is difﬁcult to evaluate. We ﬁrst restrict the result to the case where the point is at a distance that is far away with respect to thecurrent loop radius, r/greatermucha, and later will obtain the exact solution. Evaluating the denominator using the binomial theorem gives 1 R∼=⎜parenleftbigg1 r2−2arsinθcosφ⎜parenrightbigg1/2 =1 r⎜parenleftbigg 1−2a rsinθcosφ⎜parenrightbigg−1/2 ∼=1 r⎜parenleftbigg 1+a rsinθcosφ⎜parenrightbigg 60 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) Hence, we obtain Aφ∼=μ0I 2π⎜integraldisplayπ φ=0acosφ r⎜parenleftbigg 1+a rsinθcosφ⎜parenrightbigg dφ =μ0I 2πa2 r2sinθπ 2 =μ0Ia2sinθ 4r2 Taking the curl of Ain spherical coordinates (see the Appendix) to obtain B gives B=∇ ×A =1 rsinθ∂⎜parenleftbigsinθAφ⎜parenrightbig ∂θar−1 r∂⎜parenleftbigrAφ⎜parenrightbig ∂raθ =μ0a2I 4r3(2 cosθar+sinθaθ) (2.60) Themagnetic dipole moment is deﬁned in (2.27) as the product of the current and the area of the current loop: m=πa2I (2.27) Substituting (2.27) into (2.60) gives the components of Bas Br=μ0m 2πr3cosθ (2.61a) Bθ=μ0m 4πr3sinθ (2.61b) Bφ=0 (2.61c) Note that the magnetic ﬁeld of a current loop at large distances from the loop falls off inversely with the cube of distance. The exact solution of the integral in (2.59) can be obtained in terms of the complete elliptic integrals [10,11]. To do so, we make a change of variablesφ=π−2ζanddφ=−2dζso that cos φ=− cos 2ζ=2 sin 2ζ−1. Hence, (2.59) becomes Aφ=μ0Ia π⎜integraldisplayπ/2 ζ=02 sin2ζ−1⎜radicalBig (a+ρ)2+z2−4aρsin2ζdζ (2.62) Deﬁning k2=4aρ (a+ρ)2+z2(2.63) VECTOR MAGNETIC POTENTIAL 61 (2.62) reduces to Aφ=μ0I πk⎜radicalBigg a ρ⎜bracketleftBigg⎜parenleftBigg 1−k2 2⎜parenrightBigg K−E⎜bracketrightBigg (2.64) where the complete elliptic integrals of the ﬁrst and second kind are [7] K(k)=⎜integraldisplayπ/2 ζ=0dζ⎜radicalBig 1−k2sin2ζ(D773.1) and E(k)=⎜integraldisplayπ/2 ζ=0⎜radicalBig 1−k2sin2ζd ζ (D774.1) and are tabulated in Dwight, Tables 1040 and 1041 [7]. The magnetic ﬂux density is obtained in cylindrical coordinates (see the Appendix) from B=∇ ×Aas [10,11] Bρ=−∂Aφ ∂z =μ0I 2πz ρ⎜radicalBig (a+ρ)2+z2⎜bracketleftBigg −K+a2+ρ2+z2 (a−ρ)2+z2E⎜bracketrightBigg (2.65a) Bφ=0 (2.65b) Bz=1 ρ∂⎜parenleftbigρAφ⎜parenrightbig ∂ρ =μ0I 2π1⎜radicalBig (a+ρ)2+z2⎜bracketleftBigg K+a2−ρ2−z2 (a−ρ)2+z2E⎜bracketrightBigg (2.65c) Along the zaxis,ρ=0,k=0, so that K(0)=E(0)=π/2 and these general results reduce to Bρ→0 ρ=0 (2.66a) Bφ=0 ρ=0 (2.66b) Bz=μ0I 2a2 ⎜parenleftbiga2+z2⎜parenrightbig3/2ρ=0 (2.66c) 62 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) I IB B B B FIGURE 2.26. Magnetic ﬁeld of a loop. which is the result obtained in (2.24). In the plane of the loop, z=0, (2.65) reduces to Bρ=0 z=0 (2.67a) Bφ=0 z=0 (2.67b) Bz=μ0I 2π⎜parenleftbigg1 a+ρK+1 a−ρE⎜parenrightbigg =μ0Ia 2π⎜integraldisplayπ 0a−ρcosφ⎜parenleftbiga2−ρ2⎜parenrightbig⎜radicalbig a2+ρ2−2aρcosφdφ z =0 (2.67c) where k2in (2.63) becomes, for z=0, k2=4aρ (a+ρ)2z=0 (2.67d) The alternative result for Bzatz=0 in (2.67c) was obtained by substituting the change of variables φ=π−2ζanddφ=−2dζ, so that cos φ=− cos 2ζ= 2 sin2ζ−1 into KandE. At the center of the loop (z=0,ρ=0),k=0, and K=E=π/2, so that (2.67c) reduces to Bz=μ0I/2a, which is (2.25). The magnetic ﬁelds of a current loop are illustrated in Fig. 2.26. EXAMPLE Derive the Bﬁeld at any point about a current sheet of ﬁnite width Wand inﬁnite length as shown in Fig. 2.27 using the vector magnetic potential. VECTOR MAGNETIC POTENTIAL 63 to∞z y mAK (a) (b)y xK RyW WA Ato∞ Y X FIGURE 2.27. Current sheet of ﬁnite width and inﬁnite length. Again the sheet surface current can be viewed as currents I=Kd y , which are inﬁnite in length. We can assume a form for the differential contributiontoA zgiven in (2.53): dAz=−μ0Kd y 2πlnR current of inﬁnite length Here R=⎜radicalBig (y−Y)2+X2 64 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) and hence Az=−μ0K 2π⎜integraldisplayW/2 y=−W/ 2lnRd y =−μ0K 4π⎜integraldisplayW/2 y=−W/ 2ln⎜bracketleftBig (y−Y)2+X2⎜bracketrightBig dy Making a change of variables λ=y−Y, this becomes Az=−μ0K 4π⎜integraldisplayW/2−Y λ=−(W/ 2+Y)ln⎜parenleftBig λ2+X2⎜parenrightBig dλ =−μ0K 4π⎜bracketleftbigg λln⎜parenleftBig λ2+X2⎜parenrightBig −2λ+2Xtan−1λ X⎜bracketrightbiggW/2−Y −(W/ 2+Y) =−μ0K 4π⎜braceleftbigg (W/2−Y)ln⎜bracketleftBig (W/2−Y)2+X2⎜bracketrightBig −2(W/2−Y)+2Xtan−1W/2−Y X +(W/2+Y)ln⎜bracketleftBig (W/2+Y)2+X2⎜bracketrightBig −2(W/2+Y)+2Xtan−1W/2+Y X⎜bracerightbigg =−μ0K 4π⎜braceleftbigg (W/2−Y)ln⎜bracketleftBig (W/2−Y)2+X2⎜bracketrightBig +(W/2+Y)ln⎜bracketleftBig (W/2+Y)2+X2⎜bracketrightBig −2W+2Xtan−1 WX X2+Y2−(W/2)2⎜bracerightbigg (2.68) This was evaluated using integral 623 from Dwight [7]: ⎜integraldisplay ln⎜parenleftBig x2+a2⎜parenrightBig dx=xln⎜parenleftBig x2+a2⎜parenrightBig −2x+2atan−1x a(D623) We also used the trigonometric identity tan−1θ1±tan−1θ2=tan−1θ1±θ2 1∓θ1θ2θ1,θ2≥0 (2.69a) giving tan−1(x+y)+tan−1(x−y)=tan−1 2x 1−x2+y2(2.69b) VECTOR MAGNETIC POTENTIAL 65 and tan−1(x+y)−tan−1(x−y)=tan−1 2y 1+x2−y2(2.69c) This gives [x =(W/2)/Xandy=Y/X] tan−1W/2+Y X+tan−1W/2−Y X=tan−1 W/X 1−[(W/ 2)/X]2+(Y/X )2 =tan−1 WX X2+Y2−(W/2)2 The magnetic ﬁeld is determined from B=∇ ×Ain rectangular coordinates as B=∂Az ∂Yax−∂Az ∂Xay =μ0K 4π⎜braceleftBigg −ln(W/2+Y)2+X2 (W/2−Y)2+X2ax+2 tan−1 WX X2+Y2−(W/2)2ay⎜bracerightBigg (2.70) and we have used ∂ ∂utan−1u=1 1+u2(D512.4) Letting Y→0 in (2.70) gives the ﬁeld along a line perpendicular to the strip: B=μ0K 2πtan−1 WX X2−(W/2)2ay =μ0K πtan−1W 2XayY=0 (2.71) and we have used the identity in (2.69). But (2.71) is the same as (2.23) (X=d), which was derived directly for the Bﬁeld using the Biot–Savart law. This result can be derived directly from the Biot–Savart law. A cross- sectional view of the problem is shown in Fig. 2.28. The differential contribu-tion to the Bﬁeld at a general point x=Xandy=Ycan be obtained by again considering the sheet to be composed of inﬁnitely long ﬁlaments of currentsI=Kd y and using the fundamental result in (2.14): dB=μ 0Kdy 2πR(−cosθax+sinθay) where R=⎜radicalBig (Y−y)2+X2 66 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) to∞z y mAK (a) (b)y xKR yW Wto∞ Y XθθB FIGURE 2.28. Magnetic ﬁeld of a current sheet of ﬁnite width and inﬁnite length using the Biot–Savart law. and cosθ=Y−y R sinθ=X R Integrating gives the total Bﬁeld as B=⎜integraldisplayW/2 y=−W/ 2dB=Bxax+Byay VECTOR MAGNETIC POTENTIAL 67 Evaluating these gives Bx=−μ0K 2π⎜integraldisplayW/2 y=−W/ 2Y−y (Y−y)2+X2dz Making a change of variables, λ=Y−y,dλ=−dygives Bx=−μ0K 2π⎜integraldisplayY+W/ 2 λ=Y−W/ 2λ λ2+X2dλ Using integral 121.1 from Dwight [7], ⎜integraldisplayx a2+x2dx=1 2ln⎜parenleftBig a2+x2⎜parenrightBig (D121.1) gives Bx=−μ0K 4πln(Y+W/2)2+X2 (Y−W/2)2+X2 which is the xcomponent given in (2.70). The ycomponent becomes By=μ0K 2π⎜integraldisplayW/2 y=−W/ 2X (Y−y)2+X2dy Again making a change of variables, λ=Y−y,dλ=−dygives By=μ0K 2π⎜integraldisplayY+W/ 2 λ=Y−W/ 2X λ2+X2dλ Using integral 120.1 from Dwight [7], ⎜integraldisplay1 a2+x2dx=1 atan−1x a(D120.1) gives By=μ0K 2π⎜parenleftbigg tan−1Y+W/2 X−tan−1Y−W/2 X⎜parenrightbigg Using the identity in (2.69c) gives By=μ0K 2πtan−1 WX X2+Y2−(W/2)2 which is the ycomponent given in (2.70). 2.5.1 Leibnitz’s Rule: Differentiate Before You Integrate Solving for the Bﬁeld by ﬁrst obtaining the vector magnetic potential Avia (2.47) or (2.52) avoids the integration of vector quantities that occurs by a 68 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) direct solution for Busing the Biot–Savart law. However, the ﬁnal step of differentiating that to give B=∇ ×Amay involve some rather complicated differentiations. A convenient way of avoiding those complicated differentia-tions and going directly to the Bﬁeld is by using Leibnitz’s rule [12]. Leibnitz’s rule allows us to exchange the order of differentiation and integration: ∂ ∂y⎜integraldisplayb af(x, y)dx=⎜integraldisplayb a∂f(x, y) ∂ydx There are some rather mild restrictions: f(x, y)must be continuous and have continuous derivatives in a≤x≤bandy1≤y≤y2over which the result is to be obtained. For example, consider the problem of determining the Bﬁeld for a ﬁnite- length current as shown in Fig. 2.24. The vector magnetic potential was ob-tained as A z=μ0I 4π⎜integraldisplayZ+L/2 λ=Z−L/21√ λ2+r2dλ =μ0I 4πln(Z+L/2)+⎜radicalBig (Z+L/2)2+r2 (Z−L/2)+⎜radicalBig (Z−L/2)2+r2(2.57) and the Bﬁeld was obtained from B=∇ ×Aas B=−∂Az ∂raφ =μ0I 4πr⎡ ⎣Z+L/2⎜radicalBig (Z+L/2)2+r2−Z−L/2⎜radicalBig (Z−L/2)2+r2⎤ ⎦aφ(2.58) This ﬁnal step required the differentiation of a somewhat complicated natural log function. We can instead formulate this as Bφ=−∂Az ∂r =−∂ ∂r⎜bracketleftBigg μ0I 4π⎜integraldisplayZ+L/2 λ=Z−L/21√ λ2+r2dλ⎜bracketrightBigg =−μ0I 4π⎜integraldisplayZ+L/2 λ=Z−L/2∂ ∂r⎜bracketleftbigg1√ λ2+r2⎜bracketrightbigg dλ =−μ0I 4π⎜integraldisplayZ+L/2 λ=Z−L/2⎜bracketleftBigg −1 22r ⎜parenleftbigλ2+r2⎜parenrightbig3/2⎜bracketrightBigg dλ =μ0I 4πr⎜integraldisplayZ+L/2 λ=Z−L/21 ⎜parenleftbigλ2+r2⎜parenrightbig3/2dλ VECTOR MAGNETIC POTENTIAL 69 =μ0I 4πr⎜bracketleftbigg1 r2λ√ λ2+r2⎜bracketrightbiggZ+L/2 Z−L/2 =μ0I 4πr⎡ ⎣Z+L/2⎜radicalBig (Z+L/2)2+r2−Z−L/2⎜radicalBig (Z−L/2)2+r2⎤ ⎦ as obtained in (2.58) by differentiation of Azafter the integration to obtain it. We have used integral 200.03 from the table of integrals by Dwight [7]: ⎜integraldisplay1 ⎜parenleftbiga2+x2⎜parenrightbig3/2dx=x a2√ a2+x2(D200.03) Integrating Azand then obtaining Bfrom B=∇ ×Arequired the differenti- ation of a natural log function. As another example, consider the problem of a current sheet of ﬁnite width and inﬁnite length shown in Fig. 2.27. The vector magnetic potential is in thezdirection and is given by A z=−μ0K 2π⎜integraldisplayW/2 y=−W/ 2lnRd y =−μ0K 4π⎜integraldisplayW/2 y=−W/ 2ln⎜bracketleftBig (y−Y)2+X2⎜bracketrightBig dy =−μ0K 4π⎜integraldisplayW/2−Y λ=−(W/ 2+Y)ln(λ2+X2)dλ TheBﬁeld is obtained from B=∇ ×Aas B=∂Az ∂Yax−∂Az ∂Xay =μ0K 4π⎜bracketleftBigg −ln(W/2+Y)2+X2 (W/2−Y)2+X2ax+2 tan−1 WX X2+Y2−(W/2)2ay⎜bracketrightBigg (2.70) Instead of integrating to obtain Azand then differentiating to obtain B, use Leibnitz’s rule to obtain Bx=∂Az ∂Y =∂ ∂Y⎜bracketleftBigg −μ0K 4π⎜integraldisplayW/2 y=−W/ 2ln⎜bracketleftBig (y−Y)2+X2⎜bracketrightBig dy⎜bracketrightBigg =−μ0K 4π⎜integraldisplayW/2 y=−W/ 2⎜bracketleftbigg∂ ∂Yln⎜bracketleftBig (y−Y)2+X2⎜bracketrightBig⎜bracketrightbigg dy 70 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) =−μ0K 4π⎜integraldisplayW/2 y=−W/ 22(y−Y)(−1) (y−Y)2+X2dz =μ0K 2π⎜integraldisplayW/2−Y λ=−(W/ 2+Y)λ λ2+X2dλ =μ0K 2π1 2⎜bracketleftBig ln⎜parenleftBig λ2+X2⎜parenrightBig⎜bracketrightBigW/2−Y −(W/ 2+Y) =−μ0K 4πln(W/2+Y)2+X2 (W/2−Y)2+X2 as was obtained in (2.70) and we used integral 121.1 of Dwight [7]: ⎜integraldisplayx a2+x2dx=1 2ln⎜parenleftBig a2+x2⎜parenrightBig (D121.1) Similarly, the ycomponent of Bis obtained as By=−∂Az ∂X =−∂ ∂X⎜bracketleftBigg −μ0K 4π⎜integraldisplayW/2 y=−W/ 2ln⎜bracketleftBig (y−Y)2+X2⎜bracketrightBig dy⎜bracketrightBigg =μ0K 4π⎜integraldisplayW/2 y=−W/ 2⎜bracketleftbigg∂ ∂Xln⎜bracketleftBig (y−Y)2+X2⎜bracketrightBig⎜bracketrightbigg dy =μ0K 4π⎜integraldisplayW/2 y=−W/ 22X (y−Y)2+X2dy =μ0K 2πX⎜integraldisplayW/2−Y λ=−(W/ 2+Y)1 λ2+X2dλ =μ0K 2πX⎜bracketleftbigg1 Xtan−1λ X⎜bracketrightbiggW/2−Y −(W/ 2+Y) =μ0K 2π⎜bracketleftbigg tan−1W/2−Y X−tan−1−(W/2+Y) X⎜bracketrightbigg =μ0K 2πtan−1 WX X2+Y2−(W/2)2 as was obtained in (2.70) and we used integral 120.1 of Dwight [7]: ⎜integraldisplay1 a2+x2dx=1 atan−1x a(D120.1) DETERMINING THE INDUCTANCE OF A CURRENT LOOP 71 2.6 DETERMINING THE INDUCTANCE OF A CURRENT LOOP: A PRELIMINARY DISCUSSION The process of determining the inductance of a loop formed by a current- carrying conductor was discussed brieﬂy in Chapter 1. To determine the in-ductance of any loop shape we ﬁrst need to determine the Bﬁeld over the surface of the loop, s, that is bounded by the conductor. Next we must inte- grate that Bﬁeld over the loop surface with a surface integral to determine the total magnetic ﬂux through the loop surface as ψ=⎜integraldisplay sB·ds (1.6) The inductance of the loop is the ratio of this ﬂux and the current Ithat created it: L=ψ I(1.7) The circular current loop shown in Fig. 2.29 will be used to illustrate the ﬁrst part of that process: determining the Bﬁeld over the surface of the loop. The loop has a radius aand is formed by a wire of radius rw. In this section we determine the magnetic ﬂux density Bover the ﬂat surface of the loop, s, that is bounded by the interior surface of the wire by using three methods. InChapter 4, that Bﬁeld will be integrated over the loop surface via the surface Iy xa rφ s 2rwzBRφφadaIdlI= θ α FIGURE 2.29. Determining the magnetic ﬂux through the surface enclosed by a circular wire loop. 72 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) integral in (1.6) to give the total magnetic ﬂux through the loop and hence the inductance of the circular loop via (1.7). The Bﬁeld over the loop surface will, by the right-hand rule, be z-directed (out of the page), B=Bzaz, and hence the total magnetic ﬂux throught the surface via (1.6) will be obtained in Chapter 4 by further integration as ψ=⎜integraldisplaya−rw r=0⎜integraldisplay2π φ/prime=0Bzrd φ/primedr⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ds(1.6) We assume that the current is uniformly distributed over the wire cross section so that we may compute the magnetic ﬁelds from it by replacing the wire witha ﬁlament on its axis that contains the total current. Perhaps the most fundamental method for determining the magnetic ﬁeld over the surface bounded by the wire loop is by using the Biot–Savart law. Adifferential length of current produces a net magnetic ﬁeld over the surfacebounded by the wire that is in the zdirection and is therefore perpendicular to the surface of the loop. Because of symmetry, we can, without loss ofgenerality, determine the Bﬁeld in the plane of the loop, the xyplane, at a point that is located along the xaxis at a distance rfrom the center of the loop as shown in Fig. 2.29. For r<a this gives the ﬁeld inside the loop. This result will also be valid in the plane of the loop for points outside the loop, r>a . From the Biot–Savart law, the differential contribution to the B=B zazﬁeld at the point from this differential current is dBz=μ0I 4π1 R2ad φ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright dlsinθ (2.72a) where the distance Rbetween the differential current segment and the point is, using the law of cosines, R2=a2+r2−2arcosφ (2.72b) andθis the angle between the differential current vector and the vector directed from it to the point as shown in Fig. 2.29. The sine of θcan be determined in terms of the angle αasθ=90◦+αand sinθ=sin(90◦+α)=cosα Using the law of cosines again gives r2=a2+R2−2aRcosα, so that sinθ=cosα=a2+R2−r2 2aR =a−rcosφ R(2.72c) DETERMINING THE INDUCTANCE OF A CURRENT LOOP 73 The contribution to the ﬁeld at the point from the lower half of the current is the same as that from the upper half. Hence, the magnetic ﬁeld at the point isdetermined from the Biot–Savart law as B z(r)=2μ0Ia 4π⎜integraldisplayπ φ=0a−rcosφ (a2+r2−2arcosφ)3/2dφ =μ0Ia 2π⎜integraldisplayπ φ=0a−rcosφ (a2+r2−2arcosφ)3/2dφ z =0 (2.73) A second method of determining Bzis by using the vector magnetic poten- tial and performing Bz=∇ ×Aφ=(1/r)[∂(rAφ)/∂ r] (see the Appendix). The vector magnetic potential is obtained by evaluating (2.59) in the plane ofthe loop (Fig. 2.25, z=0,ρ=r,θ=0) to give A φ=μ0Ia 2π⎜integraldisplayπ φ=0cosφ⎜radicalbig a2+r2−2arcosφdφ z =0 (2.74) Hence, the magnetic ﬂux density over the loop surface is obtained from B=∇ ×A =1 r∂⎜parenleftbigrAφ⎜parenrightbig ∂raz =μ0Ia 2πr∂ ∂r⎜bracketleftBigg⎜integraldisplayπ φ=0rcosφ⎜radicalbig a2+r2−2arcosφdφ⎜bracketrightBigg az =μ0Ia 2πr⎜integraldisplayπ φ=0⎜bracketleftBigg ∂ ∂rrcosφ⎜radicalbig a2+r2−2arcosφ⎜bracketrightBigg dφaz =μ0Ia 2πr⎜integraldisplayπ φ=0acosφ(a−rcosφ) (a2+r2−2arcosφ)3/2dφ z =0 (2.75) where we have used Leibnitz’s rule to interchange the order of differentiation and integration. This result seems to be undeﬁned at the center of the loop,r=0: lim ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright r→0Bz=μ0I 2πlim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright r→0⎜integraldisplayπ φ=0a2cosφ(a−rcosφ) (a2+r2−2arcosφ)3/2dφ r =μ0I 2π⎜integraltextπ φ=0[(a3cosφ)/a3]dφ 0=0 0 74 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) However using l’H ˆopital’s rule gives a limit of lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright r→0Bz=μ0I 2πlim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright r→0∂ ∂r⎜integraldisplayπ φ=0a2cosφ(a−rcosφ) (a2+r2−2arcosφ)3/2dφ ∂ ∂r(r) =μ0I 2π⎜integraltextπ φ=0[(2 cos2φ)/a]dφ=π/a 1 =μ0I 2ar=0,z=0 which is the result derived directly by the Biot–Savart law and given in (2.25). We again used Leibnitz’s rule to interchange the order of differentiation andintegration in the numerator. The third method of obtaining the B zﬁeld in the plane of the loop is to use directly the result obtained by Smythe [10] and Weber [11] and given in(2.65). Evaluating this in the plane of the loop, z=0, gives (2.67): B z=μ0I 2π⎜parenleftbigg1 a+rK+1 a−rE⎜parenrightbigg =μ0Ia 2π⎜integraldisplayπ 0a−rcosφ⎜parenleftbiga2−r2⎜parenrightbig⎜radicalbig a2+r2−2arcosφdφ z =0 (2.67c) where K(k) andE(k) are the complete elliptic integrals of the ﬁrst and second kind, respectively [7]: K(k)=⎜integraldisplayπ/2 ζ=0dζ⎜radicalBig 1−k2sin2ζ(D773.1) E(k)=⎜integraldisplayπ/2 ζ=0⎜radicalBig 1−k2sin2ζd ζ (D774.1) and k2=4ar (a+r)2(2.67d) As a check on these results, at the center of the loop all three evaluate to Bz=μ0I 2ar=0,z=0 (2.25) which is the result obtained directly and given in (2.25). The interesting aspect of these three results for Bzis that they are all seem- ingly different! All three results have the form Bz(r)=μ0Ia 2π⎜integraldisplayπ φ=0⎜bracketleftbigIntegrand⎜bracketrightbigdφ z =0 (2.76) DETERMINING THE INDUCTANCE OF A CURRENT LOOP 75 butall three integrands are different. For example, compare the integrands of the result using the Biot–Savart law and given in (2.73): ⎜integraldisplayπ φ=0a−rcosφ (a2+r2−2arcosφ)3/2dφ=1 a2⎜integraldisplayπ φ=01−ucosφ (1+u2−2ucosφ)3/2dφ (2.77a) the result obtained by differentiating the vector magnetic potential according toB=∇ ×Aand given in (2.75): ⎜integraldisplayπ φ=0acosφ(a−rcosφ) r(a2+r2−2arcosφ)3/2dφ=1 a2⎜integraldisplayπ φ=0cosφ(1−ucosφ) u(1+u2−2ucosφ)3/2dφ (2.77b) and the result obtained by Smythe and Weber in (2.67c): ⎜integraldisplayπ 0a−rcosφ (a2−r2)⎜radicalbig a2+r2−2arcosφdφ =1 a2⎜integraldisplayπ φ=01−ucosφ (1−u2)⎜radicalbig 1+u2−2ucosφdφ (2.77c) and we have written the three integrands in terms of the ratio of the radius to the point, r, and the radius of the loop, a,a s u=r a(2.77d) Therefore, all of the results depend on the ratio of the radius to the point and the radius of the loop. For points interior to the loop, r<a andu<1, the magnetic ﬁeld should be directed out of the page, and hence the integralsshould be positive so that B z>0. For points exterior to the loop, r>a and u>1, the magnetic ﬁeld should be directed into the page and hence theintegrals should be negative, so that B z<0. This is determined using the right-hand rule. But how can these three different integrands give the same result for the Bz(r) ﬁeld over the surface enclosed by the loop? The answer is that what is important is the result of the integration, and integrands having differentcurves over the limits of the integral, 0 ≤φ≤π, are capable of enclosing the same area. Fortunately, it turns out that this is the case for the three results above: All three seemingly different integrals give the same magnetic ﬁeld B z(r), as we show next using numerical integration. Since the three integrals in (2.77) cannot be integrated in closed form, a numerical integration routine was used to perform the integration for variousvalues of the radius to the point, r, and the radius of the loop, a. Figure 2.30(a) shows the plots of the integrands for r=1 and a=2 over the range of the 76 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) Comparison of Integrands 0 < φ < π2 1.5 1 0.5 0 –0.5 0 0.5 1 1.5 2 2.5 3 3.5Value FIGURE 2.30(a). Plots of the integrands for 0 ≤φ≤π:( ) by the Biot–Savart law in (2.77a); ( ) by the vector potential method in (2.77b); and ( ) by the result from Smythe [10] and Weber [11] in (2.77c) for r=1 anda=2. All three integrals evaluate to 0.9783. integral, 0 ≤φ≤π. The curves of the three integrands are considerably dif- ferent, yet the integral evaluates to 0.9783 for all three integrands. Figure 2.30(b) shows the plots of the integrands for r=0.1 and a=2 over the range of the integral, 0 ≤φ≤π. The curve of the integrand by the vector magnetic potential method in (2.77b) is considerably different from the othertwo, yet the integral evaluates to 0.7869 for all three integrands. For r=0 all three integrals approach π/a 2=0.7854. Figure 2.30(c) shows the plots of the integrands for r=1.9 and a=2 over the range of the integral, 0 ≤φ≤π. The curves of the integrands by the Biot–Savart law in (2.77a) and by the vector potential method in (2.77b) arevirtually identical but are different from the result by Smythe and Weber in(2.77c), yet the integral evaluates to 5.6550 for all three integrands. The three integrals for B zin (2.77) are also valid for points in the plane of the loop ( z=0) which are outside the loop, r>a . Figure 2.30(d) shows the plots of the integrands for r=2 anda=1 over the range of the integral, 0≤φ≤π. The curves of the three integrands are considerably different over the range of the integral, 0 ≤φ≤π, yet the integral evaluates to −0.2709 for all three integrands. The fact that the integral is negative and Bz<0 makes sense because for points in the plane of the loop ( z=0) but lying outside the current loop, r>a , the B=Bzazﬁeld is, by the right-hand rule, in the negative zdirection (into the page). DETERMINING THE INDUCTANCE OF A CURRENT LOOP 77 Comparison of Inte grands 0 < φ < π6 4 20 –2 –4 –60 0.5 1 1.5 2 2.5 3 3.5Val ue FIGURE 2.30(b). Plots of the integrands for 0 ≤φ≤π:( ) by the Biot–Savart law in (2.77a), ( ) by the vector potential method in (2.77b), and ( ) by the result from Smythe [10] and Weber [11] in (2.77c) for r=0.1 and a=2. All three integrals evaluate to 0.7869. Comparison of Integrands 0 < φ < π120 100 8060 40 20 0 –200 0.5 1 1.5 2 2.5 3 3.5Value FIGURE 2.30(c). Plots of the integrands for 0 ≤φ≤π:( ) by the Biot–Savart law in (2.77a), ( ) by the vector potential method in (2.77b), and ( ) by the result from Smythe [10] and Weber [11] in (2.77c) for r=1.9 and a=2. All three integrals evaluate to 5.6550. 78 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) Comparison of Integrands 0 < φ < π0.4 0.2 0 –0.2 –0.4 –0.6 –0.8 –1 0 0.5 1 1.5 2 2.5 3 3.5Value FIGURE 2.30(d). Plots of the integrands for 0 ≤φ≤π:( ) by the Biot–Savart law in (2.77a), ( ) by the vector potential method in (2.77b), and ( ) by the result from Smythe [10] and Weber [11] in (2.77c) for r=2 anda=1 (points outside the loop). All three integrals evaluate to −0.2709. The curves representing the three integrands in (2.77a), (2.77b), and (2.77c) which are plotted in Fig. 2.30 are equal for only one value of φ. This equality of the three integrands occurs when in (2.77b) cosφ u=1 and when in (2.77c) 1+u2−2ucosφ 1−u2=1 Both these conditions occur when φ=cos−1uu < 1 which has values only for u≤1. For the case of r=1 and a=2 giving u=1/2 in Fig. 2.30(a), this value of φisφ=60o=1.0472 rad. For the case of r=0.1 and a=2 giving u=1 20in Fig. 2.30(b), this value of φis φ=87.134◦=1.5208 rad. For the case of r=1.9 and a=2 giving u= 0.95 in Fig. 2.30(c), this value of φisφ=18.195◦=0.3176 rad. For points outside the current loop, r>a , so that u>1, the term in the numerator of ENERGY STORED IN THE MAGNETIC FIELD 79 each integrand, (1 −ucosφ), is zero for all three integrands at φ=cos−11 uu>1 Hence for the case of r=2 anda=1 giving u=2 in Fig. 2.30(d), this value ofφwhere all three integrands are zero is φ=60◦=1.0472 rad. 2.7 ENERGY STORED IN THE MAGNETIC FIELD The electric energy stored in a region of space of volume vdue to a system of charges is [3,6] WE=1 2⎜integraldisplay vD·Edv J (2.78) If the space surrounding the charges is linear, homogeneous, and isotropic and described by a permittivity ε, then Dis related to EbyD=εEand (2.78) becomes WE=1 2ε⎜integraldisplay vE2dv J (2.79) Similarly, the magnetic energy stored in a region of space of volume vdue to a system of current loops is [3,6] WM=1 2⎜integraldisplay vB·Hdv J (2.80) If the space surrounding the currents is linear, homogeneous, and isotropic and described by a permeability μ, then Bis related to HbyB=μHand (2.80) becomes WM=1 2μ⎜integraldisplay vH2dv J (2.81) As noted in Chapter 1, these have a direct parallel with the energy stored in the ﬁelds of a capacitor (stored in its electric ﬁeld) WE=1 2CV2J (2.82) 80 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) where Vis the voltage between the capacitor plates and Cis its capacitance. Similarly, the energy stored in an inductor (stored in its magnetic ﬁeld) is WM=1 2LI2J (2.83) where Iis the current passed through the inductor and Lis its inductance. Hence, we have an alternative means of determining the capacitance or in-ductance of a structure indirectly by, instead, determining the energy storedin the electric or magnetic ﬁelds of the element: C=2W E V2(2.84a) L=2WM I2(2.84b) For some structures, (2.84b) will be a useful way of determining the inductance of the structure. 2.8 THE METHOD OF IMAGES Problems often involve a ﬂat sheet of metal that is very large in extent. Charges and/or currents exist above the sheet and it is desired to determine the electricand magnetic ﬁelds in the space above the sheet. This is a very difﬁcultproblem, because to solve it we have to determine the distribution of thecharge/currents induced on the surface of this sheet. Fortunately, with themethod of images we can replace these problems with equivalent problemsthat are much easier to solve. Although conductive metals have very largeconductivities, it is nonetheless desirable to replace them with perfect con- ductors. A perfect electric conductor is a ﬁctitious material that has an inﬁniteconductivity, σ→∞ . In the case of electric ﬁelds in the conductor, the current density in that conductor is related to the electric ﬁeld by Ohm’s law, J=σE. Forσ=∞ we must have either E=0o rJ=∞ . Having an inﬁnite current density would mean that either (1) a ﬁnite amount of charge is moved in zerotime, or (2) an inﬁnite amount of charge is moved in a ﬁnite time. Since nei-ther of these is acceptable physically, we conclude that E=0 in a perfect conductor. No charge can exist in the interior of a perfect conductor and mustexist only on its surface. Any charge in a very good conductor having a verylarge conductivity will decay to zero (move to the conductor surface) in avery short time called the relaxation time, τ=ε 0/σ[3]. Similarly, magnetic ﬁelds that vary with time cannot exist in a perfect conductor, but steady (dc)magnetic ﬁelds in a superconductor can [3]. THE METHOD OF IMAGES 81 The boundary conditions at the surface of a perfect conductor are that (1) the component of the total electric ﬁeld intensity Ethat is tangent to the surface must be zero, and (2) the component of the total magnetic ﬂux densityBnormal to the surface must be zero [3]. This means that on the surface of a perfect conductor (1) the total electric ﬁeld must be normal to it, and (2) thetotal magnetic ﬁeld must be tangent to it. The electrostatic potential function or voltage Vis deﬁned such that the neg- ative gradient of Vgives the static electric ﬁeld: E=− ∇ V(see the Appendix for the gradient function). Hence, the equipotential surfaces on which thevoltage is constant are perpendicular to the lines of the Eﬁeld and hence must be tangent to the surface of a perfect conductor. Similarly, the vectormagnetic potential Ais deﬁned such that its curl gives the magnetic ﬂux den- sity: B=∇ ×A. Since the Bﬁeld lines must be tangent to the surface of a perfect conductor with no component perpendicular to the conductor surface,the lines of Amust be tangent to the surface of a perfect conductor. Ground planes consisting of a conductor of large extent whose conduc- tance, although ﬁnite, is very large (e.g., copper) are frequently found inelectronic systems either intentionally or unintentially. The metallic frame ofan airplane fuselage acts like a ground plane to the electromagnetic ﬁeldsof the antennas that are mounted above it. Other metallic enclosures such asare used in constructing shielded rooms are intended to contain or excludeunwanted electromagnetic ﬁelds that may cause interference with sensitiveelectronic devices [5]. First consider a static charge above a perfect conductor of inﬁnite extent shown in Fig. 2.31. Although the equivalent image problem requires a perfectconductor of inﬁnite extent, in practice a reasonably good conductor of verylarge extent is usually a sufﬁcient approximation. A positive charge Qat a height habove an inﬁnite, perfectly conducting plane has, according to the boundary conditions on the electric ﬁeld at its surface, its electric ﬁeld normalto the surface of the plane [3]. If we replace the plane with a negative charge +Q +Q –Q⇔h h hE E E E E E FIGURE 2.31. Static charges above a perfect conductor and the equivalent image problem. 82 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) –Qat a depth hbelow the previous position of the plane, the electric ﬁelds above the position of the plane will be identical in either case [3]. In the case of currents, a similar imaging can be used. A foolproof way of getting the correct directions of the images of the currents is to recallthat current is the ﬂow of charge. So we can visualize a ﬁnite-length currentas having positive and negative charge being accumulated at the ends andthen image those charges as shown in Fig. 2.32. A current that is parallel tothe plane is imaged at the same depth below the plane but with the currentdirection reversed as shown in Fig. 2.32(a). A current that is perpendicular tothe plane is imaged at the same depth below the plane but with its directionthe same as the current above the plane as shown in Fig. 2.32(b). In eithercase, the magnetic ﬁeld in the space above the plane will be the same as whenthe plane is replaced by images. The total magnetic ﬁeld will be tangent to the ⇔h hI I Ih (a) I h ⇔I h Ih (b) FIGURE 2.32. Imaging currents. STEADY (DC) CURRENTS MUST FORM CLOSED LOOPS 83 plane at all points on the plane in either case, thereby satisfying the boundary conditions on the magnetic ﬁeld on the surface of a perfectly conducting plane[3]. Currents that are neither horizontal nor perpendicular to the plane can beimaged by resolving the current into its vertical and horizontal componentsand imaging those. You should show that the total magnetic ﬁeld at all points on the surface of the plane in Fig. 2.32 is tangent to the surface, and there is no componentperpendicular to the surface of the plane. Do this by replacing the plane with itsimage and then superimposing the magnetic ﬁelds due to the original currentand its image using the results for the magnetic ﬁelds of the currents (ﬁnitelength or inﬁnite length) that were derived previously. 2.9 STEADY (DC) CURRENTS MUST FORM CLOSED LOOPS Steady currents (dc currents that do not vary with time )must form closed loops (i.e., must return to their source ). This is rather simple to prove. First, we recall the law of conservation of charge: Ileaving s=⎜contintegraldisplay sJ·ds=−d dtQenclosed (2.85) The surface integral⎜contintegraltext sJ·dsgives the net current leaving the closed surface s,Jis the current density in A /m2over that surface, and Qenclosed is the net positive charge enclosed by the surface. This is illustrated in Fig. 2.33. Thismathematical statement of the law of conservation of charge is very sensiblesince it says merely that the net outﬂow of current out of a closed surface s equals the time rate of decrease of the charge enclosed by that surface . Recalling that current is the rate of ﬂow of charge, this mathematical statement closed s urface sJ JJenclosedQ FIGURE 2.33. Conservation of charge. 84 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) (a) (b)closed surface ss0enclosed = Q 0enclosed ≠ Q closed surface ssI I I FIGURE 2.34. Finite-length currents and conservation of charge. of the law of conservation of charge is elegantly obvious since it requires that if there is a netcurrent leaving the closed surface, it must be accompanied by a decrease in the net positive charge contained in that surface since charge can be neither created nor destroyed inside the closed surface s! Consider the case of a wire carrying a steady (dc) current Ias illustrated in Fig. 2.34. Surround a point along the wire with a closed surface (a nodein the vernacular of lumped-circuit theory) as shown in Fig. 2.34(a). In thecase of steady or direct (dc) currents, the right-hand side of (2.85) mustbe zero: ⎜contintegraldisplay sJ·ds=0 steady (dc) currents (2.86) According to (2.86) the dc current entering the node must equal the dc current leaving the node, and hence Kirchhoff’s current law satisﬁes conservationof charge for steady (dc) currents [1,2]. However, this also shows that ﬁnite lengths of dc currents cannot exist . This is simple to show because if we sur- round the end of a ﬁnite length of current with a closed surface as illustratedin Fig. 2.34(b), we will have current Ientering but no current leaving the closed surface, thereby violating (2.86). Hence, steady (dc) currents must form closed loops . In other words, a steady (dc) current must return to its source. STEADY (DC) CURRENTS MUST FORM CLOSED LOOPS 85 The current Jin (2.86) is conduction current, which is the ﬂow of free charge such as in a wire. When we add displacement current to Amp `ere’s law for time-varying ﬁelds (and time-varying currents) in Chapter 3 we maymake the statement that for time-varying currents, the sum of the conductionand the displacement current must form closed loops (i.e., must return to theirsource). This may be shown by taking the divergence of Amp `ere’s law for time-varying currents in point form (see Chapter 3): ∇·∇×H=∇ · ⎛ ⎜⎝J⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright conduction current⎞ ⎟⎠+∇ ·⎛ ⎜⎜⎜⎝∂D ∂t⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright displacement current⎞ ⎟⎟⎟⎠ =0 (2.87) since we have the identity ∇·∇×F=0 for any vector ﬁeld (see the Appendix). Using the divergence theorem (see the Appendix), this gives ⎜contintegraldisplay s⎜parenleftbigg J+∂D ∂t⎜parenrightbigg ·ds=0 (2.88) thereby showing that for time-varying currents, the total current must form closed loops. Where conduction current ends, displacement current takes overto complete the loop as in a circuit containing a capacitor. In several of the examples illustrating the Biot–Savart law as well as in illustrating the use of the vector magnetic potential to obtain BviaB=∇ ×A, we employed ﬁnite lengths of dc current . But if these cannot exist, of what use are those results for the magnetic ﬁelds of a current of ﬁnite length? Theanswer is that we use solutions for the ﬁelds of ﬁnite-length current segments II IBnet FIGURE 2.35. Combining magnetic ﬁelds of ﬁnite-length currents to determine the magnetic ﬁeld of a closed loop of current. 86 MAGNETIC FIELDS OF DC CURRENTS (STEADY FLOW OF CHARGE) to construct the solutions for the ﬁelds of a closed current loop of which these current elements are a part, as illustrated in Fig. 2.35. Clearly, we areusing superposition and the surrounding medium must be linear, at least with regard to its magnetic ﬁeld properties. So determining the magnetic ﬁeldsfor steady (dc) currents of ﬁnite length is useful in that regard and for thatpurpose. 3 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) In Chapter 2 we investigated the calculation of the magnetic ﬁelds produced by various conﬁgurations of static (dc) currents (the steady ﬂow of charge). We discussed in Chapter 1 how the inductance of a structure will be obtained fromthese static magnetic ﬁelds by ﬁrst obtaining the magnetic ﬂux penetratingthe surface of the current loop from (1.6): ψ= ⎜integraldisplay sB·ds (1.6) and then obtaining the inductance of the loop from (1.7): L=ψ I(1.7) This inductance parameter will then be used in lumped circuits to determine its effect in circuits in which the currents vary with time ( accelerated charge). The electromagnetics law that allows this determination is Faraday’s law ofinduction. However, we seem to have a logical inconsistency in this process:A circuit element, inductance, that was derived for static (dc) currents willbe used to evaluate its effect on time-varying currents. The ability to usea result derived for dc currents in a situation where the currents vary withtime is shown in Section 3.4 to be a valid approximation using an iterativesolution of the ﬁeld equations. This approximation will be valid for circuits Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 87 88 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) whose maximum dimensions are “electrically small,” that is, much less than a wavelength at the frequency of the driving source. The notion of electricallysmall dimensions is discussed in Section 3.3. 3.1 FARADAY’S FUNDAMENTAL LA W OF INDUCTION Faraday’s law is perhaps the most profound of the collective group of laws governing all macroscopic electromagnetic ﬁelds that are known as Maxwell’sequations. Without Faraday’s law we would not have the use of “electricity”and all its myriad implications. To state Faraday’s law in unambiguous mathematical terms, consider Fig. 3.1, which shows an open surface sthat has a contour or path csurround- ing it. With reference to Fig. 3.1, Faraday’s law can be stated in mathematicalform as [3–6] emf=−dψ dt(3.1) where the electromotive force emf around the closed loop c is obtained with aline integral as emf=⎜contintegraldisplay cE·dl (3.2) and the magnetic ﬂux that passes through the open surface s is obtained with asurface integral as ψ=⎜integraldisplay sB·ds (3.3) Hence, Faraday’s fundamental law of induction is ⎜contintegraldisplay cE·dl=−d dt⎜integraldisplay sB·ds (3.4) dsE BB B Bdl c sna FIGURE 3.1. Faraday’s law. FARADAY’S FUNDAMENTAL LAW OF INDUCTION 89 The contour cof the closed loop can be thought of as either a conducting material (as in the case of a wire) or an imaginary contour of nonconductingmaterial (as in the case of free space) and Eis the electric ﬁeld intensity vector with units of V/m along that contour. The dot product in the integrand of the emf in (3.2), E·dl, means that we take the product of the differential lengths of this contour dland the electric ﬁeld lines that are tangent to the contour. We then sum these products (with an integral) to obtain the emf around thatclosed path. Ehas a component parallel or tangent to this path and a component perpendicular to this path, and the components that are perpendicular to thispath do not contribute to the sum. Observe that the electromotive force in (3.2)has units of volts and acts like a voltage. However, the minus sign that waspresent in the deﬁnition of voltage due to a charge distribution in Chapter 1is absent here, so that instead of being a voltage produced by charge , the emf represents a form of voltage source inserted in the loop . If the electrical dimensions (in wavelengths) of the closed loop are electrically small ( /lessmuchλ), we may treat this emf as a lumped voltage source and place it anywhere in the loop. The right-hand side of Faraday’s law in (3.1) is the rate of decrease (the negative sign is referred to as Lenz’s law ) of the magnetic ﬂux ψgiven in (3.3) that passes through the surface sthat the closed loop cencloses, and Bis the magnetic ﬂux density vector with units of Wb/m 2(tesla). The result of the surface integral in (3.3), ψ, gives the net magnetic ﬂux passing through the surface that is enclosed by the contour c . The units of that ﬂux are webers. A vector differential surface of that surface is ds=dsan, where anis the unit normal to the surface. The dot product B·dsin the integrand of (3.3) means that we take the product of the differential surface areas dsand the components of Bthat are perpendicular to the surface. Then we add (with an integral) these products to give the net magnetic ﬂux ψleaving (or passing through) the open surface s. This is again sensible since Bhas two compo- nents: one perpendicular to the surface and one that is tangent to the surface.The component of Bthat is tangent to the surface does not (and should not) contribute to the net ﬂux passing through the surface. So we may interpret Faraday’s law as providing that: A time-varying magnetic ﬁeld passing through an open surface s willinduce (produce) an electric ﬁeld around the contour c that encircles the surface. In Section 3.4 we provide the rationale for saying that the magnetic ﬁeld pro-duces an electric ﬁeld rather than the reverse, although this is still somewhat 90 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) arbitrary. This is the process behind some particle accelerators that accelerate charged particles to enormous speeds and smash them into other particlesin order to break those particles into their constituent pieces. A large, time-varying magnetic ﬁeld creates an electric ﬁeld that exerts a force on electriccharge. The path here into which the electric ﬁeld is induced is an imaginarycontour in space. Faraday’s law also makes possible electric transformers andelectric motors and generators among an enormous number of other appli-cations that are absolutely essential to our daily lives and commerce. In anelectric generator, coils of wire rotate around a shaft and pass through a dcmagnetic ﬁeld, thereby causing a time-varying magnetic ﬁeld to penetrate thesurfaces enclosed by those coils of wire. Hence voltages are induced in thosecoils of wire by Faraday’s law, thereby producing electricity. The contour or path cin the general statement of Faraday’s law can be thought of as the mouth of a balloon which can be inﬂated to give differentsurfaces s, as illustrated in Fig. 3.1. All these surfaces give the same result as long as the contour cremains the same. Magnetic ﬁeld lines that enter and leave the surface and do not pass through the mouth of the balloon do notcontribute to the net ﬂux through the surface and hence do not contribute tothe induced electric ﬁeld. Only those magnetic ﬁeld lines that pass throughthe mouth of the balloon contribute to the net ﬂux exiting the balloon surface.The direction of the contour cand the direction “out of” the open surface s are again related by the right-hand rule. Placing the ﬁngers of our right hand in the direction of the contour c, our thumb will point in the direction “out of the open surface.” To simplify the discussion we choose a ﬂat surface and a circular contour enclosing that surface as shown in Fig. 3.2. Again, the components of themagnetic ﬂux density that penetrate or pass through this surface are those thatare normal (perpendicular) to the surface, B·dsandds=dsa n, where anis a unit normal to the surface. Again, this is sensible because the components ofBthat are tangent (parallel) to the surface do not “exit” the surface. Faraday’s law provides that we may replace the effect of the magnetic ﬂux density vector passing through the surface by inserting an equivalent voltage source whosevalue is V=dψ dt(3.5) into the contour of the loop that encloses the surface. To “lump” this induced emf in the loop, we will assume that the physical dimensions of this loopare electrically small ( /lessmuchλ). Furthermore, we consider the loop contour to be constructed of a conducting material such as a wire (a conductor hav-ing a circular, cylindrical cross section). We can lump these effects of the FARADAY’S FUNDAMENTAL LAW OF INDUCTION 91 dsc dtdψV=E dl sB B indBindI FIGURE 3.2. Modeling the effect of the Bﬁeld as an induced voltage source. time-changing magnetic ﬁeld through the loop into a lumped voltage source whose value is given in (3.5) and place it anywhere in the loop contour because we assume that the loop dimensions are electrically small. Getting the polarity of this induced source correct is critical. Faraday’s law essentially provides that the voltage source representing the induced emf hasa polarity such that it opposes (Lenz’s law) the rate of change of the magnetic ﬂux through the loop. A foolproof way of determining the correct polarity of the source is the following. The source should tend to induce or “push” acurrent I indaround this conducting loop in a direction such that this induced current produces another induced magnetic ﬂux Bindthatopposes any change in the original magnetic ﬁeld B. This is a very sensible result because if the magnetic ﬁeld induced by the source did not oppose the original magnetic 92 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) ﬁeld, an induced current would produce an induced magnetic ﬂux that would increase the net magnetic ﬂux through the loop, thereby increasing the valueof the induced voltage, which produces a larger induced magnetic ﬁeld, andso on without bound. As we found in Chapter 2, a current in a wire produces amagnetic ﬁeld whose direction can be obtained with the right-hand rule. That is, if we place the thumb of our right hand in the direction of the current, theﬁngers will give the direction of the induced magnetic ﬁeld about the wire. Ifthe original magnetic ﬂux through the surface enclosed by the loop is directedupward as shown in Fig. 3.2, the source should have a polarity such that ittends to push a current out of its positive terminal that circulates clockwise, thereby producing (by the right-hand rule) an induced magnetic ﬁeld that isdirected downward through the loop surface such that this induced magnetic ﬁeld opposes the original magnetic ﬁeld. Observe that the value of the inducedvoltage source Vin (3.5) depends on the time rate of change of the magnetic ﬂux. Hence, either a large Bﬁeld that is slowly varying with time (such as a 60-Hz power frequency current) or a small Bﬁeld that is rapidly varying with time (such as a 2-GHz current in a cell phone) will have a similar effect. EXAMPLE This example shows the utility of using an induced voltage source to model the effect of an incident magnetic ﬁeld through a closed loop and also showsthat the positions of the measurement leads to a voltmeter affect the reading ofthat voltmeter. Figure 3.3(a) shows a circuit where two resistors comprise thecircuit and a uniform external magnetic ﬁeld of B=5t 2Wb/m2directed out of the page threads the loop that the circuit encloses. A high-impedance voltmeterthat draws negligible current is attached across a resistor. The 2 m ×3m circuit loop encloses a total magnetic ﬂux of ψ=⎜integraldisplay sB·ds =5t2⎜parenleftBig Wb/m2⎜parenrightBig ×6⎜parenleftBig m2⎜parenrightBig =30t2(Wb) Hence, the magnitude of the voltage source induced in the loop is V=dψ dt =60t V FARADAY’S FUNDAMENTAL LAW OF INDUCTION 93 V200 Ω 100 Ω 2 m 3 m (a) 100 Ω 200 ΩV(b)200 Ω 100 ΩV0 I60tV (c) 60tV (d)200 Ω 100 ΩI60tV V0005t2 Wb/m2 FIGURE 3.3. Example showing that the position of the voltmeter leads affects its reading. 94 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) The source representing this induced emf is inserted as shown in Fig. 3.3(b). The source has the polarity shown in order to enforce Lenz’s law (it tends toproduce a current that circulates around the loop in the clockwise direction soas to produce, according to the right-hand rule, an induced Bﬁeld that tends to oppose the change in the original Bﬁeld). From that circuit we calculate a current ﬂowing around the loop in the clockwise direction of I=60t 100+200 =0.2tA Hence, the measured voltage is V=200I =40tV In Fig. 3.3(c) the voltmeter is attached to the same two points, but the voltmeter leads are routed differently. The equivalent circuit for Fig. 3.3(c) is shown inFig. 3.3(d). Observe that the voltmeter leads now also enclose the magneticﬂux, and another voltage source must be inserted in the loop formed by thosevoltmeter leads as shown. Since the impedance of the voltmeter is assumedinﬁnite, it draws neglible current and we again obtain I=60t 100+200 =0.2tA But its measured voltage is now V=200I−60t =−20tV This can also be obtained by summing KVL around the inner loop of that circuit to again obtain V=−60t+60t−100I =−20tV Hence, Faraday’s law shows that the orientation of the voltmeter leads can inﬂuence its reading signiﬁcantly. FARADAY’S FUNDAMENTAL LAW OF INDUCTION 95 EXAMPLE Consider Fig. 3.4(a), where an open-circuit loop is situated near a two-wire transmission line bearing equal but oppositely directed time-varying currentsI(t). It is assumed that the time variation of the currents is sufﬁciently slow that the loop is electrically small at the signiﬁcant spectral frequencies ofthe currents. Each current produces a component of the Bﬁeld threading the loop as shown in Fig. 3.4(b). Assuming that the currents are very longwith respect to the loop dimensions, the fundamental result in (2.14) for themagnetic ﬁeld of an inﬁnitely long current can be applied in an approximatefashion: B φ=μ0I(t) 2πr I(t) wl d s (a)s B(t)Voc=−dtdψ s d wBφ (b)I(t) I Idtdψ FIGURE 3.4. Example. 96 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) Hence, the net magnetic ﬂux threading the loop [into the page in Fig. 3.4(a)] due to both currents is (use the right-hand rule) ψ=−⎜integraldisplayl z=0⎜integraldisplays+d+w r=s+dμ0I(t) 2πrdr dz+⎜integraldisplayl z=0⎜integraldisplayd+w r=dμ0I(t) 2πrdr dz =μ0lI(t) 2π⎜parenleftbigg −lns+d+w s+d+lnd+w d⎜parenrightbigg =μ0l 2πln(d+w)(s+d) d(s+d+w)I(t) Hence, the induced voltage source in the loop has the magnitude dψ/dt and the polarity shown. Thus, the open-circuit voltage at the loop terminals withpolarity shown is V oc(t)=−dψ dt =−μ0l 2πln(d+w)(s+d) d(s+d+w)dI(t) dt EXAMPLE This ﬁnal example of Faraday’s law illustrates that since the magnitude of the induced voltage source is the time rate of change of the ﬂux through the loopand the total ﬂux through the loop is essentially the product of the Bﬁeld and the area of the loop, the induced voltage can also be produced by a constant B ﬁeld but a time-changing loop area. Consider a set of conducting rails acrosswhich a conducting shorting bar moves to the right with velocity vas shown in Fig. 3.5. The magnetic ﬁeld threading the loop is constant (independent oftime) and is uniformly distributed over the loop area. The horizontal width ofthe loop area is vt, so that the total area of the loop is area=lvt The induced voltage source has the polarity shown and a magnitude of dψ dt=Bdarea dt =Blv Hence, the open-circuit voltage is Voc=Blv FARADAY’S FUNDAMENTAL LAW OF INDUCTION 97 w = υtls Bdtdψ Voc=dtdψ υ FIGURE 3.5. Example showing Faraday’s law for a moving contour. The form of Faraday’s law in (3.4) is said to be its integral form. This form is useful for describing its meaning. The point form is useful for performing numerical solutions. It is obtained by applying Stokes’s theorem (see theAppendix) to the left-hand side to give ⎜contintegraldisplay cE·dl=⎜integraldisplay s(∇×E)·ds =−d dt⎜integraldisplay sB·ds Comparing both sides gives the point form of Faraday’s law: ∇×E=−∂B ∂t(3.6) where ∇×Egives the curl or circulation of Eat a point. Applying the general result (see the Appendix) that ∇·∇×F=0 for any general vector ﬁeld F to (3.6), we obtain Gauss’s law for the magnetic ﬁeld: ∇·B=0. Expanding the curl (see the Appendix) in a rectangular coordinate system and comparingboth sides gives ∂E z ∂y−∂Ey ∂z=−∂Bx ∂t(3.7a) ∂Ex ∂z−∂Ez ∂x=−∂By ∂t(3.7b) ∂Ey ∂x−∂Ex ∂y=−∂Bz ∂t(3.7c) 98 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) EXAMPLE A very common form of wave propagation is the uniform plane wave [3–6]. If the Eﬁeld is given by E=Emcos(ωt−βz)ax determine the corresponding Bﬁeld such that the ﬁelds satisfy Faraday’s law. Since the Eﬁeld is directed solely in the xdirection, Ey=Ez=0. Furthermore, the Eﬁeld is independent of xandy, so that ∂/∂x=∂/∂y=0. Hence, (3.7) becomes simply ∂Ex ∂z=−∂By ∂t(3.7b) Substituting the form of Egives βEmsin(ωt−βz)=−∂By ∂t Integrating this gives the Bﬁeld as B=β ωEmcos(ωt−βz)ay TheBﬁeld is in the ydirection orthogonal to the Eﬁeld. 3.2 AMP `ERE’S LA W AND DISPLACEMENT CURRENT We studied Amp `ere’s law for static (dc) ﬁelds in Chapter 2: ⎜contintegraldisplay cH·dl=⎜integraldisplay sJ·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Ienclosed AMP `ERE’S LAW AND DISPLACEMENT CURRENT 99 dsdl c snaH D Jdsna JD FIGURE 3.6. Amp `ere’s law for time-varying ﬁelds. The surface current density Jhas units of A/m2and represents current due to free charges such as electrons in a wire. For time-varying ﬁelds a term mustbe added: ⎜contintegraldisplay cH·dl=⎜integraldisplay sJ·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright conduction current+d dt⎜integraldisplay sD·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright displacement current(3.8) The vector Dis the electric ﬂux density vector with units of C/m2. The open surface sis enclosed by the closed contour cas for Faraday’s law and the directions are related by the right-hand rule. This is illustrated in Fig. 3.6. The displacement current is essentially a time-varying electric ﬁeld. For static ﬁelds (3.8) reduces to the static ﬁeld version of Amp `ere’s law. This addition of the displacement current term to the static version of Amp `ere’s law allows current to ﬂow between two plates of a capacitor, thereby completingthe current loop as shown in Fig. 3.7. Amp `ere’s law for time-varying ﬁelds in cconductionIdisplacementID D Vsin ω ts FIGURE 3.7. Displacement current ﬂows between the two plates of a capacitor. 100 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) (3.8) shows that a time-varying electric ﬁeld and its associated displacement current act exactly like conduction current, and either one can produce amagnetic ﬁeld. EXAMPLE For the capacitor circuit of Fig. 3.7, a 1- μF capacitor has a sinusoidal voltage source 10 sin ωtvolts attached across its terminals, and the frequency of the source is 1 kHz. The conduction current is Iconduction =10 V 1/ωC =62.8m A The capacitance of a parallel-plate capacitor (neglecting fringing of the ﬁelds at the edges) is C=ε(A/d ), where εis the permittivity of the dielectric between the plates, Ais the plate area, and dis the separation of the plates. The electric ﬁeld between the plates is (neglecting fringing of the ﬁelds at theedges) E=10 V/d. Hence, the Dﬁeld is D=εE =C A(10 V) =10−5 A and the displacement current is Idisplacement =d dt⎜integraldisplay sD·ds =ω10−5 AA =62.8m A Hence, the conduction and displacement currents are equal, as they must be. The form of Amp `ere’s law in (3.8) is said to be its integral form. Again, this form is useful for describing its meaning. The point form is useful for performing numerical solutions. It is obtained by applying Stokes’s theorem AMP `ERE’S LAW AND DISPLACEMENT CURRENT 101 (see the Appendix) to the left-hand side to give ⎜contintegraldisplay cH·dl=⎜integraldisplay s(∇×H)·ds =⎜integraldisplay sJ·ds+d dt⎜integraldisplay sD·ds Comparing both sides gives the point form of Amp `ere’s law: ∇×H=J+∂D ∂t(3.9) Applying the general result (see the Appendix) that ∇·∇×F=0 for any general vector ﬁeld Fto (3.9), we obtain ∇·J=−∂(∇·D)/∂ t =−∂ρ(t)/∂ t by substituting Gauss’s law so that (3.9) satisﬁes the law of conservation ofcharge. Expanding the curl in a rectangular coordinate system and comparingboth sides gives ∂H z ∂y−∂Hy ∂z=Jx+∂Dx ∂t(3.10a) ∂Hx ∂z−∂Hz ∂x=Jy+∂Dy ∂t(3.10b) ∂Hy ∂x−∂Hx ∂y=Jz+∂Dz ∂t(3.10c) EXAMPLE Again a very common form of wave propagation is the uniform plane wave [3–6]. If the Hﬁeld is given by H=Hmcos(ωt−βz)ay determine the corresponding Eﬁeld such that the ﬁelds satisfy Amp `ere’s law. We assume that the ﬁelds are in free space so that there is no conductioncurrent, J=0. Since the Hﬁeld is directed solely in the ydirection, H x= Hz=0. Furthermore, the Hﬁeld is independent of xandyso that ∂/∂x= ∂/∂y=0. Hence, (3.10) becomes simply −∂Hy ∂z=∂Dx ∂t(3.10a) Substituting the form of Hgives −βH msin(ωt−βz)=∂Dx ∂t 102 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) Integrating this gives the Dﬁeld as D=β ωHmcos(ωt−βz)ax Substituting D=ε0Egives the electric ﬁeld: E=β ε0ωHmcos(ωt−βz)ax Again, the Eﬁeld is in the xdirection orthogonal to the Hﬁeld. 3.3 WA VES, WA VELENGTH, TIME DELAY, AND ELECTRICAL DIMENSIONS We routinely model electronic circuits with a lumped-circuit model which is a particular interconnection of the lumped-circuit elements of resistance,capacitance, and inductance [1,2]. We then solve these lumped-circuit modelsfor the resulting voltages and currents of those elements using Kirchhoff’svoltage and current laws. These lumped-circuit models and the voltages and currents obtained from them are only valid as long as the largest physicaldimension of the circuit is electrically small (i.e., much less than a wavelengthat the frequency of excitation, f, of that circuit) [3–6]. A wavelength is λ=v f(3.11) where vdenotes the velocity of propagation of, for example, the currents along the connection leads attached to the elements. If the surrounding mediumis free space (for all practical purposes air), the velocity of propagation isapproximately v=3×10 8m/s. If a sinusoidal source excites the circuit and has a frequency of 300 MHz, a wavelength is 1 m, and if the excitationfrequency of the source is 3 GHz, a wavelength is 10 cm or approximately4 in. In the case of a printed circuit board the velocities of propagation of thesignals carried by the lands on the board are about 60% of that of free space,due to the interaction of the electromagnetic ﬁelds produced by those signalswith the board substrate, and hence the wavelengths are smaller than in air. In lumped circuits we can ignore the effects of the connection leads attached to the lumped elements because for the model to be valid, their physicallengths must be electrically small (i.e., /lessmuchλ). If the connection leads that are attached to an element are electrically long, currents at the two endpoints ofthis leads will not be the same but will have a phase difference between them, WA VES, WA VELENGTH, TIME DELAY , AND ELECTRICAL DIMENSIONS 103 lumped elementL (t)1i (t)2iconnection leadconnection lead (t)1i (t)2i t tυL FIGURE 3.8. Effect of element interconnection leads. as illustrated in Fig. 3.8. The current along the connection lead is, in fact, a wave. Suppose that the current and the associated wave are sinusoidal. Sucha wave can be written as a function of time, t, and position along the lead, z, as [3–6] i(z, t )=Icos(ωt −βz) (3.12) where βis the phase-shift constant in rad/m, and ω=2πf, where fis the cyclic frequency of the wave. The velocity of propagation of the wave can befound by observing that to track the movement of the wave, we must followa point on the wave. Hence, the argument of the cosine in (3.12) must be aconstant: ωt−βz=C. Differentiating this gives the velocity of propagation of the wave: v=ω β(3.13) Substituting (3.13) into (3.12) gives i(z, t )=Icos⎜parenleftbigg ω⎜parenleftbigg t−z v⎜parenrightbigg⎜parenrightbigg (3.14) Therefore, the phase shift in the frequency domain translates to a time delay ofz/vseconds in the time domain. Hence, the currents at two ends of the leads have a time delay between them of TD=L v(3.15) where L is the total length of the connection leads. 104 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) i(0, z) zλ 2λ λ (a) , z)i(t1 zυt1 (b) FIGURE 3.9. Wave propagation and wavelength. Awavelength λis the distance the wave must travel to shift phase by 2π radians or 360◦: βλ=2π as illustrated in Fig. 3.9. Substituting this into (3.13) again gives the wave- length in terms of the velocity of propagation and the frequency as λ=v f If the total length of the connection leads is one-half wavelength, these cur- rents at the endpoints of the lead will be 180◦out of phase with each other. If the length of the connection leads is only λ/100, the phase difference be- tween the two currents at the endpoints is an inconsequential 3 .6◦and can be ignored. This phase difference translates in the time domain to a time delay;the current at one end of the connection lead and the current at the other endwill have a time delay between them. For a connection lead of length L thistime delay (in seconds) can be written as T D=L/v=(L/λ)(1/f )=(L/λ)P , where P=1/fis the period of the sinusoidal waveforms. Hence, the sinu- soidal waveforms at the two ends of the connection lead will be shifted in timerelative to each other by a fraction of their period, L /λ. If the connection leads are electrically short, L /lessmuchλ, the two waveforms will be almost coincident in time and the time delay can be ignored. Otherwise, the time delay will besigniﬁcant. HOW RESULTS DERIVED USING STATIC (DC) VOLTAGES 105 3.4 HOW CAN RESULTS DERIVED USING STATIC (DC) VOLTAGES AND CURRENTS BE USED IN PROBLEMS WHERETHE VOLTAGES AND CURRENTS ARE V ARYING WITH TIME? At the beginning of this chapter we alluded to the apparent contradiction that we compute the lumped elements of capacitance and inductance using static(dc) voltages and currents, yet we use these elements to investigate the effectsof time-varying voltages and currents. How is this possible? The answer is,of course, as an approximation. In this section we look more closely at thisapproximation. Maxwell’s equations are commonly considered to be the collection of ﬁve equations: Faraday’s law, Amp `ere’s law, the two laws of Gauss, and the law of conservation of charge: ⎜contintegraldisplay cE·dl=−d dt⎜integraldisplay sB·ds (3.16a) ⎜contintegraldisplay cH·dl=⎜integraldisplay sJ·ds+d dt⎜integraldisplay sD·ds (3.16b) ⎜contintegraldisplay sD·ds=⎜integraldisplay vρvdv (3.16c) ⎜contintegraldisplay sB·ds=0 (3.16d) ⎜contintegraldisplay sJ·ds=−d dt⎜integraldisplay vρvdv (3.16e) where ρvis the volume (free) charge distribution throughout volume v. The point forms of these laws were obtained from the integral forms as ∇×E=−∂B ∂t(3.17a) ∇×H=J+∂D ∂t(3.17b) ∇·D=ρv (3.17c) ∇·B=0 (3.17d) ∇·J=−∂ρv ∂t(3.17e) 106 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) We solve these in an approximate manner by an iterative process [13]. First disregard all time derivatives giving the zero-order solutions: ∇×E0=0 (3.18a) ∇×H0=J0 (3.18b) ∇·D0=ρv0 (3.18c) ∇·B0=0 (3.18d) ∇·J0=0 (3.18e) Next, we put back the time derivatives but use the zero-order solutions in those time derivatives to obtain these ﬁrst-order solutions for other variables not contained in time derivatives: ∇×E1=−∂B0 ∂t(3.19a) ∇×H1=J1+∂D0 ∂t(3.19b) ∇·D1=ρv1 (3.19c) ∇·B1=0 (3.19d) ∇·J1=−∂ρv0 ∂t(3.19e) Similarly, we can obtain a more reﬁned solution known as the second-order solution by using the ﬁrst-order solutions in the time derivatives to obtain theother variables not contained in the time derivatives: ∇×E 2=−∂B1 ∂t(3.20a) ∇×H2=J2+∂D1 ∂t(3.20b) ∇·D2=ρv2 (3.20c) ∇·B2=0 (3.20d) ∇·J2=−∂ρv1 ∂t(3.20e) The zero-order solutions in (3.18) are the static (dc) solutions we obtained in Chapter 2. The zero-order solutions in (3.18) were used to obtain, for example,the solution for the ﬁrst-order, induced electric ﬁeld, E 1, in Faraday’s law in (3.19a) by using the zero-order solution for the Bﬁeld, B0. As we continue this process, we obtain a more accurate solution of Maxwell’s equations. Thecombination of the zero-order solutions in (3.18) and the ﬁrst-order solutions VECTOR MAGNETIC POTENTIAL FOR TIME-V ARYING CURRENTS 107 in (3.19) are usually referred to as the quasistatic solution. Generally speaking, the quasistatic solution obtained iteratively using the zero-order solutions andreﬁning them to give the ﬁrst-order solutions give adequate accuracy as longas the maximum physical dimension of the electromagnetic structure beinginvestigated is electrically small (i.e., L /lessmuchλ) [13]. This gives the rationale for using circuit elements such as capacitance and inductance which were derivedusing dc voltages and dc currents in circuits whose currents and voltages varywith time so long as the maximum dimension of the circuit is electricallysmall. It is simple to show that the sums of the partial solutions in this iterative process converge to the true solution to Maxwell’s equations: E=E 0+E1+E2+··· (3.21a) H=H0+H1+H2+··· (3.21b) D=D0+D1+D2+··· (3.21c) B=B0+B1+B2+··· (3.21d) J=J0+J1+J2+··· (3.21e) ρv=ρv0+ρv1+ρv2+··· (3.21f) Adding the zero-order equations in (3.18a), the ﬁrst-order equations in (3.19a), the second-order equations in (3.20a), and so on, gives ∇×(E0+E1+E2+··· )=0−∂ ∂t(B0+B1+B2+··· ) (3.22) Substituting (3.21a) and (3.21d) gives the ﬁrst Maxwell equation: ∇×E=−∂B ∂t The other equations of Maxwell are obtained in a similar fashion. 3.5 VECTOR MAGNETIC POTENTIAL FOR TIME-V ARYING CURRENTS In Chapter 2 we introduced the vector magnetic potential Afor determining the magnetic ﬁeld Bfor static (dc) current conﬁgurations. In this section we rederive the vector magnetic potential for time-varying currents. The main pur-pose in doing so is that the result will clearly demonstrate that the quasistaticsolutions of the ﬁeld equations, (3.18) and (3.19), are valid approximationsas long as the maximum physical dimensions of the problem are much lessthan a wavelength (L /lessmuchλ). 108 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) Because of Gauss’s law for the magnetic ﬁeld, ∇·B=0 (3.23) and the vector identity (see the Appendix) that the divergence of the curl of anyvector ﬁeld is zero; ∇·∇×A=0 (3.24) we can deﬁne the vector magnetic potential Aas B=∇ ×A (3.25) For static current distributions, Awas obtained in Chapter 2 as A=μ 4π⎜integraldisplay vJ Rdv (3.26) where Jis the current distribution (A /m2),vis the volume enclosing that current distribution, and Ris the distance between a differential volume of that current distribution containing Jdvand the point at which we wish to determine A. For time-varying currents this result obviously must be modiﬁed. To demonstrate that result, ﬁrst note that substituting (3.25) into Faraday’s lawgives ∇×⎜parenleftbigg E+∂A ∂t⎜parenrightbigg =0 This seems to imply that the sum in parentheses is zero. But we have the identity (see the Appendix) that ∇×∇φ=0 (3.27) for any scalar ﬁeld φ. Hence, we can write, in general, E=− ∇ φ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright due to charges−∂A ∂t⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright due to time−varying currents(3.28) Hence, in general, the electric ﬁeld is the result of two “sources”: the charges in the system and the time-varying currents in the system. For dc, this reducestoE=− ∇ φandφis said to be the potential function that is more commonly known as “voltage.” With these results we can now derive the result for the vector magnetic potential for time-varying currents. Proceeding in a fashion similar to that VECTOR MAGNETIC POTENTIAL FOR TIME-V ARYING CURRENTS 109 of Section 2.5 of for static ﬁelds, we substitute B=μHandD=εEinto Amp `ere’s law to yield ∇×B=μJ+με∂E ∂t Substituting the relation for Bin terms of Agiven in (3.25) yields ∇×∇×A=μJ+με∂E ∂t(3.29) Substituting the relation for Egiven in (3.28) gives ∇×∇×A=μJ+με⎜parenleftBigg −∇⎜parenleftbigg∂φ ∂t⎜parenrightbigg −∂2A ∂2t⎜parenrightBigg (3.30) But we have the vector identity [3,6] ∇×∇×A=∇ (∇·A)−∇2A (3.31) Substituting (3.31) into (3.30) and collecting terms gives ∇2A−με∂2A ∂2t=−μJ+∇⎜parenleftbigg ∇·A+με∂φ ∂t⎜parenrightbigg (3.32) Again, the complete deﬁnition of a vector quantity requires that we deﬁne both the curl and the divergence of it. We deﬁned the curl of Ain (3.25). We are free to deﬁne the divergence of A. From (3.32) a convenient way to deﬁne the divergence of Ais so that the term in (3.32) in parentheses is rendered zero: ∇·A=−με∂φ ∂t(3.33) This is commonly referred to as the Lorentz choice of gauge . Note that for static currents, this reduces to ∇·A=0, which was chosen in Chapter 2 for static (dc) currents. Hence the equation for the vector magnetic potential fortime-varying currents becomes ∇ 2A−με∂2A ∂2t=−μJ (3.34) It can be shown [3,6] that the solution to (3.34) is A=μ 4π⎜integraldisplay vJ⎜parenleftBig t−R v⎜parenrightBig Rdv (3.35a) where vis a velocity of propagation: v=1√με(3.35b) 110 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) and again, Jis the current distribution, vis the volume enclosing that current distribution, and Ris the distance between a differential volume of that current distribution containing Jdvand the point at which we wish to determine A. This shows that the vector magnetic potential at a point that is a distance R away from a current element Jdvhas a time delay of effect of R/v. This is refered to as retardation and is characteristic of all time-varying ﬁelds. If we write this result for ﬁelds and currents that are varying sinusoidally with time,the result in (3.35a) becomes the phasor vector magnetic potential ˆA[3]: ˆA=μ 4π⎜integraldisplay vˆJe−jβR Rdv (3.36a) in terms of the phasor current density ˆJwhere the phase constant βis again β=ω v =2π λ(3.36b) and the wavelength is again λ=v f =1√με f(3.36c) Again, the phase shift terme−jβRin the frequency domain amounts to a time delay in the time domain. We can expand the exponential term in (3.36a) as e−jβR=1−jβR+β2R2 2+··· (3.37a) It is this result that shows why quasistatic results can be used to approximate time-varying ﬁelds. Substituting (3.36b) for βgives e−jβR=1−j2πR λ+(2π)2 2⎜parenleftbiggR λ⎜parenrightbigg2 +··· (3.37b) Hence, the retardation term depends on powers of R/λ, which gives the physi- cal distance to the ﬁeld point in terms of its electrical distance in wavelengths.For electrically small dimensions of the problem, R/lessmuchλ, the exponential term approximates to unity, e −jβR∼=1, and the vector magnetic potential for time-varying currents in (3.36a) reduces to the static ﬁeld result for Athat was used in Chapter 2 and is given in (2.52). Also observe that in Chapter 2we chose the Coulomb gauge to deﬁne the divergence of Afor static ﬁelds: CONSERV ATION OF ENERGY AND POYNTING’S THEOREM 111 ∇·A=0. The more general Lorentz gauge for time-varying ﬁeld problems in (3.33) reduces to the Coulomb gauge for static problems. In the case of sinusoidal variation of the ﬁelds, the phasor form of Faraday’s law and Amp `ere’s law are obtained by replacing all time derivatives with jω and become [3–6] ∇×ˆE=−jωˆB (3.38a) ∇×ˆB=μˆJ+jωμε ˆE (3.38b) and we have substituted ˆB=μˆHandˆD=εˆEinto Amp `ere’s law in (3.38b). Once the phasor vector magnetic potential ˆAis obtained from (3.36a), the phasor magnetic ﬁeld is determined from ˆB=∇ ×ˆA (3.39a) The phasor electric ﬁeld is determined from Amp `ere’s law in (3.38b) in the region outside the current distribution where ˆJ=0a s ˆE=1 jωμε∇×ˆB =1 jωμε∇×∇×ˆA (3.39b) and the solution for all the ﬁelds is determined in terms of the vector magnetic potential ˆA. Using the equation for the phasor vector magnetic potential in (3.36a) and (3.37b) shows that for electrically small structures, the quasistaticﬁelds obtained from (3.39) provide reasonable approximations. 3.6 CONSERV ATION OF ENERGY AND POYNTING’S THEOREM In this section we discuss the dissipation and storage of energy in the electro- magnetic ﬁeld. The product of the units of EandHis V/m×A/m=W/m 2, representing a power density in the combined ﬁeld . Hence, it is natural to deﬁne the power density vector as S=E×H W/m2(3.40) This is referred to as the Poynting vector after the English physicist John H. Poynting. The net outﬂow of power from a point is represented by the 112 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) divergence of S. Using a vector identity [3] of ∇·(E×H )=H·(∇×E)− E·(∇×H)and substituting Faraday’s and Amp `ere’s laws gives ∇·S=H·⎜parenleftbigg −∂B ∂t⎜parenrightbigg −E·⎜parenleftbigg J+∂D ∂t⎜parenrightbigg =−E·J−E·∂D ∂t−H·∂B ∂t(3.41) Integrating this result throughout a volume vand using the divergence theorem (see the Appendix) gives −⎜contintegraldisplay sS·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright power entering surfaces=⎜integraldisplay v(E·J)dv ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright power dissipated in volume v+⎜integraldisplay v⎜parenleftbigg E·∂D ∂t⎜parenrightbigg dv ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright rate of change of stored energy in the electric ﬁeld+⎜integraldisplay v⎜parenleftbigg H·∂B ∂t⎜parenrightbigg dv ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright rate of change of stored energy in the magnetic ﬁeld(3.42) where the closed surface sencloses the volume v. This indicates the expected energy balance since the left side, which represents the total power entering the closed surface s, equals the sum of three terms. The ﬁrst term represents the ohmic power dissipation throughout the volume v, while the second and third terms represent the time rate of change of the energy stored in the electricand magnetic ﬁelds, respectively, in the volume v. The right-hand side can be rewritten, assuming that the medium is linear, homogeneous, and isotropicusing the basic relations J=σE,B=μH, and D=εE,a s − ⎜contintegraldisplay sS·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright power entering surfaces=⎜integraldisplay v⎜parenleftBig σ|E|2⎜parenrightBig dv ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright power dissipated in volume v+1 2d dt⎜integraldisplay v⎜parenleftBig ε|E|2⎜parenrightBig dv ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright rate of change of stored energy in the electric ﬁeld+1 2d dt⎜integraldisplay v⎜parenleftBig μ|H|2⎜parenrightBig dv ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright rate of change of stored energy in the magnetic ﬁeld =⎜integraldisplay v(E·J)dv+1 2d dt⎜integraldisplay v(D·E)dv+1 2d dt⎜integraldisplay v(B·H)dv(3.43) We used the1 2factor and moved the time partial derivatives outside the last two volume integrals since E·∂D ∂t=εE·∂E ∂t =ε1 2∂ ∂t|E|2 INDUCTANCE OF A CONDUCTING LOOP 113 =ε1 2∂ ∂t(E·E) =1 2∂ ∂t(D·E) H·∂B ∂t=μH·∂H ∂t =μ1 2∂ ∂t|H|2 =μ1 2∂ ∂t(H·H) =1 2∂ ∂t(B·H) because we can write for any vector ﬁeld F, using the chain rule, F·∂F ∂t=Fx∂Fx ∂t+Fy∂Fy ∂t+Fz∂Fz ∂t =1 2∂F2 x ∂t+1 2∂F2 y ∂t+1 2∂F2 z ∂t =1 2∂|F|2 ∂t =1 2∂(F·F) ∂t The result in (3.43) suggests that for linear, homogeneous, and isotropic media, the energy stored in the electric and magnetic ﬁelds inside the volume is WE=1 2⎜integraldisplay v(D·E)dv J (3.44a) and WM=1 2⎜integraldisplay v(B·H)dv J (3.44b) respectively. 3.7 INDUCTANCE OF A CONDUCTING LOOP In the remaining chapters we discuss the computation of the inductance of a closed loop that is constructed of a conductor such as a wire or a land on a 114 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) printed circuit board. This ﬁnal section will serve as a preliminary to those discussions. Faraday’s fundamental law of induction indicates that an electromotive force is induced in the perimeter of any closed loop through the enclosedsurface of which a time-varying magnetic ﬁeld passes. We have representedthat emf in Fig. 3.2 as a lumped voltage source placed at an indeterminate position in the loop: V=dψ dt(3.45) This voltage source represents the time rate of change of the total magnetic ﬂux penetrating that loop: ψ=⎜integraldisplay sB·ds (3.46) We can also write the ﬂux through the loop in an equivalent form in terms of the line integral of the vector magnetic potential Aaround the loop using the identity B=∇ × A. Hence, the magnetic ﬂux through the loop can be written as a line integral of Aaround that loop as ψ=⎜integraldisplay sB·ds =⎜integraldisplay s(∇×A)·ds =⎜contintegraldisplay cA·dl (3.47) where we have used Stokes’s theorem (see the Appendix) to convert a sur- face integral over the open surface senclosed by the conducting loop into a line integral around the contour cenclosing the surface. Hence, Faraday’s fundamental law of induction can be written in two alternative forms as ⎜contintegraltext cE·dl=−d dt⎜integraldisplay sB·ds =−d dt⎜contintegraldisplay cA·dl(3.48) INDUCTANCE OF A CONDUCTING LOOP 115 VB surface s contour cab A∫∫ === cs dtdVdtdVdtdV A•dlB•dsψ FIGURE 3.10. Conducting loop with a small gap. Consider a conducting loop composed of a perfect conductor that has a very small gap cut in it, as shown in Fig. 3.10. Along the conductor of theloop E=0, so that Faraday’s law gives ⎜contintegraldisplay cE·dl=⎜integraldisplay gapEgap·dl+⎜integraldisplay conductorEconductor⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0·dl =−d dt⎜integraldisplay sB·dl =−d dt⎜contintegraldisplay cA·dl (3.49) But the electric ﬁeld along the perfect conductor is zero, giving ⎜integraldisplay gapEgap·dl=−d dt⎜integraldisplay sB·ds =−d dt⎜contintegraldisplay cA·dl (3.50) 116 FIELDS OF TIME-V ARYING CURRENTS (ACCELERATED CHARGE) Moving the minus sign to the left-hand side gives V=−⎜integraldisplay gapEgap·dl =dψ dt =d dt⎜integraldisplay sB·ds =d dt⎜contintegraldisplay cA·dl (3.51) Hence, a voltage that is related to the time rate of change of the magnetic ﬂux appears at the terminals of the open-circuited loop. Essentially, the time-changing magnetic ﬂux through the loop induces an electric ﬁeld in the con-ductor that forces the charges (electrons) in the conductor to move to theterminals of the gap, thereby creating another electric ﬁeld due to this electricﬁeld across the gap induced by the charge that is accumulated at the terminals.The sum of this induced electric ﬁeld caused by the charges at the gap andthe original electric ﬁeld combine to give a net electric ﬁeld that is zero on thesurface of the conductor, therby satisfying the boundary conditions that thetangential electric ﬁeld on the surface of a perfect conductor must equal zero. 4 THE CONCEPT OF “LOOP” INDUCTANCE In this chapter we examine the calculation of the “loop” inductance of various conﬁgurations of closed current loops. 4.1 SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LA W OF INDUCTION Faraday’s law of induction, discussed in Chapter 3, is fundamental to the notion of inductance. For example, consider the circular loop of conductingwire shown in Fig. 4.1. Suppose that we cut a small gap in the loop and inject acurrent Iinto that gap so that the current ﬂows around the loop in the counter- clockwise direction as shown in Fig. 4.1. This current will, by the right-handrule, produce a magnetic ﬂux density Bthreading the surface sthat the current surrounds. We have shown this surface as being ﬂat to simplify the discussion,although any surface shape will give the same result as long as it is surroundedby the loop. For a current directed in the counterclockwise direction aroundthe loop, the magnetic ﬁeld is directed upward through the surface surroundedby the loop. The total magnetic ﬂux penetrating the loop is obtained as ψ=⎜integraldisplay sB·ds (4.1) Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 117 118 THE CONCEPT OF “LOOP” INDUCTANCE +–sI II I sdtdVψ= I II IdtdILdtdV ==ψdtdVψ= dtdIL V=LB B B+– +–+– FIGURE 4.1. Loop inductance by Faraday’s law. where sis the surface the current loop surrounds. If the current and asso- ciated magnetic ﬁeld varies with time, Faraday’s law of induction essen-tially provides that the time rate of change of the magnetic ﬂux through theloop will essentially induce an electromotive force (emf) around the loopcontour: emf= ⎜contintegraldisplay cE·dl=−dψ dt(4.2) where cis the contour of the loop that surrounds the surface s. If the dimensions of the loop are electrically small, we may represent this emf as a lumpedvoltage source whose value is the time rate of change of the magnetic ﬂuxthrough the loop: V=dψ dt(4.3) and place it anywhere in the loop perimeter as shown in Fig. 4.1. The exact location of the voltage source in the loop perimeter cannot be determineduniquely, nor does it need to be. SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 119 It is important to determine correctly the polarity of the induced voltage source. The minus sign in Faraday’s law in (4.2) is referred to as Lenz’slaw. The induced voltage source should induce a current, I induced, leaving its positive terminal such that this induced current will produce an induced magnetic ﬁeld, Binduced, through the loop surface that tends to oppose the rate of change of the original magnetic ﬁeld, B, produced by the original current I. Hence, the voltage source is inserted with the polarity shown in Fig. 4.1. The inductance of the current loop is deﬁned fundamentally, as the ratio of the magnetic ﬂux threading the loop and the current producing it: L=ψ I(4.4a) or ψ=LI (4.4b) If the surrounding medium is linear, homogeneous, and isotropic, the total magnetic ﬂux threading the loop is directly proportional to the current Ithat produced it, and hence the inductance of the loop is only a function of the loopshape and its dimensions as well as the material properties of the surroundingmedium. Hence, the induced voltage source is V=dψ dt =LdI dt(4.5) Figure 4.1 shows that we can replace this induced source with the usual in- ductor symbol, and the voltage induced across this inductance is given by(4.5). This voltage appears across the terminals of the loop like a Th `evenin open-circuit voltage. If the contour of the loop (represented here as a wire) hasresistance, that is represented as well by the usual resistor symbol inserted inseries with the loop, thereby giving an additional voltage drop of IRaround the loop. The process of calculating the inductance of a loop is referred to as the method of ﬂux linkages, since we compute the ﬂux that “links” the current. It is a four-step process: 1. Inject a current Iaround the closed loop. 2. Determine the magnetic ﬂux density Bover the surface of the enclosed loop by the methods of Chapter 2. 3. Compute the total magnetic ﬂux threading the loop according to (4.1).4. Divide that ﬂux by the current Iaccording to (4.4a). 120 THE CONCEPT OF “LOOP” INDUCTANCE 1I loop 1 loop 2 dtdI1Mdtdψ V 122 ==B12s1 s2+_ FIGURE 4.2. Mutual inductance between two loops. The inductance so obtained is referred to as the self inductance of the loop. Themutual inductance between two loops, one of which carries a current I1, is deﬁned with reference to Fig. 4.2 as M12=ψ2 I1(4.6) where ψ2is the ﬂux penetrating the surface of the second loop, s2, that is caused by the current of the ﬁrst loop: ψ2=⎜integraldisplay s2B12·ds (4.7) In Chapter 2 we found that the computation of the magnetic ﬂux density B could be accomplished by various methods. But they all required that we eval-uate some rather complex integrals. To complete the process of determiningthe inductance of the structure by the method of ﬂux linkages, we will furtherhave to evaluate some rather complicated integrals involving those Bﬁelds in order to determine the ﬂux through the loop via (4.1) and (4.7) or by othermeans. However, there are other methods that we will investigate to compute SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 121 the self and mutual inductances of and between current loops that avoid the direct calculation of Band the ﬂux through the loop as in (4.1) or (4.7). Nevertheless, the fundamental deﬁnition of inductance is via Faraday’s law. 4.1.1 Rectangular Loop In this section we determine the inductance of the rectangular loop shown in Fig. 4.3(a), whose length is land width is w. The conductors of the loop are (b)z II I Il wy2rw B (a) II I Iz yw2rl− w2rl+ −B(R,Z)RZs s FIGURE 4.3. Rectangular loop. 122 THE CONCEPT OF “LOOP” INDUCTANCE wires having radii rw. We assume that the current Iisuniformly distributed across the cross section of the wires , so that with regard to computing the magnetic ﬁeld from it, the current can be considered to be concentrated in aﬁlament on the axes of those wires. For isolated direct currents (dc) not inproximity to other currents, the current is, in fact, uniformly distributed overthe wire cross section. However, for a current that is in close proximity to othercurrents, the current in the wire will not be distributed uniformly over the wirecross section. Nearby currents will cause the current to be concentrated on theside of the wire nearest the neighboring current, a phenomenon known as theproximity effect . Proximity effect is usually not pronounced unless the two currents are within about four radii of each other (i.e., one wire will just ﬁtbetween the two). This is investigated in Section 4.6. High-frequency currentswill be symmetric about the wire axis but will tend to be concentrated in anannulus at the surface of thickness that is a few skin depths. High-frequencyredistribution of the current is investigated in Section 6.5. The loop through which we determine the magnetic ﬂux is the area formed by the interior edges of the wires of the loop. To determine the total ﬂux throughthat loop, we determine the ﬂux through the loop caused by the current of eachwire separately and then add the four ﬂuxes. The ﬂux through the loop causedby the left wire segment as shown in Fig. 4.3(b) can be obtained by using theresult for the Bﬁeld due to a length of wire given in equation (2.15). The B ﬁeld is perpendicular to the loop surface and directed into the page accordingto the right-hand rule: B(R,Z )=μ 0I 4πR⎡ ⎣Z+l/2⎜radicalBig (Z+l/2)2+R2−Z−l/2⎜radicalBig (Z−l/2)2+R2⎤ ⎦(4.8) Hence, the ﬂux through the loop due to the current of the left side is ψleft side =⎜integraldisplayl/2−rw Z=rw−l/2⎜integraldisplayw−r w R=r wB(R,Z )dR dZ (4.9) Then the total ﬂux through the loop is ψloop=2⎜integraldisplayl/2−rw Z=rw−l/2⎜integraldisplayw−r w R=r wB(R,Z )dR dZ +2⎜integraldisplayw/2−rw Z=rw−w/2⎜integraldisplayl−rw R=r wB(R,Z )dR dZ (4.10) SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 123 The ﬂux through the loop surface due to the left side is evaluated as follows: ψleft side =μ0I 4π⎜integraldisplayl/2−rw Z=rw−l/2⎜integraldisplayw−r w R=r w1 R⎡ ⎣Z+l/2⎜radicalBig (Z+l/2)2+R2 +l/2−Z⎜radicalBig (l/2−Z)2+R2⎤ ⎦dR dZ (4.11) Using integral 221.01 from Dwight [7], ⎜integraldisplaydx x√ x2+a2=−1 aln⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglea+√ x2+a2 x⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle (D221.01) this becomes ψleft side =μ0I 4π⎜integraldisplayl/2−rw Z=rw−l/2⎡ ⎣−ln(Z+l/2)+⎜radicalBig (Z+l/2)2+R2 R −ln(l/2−Z)+⎜radicalBig (l/2−Z)2+R2 R⎤ ⎦w−r w R=r wdZ =μ0I 4π⎜integraldisplayl/2−rw Z=rw−l/2⎜bracketleftbigg −sinh−1Z+l/2 R−sinh−1l/2−Z R⎜bracketrightbiggw−r w R=r wdZ (4.12) where we have written this result in terms of the inverse hyperbolic sine: sinh−1x=ln⎜parenleftBig x+⎜radicalbig x2+1⎜parenrightBig (D700.1) Evaluating this at the limits gives ψleft side =μ0I 4π⎜integraldisplayl/2−rw Z=rw−l/2⎜parenleftbigg −sinh−1Z+l/2 w−rw−sinh−1l/2−Z w−rw +sinh−1Z+l/2 rw+sinh−1l/2−Z rw⎜parenrightbigg dZ (4.13) To evaluate this ﬁnal integral we use a change of variables, λ=Z+l 2 dλ=dZ 124 THE CONCEPT OF “LOOP” INDUCTANCE and ζ=l 2−Z dζ=−dZ giving ψleft side =μ0I 4π⎜integraldisplayl−rw λ=r w⎜parenleftbigg −sinh−1λ w−rw+sinh−1λ rwdλ⎜parenrightbigg +μ0I 4π⎜integraldisplayl−rw ζ=rw⎜parenleftbigg −sinh−1ζ w−rw+sinh−1ζ rw⎜parenrightbigg dζ =2μ0I 4π⎜integraldisplayl−rw λ=r w⎜parenleftbigg −sinh−1λ w−rw+sinh−1λ rw⎜parenrightbigg dλ (4.14) Evaluating this using Dwight’s integral 730 [7], ⎜integraldisplay sinh−1x adx=xsinh−1x a−⎜radicalbig x2+a2a>0 (D730) gives ψleft side =μ0I 2π⎜bracketleftbigg −λsinh−1λ w−rw+⎜radicalBig λ2+(w−rw)2 +λsinh−1λ rw−⎜radicalBig λ2+(rw)2⎜bracketrightbiggl−rw λ=r w =μ0I 2π⎡ ⎢⎢⎢⎢⎣−(l−rw)sinh−1l−rw w−rw+⎜radicalBig (l−rw)2+(w−rw)2 +(l−rw)sinh−1l−rw rw−⎜radicalBig (l−rw)2+(rw)2 +rwsinh−1rw w−rw−⎜radicalBig (rw)2+(w−rw)2 −rwsinh−1rw rw⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ln (1+√ 2)+⎜radicalBig (rw)2+(rw)2 ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright√ 2rw⎤ ⎥⎥⎥⎥⎦ (4.15) Then the total ﬂux through the loop given by (4.10) is ψloop=2ψleft side (l,w,r w)+2ψtop side (w,l ,r w) (4.16) SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 125 where we simply interchange landwin (4.15) to obtain ψtop side (w,l ,r w). The inductance of the loop is Lloop=2ψleft side (l,w,r w)+ψtop side (w,l ,r w) I =μ0 π⎜bracketleftbigg −(l−rw)sinh−1l−rw w−rw−(w−rw)sinh−1w−rw l−rw +(l−rw)sinh−1l−rw rw+(w−rw)sinh−1w−rw rw +rwsinh−1rw w−rw+rwsinh−1rw l−rw +2⎜radicalBig (l−rw)2+(w−rw)2−2⎜radicalBig (w−rw)2+(rw)2 −2⎜radicalbig (l−rw)2+(rw)2−2rwln⎜parenleftbig1+√ 2⎜parenrightbig+2√ 2rw⎜bracketrightbigg (4.17) If the loop dimensions are much larger than the wire radius, l,w/greatermuchrw, the result in (4.17) simpliﬁes to Lloop∼=μ0 π⎜parenleftbigg −lsinh−1l w−wsinh−1w l +lsinh−1l rw+wsinh−1w rw+2⎜radicalbig l2+w2−2w−2l⎜parenrightbigg =μ0 π⎡ ⎣−lln⎛ ⎝l w+⎜radicalBigg⎜parenleftbiggl w⎜parenrightbigg2 +1⎞ ⎠−wln⎛ ⎝w l+⎜radicalBigg⎜parenleftbiggw l⎜parenrightbigg2 +1⎞ ⎠ +lln⎜parenleftbigg2l rw⎜parenrightbigg +wln⎜parenleftbigg2w rw⎜parenrightbigg +2√ l2+w2−2w−2l⎤ ⎦ =μ0 π⎡ ⎣−lln⎛ ⎝1+⎜radicalBigg 1+⎜parenleftbiggw l⎜parenrightbigg2⎞ ⎠−wln⎛ ⎝1+⎜radicalBigg 1+⎜parenleftbiggl w⎜parenrightbigg2⎞ ⎠ +lln2w rw+wln2l rw+2⎜radicalbig l2+w2−2w−2l⎤ ⎦ l,w/greatermuchrw (4.18) 126 THE CONCEPT OF “LOOP” INDUCTANCE This result for the inductance of a rectangular loop in 4.17 simpliﬁes con- siderably if the loop is square (i.e., l=w). The loop inductance of a square loop becomes Lsquare loop =2μ0 π⎜bracketleftbigg (l−rw)sinh−1l−rw rw−lln⎜parenleftBig 1+√ 2⎜parenrightBig +l√ 2 +rwsinh−1rw l−rw−2⎜radicalBig (l−rw)2+(rw)2⎜bracketrightbigg l=w (4.19) In practical situations, the wire radius is much smaller than the side length of the loop (i.e., l/greatermuchrw), and this simpliﬁes to Lsquare loop∼=2μ0 π⎜bracketleftbigg lsinh−1l rw−lln⎜parenleftBig 1+√ 2⎜parenrightBig +l√ 2−2l⎜bracketrightbigg ∼=2μ0 πl⎜bracketleftbigg ln⎜parenleftbigg 2l rw⎜parenrightbigg −ln⎜parenleftBig 1+√ 2⎜parenrightBig +√ 2−2⎜bracketrightbigg =2μ0 πl⎜bracketleftbigg lnl rw−0.774⎜bracketrightbigg l=w/greatermuchrw (4.20) This tedious derivation will be obtained in a simple and straightforward manner using the concept of partial inductance in Chapter 5. 4.1.2 Circular Loop Next, we determine the loop inductance of a circular loop of radius alying in thexyplane which is composed of a wire of radius rw, as shown in Fig. 4.4. Again we assume that the (dc) current is uniformly distributed over the crosssection of the wire so that for the purposes of computing the ﬂux throughthe loop surface, we can consider the current Ito be contained in a ﬁlament at the center of the wire. The magnetic ﬂux density is directed solely in thezdirection over the loop, B=B zaz, and is therefore perpendicular to the surface sthat is surrounded by the wire. Once the Bﬁeld over the surface sis computed, we next determine the total magnetic ﬂux through the surface ofthe loop with a surface integral as ψ=⎜integraldisplay sB·ds =⎜integraldisplaya−rw r=0⎜integraldisplay2π φ/prime=0Bzrd φ/primedr⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ds SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 127 Iy xaB r s 2rwφ′ FIGURE 4.4. Circular loop. Note that the integral with respect to ris from r=0 out to the inner edge of the wires at r=a−rwas with the rectangular loop. Once this is completed, the self inductance of the circular current loop is again determined from L=ψ I In Section 2.6 three methods for determining the B=Bzazﬁeld over the loop surface were evaluated. First the Biot–Savart law was the simplest methodand gave the result in (2.73): B z(r)=2μ0Ia 4π⎜integraldisplayπ φ=0a−rcosφ ⎜parenleftbiga2+r2−2arcosφ⎜parenrightbig3/2dφ =μ0Ia 2π⎜integraldisplayπ φ=0a−rcosφ ⎜parenleftbiga2+r2−2arcosφ⎜parenrightbig3/2dφ (2.73) Next, we obtained the Bﬁeld over the loop surface from the vector magnetic potential of a current loop given in (2.59). That general result in (2.59) spe-cialized for the problem of Fig. 4.4 for the ﬁeld in the plane of the loop (z=0) is A φ=μ0Ia 2π⎜integraldisplayπ φ=0cosφ⎜radicalbig a2+r2−2arcosφdφ (4.21) 128 THE CONCEPT OF “LOOP” INDUCTANCE We obtained the magnetic ﬂux density over the loop surface contained by the loop from B=∇ ×A=1/r[∂⎜parenleftbigrAφ⎜parenrightbig/∂r]azusing the result in (4.21). The magnetic ﬂux density in the plane of the loop ( z=0) is totally zdirected (out of the page within the interior of the loop and into the page outside the loop) according to the right-hand rule and is Bz=1 r∂⎜parenleftbigrAφ⎜parenrightbig ∂r =μ0I 2πr⎜integraldisplayπ φ=0a2cosφ(a−rcosφ) ⎜parenleftbiga2+r2−2arcosφ⎜parenrightbig3/2dφ (4.22) The third method for obtaining the Bﬁeld over the loop surface is to use directly the result obtained from (2.59) by Smythe [10] and Weber [11] andgiven in (2.67c). We showed in Section 2.6 that all three results give the samevalue for the Bﬁeld over the surface of the loop. So the choice of which result to use is whichever one provides the simplest integral for obtaining the totalﬂux through the loop. It is for this reason that we choose to use the resultobtained from differentiating Aand given in (4.22). The total ﬂux through the surface of the loop is ψ loop=⎜integraldisplay2π φ/prime=0⎜integraldisplaya−rw r=0Bzrd rd φ/prime =μ0I 2π⎜integraldisplay2π φ/prime=0⎜integraldisplaya−rw r=01 r⎜bracketleftBigg⎜integraldisplayπ φ=0a2cosφ(a−rcosφ) ⎜parenleftbiga2+r2−2arcosφ⎜parenrightbig3/2dφ⎜bracketrightBigg rd rd φ/prime =μ0I⎜integraldisplayπ φ=0⎜bracketleftBigg⎜integraldisplaya−rw r=0a2cosφ(a−rcosφ) ⎜parenleftbiga2+r2−2arcosφ⎜parenrightbig3/2dr⎜bracketrightBigg dφ and we have interchanged the order of integration. The interior integral can be evaluated using integrals 380.003 and 380.013 in Dwight [7]: ⎜integraldisplaydx ⎜bracketleftbigax2+bx+c⎜bracketrightbig3/2=4ax+2b ⎜parenleftbig4ac−b2⎜parenrightbig⎜bracketleftbigax2+bx+c⎜bracketrightbig1/2(D380.003) ⎜integraldisplayxd x ⎜bracketleftbigax2+bx+c⎜bracketrightbig3/2=−2bx+4c ⎜parenleftbig4ac−b2⎜parenrightbig⎜bracketleftbigax2+bx+c⎜bracketrightbig1/2(D380.013) to yield ψloop=μ0Ia(a−rw)⎜integraldisplayπ φ=0cosφ⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφdφ (4.23) SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 129 This integral cannot be evaluated in closed form, but the result can be given in terms of complete elliptic integrals of the ﬁrst and second kind [7]: K=⎜integraldisplayπ/2 θ=0dθ⎜radicalbig 1−k2sin2θ(D773.1) and E=⎜integraldisplayπ/2 θ=0⎜radicalBig 1−k2sin2θd θ (D774.1) Making a change of variables in (4.23) as φ=π−2θ,dφ=−2dθgives cosφ=2 sin2θ−1 and ψloop=2μ0Ia(a−rw)⎜integraldisplayπ/2 θ=02 sin2θ−1 (2a−rw)⎜radicalbig 1−k2sin2θdθ (4.24) where k2is deﬁned here as k2=4a(a−rw) (2a−rw)2(4.25) This can be written in terms of the complete elliptic integrals as ψloop=μ0I⎜radicalbig a(a−rw)⎜bracketleftbigg⎜parenleftbigg2 k−k⎜parenrightbigg K(k)−2 kE(k)⎜bracketrightbigg (4.26) Hence, the loop inductance is Lloop=ψloop I =μ0√a(a−rw)⎜bracketleftbigg⎜parenleftbigg2 k−k⎜parenrightbigg K(k)−2 kE(k)⎜bracketrightbigg (4.27) This result can be simpliﬁed by assuming that the loop radius is much larger than the wire radius, a/greatermuchrw. For this reasonable approximation we obtain√a(a−rw)∼=aandk2∼=1. From series expansions of the complete elliptic integrals given by Dwight [7], we obtain K(k)∼=ln⎜parenleftbigg8a rw−4⎜parenrightbigg a/greatermuchrw E(k)∼=1 a/greatermuchrw 130 THE CONCEPT OF “LOOP” INDUCTANCE Hence, the loop inductance of the circular loop approximates to Lloop∼=μ0a⎜parenleftbigg ln8a rw−2⎜parenrightbigg a/greatermuchrw (4.28) The self inductance of coils consisting of a thin wire of radius rwand the same total length, denoted as Len, are approximately independent of theirshape. For example, the circular loop of radius ahas a total circumference of Len=2πaand an inductance in (4.28) of L circular loop =μ0Len 2π⎜parenleftbigg ln4 Len rw−3.145⎜parenrightbigg whereas the square loop of equal side lengths of lhas a total circumference of Len =4land an inductance in (4.20) of Lsquare loop =μ0Len 2π⎜parenleftbigg ln4 Len rw−3.547⎜parenrightbigg 4.1.3 Coaxial Cable In this section we determine the inductance of a coaxial cable shown in Fig. 4.5(a). The cable is assumed to be inﬁnite in length (or very long com-pared with the cable radius) in order to avoid having to deal with fringingof the ﬁelds at the ends of a ﬁnite-length section. The magnetic ﬂux densityfor this cable was determined in Chapter 2. Consider a section of length 1 m.Because of the inﬁnite length and symmetry, the magnetic ﬁeld between theinner wire and the inside of the shield is circumferentially directed in the φ direction as shown in Fig. 4.5(b) and is determined in Chapter 2 as B φ=μ0I 2πrrw<r<r s (2.35a) We determine the ﬂux through a ﬂat surface that extends from the outer edge of the inner wire, r=rw, to the inner edge of the outer shield, r=rs, and is of length along the cable of 1 m. We have shown two choices for this surface. One(which we will choose) is perpendicular the inner wire surface, and the otherextends at an angle from the inner wire surface to the inner surface of the shieldas shown in Fig. 4.5(b). The best choice is the ﬁrst surface that is perpendicularto the inner wire surface and extends directly across perpendicular to the innersurface of the shield. The reason that this is preferred is that the magnetic ﬁeld, SELF INDUCTANCE OF A CURRENT LOOP FROM FARADAY’S LAW OF INDUCTION 131 wrI I (a) (b)wrφB I It tsr srto∞ to∞ top (c)sidebottom endend φB φB1 m FIGURE 4.5. Coaxial cable. Bφ, is perpendicular to that surface and hence we easily obtain the ﬂux through this surface as ψ=⎜integraldisplay sB·ds =⎜integraldisplay1m z=0⎜integraldisplayrs r=rwBφdr dz⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ds 132 THE CONCEPT OF “LOOP” INDUCTANCE =⎜integraldisplay1m z=0⎜integraldisplayrs r=rwμ0I 2πrdr dz⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ds =μ0I 2πlnrs rw Theper-unit-length inductance of the cable is the inductance of this section and is denoted as l: l=ψ I =μ0 2πlnrs rwH/m (4.29) There were two choices for the ﬂat surface through which we were to de- termine the ﬂux. Figure 4.5(c) shows this situation. Consider this as a closed,“wedge-shaped” surface. The top and bottom sides were the two choices forsurfaces. We chose the bottom surface because the magnetic ﬂux density vec-tor is perpendicular to that surface, thus allowing us to remove the dot productin the ﬂux integral and deal only with the magnitude of the ﬁeld over the sur-face. Would computing the ﬂux through the other surface, the top surface,have given a different answer? Certainly that computation would be moredifﬁcult since the magnetic ﬂux density vector would not be perpendicularto it and the dot product could not be removed from the ﬂux integral. RecallGauss’s law for the magnetic ﬁeld: ⎜contintegraldisplay sB·ds=0 In other words, the net magnetic ﬂux leaving a closed surface s is zero for the magnetic ﬁeld. Consider the closed wedge-shaped surface in Fig. 4.5(c). Applying Gauss’s law gives ⎜contintegraldisplay sB·ds=⎜integraldisplay topB·ds+⎜integraldisplay bottomB·ds+⎜integraldisplay sideB·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0+⎜integraldisplay left endB·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0 +⎜integraldisplay right endB·ds ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0=0 The ﬂux through the side of constant radius rsis zero because on that surface Bφis parallel to the surface. Similarly, the ﬂux through the left and right ends of the surface are also zero because on the surfaces Bφis also parallel to the THE CONCEPT OF FLUX LINKAGES FOR MULTITURN LOOPS 133 surfaces. Hence, we see that ⎜integraldisplay topB·ds=−⎜integraldisplay bottomB·ds But obtaining the ﬂux through the bottom surface is much easier than obtaining the ﬂux through the top surface since the magnetic ﬁeld is perpendicular tothe bottom surface. 4.2 THE CONCEPT OF FLUX LINKAGES FOR MULTITURN LOOPS Consider a single, circular current loop carrying a current I. Denote the mag- netic ﬂux through the surface of the loop due to this current Iasψ one loop . The emf voltage induced in that single loop is Vone loop =dψone loop dt =Lone loopdI dt The magnetic ﬂux ψone loop is said to linkcurrent I. Now consider a multiturn loop where we add N such identical loops that are in very close proximity (virtually on top of each other) so that all of themagnetic ﬂux that passes through one of the loops that is due to the currentof that loop, ψ one loop , also passes through all the other loop surfaces. The total are concentrically located and are tightly wound together such that theyresemble one loop carrying a current of N Iamperes as shown in Fig. 4.6. Thetotal ﬂux through each loop is therefore the sum of the ﬂuxes from all the I IN turnsINNBone loop FIGURE 4.6. A multiturn loop consisting of N loops close together and connected in series. 134 THE CONCEPT OF “LOOP” INDUCTANCE other N loops or N ψone loop . Hence, we say that each loop has N ﬂux linkages linking its current I. The emf voltage induced in each loop is therefore Vone loop =Ndψone loop dt The loops are connected in series so that each carry current Iin the same direction around the loops. Since all N loops are connected in series, the totalemf voltage at the terminals of the N loops is Vtotal=N⎜parenleftbigVone loop⎜parenrightbig =N2dψone loop dt This is an important property of N identical loops that surround a common core; the inductance is proportional to N2times the inductance of one of the loops: LN loops∝N2Lone loop 4.2.1 Solenoid For example, consider the solenoid shown in Fig. 4.7(a), consisting of N turns of wire wound in one layer on a ferromagnetic core that has a relativepermeability of μ rand a radius r. The purpose of a ferromagnetic core having a large μris to concentrate the ﬂux in that core, thereby minimizing the ﬂux that leaks out into the air , which has μr=1 [3]. Hence, if the turns of wire are closely wound on the core, there will be very little leakage of the magneticﬁeld between the adjacent turns of wire. In fact, if the solenoid is inﬁnite inlength, l→∞ , and the turns of wire are tightly wound, the magnetic ﬁeld in the core will be (1) in the zdirection parallel to the axis of the core, (2) constant along that axis, (3) uniformly distributed across the core cross section, and(4) the magnetic ﬁeld outside the solenoid will be zero. These properties ofan inﬁnite-length solenoid are also approximate properties of a solenoid ofﬁnite length on which the wires are tightly wound. To determine the Hﬁeld in the core, assume that the solenoid is inﬁnite in length (l →∞ ). Thinking of this as an inﬁnite number of current loops that are inﬁnitesimally close together shows that the Hﬁeld in the core will be in thezdirection and independent of zandr. Draw a rectangular closed contour cwhose sides are parallel to the core axis and whose ends are perpendicular to it and which encloses N turns (wires) within a length las shown in Fig. 4.7(b). If this rectangle were moved outside the core, it would enclose no current andthe line integral of Haround it must, by Ampere’s law, be zero. But this would THE CONCEPT OF FLUX LINKAGES FOR MULTITURN LOOPS 135 lr N turnsIμr c (a)cI B (b) N turns (loops) II (c)r r rI IdtdVψ= dtdVψ= dtdVψ=dtdVψ= dtdVψ= dtdVψ=I () dtt I d lr NVN V Lr ) (2 2 0out == π μ μH ()) (20t IlrN r π μ μψ=N SourcesIz z zz=0∞ ∞l + _+ − + − + −+−+−+− FIGURE 4.7. Solenoid. imply that the Hﬁeld along the sides would be constant. Hence, we conclude that the Hﬁeld outside the inﬁnite-length core is zero since the magnetic ﬁeld must go to zero as r→∞ . Therefore, the magnetic ﬁeld along the right-hand part of contour c(that passes along the outside of the coil of wire) is zero. From Amp `ere’s law and Fig. 4.7(b), we obtain ⎜contintegraldisplay cH·dl=Hl=NI (4.30) 136 THE CONCEPT OF “LOOP” INDUCTANCE since the closed contour cencloses N currents. From this result for a coil of inﬁnte length the magnetic ﬁeld intensity is H=NI/l. For a core of ﬁnite length, this result is approximately the same and relies on our assumption that(1) the turns are tightly wound, (2) the relative permeability of the core is verylarge,μ r/greatermuch1, and (3) the coil length, l, is long, l/greatermuchr. Hence, the magnetic ﬂux density in the core and parallel to the core axis is B=μrμ0H=μrμ0NI l(4.31) This result can be derived in a different fashion by using the result for the magnetic ﬁeld on the axis of a single current loop derived in Chapter 2 andgiven in (2.24). Since we assume that the coil of wires is tightly wound, thinkof the coil of wires as being a cylindrical sheet of current with a surface currentdistribution of K=NI/lA/m uniformly distributed along the core surface and directed in the circumferential direction about the core. Hence, we maythink of a section of the coil of differential length dzas being a single turn carrying a current of Kd z=NI/l dz amperes. Using (2.24) and summing the ﬁelds of these turns of differential lengths dzgives the magnetic ﬂux density on the axis of the core and midway between the two ends of the coil of wireatz=0a s B=μ rμ0 2NI l⎜integraldisplayl/2 z=−l/2r2 ⎜parenleftbigr2+z2⎜parenrightbig3/2dz =μrμ0 2NI l⎜bracketleftbiggz√ r2+z2⎜bracketrightbiggl/2 z=−l/2 =μrμ0 2NI l⎡ ⎣l/2⎜radicalBig r2+(l/2)2+l/2⎜radicalBig r2+(l/2)2⎤ ⎦ =μrμ0NI√ 4r2+l2(4.32) and we have used integral 200.03 from Dwight [7]: ⎜integraldisplay1 ⎜parenleftbigx2+a2⎜parenrightbig3/2dx=x a2√ x2+a2(D200.03) For a very long coil length with respect to the radius, l/greatermuchr, (4.32) reduces to (4.31) derived by the previous method using Amp `ere’s law. Since the ﬁeld for a very long coil, l/greatermuchr, is (approximately) uniformly distributed over the core cross section, which has an area of πr2, the magnetic ﬂux through each turn of the solenoid is ψeach loop =μrμ0NI lπr2(4.33) THE CONCEPT OF FLUX LINKAGES FOR MULTITURN LOOPS 137 Hence, the emf voltage induced in each loop is Veach loop =Nμrμ0πr2 l⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Leach loopdI(t) dt(4.34) If we visualize the entire coil of wire as being the N loops connected in series as shown in Fig. 4.7(c), the emf voltage sources of each loop are connectedin series so that the voltage across the terminals of the entire coil is V= NV each loop . Hence, the total inductance of the solenoid is L=N2μrμ0πr2 l(4.35) 4.2.2 Toroid Next, consider the toroid shown in Fig. 4.8(a). The toroid consists of N turns of wire wound tightly around a toroidal core of ferromagnetic material havingrelative permeability of μ r, an inner radius a, and an outer radius b. The cross section of the toroid is usually rectangular with thickness tand width w=b−a, as shown in Fig. 4.8(b). If we assume that the turns are tightly wound on the core and μr/greatermuch1 so that there is no signiﬁcant leakage of the I Iμr a btcontour c r N turns (a) b atφB φB φB (b)w FIGURE 4.8. Toroid. 138 THE CONCEPT OF “LOOP” INDUCTANCE magnetic ﬁeld outside the core, we may assume as an approximation that the magnetic ﬁeld is contained within the core and is in the circumferential or φ direction. Alternatively, we can view the toroid as a ﬁnite-length solenoid thatis formed into a circle. To determine that magnetic ﬁeld, we choose a circular contour cof radius rin the core as shown in Fig. 4.8(a) and write Amp `ere’s law as ⎜contintegraldisplay cH·dl=Hφ(2πr)=NI (4.36a) since the contour csurrounds NI currents. Hence, the magnetic ﬁeld intensity in the core is Hφ=NI/2πr, and the ﬂux density in the core is Bφ=μrμ0Hφ =Nμrμ0 2πrI a<r<b (4.36b) Expanding the contour to a radius r>b encloses zero net current, and hence theHﬁeld outside the toroid is zero, as is the ﬁeld for r<a . If the core cross section is rectangular with width wand thickness tas shown in Fig. 4.8(b), we can determine the total magnetic ﬂux through each loop as ψeach loop =⎜integraldisplay sB·ds =⎜integraldisplayt z=0⎜integraldisplayb r=aNIμrμ0 2πrdr dz =NIμrμ0 2πtlnb a(4.37) where surface sis the rectangular ﬂat surface of a cross section of the core. The emf voltage induced in each turn is Veach loop =Nμrμ0 2πtlnb a⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Leach loopdI(t) dt(4.38) Since the loops are connected in series, the total inductance is L=NLeach loop =N2μrμ0 2πtlnb a(4.39) LOOP INDUCTANCE USING THE VECTOR MAGNETIC POTENTIAL 139 This expression can be simpliﬁed for cores of rectangular cross section where the width, w=b−a, is much less than the inner radius, w/lessmucha,b y using the approximation of the natural logarithm: lnb a=ln⎜parenleftbiggw a+1⎜parenrightbigg ∼=w aw/lessmucha (D601) Evaluating (4.39) gives L∼=N2μrμ0 2πatw (4.40a) Since the cross-sectional area of the core is A=tw, we can write a general relation for the inductance of a toroid as L∼=μrμ0N2A 2πa(4.40b) 4.3 LOOP INDUCTANCE USING THE VECTOR MAGNETIC POTENTIAL The inductance of a current loop is deﬁned fundamentally by Faraday’s law as the ratio of the magnetic ﬂux penetrating the open surface sthat is surrounded by the current and the current Ias illustrated in Fig. 4.9: L=ψ=⎜integraltext sB·ds I(4.41) In the previous examples we evaluated this by ﬁrst computing the magnetic ﬂux density Band then evaluating (4.41) by computing the ﬂux ψthrough the surface that is surrounded by the current loop. This required the evaluation oftwo integrals: one to obtain B(by the Biot–Savart law or Amp `ere’s law) and two[since (4.41) is a surface integral] to obtain ψ. There is another way of obtaining this result by using the vector magnetic potential Arather than using B. To obtain this alternative result, recall from Chapter 2 that Ais deﬁned by B=∇ ×A (4.42) 140 THE CONCEPT OF “LOOP” INDUCTANCE B AIc s FIGURE 4.9. Using the vector magnetic potential Ato obtain the magnetic ﬂux through an open surface s. Hence, the magnetic ﬂux through surface scan alternatively be obtained in terms of Aas ψ=⎜integraldisplay sB·ds =⎜integraldisplay s(∇×A)·ds =⎜contintegraldisplay cA·dl (4.43) where we have used Stokes’s theorem (see the Appendix) and cis the closed contour that surrounds the open surface s . Hence, to obtain the to- tal magnetic ﬂux penetrating the open surface swe only need to obtain A (which is usually easier to obtain than B) and then integrate with only one integral, a line integral, the component of Athat is tangent to the contour caround the perimeter of that open surface, as illustrated in Fig. 4.9. The inductance calculation becomes L=ψ=⎜integraldisplay sB·ds I =⎜contintegraldisplay cA·dl I(4.44) LOOP INDUCTANCE USING THE VECTOR MAGNETIC POTENTIAL 141 z II I Il wy2rw2rwl− w2rl+ −Aleft ArightAtop Abottom FIGURE 4.10. Determining the inductance of a rectangular loop by using the vector magnetic potential A. 4.3.1 Rectangular Loop We now apply this to the calculation of the self inductance of a rectangular loop composed of four wires of radii rwhaving lengths wandlas shown in Fig. 4.3. The basic idea is to integrate the line integral of Aalong the interior edge of the wire of one side as illustrated in Fig. 4.10 and then repeat this for the otherthree sides. The total magnetic ﬂux threading the loop, according to (4.43), is ψ loop=2⎜integraldisplay left sideA·dl+2⎜integraldisplay top sideA·dl (4.45) It is very important to realize that the total vector magnetic potential tangent toeach side has contributions from the current of that side andthe currents of the other three sides. This is illustrated for the left side in Fig. 4.10. Againwe assume that the currents are dc and are uniformly distributed over the wirecross sections so that they can be represented by ﬁlaments on the axes of thewires. Two of these contributions, A left(that is due to the current in the left side) and Aright(that is due to the current in the right side), are parallel to the left side and are oppositely directed. Aleftis larger in magnitude than Aright since the current of the right side is further away. The other contributions along the left side, AtopandAbottom , are due to the currents in the top and bottom sides and are perpendicular to the left side since the vector magneticpotential is in the direction of the current producing it. Hence, the line integral 142 THE CONCEPT OF “LOOP” INDUCTANCE of the total vector magnetic ﬁeld along the left side is ψleft side =⎜integraldisplay left sideA·dl =⎜integraldisplayl/2−rw z=r w−l/2Aleft·dl+⎜integraldisplayl/2−rw z=r w−l/2Aright ·dl +⎜integraldisplayl/2−rw z=r w−l/2Atop·dl ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright 0+⎜integraldisplayl/2−rw z=r w−l/2Abottom ·dl ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright 0(4.46) The contributions to AleftandArightare derived in Chapter 2 and given in (2.57) with respect to Fig. 2.24: Az=μ0I 4π⎜parenleftbigg sinh−1Z+L/2 r+sinh−1L/2−Z r⎜parenrightbigg (2.57) Hence, the contribution to the magnetic ﬂux through the loop surface inte- grated along the left side is the same as obtained in Section 4.1.1 and givenin (4.13): ψ left side =μ0I 4π⎜integraldisplayl/2−rw Z=rw−l/2⎜parenleftbigg sinh−1Z+l/2 rw+sinh−1l/2−Z rw −sinh−1Z+l/2 w−rw−sinh−1l/2−Z w−rw⎜parenrightbigg dZ (4.13) Notice that the vector magnetic potential in (2.57) is evaluated over the left wire surface, giving Aleft=Az|r=rw(4.47a) and Aright=− Az|r=w−r w(4.47b) since Aleftalong the left side is at a distance r=rwfrom the current of that side, and Arightalong the left side is at a distance r=w−rwfrom the current LOOP INDUCTANCE USING THE VECTOR MAGNETIC POTENTIAL 143 of the right side that produces it. The integral of (4.13) was evaluated in Section 4.1.1, giving ψleft side =μ0I 2π⎡ ⎢⎢⎢⎢⎢⎣−(l−rw)sinh−1l−rw w−rw+⎜radicalBig (l−rw)2+(w−rw)2 +(l−rw) sinh−1l−rw rw−⎜radicalBig (l−rw)2+(rw)2 +rwsinh−1rw w−rw−⎜radicalBig (rw)2+(w−rw)2 −rwsinh−1⎜parenleftbiggrw rw⎜parenrightbigg ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ln⎜parenleftbig 1+√ 2⎜parenrightbig+⎜radicalBig (rw)2+(rw)2 ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright√ 2rw⎤ ⎥⎥⎥⎥⎥⎦(4.15) Similarly, we obtain the contributions to the magnetic ﬂux through the surface from the right side and the top and bottom sides by integratingAalong those remaining three sides of the loop (ψ left side =ψright side and ψtop side =ψbottom side ). The total ﬂux through the loop is ψloop=2ψleft side (l,w,r w)+2ψtop side (w,l ,r w) (4.48) where we simply interchange land winψleft side (l,w,r w)to obtain ψtop side (w,l ,r w). Since the result is identical to that obtained in Section 4.1.1 using B, the inductance of the loop is identical to that obtained in Section 4.1.1: Lloop=2ψleft side (l,w,r w)+ψtop side (w,l ,r w) I =μ0 π⎜bracketleftbigg −(l−rw)sinh−1l−rw w−rw−(w−rw)sinh−1w−rw l−rw +(l−rw)sinh−1l−rw rw+(w−rw)sinh−1w−rw rw +rwsinh−1rw w−rw+rwsinh−1rw l−rw 144 THE CONCEPT OF “LOOP” INDUCTANCE +2⎜radicalBig (l−rw)2+(w−rw)2−2⎜radicalBig (w−rw)2+(rw)2 −2⎜radicalBig (l−rw)2+(rw)2−2rwln⎜parenleftBig 1+√ 2⎜parenrightBig +2√ 2rw⎜bracketrightbigg (4.17) The remaining results in (4.18), (4.19), and (4.20) for a square loop and for loop side lengths greater than the wire radius are obtained from (4.17) and areidentical to those obtained with this method. But the method of this sectionis much simpler since it avoids having to integrate Bover the surface of the loop, thereby eliminating one integration. 4.3.2 Circular Loop The circular loop of radius ais composed of a wire having radius r wand shown in Fig. 4.4 and is illustrated for this problem in Fig. 4.11. Again weassume that the current is dc and is uniformly distributed over the wire crosssection so that it can be represented by a ﬁlament on the axis of the wire. Toobtain the magnetic ﬂux through the loop enclosed by the wire surface using Iy x a 2r w(a-r w)), ( φ φ wr a− A φ′ FIGURE 4.11. Determining the inductance of a circular loop by using the vector magnetic potential A. NEUMANN INTEGRAL FOR SELF AND MUTUAL INDUCTANCES 145 the vector magnetic potential method in (4.44), we ﬁrst obtain the vector magnetic potential along the inner surface of the wire at r=a−rw. The vector magnetic potential for a circular current loop was obtained in Chapter 2and given in (2.59) with reference to Fig. 2.25. That result is used to give themagnetic vector potential over the loop surface given in (4.21): A φ=μ0Ia 2π⎜integraldisplayπ φ=0cosφ⎜radicalbig a2+r2−2arcosφdφ (4.21) Evaluating (4.21) at r=a−rwgives the vector magnetic potential along the inner wire surface as Aφ(a−rw,φ)=μ0Ia 2π⎜integraldisplayπ φ=0cosφ⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφdφ (4.49) Then we obtain the result for the total ﬂux through the loop as ψloop=⎜contintegraldisplay cA·dl =⎜integraldisplay2π φ/prime=0Aφ⎜vextendsingle⎜vextendsingle r=a−rw(a−rw)dφ/prime ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright rd φ/prime =μ0Ia 2π⎜integraldisplay2π φ/prime=0⎜integraldisplayπ φ=0(a−rw)cosφ⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφdφ dφ/prime =μ0Ia(a−rw)⎜integraldisplayπ φ=0cosφ⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφdφ (4.50) But this is identical to the result obtained by integrating Bin Section 4.1.2 and given in (4.23). Hence, the remaining results in Section 4.1.2 and theinductance of the loop obtained in (4.27) and (4.28) are identical to thoseobtained by this method. But the method of this section is much simpler sinceit avoids having to integrate Bover the surface of the loop, thereby eliminating one (difﬁcult) integration. 4.4 NEUMANN INTEGRAL FOR SELF AND MUTUAL INDUCTANCES BETWEEN CURRENT LOOPS Mutual inductance between two current loops was discussed at the beginning of this chapter with reference to Fig. 4.2. With the ﬁrst loop carrying a current 146 THE CONCEPT OF “LOOP” INDUCTANCE I1, the mutual inductance between the two loops is M12=ψ2 I1(4.6) where ψ2is the ﬂux penetrating the surface of the second loop, s2, that is caused by the current of the ﬁrst loop: ψ2=⎜integraldisplay s2B12·ds (4.7) andB12is the magnetic ﬂux density through loop 2 that is due to the current I1of loop 1. This result can be put into a more compact form by recalling that the magnetic ﬂux through the second loop can be written in terms of thevector magnetic potential around the perimeter of that loop (the interior edgeof the wire), A 12,a s ψ2=⎜contintegraldisplay c2A12·dl2 (4.51) andc2is the contour surrounding the surface of the second loop, s2. But A12 is the magnetic vector potential around contour c2of loop 2 that is due to the current of loop 1 as A12=μ0 4π⎜contintegraldisplay c1I1 R12dl1 (4.52) andc1is the contour of the current of loop 1. The distance R12is the distance from a “chunk” of current I1dl1of loop 1 to the point on the contour of loop 2,c2, where we are evaluating the integral in (4.51). Substituting (4.52) into (4.51) yields ψ2=μ0I1 4π⎜contintegraldisplay c1⎜contintegraldisplay c2dl1·dl2 R12(4.53) Hence, the mutual inductance between the two loops is M12=ψ2 I1 =μ0 4π⎜contintegraldisplay c1⎜contintegraldisplay c2dl1·dl2 R12(4.54) This result is called the Neumann integral. It shows that the mutual inductance between two loops is only a function of the shapes of the two loops and theirorientation with respect to each other. It is also important to remember that ifthe currents are not ﬁlamentary but are uniformly distributed over the crosssections of wires of radii r w1in loop 1 and rw2in loop 2, contour c1is along the ﬁlamentary current I1but contour c2is along the interior surface of the second NEUMANN INTEGRAL FOR SELF AND MUTUAL INDUCTANCES 147 wire, which bounds the surface s2that is enclosed by that wire. The order of integration is immaterial. This important result shows that M12=M21simply by interchanging the roles of the two loops in (4.54). The Neumann integral for mutual inductance between two current loops in (4.54) can also be used to determine the self inductance of a loop by lettingthe two loops be coincident: L=ψ I =μ0 4π⎜contintegraldisplay c/prime⎜contintegraldisplay cdl·dl/prime R(4.55) Contour c/primeis along the ﬁlamentary current bearing current Iat the center of the wire, and contour cis along the interior edge of that wire that bounds the surface of the loop through which we desire to compute the ﬂux through thatloop. 4.4.1 Mutual Inductance Between Two Circular Loops Consider two coaxial loops having N 1and N 2turns, respectively, that are tightly wound, as shown in Fig. 4.12. The two loops are parallel, have radii a andb, and are separated by distance d. First we ﬁx the point on the second loop b φz d x IN1 turnsN2 turns dl1dl2 aR12 FIGURE 4.12. Concentric, coaxial loops. 148 THE CONCEPT OF “LOOP” INDUCTANCE and vary the angle φof the ﬁrst loop. Using the law of cosines, the distance between the two differential arc lengths is R12=⎜radicalBig a2+b2+d2−2abcosφ (4.56) First, we perform the calculation for one turn in each loop and then we mul- tiply the result by the square of the number of turns, N2 1and N2 2, as discussed previously. The dot product in the Neumann integral depends on the dot prod-uctdl 1·dl2=cosφd l 1dl2anddl1=ad φ anddl2=bd φ/prime. Once we integrate with respect to φfromφ=0t oφ=2π, we ﬁnally integrate with respect to the angle of loop 2: φ/prime=0t oφ/prime=2π, giving the Neumann integral as M12=μ0ab 4π⎜integraldisplay2π φ/prime=0⎜integraldisplay2π φ=0cosφ⎜radicalbig a2+b2+d2−2abcosφdφ dφ/prime =μ0ab 2⎜integraldisplay2π φ=0cosφ⎜radicalbig a2+b2+d2−2abcosφdφ (4.57) Making a change of variables to φ=2θso that cos φ=cos 2θ=2 cos2θ−1 anddφ=2dθgives M12=μ0ab 2⎜integraldisplayπ θ=02 cos 2 θ⎜radicalBig (a+b)2+d2−4abcos2θdθ =μ0√ ab 2⎜integraldisplayπ θ=0kcos 2θ√ 1−k2cos2θdθ (4.58a) where k2=4ab (a+b)2+d2 =4(a/d)(b/d) (a/d+b/d)2+1(4.58b) and the factor kdepends on the ratios of the circle radii to their separation. But the numerator of the integrand of (4.58a) can be written as kcos 2θ=k⎜parenleftBig 2 cos2θ−1⎜parenrightBig =⎜parenleftbigg2 k−k⎜parenrightbigg −2 k⎜parenleftBig 1−k2cos2θ⎜parenrightBig Note that ⎜integraldisplayπ θ=0⎜radicalbig 1−k2cos2θd θ=2⎜integraldisplayπ/2 θ=0⎜radicalbig 1−k2cos2θd θ =2⎜integraldisplayπ/2 θ=0⎜radicalBig 1−k2sin2θd θ NEUMANN INTEGRAL FOR SELF AND MUTUAL INDUCTANCES 149 ⎜integraldisplayπ θ=01√ 1−k2cosθdθ=2⎜integraldisplayπ/2 θ=01√ 1−k2cosθdθ =2⎜integraldisplayπ/2 θ=01√ 1−k2sinθdθ Hence, (4.58) can be written as M12=μ0√ ab 2⎜integraldisplayπ θ=0kcos 2θ√ 1−k2cos2θdθ =μ0√ ab⎜integraldisplayπ/2 θ=0⎜bracketleftbigg⎜parenleftbigg2 k−k⎜parenrightbigg1√ 1−k2cos2θ−2 k⎜radicalbig 1−k2cos2θ⎜bracketrightbigg dθ =μ0√ ab⎜integraldisplayπ/2 θ=0⎜bracketleftBigg⎜parenleftbigg2 k−k⎜parenrightbigg1⎜radicalbig 1−k2sin2θ−2 k⎜radicalBig 1−k2sin2θ⎜bracketrightBigg dθ =μ0√ ab⎜bracketleftbigg⎜parenleftbigg2 k−k⎜parenrightbigg K(k)−2 kE(k)⎜bracketrightbigg (4.59) where K(k) andE(k) are the complete elliptic integrals of the ﬁrst and second kind that are tabulated by Dwight [7]: K(k)=⎜integraldisplayπ/2 θ=0dθ⎜radicalbig 1−k2sin2θ(D773.1) E(k)=⎜integraldisplayπ/2 θ=0⎜radicalBig 1−k2sin2θd θ (D774.1) This was ﬁrst obtained by Maxwell [23]. (In [23] 4 πdenotes μ0, the units of length taken to be 107m.) Setting d=0 and b=a−rwgives the self inductance of a loop of radius acomposed of a wire of radius rwgiven in (4.27). If the separation between the two coils, d, is much larger than the radii (i.e., d/greatermucha, b),R12in (4.56) approximates to 1 R12=1⎜radicalbig a2+b2+d2−2abcosφ =⎜parenleftBig a2+b2+d2−2abcosφ⎜parenrightBig−1/2 ∼=1 d⎜parenleftbigg 1−2abcosφ d2⎜parenrightbigg−1/2 ∼=1 d⎜parenleftbigg 1+abcosφ d2⎜parenrightbigg d/greatermucha, b (4.56) 150 THE CONCEPT OF “LOOP” INDUCTANCE and we used the binomial theorem: (1−x)−1/2∼=1+1 2x+··· (D1) Using this result, the integral in (4.57) approximates to M12=N2 1N2 2μ0ab 4π⎜integraldisplay2π φ/prime=0⎜integraldisplay2π φ=0cosφ⎜radicalbig a2+b2+d2−2abcosφdφ dφ/prime =N2 1N2 2μ0ab 2⎜integraldisplay2π φ=0cosφ⎜radicalbig a2+b2+d2−2abcosφdφ ∼=N2 1N2 2μ0ab 2d⎜integraldisplay2π φ=0⎜parenleftbigg 1+abcosφ d2⎜parenrightbigg cosφd φ =N2 1N2 2μ0πa2b2 2d3d/greatermucha, b (4.60) and we have multiplied by the squares of the number of turns in each coil for multiturn coils. 4.4.2 Self Inductance of the Rectangular Loop Figure 4.13 shows a rectangular loop for computing the self inductance using the Neumann integral in (4.55): L=ψ I =μ0 4π⎜contintegraldisplay c/prime⎜contintegraldisplay cdl·dl/prime R(4.55) The differential element along the ﬁlament of current at the center of the wires is denoted as dl/prime, and the differential element along the inside of the wire (bounding the surface of the loop) is denoted as dl. For the left and right sides of the loop these become dl/prime left=dz/primeaz,dlleft=dzazanddl/prime right=−dz/primeaz, dlright=−dzaz. For the top and bottom sides of the loop these become dl/prime top= dy/primeay,dltop=dyayanddl/prime bottom=−dy/primeay,dlbottom=−dyay. We ﬁrst compute the integral along the inside surface of the left wire (at y= rw),dlleft, which is due to the currents of each of the four sides, dl/prime, according to (4.55) to give the contribution of that side to the total loop inductance ofthe loop, L left. Then repeat this for the contributions to the inductance of the loop for each of the other three sides that are due to the currents of each of thefour sides according to (4.55): L top,Lright, andLbottom . The dot product dl/prime·dl equals dz/primedzalong the left side and −dz/primedzalong the right side but is zero along the top and bottom sides since dl/primeanddlare orthogonal to each other NEUMANN INTEGRAL FOR SELF AND MUTUAL INDUCTANCES 151 z II I Il wyrw2w2rl− w2rl+ −12R12R12R12R leftl′dleftldtopl′d rightl′d bottoml′d FIGURE 4.13. Neumann integral and the rectangular loop. along those sides. So the contributions along the left side due to the currents of each of the other four sides is Lleft=μ0 4π⎜integraldisplayl/2−rw z=r w−l/2⎡ ⎢⎢⎢⎢⎢⎣⎜integraldisplay l/2 z/prime=−l/21⎜radicalBig (z/prime−z)2+r2wdz/prime ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright left −⎜integraldisplayl/2 z/prime=−l/21⎜radicalBig (z/prime−z)2+(w−rw)2dz/prime ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright right⎤ ⎥⎥⎥⎥⎥⎥⎦dz =μ0 4π⎜integraldisplayl/2−rw z=r w−l/2⎡ ⎣⎜integraldisplayl/2−z λ=−l/2 −z1⎜radicalBig λ2+r2wdλ −⎜integraldisplayl/2−z λ=−l/2 −z1⎜radicalBig λ2+(w−rw)2dλ⎤ ⎦dz 152 THE CONCEPT OF “LOOP” INDUCTANCE =μ0 4π⎜integraldisplayl/2−rw z=r w−l/2⎜braceleftBigg⎜bracketleftbigg ln⎜parenleftbigg λ+⎜radicalBig λ2+r2w⎜parenrightbigg⎜bracketrightbiggl/2−z λ=−l/2 −z −⎜bracketleftbigg ln⎜parenleftbigg λ+⎜radicalBig λ2+(w−rw)2⎜parenrightbigg⎜bracketrightbiggl/2−z λ=−l/2 −z⎜bracerightBigg dz =μ0 4π⎜integraldisplayl/2−rw z=r w−l/2⎡ ⎣ln(l/2−z)+⎜radicalBig (l/2−z)2+r2w (−l/2−z)+⎜radicalBig (−l/2−z)2+r2w −ln(l/2−z)+⎜radicalBig (l/2−z)2+(w−rw)2 (−l/2−z)+⎜radicalBig (−l/2−z)2+(w−rw)2⎤ ⎦dz (4.61) where we used a change of variables λ=z/prime−zand integral 200.01. Using sinh−1x=ln⎜parenleftBig x+⎜radicalbig x2+1⎜parenrightBig =− sinh−1(−x) =− ln⎜parenleftBig −x+⎜radicalbig x2+1⎜parenrightBig (D700.1) (4.61) reduces to Lleft=μ0 4π⎜integraldisplayl/2−rw z=r w−l/2⎜parenleftbigg sinh−1z+l/2 rw+sinh−1l/2−z rw −sinh−1z+l/2 w−rw−sinh−1l/2−z w−rw⎜parenrightbigg dz (4.62) Essentially, we have rederived the equation for Azin (2.57) and then repeated the derivation using Azin Section 4.3.1. The integral of (4.62) was evaluated in (4.13)–(4.15), giving Lleft=μ0 2π⎡ ⎢⎢⎢⎢⎢⎣−(l−rw)sinh−1l−rw w−rw+⎜radicalBig (l−rw)2+(w−rw)2 +(l−rw)sinh−1l−rw rw−⎜radicalBig (l−rw)2+(rw)2 NEUMANN INTEGRAL FOR SELF AND MUTUAL INDUCTANCES 153 +rwsinh−1rw w−rw−⎜radicalBig (rw)2+(w−rw)2 −rwsinh−1rw rw⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ln⎜parenleftbig 1+√ 2⎜parenrightbig+⎜radicalBig (rw)2+(rw)2 ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright√ 2rw⎤ ⎥⎥⎥⎥⎥⎦(4.63) Repeating this for the top, right, and left sides gives Lloop=2Lleft(l,w,r w)+2Ltop(w,l ,r w) (4.64) where we simply interchange landwinLleft(l,w,r w) to obtain Ltop(w,l ,r w). Since the result is identical to that obtained in Section 4.1.1 using B, the inductance of the loop is identical to that obtained in Section 4.1.1: Lloop=2Lleft side (l,w,r w)+2Ltop side (w,l ,r w) =μ0 π⎜bracketleftbigg −(l−rw)sinh−1l−rw w−rw−(w−rw)sinh−1w−rw l−rw +(l−rw)sinh−1l−rw rw+(w−rw)sinh−1w−rw rw +rwsinh−1rw w−rw+rwsinh−1rw l−rw +2⎜radicalBig (l−rw)2+(w−rw)2−2⎜radicalBig (w−rw)2+(rw)2 −2⎜radicalBig (l−rw)2+(rw)2−2rwln⎜parenleftBig 1+√ 2⎜parenrightBig +2√ 2rw⎜bracketrightbigg (4.65) The remaining results in (4.18), (4.19), and (4.20) for a square loop and for loop side lengths greater than the wire radius are obtained from (4.65) and areidentical to those obtained with this method. 4.4.3 Self Inductance of the Circular Loop Applying the Neumann integral in (4.55) to the circular loop in Fig. 4.14 yields L loop=μ0 4π⎜integraldisplay2π φ/prime=0⎜bracketleftBigg⎜integraldisplay2π φ=0cosφ R12(a−rw)dφ⎜bracketrightBigg ad φ/prime(4.66a) 154 THE CONCEPT OF “LOOP” INDUCTANCE where R12=⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφ (4.66b) and the dot product in (4.55) is dl·dl/prime=(a−rw)dφ a dφ/primecosφ (4.66c) Substituting gives Lloop=μ0a(a−rw) 4π⎜integraldisplay2π φ/prime=0⎜integraldisplay2π φ=0cosφ⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφdφ dφ/prime =μ0a(a−rw) 2⎜integraldisplay2π φ=0cosφ⎜radicalBig a2+(a−rw)2−2a(a−rw)cosφdφ(4.67) But this is identical to the result in (4.57) for the mutual inductance between two coaxial loops obtained in Section 4.4.1 if we let the two loops in Fig.4.12 be coincident, d=0, and the radius of the second loop be b=a−r w. Hence, the results for that case given in (4.59) yields the self inductance ofthe loop in Fig. 4.14 and substitute d=0 and b=a−r win that result. But that is identical to (4.27), which, for thin wires, rw/lessmucha, reduces to (4.28). Iy x rwφ 2φ′a12R wr a− φφ ′′ =′ a l da d() φφa l dr a d w− = FIGURE 4.14. Neumann integral and the circular loop. INTERNAL INDUCTANCE VS. EXTERNAL INDUCTANCE 155 4.5 INTERNAL INDUCTANCE VS. EXTERNAL INDUCTANCE Thus far, we have determined the external inductance of a current loop: that is, the inductance due to the magnetic ﬂux that is external to the wires. In Section2.4 we determined the magnetic ﬂux density both external and internal to awire of radius r wthat carries a dc current Ithat is uniformly distributed over the wire cross section. Those results are given in (2.33a) for r>r wand in (2.33b) for r<r w. The magnetic ﬂux internal to the wire also links a portion of the current and gives rise to an internal inductance . Hence, the internal inductance of the wire should be added to the external self-inductances of the current loops that were determined in the previous examples of this chapter as Lloop=Lexternal +Linternal (4.68) We next show that the internal inductance of a wire carrying a dc current I that is uniformly distributed over the wire cross section is Linternal=μ0 8π×wire length (4.69) Hence the per-unit-length internal inductance is μ0/8π=0.5×10−7H/m= 50 nH/m=1.27 nH /in. Usually, this is inconsequential compared to the external inductance. For a wire formed into a circular loop of radius r=a such as those shown in Figs. 4.4, 4.11, and 4.14, the total internal inductanceis (approximately) ( μ 0/8π)2πa. For currents whose frequency is not zero (dc), the current tends to be concentrated increasingly in an annulus of a skindepth, δ, at the surface, where the skin depth is δ=1 √πfμ 0σ where σis the conductivity of the wire material. As f→∞ , the current tends to reside on the surface of the wire and Linternal→0 since no internal current is linked by the ﬁeld. To determine the internal inductance of a wire, consider the cross sec- tion shown in Fig. 4.15. The magnetic ﬂux density internal to the wire wasdetermined in Chapter 2, using Amp `ere’s law, to be B φ=μ0 2πrI⎜parenleftBigg πr2 πr2w⎜parenrightBigg =μ0Ir 2πr2wr<r w (2.33b) 156 THE CONCEPT OF “LOOP” INDUCTANCE rrw IIdrφB FIGURE 4.15. Determining the internal inductance of a wire. An annulus of radius rand thickness drhas a total ﬂux through it, for a unit length along the wire axis, of dψ=μ0Ir 2πr2wdr But this ﬂux links only a portion of the total wire current of Iπr2 πr2w so that the ﬂux linkages for that annulus are dψ=μ0Ir 2πr2wr2 r2wdr =μ0Ir3 2πr4wdr Hence, the total ﬂux linkage per unit length of the wire is ψ=⎜integraldisplayrw r=0μ0Ir3 2πr4wdr =μ0I 8π INTERNAL INDUCTANCE VS. EXTERNAL INDUCTANCE 157 and the per-unit-length internal inductance of the wire is linternal=ψ I =μ0 8πH/m (4.70) EXAMPLE Consider the coaxial cable shown in Fig. 4.5. The inner wire has radius rw and an internal inductance per unit length along the cable of linternal ,wire=μ0 8πH/m (4.71a) as determined above. The internal inductance of the shield which is of interior radius rsand thickness tis determined as follows. In Chapter 2 we determined the magnetic ﬂux density in the shield as Bφ=μ0I 2πr(rs+t)2−r2 (rs+t)2−r2srs<r<r s+t (2.35c) and the dc return current in the shield, – I, is distributed uniformly over the cross section of the shield. Again constructing an annulus at radius rand thickness drwithin the shield, the magnetic ﬂux through the annulus per unit of cable length is dψ=μ0I 2πr(rs+t)2−r2 (rs+t)2−r2sdr r s<r<r s+t But this links only a portion of the total cable current of I−Iπr2−πr2 s π⎜bracketleftBig (rs+t)2−r2s⎜bracketrightBig=I(rs+t)2−r2 (rs+t)2−r2sA rs<r<r s+t Hence, the total ﬂux linkages for this annulus are dψ=μ0I 2πr⎜bracketleftBigg (rs+t)2−r2 (rs+t)2−r2s⎜bracketrightBigg2 dr r s<r<r s+t 158 THE CONCEPT OF “LOOP” INDUCTANCE The total per-unit-length internal inductance of the shield is therefore linternal ,shield =ψ I =⎜integraldisplayrs+t r=rsμ0 2πr⎜bracketleftbigg(rs+t)2−r2 (rs+t)2−r2s⎜bracketrightbigg2 dr =μ0 2π⎜bracketleftbig (rs+t)2−r2s⎜bracketrightbig2⎜integraldisplayrs+t r=rs⎜bracketleftbigg (rs+t)41 r−2r(rs+t)2+r3⎜bracketrightbigg =μ0 2π⎜bracketleftBigg (rs+t)4ln[(r s+t)/rs]−(rs+t)2⎜bracketleftbig (rs+t)2−r2 s⎜bracketrightbig +1 4⎜bracketleftbig (rs+t)4−r4 s⎜bracketrightbig ⎜bracketleftbig (rs+t)2−r2s⎜bracketrightbig2⎜bracketrightBigg H/m (4.71b) To this is added the per-unit-length external inductance determined in (4.29): lexternal =ψ I =μ0I 2πlnrs rwH/m (4.29) Hence the total per-unit-length inductance of the coaxial cable is l=linternal ,wire+lexternal +linternal ,shield H/m (4.72) 4.6 USE OF FILAMENTARY CURRENTS AND CURRENT REDISTRIBUTION DUE TO THE PROXIMITY EFFECT Throughout this chapter and in Chapter 2 we have assumed that the (dc) current Iin a wire was uniformly distributed over the wire cross section. Hence, we were able to represent the wire current as a ﬁlament of current on the axis ofthe wire, thereby simplifying the computations. If there are no other currents in close proximity, this will be the case. However, if another current is within a few radii of this wire, the currentover the wire cross section will not be distributed uniformly but will tend tobe concentrated toward the side facing the other wire. This phenomenon iscalled the proximity effect . If that is the case, the previous results for the B ﬁeld in Chapter 2 as well as the inductances associated with the wire (both USE OF FILAMENTARY CURRENTS AND CURRENT REDISTRIBUTION 159 external and internal inductances) in this chapter will be only approximately correct and will become less so the closer the wires are spaced. Typically, theproximity effect does not substantially alter the results that were obtained byassuming that the current is uniformly distributed over the wire cross sectionif the separation of the two wires is greater than approximately four wire radii,as we will see. In other words, one wire of the same radius as the other twocould be placed exactly between the two wires. 4.6.1 Two-Wire Transmission Line To demonstrate this dependence, consider a two-wire transmission line con- sisting of two wires of equal radii, r w, carrying equal but oppositely directed currents and separated by a distance (center to center) of sas illustrated in Fig. 4.16(a). The wires are considered inﬁnitely long (or at least very longcompared with their radii) so that we will not have to deal with fringing ofthe ﬁelds at the endpoints of ﬁnite-length wires. For the widely spaced wires wr wr sI– I (a) D (b)B B I –I FIGURE 4.16. Proximity effect in a two-wire transmission line. 160 THE CONCEPT OF “LOOP” INDUCTANCE shown in Fig. 4.16(a), the current is distributed uniformly over the wire cross sections so that the current may be replaced by ﬁlaments on the wire axes.Hence, the Bﬁelds of each wire form circles that are centered on the centers of the respective wires. The magnetic ﬂux density about each wire is B φ=μ0I 2πr where ris measured from the center of each wire. Constructing a ﬂat surface between the adjacent surfaces, the total ﬂux through the surface per unit of itslongitudinal length is ψ=2⎜integraldisplays−rw rwμ0I 2πrdr =μ0I πlns−rw rw ∼=μ0I πlns rw We have made the approximation that s−rw∼=ssince the wires are assumed to be widely spaced. Hence, the approximate per-unit-length inductance ofthe line for widely spaced wires which assumes a uniform current distributionover the wire cross sections is lapproximate =μ0 πlns rwH/m (4.73) If the wires are closely spaced as shown in Fig. 4.16(b), the currents will be concentrated toward the facing sides, and the result above for the per-unit-length inductance in (4.73) is an approximation since that relied on the currentsbeing uniformly distributed over the wire cross sections. It can be shown (see[3,8]) that the magnetic ﬁelds are as though the total currents are concentratedas ﬁlaments but separated by a distance D≤sas shown in Fig. 4.16(b). The exact per-unit-length inductance for this result can be shown to be [3,8] lexact=μ0 πln⎡ ⎣s 2rw+⎜radicalBigg⎜parenleftbiggs 2rw⎜parenrightbigg2 −1⎤ ⎦ H/m (4.74) Observe that (4.74) reduces to (4.73) if s/greatermuch2rw. USE OF FILAMENTARY CURRENTS AND CURRENT REDISTRIBUTION 161 Effect of wire separation on nonuniform current distributionRatio= approx/exact s/rw2.5 2 1.5 12 3 4 5 6 7 8 FIGURE 4.17. Ratio of the approximate and exact per-unit-length inductances of a two-wire transmission line as a function of the ratio of separation to wire radius. Figure 4.17 shows a plot of the ratio R=lapproximate lexact =ln(s/rw) ln⎜bracketleftbigg (s/2rw)+⎜radicalBig (s/2rw)2−1⎜bracketrightbigg for ratios of wire separation to wire radius between 2.1 and 8: 2 .1< s/r w≤8. (Note: For a ratio of s/rw=2, the two wires would be touching.) For a ratio ofs/rw=4, the error is 5.3%. 4.6.2 One Wire Above a Ground Plane The results for the two-wire transmission line can be extended rather easily to cover the case of a transmission line consisting of one wire at a height habove an inﬁnite and perfectly conducting ground plane, as shown in Fig. 4.18(a).The basic idea is to use the method of images discussed in Section 2.7 toreplace the ground plane with the image of the current as shown in Fig. 4.18(b).The image of the current above the ground plane is the same but with thecurrent direction reversed and at a distance hbelow the position of the ground plane but with the ground plane removed. All the ﬁelds above the position ofthe ground plane remain the same in the image problem of Fig. 4.18(b). Note 162 THE CONCEPT OF “LOOP” INDUCTANCE wr wrh hground planeI –I (a) ground plane (b)Loop AreahI h –IφB FIGURE 4.18. Transmission line consisting of one wire above and inﬁnite and perfectly conducting ground plane. that the method of images also applies to the case where the current is not distributed uniformly over the wire cross section. Now we have an equivalent problem of a two-wire transmission line where the separation between the two wires is s=2h. In the original problem of one wire above a ground plane, the loop area for a 1-m length of the line is betweenthe surface of the wire and the ground plane, whereas the equivalent surface forthe image problem is between the surfaces of the wire and its image. Hence,the per-unit-length inductance of the problem of one wire above a groundplane is one-half the value of the per-unit-length inductance of the two-wirebut with sreplaced by s=2h. Therefore, the per-unit-length inductance of ENERGY STORAGE METHOD FOR COMPUTING LOOP INDUCTANCE 163 the transmission line consisting of one wire above a ground plane is lapproximate =μ0 2πln2h rwH/m (4.75) and lexact=μ0 2πln⎡ ⎣h rw+⎜radicalBigg⎜parenleftbiggh rw⎜parenrightbigg2 −1⎤ ⎦ H/m (4.76) Figure 4.17, demonstrating the impact of proximity effect and current redis- tribution, applies to this case, but the horizontal axis is h/r w, which varies f r o m1t o4 . The internal inductances of the wires can then be added to all these external inductances to give the total per-unit-length inductances of the lines. In thecase of the two-wire line, we add l internal=2(μ 0/8π)=μ0/4πH/m, and in the case of one wire above a ground plane we add linternal=μ0/8π H/m. 4.7 ENERGY STORAGE METHOD FOR COMPUTING LOOP INDUCTANCE In Section 3.6 we obtained the result that the magnetic energy stored in the magnetic ﬁeld is WM=1 2⎜integraldisplay vB·Hdv =μ0 2⎜integraldisplay vH2dv =1 2μ0⎜integraldisplay vB2dv (4.77) andvis the volume of space containing the magnetic ﬁeld. From a circuit analysis standpoint, the energy stored in an inductance is WM=1 2LI2(4.78) 164 THE CONCEPT OF “LOOP” INDUCTANCE Hence, we can determine the inductance in terms of the stored magnetic ﬁeld from L=2WM I2 =μ0 I2⎜integraldisplay vH2dv =1 μ0I2⎜integraldisplay vB2dv(4.79) 4.7.1 Internal Inductance of a Wire In Chapter 2 we determined the magnetic ﬁelds both inside and outside a wire of radius rwthat contained a current Ithat is distributed uniformly over the wire cross section. Hence we assumed that the current is dc and there are noother currents in close proximity to disturb this uniform distribution. Thosemagnetic ﬁelds at a radius rare directed circumferentially in the φdirection about the wire axis: B φ=⎧ ⎪⎪⎨ ⎪⎪⎩μ0Ir 2πr2w0<r<r w (2.33b) μ0I 2πrrw<r (2.33a) The per-unit-length internal inductance of the wire is obtained from (4.79) using (2.33b) by integrating throughout a cylindrical volume of unit lengthwithin the wire as l internal=1 μ0I2⎜integraldisplay1m z=0⎜integraldisplay2π φ=0⎜integraldisplayrw r=0⎜parenleftBigg μ0Ir 2πr2w⎜parenrightBigg2 rd φd rd z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright dv =2πμ0 4π2r4w⎜integraldisplayrw r=0r3dr =μ0 2πr4w⎜bracketleftBigg r4 4⎜bracketrightBiggr=rw r=0 =μ0 8πH/m (4.80) which was obtained directly by the ﬂux linkage method in Section 4.5. ENERGY STORAGE METHOD FOR COMPUTING LOOP INDUCTANCE 165 4.7.2 Two-Wire Transmission Line The per-unit-length external inductance of a two-wire transmission line con- sisting of two identical wires of radii rwwith center-to-center separation sas shown in Fig. 4.16(a) was obtained in Section 4.6.1. Again we assume thatthe wire separation is sufﬁciently large, s/greatermuchr w, so that the current of each wire remains distributed uniformly (or approximately so) over the wire crosssection. We obtain the total per-unit-length inductance by integrating (4.79)throughout a cylindrical volume of unit length using the results for the Bﬁelds inside and outside the wires given in (2.33a) and (2.33b): l=21 μ0I2⎜integraldisplay1m z=0⎜integraldisplay2π φ=0⎜integraldisplayrw r=0⎜parenleftBigg μ0Ir 2πr2w⎜parenrightBigg2 rdφ dr dz⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright dv⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright linternal +21 μ0I2⎜integraldisplay1m z=0⎜integraldisplay2π φ=0⎜integraldisplays−rw r=rw⎜parenleftbiggμ0I 2πr⎜parenrightbigg2 rdφ dr dz⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright dv⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright lexternal =22πμ 0 4π2r4w⎜integraldisplayrw r=0r3dr ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright linternal+22πμ0 4π2⎜integraldisplays−rw r=rw1 rdr ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright lexternal =2μ0 2πr4w⎜bracketleftBigg r4 4⎜bracketrightBiggrw r=0⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright linternal+μ0 π[lnr]s−rw r=rw ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright lexternal =2μ0 8π⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright linternal+μ0 πlns−rw rw⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright lexternalH/m (4.81) 4.7.3 Coaxial Cable Finally, we obtain the per-unit-length inductance of a coaxial cable shown in Fig. 2.19 consisting of an inner wire of radius rwand an overall shield of inner radius rsand thickness t. The dc current of the inner wire returns in the shield. Observe that because of symmetry there is no proximity effect regardless ofthe spacing of the conductors. The magnetic ﬁelds of the cable were derived 166 THE CONCEPT OF “LOOP” INDUCTANCE in Chapter 2 and are Bφ=⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎩μ0Ir 2πr2wr<r w (2.35b) μ0I 2πrrw<r<r s (2.35a) μ0I 2πr(rs+t)2−r2 (rs+t)2−r2srs<r<r s+t (2.35c) Integrating (4.79) over a differential volume of unit length, dv=rd φd rd z , throughout the appropriate regions gives linternal wire =μ0 8πH/m (4.82a) lexternal =1 μ0I2⎜integraldisplay1m z=0⎜integraldisplay2π φ=0⎜integraldisplayrs r=rw⎜parenleftbiggμ0I 2πr⎜parenrightbigg2 rd φ d r d z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright dv =2πμ0 4π2⎜integraldisplayrs r=rw1 rdr=μ0 2π[lnr]rs r=rw =μ0 2πlnrs rwH/m (4.82b) linternal ,shield=1 μ0I2⎜integraldisplay1m z=0⎜integraldisplay2π φ=0⎜integraldisplayrs+t r=rs⎜parenleftBigμ0I 2πr⎜parenrightBig2⎜bracketleftbigg (rs+t)2−r2 (rs+t)2−r2 s⎜bracketrightbigg2 rd φ d r d z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright dv =2πμ 0 4π2⎜integraldisplayrs+t r=rs1 r⎜bracketleftbigg (rs+t)2−r2 (rs+t)2−r2 s⎜bracketrightbigg2 dr =μ0 2π⎜bracketleftBigg (rs+t)4ln[(r s+t)/rs]−(rs+t)2⎜bracketleftbig (rs+t)2−r2 s⎜bracketrightbig +1 4⎜bracketleftbig (rs+t)4−r4 s⎜bracketrightbig ⎜bracketleftbig (rs+t)2−r2 s⎜bracketrightbig2⎜bracketrightBigg H/m (4.82c) But the result in (4.82c) is the same result as obtained in (4.71b) by the method of ﬂux linkages. LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS 167 4.8 LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS Figure 4.19(a) shows ncurrent-carrying loops which are in close proximity such that their magnetic ﬁelds interact with each other so that the loops aresaid to be coupled . This structure can be characterized by self and mutual I1 Ii Ins1 si sn (a) I1 Ii In (b)11L iiL nnLiM1 nM1inM FIGURE 4.19. Multiloop coupled structure. 168 THE CONCEPT OF “LOOP” INDUCTANCE inductances as shown in Fig. 4.19(b). Denote the ﬂuxes through each loop asψ1,...,ψ i,...,ψ n. These ﬂuxes are related to the currents of each loop, I1,...,I i,...,I nas ψ1=L11I1+···+ M1iIi+···+ M1nIn ... ψn=Mn1I1+···+ MniIi+···+ LnnIn(4.83) andMij=Mji. The terms Liiare the self inductances of the current loops, and the Mijare the mutual inductances between the current loops. These individual inductances can be obtained with the methods of this chapter bysetting all but one of the currents in each equation of (4.83) to zero: Lii=ψi Ii⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I i−1=Ii+1=···=I n=0(4.84a) Mij=ψi Ij⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I j−1=Ij+1=···=I n=0(4.84b) The equations in (4.83) can be placed in matrix form as ψ=LI (4.85a) where ψ=⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣ψ1 ... ψi ... ψn⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (4.85b) L=⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣L11···M1i···M1n ............... M i1...Lii...Min ............... M n1···Mni···Lnn⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (4.85c) LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS 169 I=⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣I1 ... Ii ... In⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦(4.85d) Then×nmatrix Lis said to be the inductance matrix for this coupled set of current loops. Assuming that the entire structure is electrically small at thefrequencies of the currents, the Faraday law voltages induced into each loop(see Fig. 4.1) are obtained by differentiating (4.85) to give V(t)=LdI(t) dt(4.86a) where the n×1 vector of induced voltages is V=⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣V1 ... Vi ... Vn⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (4.86b) 4.8.1 Dot Convention It is important to review the dot convention for computing the contributions to the voltages across each inductance that are due to the mutual inductances be-tween that loop and the other loops [1,2]. Figure 4.20 illustrates this. The dotsare placed on the individual inductors in order to give the relative orientationsof the loops with respect to each other. For example, consider the two coupledcurrent-carrying loops (shown for simplicity as being rectangular) shown inFig. 4.20. The currents around each loop, I 1andI2, are arbitrarily chosen to be in the clockwise direction around those loops, and the directions of the ﬂuxesthrough each loop, ψ 1andψ2, are arbitrarily chosen to be into the page for loop 1 and out of the page for loop 2. Using the right-hand rule, we see that ψ1=L11I1−M12I2 (4.87a) and ψ2=−L22I2+M12I1 (4.87b) (which you should verify using the right-hand rule) where L11,L22,and M12=M21are positive numbers. 170 THE CONCEPT OF “LOOP” INDUCTANCE 21 12M M= 1s2s11L 22L1I2I 1V2V1ψ2ψ dtdIMdtdIL V2 121 11 1 − =1 12 2 22 2I M I L + − = ψ dtdIMdtdIL V1 122 22 2 + − =2 12 1 11 1I M I L − = ψ2I 1ILoop 1 Loop 2 FIGURE 4.20. Dot convention. Next, we label the inductors with dots and choose the voltage polarities for the voltages, V1andV2, across those inductances as shown. These voltages are the Faraday law voltage sources that are induced into each loop (seeFig. 4.1). Each current contributes to each voltage, so we initially set up therelation V 1=(?)L11I1+(?)M12I2 V2=(?)M12I1+(?)L2I2(4.88) and (?) denotes the signs that are to be determined. We determine the signs of each of the terms according to the following rules. For each loop we write theequation for the induced voltage in that loop as the sum of a self term and amutual term, with the signs of each being determined by the following [1,2]: 1. The sign of the self-inductance term is positive if the current of that loop enters the assumed positive or +terminal of the voltage for that loop (the passive sign convention [1,2]). For loop 1, I 1enters the +terminal ofV1, so the sign of the self inductance contribution, L11I1, is positive. For loop 2, I2enters the negative or −terminal of V2, so the sign of the self inductance contribution, L22I2,i sn e g a t i v e . 2. The mutual inductance contribution to the voltage of a loop is positive at the dotted end of that loop inductance if the current of the other loop enters the dotted end of the inductance of that loop. Otherwise, it isnegative. For example, the current of loop 2, I 2, enters the undotted LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS 171 end of its inductance. Hence, it produces a contribution to the voltage of the ﬁrst loop, M12I2, that is positive at the undotted end of V1and is therefore entered as a negative contribution to the equation for V1. Hence, the equation for the voltage of loop 1 is V1=L11I1−M12I2 (4.89a) The current of loop 1 enters the dotted end of its inductance. Hence, it produces a contribution to the voltage of the second loop, M12I1, that is positive at the dotted end of V2and is entered as a positive contribution to the equation for V2. Hence, the equation for the voltage of loop 2 is V2=−L22I1+M12I2 (4.89b) 4.8.2 Multiconductor Transmission Lines These concepts are very useful in constructing transmission-line equations characterizing multiconductor transmission lines (MTLs) [8]. Solving thoseMTL equations allows the prediction of crosstalk, which is the unintended coupling of a signal from one current loop into another current loop [8]. For example, consider the MTL shown in Fig. 4.21(a) consisting of n+1 parallel conductors of inﬁnite (or very long) length. The ( n+1)st conduc- tor serves as the return for all the other currents. The currents of all con- ductors are directed to the right (in the zdirection parallel to the conductor axes). The current of the ( n+1)st conductor is therefore I n+1=−⎜summationtextn i=1Ii= −(I1+···+ Ii+···+ In). Each of the ncurrents therefore forms a loop between that current and the (n +1)st conductor. Hence, the ﬂuxes of each loop,ψi(assumed arbitrarily to be directed into the page) can be related to the currents with a per-unit-length inductance matrix as ψ=LI (4.90a) where the n×nmatrix of per-unit-length inductances (denoted as lowercase) is L=⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣l11···m1i···m1n ............... m i1...lii...min ............... m n1···mni···lnn⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ H/m (4.90b) 172 THE CONCEPT OF “LOOP” INDUCTANCE 1I nI ∑ =n iiI 1 (a) 1I nI ∑ =n iiI 1 (b)11lzΔ Δz zΔ zΔ nnl nsnm1z nψ FIGURE 4.21. Multiconductor transmission line. andmij=mji. The per-unit-length equivalent circuit of a /Delta1zlength of line is shown in Fig. 4.21(b). From this we can determine the voltages across eachof the ninductors with the +terminal assumed at the dotted end as V(t)=L/Delta1zdI(t) dt(4.91) To this circuit are added the per-unit-length self and mutual capacitances between the n+1 conductors from which the MTL equations are derived LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS 173 and whose solution can be used to predict crosstalk between the ncircuits (loops) [8]. In the following subsections we determine approximate relations for the per-unit-length self and mutual inductances of MTLs that are composed of n wires. To make our calculations feasible, we assume that all wires are “widelyspaced,” so that the current of each wire is distributed uniformly over the wirecross section. In other words, the wires are separated sufﬁciently, so that theproximity effect is not pronounced. As we saw in Section 4.6, this will be agood approximation as long as the ratio of wire separation to wire radius islarger than about 4 : 1. This is not an unduly restrictive assumption since itmeans that one wire can just be placed between two other wires of the sameradii, so that wires separated by this ratio are rather “closely spaced.” Withthis assumption of “widely spaced” wires we can replace the current of eachwire with a ﬁlament on its axis containing the total current of the wire. The n wires are assumed to be inﬁnite in length to avoid having to deal with fringingof the ﬁeld at the endpoints of a ﬁnite-length line. Therefore, the magnetic ﬂuxdensity of each wire is in the circumferential or φdirection about the wire, and we obtain the familiar result for the magnetic ﬂux density at a radius r about an inﬁnitely long wire that has a uniform current distribution over itscross section that was obtained in Chapter 2: B φ=μ0I 2πr(2.14) In determining the per-unit-length inductances of the line, we determine the total magnetic ﬂux through a surface by using superposition to give the totalcontribution from all wires of the line. Using the result in (2.14), we can develop a useful wide-separation approx- imation for calculating the total magnetic ﬂux through a surface. Consider theproblem shown in Fig. 4.22(a) of an isolated wire where we wish to calculatethe total magnetic ﬂux through a tilted surface swhose edges are at radii R 1andR2from the wire axis with R2>R 1. The total per-unit-length ﬂux through this surface for R2>R 1is in the direction indicated through surface sand is obtained from ψ=⎜integraldisplay sB·ds But as shown in the ﬁgure, this is a difﬁcult calculation because the magnetic ﬂux density Bφis not perpendicular to the surface s, so that the dot product cannot be removed from the integrand. However, consider the closed “wedge- shaped” surface that is1mi nlength into the page and has a side s(the original side through which the ﬂux is desired), a side s2that is at a constant radius 174 THE CONCEPT OF “LOOP” INDUCTANCE φB ψ 2R1R Iwr1s2s sφB (a) Id Iw1r w2r 1ψ2ψ (b) FIGURE 4.22. Fundamental problem for determining ﬂux through a surface. r=R2from the wire, a side s1that extends radially fromr=R1tor=R2, and two “end caps.” Gauss’s law provides that the total ﬂux leaving thisclosed surface is ⎜contintegraltextB·ds=⎜integraldisplay sB·ds+⎜integraldisplay s1B·ds+⎜integraldisplay s2B·ds+⎜integraldisplay end capsB·ds =0 From the ﬁgure we see that ⎜integraldisplay s2B·ds=0 because the magnetic ﬂux density Bφis tangent to this side. Similarly, we see that ⎜integraldisplay end capsB·ds=0 because the magnetic ﬂux density Bφis tangent to these sides. Hence, we obtain the important result that ⎜integraldisplay sB·ds=−⎜integraldisplay s1B·ds =⎜integraldisplay1m z=0⎜integraldisplayR2 r=R 1Bφdr LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS 175 =⎜integraldisplay1m z=0⎜integraldisplayR2 r=R 1μ0I 2πrdr =μ0I 2πlnR2 R1(4.92) and the dot product may be removed from the integrand since Bφis perpen- dicular to surface s1. In addition, when the surface is between two wires that are separated center to center by distance d, we would integrate from r=rwtor=d−rw∼=d, which results from our assumption that d/greatermuchrw. This is illustrated in Fig. 4.22(b). By superposition the total ﬂux through the ﬂat surface between theinterior edges of the two wires is the sum of the ﬂuxes due to each current: ψ=ψ 1+ψ2 =μ0I 2πlnd−rw2 rw1+μ0I 2πlnd−rw1 rw2 ∼=μ0I 2πlnd rw1+μ0I 2πlnd rw2 =μ0I 2πlnd2 rw1rw2 since we must assume that d/greatermuchrw1,rw2in order for the current to be uniformly distributed over the wire cross sections and for (2.14) to apply. Lines Composed of n+1Wires Consider the case of n+1 wires of radii rwithat are parallel to each other as shown in cross section in Fig. 4.23(a). The (n+1)st conductor through which the other ncurrents “return” is denoted as the zeroth conductor. Figures 4.23(b) and (c) show the calculation of theper-unit-length self and mutual inductances of the line: lii=ψi Ii⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I i−1=Ii+1=···=I n=0 =μ0 2πlndi0 rw0+μ0 2πlndi0 rwi =μ0 2πlnd2 i0 rw0rwi(4.93a) 176 THE CONCEPT OF “LOOP” INDUCTANCE 0w0ri 0idijd irw j jjrw (a) 00 wri 0idirw jrw (b) w0r0idijd irw jrw (c)0=jI iI iI jI jI0=iIiψ iψ 0jdi j 0 FIGURE 4.23. (n+1) wires. LOOP INDUCTANCE MATRIX FOR COUPLED CURRENT LOOPS 177 and lij=lji=ψi Ij⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I j−1=Ij+1=···=I n=0 =μ0 2πlndj0 dij+μ0 2πlndi0 rw0 =μ0 2πlndi0dj0 dijrw0 (4.93b) Lines Composed of nWires Above an Inﬁnite Ground Plane Figure 4.24 shows the case of nparallel wires of radii rwisituated at heights hiabove an inﬁnite “ground plane” which is designated as the zeroth conductor throughwhich all the other ncurrents “return.” Replacing the ground plane with the images of the currents according to Section 2.8 allows calculation of the irwjrw ih ihjh jhijs s′ s′s′ ′0=jI iψjψ 0iI iIij FIGURE 4.24. nwires above a ground plane. 178 THE CONCEPT OF “LOOP” INDUCTANCE per-unit-length self and mutual inductances as shown: lii=ψi Ii⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I i−1=Ii+1=···=I n=0 =μ0 2πlnhi rwi+μ0 2πln2hi hi =μ0 2πln2hi rwi(4.94a) and lij=lji=ψj Ii⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I i−1=Ii+1=···=I n=0 =μ0 2πlns/prime sij+μ0 2πlns/prime/prime s/prime =μ0 2πln⎜radicalBig s2 ij+4hihj sij =μ0 4πln⎜bracketleftBigg 1+4hihj s2 ij⎜bracketrightBigg (4.94b) Lines Composed of nWires Within an Overall Shield Figure 4.25(a) shows the case of nparallel wires of radii rwiwithin an overall circular shield of interior radius rswhich is designated as the zeroth conductor through which all the other ncurrents “return.” We can replace the shield with the wire images that are located at radii r2 s/difrom the axis of the shield [10]. This allows calculation of the per-unit-length self and mutual inductances as shownin Fig. 4.25(b) [8]: lii=ψi Ii⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I i−1=Ii+1=···=I n=0 =μ0 2πlnrs−di rwi+μ0 2πlnr2 s/di−di r2s/di−rs =μ0 2πlnr2 s−d2 i rsrwi(4.95a) LOOP INDUCTANCES OF PRINTED CIRCUIT BOARD LANDS 179 irw isr jrw j ijθiψ jψid dj (a) irwsr jrw jiψ jψid dj (b)irwis dr2 js dr2 iI iI0 0 jrwi FIGURE 4.25. nwires within an overall shield. and lij=lji=ψj Ii⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle I1=···=I i−1=Ii+1=···=I n=0 =μ0 2πln⎛ ⎝dj rs⎜radicaltp⎜radicalvertex⎜radicalvertex⎜radicalbt⎜parenleftbigdidj⎜parenrightbig2+r4 s−2didjr2 scosθij⎜parenleftbigdidj⎜parenrightbig2+d4 j−2did3 jcosθij⎞ ⎠(4.95b) 4.9 LOOP INDUCTANCES OF PRINTED CIRCUIT BOARD LANDS So far we have concentrated on conductors having circular, cylindrical cross sections (i.e., wires). Finally, we turn our attention to transmission lines that are 180 THE CONCEPT OF “LOOP” INDUCTANCE constructed of conductors that have rectangular cross sections. These appear on printed circuit boards (PCBs) and are referred to as lands . The calcula- tion of the loop inductances of structures composed of lands is considerablymore difﬁcult than for wires, and only approximate relations are generallyobtained. Figure 4.26 shows three common conﬁgurations used in constructing PCBs. Figure 4.26(a) shows the stripline that appears in PCBs that contain inner- planes. Innerplanes are layers of conductors sandwiched at various levelswithin the board substrate, which is glass epoxy, a dielectric with ε r∼=4.7 andμr=1. Hence the board substrate is not ferromagnetic and does not affect the magnetic ﬁelds (but does affect the electric ﬁelds). A land of thick-ness tis situated between two “ground planes.” The thickness of the land is commonly that of 1-oz copper, which is t=1.4 mils =0.036 mm (1 mil = 0.001 in). However, in the following results it is assumed that t=0. The sep- aration between the two surrounding ground planes is denoted as s, and the land is situated midway between the two ground planes (as is common). The FIGURE 4.26. (a) Stripline; (b) microstrip line; (c) PCB lands. LOOP INDUCTANCES OF PRINTED CIRCUIT BOARD LANDS 181 per-unit-length loop inductance of the stripline is [8] l=⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩30 v0ln⎜bracketleftBigg 21+√ k 1−√ k⎜bracketrightBigg 1√ 2≤k≤1 30π2 v0ln⎜bracketleftBig 2⎜parenleftBig 1+√ k/prime⎜parenrightBig /⎜parenleftBig 1−√ k/prime⎜parenrightBig⎜bracketrightBig 0≤k≤1√ 2H/m (4.96a) where k=1 cosh (πw/2s)(4.96b) andk/prime=√ 1−k2. This can be approximated as [8] l∼=30π v01 we/s+0.441H/m (4.96c) and the effective width of the conductor is we s=⎧ ⎪⎪⎨ ⎪⎪⎩w sw s≥0.35 w s−⎜parenleftbigg 0.35−w s⎜parenrightbigg2w s≤0.35(4.96d) The speed of light is denoted as v0∼=3×108m/s. Themicrostrip line shown in Fig. 4.26(b) is typical of the outer layers of a PCB that has innerplanes. A land of thickness tlies on top of a dielectric substrate of thickness h, and a ground plane (representing an adjacent inner- plane) is below the substrate. Assuming that the thickness of the land is zero,t=0, approximate relations for the per-unit-length loop inductance are [8] l=⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩60 v0ln⎜parenleftbigg8h w+w 4h⎜parenrightbiggw h≤1 120π v0⎜bracketleftbiggw h+1.393+0.667 ln⎜parenleftbiggw h+1.444⎜parenrightbigg⎜bracketrightbigg−1w h≥1H/m (4.97) 182 THE CONCEPT OF “LOOP” INDUCTANCE Finally, the case of two lands on the surface of a PCB is shown in Fig. 4.26(c). The approximate per-unit-length loop inductance is [8] l=⎧ ⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎩120 v0ln⎜parenleftBigg 21+√ k 1−√ k⎜parenrightBigg 1√ 2≤k≤1 120π2 v0ln⎜bracketleftBig 2(1+√ k/prime)/(1−√ k/prime)⎜bracketrightBig 0≤k≤1√ 2H/m (4.98a) where k=s s+2w(4.98b) and k/prime=⎜radicalbig 1−k2 (4.98c) 4.10 SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE There are several methods for computing the Bﬁeld of currents: the Biot– Savart law, Amp `ere’s law, the vector magnetic potential A, and the method of images for problems with ground planes. There are also several methods forcomputing the loop inductance of a closed current loop, but all these methodsfundamentally require computation of the ﬂux through the open surface sthat is enclosed by the closed current loop: ψ=⎜integraldisplay sB·ds (4.99a) The loop inductance is computed from this result as L=ψ I(4.99b) We investigated several methods in this chapter for calculating Leither directly or indirectly. The direct method is to use the Biot-Savart law: B=μ0I 4π⎜integraldisplay ldl×aR R2(4.100a) SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE 183 or Amp `ere’s law for problems with symmetry: ⎜contintegraldisplay cB·dl=μ0I (4.100b) to compute Bover the surface enclosed by the current loop, and then to compute the inductance of the loop via (4.99a) and (4.99b). The next method is to compute the vector magnetic potential Adirectly from A=μ0I 4π⎜integraldisplay l1 Rdl (4.101a) Substituting B=∇ ×A (4.101b) into (4.99) yields L=⎜contintegraltext cA·dl I(4.101c) where cis the contour around the open surface that the current loop surrounds. The third method is via the Neumann integral. Substituting (4.101a) into (4.101c) yields L=μ0 4π⎜contintegraldisplay c⎜contintegraldisplay c/primedl·dl/prime R12(4.102) A fourth indirect method is by computing the energy stored in the magnetic ﬁeld: WM=1 2LI2 =1 2⎜integraldisplay vB·Hdv (4.103a) giving L=1 μ0I2⎜integraldisplay vB2dv (4.103b) Generally, this energy method works best for closed structures where the magnetic ﬁeld is contained within a ﬁnite region of space as with a coaxialcable or in computing the internal inductance of a wire. 184 THE CONCEPT OF “LOOP” INDUCTANCE For some structures, such as the circular current loop, the vector magnetic potential method in (4.101) and the Neumann integral in (4.102) are easiest,whereas for rectangular current loops, the direct method of computing themagnetic ﬂux through the loop in (4.99) is somewhat simpler. All of thesemethods can be applied in a similar fashion to the computation of the mutualinductance between two closed current loops: M12=ψ2 I1(4.104) 4.10.1 Mutual Inductance Between Two Rectangular Loops We ﬁnally illustrate all three methods by computing the mutual inductance between two rectangular loops that lie in the same plane and whose sidesare either parallel or perpendicular as shown in Fig. 4.27. We consider theloops to be composed of wires which can be approximated by ﬁlaments onthe axes of those wires on the assumption that the currents of the wires are 1l2l 1m2m ms ls 1 LoopLoop 2 1I1I 1I 1I FIGURE 4.27. Computation of the mutual inductance between two rectangular loops. SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE 185 uniformly distributed over their cross sections (i.e., all of the parallel wires are “widely spaced”). Loop 1 carries a current I1that circulates about that loop in the clockwise direction. We ﬁrst compute the mutual inductance between thetwo loops from the fundamental deﬁnition of mutual inductance in (4.104) bycomputing the total magnetic ﬂux penetrating the surface enclosed by loop 2: ψ 2=⎜integraldisplay s2B·ds (4.105) where s2is the open surface enclosed by loop 2. Using the Biot–Savart law, we see that the magnetic ﬂux density Bis perpendicular to the surface of loop 2, and hence the dot product in (4.105) can be removed: ψ2=⎜integraldisplay s2Bd s (4.106) Note that the total magnetic ﬂux through loop 2 can be obtained as the super- position of the ﬂuxes through that loop, due to each of the four currents ofthe four sides comprising loop 1. Hence, we essentially need ﬁrst to solve thefundamental problem shown in Fig. 4.28 of determining the total magneticﬂux through a rectangular loop due to a current ﬁlament of length Land then use that fundamental result and the right-hand rule to superimpose the ﬂuxesdue to the currents of the four sides of loop 1. The magnetic ﬂux density ofa line current was determined in (2.15). This was for the origin located at the B r2r1 1Z2Z I Lz FIGURE 4.28. Fundamental subproblem. 186 THE CONCEPT OF “LOOP” INDUCTANCE midpoint of the current. We modify that result for the origin at the lower end of the current ﬁlament, giving B=μ0I 4πr⎡ ⎣Z√ Z2+r2−Z−L⎜radicalBig (Z−L)2+r2⎤ ⎦ (4.107) Hence, the magnetic ﬂux through the loop due to this current ﬁlament is ψ=μ0I 4π⎜integraldisplayr2 r=r1⎜integraldisplayZ2 Z=Z 11 r⎡ ⎣Z√ Z2+r2−Z−L⎜radicalBig (Z−L)2+r2⎤ ⎦dZ dr (4.108) The inner integral with respect to Zis evaluated as (I)=⎜integraldisplayZ2 Z=Z 1⎡ ⎣Z√ Z2+r2−Z−L⎜radicalBig (Z−L)2+r2⎤ ⎦dZ Using an integral from Dwight [7], ⎜integraldisplayx√ x2+a2dx=⎜radicalbig x2+a2 (D201.01) gives (I)=⎜radicalBig Z2 2+r2−⎜radicalBig Z2 1+r2−⎜radicalBig (Z2−L)2+r2+⎜radicalBig (Z1−L)2+r2 where we have used a change of variables λ=Z−L,dλ=dZin the second part of the integral. The second integral with respect to ris (II)=⎜integraldisplayr2 r=r1⎡ ⎣⎜radicalBig Z2 2+r2 r−⎜radicalBig Z2 1+r2 r −⎜radicalBig (Z2−L)2+r2 r+⎜radicalBig (Z1−L)2+r2 r⎤ ⎦dr This can be evaluated using an integral from Dwight [7]: ⎜integraldisplay√ x2+a2 xdx=⎜radicalbig x2+a2−alna+√ x2+a2 x =⎜radicalbig x2+a2−aln⎜parenleftBig a+⎜radicalbig x2+a2⎜parenrightBig +alnx(D241.01) SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE 187 giving the ﬂux through the loop as ψ=μ0I 4π⎜bracketleftbigg⎜radicalBig Z2 2+r2 2−Z2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 2⎜parenrightbigg −⎜radicalBig Z2 2+r2 1 +Z 2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 1⎜parenrightbigg −⎜radicalBig Z2 1+r2 2+Z1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 2⎜parenrightbigg +⎜radicalBig Z2 1+r2 1−Z1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 1⎜parenrightbigg −⎜radicalBig (Z2−L)2+r2 2 +(Z2−L)ln⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 2⎜parenrightbigg +⎜radicalBig (Z2−L)2+r2 1 −(Z2−L)ln⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 1⎜parenrightbigg +⎜radicalBig (Z1−L)2+r2 2 −(Z1−L)ln⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 2⎜parenrightbigg −⎜radicalBig (Z1−L)2+r2 1 +(Z1−L)ln⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 1⎜parenrightbigg⎜bracketrightbigg (4.109a) This can be written more compactly as ψ=μ0I 4πK(Z1,Z2,r1,r2,L) (4.109b) where K(Z1,Z2,r1,r2,L)=⎜summationdisplay 2 i=1⎜summationdisplay 2j=1(−1)i+j⎜bracketleftbigf⎜parenleftbigZi,rj,0⎜parenrightbig−f⎜parenleftbigZi,rj,L⎜parenrightbig⎜bracketrightbig (4.109c) and f(Z, r, L )=⎜radicalBig (Z−L)2+r2−(Z−L)ln⎜bracketleftbigg (Z−L)+⎜radicalBig ((Z−L)2+r2⎜bracketrightbigg (4.109d) Hence, by superimposing the magnetic ﬂuxes through loop 2 in Fig. 4.27 due to each of the four sides of loop 1 (using the right-hand rule and matchingeach case to Fig. 4.28), we obtain the mutual inductance between the two 188 THE CONCEPT OF “LOOP” INDUCTANCE rectangular loops in Fig. 4.27 as M12=μ0 4π[K(m1+sm,m1+sm+m2,l1+sl,l1+sl+l2,m1) −K(m1+sm,m1+sm+m2,sl,sl+l2,m1) −K(l1+sl,l1+sl+l2,sm,sm+m2,l1) +K(l1+sl,l1+sl+l2,m1+sm,m1+sm+m2,l1)] (4.110) Note that the smandslmay be negative. If both loops are identical andsquare (i.e.,l1=l2=m1=m2=L), this result simpliﬁes to M12=μ0 4π[K(L+sm,2L+sm,L+sl,2L+sl,L) −K(L+sm,2L+sm,sl,L+sl,L) −K(L+sl,2L+sl,sm,L+sm,L) +K(L+sl,2L+sl,L+sm,2L+sm,L)] (4.111) Next, we obtain the mutual inductance by determining the total magnetic ﬂux through the second loop using the vector magnetic potential, A,a s ψ2=⎜integraldisplay s2B·ds =⎜contintegraldisplay c2A·dl (4.112) This requires that we solve the fundamental subproblem shown in Fig. 4.29. Once this is done, we superimpose the result around the four sides of loop2 from each of the four currents in loop 1. Since the current Iand the left segment of loop 2 are parallel, the vector magnetic potential Ais tangent to the conductor and the dot product in (4.112) can be removed. The vectormagnetic potential for the case in Fig. 4.29 was derived in equation (2.57).That was derived for the origin at the midpoint of the current. Rederiving thatfor the origin at the bottom of the current as in Fig. 4.29 gives A=μ 0I 4π⎜bracketleftbigg ln⎜parenleftBig Z+⎜radicalbig Z2+r2⎜parenrightBig −ln⎜parenleftbigg (Z−L)+⎜radicalBig (Z−L)2+r2⎜parenrightbigg⎜bracketrightbigg (4.113) SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE 189 2r1r 1Z2Z I Lz A 2c2c 2c 2cA A A FIGURE 4.29. Another fundamental subproblem. Integrating this along the left side of the second loop gives ⎜integraldisplayZ2 Z=Z1Ad Z=μ0I 4π⎜integraldisplayZ2 Z=Z1⎜bracketleftBig ln⎜parenleftBig Z+⎜radicalbig Z2+r2 1⎜parenrightBig −ln⎜parenleftbigg (Z−L)+⎜radicalBig (Z−L)2+r2 1⎜parenrightbigg⎜bracketrightbigg dZ =μ0I 4π⎜bracketleftBig Z2ln⎜parenleftBig Z2+⎜radicalbig Z2 2+r2 1⎜parenrightBig −⎜radicalbig Z2 2+r2 1 −Z 1ln⎜parenleftBig Z1+⎜radicalbig Z2 1+r2 1⎜parenrightBig +⎜radicalbig Z2 1+r2 1 −(Z2−L)ln⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 1⎜parenrightbigg +⎜radicalBig (Z2−L)2+r2 1 +(Z1−L)ln⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 1⎜parenrightbigg −⎜radicalBig (Z1−L)2+r2 1⎜bracketrightbigg (4.114) where we have used an integral from Dwight [7]: ⎜integraldisplay ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig dx=xln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig −⎜radicalbig x2+a2 (D625) 190 THE CONCEPT OF “LOOP” INDUCTANCE and have made a change of variables in the second half of the integral of λ= Z−L,dλ=dZ. Realizing that the vector magnetic potential is perpendicular to the top and bottom sides of the loop in Fig. 4.29 and contribute nothing tothe line integral around the loop, the contribution to integration around loop 2in (4.112) due to the current in the left side of loop 1 as in Fig. 4.28 is obtainedas ψ=μ 0I 4π⎜bracketleftbigg Z2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 1⎜parenrightbigg −⎜radicalBig Z2 2+r2 1 −Z 1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 1⎜parenrightbigg +⎜radicalBig Z2 1+r2 1 −(Z2−L)ln⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 1⎜parenrightbigg +⎜radicalBig (Z2−L)2+r2 1 +(Z1−L)ln⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 1⎜parenrightbigg −⎜radicalBig (Z1−L)2+r2 1 −Z 2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 2⎜parenrightbigg +⎜radicalBig Z2 2+r2 2 +Z 1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 2⎜parenrightbigg −⎜radicalBig Z2 1+r2 2 +(Z2−L)ln⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 2⎜parenrightbigg −⎜radicalBig (Z2−L)2+r2 2 −(Z1−L)ln⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 2⎜parenrightbigg +⎜radicalBig (Z1−L)2+r2 2⎜bracketrightbigg (4.115) giving the same result as in (4.109). Using this result and superimposing the ﬂuxes through loop 2 due to the the top, right and bottom currents of loop 1gives the same result as the previous direct computation of the ﬂux throughloop 2 and given in (4.110). Using the Neumann integral we compute the mutual inductance between loops 1 and 2 directly from M 12=μ0 4π⎜contintegraldisplay c2⎜contintegraldisplay c1dl·dl2 R12(4.116) where c1andc2are the contours of loops 1 and 2, respectively, and R12is the distance between a point on loop 1 and a point on loop 2. This was derived bysubstituting the expression for the vector magnetic potential produced alongthe contour of loop 2 by the current of loop 1: A 12=μ0I1 4π⎜contintegraldisplay c11 R12dl1 (4.117a) SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE 191 into the basic expression for the magnetic ﬂux through loop 2: ψ12=⎜contintegraldisplay s2B12·ds =⎜contintegraldisplay c2A12·dl (4.117b) The mutual inductance is obtained by dividing the ﬂux by the current I1 according to the basic deﬁnition in (4.104). Once again we need to solve a basic subproblem shown in Fig. 4.30. This represents the contribution to theNeumann integral along the left side of loop 1 and the left side of loop 2. Theportion of the Neumann integral due to the left side of loop 1 and the left sideof loop 2 represented in Fig. 4.30 beomes Int left-left=μ0 4π⎜integraldisplay c2⎜integraldisplay c11 R12dl1dl2 =μ0 4π⎜integraldisplayZ2 z2=Z 1⎜integraldisplayL z1=01⎜radicalbig (z2−z1)2+r2 1dz1dz2 (4.118) 2r1r 1Z2Z Lz 2c2c 2c 2c 1c12R 1dl2dl FIGURE 4.30. Basic subproblem for the Neumann integral. 192 THE CONCEPT OF “LOOP” INDUCTANCE The inner integral with respect to z1is integrated as (I)=⎜integraldisplayL z1=01⎜radicalBig (z2−z1)2+r2 1dz1 =⎜integraldisplayz2 λ=z 2−L1⎜radicalBig λ2+r2 1dλ =ln⎜parenleftbigg z2+⎜radicalBig z2 2+r2 1⎜parenrightbigg −ln⎜bracketleftbigg (z2−L)+⎜radicalBig (z2−L)2+r2 1⎜bracketrightbigg and we have used an integral from Dwight [7]: ⎜integraldisplaydx√ x2+a2=ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig (D200.01) and a change of variables λ=z2−z1,d λ=−dz1. Integrating with respect toz2gives Intleft-left=μ0 4π⎜integraldisplayZ2 z2=Z 1⎜bracketleftbigg ln⎜parenleftbigg z2+⎜radicalBig z2 2+r2 1⎜parenrightbigg −ln⎜parenleftbigg (z2−L)+⎜radicalBig (z2−L)2+r2 1⎜parenrightbigg⎜bracketrightbigg dz2 Using a change of variables λ=z2−L,dλ =dz2gives Intleft-left=μ0 4π⎜integraldisplayZ2 z2=Z 1⎜bracketleftbigg ln⎜parenleftbigg z2+⎜radicalBig z2 2+r2 1⎜parenrightbigg⎜bracketrightbigg dz2 −μ0 4π⎜integraldisplayZ2−L λ=Z 1−L⎜bracketleftbigg ln⎜parenleftbigg λ+⎜radicalBig λ2+r2 1⎜parenrightbigg⎜bracketrightbigg dλ Integrating this using an integral from Dwight [7], ⎜integraldisplay ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig dx=xln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig −⎜radicalbig x2+a2(D625) SUMMARY OF METHODS FOR COMPUTING LOOP INDUCTANCE 193 yields Intleft-left=μ0 4π⎜bracketleftbigg Z2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 1⎜parenrightbigg −⎜radicalBig Z2 2+r2 1 −Z 1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 1⎜parenrightbigg +⎜radicalBig Z2 1+r2 1 −(Z2−L)l n⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 1⎜parenrightbigg +⎜radicalBig (Z2−L)2+r2 1 +(Z1−L)l n⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 1⎜parenrightbigg −⎜radicalBig (Z1−L)2+r2 1⎜bracketrightbigg The contribution from the left side of loop 1 to the right side of loop 2 is the same but negated and with r1replaced with r2. Realizing that dl1·dl2=0 along the top and bottom sides of loop 2 gives the total contribution to theNeumann integral due to the left side of loop 1 as Int left=μ0 4π⎜bracketleftbigg Z2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 1⎜parenrightbigg −⎜radicalBig Z2 2+r2 1 −Z 1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 1⎜parenrightbigg +⎜radicalBig Z2 1+r2 1 −(Z 2−L)l n⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 1⎜parenrightbigg +⎜radicalBig (Z2−L)2+r2 1 +(Z 1−L)l n⎜parenleftbigg (Z2−L)+⎜radicalBig (Z1−L)2+r2 1⎜parenrightbigg −⎜radicalBig (Z1−L)2+r2 1 −Z 2ln⎜parenleftbigg Z2+⎜radicalBig Z2 2+r2 2⎜parenrightbigg +⎜radicalBig Z2 1+r2 2 +Z 1ln⎜parenleftbigg Z1+⎜radicalBig Z2 1+r2 2⎜parenrightbigg −⎜radicalBig Z2 1+r2 2 +(Z 2−L)l n⎜parenleftbigg (Z2−L)+⎜radicalBig (Z2−L)2+r2 2⎜parenrightbigg −⎜radicalBig (Z2−L)2+r2 2 −(Z 1−L)l n⎜parenleftbigg (Z1−L)+⎜radicalBig (Z1−L)2+r2 2⎜parenrightbigg +⎜radicalBig (Z1−L)2+r2 2⎜bracketrightbigg (4.119) 194 THE CONCEPT OF “LOOP” INDUCTANCE which is the same as (4.115) with the current Iremoved from that expression. Using this result and superimposing the contributions to the Neumann integralaround loop 2 due to the top, right, and bottom segments of loop 1 gives thesame result as the previous direct computation of the ﬂux through loop 2 andgiven in (4.110). 5 THE CONCEPT OF “PARTIAL” INDUCTANCE In the preceding chapters we discussed the meaning and calculation of the “loop” inductance of various conducting structures that support a closed loopof current. This “loop” inductance is calculated fundamentally for steady (dc)currents which we showed in Section 2.9 must form closed loops. If we openthe loop at a point with a small gap, the loop inductance of that current loop isseen as an inductance Lat these input terminals. When we pass a time-varying current around the loop via these terminals a voltage, V(t)=LdI(t)/dt,i s developed across the terminals. This voltage is essentially the Faraday’s lawvoltage induced into the loop. For electrically small loop dimensions, thislumped inductance and the voltage across its terminals can be represented asa lumped voltage source and placed anywhere in the loop perimeter (see Fig. 4.1). It is important, however, to remember that neither this lumped inductance nor the equivalent voltage source it represents can be placed in aunique position in the loop! This loop inductance is a property of the entireloop and its use is valid only at the input terminals of the loop. Hence, it isnot possible to associate the loop inductance with any particular segment ofthe loop. However, there are numerous situations, some of which were described in Chapter 1, where it is useful to develop a lumped-circuit model of a closedcurrent loop wherein the segments of the perimeter of the loop are represented Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 195 196 THE CONCEPT OF “PARTIAL” INDUCTANCE with a self inductance as well as mutual inductances between that segment and other segments of this and other adjacent current loops . The concept of “partial” inductance allows us to do that in a unique way. It has been said that “you cannot ascribe the properties of inductance to an isolated piece of wire.” Of course you can’t because an isolated piece ofwire is not capable of supporting a dc current, which must form a closed loop(i.e., it must return to its source). This is therefore a misleading statement.The proper question is: Can you ascribe the properties of inductance uniquely to a segment of a closed loop of current? The answer to this question is yes, and the method for doing so is with “partial” inductances. There are three signiﬁcant references regarding partial inductance. Those by Grover [14] and Ruehli [15] are excellent general references, and the paperby Hoer and Love [16] gives results for the partial inductances of conductorsof rectangular cross section [e.g., printed circuit board (PCB) lands]. 5.1 GENERAL MEANING OF PARTIAL INDUCTANCE Consider a closed physical loop constructed of a conductor such as a wire, PCB land, and so on, that supports a dc current I. The “loop” inductance of this current loop is deﬁned fundamentally in previous chapters as L=ψ I(5.1a) where ψ=⎜integraldisplay sB·ds (5.1b) is the total magnetic ﬂux that penetrates the open surface sthat is surrounded by the closed contour of the loop, c, and Bis the magnetic ﬂux density (caused by current I) through the surface s. In Chapter 2 we calculated Bfor various conﬁgurations of loop shapes. In Chapter 4 we calculated the ﬂux ψand hence the inductance according to (5.1) for various loop shapes. Faraday’sfundamental law of induction (Chapter 3) gives the induced voltage appearingat the terminals of the loop as V=dψ dt =LdI dt(5.2) where the current Iis now allowed to be “slowly varying with time,” as demon- strated in Section 3.4. Essentially, the condition “slowly varying with time”is satisﬁed approximately as long as the physical dimensions of the loop aremuch less than a wavelength (e.g., <λ / 10, where the wavelength is λ=v/f, GENERAL MEANING OF PARTIAL INDUCTANCE 197 fis the highest signiﬁcant frequency in the waveform of the current I, and v is the velocity of propagation of the current. In Chapter 4 we developed an alternative means of calculating the induc- tance by using the vector magnetic potential A, which is deﬁned by B=∇ ×A (5.3) Hence, the total magnetic ﬂux through the surface sis ψ=⎜integraldisplay sB·ds =⎜integraldisplay s(∇×A)·ds =⎜contintegraldisplay cA·dl (5.4) where we have used Stokes’s theorem (see the Appendix) to convert the surface integral over surface sto a line integral around contour cthat encloses the surface. This gives an alternative way of calculating the ﬂux through the loop,ψ, in terms of A. Hence, an alternative way of calculating the inductance of the current loop is L=⎜contintegraltext cA·dl I(5.5) where cis the closed contour that bounds the open surface s. Hence, we can compute the inductance of a loop by integrating, with a line integral, theproduct of the differential path lengths around the contour cthat surrounds the open surface s and the components of the vector magnetic potential A that are tangent to that closed path. But (5.5) can be decomposed into the lineintegral along unique segments of the closed loop as L=⎜contintegraltext cA·dl I =⎜integraltext c1A1·dl I+⎜integraltext c2A2·dl I+···+⎜integraltext cnAn·dl I (5.6) where the closed path cis segmented into n contiguous segments ciso that c=c1+c2+···+ cnandAiis the total Aalong contour cithat is due to the current of that segment as well as the currents of the other segments of cor of some other current loop. This allows us to uniquely associate an inductance contribution to each segment of the closed loop as Li=⎜integraltext ciAi·dl I(5.7a) 198 THE CONCEPT OF “PARTIAL” INDUCTANCE z II I Il wy2rww2rl− w2rl+ −Aleft ArightAtop Abottomc c ccs FIGURE 5.1. Rectangular loop. so that the total loop inductance is the sum of these parts: L=L1+L2+···+ Ln (5.7b) For example, in Fig. 5.1 we have shown Fig. 4.10, where in Section 4.3.1 we detailed the calculation of the inductance of a rectangular loop using thevector magnetic potential according to (5.5) Essentially, we are indirectlycomputing the total magnetic ﬂux threading the loop, which is the regionsurrounded by the interior surfaces of the wires whose radii are r w. This contour surrounding the open surface sis denoted as contour cin Fig. 5.1. As discussed in Sections 4.5 and 4.6, two important assumptions in computing theBﬁeld (and the subsequent calculation of the Aﬁeld) are that (1) the current Iisdistributed uniformly over the wire cross section so that the current I can be represented as a ﬁlament on the wire axis (as it is for dc currents), and (2) there are no other currents in close enough proximity to this wire to upset thisuniform current distribution over its cross section (i.e., the “proximity effect”is not pronounced). The total vector magnetic potential along the left side ofthe loop, A 1, is the sum of the vector magnetic potentials along that side that are due to the current of that side, Aleft, and those that are due to the currents of the other three sides of the loop, Aright,Atop, and Abottom : A1=Aleft+Aright+Atop+Abottom (5.8a) GENERAL MEANING OF PARTIAL INDUCTANCE 199 Hence, the portion of the loop inductance uniquely attributable to the left sideis L1=⎜integraltext left sideA1·dl I =⎜integraltext left sideAleft·dl I+⎜integraltext left sideAright ·dl I+⎜integraltext left sideAtop·dl I +⎜integraltext left sideAbottom ·dl I(5.8b) Observe that AtopandAbottom are in the directions of the currents of those sides and hence are both orthogonal to the left side and do not contribute tothe line integral for L 1along the left side. In a similar fashion we obtain the inductances attributable to the other three sides, L2,L3, andL4. Hence, the rectangular loop can be represented uniquely by the lumped equivalent circuit shown in Fig. 5.2. Observe that the total vector magnetic potential along the left side ,A1,i n (5.8a) has contributions due to its own current as well as the currents of theother three sides. So this leads us to break the inductance of the left side, L 1, into four distinct pieces according to (5.8b): L1=Lp1+Mp12+Mp13+Mp14 (5.9) I I II1c2c 3c 4cs 1L2L 3L L4I III BA 2 A 1 A 3 A4 FIGURE 5.2. Uniquely attributing inductances to the sides of the rectangular loop of Fig. 5.1. 200 THE CONCEPT OF “PARTIAL” INDUCTANCE The ﬁrst contribution is the self partial inductance of the left side: Lp1=⎜integraltext left sideAleft·dl I(5.10a) which is due to the current of the left side. The other three contributions are due to the currents of the other three sides and are referred to as the mutual partial inductances between the other three sides and the left side: Mp12=⎜integraltext left sideAtop·dl I(5.10b) Mp13=⎜integraltext left sideAright ·dl I(5.10c) Mp14=⎜integraltext left sideAbottom ·dl I(5.10d) Hence, the more complete equivalent circuit of the rectangular loop in terms of the partial inductances is shown in Fig. 5.3. According to the dot convention described in Section 4.8.1, the total voltage across the left conductor is V1=Lp1dI dt+Mp12dI dt+Mp13dI dt+Mp14dI dt =⎜parenleftbigLp1+Mp12+Mp13+Mp14⎜parenrightbig ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright L1dI dt(5.11) Observe that because sides 2 and 4 are orthogonal to side 1, AtopandAbottom are orthogonal to the left side, so that Mp12=Mp14=0. Also, because the direction of Arightis opposite the direction of the contour calong the left side, I I II Lp1Lp2 Lp3 Lp4Mp13 V1Mp12 Mp14 FIGURE 5.3. Rectangular loop equivalent circuit in terms of the partial inductances. PHYSICAL MEANING OF PARTIAL INDUCTANCE 201 Mp13in (5.10c) is negative. The effective inductance of the left side of the loop,L1, in (5.9) is referred to as the net partial inductance but has little value or use. Separating the net partial inductance into its constituent parts as in (5.9)gives more information about the contributions of all the other side currents. 5.2 PHYSICAL MEANING OF PARTIAL INDUCTANCE The self partial inductance of the ith segment of a current loop is Lpi=⎜integraltext ciAi·dl Ii(5.12a) andAiis the portion of Aalong cithat is produced by the current Iiof that segment. The voltage developed across that self partial inductance is Vi=LpidIi dt(5.12b) as shown in Fig. 5.4. Although (5.12a) gives the mathematical deﬁnition of self partial induc- tance, we now investigate the physical meaning of self partial inductance.Consider a segment c iof a current loop carrying current Iias shown in Fig. 5.5(a). Draw a surface extending from the segment to inﬁnity with sidesthat are perpendicular to the current segment. Now determine the magneticﬂux through that surface: ψ ∞ Ii=⎜integraltext sB·ds Ii=⎜contintegraltext cA·dl Ii =⎜integraltext ciAi·dl Ii⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ci+⎜integraltext cA·dl Ii⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright left side+⎜integraltext cA·dl Ii⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright right side+⎜integraltext cA·dl Ii⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ∞ =⎜integraltext ciAi·dl Ii⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ci =Lpi (5.13) iciIiI iV iApiL FIGURE 5.4. Self partial inductance of the ith segment of a current loop. 202 THE CONCEPT OF “PARTIAL” INDUCTANCE ∞ ic ic∞iI(a)iAAA 0=A iI (b)iBc s s FIGURE 5.5. Physical meaning of self partial inductance. The line integrals along the left and right sides are zero since the vector mag- netic potential Ais parallel to the current Iithat produces it and is therefore per- pendicular to the left and right sides of the closed contour. The vector magneticpotential from a line current goes to zero at inﬁnity [see (2.57)], so that the lineintegral along this portion of the closed contour at inﬁnity is also zero. Hence,we are left with the partial inductance given in (5.12a) and the observationthat: The self partial inductance of a segment of a current loop is the ratioof the magnetic ﬂux between the current segment and inﬁnity and thecurrent of that segment. This is illustrated in cross section in Fig. 5.5(b). The mutual partial inductance between two segments ciandcj(which may be parts of the same current loop or different current loops) isdeﬁned by Mpij=⎜integraltext ciAij·dl Ij(5.14a) PHYSICAL MEANING OF PARTIAL INDUCTANCE 203 cicj Ij Ij ViLpiLpj Mpij Aij FIGURE 5.6. Mutual partial inductance between two current loop segments ciandcj. where Aijis along contour ciand is due to the current of another segment, Ij. The voltage developed across that self partial inductance is Vi=MpijdIj dt(5.14b) as shown in Fig. 5.6. The physical meaning of mutual partial inductance is illustrated in Fig. 5.7. Consider a current loop and two segments of that loop, ciandcj, as shown in Fig. 5.7(a). Again draw a surface sextending from the jth segment (car- rying the current) to inﬁnity with sides that are perpendicular to that current (b)(a) ijBci cicj cjIj IjAA A = 0Aij cs s FIGURE 5.7. Physical meaning of mutual partial inductance. 204 THE CONCEPT OF “PARTIAL” INDUCTANCE segment. Now determine the magnetic ﬂux through the surface sbetween the ith segment and inﬁnity. Carrying through a development similar to that in(5.13) we see that the line integrals along the left and right sides are zero sincethe vector magnetic potential Ais parallel to the current I jthat produces it and is therefore perpendicular to the left and right sides of the contour cthat surrounds surface s. Also, the vector magnetic potential from a line current goes to zero at inﬁnity so that the line integral along the portion of the contourat inﬁnity is also zero. Hence, we are left with the mutual partial inductancegiven in (5.14a) and the observation that The mutual partial inductance between two segments of the same ordifferent current loops is the ratio of the magnetic ﬂux (produced by the current of the ﬁrst segment )that penetrates the surface between the second segment and inﬁnity and the current of the ﬁrst segment. This is illustrated in cross section in Fig. 5.7(b). Although Fig. 5.7 shows the result for the mutual partial inductance of twoparallel conductors, the result also obtains for two conductors at any angle to each other as shown in Fig. 5.8. Again draw two lines to inﬁnity ∞ 0=A A A ijA Ijs c ic ijRidl dlj FIGURE 5.8. Mutual partial inductance for conductors at any angle to each other. SELF PARTIAL INDUCTANCE OF WIRES 205 that are perpendicular (shown as small rectangles) to current Ijand which enclose the open surface sthat lies between those parallel lines and between the skewed conductor and inﬁnity. Integrating the line integral of the vectormagnetic potential around the closed contour csurrounding this surface to inﬁnity again gives M pij=ψ∞ Ij =⎜contintegraltext cA·dl Ij =⎜integraltext ciAij·dl Ij(5.15) This is obtained again since Ais parallel to Ijat all points in space, so that A is perpendicular to the left and right sides of sand contribute nothing to the line integral along those sides, and Agoes to zero at inﬁnity. Observe that the same result is obtained even if the two conductors do not lie in the same plane , since Awill still be orthogonal to the two sides of the open surface because they were constructed perpendicular to the current Ijand will also go to zero at inﬁnity. (Again draw two lines for the sides of sthat are perpendicular to conductor cj.) The mutual partial inductance can also be obtained from the Neumann integral by substituting the explicit equation for Aijinto (5.15): Mpij=μ0 4π⎜integraldisplay ci⎜integraldisplay cj1 Rijdli·dlj (5.16) where cjis the contour along the conductor carrying current Ij, andRijis the distance between differential segments dlialong contour cianddljalong contour cj, as shown in Fig. 5.8. 5.3 SELF PARTIAL INDUCTANCE OF WIRES In this section we derive some fundamental results for the self partial induc- tance of wires having radii rw. Again we assume that the current of the wire, I, is distributed uniformly over the wire cross section so that for the purpose ofcomputing the BandAﬁelds, we can concentrate the current Ias a ﬁlament on the axis of the wire. The fundamental problem for computing the self partial inductance of a wire is a wire of length lcarrying a current Ias shown in Fig. 5.9. We de- termine the self partial inductance of this segment of wire by integrating themagnetic ﬂux density through the surface sbetween the wire surface, y=r w, 206 THE CONCEPT OF “PARTIAL” INDUCTANCE l Iz yZr s2lz= 2lz–=B A2rw FIGURE 5.9. Determination of the self partial inductance of a wire. and inﬁnity, y→∞ . The magnetic ﬂux density was derived in Chapter 2 and given in (2.15): B=μ0I 4πr⎜bracketleftBigg Z+l/2⎜radicalbig (Z+l/2)2+r2−Z−l/2⎜radicalbig (Z−l/2)2+r2⎜bracketrightBigg aφ (2.15) The total ﬂux through the surface sis ψ∞=⎜integraldisplay∞ r=rw⎜integraldisplayl/2 Z=−l/2BφdZ dr =μ0I 4π⎜integraldisplay∞ r=rw1 r⎜integraldisplayl/2 Z=−l/2⎜bracketleftBigg Z+l/2⎜radicalbig (Z+l/2)2+r2 −Z−l/2⎜radicalbig (Z−l/2)2+r2⎜bracketrightBigg dZ dr =2μ0I 4π⎜integraldisplay∞ r=rw1 r⎜integraldisplayl λ=0λ√ λ2+r2dλ dr =μ0I 2π⎜integraldisplay∞ r=rw1 r⎜bracketleftBig⎜radicalbig λ2+r2⎜bracketrightBigl λ=0dr =μ0I 2π⎜integraldisplay∞ r=rw1 r(⎜radicalbig l2+r2−r)dr (5.17a) SELF PARTIAL INDUCTANCE OF WIRES 207 and we have used a change of variables, λ=Z±l/2,d λ=dZ, and integral 201.01 of Dwight [7]: ⎜integraldisplayx√ x2+a2dx=⎜radicalbig x2+a2 (D201.01) Further integration with respect to ryields ψ∞=μ0I 2π⎜integraldisplay∞ r=rw⎜bracketleftBigg√ l2+r2 r−1⎜bracketrightBigg dr =μ0I 2π⎜bracketleftBigg⎜radicalbig l2+r2−llnl+√ l2+r2 r−r⎜bracketrightBiggr→∞ r=rw =−μ0I 2πl⎡ ⎣ln⎛⎝ l r+⎜radicalBigg⎜parenleftbiggl r⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggr l⎜parenrightbigg2 +r l⎤ ⎦r→∞ r=rw =μ0I 2πl⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ (5.17b) and we have used integral 241.01 of Dwight [7]: ⎜integraldisplay√ x2+a2 xdx=⎜radicalbig x2+a2−alna+√ x2+a2 x(D241.01) Hence, the self partial inductance is Lp=ψ∞ I =μ0 2πl⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ =2×10−7l⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ (5.18a) and we have substituted μ0/2π=2×10−7. Using the inverse hyperbolic sine, sinh−1x a=ln⎛ ⎝x a+⎜radicalBigg⎜parenleftbiggx a⎜parenrightbigg2 +1⎞ ⎠ =− sinh−1⎜parenleftbigg −x a⎜parenrightbigg (D700.1) 208 THE CONCEPT OF “PARTIAL” INDUCTANCE gives an alternative form of the result: Lp=μ0 2πl⎡ ⎣sinh−1l rw−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ (5.18b) In a practical case, the length of the segment is usually much larger than the wire radius, l/greatermuchrw, so we have the following approximations: ln⎡ ⎣l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎤ ⎦=ln2l rw+1 4⎜parenleftbiggrw l⎜parenrightbigg2 −3 32⎜parenleftbiggrw l⎜parenrightbigg4 +···l rw/greatermuch1 (D602.1) and ⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 =1+1 2⎜parenleftbiggrw l⎜parenrightbigg2 −1 8⎜parenleftbiggrw l⎜parenrightbigg4 +···rw l≤1 (D5.3) so that (5.18a) approximates to Lp=μ0 2πl⎜bracketleftBigg ln2l rw−1+rw l−1 4⎜parenleftbiggrw l⎜parenrightbigg2 +···⎜bracketrightBigg ∼=2×10−7l⎜parenleftbigg ln2l rw−1⎜parenrightbigg l/greatermuchrw (5.18c) Alternatively, we can determine the self partial inductance by integrating the vector magnetic potential along the wire surface also shown in Fig. 5.9.The vector magnetic potential Afor this case was determined in Chapter 2 and given in (2.57): A z=μ0I 4π⎜parenleftbigg sinh−1Z+l/2 r−sinh−1Z−l/2 r⎜parenrightbigg (2.57) Hence, we set up the integral Lp=⎜integraldisplayl/2 Z=−l/2Az|r=rwdZ I =μ0 4π⎜integraldisplayl/2 Z=−l/2⎜parenleftbigg sinh−1Z+l/2 rw−sinh−1Z−l/2 rw⎜parenrightbigg dZ =2μ0 4π⎜integraldisplayl λ=0⎜parenleftbigg sinh−1λ rw⎜parenrightbigg dλ MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 209 =2μ0 4π⎜bracketleftbigg λsinh−1λ rw−⎜radicalBig λ2+r2w⎜bracketrightbiggl λ=0 =μ0 2π⎜parenleftbigg lsinh−1l rw−⎜radicalBig l2+r2w+rw⎜parenrightbigg =μ0 2πl⎡ ⎣sinh−1l rw−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ =μ0 2πl⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ (5.19) which is the same as (5.18a). We have used a change of variables, λ=Z± l/2,dλ=dZ, integral 730 of Dwight [7], ⎜integraldisplay sinh−1x adx=xsinh−1x a−⎜radicalbig x2+a2 (D730) and the identity for inverse hyperbolic sine, sinh−1x a=ln⎛ ⎝x a+⎜radicalBigg⎜parenleftbiggx a⎜parenrightbigg2 +1⎞ ⎠ =− sinh−1⎜parenleftbigg −x a⎜parenrightbigg (D700.1) 5.4 MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES Next, we determine another fundamental result: the mutual partial inductance between two parallel wires shown in Fig. 5.10. We ﬁrst assume that both wiresare of the same length and their endpoints are aligned. In the next section wederive the result for this situation but with the wires offset and their lengthsdifferent. The only difference between this computation and those for the selfpartial inductance of Section 5.3 is that here we integrate from y=d+r wto y→∞ rather than from the surface of the ﬁrst wire. Hence, the integral in (5.17a) becomes ψ∞=⎜integraldisplay∞ r=d+rw⎜integraldisplayl/2 Z=−l/2BφdZ dr (5.20) 210 THE CONCEPT OF “PARTIAL” INDUCTANCE lIz yZr s2lz= 2lz− =B A dwr2 FIGURE 5.10. Determination of the mutual partial inductance between parallel wires. It is easy to see that we only need to replace rwwithd+rwin the previous derivation for the self partial inductance in (5.17a)–(5.17b) and obtain Mp=ψ∞ I =μ0 2πl⎡ ⎣ln⎛⎝ l d+rw+⎜radicalBigg⎜parenleftbiggl d+rw⎜parenrightbigg2 +1⎞ ⎠ −⎜radicalBigg 1+⎜parenleftbiggd+rw l⎜parenrightbigg2 +d+rw l⎤ ⎦ ∼=2×10−7l⎡ ⎣ln⎛⎝ l d+⎜radicalBigg⎜parenleftbiggl d⎜parenrightbigg2 +1⎞ ⎠ −⎜radicalBigg 1+⎜parenleftbiggd l⎜parenrightbigg2 +d l⎤ ⎦ d/greatermuchrw (5.21a) Using the inverse hyperbolic sine, sinh−1x a=ln⎛ ⎝x a+⎜radicalBigg⎜parenleftbiggx a⎜parenrightbigg2 +1⎞ ⎠ =− sinh−1⎜parenleftbigg −x a⎜parenrightbigg (D700.1) MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 211 gives an alternative form of the result: Mp=μ0 2πl⎡ ⎣sinh−1l d−⎜radicalBigg 1+⎜parenleftbiggd l⎜parenrightbigg2 +d l⎤ ⎦ d/greatermuchrw(5.21b) For wires that are very long compared to their separation, l/d/greatermuch1, or, equivalently, separations much smaller than their length, d/l/lessmuch1, the result in (5.21a) can be approximated by using ln⎡ ⎣l d+⎜radicalBigg⎜parenleftbiggl d⎜parenrightbigg2 +1⎤ ⎦=ln2l d+1 4⎜parenleftbiggd l⎜parenrightbigg2 −3 32⎜parenleftbiggd l⎜parenrightbigg4 +···l d>1 (D602.1) ⎜radicalBigg 1+⎜parenleftbiggd l⎜parenrightbigg2 =1+1 2⎜parenleftbiggd l⎜parenrightbigg2 −1 8⎜parenleftbiggd l⎜parenrightbigg4 +···d l≤1 (D5.3) giving Mp=μ0 2πl⎜bracketleftBigg ln2l d−1+d l−1 4⎜parenleftbiggd l⎜parenrightbigg2 +1 32⎜parenleftbiggd l⎜parenrightbigg4 −···⎜bracketrightBigg ∼=μ0 2πl⎜parenleftbigg ln2l d−1⎜parenrightbigg l/greatermuchd (5.21c) For wires that are very short compared to their separation, l/d/lessmuch1, or, equivalently, separations much greater than their length, d/l/greatermuch1, the result in (5.21a) can be approximated by using ln⎡ ⎣l d+⎜radicalBigg⎜parenleftbiggl d⎜parenrightbigg2 +1⎤ ⎦=l d−1 6⎜parenleftbiggl d⎜parenrightbigg3 +3 40⎜parenleftbiggl d⎜parenrightbigg5 −···l d<1 (D602.1) ⎜radicalBigg 1+⎜parenleftbiggd l⎜parenrightbigg2 =d l⎜radicalBigg⎜parenleftbiggl d⎜parenrightbigg2 +1 =d l+1 2⎜parenleftbiggl d⎜parenrightbigg −1 8⎜parenleftbiggl d⎜parenrightbigg3 +1 16⎜parenleftbiggl d⎜parenrightbigg5 −···l d≤1 (D5.3) 212 THE CONCEPT OF “PARTIAL” INDUCTANCE giving Mp=μ0 2πl 2d⎜bracketleftBigg 1−1 12⎜parenleftbiggl d⎜parenrightbigg2 +1 40⎜parenleftbiggl d⎜parenrightbigg4 −···⎜bracketrightBigg l/lessmuchd (5.21d) We can obtain the same result as in (5.21a) from the vector magnetic potential A: Mp=⎜integraldisplayl/2 Z=−l/2Az|r=d+rwdZ I(5.22) and evaluating Azalong the second wire at y=d+rw. Carrying through the same integration in (5.19) but with r=rwreplaced by r=d+rwagain gives (5.21a). Finally, we show that the mutual partial inductance in (5.21a) can also be derived from the Neumann integral in (5.16): Mp=μ0 4π⎜integraldisplay c1⎜integraldisplay c2dl1·dl2 R12 =μ0 4π⎜integraldisplayl/2 z2=−l/2dz2⎜integraldisplayl/2 z1=−l/21⎜radicalBig (d+rw)2+(z1−z2)2dz1 =μ0 4π⎜integraldisplayl/2 z2=−l/2dz2⎜integraldisplayl/2−z2 λ=−l/2 −z21⎜radicalBig (d+rw)2+λ2dλ =μ0 4π⎜integraldisplayl/2 z2=−l/2⎜bracketleftbigg ln⎜parenleftbigg λ+⎜radicalBig (d+rw)2+λ2⎜parenrightbigg⎜bracketrightbiggl/2−z2 λ=−l/2 −z2dz2 =μ0 4π⎜integraldisplayl/2 z2=−l/2⎜parenleftbigg sinh−1l/2−z2 d+rw+sinh−1l/2+z2 d+rw⎜parenrightbigg dz2 =μ0 4π⎜integraldisplayl ζ=0⎜parenleftbigg 2 sinh−1ζ d+rw⎜parenrightbigg dζ =2μ0 4π⎜bracketleftbigg ζsinh−1ζ d+rw−⎜radicalBig ζ2+(d+rw)2⎜bracketrightbiggl ζ=0 =μ0 2π⎜bracketleftbigg lsinh−1l d+rw−⎜radicalBig l2+(d+rw)2+(d+rw)⎜bracketrightbigg =μ0 2πl⎡ ⎣sinh−1l d+rw−⎜radicalBigg 1+⎜parenleftbiggd+rw l⎜parenrightbigg2 +d+rw l⎤ ⎦(5.23) MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 213 and the differential lengths dl1anddl2are parallel so that the dot product goes away: dl1·dl2=dl1dl2=dz1dz2. But (5.23) is the same as (5.21a). We have substituted a change of variables in the inner integral: λ=z1−z2,dλ=dz1, and a change of variables in the outer integral: ζ=l/2±z2,dζ=±dz2, and have again used the integrals ⎜integraldisplay1√ x2+a2dx=ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig (D200.01) and⎜integraldisplay sinh−1x adx=xsinh−1x a−⎜radicalbig x2+a2 (D730) We also used the important identity lna+√ x2+a2 −b+√ x2+b2=sinh−1a x−sinh−1⎜parenleftbigg −b x⎜parenrightbigg =sinh−1a x+sinh−1b x From these results we see that the self partial inductance Lpcan be obtained from the mutual partial inductance simply by replacing d+rwinMp, with rw, and vice versa. In other words, Lp=Mp⎜vextendsingle⎜vextendsingle d+rw→r w(5.24) Using Mpto get Lpin this way presupposes that both wires are of the same length and radii, and their endpoints are aligned. 5.5 MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES THAT ARE OFFSET Consider the case of two offset, parallel wires whose lengths are landmshown in Fig. 5.11. The two wires are parallel to the zaxis, have a center-to-center separation of d, and their endpoints are offset by a distance s. The radius of the second wire of length lisrw. The radius of the ﬁrst wire of length mcarrying the current Iwhich produces the magnetic ﬁeld is immaterial since we assume that the current Iis distributed uniformly over the cross section of that wire so that this current can be concentrated as a ﬁlament on the axis of the wire.The ﬁrst wire carrying the current Ihas its lower end at the origin of the coordinate system, z=0. The two ends of the other wire of length lare at positions z=z 1andz=z2. In all such problems of determining the mutual partial inductance between two parallel but offset wires using the result derivedin this section, it is important to determine these wire lengths and positions,z=0,z 1, andz2, for each particular problem. 214 THE CONCEPT OF “PARTIAL” INDUCTANCE z IZd 0=z1z2z d r d r≅ + =w 21A 2rw sl mm z= FIGURE 5.11. Mutual partial inductance between offset wires. Again we have three methods for calculating the mutual partial inductance between the two wire segments: the magnetic ﬂux linkage method using B, the vector magnetic potential method using A, and the Neumann integral. For this problem we choose to use the vector magnetic potential method using A. We must integrate the vector magnetic potential due to the current Iof the ﬁrst wire of length malong the surface of the second wire of length lwith a MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 215 line integral: Mp=⎜integraldisplay lA21|r=d+rw∼=d·dl I =⎜integraldisplayz2 Z=z1A21(Z, r)|r=ddZ I(5.25) where A21is the vector magnetic potential along the surface of the second wire that is produced by the current Iof the ﬁrst wire. Hence, we need the result for the vector magnetic potential from a wire of length mcarrying a current I. This was derived in Chapter 2 from Fig. 2.24 and given in (2.57). Note in Fig. 2.24 that the origin of the coordinate system at z=0 was located at the midpoint of the wire. We must modify that result to ﬁt Fig. 5.11 byrederiving the result for the case where the lower end of the wire is at z=0. Carrying through the development that led to (2.57) yields for this case A 21(Z, r)=μ0I 4π⎜braceleftBig ln⎜parenleftBig Z+⎜radicalbig Z2+r2⎜parenrightBig −ln⎜bracketleftbigg (Z−m)+⎜radicalBig (Z−m)2+r2⎜bracketrightbigg⎜bracerightbigg =μ0I 4π⎜parenleftbigg sinh−1Z r−sinh−1Z−m r⎜parenrightbigg (5.26) and we have again used the identity sinh−1x a=− sinh−1⎜parenleftbigg −x a⎜parenrightbigg =ln⎡ ⎣x a+⎜radicalBigg⎜parenleftbiggx a⎜parenrightbigg2 +1⎤ ⎦ =ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig −lna (D700.1) Hence, (5.25) becomes Mp=⎜integraldisplayz2 Z=z1A21|r=ddZ I =μ0 4π⎜integraldisplayz2 Z=z1⎜parenleftbigg sinh−1Z d−sinh−1Z−m d⎜parenrightbigg dZ (5.27) 216 THE CONCEPT OF “PARTIAL” INDUCTANCE Carrying through with the integration of (5.27) gives Mp=μ0 4π⎜integraldisplayz2 Z=z1⎜parenleftbigg sinh−1Z d−sinh−1Z−m d⎜parenrightbigg dZ =μ0 4π⎜parenleftbigg⎜integraldisplayz2 Z=z1sinh−1Z ddZ−⎜integraldisplayz2−m λ=z 1−msinh−1λ ddλ⎜parenrightbigg =μ0 4π⎜bracketleftbigg z2sinh−1z2 d−z1sinh−1z1 d−(z2−m)sinh−1z2−m d +(z1−m)sinh−1z1−m d−⎜radicalBig z2 2+d2+⎜radicalBig z2 1+d2 +⎜radicalBig (z2−m)2+d2−⎜radicalBig (z1−m)2+d2⎜bracketrightbigg (5.28) where we have used a change of variables, λ=Z−m,dλ=dZ, in the second integral and have used integral 730 of Dwight [7]: ⎜integraldisplay sinh−1x adx=xsinh−1x a−⎜radicalbig x2+a2 (D730) In the case where the two wires lie on the zaxis,d=0, as shown in Fig. 5.12, we could reintegrate (5.27) for r=rwor simply substitute d=rw into (5.28) to give Mp(d=rw)=μ0 4π⎜bracketleftbigg z2sinh−1z2 rw−z1sinh−1z1 rw−(z2−m)sinh−1z2−m rw +(z1−m)sinh−1⎜parenleftbiggz1−m rw⎜parenrightbigg −⎜radicalBig z2 2+r2w+⎜radicalBig z2 1+r2w +⎜radicalBig (z2−m)2+r2w−⎜radicalBig (z1−m)2+r2w⎜bracketrightbigg (5.29a) MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 217 z I 0=z1z2z 2rw21A sZwr r= l mm z= FIGURE 5.12. Aligned but offset wires. 218 THE CONCEPT OF “PARTIAL” INDUCTANCE Substituting the dimensions gives Mp(d=rw)=μ0 4π⎜bracketleftbigg (l+s+m)sinh−1l+s+m rw−(m+s)sinh−1m+s rw −(l+s)sinh−1l+s rw+ssinh−1s rw−⎜radicalBig (l+s+m)2+r2w +⎜radicalBig (m+s)2+r2w+⎜radicalBig (l+s)2+r2w−⎜radicalBig s2+r2w⎜bracketrightbigg ∼=μ0 4π⎜braceleftbigg (l+s+m)⎜bracketleftbigg ln⎜parenleftbigg2(l+s+m) rw⎜parenrightbigg −1⎜bracketrightbigg −(m+s)⎜bracketleftbigg ln⎜parenleftbigg2(m+s) rw⎜parenrightbigg −1⎜bracketrightbigg −(l+s)⎜bracketleftbigg ln⎜parenleftbigg2(l+s) rw⎜parenrightbigg −1⎜bracketrightbigg +s⎜bracketleftbigg ln⎜parenleftbigg2s rw⎜parenrightbigg −1⎜bracketrightbigg⎜bracerightbigg (5.29b) In terms of the self partial inductances of a wire of radius rwand length l obtained in (5.18b), Ll=μ0 2π⎜parenleftbigg lsinh−1l rw−⎜radicalBig l2+r2w+rw⎜parenrightbigg (5.18b) the result for aligned but offset wires in (5.29a,b) can be written as 2Mp(d=rw)=⎜parenleftbigLz2+Lz1−m⎜parenrightbig−⎜parenleftbigLz2−m+Lz1⎜parenrightbig =(Ll+s+m+Ls)−(Ll+s+Lm+s)(5.29c) Notice that (5.29c) gives 2 Mpsince the self partial inductance Llin (5.18b) is multiplied by μ0/2π, whereas the result for Mpin (5.29a,b) is multiplied byμ0/4π. There is a simple explanation for why the result for the mutual partial inductance between two aligned but offset wires can be written in terms ofthe self partial inductances of wires of various lengths obtained previously, asin (5.29c). Recall that the self partial inductance of a wire is the ratio of themagnetic ﬂux between that wire and inﬁnity, ψ l, and the current of that wire: Ll=ψl I Figure 5.13 shows that a current on each wire segment produces not only ﬂux between that segment and inﬁnity but also between each of the other segmentsand inﬁnity. For example, observe from Fig. 5.13 that superimposing the ﬂuxes MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 219 s l mI I I(a) (b) (c)mmmsml ss smll lssl lms l m s l m FIGURE 5.13. Mutual partial inductance for aligned but offset wires in terms of ﬂuxes to inﬁnity. opposite each segment that are due to currents on the other three segments gives ψl=ψll+ψls+ψlm (5.30a) ψs=ψsl+ψss+ψsm (5.30b) ψm=ψml+ψms+ψmm (5.30c) 220 THE CONCEPT OF “PARTIAL” INDUCTANCE where the notation ψijdenotes the ﬂux to inﬁnity opposite segment idue to a current I only on segment j. Keep in mind that these mutual inductances are reciprocal (i.e., Mij=Mji). But if all three segments have current Ion them, they produce the total ﬂux ψl+s+m. Hence, the self inductance of a wire of total length l+s+mcan be written as Ll+s+m=ψl+s+m I =ψl I+ψs I+ψm I =ψll I+ψls I+ψlm I +ψsl I+ψss I+ψsm I +ψml I+ψms I+ψmm I(5.31) The key to simplifying this and writing it in the form of (5.29c) is to write the result in terms of the self partial inductances of segments of a single lengthso that we can use the result derived in (5.18a,b) without having to rederive a new result [which we have already done in (5.29)]. To do this, note that the total ﬂuxes given by (5.31) can be written as L l+s+m=ψll I+ψls I+ψsl I+ψss I⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright Ll+s+ψmm I+ψms I+ψsm I+ψss I⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright Lm+s −ψss I⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Ls+ψlm I+ψml I⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 2Mp(5.32a) Solving this gives the result in (5.29c) since 2Mp=Mlm+Mml I =(Ll+s+m+Ls)−(Ll+s+Lm+s) (5.32b) This gives a very basic principle for adding inductors in series where the inductors have not only their self inductance but also mutual inductancesbetween each other: L1+2+3=L1+M12+M13+L2+M12+M23+L3+M13+M23 =L1+L2+L3+2M12+2M13+2M23 (5.33) MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 221 VL1 L2 L3M12M13 M23 I I FIGURE 5.14. Adding inductors in series. This can be veriﬁed from the electric circuit diagram in Fig. 5.14 by determin- ing the total voltage across the series combination using the dot convention[1,2]. The basic result in (5.28) for parallel, offset wires with d/=0 can be written similarly in terms of the result in Section 5.4 for the mutual partial inductancebetween two identical wires of lengths land separation dwhose endpoints coincide as shown in Fig. 5.10 and given in (5.21b): M l=μ0 2π⎜parenleftbigg lsinh−1l d−⎜radicalbig l2+d2+d⎜parenrightbigg d/greatermuchrw (5.21b) Hence, the result in (5.28) can be written in terms of (5.21b) as 2Mp=⎜parenleftbigMz2+Mz1−m⎜parenrightbig−⎜parenleftbigMz1+Mz2−m⎜parenrightbig =(Ml+s+m+Ms)−(Mm+s+Ml+s)(5.34) Notice again that (5.34) gives 2 MpsinceMlin (5.21b) is multiplied by μ0/2π, whereas the result for Mpin (5.28) is multiplied by μ0/4π. Note from (5.21b) that M0=0 (5.35a) and M−l=Ml (5.35b) with (5.35b) resulting from the identity sinh−1(−x)=− sinh−1x. If the wires overlap, replace swith−sin (5.34). We can easily determine the mutual partial inductance between the various offset structures shown in Fig. 5.15 by using the basic result in (5.34) and comparing each of these structures to Fig. 5.11, from which (5.34) was derivedin order to (1) determine the location point of z=0 on those structures, and (2) hence to determine the values of z 1andz2in (5.34). For example, 222 THE CONCEPT OF “PARTIAL” INDUCTANCE s (a)d (d)d(c)d(b)d2z 2z 2z 2z1z 1z 1z 1zq pp0=z 0=z 0=z 0=zml m m ml l ls = 0 ) (p l s− − = ) ( ) (q m p l s+ − = − − = FIGURE 5.15. Using the basic relation in (5.34) to determine the mutual inductance for other offset structures. in Fig. 5.15(a) we identify z2=l+s+mandz1=m+s. Hence, for the structure in Fig. 5.15(a) we obtain 2Mp=(Ml+s+m+Ms)−(Mm+s+Ms+l) Similarly, for the case in Fig. 5.15(b) we identify z2=l+mandz1=mor, equivalently, s=0. Hence, for the structure in Fig. 5.15(b) we obtain 2Mp=(Ml+m+M0)−(Mm+Ml) =Ml+m−Mm−Ml For the case in Fig. 5.15(c) we identify z2=p+mandz1=p+m−lor, equivalently, s=− (l−p). Hence, for the structure in Fig. 5.15(c) we obtain 2Mp=⎜parenleftbigMp+m+Mp−l⎜parenrightbig−⎜parenleftbigMp+m−l +Mp⎜parenrightbig =⎜parenleftbigMp+m+Ml−p⎜parenrightbig−⎜parenleftbigMp+m−l +Mp⎜parenrightbig MUTUAL PARTIAL INDUCTANCE BETWEEN PARALLEL WIRES 223 For the case in Fig. 5.15(d) we identify z2=p+mandz1=−q=p+ m−lor, equivalently, s=− (l−p)=− (m+q). Hence, for the structure in Fig. 5.15(d) we obtain 2Mp=⎜parenleftbigMp+m+M−q−m⎜parenrightbig−⎜parenleftbigM−q+Mp⎜parenrightbig =⎜parenleftbigMp+m+Mq+m⎜parenrightbig−⎜parenleftbigMq+Mp⎜parenrightbig Figure 5.16 shows how we could have easily obtained the basic result in (5.34) by using lumped-circuit analysis principles and the dot convention d s A aB bC c IMAa MBb MCcMBa IV Vlm l s m M M + + =net FIGURE 5.16. Combining mutual partial inductances that are in series. 224 THE CONCEPT OF “PARTIAL” INDUCTANCE [1,2]. We have shown an equivalent circuit for two parallel conductors of total length m+s+lalong with the mutual inductances between the three segments of lengths m,s, and l. Denoting the voltage between the endpoints of the ends of the top conductor as V, and passing a current Ithrough the lower conductor, the total contribution to Vdue to the mutual inductances between all segments is V=MnetdI dt Using the dot convention [1,2] and analyzing this circuit for the total voltage contributed to Vby the mutual inductances between the segments, the net mutual inductance between the entire lengths is Mnet=MAa+MAb+MAc+MBa+MBb+MBc+MCa+MCb+MCc But Mnet=Mm+s+l Ms+l=MAa+MAb+MBa+MBb Mm+s=MBb+MBc+MCb+MCc Ms=MBb Hence, we can write Mm+s+l=Ms+l+Mm+s−Ms+MAc+MCa However, the mutual inductance we desire between the two conductors of lengths landmis 2Mp=MAc+MCa Solving the last two relations gives the basic relation in (5.34), which we derived through a lengthy integration! 5.6 MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES AT AN ANGLE TO EACH OTHER We ﬁrst consider a special case of two straight wires of lengths landmthat are inclined with respect to each other at an angle θand joined at one end (or at least inﬁnitesimally close) as shown in Fig. 5.17. The solution for the mutualpartial inductance for this special case can be adapted to give the solution fora large class of similar problems, as we will see. This will be very similar MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 225 I 1dl2dl 1l2l 02 1= =l l m l=1l l=2 l mR 12R FIGURE 5.17. Wires inclined at an angle to each other. to our recognizing that the mutual partial inductance for the case for two parallel but offset wires shown in Fig. 5.11 could be obtained in terms of thesolution for two equal-length wires whose endpoints are aligned and shownin Fig. 5.10. This adaptation is given in (5.34) in terms of the mutual partialinductance of two equal-length parallel wires whose endpoints are alignedgiven in (5.21a,b). We obtain the mutual partial inductance for the conﬁguration in Fig. 5.17 using the Neumann integral: M p=μ0 4π⎜integraldisplay l2⎜integraldisplay l1dl1·dl2 R12 =μ0 4πcosθ⎜integraldisplay l2⎜integraldisplay l11 R12dl1dl2 (5.36a) where l1andl2are the contours along the axes of the two wires, and R12is the distance between the differential segments dl1anddl2given by R12=⎜radicalBig l2 1+l2 2−2l1l2cosθ (5.36b) and we have used the law of cosines. The dot product of the vector differential segments becomes dl1·dl2=cosθd l 1dl2. 226 THE CONCEPT OF “PARTIAL” INDUCTANCE We can place this integral in an integrable form using the following tech- nique [17]. We can show that (5.36a) can be written as Mp=μ0 4πcosθ⎜integraldisplay l2⎜integraldisplay l11 R12dl1dl2 =μ0 4πcosθ⎜integraldisplay l2⎜integraldisplay l1⎜bracketleftbiggd dl1⎜parenleftbiggl1 R12⎜parenrightbigg +d dl2⎜parenleftbiggl2 R12⎜parenrightbigg⎜bracketrightbigg dl1dl2 =μ0 4πcosθ⎜bracketleftbigg l1⎜integraldisplay l21 R12dl2+l2⎜integraldisplay l11 R12dl1⎜bracketrightbigg (5.37) This ﬁrst equivalence in (5.37) can be shown, using R12from (5.36b), to give d dl1⎜parenleftbiggl1 R12⎜parenrightbigg =R12−l1R−1 12(l1−l2cosθ) R2 12 =1 R12−l1(l1−l2cosθ) R3 12(5.38a) d dl2⎜parenleftbiggl2 R12⎜parenrightbigg =R12−l2R−1 12(l2−l1cosθ) R2 12 =1 R12−l2(l2−l1cosθ) R3 12(5.38b) and we have used d⎜parenleftbigu v⎜parenrightbig dx=vdu dx−udv dx v2(D65) Hence, d dl1⎜parenleftbiggl1 R12⎜parenrightbigg +d dl2⎜parenleftbiggl2 R12⎜parenrightbigg =1 R12(5.39) The second equivalence in (5.37) can easily be shown from ⎜integraldisplay l2⎜integraldisplay l1⎜bracketleftbiggd dl1⎜parenleftbiggl1 R12⎜parenrightbigg⎜bracketrightbigg dl1dl2=⎜integraldisplay l2⎜bracketleftbigg⎜integraldisplay l1d dl1⎜parenleftbiggl1 R12⎜parenrightbigg dl1⎜bracketrightbigg dl2 =⎜integraldisplay l2l1 R12dl2 =l1⎜integraldisplay l21 R12dl2 (5.40a) MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 227 ⎜integraldisplay l1⎜integraldisplay l2⎜bracketleftbiggd dl2⎜parenleftbiggl2 R12⎜parenrightbigg⎜bracketrightbigg dl2dl1=⎜integraldisplay l1⎜bracketleftbigg⎜integraldisplay l2d dl2⎜parenleftbiggl2 R12⎜parenrightbigg dl2⎜bracketrightbigg dl1 =⎜integraldisplay l1l2 R12dl1 =l2⎜integraldisplay l11 R12dl1 (5.40b) and we have obtained the equivalence in (5.37). But the last result in (5.37) can easily be integrated using integral 380.001 from Dwight [7]: ⎜integraldisplaydx√ x2+bx+c=ln⎜parenleftBig 2⎜radicalbig x2+bx+c+2x+b⎜parenrightBig (D380.001) and the equation for R12in (5.36b) to give ⎜integraldisplay li1 R12dli=⎜integraldisplay li1⎜radicalBig l2 i+l2 j−2liljcosθdli =ln⎜parenleftbigg 2⎜radicalBig l2 i+l2 j−2liljcosθ+2li−2ljcosθ⎜parenrightbigg =ln⎜parenleftbigg⎜radicalBig l2 i+l2 j−2liljcosθ+li−ljcosθ⎜parenrightbigg (5.41) The factor of 2 cancels out when we evaluate at the upper and lower limits of the integral. Now we apply this result to the problem of Fig. 5.17. For economy of notation we denote the mutual partial inductance between the two segmentsas M p=μ0 4πN (5.42) andNbecomes, in terms of the limits of the integrals, N=⎜integraldisplayB l2=A⎜integraldisplayb l1=a1 R12dl1dl2 =l1⎜integraldisplay l21 R12dl2+l2⎜integraldisplay l11 R12dl1 =⎜bracketleftbigg l1⎜integraldisplayB l2=A1 R12dl2⎜bracketrightbiggb l1=a+⎜bracketleftBigg l2⎜integraldisplayb l1=a1 R12dl1⎜bracketrightBiggB l2=A =b⎜braceleftBig ln⎜bracketleftBig⎜radicalbig b2+B2−2bBcosθ+B−bcosθ⎜bracketrightBig 228 THE CONCEPT OF “PARTIAL” INDUCTANCE −ln⎜bracketleftBig⎜radicalbig b2+A2−2bAcosθ+A−bcosθ⎜bracketrightBig⎜bracerightBig −a⎜braceleftBig ln⎜bracketleftBig⎜radicalbig a2+B2−2aBcosθ+B−acosθ⎜bracketrightBig −ln⎜bracketleftBig⎜radicalbig a2+A2−2aAcosθ+A−acosθ⎜bracketrightBig⎜bracerightBig +B⎜braceleftBig ln⎜bracketleftBig⎜radicalbig b2+B2−2bBcosθ+b−Bcosθ⎜bracketrightBig −ln⎜bracketleftBig⎜radicalbig a2+B2−2aBcosθ+a−Bcosθ⎜bracketrightBig⎜bracerightBig −A⎜braceleftBig ln⎜bracketleftBig⎜radicalbig b2+A2−2bAcosθ+b−Acosθ⎜bracketrightBig −ln⎜bracketleftBig⎜radicalbig a2+A2−2aAcosθ+a−Acosθ⎜bracketrightBig⎜bracerightBig =blnRbB+B−bcosθ RbA+A−bcosθ−alnRaB+B−acosθ RaA+A−acosθ +BlnRbB+b−Bcosθ RaB+a−Bcosθ−AlnRAb+b−Acosθ RaA+a−Acosθ(5.43) The beginning and ending coordinates of the two lines are denoted as a,bfor l1andA,Bforl2. The distances Rijare the distances between the endpoints of the segments. For the problem in Fig. 5.17, we obtain N=⎜bracketleftbigg l1⎜integraldisplay l21 R12dl2+l2⎜integraldisplay l11 R12dl1⎜bracketrightbigg =⎜braceleftBigg⎜bracketleftBigg l1⎜integraldisplayl l2=01 R12dl2⎜bracketrightBiggm l1=0+⎜bracketleftbigg l2⎜integraldisplaym l1=01 R12dl1⎜bracketrightbiggl l2=0⎜bracerightBigg =mlnRml+l−mcosθ Rm0+0−mcosθ−0l nR0l+l−0 cosθ R00+0−0 cosθ +llnRml+m−lcosθ R0l+0−lcosθ−0l nR0m+m−0 cosθ R00+0−0 cosθ =llnR+m−lcosθ l−lcosθ+mlnR+l−mcosθ m−mcosθ(5.44) and, by using l’H ˆopital’s rule, lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright x→0xln(x)=0 (D605) MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 229 The distance between the endpoints is denoted as R=Rml =⎜radicalbig l2+m2−2lmcosθ (5.45) andRm0=mandR0l=l. Substituting the result in (5.44) into (5.42) gives the mutual partial inductance between the two segments in Fig. 5.17: Mp=μ0 4πcosθ⎜parenleftbigg llnR+m−lcosθ l−lcosθ+mlnR+l−mcosθ m−mcosθ⎜parenrightbigg (5.46a) But this result can be put into an equivalent form as [14] Mp=μ0 4πcosθ⎜parenleftbigg llnR+m+l R+l−m+mlnR+l+m R+m−l⎜parenrightbigg (5.46b) To demonstrate the equivalence between the two forms of the result in (5.46) we need to show that R+m−lcosθ l−lcosθ=R+m+l R+l−m(5.47) This can be shown directly by multiplying it out to give R2+R(l−lcosθ)+(m−lcosθ)(l−m) ?=R(l−lcosθ)+(m+l)(l−lcosθ) or R2?=(m+l)(l−lcosθ)−(m−lcosθ)(l−m) =l2+m2−2mlcosθ which is satisﬁed. In fact, a more general result can be proven which will be useful for other situations. Consider the triangles shown in Fig. 5.18. Each triangle is com-posed of two sides labeled RandR /primewith included angles θandθ/primewith respect to the horizontal axes. These sides RandR/primemake projections on the hori- zontal axes of PandP/prime, respectively, where P=RcosθandP/prime=R/primecosθ/prime. The total length on the horizontal axis between the intersections of each linewith the horizontal axis is denoted as T. We can prove the following important 230 THE CONCEPT OF “PARTIAL” INDUCTANCE R R' h θ′ θh (a)( b)R' R θ′ θ P' P′P P P P T′+ =P P T−′= FIGURE 5.18. Important theorem. equivalences. For the left triangle in Fig. 5.18(a) we have lnR+P R/prime−P/prime=lnR/prime+P/prime R−P =lnR+R/prime+T R+R/prime−T =2 tanh−1T R+R/prime(5.48a) and for the right triangle in Fig. 5.18(b) we have lnR−P R/prime−P/prime=lnR/prime+P/prime R+P =lnR+R/prime+T R+R/prime−T =2 tanh−1T R+R/prime(5.48b) The conversion of (5.48a) to (5.48b) is accomplished simply by replacing Pin (5.48a) with −P. This is somewhat evident since Pin Fig. 5.18(a) adds to P/prime to give the total length between the endpoints of RandR/prime, which is denoted as T=P/prime+P, whereas in Fig. 5.18(b) Psubtracts from P/primeto give T=P/prime−P. The identity for Fig. 5.17 in (5.47) follows from the identity in (5.48a). The proofs of (5.48) are fairly simple by comparing the arguments of the log functions. For example, (5.48a) gives R+P R/prime−P/prime?=R/prime+P/prime R−P?=R+R/prime+T R+R/prime−T MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 231 Multiplying these out gives (R+P)(R−P)?=⎜parenleftbigR/prime+P/prime⎜parenrightbig⎜parenleftbigR/prime−P/prime⎜parenrightbig ButR2=P2+h2andR/prime2=P/prime2+h2. Substituting T=P+P/prime, we need to show that R+P R/prime−P/prime?=R+R/prime+⎜parenleftbigP+P/prime⎜parenrightbig R+R/prime−(P+P/prime) Multiplying this out and canceling common terms gives R2−P2?=R/prime2−P/prime2 which is satisﬁed. The results in (5.48b) can be veriﬁed similarly. The last results in (5.48) are veriﬁed using the identity for the inverse hyperbolic tan-gent: tanh −1x=1 2ln1+x 1−xx2<1 (D702) Hence, a further equivalent form for the result in (5.46) for Fig. 5.17 can be obtained in terms of the inverse hyperbolic tangent as Mp=μ0 4πcosθ⎜parenleftbigg llnR+m+l R+l−m+mlnR+l+m R+m−l⎜parenrightbigg =μ0 2πcosθ⎜parenleftbigg ltanh−1m R+l+mtanh−1l R+m⎜parenrightbigg (5.46c) This solution process for the conﬁguration of Fig. 5.17 can readily be adapted to obtain the mutual partial inductance between two segments that donot physically join at a common point but are inclined at an angle θto each other as shown in Fig. 5.19. Extend the segments of lengths landmto a point where they join, thereby generating the extension lengths αandβ. Adapting θl mα β1R 2R 3R4R FIGURE 5.19. More general case of Fig. 5.17. 232 THE CONCEPT OF “PARTIAL” INDUCTANCE the result in (5.43) gives the result for Fig. 5.19 as N=l1⎜integraldisplay l21 R12dl2+l2⎜integraldisplay l11 R12dl1 =⎧ ⎨ ⎩⎜bracketleftBigg l1⎜integraldisplayα+l l2=α1 R12dl2⎜bracketrightBiggβ+m l1=β+⎜bracketleftbigg l2⎜integraldisplayβ+m l1=β1 R12dl1⎜bracketrightbiggα+l l2=α⎫ ⎬ ⎭ =(β+m)lnR(β+m)(α+l)+(α+l)−(β+m)cosθ R(β+m)α+α−(β+m)cosθ −βlnRβ(α+l)+(α+l)−βcosθ Rβα+α−βcosθ +(α+l)lnR(β+m)(α+l)+(β+m)−(α+l)cosθ Rβ(α+l)+β−(α+l)cosθ −αlnRα(β+m)+(β+m)−αcosθ Rβα+β−αcosθ(5.49) Denoting the distances between the endpoints of the lines as shown in Fig. 5.19 gives R1=R(α+l)(β+m) =⎜radicalBig (α+l)2+(β+m)2−2(α+l)(β+m)cosθ (5.50a) R2=R(α+l)β =⎜radicalBig (α+l)2+β2−2(α+l)βcosθ (5.50b) R3=Rαβ =⎜radicalBig α2+β2−2αβcosθ (5.50c) R4=Rα(β+m) =⎜radicalBig α2+(β+m)2−2α(β+m)cosθ (5.50d) MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 233 Hence, the mutual partial inductance between the two segments of Fig. 5.19 becomes Mp=μ0 4πN =μ0 4π⎜bracketleftbigg (β+m)l nR1+(α+l)−(β+m)cosθ R4+α−(β+m)cosθ −βlnR2+(α+l)−βcosθ R3+α−βcosθ +(α+l)lnR1+(β+m)−(α+l)cosθ R2+β−(α+l)cosθ −αlnR4+(β+m)−αcosθ R3+β−αcosθ⎜bracketrightbigg (5.51a) Using the identities in (5.48) and the inverse hyperbolic tangent identity in (D702) gives equivalent forms as Mp=μ0 4πN =μ0 4π⎜bracketleftbigg (β+m)lnR1+R4+l R1+R4−l−βlnR2+R3+l R2+R3−l +(α+l)lnR1+R2+m R1+R2−m−αlnR4+R3+m R4+R3−m⎜bracketrightbigg =μ0 2π⎜bracketleftbigg (β+m)tanh−1 l R1+R4−βtanh−1 l R2+R3 +(α+l)tanh−1 m R1+R2−αtanh−1 m R4+R3⎜bracketrightbigg (5.51b) We can then use the previous result for the mutual partial inductance between two segments of lengths xandythat are joined at a common point that was derived for Fig. 5.17 and given in (5.46c) to obtain the mutual partialinductance for Fig. 5.19 indirectly. Denote the result for Fig. 5.17 as M x,y=μ0 4πcosθ⎜parenleftbigg xlnR+x+y R+x−y+ylnR+y+x R+y−x⎜parenrightbigg =μ0 2πcosθ⎜parenleftbigg xtanh−1y R+x+ytanh−1x R+y⎜parenrightbigg (5.52a) 234 THE CONCEPT OF “PARTIAL” INDUCTANCE where θis the included angle where they are joined and R=⎜radicalBig x2+y2−2xycosθ (5.52b) Visualize the structure of Fig. 5.19 as consisting of four such structures, each consisting of the following lengths, with each pair being joined at a commonpoint: (1) x=α+l,y=β+m, (2) x=α,y=β, (3)x=α+l,y=β, and (4) x=α, and y=β+m. We can then obtain the total mutual partial inductance for the structure in Fig. 5.19 of overall lengths α+landβ+mto give, in a fashion similar to that of Fig. 5.16, M α+l,β +m=Mα,β+Mα,m+Ml,β+Ml,m The desired result is Ml,mgiving Ml,m=Mα+l,β +m−Mα,β−Mα,m−Ml,β But the result for Fig. 5.17 does not apply to generating Mα,morMl,βsince the two lengths in each of these are not joined at a common point. So we writethis as M l,m=Mα+l,β +m−Mα,β−⎜parenleftbigMα,m+Mα,β⎜parenrightbig ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Mα,β+m−⎜parenleftbigMl,β+Mα,β⎜parenrightbig ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Mα+l,β+2Mα,β giving Mp=⎜parenleftbigMα+l,β +m+Mαβ⎜parenrightbig−⎜parenleftbigMα+l,β+Mβ+m,α⎜parenrightbig(5.53) Using the result for two segments joined at one end in (5.52) gives the result as Mp=μ0 4πcosθ⎜bracketleftbigg (α+l)lnR1+(α+l)+(β+m) R1+(α+l)−(β+m) +(β+m)lnR1+(β+m)+(α+l) R1+(β+m)−(α+l) +αlnR3+α+β R3+α−β+βlnR3+β+α R3+β−α −(α+l)lnR2+(α+l)+β R2+(α+l)−β−βlnR2+β+(α+l) R2+β−(α+l) −(β+m)lnR4+(β+m)+α R4+(β+m)−α−αlnR4+α+(β+m) R4+α−(β+m)⎜bracketrightbigg (5.54a) MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 235 This result can be simpliﬁed to Mp=μ0 4πcosθ⎜bracketleftbigg (α+l)lnR1+R2+m R1+R2−m+(β+m)lnR1+R4+l R1+R4−l −αlnR3+R4+m R3+R4−m−βlnR2+R3+l R2+R3−l⎜bracketrightbigg (5.54b) which agrees with (5.51b). The equivalence of (5.54a) and (5.54b) can be shown with the following important identity for triangles. Consider the three triangles shown inFig. 5.20. Triangle T 1has sides of a,b, and R1. Triangle T2has sides of a,c, and R2. Triangle T3has sides of R1,b−c, andR2and is formed from tri- angles T1andT2asT3=T1−T2. It is a simple matter to prove the following (a ( ) b)1T 2Tc2R θa 1R bθa 1R c2R θa b – c (c)1R 2R 2 1 3T T T− = FIGURE 5.20. Important identity for triangles. 236 THE CONCEPT OF “PARTIAL” INDUCTANCE identity for these three related triangles: lnR1+a+b R1+a−b−lnR2+a+c R2+a−c=lnR1+R2+(b−c) R1+R2−(b−c)(5.55a) andR1andR2are given by the law of cosines: R2 1=a2+b2−2abcosθ (5.55b) R2 2=a2+c2−2accosθ (5.55c) The important identity in (5.55a) can easily be veriﬁed by multiplying out the arguments of the logarithms as ln A−lnB=lnC⇒A/B=C. Applying (5.55a) to (5.54a) gives the equivalence to (5.54b). Figure 5.21 shows the general conﬁguration for skewed and displaced con- ductors. The general result for this was derived by G.A. Campbell in 1915[17]. This ﬁgure is modeled after that of Grover [14], pp. 56, who clearlyexplained the general result obtained by Campbell. The ﬁrst conductor is oflength land its endpoints are denoted as AandB. It is shown as lying in a plane. The second conductor is of length mand its endpoints are denoted as aandb. It is shown as lying in another plane. These two planes containing the two conductors are parallel and separated by distance dbetween the two θ R1R2 R3R4A B l αP pd ma bCPlane containing AB βR12 FIGURE 5.21. General conﬁguration for skewed and displaced conductors. MUTUAL PARTIAL INDUCTANCE BETWEEN WIRES 237 planes. The line Ppbetween the two planes is of length dand is mutually perpendicular to the two planes containing the two conductors. Hence, Ppis said to be the common perpendicular to the two conductors. The endpoints of the conductors, Aanda, are displaced from points Pandpby distances α andβ, respectively. The line PClying in the plane containing ABis parallel to the line abrepresenting the second conductor and is at an angle θto the ﬁrst conductor AB. This is what is meant by the two conductors ABandab having an angle of inclination of θwith respect to each other. If the displace- ment between the planes, d, is zero, d=0, then the angle θbetween the two conductors is the same as in the previous results. The Neumann integral in (5.36) remains the same for this case: Mp=μ0 4πcosθ⎜integraldisplay l1⎜integraldisplay l11 R12dl1dl2 (5.56a) where l1andl2again denote the contours along the two conductors of lengths mandl, respectively, and R12=⎜radicalBig d2+l2 1+l2 2−2l1l2cosθ (5.56b) Carrying through with a similar development as before gives Mp=μ0 4πcosθ⎜integraldisplay l2⎜integraldisplay l11 R12dl1dl2 =μ0 4πcosθ⎜integraldisplay l2⎜integraldisplay l1⎜bracketleftBigg d dl1⎜parenleftbiggl1 R12⎜parenrightbigg +d dl2⎜parenleftbiggl2 R12⎜parenrightbigg −d2 R3 12⎜bracketrightBigg dl1dl2 =μ0 4πcosθ⎜parenleftbigg l1⎜integraldisplay l21 R12dl2+l2⎜integraldisplay l11 R12dl1 −d sinθ⎜integraldisplay l2⎜integraldisplay l1dsinθ R3 12dl1dl2⎜parenrightBigg (5.57a) and one can similarly show using R12in (5.56b), as was done previously for d=0, that d dl1⎜parenleftbiggl1 R12⎜parenrightbigg +d dl2⎜parenleftbiggl2 R12⎜parenrightbigg −d2 R3 12=1 R12(5.57b) 238 THE CONCEPT OF “PARTIAL” INDUCTANCE Again, the last result in (5.57a) can be integrated, using (D380.001), to yield Mp=μ0 4πcosθ⎜parenleftBigg l1⎜integraldisplay l21 R12dl2+l2⎜integraldisplay l11 R12dl1 −d sinθ⎜integraldisplay l2⎜integraldisplay l1dsinθ R3 12dl1dl2⎜parenrightBigg =μ0 4πcosθ⎜bracketleftBigg⎜bracketleftbigg l1ln(R12+l2−l1cosθ)+l2ln(R12+l1−l2cosθ) −/Omega1d sinθ⎜bracketrightbiggl2=PB l2=PA⎜bracketrightBiggl1=pb l1=pa =μ0 2π⎜parenleftbigg pB/primetanh−1 ab aB+Bb−pA/primetanh−1 ab aA+Ab +Pb/primetanh−1AB Ab+bB−Pa/primetanh−1AB Aa+aB−/Omega1d tanθ⎜parenrightbigg (5.58a) where the solid angle /Omega1is /Omega1=tan−1⎜parenleftbiggPp Bbcotθ+PB Pppb Bbsinθ⎜parenrightbigg −tan−1⎜parenleftbiggPp Bacotθ+PB Pppa Basinθ⎜parenrightbigg −tan−1⎜parenleftbiggPp Abcotθ+PA Pppb Absinθ⎜parenrightbigg +tan−1⎜parenleftbiggPp Aacotθ+PA Pppa Aasinθ⎜parenrightbigg (5.58b) In (5.58a) primes denote the projection of the point on one conductor perpen- dicular to and onto the other conductor, and the inverse hyperbolic tangent isagain deﬁned in terms of the natural logarithm as tanh −1x=1 2ln1+x 1−xx2<1 (D702) Grover [14] simpliﬁed this in terms of the quantities in Fig. 5.21 and the result becomes Mp=μ0 2πcosθ⎜bracketleftbigg (α+l)tanh−1 m R1+R2+(β+m)tanh−1 l R1+R4 −αtanh−1 m R3+R4−βtanh−1 l R2+R3⎜bracketrightbigg −μ0 4π/Omega1d tanθ (5.59a) NUMERICAL V ALUES OF PARTIAL INDUCTANCES AND SIGNIFICANCE 239 where the solid angle /Omega1is /Omega1=tan−1d2cosθ+(α+l)(β+m)sin2θ dR1sinθ −tan−1d2cosθ+(α+l)βsin2θ dR2sinθ +tan−1d2cosθ+αβsin2θ dR3sinθ −tan−1d2cosθ+α(β+m)sin2θ dR4sinθ(5.59b) The distances between the ends of the two conductors are shown in Fig. 5.21 and are R1=Bb,R2=Ba,R3=Aa, andR4=Ab. Using the law of cosines, these distances are R2 1=d2+(α+l)2+(β+m)2−2(α+l)(β+m)cosθ(5.60a) R2 2=d2+(α+l)2+β2−2β(α+l)cosθ (5.60b) R2 3=d2+α2+β2−2αβcosθ (5.60c) R2 4=d2+α2+(β+m)2−2α(β+m)cosθ (5.60d) The only difference between this result and the result for Fig. 5.19 given in (5.51b) is the solid angle /Omega1, which goes away for d=0. 5.7 NUMERICAL V ALUES OF PARTIAL INDUCTANCES AND SIGNIFICANCE OF INTERNAL INDUCTANCE It is helpful to obtain some representative values of the self and mutual partial inductances for typical conﬁgurations. The self partial inductance of a wireof radius r wand length lis obtained as Lp=2×10−7l⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠ −⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ (5.18a) Observe that this depends on the ratio of the wire length and the wire radius: l/rw. It is typical to specify wire radii rwin mils (1000 mils =1 in. and 1 in.=2.54 cm). Also observe that the length of the wire, l, also appears outside 240 THE CONCEPT OF “PARTIAL” INDUCTANCE the equation. Hence, it is not possible to speak of an absolute per-unit-length inductance as is the case for a two-wire transmission line of inﬁnite length.Nevertheless, we can divide both sides of (5.18a) by the wire length and obtaina universal plot of the ratio of self partial inductance per unit length, L p/l, versus the ratio l/rwas Lp l=5.08⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠ −⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ nH/in. (5.61) This is shown in Fig. 5.22 for ratios of 10 ≤l/rw≤500. For example, a No. 20 gauge (AWG) wire is a common wire size and has a radius of 16 mils.Hence, the last ratio of 500 plotted represents a wire length of 8 in. for a No. 20gauge wire (30.02 nH/in.), and a ratio of 10 represents a length of 0.16 in.,or about 3/16 in. (10.63 nH/in.). Because the wire length lappears outside the result, it is not possible to state a single per-unit-length value of the selfpartial inductance. But the plot in Fig. 5.22 indicates that a reasonable rule ofthumb for practical wire sizes and wire lengths is a value of between 15 and30 nH/in. Self partial inductance (nH/inch) Ratio of wire length to wire radius, l/r wLp/l (nH/inch)35 30252015 100 100 200 300 400 500 FIGURE 5.22. Plot of Lp/l(nH/in.) vs. the ratio of wire length to wire radius, l/rw. NUMERICAL V ALUES OF PARTIAL INDUCTANCES AND SIGNIFICANCE 241 We obtained the dc per-unit-length value of the internal inductance of a wire (which is independent of the wire radius) as linternal=μ0 8π =0.5×10−7H/m =1.27nH /in. (5.62) Technically, this should be multiplied by the wire length and added to the ex- ternal self partial inductance in (5.18a) to give the total self partial inductance: Lp,total=Lp,(5.18a) +linternal×l (5.63) But we see from Fig. 5.22 that for practical situations the internal inductance of the wire can generally be neglected. Furthermore, the value for the internalinductance in (5.62) is its value at dc. As frequency is increased from zero, thecurrent tends to move toward the surface of the wire, and hence the internalinductance goes to zero. This gives further support to the observation that theinternal inductance can generally be neglected. The mutual partial inductance between two wires of common length land separation dis obtained as M p=2×10−7l⎡ ⎣ln⎛⎝ l d+⎜radicalBigg⎜parenleftbiggl d⎜parenrightbigg2 +1⎞ ⎠ −⎜radicalBigg 1+⎜parenleftbiggd l⎜parenrightbigg2 +d l⎤ ⎦ (5.21a) As was the case for self partial inductance, notice that this depends on the ratio of wire length to wire separation, l/d. But the wire length, l, also appears outside the result, so it is not possible to speak of an absolute value of per-unit-length mutual inductance, as is the case for a transmission line of inﬁnitelength. Nevertheless, we can divide both sides of (5.21a) by the wire lengthand obtain a universal plot of the ratio of the per-unit-length mutual partialinductance, M p/l, versus the ratio l/das Mp l=5.08⎡ ⎣ln⎛⎝ l d+⎜radicalBigg⎜parenleftbiggl d⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggd l⎜parenrightbigg2 +d l⎤ ⎦ nH/in. (5.64) This is plotted in Fig. 5.23 for ratios of 1 ≤l/d≤100. For example, a ratio of 80 would apply to two wires of length 5 in. and a separation between themof 0.0625 in., or 1/16 in. (20.77 nH/in.), and a ratio of 10 would apply to 242 THE CONCEPT OF “PARTIAL” INDUCTANCE Mutual partial inductance (nH/inch) Ratio of wire length to wire separation, l /d0 20 40 60 80 10025 20 1510 5 0 –5Mp/l (nH/inch) FIGURE 5.23. Plot of Mp/lin (nH/in.) vs. the ratio of wire length to wire separation l/d. two wires of length 5 in. and a separation between them of 1/2 in. (10.63 nH/in.). Observe that as the wire separation increases without bound (i.e., theratio goes to zero), the mutual partial inductance goes to zero: an expectedresult. Similarly, as the wire separation goes to zero (approaches the radii ofthe wires) (i.e., the ratio increases), the mutual partial inductance approachesthe self partial inductance shown in Fig. 5.22: again, an expected result. 5.8 CONSTRUCTING LUMPED EQUIV ALENT CIRCUITS WITH PARTIAL INDUCTANCES Unlike the case of loop inductances, for current loops whose borders are bounded by piecewise-linear segments of wires, there are no further partialinductances to be derived. We simply “put together” the partial inductances(self and mutual) derived previously in this chapter and “turn the crank.”We construct an equivalent lumped-circuit model that can be solved with,for example, the SPICE circuit analysis computer program [2]. To do so,we ﬁnally need to discuss the allocation of the dots in an inductor equivalentcircuit of the segments. The key to doing so is to be able to determine correctlythe total voltage developed across the segment, magnitude and polarity, byusing the dot convention that replicates the derivation of that inductance in thischapter. The self partial inductance of a segment determines the voltage across CONSTRUCTING LUMPED EQUIV ALENT CIRCUITS 243 Ij Ij ViLpiLpj Mpij Aij FIGURE 5.24. Mutual partial inductance between pairs of segments. it that is due to the current through the element according to the passive sign convention: Current entering one end of the inductance produces a voltageacross the element that is positive at that end. The dots do not have anythingto do with this self voltage. Let us ﬁrst address the situation for a pair of segments shown in Fig. 5.24. The dots are placed on the ends of two elements so that the magnitude andpolarity of the contribution to the voltage across one of the elements that isdue to the current through the other element via the mutual partial inductancebetween the associated segments will be determined correctly. The key to doing so is to replicate the situation for which the mutual partial inductancebetween two segments was as derived in this chapter. Note that if a current I on one segment enters the dotted end of that segment, a voltage M pdI/dt will be developed along the other segment that is positive at the dotted end of thatsegment: V i=MpijdIj dt The key to getting the dots placed correctly on a pair of segments is observed to be in the relation between the current in one segment and the direction ofthe vector magnetic potential Aalong the other segment, which was used in the derivations of the mutual partial inductance. The vector magnetic potentialAis everywhere parallel to the current that produced it. Hence, the positive terminal of the induced voltage is on the end of the segment that Aenters, as shown in Fig. 5.24. In other words, Apoints from the positive terminal of the induced voltage tothe negative terminal of the induced voltage. This can be done easily for a pair of elements. For more than two cou- pled segments we must arbitrarily place the dots on the ends of the inductor symbols for each segment. But some of the mutual partial inductances mayturn out to be negative for that placement. A good example of this is the rect- angular loop shown in Fig. 5.1. The inductive equivalent circuit is shown inFig. 5.3 and the dots are assigned arbitrarily. Observe that a current directed 244 THE CONCEPT OF “PARTIAL” INDUCTANCE 1 2 3 (a)1I 12A 13A (b)1V I2 3IMp12−Mp13 Lp1 Lp2 Lp3 FIGURE 5.25. Assigning dots to the segments. down through the right segment of the loop produces a vector magnetic po- tential along the left side of the loop that is also directed downward. But thisis opposite the assigned dotted terminal of the left inductor. Hence, M 13here is negative. Figure. 5.25 shows an example of this. Assigning the dots arbitrarily gives the inductive equivalent circuit in Fig. 5.25(b). The voltage across the ﬁrstinductor is assigned the polarity of positive at its dotted end. Directing acurrent I 1through the ﬁrst inductor that enters the assigned dotted end of it as shown in Fig. 5.25(a) generates vector magnetic potentials along the othersegments that enters the dotted end of the third segment (assigned arbitrarily)so that the mutual inductance between the ﬁrst and third inductors is positiveasM p13. But current I1that enters the dotted terminal of the ﬁrst inductor (arbitrarily assigned) generates a vector magnetic potential that enters theundotted end of the second segment (assigned arbitrarily) so that the mutualpartial inductance between that pair is −M p12, where Mp12here has a positive CONSTRUCTING LUMPED EQUIV ALENT CIRCUITS 245 value. Hence, using the dot convention, the voltage generated across the ﬁrst inductor that is due only to the mutual inductances is V1=−⎜parenleftbig−Mp12⎜parenrightbigdI2 dt−Mp13dI3 dt =+Mp12dI2 dt−Mp13dI3 dt Computer-aided circuit analysis programs such as SPICE require that all self inductances such as Lphere must be positive. However, there are no restric- tions on the signs of any of the mutual inductances: Some may have negativevalues. 6 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULARCROSS SECTION In this chapter we obtain the self and mutual partial inductances for conductors of rectangular cross section, referred to here as printed circuit board (PCB)lands. Figure 6.1 shows this type of conductor. The width is denoted as w, the length is denoted as l, and the thickness is denoted as t. In previous chapters we have detailed the computation of inductances for conductors having circular, cylindrical cross sections (i.e., wires). The com-putation of the partial inductances of and between wires is fairly simple, foran important reason. We consistently made the assumption that the current carried by a wire is uniformly distributed over the cross section of the wirewhich is true for dc and widely spaced wires. In this case, for the purposes of computing the magnetic ﬁelds from that wire ,we can replace the wire with a ﬁlament containing the total current I=JA, where Jis the uniform current density distribution and Ais the area of the wire cross section. This is an extraordinarily important simplifying assumption, for a number of reasons.First, we can equate the self partial inductance of a wire having a uniformcurrent distribution over its cross section to the total magnetic ﬂux threading the surface formed between the surface of the wire and inﬁnity per unit of that current. No matter what radial direction about the wire we choose togo to inﬁnity, the result is the same since the magnetic ﬁeld is symmet-ric about the wire. This provides important alternative methods of directly Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 246 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION 247 t wl FIGURE 6.1. Printed circuit board land. computing the self and mutual partial inductances of wires. We can directly compute the magnetic ﬂux through that surface from the surface integral of themagnetic ﬂux density vector B, or we could integrate via a line integral the vector magnetic potential Aalong the surface of the wire, which leads to the third method, the Neumann integral. Now consider the case of the PCB land. Even if we assume that the current is distributed uniformly over the cross section of the land (as we do in thischapter), the magnetic ﬁelds about the land do not form concentric circles,and hence we cannot replace the land with an equivalent ﬁlament containingthe total current for the purposes of computing the magnetic ﬁelds due toit. [See Chapter 2 for the ﬁelds around an inﬁnitely long, ﬂat conductor ofwidth Wgiven in (2.70).] Hence, for lands, we can no longer relate the partial inductances to the magnetic ﬂux between the land surface and inﬁnity: Fromwhich point on the land shall we draw the boundaries of the surface? Althoughthe magnetic ﬁelds about a land will appear at very large distances as thoughthey are due to the current from a ﬁlament, the computation of the ﬂux atnearer distances is dominant. We need another way of meaningfully formulating the partial inductances of and between lands. In the next section we derive that formulation in termsof stored energy in the magnetic ﬁeld. A fundamental assumption in thatderivation is again that the current is distributed uniformly over the crosssection of the land. This assumption is also true for dc currents and will providea simpliﬁcation in the computation, as we will see. When is this assumptionof a uniform current distribution over the land cross section invalidated? Asin the case of wires, adjacent land currents will cause the currents to migratetoward the facing surfaces of the lands in the same way that closely spacedwires will cause their current distributions to move toward the facing surfacesof the wires. This is again the phenomenon of “proximity effect.” For wires, 248 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION this was not pronounced enough to invalidate the replacement of the wire with a ﬁlament as long as the ratio of wire separation to wire radius was largerthan about 4 (i.e., for two identical wires, another identical wire would justﬁt between the two). So this assumption of uniform current distribution overthe cross section of the wire and the subsequent replacement of the wire witha ﬁlament is not a limiting assumption for wires having typical separations.For two lands this region of separation for the current to be approximatelyuniformly distributed over the cross section cannot easily be determined, butwe will make the assumption that the proximity effect is not pronounced inorder to make computation of the self and mutual partial inductances feasible.We provide numerical computations in Section 6.5 to give us some feel forwhat “too close” means for this problem. Recall that the fundamental computation of inductance is for dc currents! If a wire or a PCB land carries a dc current and is isolated from other wiresor PCB lands, the current will be distributed uniformly over the wire or landcross section. For wires, as the frequency of the current increases from dc, thecurrent will crowd to the surface, lying in a region of the surface of thicknesson the order of a skin depth, but the current will remain symmetric about the wire axis and can still be replaced by a ﬁlament as long as the wire is not “too close” to other wires. The internal inductance will go to zero but the externalinductances will remain the same. In the case of a PCB land, as the frequencyof the current also increases from dc, the current will migrate toward thesurface of the land but it will also peak at the sharp corners of the land ,a s numerical computations in Section 6.5 will show. Hence, the advantage of awire current of increasing frequency remaining symmetric about the wire axisis not shared by the PCB land, even if the proximity effect is not pronounced. This prior discussion illustrates that although the computation of loop and partial inductances for wires was rather straightforward, the same computa- tions for PCB lands will be much more difﬁcult . Even using the assumption that the current is uniformly distributed over the cross section of the land, thecomputation of its self and mutual partial inductances is considerably moredifﬁcult than for wires. Hoer and Love [16] provide general formulas for theself and mutual inductances of PCB lands. These formulas are not derived in[16], but in the following we provide detailed derivations of them. Again thederivations of these formulas for PCB lands are very complicated, as are theresulting formulas themselves. 6.1 FORMULATION FOR THE COMPUTATION OF THE PARTIAL INDUCTANCES OF PCB LANDS As indicated previously, we need to determine a suitable method for computing the self and mutual partial inductances of PCB lands. The method we use is in FORMULATION FOR THE COMPUTATION 249 terms of the energy stored in the magnetic ﬁeld. First recall that the magnetic energy stored in the magnetic ﬁeld is given by WM=1 2⎜integraldisplay all spaceB·Hdv J (6.1) Recall that we deﬁne the vector magnetic potential Ain terms of the magnetic ﬂux density vector BasB=∇ ×A. Substituting into 6.1 gives WM=1 2⎜integraldisplay all space(∇×A)·Hdv (6.2) Substituting the vector identity [3] (∇×A)·H=∇ ·(A×H)+A·(∇×H) (6.3) into (6.2) and using Amp `ere’s law for dc currents, ∇×H=J,g i v e s WM=1 2⎜integraldisplay all space∇·(A×H)dv+1 2⎜integraldisplay all spaceA·Jdv (6.4) Applying the divergence theorem (see the Appendix) to the ﬁrst integral gives WM=1 2⎜contintegraldisplay s∞(A×H)·ds+1 2⎜integraldisplay all spaceA·Jdv (6.5) where s ∞is the closed surface at inﬁnity. Now recall the Biot–Savart law in (2.11) and the equation for computing the vector magnetic potential in(2.47a). These show that at points far from a ﬁnite current distribution, themagnitude of Awill decrease at a rate greater than or equal to the inverse distance (1/r ), and the magnitude of Bwill decrease at a rate greater than or equal to the inverse distance squared (1 /r 2). The AandBﬁelds of some current distributions, such as the current loop in Fig. 2.25 whose ﬁelds werederived in Chapter 2, decrease at large distances from the loop at a greater rate:1/r 2and 1/r3, respectively. In spherical coordinates, the differential surface is ds=r2sinθd θd φ . Since the product of the magnitudes of AandHdecrease at a rate no less than 1 /r3, the ﬁrst integral in (6.5) over the surface at inﬁnity, s∞, will go to zero. The remaining integral is over all space, but the current density Jis zero except where the current is located. Hence, the result for the stored energy is an integral only over the volume containing the current: WM=1 2⎜integraldisplay throughout the volume containing the currentA·Jdv (6.6) Now consider applying the result in (6.6) to the case of two lands. Each land carries a total current I1andI2, respectively. The total vector magnetic potential in the space around the two lands is A=A1+A2, where A1is due 250 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION to current I1that is carried by land 1, and A2is due to current I2that is carried by land 2. The total magnetic energy in the ﬁeld surrounding the lands is WM=1 2⎜integraldisplay v1A·J1dv1+1 2⎜integraldisplay v2A·J2dv2 (6.7) where volumes v1andv2are the volumes of the respective lands that en- close the respective current densities, J1andJ2. Substituting the total vector magnetic potential in the space surrounding the two lands, A=A1+A2, gives WM=1 2⎜integraldisplay v1A1·J1dv1+1 2⎜integraldisplay v2A1·J2dv2+1 2⎜integraldisplay v2A2·J2dv2 +1 2⎜integraldisplay v1A2·J1dv1 (6.8a) Our intent is to derive self and mutual partial inductances that can be used to model these two lands, as illustrated in Fig. 5.6. The total magnetic energy inthe ﬁeld is represented in terms of self and mutual partial inductances of thetwo lands as W M=1 2Lp1I2 1+1 2Mp12I1I2+1 2Lp2I2 2+1 2Mp21I2I1 (6.8b) and of course the mutual partial inductances are reciprocal (i.e., Mp12= Mp21). Comparing (6.8a) and (6.8b), the various self and mutual partial in- ductances can be found from Lp1=1 I2 1⎜integraldisplay v1A1·J1dv1 (6.9a) Lp2=1 I2 2⎜integraldisplay v2A2·J2dv2 (6.9b) Mp12=1 I1I2⎜integraldisplay v2A1·J2dv2 (6.9c) Mp21=1 I1I2⎜integraldisplay v1A2·J1dv1 (6.9d) Next we substitute the equation for the vector magnetic potential given in (2.47a), A=μ0 4π⎜integraldisplay vJdv R(2.47a) FORMULATION FOR THE COMPUTATION 251 giving Lp1=1 I2 1⎜integraldisplay v1⎜integraldisplay v1μ0 4πJ1·J/prime1 Rdv/prime 1dv1 (6.10a) Lp2=1 I2 2⎜integraldisplay v2⎜integraldisplay v2μ0 4πJ2·J/prime2 Rdv/prime 2dv2 (6.10b) Mp12=1 I1I2⎜integraldisplay v1⎜integraldisplay v2μ0 4πJ/prime1·J2 Rdv2dv/prime 1 (6.10c) Mp21=1 I1I2⎜integraldisplay v2⎜integraldisplay v1μ0 4πJ/prime2·J1 Rdv1dv/prime 2 (6.10d) where the term Ris the distance between two differential chunks of current JidviandJ/primejdv/prime j. We have denoted one current density with a prime to denote that chunk of current J/primedv/primeas being the cause of Aand it lies in the differ- ential volume dv/prime, whereas the differential chunk of current Jdvlies in the differential volume dv. It is important to point out the following observation about the internal inductances of lands: Since the formulations for determining the self and mutual partial in-ductances in (6.10) result from the integration of the magnetic energy density over all space (including that internal to the lands ), the results for the self partial inductances in (6.10a) and(6.10b) include the inter- nal self inductances of the lands due to the magnetic ﬁelds internal tothem. As the frequency of the current increases from dc, the current distributionover the land cross sections migrate to the outer edges of the lands, and hencethe internal inductances go to zero, so that ( 6.10a) and(6.10b) for higher frequencies represent the external self partial inductances . This will be shown through numerical computations later. However, it would not be simple todetermine the external self partial inductances by integrating (6.10) only overa thin volume near the land surfaces that contains the total land current, sincethe high-frequency current distribution peaks at the corners of the lands, andthis would generate a very difﬁcult computation. It turns out that Hollowayand Kuester have recently derived the result for the internal self inductanceof a PCB land [18]. Their result for a square land was 48 .3nH/m, which had been conﬁrmed with numerical computation [19]. This is on the order of the 252 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION internal inductance of a circular wire (50nH /m), which is independent of the wire radius. Now we make two crucial assumptions in order to make the results in (6.10) useful for computing the self and mutual partial inductances of PCB lands. 1. Assume that the currents Iwith density Jin volume vthat are carried by the lands are uniformly distributed over the cross sections of those lands . Hence, we can simply write the magnitudes of the current distributionsasJ=I/A, where Adenotes the cross-sectional area of the respective lands. 2. Assume that the currents Jinvthat are carried by the lands are also uniformly distributed along the lengths of the lands . This agrees with the fundamental limitation that the conductors must be electrically shortfor currents of nonzero frequency in order to represent those conductorswith lumped-circuit elements such as an inductance. Hence, we canwrite Jdv=I AdA dl With these two assumptions the magnitudes of the current densities, J, are constants (independent of the longitudinal as well as the cross-sectionalvariables of the lands) and can therefore be removed from the integrals in(6.10). Hence, the results in (6.10) simplify considerably. The self partialinductances are then computed from Lp=1 AA/prime⎜integraldisplay A⎜integraldisplay A/primeMfdA/primedA (6.11a) where AandA/primeare over the same land, and Mfis the Neumann integral de- noting the mutual partial inductance between two ﬁlamentary currents (withinthe same land) that are separated by distance R: Mf=μ0 4π⎜integraldisplay l⎜integraldisplay ldl/prime·dl R(6.12) In the case of two different lands, the mutual partial inductance between them is obtained from Mp=1 AA/prime⎜integraldisplay A⎜integraldisplay A/primeMfdA/primedA (6.11b) FORMULATION FOR THE COMPUTATION 253 where AandA/primeare over different lands, and Mfin (6.12) denotes the mutual partial inductance between two ﬁlamentary currents (in different lands) thatare separated by distance R. So the computation of the self and mutual partial inductances essentially involves representing the currents of each land as being composed of ﬁlaments.We obtain the total self and mutual partial inductances of the lands as thesummation, over the cross-sectional areas of the lands, of these mutual partialinductances between the ﬁlaments. Note that (6.11a) and (6.11b) for the selfand mutual partial inductances seem to indicate an “averaging” over the cross-sectional areas of the lands. This averaging is not in the derivation of the resultfrom the outset (i.e., it is not an approximation, but just comes out of the formalderivation). The computations in (6.11a) and (6.11b) involve sixfold integrals:two over the ﬁlament lengths and four over the cross-sectional areas of thelands. But the mutual partial inductance between ﬁlaments was derived inChapter 5 and given in Section 5.5. For parallel but offset ﬁlaments of lengthlandmseparated by distance dand whose endpoints are offset by distance swith reference to Fig. 5.11, the mutual partial inductance between the two parallel but offset ﬁlaments is given as M f=μ0 4π⎜bracketleftbigg (l+s+m)sinh−1l+s+m d−(s+m)sinh−1s+m d −(l+s)sinh−1l+s d+ssinh−1s d−⎜radicalBig (l+s+m)2+d2 +⎜radicalBig (s+m)2+d2+⎜radicalBig (l+s)2+d2−⎜radicalbig s2+d2⎜bracketrightbigg (5.28) This can be put into an alternative form by using the identity for the inverse hyperbolic sine: sinh−1z d=ln⎡ ⎣z d+⎜radicalBigg⎜parenleftbiggz d⎜parenrightbigg2 +1⎤ ⎦ =ln⎜parenleftBig z+⎜radicalbig z2+d2⎜parenrightBig −lnd (D700.1) as Mf=μ0 4π⎜bracketleftbigf(z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle(l+s+m),s (z) (s+m),(l+s) =μ0 4π⎜bracketleftbigf(l+s+m)−f(s+m)+f(s)−f(l+s)⎜bracketrightbig(6.13a) 254 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION where f(z)=zln⎜parenleftBig z+⎜radicalbig z2+d2⎜parenrightBig −⎜radicalbig z2+d2 (6.13b) Note that the −lndterm in the identity in (D700.1) cancels out because we add and subtract this term twice in (5.28). 6.2 SELF PARTIAL INDUCTANCE OF PCB LANDS For self partial inductance calculations, the mutual partial inductance be- tween two parallel ﬁlaments in (6.13) simpliﬁes since the two ﬁlaments areofidentical length l and their endpoints coincide, giving m=lands=−l. Hence, (6.13) reduces to Mf=μ0 4π⎜bracketleftbigf(z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglel,−l (z) 0,0 =μ0 4π⎜bracketleftbigf(l)−f(0)+f(−l)−f(0)⎜bracketrightbig =μ0 4π⎜bracketleftBig lln⎜parenleftBig l+⎜radicalbig l2+d2⎜parenrightBig −lln⎜parenleftBig −l+⎜radicalbig l2+d2⎜parenrightBig −2√ l2+d2+2d⎜bracketrightBig =μ0 2π⎜parenleftbigg lsinh−1l d−⎜radicalbig l2+d2+d⎜parenrightbigg m=l, s=−l (6.14) which agrees with (5.21b), which was derived directly in Chapter 5. We ﬁrst compute the self partial inductance for a land of zero thickness, t=0. Hence, we integrate the mutual partial inductances of ﬁlaments in (6.14) that are separated by distance d2=(x2−x1)2ﬁrst from x1=0t ox1=wand then integrate that result from x2=0t ox2=w. Hence, we integrate (6.11a) over the same cross-section land to give Lp(t=0)=μ0 4π1 w2⎜integraldisplayw x2=0⎜integraldisplayw x1=0Mfdx1dx2t=0 (6.15) as illustrated in Fig. 6.2. The inner integral is evaluated ﬁrst: (I)=⎜integraldisplayw x1=0⎜bracketleftBig zln⎜parenleftBig z+⎜radicalbig z2+d2⎜parenrightBig −⎜radicalbig z2+d2⎜bracketrightBig dx1 andd2=(x1−x2)2. Since zhere is treated as a constant, we evaluate f(z) in the integrand at the four limits of z=l,−l,0,0 as in (6.14) after we SELF PARTIAL INDUCTANCE OF PCB LANDS 255 xx1x2 d w FIGURE 6.2. Computation of the self partial inductance for a land of zero thickness ( t=0). have ﬁnished the integrations. Making a change of variables as λ=x1−x2, dλ=dx1gives (I)=⎜integraldisplayw−x2 λ=−x 2⎜bracketleftBig zln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig −⎜radicalbig z2+λ2⎜bracketrightBig dλ which is equivalent to (I)=⎜integraldisplayx2 λ=x 2−w⎜bracketleftBig zln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig −⎜radicalbig z2+λ2⎜bracketrightBig dλ This can be integrated using integrals from Dwight [7]: ⎜integraldisplay ln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig dλ=λln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig −λ+zln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig (D740) ⎜integraldisplay⎜radicalbig z2+λ2dλ=z2 2ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig +λ 2⎜radicalbig z2+λ2 (D230.01) Hence, the inner integral evaluates to (I)=⎜bracketleftBigg λzln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig −λz+z2 2ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig −λ 2⎜radicalbig z2+λ2⎜bracketrightBiggx2 λ=x 2−w Note that this result contains a term −λz. But when this is evaluated for the four values of z,z=l,−l,0,0, it will cancel out: ⎜bracketleftbigf(z)⎜bracketrightbigl,l (z) 0,0=⎜bracketleftbigf(l)−f(0)+f(−l)−f(0)⎜bracketrightbig =−λl+λl =0 256 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION Hence, we ignore this term. This gives (I)=⎜bracketleftBigg λzln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig +z2 2ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig −λ 2⎜radicalbig z2+λ2⎜bracketrightBiggx2 λ=x 2−w =⎜bracketleftBigg x2zln⎜parenleftbigg z+⎜radicalBig z2+x2 2⎜parenrightbigg +z2 2ln⎜parenleftbigg x2+⎜radicalBig z2+x2 2⎜parenrightbigg −x2 2⎜radicalBig z2+x2 2⎜bracketrightBigg −⎜bracketleftbigg (x2−w)zln⎜parenleftbigg z+⎜radicalBig z2+(x2−w)2⎜parenrightbigg +z2 2ln⎜parenleftbigg (x2−w)+⎜radicalBig z2+(x2−w)2⎜parenrightbigg −(x2−w) 2⎜radicalBig z2+(x2−w)2⎜bracketrightbigg Next, we integrate (I)fromx2=0t ox2=w: (II)=⎜integraldisplayw x2=0(I)dx2 giving (II)=⎜integraldisplayw x2=0⎜bracketleftBigg x2zln⎜parenleftbigg z+⎜radicalBig z2+x2 2⎜parenrightbigg +z2 2ln⎜parenleftbigg x2+⎜radicalBig z2+x2 2⎜parenrightbigg −x2 2⎜radicalBig z2+x2 2⎜bracketrightBigg dx2 +⎜integraldisplay−w λ=0⎜bracketleftBigg λzln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig +z2 2ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig −λ 2⎜radicalbig z2+λ2⎜bracketrightBigg dλ and we have used a change of variables λ=x2−w,dλ=dx2in the second integral. These integrals can be evaluated using Dwight [7]: ⎜integraldisplay λln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig dλ=−λ2 4+z 2⎜radicalbig z2+λ2 +λ2 2ln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig (D602.5) ⎜integraldisplay ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig dλ=λln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig −⎜radicalbig z2+λ2 (D625) SELF PARTIAL INDUCTANCE OF PCB LANDS 257 ⎜integraldisplay λ⎜radicalbig z2+λ2dλ=1 3⎜parenleftBig z2+λ2⎜parenrightBig3/2(D231.01) to give (II)=⎜bracketleftBigg −λ2z 4+zλ2 2ln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig +z2λ 2ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig −1 6⎜parenleftBig z2+λ2⎜parenrightBig3/2⎜bracketrightBiggw,−w λ=0,0 Once again this contains a term −λ2z/4, which will be canceled out when this is evaluated at the four limits of z,z=l,−l,0,0, so it will also be ignored. Hence, the result gives the self partial inductance of a land of zero thickness (t=0)as Lp(t=0)=μ0 4π1 w2⎜bracketleftbigf(λ,z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglew,−w (λ) 0,0⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglel,−l (z) 0,0t=0 (6.16a) where f(λ,z)=zλ2 2ln⎜parenleftBig z+⎜radicalbig z2+λ2⎜parenrightBig +z2λ 2ln⎜parenleftBig λ+⎜radicalbig z2+λ2⎜parenrightBig −1 6⎜parenleftBig z2+λ2⎜parenrightBig3/2(6.16b) Evaluating this gives Lp(t=0)=μ0 4π1 w2⎜bracketleftbigf(λ,z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglew,−w (λ) 0,0⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglel,−l (z) 0,0t=0 =μ0 4π1 w2⎡ ⎢⎢⎢⎢⎣+f(w,l)−f(w,0)+f(w,−l)−f(w,0) −f(0,l)+f(0,0)−f(0,−l)+f(0,0) +f(−w,l )−f(−w, 0)+f(−w,−l)−f(−w, 0) −f(0,l)+f(0,0)−f(0,−l)+f(0,0)⎜bracketrightbig⎤ ⎥⎥⎥⎥⎦ (6.17) Note that f(0,0)=0. In the sum in (6.17), the following identity may be used: ln⎜parenleftBig a+⎜radicalbig a2+b2⎜parenrightBig −ln⎜parenleftBig −a+⎜radicalbig a2+b2⎜parenrightBig =2l n⎡ ⎣a b+⎜radicalBigg⎜parenleftbigga b⎜parenrightbigg2 +1⎤ ⎦ This identity can be conﬁrmed by writing the left side as ln A−lnB= ln(A/B )and comparing the arguments of the natural logarithms on both sides. Caution should be observed in evaluating the term⎜parenleftbigz2+λ2⎜parenrightbig3/2. When 258 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION either z=0o rλ=0, the absolute values of zorλshould be used: ⎜bracketleftbigg⎜parenleftBig z2+λ2⎜parenrightBig3/2⎜bracketrightbigg z=0o r λ=0=|λ|3or|z|3 Finally, in evaluating the terms in (6.17) we should note that lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright x→0xln[ax]=0 as can be proven using l/primeHˆopital’s rule. This gives the self partial inductance of a PCB land of zero thickness, t=0, as Lp(t=0)=μ0 2π1 w2⎡ ⎣lw2ln⎛ ⎝l w+⎜radicalBigg⎜parenleftbiggl w⎜parenrightbigg2 +1⎞ ⎠ +l2wln⎛ ⎝w l+⎜radicalBigg⎜parenleftbiggw l⎜parenrightbigg2 +1⎞ ⎠ +1 3⎜parenleftBig l3+w3⎜parenrightBig −1 3⎜parenleftBig l2+w2⎜parenrightBig3/2⎤ ⎦ t=0(6.18) which agrees with the formula given by Hoer and Love [16]. The result for inﬁnitesimally thin lands in (6.18) can be written compactly in terms of the “aspect ratio” of the land as the ratio of the land length to landwidth, u=l/w: Lp(t=0) l=μ0 2π⎡ ⎣ln⎜parenleftBig u+⎜radicalbig u2+1⎜parenrightBig +uln⎛ ⎝1 u+⎜radicalBigg⎜parenleftbigg1 u⎜parenrightbigg2 +1⎞ ⎠ +1 3⎜parenleftBigg u2+1 u−⎜parenleftbigu2+1⎜parenrightbig3/2 u⎜parenrightBigg⎤ ⎦ u=l w,t=0 (6.19) This can be simpliﬁed for extreme values of the “aspect ratio” of the lands: u/greatermuch1 (very “long” lands, l/greatermuchw)o ru/lessmuch1 (very “wide” lands, l/lessmuchw). We have the following identities [7]: ln(u+⎜radicalbig u2+1)=⎧ ⎪⎪⎨ ⎪⎪⎩ln 2u+1 4u2−3 32u4+··· u> 1 u−1 6u3+3 40u5+··· u< 1(D602.1) SELF PARTIAL INDUCTANCE OF PCB LANDS 259 ln⎡ ⎣1 u+⎜radicalBigg⎜parenleftbigg1 u⎜parenrightbigg2 +1⎤ ⎦=⎧ ⎪⎪⎨ ⎪⎪⎩1 u−1 6u3+3 40u5+··· u> 1 ln2 u+u2 4−3u4 32+··· u< 1(D602.5) The result in (D602.5) follows from (D602.1) by substituting u=1/xinto (D602.1). In addition, we have the series expansion (1+x)n=1+nx+n(n−1) 2!x2+n(n−1)(n−2) 3!+··· x2≤1 (D1) Using (D1) we obtain approximations for the term⎜parenleftbigu2+1⎜parenrightbig3/2foru≤1: ⎜parenleftbigu2+1⎜parenrightbig3/2=1+3 2u2+3 8u4+··· u≤1 ∼=1+3 2u2u/lessmuch1 and for u≥1: ⎜parenleftBig u2+1⎜parenrightBig3/2=u3⎜parenleftbigg1 u2+1⎜parenrightbigg3/2 =u3⎜parenleftbigg 1+3 21 u2+3 81 u4+···+⎜parenrightbigg u≥1 ∼=u3+3 2uu /greatermuch1 Hence the last term in (6.19) becomes 1 3⎜parenleftBigg u2+1 u−⎜parenleftbigu2+1⎜parenrightbig3/2 u⎜parenrightBigg ∼=⎧ ⎪⎪⎨ ⎪⎪⎩−1 2+1 3uu/greatermuch1 −1 2u+1 3u2u/lessmuch1 Hence, (6.19) should approach Lp(t=0) l∼=μ0 2π⎧ ⎪⎪⎨ ⎪⎪⎩ln 2u+1 2+1 3uu/greatermuch1,l/greatermuchw u⎜parenleftbigg ln2 u+1 2+u 3⎜parenrightbigg u/lessmuch1,l/lessmuchw(6.20) The following table summarizes some computed data comparing (6.19) and (6.20). u (6.19) (6.20) 10 17.925 nH /in. 17.928 nH /in. 100 29.472 nH /in. 29.473 nH /in. 1 101.7927 nH /in. 1.7928 nH /in. 1 1000.2946 nH /in. 0.2947 nH /in. 260 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION xx1x2 d wy y1y2t FIGURE 6.3. Computing the self partial inductance of a PCB land of nonzero thickness. For very large aspect ratios, l/greatermuchw, the self partial inductance approaches a variation of 5 .08⎜bracketleftBig ln(2l/w)+1 2⎜bracketrightBig nH/in. For this to give the same result as in (5.18c) for a “long” wire would require the land to be replaced by a wire whose diameter is 0.446 w. This tends to support a commonly held design rule that rectangular straps have less “inductance” than do comparable sized wires. Now we turn our attention to PCB lands with nonzero thicknesses, as illustrated in Fig. 6.3. In this case we must integrate (6.11b) over the entirecross section: M p=1 AA/prime⎜integraldisplay A⎜integraldisplay A/primeMfdA/primedA (6.11b) The mutual inductance between the ﬁlaments, Mf, is the same as in (6.14) but with d2=(x2−x1)2+(y2−y1)2(6.21) Hence, the integral we must evaluate is Lp=1 t2w2⎜integraldisplayt y2=0⎜integraldisplayt y1=0⎜integraldisplayw x2=0⎜integraldisplayw x1=0Mfdx1dx2dy1dy2 (6.22) and, since zinMfis treated here essentially as a constant, we evaluate the result at the limits of z=l,−l,0,0 according to (6.14) after we ﬁnish the integration. The interior integrals with respect to xrepresent the self partial inductances of two identical lands of zero thickness that we integrated before,but we must repeat that because d 2inMfis no longer just (x2−x1)2but is given now by (6.21). The ﬁrst two integrals with respect to x1andx2are fairly simple to integrate, and the process is very similar to what was done to obtainthe result for a land of zero thickness in (6.15). Once these are performed,we are left with the integrals with respect to y 1andy2. The entire process, SELF PARTIAL INDUCTANCE OF PCB LANDS 261 although straightforward, is exceedingly tedious. Hoer and Love give that result [16] as Lp=μ0 4π1 t2w2⎜bracketleftbigg⎜bracketleftbigg⎜bracketleftbigf(x, y, z )⎜bracketrightbigw (x) 0⎜bracketrightbiggt (y) 0⎜bracketrightbiggl (z) 0(6.23a) where ⎜bracketleftBigg⎜bracketleftBigg ⎜bracketleftbigf(x, y, z )⎜bracketrightbigq1 (x) q2⎜bracketrightBigg r1(y) r2⎜bracketrightBigg s1(z) s2=2⎜summationdisplay i=12⎜summationdisplay j=12⎜summationdisplay k=1(−1)i+j+k+1f⎜parenleftbigqi,rj,sk⎜parenrightbig (6.23b) andf(x, y, z )is given by f(x, y, z )=⎜parenleftBigg y2z2 4−y4 24−z4 24⎜parenrightBigg xlnx+⎜radicalbig x2+y2+z2 ⎜radicalbig y2+z2 +⎜parenleftBigg x2z2 4−x4 24−z4 24⎜parenrightBigg ylny+⎜radicalbig x2+y2+z2 √ x2+z2 +⎜parenleftBigg x2y2 4−x4 24−y4 24⎜parenrightBigg zlnz+⎜radicalbig x2+y2+z2 ⎜radicalbig x2+y2 +1 60⎜parenleftBig x4+y4+z4−3x2y2−3y2z2−3x2z2⎜parenrightBig⎜radicalBig x2+y2+z2 −xyz3 6tan−1 xy z⎜radicalbig x2+y2+z2 −xy3z 6tan−1 xz y⎜radicalbig x2+y2+z2 −x3yz 6tan−1 yz x⎜radicalbig x2+y2+z2 (6.23c) Ruehli shows [15] that the general result in (6.23) can be written solely in terms of u=l/wandv=t/w. One would expect that the self partial induc- tance should vary in a smooth fashion as a function of u=l/wandv=t/w. Hence, it should be possible to obtain a general formula that is much simplerthan (6.23) by curve ﬁtting to computed data from (6.23) or Ruehli’s versionof it. Ruehli in [15] also points out that for extreme values of the aspect ratiosofu=l/wandv=t/w, the general result in (6.23) may involve subtraction of terms of similar magnitude, thereby giving numerical errors. He gives a morestable form of this result in [15]. His computations also show that the result 262 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION for a land of zero thickness in (6.19) gives reasonable accuracy for v<0.1. In addition, his results show that the zero-thickness land result in (6.19) givesapproximately the same result as (6.23) for all values of vandu>100 (i.e., for “very long” lands). Hoer and Love [16] give an approximation for thegeneral result in (6.23) for v=t/w≤0.1 using the zero-thickness land result in (6.19) as Lp∼=Lp(t=0)−2×10−7t wl w≥10t (6.24) 6.3 MUTUAL PARTIAL INDUCTANCE BETWEEN PCB LANDS Calculating the mutual partial inductance between lands follows a pattern similar to that for self partial inductance. Treat each land as a set of ﬁlamentsand use the basic result M p=1 AA/prime⎜integraldisplay A⎜integraldisplay A/primeMfdA/primedA (6.11b) In this case the lands may be offset from each other and hence we use the general relation for the mutual partial inductances between two ﬁlaments oflengths landmwhich are offset from each other by a distance sas shown in Fig. 5.11. Hence, the mutual partial inductance between the two ﬁlaments isgiven by (5.28), which may be written as M f=μ0 4π⎜bracketleftbigf(z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle(l+s+m),s (z) (s+m),(l+s) =μ0 4π⎜bracketleftbigf(l+s+m)−f(s+m)+f(s)−f(l+s)⎜bracketrightbig(6.13a) where f(z)=zln⎜parenleftBig z+⎜radicalbig z2+d2⎜parenrightBig −⎜radicalbig z2+d2 (6.13b) If the lands (and their associated ﬁlaments) overlap, swill be negative by the amount of overlap. We ﬁrst obtain the mutual partial inductance between two lands of zero thickness whose lengths in the zdirection are landmand whose surfaces are parallel to each other (but perhaps offset by distance s) as shown in Fig. 6.4(a). For this case, din (6.13b) is the distance in the xyplane between the ﬁlaments composing each land and is given by d2=(x2−x1)2+b2(6.25) MUTUAL PARTIAL INDUCTANCE BETWEEN PCB LANDS 263 x x1 x2y d w1w2 a ba b m sx zy l (a) (b)w1 w2 FIGURE 6.4. Computing the mutual partial inductance between two PCB lands of zero thickness. where bis the vertical separation in the xyplane between the lands as shown in Fig. 6.4. Hence, the integral we must evaluate is Mp=1 w1w2⎜integraldisplaya+w2 x2=a⎜integraldisplayw1 x1=0Mfdx1dx2 (6.26) as illustrated in Fig. 6.4(b), and since zinMfis essentially treated as a constant here, we evaluate the result at the limits of z=(l+s+m),(s+m),s ,(l +s) according to (6.13a) after we ﬁnish the integration. Hence, we ﬁrst integrate (I)=⎜integraldisplayw1 x1=0⎜bracketleftbigg zln⎜parenleftbigg z+⎜radicalBig z2+b2+(x2−x1)2⎜parenrightbigg −⎜radicalBig z2+b2+(x2−x1)2⎜bracketrightbigg dx1 264 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION Making a change of variables as λ=(x2−x1),dλ=−dx1gives (I)=⎜integraldisplayx2 λ=x 2−w1⎜bracketleftBig zln⎜parenleftBig z+⎜radicalbig z2+b2+λ2⎜parenrightBig −⎜radicalbig z2+b2+λ2⎜bracketrightBig dλ This can be integrated as [16] (I)=⎡ ⎣zx2ln⎜parenleftbigg z+⎜radicalBig z2+b2+x2 2⎜parenrightbigg −zbtan−1 zx2 b⎜radicalBig z2+b2+x2 2 +z2−b2 2ln⎜parenleftbigg x2+⎜radicalBig z2+b2+x2 2⎜parenrightbigg −x2 2⎜radicalBig z2+b2+x2 2⎤ ⎦ −⎜bracketleftbigg z(x2−w1)ln⎜parenleftbigg z+⎜radicalBig z2+b2+(x2−w1)2⎜parenrightbigg −zbtan−1 z(x2−w1) b⎜radicalBig z2+b2+(x2−w1)2 −z2−b2 2ln⎜parenleftbigg (x2−w1)+⎜radicalBig z2+b2+(x2−w1)2⎜parenrightbigg −x2−w1 2⎜radicalBig z2+b2+(x2−w1)2⎜bracketrightbigg Hence, it can be written symbolically as (I)=⎜bracketleftbigg zλln⎜parenleftBig z+⎜radicalbig z2+b2+λ2⎜parenrightBig −zbtan−1 zλ b√ z2+b2+λ2 +z2−b2 2ln⎜parenleftBig λ+⎜radicalbig z2+b2+λ2⎜parenrightBig −λ 2⎜radicalbig z2+b2+λ2⎜bracketrightBiggx2 λ=x 2−w1 which, when multiplied by ( μ0/4π)( 1/w1) and evaluated at the four limits of zaccording to (6.13a), represents the mutual partial inductance between the ﬁrst land and a ﬁlament in the second land at (y, x)=(b, x 2)[16]. The second integral in (6.26) becomes (II)=⎜integraldisplaya+w2 x2=a(I)dx2 MUTUAL PARTIAL INDUCTANCE BETWEEN PCB LANDS 265 Making a change of variables in the second half of the result for the ﬁrst integral of λ=(x2−w1),dλ=dx2gives (II)=⎜integraldisplaya+w2 x2=a⎡ ⎣zx2ln⎜parenleftbigg z+⎜radicalBig z2+b2+x2 2⎜parenrightbigg −zbtan−1 zx2 b⎜radicalBig z2+b2+x2 2 +z2−b2 2ln⎜parenleftbigg x2+⎜radicalBig z2+b2+x2 2⎜parenrightbigg −x2 2⎜radicalBig z2+b2+x2 2⎤ ⎦dx2 +⎜integraldisplaya−w1 λ=a+w2−w1⎡ ⎣zλln⎜parenleftBig z+⎜radicalbig z2+b2+λ2⎜parenrightBig −zbtan−1 zλ b√ z2+b2+λ2 +z2−b2 2ln⎜parenleftBig λ+⎜radicalbig z2+b2+λ2⎜parenrightBig −λ 2⎜radicalbig z2+b2+λ2⎤ ⎦dx2 These integrals can be evaluated, giving the mutual partial inductance between two lands of zero thickness as [16] Mp=μ0 4π1 w1w2⎜bracketleftbigf(x, z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglea+w2,a−w1(x) a,a+w2−w1⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle(l+s+m),s (z) (s+m),(l+s)(6.27a) where ⎜bracketleftbigf(x, z)⎜bracketrightbig⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingleq1,q3 (x) q2,q4⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingles1,s3(z) s2,s4=4⎜summationdisplay i=14⎜summationdisplay j=1(−1)i+jf(qi,sj) (6.27b) and f(x, z)=x2−b2 2zln⎜parenleftBig z+⎜radicalbig z2+b2+x2⎜parenrightBig −1 6⎜parenleftBig z2−2b2+x2⎜parenrightBig⎜radicalbig z2+b2+x2 +z2−b2 2xln⎜parenleftBig x+⎜radicalbig z2+b2+x2⎜parenrightBig −zbx tan−1zx b√ z2+b2+x2(6.27c) 266 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION The mutual partial inductance between two parallel lands of widths w1,w2 and corresponding thicknesses of t1,t2is extraordinarily complicated and becomes [16] Mp=μ0 4π1 w1t1w2t2⎜bracketleftBigg⎜bracketleftbigg⎜bracketleftbigf(x, y, z )⎜bracketrightbiga−w 1,a+w 2(x) a+w 2−w 1,a⎜bracketrightbiggb−t1,b+t2(y) b+t2−t1,b⎜bracketrightBigg (l+s+m),s (z) (s+m),(l+s) (6.28a) where ⎜bracketleftBigg⎜bracketleftBigg ⎜bracketleftbigf(x, y, z )⎜bracketrightbigq1,q3 (x) q2,q4⎜bracketrightBigg r1,r3(y) r2,r4⎜bracketrightBigg s1,s3(z) s2,s4=4⎜summationdisplay i=14⎜summationdisplay j=14⎜summationdisplay k=1(−1)i+j+k+1f⎜parenleftbigqi,rj,sk⎜parenrightbig (6.28b) andf(x, y, z ) is given in (6.23c). 6.4 CONCEPT OF GEOMETRIC MEAN DISTANCE The concept of the geometric mean distance (GMD) between two objects gives a method for obtaining simpliﬁed (but approximate) calculations of themutual partial inductance between those objects [14,20–22]. The origin ofthe name is illustrated in Fig. 6.5, where we have shown a point Pand a line. Distances d iare drawn from the point to the line. The geometric mean distance Dbetween the point and the line is the nth root of the product of the distances Pd1 d2 di dn FIGURE 6.5. Concept of geometrical mean distance. CONCEPT OF GEOMETRIC MEAN DISTANCE 267 (their geometric mean) as the number of distances nincreases without bound: D=lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright n→∞(d1d2···d i···d n)1/n(6.29) Taking the natural logarithm of (6.29) gives lnD=lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright n→∞lnd1+lnd2+···+ lndi+···+ lndn n(6.30) Hence, the natural logarithm of the geometric mean distance between the point and the line is the arithmetic mean of the natural logarithms of the distances from the point to the line. The concept of geometric mean distance has a particularly beneﬁcial ap- plication in computing the mutual partial inductance between a pair of two-dimensional shapes representing the cross sections of two conductors whenthe lengths of the two conductors are much greater than their separation . The concept of geometric mean distance was originally developed by Maxwellin the late nineteenth century [23]. It is used routinely in the electric powerdistribution area to compute the self inductance of a bundle of wires, as well asthe mutual inductances between sets of multiphase, high-voltage power trans-mission lines [24]. A very large number of formulas for the GMD of variousshapes was published by Rosa and his colleagues in the Bulletin of the Na- tional Bureau of Standards in the period 1900–1910 [25–28]. A magniﬁcent book by Andrew Gray gives a very thorough discussion of GMD, and it waswritten in 1893 [29]! We computed the mutual partial inductance by considering the conductors to be composed of parallel current ﬁlaments via (6.11b): M p=1 AA/prime⎜integraldisplay A⎜integraldisplay A/primeMfdA/primedA (6.11b) where Mfis the mutual partial inductance between two ﬁlaments of current within the cross-sectional areas AandA/primegiven in (5.21a). The basic idea here is to treat the currents (assumed to be uniformly distributed over their crosssections) as being composed of ﬁlaments of current, sweep the ﬁlaments overthe two cross sections, and then average the result over the cross sections asillustrated in Fig. 6.6. The basic result for the mutual partial inductance between two ﬁlaments of current of length l(into the page) separated by distance dis given in (5.21a). If the lengths of the ﬁlaments are much greater that their separation, (5.21a) 268 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION d 1t 1w 2w2t1S 2S D FIGURE 6.6. Computing the mutual partial inductance using the GMD between two shapes. approximates to Mf=μ0 2πl⎜bracketleftBigg ln2l d−1+d l−1 4⎜parenleftbiggd l⎜parenrightbigg2 +···⎜bracketrightBigg ∼=μ0 2πl⎜parenleftbigg ln2l d−1⎜parenrightbigg l/greatermuchd (5.21c) Note that only dwill vary as we sweep the ﬁlaments over the surfaces as in (6.11b). Substituting (5.21c) into (6.11b) gives Mp∼=1 AA/prime⎜integraldisplay A⎜integraldisplay A/primeμ0 2πl⎜parenleftbigg ln2l d−1⎜parenrightbigg dA/primedA =μ0 2πl1 AA/prime⎜integraldisplay A⎜integraldisplay A/prime(ln 2l−lnd−1)dA/primedA =μ0 2πl(ln 2l−1)−μ0 2πl1 AA/prime⎜integraldisplay A⎜integraldisplay A/prime(lnd)dA/primedA (6.31) The goal here is to determine an equivalent distance between two ﬁlaments, D, which will have the same mutual partial inductance between them as betweenthe two surfaces, as illustrated in Fig. 6.6. Note that for very long ﬁlamentsl/greatermuchd, the only parameter in M fin (5.21c) that varies is d. Hence, we may instead determine a D(the GMD between the two surfaces) such that lnD=1 AA/prime⎜integraldisplay A⎜integraldisplay A/prime(lnd)dA/primedA (6.32) If this computation is carried out, two (very long) ﬁlaments of length lspaced a distance equal to the geometric mean distance between the two cross-sectional CONCEPT OF GEOMETRIC MEAN DISTANCE 269 shapes of the two conductors, D, will have a mutual partial inductance between those conductors of Mp∼=μ0 2πl⎜parenleftbigg ln2l D−1⎜parenrightbigg l/greatermuchD (6.33) which will give the same mutual partial inductance as between the (very long) conductors originally desired and obtained with (6.11b). If we include the third term of (5.21c), d/l, the dshould properly be the arithmetic mean distance between the ﬁlaments used to represent the conductors. But usually this term is inconsequential and will be neglected. So (6.33) is an approximatesolution to (6.11b) which is reasonably valid only for “very long” conductors(i.e.,D/lessmuchl). Of course, the work in computing the GMD in (6.32) can be as tedious as that in directly computing the original integral in (6.11b), butthe GMD for various shapes has been tabulated over the years in variouspublications [14,20,21]. The discussion above regarding the use of the GMD in computing the mutual partial inductance applies to computation of the self partial in-ductance in (6.11a). The GMD here is said to be between the shape anditself. Another interpretation of the utility of the GMD is computation of the vector magnetic potential due to a very long conductor of rectangular crosssection whose current Iisdistributed uniformly over the cross section of the conductor as illustrated in Fig. 6.7 [22]. The vector magnetic potential of a current ﬁlament of inﬁnite length (pointing in the zdirection, into the page) is given as A z=−μ0I 2πlnd (2.53) where dis the distance between that ﬁlament of current and the point P(X, Y ) at which we desire to determine the vector magnetic potential. This is uniquewithin a constant. Assuming that the current I of the rectangular bar is dis- tributed uniformly over the bar cross section, the current density over thebar is J=I/A, where Ais the cross-sectional area of the bar. Hence, the current over the bar can be concentrated into ﬁlaments, giving differentialcontributions to A z(X, Y ) at point Plocated at x=Xandy=Yas dAz=−μ0J 2πlndd A (6.34) 270 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION yx P(X,Y) Δx Δy(X−x)2 + (Y−y)2d=Az (X,Y) (x,y) FIGURE 6.7. Another interpretation of geometric mean distance between a point and a surface. anddAis a differential area of the cross section. Hence, the magnetic vector potential at point Pis obtained as Az(X, Y )=−μ0J 2π⎜integraldisplay y⎜integraldisplay xln⎜bracketleftbigg⎜radicalBig (X−x)2+(Y−y)2⎜bracketrightbigg dx dy =−μ0I 2π1 A⎜integraldisplay y⎜integraldisplay xln⎜bracketleftbigg⎜radicalBig (X−x)2+(Y−y)2⎜bracketrightbigg dx dy (6.35a) where the distance between the ﬁlament and the point is d=⎜radicalBig (X−x)2+(Y−y)2(6.35b) and the integral is to be taken over the coordinates of the conductor cross section. We can evaluate the integral in (6.35) in an approximate fashion via numerical means simply by dividing the cross section into Nsubrectangles of area /Delta1x /Delta1y as illustrated in Fig. 6.7, performing (6.35) over the individual CONCEPT OF GEOMETRIC MEAN DISTANCE 271 subrectangles, and summing the results: Az(X, Y )∼=−μ0I 2π1 AN⎜summationdisplay i=1⎜integraldisplay yi⎜integraldisplay xiln⎜bracketleftbigg⎜radicalBig (X−xi)2+(Y−yi)2⎜bracketrightbigg dxidyi (6.36) where xiandyiare the coordinates of the subrectangles. If this division is such that the dimensions of the subrectangles are sufﬁciently small, the integral in(6.36) over each subrectangle can be further approximated as ⎜integraldisplay yi⎜integraldisplay xiln⎜bracketleftbigg⎜radicalBig (X−xi)2+(Y−yi)2⎜bracketrightbigg dxidyi ∼=/Delta1x /Delta1y ln⎜bracketleftbigg⎜radicalBig (X−xn)2+(Y−yn)2⎜bracketrightbigg (6.37) This amounts to replacing the smooth and uniform current distribution by ﬁlaments at the centers of the subrectangles located at ( xn,yn). Hence, (6.35) is approximated as (the area of the conductor cross section is A=N/Delta1x /Delta1y ) Az(X, Y )∼=−μ0I 2π1 NN⎜summationdisplay n=1ln⎜bracketleftbigg⎜radicalBig (X−xn)2+(Y−yn)2⎜bracketrightbigg (6.38) By ﬁnely dividing the cross section, we can obtain a reasonably accurate approximation to (6.35). The computation in (6.38) requires an excessive number of operations to (1) take the square root, (2) take the natural logarithm of that result, and (3)sum all of these Nresulting contributions. Alternatively, we can write this result in a more computationally efﬁcient form as A z(X, Y )∼=−μ0I 2π1 NN⎜summationdisplay n=1ln⎜bracketleftbigg⎜radicalBig (X−xn)2+(Y−yn)2⎜bracketrightbigg =−μ0I 4π1 Nln⎜braceleftBiggN⎜productdisplay n=1⎜bracketleftBig (X−xn)2+(Y−yn)2⎜bracketrightBig⎜bracerightBigg (6.39) and we have used the result that ln x1+lnx2+···+ lnxN=ln(x1x2···x N) along with ln√a=1 2lna. Hence, the result in (6.35) can be approximated as Az(X, Y )∼=−μ0I 2πlnD (6.40a) 272 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION where the geometric mean distance D between the point P and the rectangle is lnD=1 NN⎜summationdisplay n=1ln⎜bracketleftbigg⎜radicalBig (X−xn)2+(Y−yn)2⎜bracketrightbigg =1 2Nln⎜braceleftBiggN⎜productdisplay n=1⎜bracketleftBig (X−xn)2+(Y−yn)2⎜bracketrightBig⎜bracerightBigg (6.40b) or D=2N⎜radicaltp⎜radicalvertex⎜radicalvertex⎜radicalbtN⎜productdisplay n=1⎜bracketleftBig (X−xn)2+(Y−yn)2⎜bracketrightBig (6.40c) So the vector magnetic potential at a point Pfrom a conductor of rectangular cross section and very long length compared to D,l/greatermuchD, can be computed alternatively as being the same as that due to a ﬁlament containing the totalcurrent Ithat is separated from the point Pby the distance D, which is the geometric mean distance between the rectangle and the point. This works well for points Poutside the conductor. In using it to determine A z(X, Y )at points within the conductor, we run into an obvious problem when the desired point is at the center of a subrectangle, X=xn,Y=yn.F o r this particular “self term” in (6.38), we integrate over the subrectangle andaverage to give 1 /Delta12x/Delta12y⎜integraldisplay y⎜integraldisplay x⎜integraldisplay η⎜integraldisplay ξln⎜bracketleftbigg⎜radicalBig (x−ξ)2+(y−η)2⎜bracketrightbigg dξ dη dx dy (6.41) The GMD between the subrectangle and itself is proportional to the perimeter of the subrectangle and evaluates to ln Dn=ln[0.223525 (W +H)][14,20]. This represents a special case of the geometric mean distance between a shape and itself. The numerical solution above was used by Antonini et al. [19] to compute the internal self partial inductance of a conductor of rectangular cross section.We discuss those results in Section 6.5. The GMD of the combination of two or more surfaces S 1,S2,... with another surface Scan be found from lnDS=A1lnDS1+A2lnDS2+··· A1+A2+···(6.42) which follows from the deﬁnition of the GMD. The notation DSidenotes the GMD from surface Sito the desired surface S, and Aidenotes the cross- sectional area of surface Si. CONCEPT OF GEOMETRIC MEAN DISTANCE 273 6.4.1 Geometrical Mean Distance Between a Shape and Itself and the Self Partial Inductance of a Shape The GMD between a shape and itself is given by (6.32): ln(D)=1 AA/prime⎜integraldisplay A⎜integraldisplay A/prime(lnd)dA/primedA (6.32) where AandA/primeare the same surface area. When the conductor length is much greater than the GMD given by (6.32), the self partial inductance isapproximately L p∼=μ0 2πl⎜parenleftbigg ln2l D−1⎜parenrightbigg l/greatermuchD (6.33) Since the integration for the GMD in (6.32) is over the entire cross section of the conductor, the self partial inductance in (6.33) includes the internal self partial inductance of the conductor due to magnetic ﬂux internal to it. Althoughthere are innumerable shapes having a GMD with itself, we consider only theGMDs of the common and useful shapes shown in Fig. 6.8. GMDs of othershapes may be found in Grover [14], Rosa and Grover [20], Gray [29], andHiggins [30]. wr (a) w (b) wt (c) FIGURE 6.8. GMDs of various shapes from which the self partial inductance can be obtained. 274 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION GMD of a Circular Shape from Itself The ﬁrst shape is the circle shown in Fig. 6.8(a), representing the cross section of a wire. The self partial inductancedue to magnetic ﬂux external to the wire was computed directly in Chapter 4 by replacing the wire with a ﬁlament on its axis carrying the total wire currentI. This was permissible by the fundamental assumption that the current of the wire is distributed uniformly over its cross section . Hence, the external self partial inductance of a wire of length land radius r wwas obtained in Chapter 5 and is given by Lp,external =ψ∞ I =μ0 2πl⎡ ⎣ln⎛⎝ l rw+⎜radicalBigg⎜parenleftbiggl rw⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggrw l⎜parenrightbigg2 +rw l⎤ ⎦ (5.18a) For a long wire such that l/greatermuchrw(a very reasonable assumption) this approx- imates to Lp,external =μ0 2πl⎜bracketleftBigg ln2l rw−1+rw l−1 4⎜parenleftbiggrw l⎜parenrightbigg2 +···⎜bracketrightBigg ∼=μ0 2πl⎜parenleftbigg ln2l rw−1⎜parenrightbigg l/greatermuchrw (5.18c) But this only gives the self partial inductance due to the magnetic ﬂux external to the wire. The contribution to the self partial inductance due to the magneticﬂux internal to the wire was obtained as L p,internal =μ0 8πl (4.70) Adding the external self partial inductance in (5.18c) to the internal self partial inductance in (4.70) gives the total self partial inductance of a length of wirethat is due to magnetic ﬂux both external to the wire and internal to the wireas L p=Lp,external +Lp,internal =μ0 2πl⎜parenleftbigg ln2l rw−1+1 4⎜parenrightbigg =μ0 2πl⎜parenleftbigg ln2l rw−3 4⎜parenrightbigg (6.43a) CONCEPT OF GEOMETRIC MEAN DISTANCE 275 But this can be written as Lp=Lp,external +Lp,internal =μ0 2πl⎜parenleftbigg ln2l rw−1+1 4⎜parenrightbigg =μ0 2πl⎜parenleftbigg ln2l rwe−1/4−1⎜parenrightbigg (6.43b) Hence, the GMD of a circular area of radius rw(a wire cross section) from itself is D=rwe−1/4 =0.7788 rw(6.44a) or lnD=lnrw−1 4(6.44b) and the self partial inductance of a cylinder (a wire) of radius rwand length l is Lp∼=μ0 2πl⎜parenleftbigg ln2l D−1⎜parenrightbigg =μ0 2πl⎜parenleftbigg ln2l rw−3 4⎜parenrightbigg l/greatermuchD (6.44c) Note that the self partial inductance here includes the internal inductance of the wire due to magnetic ﬂux internal to the wire. To compute the GMD of a circular area with itself directly, we ﬁrst compute the GMD between a point Pand a circular area of radius rwas shown in Fig. 6.9. The GMD becomes, from (6.32), lnD=1 πr2w⎜integraldisplayrw r=0⎜integraldisplay2π θ=0ln⎜parenleftBig⎜radicalbig R2+r2−2rRcosθ⎜parenrightBig rd θd r =1 21 πr2w⎜integraldisplayrw r=0⎜integraldisplay2π θ=0ln⎜parenleftBig R2+r2−2rRcosθ⎜parenrightBig rd θd r (6.45a) and we have used the law of consines to write d2=R2+r2−2rRcosθ (6.45b) 276 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION wrd Pθr R FIGURE 6.9. Computing the GMD between a point and a circular area. The integral with respect to θcan be written as (I)=⎜integraldisplay2π θ=0ln⎜parenleftBig R2+r2−2rRcosθ⎜parenrightBig dθ =⎜integraldisplay2π θ=0ln⎜bracketleftBigg 1+⎜parenleftbiggr R⎜parenrightbigg2 −2⎜parenleftbiggr R⎜parenrightbigg cosθ⎜bracketrightBigg dθ+2l nR⎜integraldisplay2π θ=0dθ =⎜integraldisplay2π θ=0ln⎜bracketleftBigg 1+⎜parenleftbiggr R⎜parenrightbigg2 −2⎜parenleftbiggr R⎜parenrightbigg cosθ⎜bracketrightBigg dθ+4πlnR (6.46) This integral can be evaluated using an integral from Dwight [7]: ⎜integraldisplay2π θ=0ln⎜parenleftBig 1+a2−2acosx⎜parenrightBig dx=⎜braceleftBigg 4πlnaa > 1 0 a<1(D865.73c) giving (I) =4πln(r/R )+4πlnR=4πlnrfor the point inside the circle, R<r , and (I) =4πlnRfor the point outside the circle, R>r . Evaluating the second integral with respect to ryields lnD=4π 2πr2w⎜integraldisplayrw r=0ln (r)rd r R<r w =4π 2πr2w⎜bracketleftBigg r2 2lnr−r2 4⎜bracketrightBiggrw r=0 =⎜parenleftbigg lnrw−1 2⎜parenrightbigg R<r w (6.47a) CONCEPT OF GEOMETRIC MEAN DISTANCE 277 and lnD=4π 2πr2wlnR⎜integraldisplayrw r=0rd r R>r w =4π 2πr2wlnR⎜bracketleftBigg r2 2⎜bracketrightBiggrw r=0 =lnR R>r w (6.47b) and we have used an integral from Dwight [7]: ⎜integraldisplay xlnxd x=x2 2lnx−x2 4(D610.1) in (6.47a). The result in (6.47a) shows that the natural logarithm of the GMD from a circular area to any point inside it is the natural logarithm of the radiusof the circular area, r w, minus1 2. The result in (6.47b) shows that the GMD from a circular area to any point P outside it is simply the distance betweenthe point and the center of the circular area. Now consider an annulus of differential thickness drand radius rand a point Pwithin the annulus as shown in Fig. 6.10. The total circular surface is divided into a part internal to the point and the annulus, and a part externalto the point and the annulus. The GMD of the combination of two or moresurfaces S 1,S2,...with another surface Scan be found from lnDS=A1lnDS1+A2lnDS2+··· A1+A2+···(6.42) which follows from the deﬁnition of the GMD. The notation DSidenotes the GMD from surface Sito the desired surface S, and Aidenotes the wrP rdr FIGURE 6.10. GMD of a circle from itself. 278 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION cross-sectional area of surface Si. Hence, the GMD of the entire circle to point Pis lnDP=πr2 w⎜parenleftBig lnrw−1 2⎜parenrightBig −πr2⎜parenleftBig lnr−1 2⎜parenrightBig +πr2lnr πr2w =⎜parenleftbigg lnrw−1 2⎜parenrightbigg +r2 r2w1 2(6.48) Hence, the GMD of the circle from itself is lnD=1 πr2w2π⎜integraldisplayrw r=0ln (D P)rd r =2 r2w⎜integraldisplayrw r=0⎜bracketleftBigg⎜parenleftbigg lnrw−1 2⎜parenrightbigg +r2 r2w1 2⎜bracketrightBigg rd r =⎜parenleftbigg lnrw−1 2⎜parenrightbigg +1 4 =lnrw−1 4(6.49) as before. GMD of a Line from Itself Next we determine the GMD of a line of width wand zero thickness from itself as illustrated in Fig. 6.11. The GMD of a line from itself is again obtained from the basic deﬁnition in (6.32) as lnD=1 w2⎜integraldisplayw x2=0⎜integraldisplayw x1=0ln|x2−x1|dx1dx2 =1 2w2⎜integraldisplayw x2=0⎜integraldisplayw x1=0ln(x2−x1)2dx1dx2 (6.50) wx1x2x FIGURE 6.11. GMD of a line from itself. CONCEPT OF GEOMETRIC MEAN DISTANCE 279 The integral with respect to x1becomes (I)=⎜integraldisplayw x1=0ln(x2−x1)2dx1 =⎜integraldisplayx2 λ=x 2−wlnλ2dλ =⎜bracketleftBig λlnλ2−2λ⎜bracketrightBigx2 λ=x 2−w =x2lnx2 2−2x2−(x2−w)ln(x2−w)2+2(x2−w)(6.51) and we have used the change of variables λ=x2−x1,dλ=−dx1and an integral from Dwight [7]: ⎜integraldisplay ln⎜parenleftBig x2+a2⎜parenrightBig dx=xln⎜parenleftBig x2+a2⎜parenrightBig −2x+2atan−1x a(D623) The integral with respect to x2becomes (II)=⎜integraldisplayw x2=0(I)dx2 =⎜bracketleftBigg λ2 2lnλ2−λ2 2−λ2⎜bracketrightBiggw λ=0+⎜bracketleftBigg λ2 2lnλ2−λ2 2−λ2⎜bracketrightBigg−w λ=0 =w2lnw2−3w2(6.52) and we have used a change of variables λ=x2−w,dλ=dx2in the second half of the integral and an integral from Dwight [7]: ⎜integraldisplay xln⎜parenleftBig x2+a2⎜parenrightBig dx=1 2⎜parenleftBig x2+a2⎜parenrightBig ln⎜parenleftBig x2+a2⎜parenrightBig −1 2x2(D623.1) Hence, dividing (6.52) by 2 w2according to (6.50) gives the GMD of a line with itself as lnD=lnw−3 2(6.53a) or D=0.22313w (6.53b) and the self partial inductance of a line is Lp∼=μ0 2πl⎜parenleftbigg ln2l D−1⎜parenrightbigg =μ0 2πl⎜parenleftbigg ln2l w+1 2⎜parenrightbigg l/greatermuchD(6.53c) 280 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION Compare this result to (6.20) for u=l/w/greatermuch1: Lp(t=0) l∼=μ0 2π⎧ ⎪⎪⎨ ⎪⎪⎩ln 2u+1 2+1 3uu/greatermuch1,l/greatermuchw u⎜parenleftbigg ln2 u+1 2+u 3⎜parenrightbigg u/lessmuch1,l/lessmuchw(6.20) Foru=1(l=w)the exact result in (6.19) gives Lp(t=0)=7.552 nH /in., whereas (6.53) using the GMD gives Lp(t=0)=6.061nH /in., a difference of 24.6%. For u=10(l=10w)the exact result in (6.19) gives Lp(t=0)= 17.926 nH /in., whereas (6.53) using the GMD gives Lp(t=0)=17.758 nH /in., a difference of 0.9%. For a typical PCB land of width w=8 mils, a ratio of l/w=10 gives the length of the land as 0.08 in., which is not an unreasonably long length for a typical PCB land. For u=100(l=100w)the exact result in (6.19) gives Lp(t=0)=29.472 nH /in., whereas (6.53) using the GMD gives Lp(t=0)=29.455 nH /in., a difference of 0.06%. For a typical PCB land of width w=8 mils, a ratio of l/w=100 gives the length of the land as 0.8 in., which is still not an unreasonably long length of a typical PCB land. Thisreinforces the restriction on the validity of the GMD concept to situationswhere the length of the land is much greater than its width. But the computationof the self partial inductance of a thin land via the GMD is considerably simplerthan the exact formula given in (6.18) or (6.19). GMD of a Rectangular Shape from Itself The GMD of the rectangle shown in Fig. 6.8(c) can be derived in a fashion similar to that from the basic deﬁnitionin (6.32) and is given in [19,30] as lnD=−25 12+1 2t2w24⎜summationdisplay i=14⎜summationdisplay j=1(−1)i+jf⎜parenleftbigqi,rj⎜parenrightbig(6.54a) where f(q, r)=⎜parenleftBigg q2r2 4−q4 24−r4 24⎜parenrightBigg ln⎜parenleftbigq2+r2⎜parenrightbig +q3r 3tan−1r q+qr3 3tan−1q r(6.54b) CONCEPT OF GEOMETRIC MEAN DISTANCE 281 and q1=−w q2=q4=0 q3=w r1=−t r2=r4=0 r3=t(6.54c) Writing this out and using the fact that f(0,0)=0,f(±w,±t)=f(w,t), f(−w, 0)=f(w,0), andf(−t,0)=f(t,0)gives a clearer result as lnD=−25 12+1 2ln⎜parenleftBig w2+t2⎜parenrightBig −1 12w2 t2ln⎜parenleftBigg 1+t2 w2⎜parenrightBigg −1 12t2 w2ln⎜parenleftBigg 1+w2 t2⎜parenrightBigg +2 3w ttan−1t w+2 3t wtan−1w t(6.55a) which is equivalent to the result given by Gray [29], p. 302, eq. (114). If the rectangle is square, w=t, its GMD is lnD=−25 12+1 2ln⎜parenleftBig 2w2⎜parenrightBig −1 12ln(2)−1 12ln(2)+2 3π 4+2 3π 4 =−25 12+lnw+1 3ln(2)+π 3 =lnw−0.80509 w=t (6.55b) It turns out that the GMD of a rectangle shape (a bar) can be represented approximately in terms of the dimensions of its perimeter, its width wand its thickness t, as [14,20,29] logD∼=ln(w+t)−3 2(6.56a) and the GMD is D∼=(0.2235 )(w+t) (6.56b) 282 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION If the bar is square, w=t, this reduces approximately to (6.55b). So the self partial inductance of a conductor of rectangular cross section is approximately Lp∼=μ0 2πl⎜parenleftbigg ln2l w+t+1 2⎜parenrightbigg l/greatermuchD (6.56c) which reduces to (6.53c) for t=0. Note that the self partial inductance here includes the internal inductance of the bar due to magnetic ﬂux internal tothe bar. The result for the GMD of a rectangle given in (6.54) can be obtained by direct integration using the basic result in (6.32) as lnD=1 (wt)2⎜integraldisplayt y2=0⎜integraldisplayt y1=0⎜integraldisplayw x2=0⎜integraldisplayw x1=0 ln⎜bracketleftbigg⎜radicalBig (y2−y1)2+(x2−x1)2⎜bracketrightbigg dx1dx2dy1dy2 =1 (wt)21 2⎜integraldisplayt y2=0⎜integraldisplayt y1=0⎜integraldisplayw x2=0⎜integraldisplayw x1=0 ln⎜bracketleftBig (y2−y1)2+(x2−x1)2⎜bracketrightBig dx1dx2dy1dy2 (6.57) The ﬁrst integral with respect to x1is evaluated as (I)=⎜integraldisplayw x1=0ln⎜bracketleftBig (y2−y1)2+(x2−x1)2⎜bracketrightBig dx1 =⎜integraldisplayx2 λ=x 2−wln⎜bracketleftBig (y2−y1)2+λ2⎜bracketrightBig dλ where we have used a change of variables λ=x2−x1,dλ=−dx1. This can be evaluated using Dwight [7]: ⎜integraldisplay ln⎜parenleftBig a2+x2⎜parenrightBig dx=xln⎜parenleftBig a2+x2⎜parenrightBig −2x+2atan−1x a(D623) to give (I)=⎜bracketleftbigg λln⎜parenleftBig (y2−y1)2+λ2⎜parenrightBig −2λ+2(y2−y1)tan−1λ y2−y1⎜bracketrightbiggx2 λ=x 2−w =⎜bracketleftbigg x2ln⎜parenleftBig (y2−y1)2+x22⎜parenrightBig −2x2+2(y2−y1)tan−1x2 y2−y1⎜bracketrightbigg CONCEPT OF GEOMETRIC MEAN DISTANCE 283 −⎜bracketleftbigg (x2−w)ln⎜parenleftBig (y2−y1)2+(x2−w)2⎜parenrightBig −2⎜parenleftbigg x2−w⎜parenrightbigg +2(y2−y1)tan−1x2−w y2−y1⎜bracketrightbigg The next integral with respect to x2is evaluated as (II)=⎜integraldisplayw x2=0(I)dx2 =⎜integraldisplayw x2=0⎜bracketleftbigg x2ln⎜parenleftBig (y2−y1)2+x2 2⎜parenrightBig −2x2+2(y2−y1)tan−1x2 y2−y1⎜bracketrightbigg dx2 +⎜integraldisplay−w λ=0⎜bracketleftbigg λln⎜parenleftBig (y2−y1)2+λ2⎜parenrightBig −2λ+2(y2−y1)tan−1λ y2−y1⎜bracketrightbigg dλ where we have made a change of variables λ=x2−w,dλ=dx2in the second integral. These can be evaluated using integrals from Dwight [7]: ⎜integraldisplay xln⎜parenleftBig a2+x2⎜parenrightBig dx=1 2⎜parenleftBig a2+x2⎜parenrightBig ln⎜parenleftBig a2+x2⎜parenrightBig −x2 2(D623.1) and ⎜integraldisplay tan−1x adx=xtan−1x a−a 2ln⎜parenleftBig a2+x2⎜parenrightBig (D525) to give (II)=⎜bracketleftBigg λ2 2ln⎜parenleftBig (y2−y1)2+λ2⎜parenrightBig −3λ2 2 +2(y2−y1)λtan−1λ y2−y1−1 2(y2−y1)2ln⎜parenleftBig (y2−y1)2+λ2⎜parenrightBig⎜bracketrightBigg w,−w (λ) 0,0 where the notation is the same as in Sections 6.2 and 6.3. The third integral with respect to y1can be similarly integrated as (III)=⎜integraldisplayt y1=0(II)dy1 =⎜integraldisplayt y1=0⎜bracketleftBigg λ2 2ln⎜parenleftBig (y2−y1)2+λ2⎜parenrightBig −3λ2 2 +2(y2−y1)λtan−1λ y2−y1 −1 2(y2−y1)2ln⎜parenleftBig (y2−y1)2+λ2⎜parenrightBig⎜bracketrightBigg w,−w (λ) 0,0dy1 284 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION =⎜integraldisplayy2 ξ=y2−t⎜bracketleftBigg λ2 2ln⎜parenleftBig ξ2+λ2⎜parenrightBig −3λ2 2+2ξλtan−1λ ξ −1 2ξ2ln⎜parenleftBig ξ2+λ2⎜parenrightBig⎜bracketrightbiggw,−w (λ) 0,0dξ and we have used a change of variables ξ=y2−y1,dξ=−dy1. These can be evaluated using integrals from Dwight [7]: ⎜integraldisplay ln⎜parenleftBig a2+x2⎜parenrightBig dx=xln⎜parenleftBig a2+x2⎜parenrightBig −2x+2atan−1x a(D623) ⎜integraldisplay x2ln⎜parenleftBig a2+x2⎜parenrightBig dx=x3 3ln⎜parenleftBig a2+x2⎜parenrightBig −2 9x3+2 3xa2−2 3a3tan−1x a (D623.2) ⎜integraldisplay xtan−1a xdx=ax 2+x2+a2 2tan−1a x(D528.1) giving (III)=⎜bracketleftBigg⎜bracketleftBigg λ2ξ 2ln⎜parenleftBig ξ2+λ2⎜parenrightBig −ξ3 6ln⎜parenleftBig ξ2+λ2⎜parenrightBig −λ2ξ 3−3 2λ2+ξ2 9 +4 3λ3tan−1⎜parenleftbiggξ λ⎜parenrightbigg +λ⎜parenleftBig ξ2+λ2⎜parenrightBig tan−1λ ξ⎜bracketrightBigg w,−w (λ) 0,0⎜bracketrightBiggy2 ξ=y2−t =⎜bracketleftBigg λ2y2 2ln⎜parenleftBig y2 2+λ2⎜parenrightBig −y3 2 6ln⎜parenleftBig y2 2+λ2⎜parenrightBig −λ2y2 3−3 2λ2+y2 2 9 +4 3λ3tan−1y2 λ+λ⎜parenleftBig y2 2+λ2⎜parenrightBig tan−1λ y2⎜bracketrightBigg w,−w (λ) 0,0 −⎜bracketleftBigg λ2(y2−t) 2ln⎜parenleftBig (y2−t)2+λ2⎜parenrightBig −(y2−t)3 6ln⎜parenleftBig (y2−t)2+λ2⎜parenrightBig −λ2(y2−t) 3−3 2λ2 +(y2−t)2 9+4 3λ3tan−1y2−t λ+λ⎜parenleftBig (y2−t)2 +λ2⎜parenrightBig tan−1λ y2−t⎜bracketrightBigg w,−w (λ) 0,0 CONCEPT OF GEOMETRIC MEAN DISTANCE 285 or (III)=⎜bracketleftBigg⎜bracketleftBigg λ2η 2ln⎜parenleftBig η2+λ2⎜parenrightBig −η3 6ln⎜parenleftBig η2+λ2⎜parenrightBig −λ2η 3−3 2λ2+η2 9 +4 3λ3tan−1η λ+λ⎜parenleftBig η2+λ2⎜parenrightBig tan−1λ η⎜bracketrightBigg w,−w (λ) 0,0⎜bracketrightBigg y2 (η) y2−t In a similar fashion, the fourth integral can be evaluated as (IV)=⎜integraldisplayt y2=0(III)dy2 =⎜integraldisplayt η=0⎜bracketleftBigg λ2η 2ln⎜parenleftBig η2+λ2⎜parenrightBig −η3 6ln⎜parenleftBig η2+λ2⎜parenrightBig −λ2η 3−3 2λ2+η2 9 +4 3λ3tan−1η λ+λ⎜parenleftBig η2+λ2⎜parenrightBig tan−1λ η⎜bracketrightBigg w,−w (λ) 0,0dη +⎜integraldisplay−t η=0⎜bracketleftBigg λ2η 2ln⎜parenleftBig η2+λ2⎜parenrightBig −η3 6ln⎜parenleftBig η2+λ2⎜parenrightBig −λ2η 3−3 2λ2+η2 9 +4 3λ3tan−1η λ+λ⎜parenleftBig η2+λ2⎜parenrightBig tan−1λ η⎜bracketrightBigg w,−w (λ) 0,0dη and using Dwight [7]: ⎜integraldisplay xln⎜parenleftBig a2+x2⎜parenrightBig dx=1 2⎜parenleftBig a2+x2⎜parenrightBig ln⎜parenleftBig a2+x2⎜parenrightBig −x2 2(D623.1) ⎜integraldisplay x3ln⎜parenleftBig x2+a2⎜parenrightBig dx=x4−a4 4ln⎜parenleftBig x2+a2⎜parenrightBig −x4 8+x2a2 4(D623.3) ⎜integraldisplay tan−1x adx=xtan−1x a−a 2ln⎜parenleftBig a2+x2⎜parenrightBig (D525) ⎜integraldisplay tan−1a xdx=xtan−1a x+a 2ln⎜parenleftBig x2+a2⎜parenrightBig (D528) ⎜integraldisplay x2tan−1a xdx=x3 3tan−1a x+ax2 6−a3 6ln⎜parenleftBig x2+a2⎜parenrightBig (D528.2) giving the result in (6.54). 6.4.2 Geometrical Mean Distance and Mutual Partial Inductance Between Two Shapes In this subsection we obtain the geometrical mean distances between the common shapes shown in Fig. 6.12. These represent (a) the cross sections of 286 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION sw1rw2r (a) (b) (c)1w 1w2w 2w1t 2th hs s FIGURE 6.12. Computing the GMDs between common shapes. two wires of radii rw1andrw2that are separated a distance s, (b) two strips of widths w1andw2and zero thickness, and (c) two rectangular bars of widths w1andw2and thicknesses t1andt2that represent the cross sections of PCB lands. Once again we stress that the GMDs for these shapes will be valid onlyas long as the lengths of the conductors whose cross sections these shapesrepresent are much longer than the separation between them. CONCEPT OF GEOMETRIC MEAN DISTANCE 287 GMD Between Two Circular Shapes The GMD between two circular shapes representing the cross sections of two wires shown in Fig. 6.12(a)is simply the distance between their centers, s: lnD=lns This follows from the fact that the mutual partial inductance between be- tween two parallel wires of equal length lhaving currents that are uniformly distributed over their cross sections can be obtained by replacing them with two ﬁlaments located on the axes of the conductors . This result is the basic result for the mutual partial inductance between two ﬁlaments separated adistance sand given as M p=μ0 2πl⎜bracketleftBigg ln2l s−1+s l−1 4⎜parenleftbiggs l⎜parenrightbigg2 +1 32⎜parenleftbiggs l⎜parenrightbigg4 −···⎜bracketrightBigg ∼=μ0 2πl⎜parenleftbigg ln2l s−1⎜parenrightbigg l/greatermuchs (5.21c) GMD Between Two Lines The GMD between two parallel lines having widths w1andw2and zero thickness with vertical separation hand horizontal separation sbetween their centers as shown in Fig. 6.12(b) can again be obtained from the basic deﬁnition of the GMD in (6.32): lnD=1 w1w2⎜integraldisplays+w 1/2+w 2/2 x2=s+w 1/2−w 2/2⎜integraldisplayw1 x1=0ln⎜bracketleftbigg⎜radicalBig (x2−x1)2+h2⎜bracketrightbigg dx1dx2 =1 w1w21 2⎜integraldisplays+w 1/2+w 2/2 x2=s+w 1/2−w 2/2⎜integraldisplayw1 x1=0ln⎜bracketleftBig (x2−x1)2+h2⎜bracketrightBig dx1dx2(6.58) The integral with respect to x1becomes (I)=⎜integraldisplayw1 x1=0ln⎜bracketleftBig (x2−x1)2+h2⎜bracketrightBig dx1 =⎜integraldisplayx2 λ=x 2−w 1ln⎜parenleftBig λ2+h2⎜parenrightBig dλ =⎜bracketleftbigg λln⎜parenleftBig λ2+h2⎜parenrightBig −2λ+2htan−1λ h⎜bracketrightbiggx2 λ=x 2−w 1 =x2ln⎜parenleftBig x2 2+h2⎜parenrightBig −2x2+2htan−1x2 h −(x2−w1)ln⎜parenleftBig (x2−w1)2+h2⎜parenrightBig +2(x2−w1)−2htan−1x2−w1 h (6.59) 288 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION and we have used the change of variables λ=x2−x1,dλ=−dx1and an integral from Dwight [7]: ⎜integraldisplay ln⎜parenleftBig x2+a2⎜parenrightBig dx=xln⎜parenleftBig x2+a2⎜parenrightBig −2x+2atan−1x a(D623c) The integral with respect to x2becomes (II)=⎜integraldisplays+w 1/2+w 2/2 x2=s+w 1/2−w 2/2(I)dx2 =⎜bracketleftBigg x2 2−h2 2ln⎜parenleftBig x2 2+h2⎜parenrightBig −3 2x2 2+2hx2tan−1x2 h⎜bracketrightBiggs+w 1/2+w 2/2 x2=s+w 1/2−w 2/2 −⎜bracketleftBigg λ2−h2 2ln⎜parenleftBig λ2+h2⎜parenrightBig −3 2λ2+2hλtan−1λ h⎜bracketrightBiggs−w 1/2+w 2/2 λ=s−w 1/2−w 2/2 (6.60) and we have used a change of variables λ=x2−w1,dλ=dx2in the second half of the integral and integrals from Dwight [7]: ⎜integraldisplay xln⎜parenleftBig x2+a2⎜parenrightBig dx=x2+a2 2ln⎜parenleftBig x2+a2⎜parenrightBig −x2 2(D623.1) ⎜integraldisplay tan−1x adx=xtan−1x a−a 2ln⎜parenleftBig x2+a2⎜parenrightBig (D525) Hence, dividing (6.60) by 2 w1w2according to (6.58) gives the GMD between the two lines. Although the integration is complete, this gives a complicatedexpression which we obtain in the next section. A more interesting and simpler result is for the case where the line widths are identical, w 1=w2=w, and the two lines are on line with each other, h=0. The results above simplify to give the GMD as 2w2lnD=(s+w)2ln(s+w)−3 2(s+w)2−2s2lns +3s2+(s−w)2ln(s−w)−3 2(s−w)2h=0,w1=w2=w (6.61) We can simplify this further. Suppose that the lines lie in the same plane, h=0, and the separation between the lines is s=nw. The lines are touching whenn=1 and are separated edge to edge by one line width wwhenn=2. CONCEPT OF GEOMETRIC MEAN DISTANCE 289 Substituting s=nwinto (6.61) gives lnD=(n+1)2 2ln[(n+1)w]−3 4(n+1)2−n2lnnw +3 2n2+(n−1)2 2ln[(n−1)w]−3 4(n−1)2 =(n+1)2 2ln[(n+1)w]−n2lnnw +(n−1)2 2ln[(n−1)w]−3 2s=nw (6.62) This matches the result given by Rosa [25], p. 164, eq. (11). Rosa also gives a convenient series expansion for this case of h=0,w1=w2=w, and s=nw: lnD=lnnw−⎜parenleftbigg1 12n2+1 60n4+1 168n6+1 360n8+1 660n10+···⎜parenrightbigg (6.63) which converges very rapidly. In either case, the mutual partial inductance between the conductors is Mp∼=μ0 2πl⎜parenleftbigg ln2l D−1⎜parenrightbigg l/greatermuchD (6.33) GMD Between Two Rectangles Finally, we obtain the GMD for the general case shown in Fig. 6.12(c) of two parallel rectangles with widths w1,w2, thicknesses t1,t2, and vertical separation hand horizontal separation center to center of s. The result is given in [19,30] as lnD=−25 12+1 2(t1w1)(t2w2)4⎜summationdisplay i=14⎜summationdisplay j=1(−1)i+jf⎜parenleftbigqi,rj⎜parenrightbig(6.64a) where f(q, r)=⎜parenleftBigg q2r2 4−q4 24−r4 24⎜parenrightBigg ln⎜parenleftBig q2+r2⎜parenrightBig +q3r 3tan−1r q+qr3 3tan−1q r(6.64b) 290 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION and q1=s−w1 2−w2 2 q2=s+w1 2−w2 2 q3=s+w1 2+w2 2 q4=s−w1 2+w2 2 r1=h−t1 2−t2 2 r2=h+t1 2−t2 2 r3=h+t1 2+t2 2 r4=h−t1 2+t2 2(6.64c) Once again, for “long conductors,” the mutual partial inductance between them is Mp∼=μ0 2πl⎜parenleftbigg ln2l D−1⎜parenrightbigg l/greatermuchD (6.33) Grover [14] gives several tables for determining the GMD of two rectangles for the cases of Fig. 6.12(c) with various ratios of width to thickness and lengthto width for h=0. Ruehli [15] has computed results for the mutual partial inductance between two parallel lands of equal width and equal thickness forvarious values of u=l/wandv=t/w. He shows that there is little error be- tween the exact result and the mutual partial inductance obtained by replacingthe lands with one ﬁlament at the center of each land (i.e., M p∼=Mf) as long as the lands are not very close to each other. This correlates with the exactsolution for round wires discussed in Section 4.6, in that proximity effect andthe associated redistribution of the current over the cross section to the facingsurfaces of the wires is not signiﬁcantly pronounced unless the wires are closeenough that they are separated by a distance such that one wire will exactlyﬁt between them. This is somewhat remarkable since as the lands are broughtclose together, their currents will no longer be distributed uniformly over their cross sections but will migrate toward the facing sides , as we show with nu- merical computations in the next section. It should be reiterated that all our previous formulas were derived assuming that the current remains uniformlydistributed over the conductor cross section . Considering the nonuniform COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 291 distribution of the current over the cross section of a PCB land does not seem to be feasible except with approximate numerical solutions which we addressin the next section. 6.5 COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES OF LANDS AND NUMERICAL METHODS We have seen that computation of the self and mutual partial inductances of conductors having rectangular cross sections (PCB lands) is very com-plicated and the results are quite tedious. This is why the earlier textbooksand publications contained extensive tables for the calculation of these par-tial inductances; digital computers had not been invented. Today we enjoythe enormous computing power of computers, and involved formulas are nolonger the obstacle they used to be. Still there remain many problems in partialinductance for which there are no feasible closed-form solutions. An exampleis the nonuniform current distribution over the conductor cross section causedeither by (1) proximity effect, or (2) frequencies of excitation other than dc.In this section we develop methods for numerically computing the self andpartial inductances of conductors of rectangular cross section that can be usedto solve those difﬁcult problems. The primary restriction on all the previous results of this book is that the cur- rent is assumed to be uniformly distributed over the conductor cross sections . This condition is not satisﬁed for either proximity effect or high-frequencyexcitation. However, we can approximate a nonuniform current distribution ina discrete fashion, that is, by approximating the current distribution over theconductor cross section in a piecewise-constant or piecewise-linear manner.The actual parameters in that distribution are unknown but will be determinedby enforcing the constraints that the currents must satisfy, resulting in the si-multaneous solution of a large set of simultaneous equations which computerscan readily handle. For example, consider breaking the cross section of a rectangular con- ductor into individual “subbars” of rectangular cross section, as illustrated inFig. 6.13(a). The number of divisions along the width is NW and the number ofdivisions along the thickness is NT. Each subbar dimension is /Delta1t=t/NT and /Delta1w=w/NW. The total number of subbars is therefore N =NW×NT. The bar can be modeled as an equivalent circuit shown in Fig. 6.13(b). Representeach subbar with its dc resistance R i=l/σ /Delta1w /Delta1t, where σis the conductiv- ity of the metal, which we will assume is copper (having σCu=5.8×107). TheLpiare the exact self partial inductances of the subbar, given by Hoer and 292 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION l (a)12 11Rp1L 2 NN N = (NW)(NT) 2R NRp2L pNLp12M p1NM p2NM (b)I I V1I 2I NI+– FIGURE 6.13. Approximating a conductor of rectangular cross section as a set of parallel “subbars” of rectangular cross section. Love in (6.23), which Ruehli ([15], eq. (15)) put into a more stable numerical form. The mutual partial inductances between the subbars is approximated asbeing between ﬁlaments at the centers of the subbars and is given by M pij=μ0 2πl⎡ ⎢⎣ln⎛ ⎜⎝l dij+⎜radicaltp⎜radicalvertex⎜radicalvertex⎜radicalbt⎜parenleftBigg l dij⎜parenrightBigg2 +1⎞ ⎟⎠−⎜radicalBigg 1+⎜parenleftbiggdij l⎜parenrightbigg2 +dij l⎤ ⎥⎦(5.21a) where dijis the distance between the centers of subbars iandj. COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 293 All the subbars are connected in parallel so that the voltage across each subbar is the same and is the voltage between the ends of the entire conductorand denoted as V. The total current drawn by the conductor is denoted as I, and we determine and plot over the conductor cross section the individualcurrents of the subbars, I i, fori=1,2,..., N where N =NW×NT. This demonstrates two important aspects of this problem. For increasing frequen-cies of excitation (1) the currents crowd to the edges of the conductor, and (2)the currents peak at the corners of the conductor, giving the so-called “bed-post” distribution over the conductor cross section. The equivalent circuit ofthe conductor and its subbars is shown in Fig. 6.13(b). The voltage acrosseach subbar, V i, is related to the current through it, Ii,a s Vi=⎜parenleftbigRi+jω L pi⎜parenrightbigIi+jωN⎜summationdisplay j=1 j/=iMpijIj (6.65a) In matrix notation this becomes V=ZI (6.65b) where VandIare vectors of N rows and one column containing the voltages and currents of the N individual subbars as V=⎡ ⎢⎢⎢⎢⎢⎣V1 V2 ... VN⎤ ⎥⎥⎥⎥⎥⎦,I=⎡ ⎢⎢⎢⎢⎢⎣I1 I2 ... IN⎤ ⎥⎥⎥⎥⎥⎦(6.65c) The “impedance matrix” Zis square with N =NW×NT rows and N = NW×NT columns and contains the self impedances of the individual sub- bars and the mutual impedances between the subbars. This matrix of subbarimpedances can be separated into two parts: Z=Z s+Zm (6.65d) The “self impedance” matrix is a diagonal matrix as Zs=⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣Zs100 ··· 0 0Zs20··· 0 00...···... ............0 00 ··· 0Z sN⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (6.65e) 294 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION where Zsi=Ri+jω L pi (6.65f) The “mutual impedance” matrix is Zm=jω⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣0Mp12Mp13···Mp1N Mp12 0Mp23···Mp2N Mp13Mp23...···... ............... M p1NMp2N ··· ··· 0⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (6.65g) We solve these equations by inverting (6.65b) to give, at each frequency of excitation, I=Z−1V (6.66) But the subbars are all connected in parallel, so that V=V1=V2=···= VN. Hence, (6.66) becomes ⎡ ⎢⎢⎢⎢⎢⎣I1 I2 ... IN⎤ ⎥⎥⎥⎥⎥⎦=Z−1⎡ ⎢⎢⎢⎢⎢⎣1 1 ... 1⎤ ⎥⎥⎥⎥⎥⎦V (6.67) Hence, we sum the columns ofZ−1to obtain the individual subbar currents: Ii=N⎜summationdisplay j=1⎜bracketleftBig Z−1⎜bracketrightBig ijV (6.68) Choosing V=1V gives the explicit currents of the subbars. We can also obtain the overall impedance (resistance and self partial inductance) of the entireconductor by noting that the total conductor current is the sum of the subbarcurrents computed in (6.68) (i.e., I=I 1+I2+···+ IN). Hence, we sum the rows andcolumns (the sum of all entries) of Z−1to give the relationship between the total current through the conductor, I, and the voltage across the conductor, V,t og i v e I=⎛ ⎝N⎜summationdisplay i=1N⎜summationdisplay j=1⎜bracketleftBig Z−1⎜bracketrightBig ij⎞ ⎠V (6.69) COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 295 We obtain the effective impedance of the entire conductor by inverting (6.69) to give Ztotal=1 ⎜summationtextN i=1⎜summationtextNj=1[Z]ij =Rtotal+jω L p,total (6.70) where the real part of Ztotalis the equivalent resistance of the land, Rtotal, and the imaginary part of Ztotalis the product of ω=2πfand the equivalent self partial inductance of the land, Lp,total . Note that Lp,total is the sum of the external self partial inductance and the internal self partial inductance.As the frequency increases and the current crowds to the surface of the land,the magnetic ﬂux internal to the land goes to zero, so the internal inductanceshould also go to zero, leaving the external self partial inductance of the landas the high-frequency inductance. We will show computed results for typical lands. The thickness of a PCB land is 1.4 mils, where a mil is one-thousandth of an inch. The lands aretypically etched from a copper-cladded board that is made from glass–epoxyor FR-4 material. The glass–epoxy substrate has a relative permeability ofμ r=1 (is not magnetic) and hence does not affect the inductance. It has a relative permittivity of about εr=4.7, which does affect the capacitance. The copper cladding is said to be “1 ounce” since 1 ft2of this thickness weighs 1 oz. Typical land widths range from 5 to 30 mils. The following shows theresult for a PCB land whose thickness is 1.4 mils, whose width is 15 mils,and whose length is 10 in. The current distribution over the land cross sec-tion will be shown for four frequencies: 100 kHz, 10 MHz, 100 MHz, and 1GHz. The results for this case were obtained with NT =16 and NW =172. The width dimension of w=15 mils is one skin depth ( δ=1/√ πfμ 0σ)a t f1δ=30 kHz and two skin depths at f2δ=120.34 kHz. The thickness dimen- sion of t=1.4 mils is one skin depth at f1δ=3.45 MHz and two skin depths at f2δ=13.8 MHz. For the discretization of each conductor, the widths, /Delta1w, and the thicknesses, /Delta1t, of each subbar should be less than two skin depths in order that the current over each subbar will be approximately uniformly distributedover it, which was the basic assumption in the derivation of the subbar resis-tances and partial inductances. Hence, we should have /Delta1w=w/NW <2δ and/Delta1t=t/NT<2δ. For NT =16 and NW =172,/Delta1t=/Delta1w =2.22/H9262m. (This was the reason for choosing the NT and NW as we did.) Each dimen-sion of the subbars, /Delta1wand/Delta1t, is two skin depths at 3.5 GHz. Hence, at the largest frequency of 1 GHz the current will be approximately uniformly dis-tributed over each subbar. The distribution of the current over the conductorcross section is shown at the four frequencies in Fig. 6.14(a) through (d). 296 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION FIGURE 6.14(a). Current distribution over the cross section of a 1 .4 mil ×15 mil land at 100 kHz. FIGURE 6.14(b). Current distribution over the cross section of a 1 .4 mil ×15 mil land at 10 MHz. FIGURE 6.14(c). Current distribution over the cross section of a 1 .4 mil ×15 mil land at 100 MHz. FIGURE 6.14(d). Current distribution over the cross section of a 1 .4 mil ×15 mil land at 1 GHz. COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 297 FIGURE 6.15. Current concentrating at the land surface in a thickness of dimension on the order of one skin depth. Figure 6.14(a) shows that the current is uniformly distributed over the cross section at 100 kHz. We would expect that as the frequency of excitationincreases, the current will move toward the surface, eventually lying predom-inantly in a thickness at the surface of dimension on the order of a skin depth,as illustrated in Fig. 6.15. Consequently, we would expect that the currentwould start to exhibit this concentration at the surface when the width or thethickness becomes greater than two skin depths: 2 δ<wor 2δ<t , whichever occurs ﬁrst. For the width dimension of w=15 mils, f 2δ=120.34 kHz, and for the thickness dimension of t=1.4 mils, f2δ=13.8 MHz. Hence, 100 kHz is such that the current distribution should remain uniformly distributed overthe cross section. Since the width is much larger than the thickness, the ﬁrstdeparture from a uniform distribution should occur when the width becomeson the order of two skin depths. Figure 6.14(b) for 10 MHz clearly showsthat since the frequency is well above that for which the width is two skindepths but has not reached the point where the thickness is two skin depths orf 2δ=13.8 MHz, the current is crowding to the ends of the width dimension. Similarly, Fig. 6.14(c) for 100 MHz is above the point where the thicknessis two skin depths, so crowding of the current is beginning to occur alongthe thickness dimension. Finally, Fig. 6.14(d) for 1 GHz shows the classic“bedpost” pattern, where the frequency is high enough that the current peakssharply at all four corners of the land cross section. We next show the frequency behavior of the net resistance and self partial inductance of the 1.4-mil ×15-mil land computed from (6.70). Figure 6.16(a) shows the behavior of the net resistance versus frequency. Figure 6.16(b)shows the internal inductance of the land computed from the current data via the method of Antonini et al. [19]. As the frequency increases such that the current lies in a thickness of ap- proximately one skin depth at all four surfaces as in Fig. 6.15(b), we would ex-pect the per-unit-length high-frequency resistance to asymptotically approach r hf=1 σ(2δt+2δw) =1 2σδ(w+t)/Omega1/m (6.71a) 298 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION FIGURE 6.16(a). Net resistance of the 1.4 mil ×15 mil land vs. frequency. The dc resistance, rdc=1/σwt, and this high-frequency asymptote join at 2δ=wt w+t(6.71b) Hence, the high-frequency resistance should asymptotically approach a√fincrease since the skin depth δ=1/√πfμ 0σdecreases as√f. Figure 6.16(a) exhibits this behavior. Similarly, as the frequency increasesand the current crowds to the surface of the land, the magnetic ﬂux internal to FIGURE 6.16(b). Internal inductance of the 1.4 mil ×15 mil land vs. frequency. COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 299 the land goes to zero as√f, so the internal inductance should also go to zero as√f. Figure 6.16(b) also exhibits this behavior. Because of the “peaking” of the current at the corners of the land, the high-frequency resistance in(6.71) is somewhat less than the actual high-frequency resistance [8]. Next, we investigate two identical lands having identical cross-sectional dimensions of w=t=50/H9262m∼=2 mils with various separations between the two lands. The total length of the two lands is 2.5 cm (about 1 in.). We computethe total resistance and partial inductance of each land from 1 to 100 MHz.We also plot the current distributions across the conductor cross sections for(1) an isolated conductor and (2) two identical conductors having variousspacings between them when the two conductors carry (a) differential-modecurrents and (b) common-mode currents. This demonstrates the proximityeffect. The total resistance and partial inductance of each land will also beplotted for various land separations. We subdivide each land cross section intosubbars of thickness /Delta1t=/Delta1w=t/NT=w/NW for NT =NW=20. For the discretization of each conductor, the widths, /Delta1w, and the thicknesses, /Delta1t, of each subbar should again be less than two skin depths in order that the current over each subbar will be approximately uniformly distributed over it , which was the basic assumption in the derivation of the subbar resistances andpartial inductances (see Fig. 6.15). Hence, we should have /Delta1w=w/NW <2δ and/Delta1t=t/NT<2δ. For NT =NW=20,/Delta1t=/Delta1w =2.5/H9262m and each dimension of the subbars, /Delta1wand/Delta1t, is two skin depths at 2.8 GHz. Each subbar will again be represented as shown in Fig. 6.13(b). The self partialinductances and the mutual partial inductances between all subbars will bothbe calculated using results from Hoer and Love [16] and given in (6.23) and(6.28), respectively. The computations for an isolated conductor are as described before. For the case of two conductors, we write the relations between the impedances ofthe subbars of the two conductors as ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣V1 V1 ... V1 ··· V2 V2 ... V2⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ =⎡ ⎢⎢⎢⎣Z1...jωM12 ··· ··· ··· jωM12... Z2⎤ ⎥⎥⎥⎦ ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright Z⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣I11 I12 ... I1N ··· I21 I22 ... I2N⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (6.72) 300 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION where N =NW×NT. The Z1andZ2are each square with N rows and N columns each and contain the self impedances Z=R+jω L pof the subbars of that conductor as well as the mutual partial inductances jω M pbetween the subbars of that conductor. The M12matrix is square with N rows and N columns and contains the mutual partial inductances between subbars of the two conductors. We ﬁrst invert Zin (6.72), giving ⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣I11 I12 ... I1N ··· I21 I22 ... I2N⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ =⎜bracketleftBigg Y11Y12 Y12Y22⎜bracketrightBigg ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright Y=Z−1⎡ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣V1 V1 ... V1 ··· V2 V2 ... V2⎤ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (6.73) Since the total current of each conductor is I1=I11+···I 1NandI2=I21+··· I2Nand all subbars of each conductor are connected in parallel, we then sum the rows and columns of each of the four blocks of Y=Z−1to give a relation between the total currents and voltages of the two conductors as ⎜bracketleftBigg I1 I2⎜bracketrightBigg =⎜bracketleftBigg Y11Y12 Y12Y22⎜bracketrightBigg⎜bracketleftBigg V1 V2⎜bracketrightBigg (6.74a) where each Yijis a scalar: Yij=N⎜summationdisplay n=1N⎜summationdisplay m=1⎜bracketleftbigYij⎜bracketrightbig mn(6.74b) Equation (6.74a) is inverted to give ⎜bracketleftBigg V1 V2⎜bracketrightBigg =⎜bracketleftBigg Z11Z12 Z12Z22⎜bracketrightBigg⎜bracketleftBigg I1 I2⎜bracketrightBigg (6.75) We have two cases to consider: (1) differential-mode currents where the total currents through the two conductors are related as I2=−I1, and (2) common- mode currents where the total currents through the two conductors are related COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 301 asI2=I1. From this we can determine the voltages across the two conductors forI1=1Aa s V1,DM=(Z11−Z12)I1 V2,DM=(Z22−Z12)I2 V1,CM=(Z11+Z12)I1 V2,CM=(Z22+Z12)I2(6.76) where DM and CM denote differential- and common-mode voltages across the conductors, respectively. This gives the voltages across each of the twoconductors for DM and CM excitation for a total current of 1 A througheach of the two conductors. Hence the resistance and partial inductance ofeach conductor can be determined for DM and CM excitation as the real andimaginary parts of (6.76). (The self partial inductance of the conductor is theimaginary part divided by ω.) The voltages determined in (6.76) for 1-A DM and CM excitation can then be substituted into (6.73) to determine and plot thecurrents of the subbars for each of the conductors for DM and CM excitation. Figure 6.17(a) shows the current distribution over the cross sections of the two conductors for a separation (edge to edge) of s=50/H9262m∼=2 mils, differential-mode excitation, and an excitation frequency of 1 MHz, whileFig. 6.17(b) shows this for an excitation frequency of 100 MHz. Figure 6.18repeats this for common-mode excitation. These plots show the expectedcrowding of the current toward the surfaces of the conductors when the cross-sectional dimensions become on the order of two skin depths. They also show FIGURE 6.17(a). Current distribution over the conductor cross sections for differential-mode current and s=50/H9262m and 1 MHz. 302 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION FIGURE 6.17(b). Current distribution over the conductor cross sections for differential-mode current and s=50/H9262m and 100 MHz. the proximity effect. For differential-mode excitation shown in Fig. 6.17(b) the currents tend to concentrate on the facing sides as is the case for wires.For common-mode excitation shown in Fig. 6.18(b), the currents tend to con-centrate on opposide sides. FIGURE 6.18(a). Current distribution over the conductor cross sections for common-mode current and s=50/H9262m and 1 MHz. COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 303 FIGURE 6.18(b). Current distribution over the conductor cross sections for common-mode current and s=50/H9262m and 100 MHz. Figure 6.19(a) shows the resistance for (1) an isolated conductor and (2) two conductors. Also shown is the high-frequency approximation in (6.71a). Thedc and high-frequency asymptotes join at a frequency where 2δ =wt/(w+t). Observe that for this close separation of the two conductors of s=50/H9262m, FIGURE 6.19(a). Total resistance for an isolated conductor and for two conductors. 304 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION FIGURE 6.19(b). Total partial inductance for an isolated conductor and for two conductors. exactly one conductor will ﬁt between the two. Figure 6.19(b) shows the net self partial inductance of the isolated conductor as well as the two conductors.In addition, the dc and high-frequency limits of the self partial inductance areshown. FIGURE 6.20(a). Total resistance for an isolated conductor and for two conductors for a separation of s=200/H9262m. COMPUTING THE HIGH-FREQUENCY PARTIAL INDUCTANCES 305 FIGURE 6.20(b). Total partial inductance for an isolated conductor and for two conductors for a separation of s=200/H9262m. Figure 6.20 shows the total resistance and partial inductance for an isolated conductor and for two conductors for a separation (edge to edge) of s= 200/H9262m. For these wide separations (four conductors will ﬁt between the two) the resistance and inductance are not affected appreciably by the presence ofthe other conductor, as we would expect. FIGURE 6.21(a). Total resistance for an isolated conductor and for two conductors for a separation of s=10/H9262m. 306 PARTIAL INDUCTANCES OF CONDUCTORS OF RECTANGULAR CROSS SECTION FIGURE 6.21(b). Total partial inductance for an isolated conductor and for two conductors for a separation of s=10/H9262m. Figure 6.21 shows the total resistance and partial inductance for an iso- lated conductor and for two conductors for a separation (edge to edge) ofs=10/H9262m. For this very close separation, the resistance and inductance are affected signiﬁcantly by the presence of the other conductor more than fors=50/H9262m. 7 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE In the preceding chapters we have detailed the concept and calculation of “loop” inductance and “partial” inductance for various current-carrying struc-tures consisting of conductors of circular, cylindrical cross section (wires), aswell as conductors of rectangular cross section (PCB lands). In this ﬁnalchapter we summarize the advantages and disadvantages of characterizingthese structures with “loop” inductance or with “partial” inductance and giveexamples of the applications of partial inductance. 7.1 LOOP INDUCTANCE VS. PARTIAL INDUCTANCE: INTENTIONAL INDUCTORS VS. NONINTENTIONALINDUCTORS An important question that this book intends to resolve is: When should loop inductance be used to characterize a current-carrying, conductive structure,and when should partial inductance be used? A related question to be an-swered is: What are the advantages and disadvantages of loop inductancevs. partial inductance? There exists considerable misunderstanding through-out the electrical engineering community regarding “partial” inductance andwhere it is appropriate to use the concept to characterize the inductance ef-fects of current-carrying structures. “Loop” inductance is a standard topic in Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 307 308 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE undergraduate electrical engineering textbooks, but these textbooks do not contain any reference to “partial” inductance. Hence, electrical engineers arewell trained in the understanding and calculation of loop inductance, but theyhave little or no understanding of the concept and uses of partial inductance.This unfortunate deﬁciency in the training of electrical engineers causes themerroneously to use formulas for loop inductance that do not apply to theirsituation of interest when they should instead use partial inductance formu-las.If one models a section of wire or PCB land with an inductance, they are inherently using partial inductance, whether they know it or not . Loop inductance cannot be used to characterize a section of a conductor becauseas we have discussed, the induced emf in Faraday’s law of induction that theloop inductance represents cannot be placed uniquely in any speciﬁc place in a closed current loop. The literature, both trade magazines and scholarlyjournals, is replete with examples of this misunderstanding and misuse ofinductance, wherein the symbol for an inductor is used to model a section ofwire or PCB land, yet a formula for “loop” inductance is used erroneously tocompute the value of that inductance. The inductive effects of a time-varying current inherent in Faraday’s law of induction represent one of the most important parameters that determinethe performance of today’s electrical circuits and systems. Digital circuitsand systems today have clock and data rates in the GHz range. The spec-tral content of these digital waveforms of trapezoidal pulse shape generallyextends at least to the ﬁfth harmonic of the repetition rate and depends onthe pulse rise/fall times [5]. Frequencies of analog systems have also movedsteadily into the GHz range. Interconnects such as wires and PCB lands wereelectrically short some 10 years ago and could be ignored in an analysis ofthe system performance. Today, the physical lengths of those same intercon-nects have not changed substantially but their electrical lengths have becomea signiﬁcant portion of a wavelength, and hence can no longer be ignored.These interconnects represent one of the most important parameters affectingdigital as well as analog system performance. Most power distribution andsignal interconnects in today’s digital and analog systems must be modeledto predict the true performance of the system. To understand the distinction in use between “loop” inductance and “par- tial” inductance, it is important to focus on “intentional” and “nonintentional”inductors. The solenoid and the toroid covered in Sections 4.2 are examples of“intentional” inductors. They are constructed intentionally to take advantageof the inductive effects inherent in Faraday’s law of induction. One of theprimary uses of intentional inductors such as the solenoid and the toroid areto block high-frequency signals while passing lower-frequency signals suchas dc power. They are also used, along with capacitors, to construct bandpassﬁlters that are so essential to radio communication. Lowpass and highpass TO COMPUTE “LOOP” INDUCTANCE, THE “RETURN PATH” 309 ﬁlters are constructed as well using inductors in combination with capacitors. Similarly, bandreject ﬁlters remove unwanted signals. Shorted stubs consist-ing of two parallel lands shorted together at the far end are used to constructinductors that are suitable for use at microwave frequencies, where the para-sitic effects of interwinding capacitance in conventional inductors would shortout the inductor at these very high frequencies. These are examples of “in-tentional” inductors. For these structures we are interested only in the voltageat the terminals of the structure and are not interested in the voltages gener-ated at points internal to the structure. Hence, “loop” inductance is useful incharacterizing these structures for their typical uses. Certain “transmission lines,” such as the coaxial cable (Sections 4.1.3, and 4.7.3); the two-wire transmission line (Sections 4.6.1 and 4.7.2); one wireabove a ground (Section 4.6.2); transmission lines composed of conductorsof rectangular cross section such as the stripline, the microstrip line, and thePCB (Section 4.9); and multiconductor transmission lines (Section 4.8.2) aresuitably characterized by per-unit-length “loop” inductances. Again, for thesetransmission-line structures we are only interested in the voltages at the termi-nals of the structure and are not interested in the voltages generated at pointsinternal to the structure. Hence, “loop” inductance is useful in characterizingthese structures for analyzing their typical use. However, “nonintentional” inductances are generally undesired induc- tances and represent detrimental effects that must be incorporated into ananalysis of the overall system to determine its performance degradation. Itis generally not feasible or useful to attempt to characterize nonintentionalinductances with “loop” inductance for a number of reasons, outlined in thefollowing sections. Hence, partial inductance is the most appropriate charac-terization for nonintentional inductances. 7.2 TO COMPUTE “LOOP” INDUCTANCE, THE “RETURN PATH” FOR THE CURRENT MUST BE DETERMINED Dc currents must form closed loops along conductors. The “loop” inductance characterizes that complete current loop at its terminals. For intentional in-ductors, the complete current loop is evident virtually “by design.” Hence, thecomplete current loop for calculating the loop inductance is evident. However, consider nonintentional inductors such as lands on a PCB that interconnect a source and a load. The “going down” path for a current fromthe source to the load is fairly easy to determine. But the return path for thecurrent back to the source can take a number of alternative routes that are farfrom obvious. In today’s highly dense and complicated PCBs it is virtuallyan impossible task to identify the return path for most currents. If one cannot 310 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE identify the complete path of the current loop, the loop inductance cannot be computed. We therefore have no other recourse than to use partial inductance for characterizing the Faraday law inductive effect of a segment of the currentloop such as a wire or a PCB land. We ascribe the property of “inductance” to an intentional inductor such as a toroid or a solenoid, whether or not that inductor has any current ﬂowingthrough it. But its utility exists only if a current passes through it. Similarly,when we ascribe a partial inductance to a segment of a conductor, that partialinductance is effective only when it has a current passing through it. Thatcurrent must form a closed loop by some path that may not yet be obvious,nor do we need to determine “that path” when computing the partial inductanceof a segment of the path. We must be assured that a return path for the currenthas been provided by the designer in some fashion that may not be readilyobvious. If, by some omission in the circuit’s physical construction, no pathis provided for the current to return to its source, the partial inductance of asegment has no effect, for the same reason that an intentional inductor wouldhave no effect: No current passes through it. There is an important case where the return path of the current on a PCB is somewhat more obvious. If the current is allowed to return through one of theinnerplanes buried in the board (either a “ground” distribution innerplane or apower distribution innerplane), it is well known that the current in that planewill tend to concentrate directly beneath the “going down” path. The returncurrent will peak beneath the “going down” current and spread out in (on thesurface of) the adjacent plane, giving a current density on the plane of [5] J s(x)=I πh⎜parenleftBig 1+(x/h)2⎜parenrightBig A/m where his the height of the “going down” current above the plane and xis the horizontal position along the plane, with x=0 being directly beneath the “going down” current. This result was derived for a very ideal situation ofa ﬁlamentary current above an inﬁnite, perfectly conducting ground plane.Innerplanes in PCBs have various discontinuities in them and are of ﬁnitedimensions, so they are represented only approximately by this ideal case.For example, lands on a PCB that are above but near the edges of the in-nerplanes have substantial fringing ﬁelds and probably do not represent thecase of an inﬁnite ground plane. Similarly, the innerplanes may have gaps andother discontinuities in them. A power plane must have isolated sections toaccommodate the various dc voltages of the system. Even the ground inner-planes may have gaps cut into them for various reasons. It is generally notrecommended and is unnecessary to cut gaps in a ground innerplane, as thisdisrupts the return current paths [5]. But even for the ideal situation, which GENERALLY , THERE IS NO UNIQUE RETURN PATH FOR ALL FREQUENCIES 311 resembles a stripline or a microstrip line, computation of the loop inductance is a formidable task. Holloway and Kuester have made this calculation for themicrostrip line [31]. It is worth noting that they do this using partial inductanceconcepts. We can model all the conductor segments of a structure with their partial inductances (self and mutual). We can then solve the resulting lumped circuitusing, for example, SPICE and hence determine the return paths for the cur-rents without having to guess about the return path for a current. If the modeldoes not have provision for at least one return path back to its source for aconductor segment, assigning a partial inductance to that segment will haveno effect in the same fashion that not passing a current around the closed loopof an intentional inductor will eliminate any effect of that loop inductance.But to compute, a priori, a loop inductance, we must ﬁrst determine the com- plete path for the loop current. To compute, a priori, the partial inductance ofa conductor segment, we do notneed to determine the complete current loop path. 7.3 GENERALLY, THERE IS NO UNIQUE RETURN PATH FOR ALL FREQUENCIES, THEREBY COMPLICATINGTHE CALCULATION OF A “LOOP” INDUCTANCE Circuit designers provide only one path for the “going down” current. How- ever, they tend to “leave it to the current” to select a return path back to thesource. At dc and low frequencies, a current will return to its source along thepath of lowest resistance. At higher frequencies, the current will return to its source along the path of lowest impedance, which is generally the path of low- estinductance: Resistance is an insigniﬁcant portion of the total impedance of a conductor at the higher frequencies [5]. A good example of this is theshielded wire, where the shield is above and “grounded to” a “ground” planethat was discussed in Chapter 1 and shown in Fig. 1.2. At dc and low fre-quencies, the “going down” current Itakes its return path, I G, through the massive ground plane, which obviously has a much lower dc resistance thanother possible return paths. However, at higher frequencies, the current Itakes its return path up through the shield, I S, thereby minimizing the area and in- ductive impedance between the “going down” path and the return path [5].Therefore, the return paths and hence the complete current loops are differentfor different frequencies for this structure. Another example of this is the popular “gridded ground” system on a PCB shown in Fig. 7.1, where a “grid” of conductors is interconnected so as to pro-vide a number of possible paths for the current to return to its source along [5]. 312 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE I SVSR LR FIGURE 7.1. Ground grid for reducing the loop area of the loop current path. This is done to avoid the radiated emissions of large current loops and is used in low-cost products to avoid more costly boards having innerplanes. It is alsoused even on PCBs with innerplanes, to minimize the (partial) inductanceand associated “ground bounce” as well as common-mode currents generatedby return currents that do not take the innerplane route [5]. At dc and lowfrequencies the “going down” current Ireturns along the path of lowest resistance using the simple “current-division” principle [1,2]. Simply modeleach wire segment of the grid with its dc resistance and compute (usingSPICE or, if the circuit is simple, by hand) the path of lowest resistance. Asthe frequency of the current increases, the impedance of the complete path isgoverned by the inductance of the entire loop. From our previous calculationswe know that the inductance of a current loop is related directly to its looparea. Hence, the return current “chooses” the path to minimize the total looparea giving the path of lowest loop impedance (inductance) nearest the “goingdown” current, as shown in Fig. 7.1. Modeling all the conductor segments ofthis structure with their resistances as well as their partial inductances (selfand mutual), we can solve the resulting lumped-circuit model using SPICEand hence determine the lowest-impedance return current path. This wasaccomplished by Paul and Smith [32]. 7.4 COMPUTING THE “GROUND BOUNCE” AND “POWER RAIL COLLAPSE” OF A DIGITAL POWER DISTRIBUTIONSYSTEM USING “LOOP” INDUCTANCES Signal integrity has become a paramount design consideration in today’s high- speed digital systems. Signal integrity has to do with ensuring that the wave-shape and levels of those signals are maintained within tightly controlledlimits to avoid logic errors and false switching when these levels rise or fall COMPUTING THE “GROUND BOUNCE” AND “POWER RAIL COLLAPSE” 313 V 5+VPR VPR VGB VGBI I I I LGBLPR power supply ground FIGURE 7.2. “Ground bounce” and “power rail collapse” in digital circuits. into gray regions. Dc voltages are supplied to the modules in a digital system by lands routed on and within a PCB. For example, consider a two-conductorpower distribution circuit for supplying dc voltages from a power supply toCMOS inverters as shown in Fig. 7.2. When the left inverter is in the highstate, the right inverter is in the low state and current passes along the powerdistribution lands from the dc power supply to the power pin of the left in-verter, into the input of the right inverter, and returns to the power supplyalong the “ground” lands. When the left inverter is switched to the low state,the current of the power supply passes down through the right inverter andreturns to the power supply. We have modeled the PCB lands connecting thepower and the ground pins of the inverter modules to each other and to thedc power supply with inductances labeled as L PRandLGB. As the inverters switch, the currents through these lands go to or increase from zero or changedirection, thereby inducing voltages across these inductors that are related tothe rate of change of the currents through them. V oltages V PRare developed across the inductors LPRthat may cause the voltages of the power pins of the modules to drop signiﬁcantly, which is called power rail collapse . Similarly, voltages VGBare developed across the inductors LGB, causing the voltages of the two “ground” pins of the inverters to differ signiﬁcantly, which is referredto as ground bounce. Both of these phenomena may cause logic errors, thereby degrading the signal integrity of the system. There is considerable evidence that these voltages exist [33], but the essen- tial question is: What do we mean by these inductances? They certainly arenot “loop” inductances since we know that an inductance of a closed currentloop cannot be assigned uniquely to any place in that loop. These inductancesare, in fact, partial inductances. Although not shown in this diagram, there 314 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE are also mutual partial inductances between these self partial inductances that should be included if the conductors are close enough to each other. A diagramsuch as Fig. 7.2 is seen throughout the literature not only in “trade” magazinesbut also in scholarly journals. A closer inspection of those articles shows thatmany of the authors do not know how to correctly calculate the values forthese inductances and erroneously use formulas for “loop” inductance. If weaccept the fact that ground bounce and power rail collapse actually exist indigital circuits and are a severe problem, we have no recourse but to admitthat “loop” inductance does not explain the phenomenon, and we must ac-cept the utility of partial inductance concepts. A means for measuring theseground bounce and power rail collapse voltages using the concept of partialinductance was given in [33,34]. 7.5 WHERE SHOULD THE “LOOP” INDUCTANCE OF THE CLOSED CURRENT PATH BE PLACED WHEN DEVELOPINGA LUMPED-CIRCUIT MODEL OF A SIGNAL OR POWERDELIVERY PATH? As we know from previous discussions, one cannot place the loop inductance uniquely in any speciﬁc position in the loop. You can only attribute partialinductances to speciﬁc segments of the closed current loop. For example,consider the parallel-wire transmission line shown in Fig. 7.3. In transmission-line analyses, we are only interested in the terminal voltages at the endpointsof the line. Hence, we can place the “loop” inductance in either conductor, asshown in Fig. 7.3, and obtain the same result for these terminal voltages. It isclear from Fig. 7.3 that the ground bounce and power rail collapse voltagescannot be determined uniquely using “loop” inductance. Throughout the literature one sees equivalent-circuit models with lumped “inductances” representing segments of conductors. Upon closer scrutiny it loopL I II Lloop I dtdIL V 2V2 V1V1 V2 V1 loop= = FIGURE 7.3. Loop inductance and the transmission line. WHERE SHOULD THE “LOOP” INDUCTANCE OF THE CLOSED CURRENT 315 sI I I IMp VGBVPRLp Lp FIGURE 7.4. Modeling a two-wire transmission line with partial inductances. is found that formulas for “loop” inductance are used to compute the values of those inductances. We can compute the “loop” inductance using partialinductances of the loop segments, but the reverse is not true. For example, wecan model the two-wire transmission line in Fig. 7.3 using partial inductancesas shown in Fig. 7.4. Using the dot convention [1,2] we obtain V GB=VPR=LpdI dt−MpdI dt =⎜parenleftbigLp−Mp⎜parenrightbigdI dt(7.1) The total voltage drop around the transmission line loop is the product of the “loop” inductance of the transmission line loop and the time derivativeof the current. Hence, the total voltage around the loop is twice (7.1). There-fore, the transmission line “loop” inductance can be obtained from the partialinductances as L loop=2⎜parenleftbigLp−Mp⎜parenrightbig(7.2) 316 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE These self and mutual partial inductances for wires were obtained in Sec- tions 5.3 and 5.4: Lp=ψ∞ I ∼=μ0 2π/Delta1z⎜parenleftbigg ln2/Delta1z rw−1⎜parenrightbigg /Delta1z/greatermuchrw (7.3a) and Mp=ψ∞ I =μ0 2π/Delta1z⎡ ⎣ln⎛⎝ /Delta1z s+⎜radicalBigg⎜parenleftbigg/Delta1z s⎜parenrightbigg2 +1⎞ ⎠−⎜radicalBigg 1+⎜parenleftbiggs /Delta1z⎜parenrightbigg2 +s /Delta1z⎤ ⎦ ∼=μ0 2π/Delta1z⎜parenleftbigg ln2/Delta1z s−1⎜parenrightbigg /Delta1z/greatermuchs (7.3b) Combining these gives the loop inductance of the transmission line: Lloop=2⎜parenleftbigLp−Mp⎜parenrightbig =2μ0 2π/Delta1z⎜parenleftbigg ln2/Delta1z rw−1⎜parenrightbigg −2μ0 2π/Delta1z⎜parenleftbigg ln2/Delta1z s−1⎜parenrightbigg =μ0 π/Delta1zlns rw(7.4) which was obtained directly by computing the magnetic ﬂux threading the loop between the two wires in (4.73) in Section 4.6.1. This result for the loop inductance of the transmission line in terms of partial inductances in (7.2) is rather obvious if we recall the physical meaningof the partial inductances as being the ratios of the magnetic ﬂux between awire and inﬁnity and the current producing that ﬂux. The quantity ⎜parenleftbigLp−Mp⎜parenrightbig is the net magnetic ﬂux through the loop formed by the two transmission-line conductors per unit of current as shown in Fig. 7.5. Adding the ﬂuxes throughthe loop due to the two currents (equal but oppositely directed) gives the resultin (7.2). This partial inductance model of the transmission line in Fig. 7.4 also clearly shows an interesting observation that is not obtained from the transmission-line “loop” inductance circuit of Fig. 7.3. As we move the two wires closer, thevalue of the mutual partial inductance approaches the value of the self partialinductance (i.e., M p→Lpass→0), and hence the ground bounce and power rail collapse voltages approach zero (i.e., VGB,VPR→0a ss→0). This HOW CAN A LUMPED-CIRCUIT MODEL OF A COMPLICATED SYSTEM 317 cjci I loopψ (a) ci Icj B B sjsiloopB Lp Mp (b) FIGURE 7.5. Loop inductance of a transmission line from partial inductances. shows a routinely observed design rule that in order to reduce ground bounce and power rail collapse, the “going down” and return conductors should beplaced as close as possible to each other. This useful design rule could not bedetermined using “loop” inductance, but it is frequently used without under-standing the distinction between “loop” and “partial” inductances. 7.6 HOW CAN A LUMPED-CIRCUIT MODEL OF A COMPLICATED SYSTEM OF A LARGE NUMBER OF TIGHTLYCOUPLED CURRENT LOOPS BE CONSTRUCTED USING“LOOP” INDUCTANCE? The electromagnetic ﬁelds of all neighboring currents interact with each other to some degree, and to include all their effects, this coupling should be includedin each circuit loop representation. Consider Fig. 7.6, where we have shown 318 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE MpI I VGBVPRIother M1 M2Lother Lp2Lp1 FIGURE 7.6. Including the effects of other currents. a power distribution current loop in a digital system carrying current I, and also a conductor of a neighboring current loop on the PCB carrying currentI other. The neighboring current Iother as well as other neighboring currents will affect the ground bounce and power rail collapse voltages. This effectof neighboring currents could not be determined using “loop” inductances.However, their effect can be determined easily by modeling this situationwith the self and mutual partial inductances of the conductors. To include theeffect of the other current, we simply write the circuit equations using the dotconvention as V GB=Lp2dI dt−MpdI dt−M2dIother dt VPR=Lp1dI dt−MpdI dt+M1dIother dt With this circuit model we simply “turn the crank” and compute (perhaps with SPICE) the “ground bounce” voltage VGBand the “power rail collapse” voltage VPR, which are of considerable interest in signal integrity analyses for today’s high-speed digital systems [5]. This would be a formidable if notimpossible task using only “loop” inductances. 7.7 MODELING VIAS ON PCBS Avia(pronounced “veeya”) is a hole drilled through a PCB to interconnect lands on the top and bottom surfaces as well as on innerplane layers within the MODELING VIAS ON PCBS 319 d hpad land landpadbarrel FIGURE 7.7. Via for interconnecting lands on a PCB that are on different layers. PCB, as illustrated in Fig. 7.7. Circular “pads” attach the barrel to the lands. The via is an important feature in keeping the physical size of the PCBs frombecoming prohibitive. However, it is also a signiﬁcant factor affecting signalintegrity since it represents a discontinuity in the transmission lines that areconnected by it, thereby causing reﬂections that degrade the waveshape of thevoltages and currents on the lands. A particularly simple inductance model ofthe via is as a simple wire of diameter drepresenting the barrel. Hence, the inductance of the via is simply the self partial inductance of a wire of diameterdand length h: L via h=μ0 2π⎜parenleftbigg ln2h rvia−1⎜parenrightbigg =μ0 2π⎜parenleftbigg ln4h d−1⎜parenrightbigg H/m (7.5a) It is common to write this in nH and to use the dimensions for dandhin inches. Hence, the constant becomes μ0/2π→5.08 and we obtain Lvia h=5.08⎜parenleftbigg ln4h d−1⎜parenrightbigg nH/in. (7.5b) For example, for a board of standard thickness 62 mils and a via of radius 16 mils, which is equivalent to a No. 20 gauge wire, the inductance of the via is 320 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE 5.32 nH/in., for a total via inductance of 0.33 nH. Mutual partial inductance between neighboring vias should be included in this model. A popular signal integrity book gives the result as Lvia=5.08h⎜bracketleftbigln(4h/d)+1⎜bracketrightbig. Observe that there is a +1 in this result, whereas the correct partial inductance result in (7.5) has a −1 in it. For the previous dimensions, this gives a via inductance of 15.5 nH/in., a factor of 3 larger than the cor-rect result in (7.5b). The authors of that book argued that their result wasobtained using the per-unit-length inductance of a coaxial cable derived in(4.29), L via=(μ0/2π)hln(rs/rw), where the barrel of the via represents the inner wire of the cable of radius rwand the “shield” (the “return path” for the current) is at a distance of rs=2ehcylindrically about the barrel and e=2.71828 .... Substituting rs=2ehinto the equation for the coaxial cable gives their result. A more defensible result is obtained with partial inductanceconcepts, and we would not need to determine a ﬁctitious “return path” forthe via current. 7.8 MODELING PINS IN CONNECTORS Another aspect of system design that has the potential for degrading signal integrity are the numerous connectors in the system that make the inevitableconnection between an off-board cable and the lands on the PCB. Theseconnectors have numerous pins in them of radius r pinand length lpinthat are inserted into a receptacle on the PCB. These essentially insert inductances intothe signal propagation path that have the potential for degrading the qualityof the signals being transferred through the connector. How shall we modelthese connector pins? The obvious choice is with partial inductances. Usingthe result for the self partial inductance of a wire in (7.3a) gives L pin=μ0 2πlpin⎜parenleftBigg ln2lpin rpin−3 4⎜parenrightBigg H (7.6a) This was cited in a textbook without recognition being given to it being a “partial” inductance. Where does the factor of 3 /4 arise? If we add the internal inductance of the wire, μ0/8π, to the self partial inductance in (7.3a), we obtain a total self partial inductance of Lpin=μ0 2πlpin⎜parenleftBigg ln2lpin rpin−1⎜parenrightBigg +μ0 8πlpin =μ0 2πlpin⎜parenleftBigg ln2lpin rpin−3 4⎜parenrightBigg (7.6b) NET SELF INDUCTANCE OF WIRES IN PARALLELAND IN SERIES 321 Again mutual partial inductaries between neighbours pins should be included. As we computed earlier, the internal inductance is usually a negligible termand goes to zero as the frequency increases. This was not explained in thetextbook. If all this were explained and the use of “partial” inductance of awire were mentioned, the result would not seem to be “magic” and of unknownorigin. For pins of rectangular cross section the textbook gives the formula L pin=μ0 2πlpin⎜parenleftbigg ln4lpin P+1 2⎜parenrightbigg (7.7) where Pis the perimeter of the rectangular cross section of the pin: P= 2(w+t). This result was given in equation (6.56c). Neither of these “mys- terious formulas” in (7.6) and (7.7) were described as being “partial” induc-tances that we derived previously and were originally published by Grover[14] in 1946. In addition, the mutual inductance between two pins separatedby a distance sis given as M p=μ0 2πlpin⎡ ⎢⎣ln⎛ ⎝lpin s+⎜radicalBigg⎜parenleftbigglpin s⎜parenrightbigg2 +1⎞ ⎠−⎜radicaltp⎜radicalvertex⎜radicalvertex⎜radicalbt 1+⎜parenleftBigg s lpin⎜parenrightBigg2 +s lpin⎤ ⎥⎦ ∼=μ0 2πlpin⎜bracketleftbigg ln⎜parenleftbigg2lpin s⎜parenrightbigg −1⎜bracketrightbigg lpin/greatermuchs (7.3b) But, of course, this is simply the mutual partial inductance between the two current ﬁlaments that is derived in Chapter 5 and given in (5.21) and was alsooriginally given by Grover [14]. So once again there has been widespread useof the concept of partial inductance without apparently knowing it or under-standing the distinction between “loop” inductance and “partial” inductance. 7.9 NET SELF INDUCTANCE OF WIRES IN PARALLEL AND IN SERIES Consider two wires of equal radii r wthat are connected in series as shown in Fig. 7.8. The lengths of the wires are l1andl2and their adjacent ends are separated by a distance of s. The equivalent circuit is also shown in Fig. 7.8. Summing the voltages developed across the two inductors of the equivalentcircuit and using the dot convention gives V=L p1dI dt+MpdI dt+Lp2dI dt+MpdI dt =⎜parenleftbigLp1+Mp+Lp2+Mp⎜parenrightbigdI dt(7.8) 322 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE I Mp Lp2I IILp1 Lp netI II1l2ls FIGURE 7.8. Two wires in series. Hence, the net self partial inductance of the combination is Lpnet=Lp1+Lp2+2Mp (7.9) where the self partial inductances are obtained in Section 5.3 as Lpi∼=μ0 2πli⎜parenleftbigg ln2li rw−1⎜parenrightbigg li/greatermuchrw (5.18c) The mututal partial inductance between two wires that are aligned but are offset by a distance swas obtained in Section 5.5 and Fig. 5.12 as 2Mp=Lp(l2+s+l1)−Lp(l1+s)−Lp(l2+s)+Lps (5.28d) In the special case where the two conductors are joined together, s=0, the mutual inductance becomes 2Mp=Lp(l2+l1)−Lpl2−Lpl1s=0 (5.28d) Combining this result for s=0 with (7.9) gives Lpnet=Lp(l2+l1), which makes sense. Figure 7.9 shows two wires (possibly of different radii) connected in par- allel. From the equivalent circuit and using the dot convention, we obtain the NET SELF INDUCTANCE OF WIRES IN PARALLELAND IN SERIES 323 I MpLp1 Lp21I 2II I I II1I I2 netpL FIGURE 7.9. Two wires in parallel. voltage across each conductor as V=Lp1dI1 dt+MpdI2 dt =Lp2dI2 dt+MpdI1 dt(7.10) Since the endpoints of the wires are connected, the two voltages across each wire must be equal. Writing this in matrix form gives V⎜bracketleftBigg 1 1⎜bracketrightBigg =s⎜bracketleftBigg Lp1Mp MpLp2⎜bracketrightBigg⎜bracketleftBigg I1 I2⎜bracketrightBigg (7.11) andsdenotes the Laplace transform variable (essentially, the derivative oper- ator here). Inverting this gives ⎜bracketleftBigg I1 I2⎜bracketrightBigg =1 s1 Lp1Lp2−M2p⎜bracketleftBigg Lp2−Mp −MpLp1⎜bracketrightBigg⎜bracketleftBigg 1 1⎜bracketrightBigg V (7.12) 324 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE Solving gives ⎜bracketleftBigg I1 I2⎜bracketrightBigg =1 s1 Lp1Lp2−M2p⎜bracketleftBigg Lp2−Mp Lp1−Mp⎜bracketrightBigg V (7.13) Adding the rows gives I=I1+I2 =1 sLp1+Lp2−2Mp Lp1Lp2−M2pV (7.14) Inverting this result gives the net partial inductance of the parallel combina- tion: Lnet=Lp1Lp2−M2 p Lp1+Lp2−2Mp(7.15) If the two wires have identical lengths and radii, Lp1=Lp2=Lp, (7.15) reduces to Lnet=Lp+Mp 2(7.16) It is generally thought that placing two wires in parallel gives a net induc- tance of the combination that is half that of one wire alone, since in electriccircuit analysis, inductors in parallel combine like resistors in parallel [1,2].This is not necessarily true because that usual assumption neglects to considerthe mutual partial inductance between the two wires. The result in (7.16) showsthat unless the two wires are placed relatively far apart, the net partial induc-tance of the combination will not equal half that of one wire. Placing the wiresrelatively far apart means that the mutual partial inductance approaches zero,M p→0, as we have seen, and the result in (7.16) approaches Lpnet→Lp/2. On the other hand, moving the two wires closer together causes the mutual par-tial inductance to approach the value of the self partial inductance, M p→Lp, and the result in (7.16) approaches that of one wire, Lpnet→Lp. Hence, plac- ing two wires in parallel and close together provides a net partial inductancethat is not signiﬁcantly less than using only one wire! 7.10 COMPUTATION OF LOOP INDUCTANCES FOR V ARIOUS LOOP SHAPES With the concept of partial inductances and the results derived previously for the self and mutual partial inductances of wires, it is a simple matter to derivethe loop inductances of loops of various shapes as long as their perimetersconsist of piecewise-linear segments. For example, consider the rectangular COMPUTATION OF LOOP INDUCTANCESFOR V ARIOUS LOOP SHAPES 325 II I Il w2rw I Il wI IpwL pwLplL plLM = 0 M = 0plwMpwlM FIGURE 7.10. Rectangular loop. loop composed of wires of equal radii rwand side lengths of landwshown in Fig. 7.10. Summing the voltages across the inductances in the equivalentcircuit and using the dot convention for the mutuals gives the inductance ofthe loop as L loop=2⎜parenleftbigLpw−Mplw⎜parenrightbig+2⎜parenleftbigLpl−Mpwl⎜parenrightbig(7.17) 326 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE Notice the placement of the dots on the inductors. Parallel conductors should have the dots on the same ends since the self and mutual partial inductancesgive the magnetic ﬂuxes between each conductor and inﬁnity. To obtain themagnetic ﬂux threading the loop between the two conductors, the dots shouldbe on the same ends of parallel wires, thereby giving the magnetic ﬂux throughthe surface between the two conductors carrying oppositely directed cur-rents as the difference between the self and mutual partial inductances (seeFigure 7.5 as well as Section 5.8 and Figures 5.5, 5.6, and 5.7). Substitutingthe self and mutual partial inductances from (5.18c) and (5.21b) into (7.17)yields, for l,w/greatermuchr w: Lloop=μ0 π⎜parenleftbigg wln2w rw−w−wsinh−1w l+⎜radicalbig w2+l2−l +lln2l rw−l−lsinh−1l w+⎜radicalbig l2+w2−w⎜parenrightbigg (7.18) Simplfying this gives the same loop inductance obtained in Chapter 4 after a lengthy integration of the Bﬁeld over the loop surface: Lloop=μ0 π⎜parenleftbigg wln2w rw+lln2l rw−wsinh−1w l−lsinh−1l w +2⎜radicalbig l2+w2−2(w+l)⎜parenrightbigg (4.18) In the case of a square loop, l=w, (4.18) simpliﬁes to the result obtained in Chapter 4 that was obtained after some tedious integration of the Bﬁeld over the loop surface: Lsquare loop =μ0 π⎡ ⎢⎢⎢⎣2lln2l rw−2lsinh−1(1)⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ln⎜parenleftbig 1+√ 2⎜parenrightbig+2l√ 2−4l⎤ ⎥⎥⎥⎦ =2μ0 πl⎜bracketleftbigg ln2l rw−ln⎜parenleftBig 1+√ 2⎜parenrightBig +√ 2−2⎜bracketrightbigg =2μ0 πl⎜bracketleftbigg lnl rw−0.774⎜bracketrightbigg l=w/greatermuchrw (4.20) which matches Grover’s result [14]. To these results we may add the internal inductances of the wire if necessary: Lloop, internal=2μ0 8πl+2μ0 8πw (7.19) Next, consider the equilateral triangle shown in Fig. 7.11 (see the discussion on placement of the dots in Section 5.8). Writing the voltage Vacross one of COMPUTATION OF LOOP INDUCTANCESFOR V ARIOUS LOOP SHAPES 327 60º 60º 60ºl l lLpLp LpMp I II V FIGURE 7.11. Equilateral triangle. the inductors gives V=LpdI dt−2MpdI dt =⎜parenleftbigLp−2Mp⎜parenrightbigdI dt(7.20) The total voltage around the loop is three times (7.20), accounting for the voltages of all three sides. Hence, the net loop inductance as Lloop=3⎜parenleftbigLp−2Mp⎜parenrightbig(7.21) Substituting the self partial inductances of the three wires from Chapter 5, Lp∼=μ0 2πl⎜parenleftbigg ln2l rw−1⎜parenrightbigg l/greatermuchrw (5.18c) and the mutual partial inductances between two inclined wires of equal length from Section 5.6, equation (5.46b), Mp=μ0 2πcos⎜parenleftbig60o⎜parenrightbig⎜parenleftbigg llnl+2l l⎜parenrightbigg =μ0 2πl(0.549) (5.46b) gives Lloop=3⎜parenleftbigLp−2Mp⎜parenrightbig =3μ0 2πl⎜parenleftbigg ln2l rw−1−2×0.549⎜parenrightbigg =3μ0 2πl⎜parenleftbigg lnl rw+ln 2−1−2×0.549⎜parenrightbigg =3μ0 2πl⎜parenleftbigg lnl rw−1.405⎜parenrightbigg (7.22) 328 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE which matches Grover’s result [14]. To this result we may add the internal inductances of the wire if necessary: Lloop, internal=3μ0 8πl (7.23) 7.11 FINAL EXAMPLE: USE OF LOOP AND PARTIAL INDUCTANCE TO SOLVE A PROBLEM In this ﬁnal section of the book we examine a typical example of computing the inductive coupling between two loops using (a) loop inductances and (b)partial inductances. Figure 7.12 shows the example dimensions. To simplifythe numbers, each loop is chosen to be square with side dimensions 1 m ×1m , and the two loops are offset by 1 m in the vertical dimension and by 1 m in thehorizontal dimension. The loops are constructed of No. 20 gauge wires havingradii of r w=16 mils. The ﬁrst loop is driven by a 1-V sinusoidal source of frequency 10 MHz having a source resistance of 10 /Omega1. At a frequency of 10 MHz, a wavelength (in free space) is 30 m. Hence the dimensions of theloops and their separation can be considered to be electrically small, therebyallowing us to treat this problem as a lumped-circuit problem. The secondloop also has a 10-/Omega1 resistor inserted in it, and it is desired to compute the voltage induced across the terminals of that resistor, V out(t). 1m +– 1m1m 1m1m 1m 10 Ω10 Ω 10 MHz=fVS(t) = 1sin ωt VVout (t)+ – FIGURE 7.12. Example comparing loop inductance to partial inductance. FINAL EXAMPLE: USE OF LOOP AND PARTIAL INDUCTANCE 329 10 Ω 10 MHz+–V sin1 (t) == ft VS ω10 Ω Vout (t) 5.627 μH 5.627 μH I12I 13 2 0 04.901 nH –+ FIGURE 7.13. Example of Fig. 7.12 modeled with loop inductances. We ﬁrst compute the output voltage by modeling each loop as its loop self inductance and a mutual inductance between the two loops as shownin Fig. 7.13. The self inductance of each loop is computed from equation(4.20) as L loop=5.627μH. The mutual inductance between the two loops is computed from equation (4.111) as M12=4.901 nH. The complete model for the example using loop inductances is shown in Fig. 7.13. This can be solvedby writing the (phasor) mesh current equations around the two loops giving[1,2] ˆV S=1∠0o=⎜parenleftBig 10+jω5.627×10−6⎜parenrightBig ˆI1−jω4.901×10−9ˆI2 0=−jω4.901×10−9ˆI1+⎜parenleftBig 10+jω5.627×10−6⎜parenrightBig ˆI2 (7.24) Substituting ω=2πf=2π×107gives the phasor equations as 1∠0o=(10+j353.58 )ˆI1−j0.30795 ˆI2 0=−j0.30795 ˆI1+(10+j353.58 )ˆI2 (7.25) Solving this gives ˆI2=2.461×10−6∠−86.76oandˆVout=10ˆI2=2.461× 10−5∠−86.76oV. A simpler way to compute this result is by using PSPICE [2]. The nodes are numbered as shown in Fig. 7.13. The PSPICE program is EXAMPLE V S10A C10R S121 0L1 2 0 5.6273UL2 3 0 5.6273UK12 L1 L2 8.7097E-4R L301 0.AC DEC 1 10MEG 10MEG.PRINT AC VM(3) VP(3).END 330 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE 10 Ω 10 ΩVout (t)13 024 5 6 7 80V1∠ 0 9Lp1Lp2 Lp3 Lp4Lp5Lp6 Lp7 Lp8 FIGURE 7.14. Modeling the example in Fig. 7.12 using partial inductances. Note that PSPICE requires the description of mutual inductances in terms of their “coupling coefﬁcients” as k12=M12/√L1L2=8.7097 ×10−4. The result is, as by hand calculation, ˆVout=V(3)=2.461×10−5∠−86.76◦V. Next, we compute this result using partial inductances to model the seg- ments of the loops and their interaction. The equivalent circuit is shown inFig. 7.14. All of the self partial inductances are equal since the lengths ofthe sides of the loops are identical and equal to 1 m. These self partial in-ductances of each of the four sides of the two loops are computed fromequation (5.18a) or approximately from (5.18c) and yield L p=1.5μH. The inductances are labeled and have even or odd numbers. Mutual partial in-ductances between orthogonal segments are zero and hence there are mu-tual inductances only between even-numbered segments and only betweenodd-numbered segments. First, we compute the mutual partial inducancesbetween parallel segments in the same loop using (5.21a): M p13=Mp24= Mp57=Mp68=93.432 nH. Next, we compute the mutual partial inductances between the vertical segments and between the horizontal segments of the twoseparate loops that are parallel but offset. We use (5.28) to perform that com-putation: M p15=Mp37=Mp48=Mp26=35.524 nH. Similarly, we obtain Mp17=Mp46=27.7175 nH and Mp28=Mp35=45.7816 nH. The PSPICE program is EXAMPLE V S10A C10R S121 0 FINAL EXAMPLE: USE OF LOOP AND PARTIAL INDUCTANCE 331 L1 3 2 1.5U L2 3 4 1.5U L3 4 9 1.5U L4 0 9 1.5UL5 5 0 1.5UL6 5 6 1.5UL7 6 7 1.5UL8 0 8 1.5UR L781 0K13 L1 L3 0.062275K15 L1 L5 0.023678K17 L1 L7 0.018474K35 L3 L5 0.030515K37 L3 L7 0.023678K57 L5 L7 0.062275K24 L2 L4 0.062275K68 L6 L8 0.062275K26 L2 L6 0.023678K48 L4 L8 0.023678K28 L2 L8 0.030515K46 L4 L6 0.018474.AC DEC 1 10MEG 10MEG.PRINT AC VM(7,8) VP(7,8).END The result is ˆV(7,8)=ˆV out=2.461×10−5∠−86.76oV, which is precisely the same result as was obtained by using loop inductances! If we examine the values of the mutual inductances for this problem, we ﬁnd a seemingly curious result. All coupling between the two loops is trans-ferred only through the mutual inductances, loop or partial. In the case ofloop inductances in Fig. 7.13, this is solely through the mutual inductancebetween the two loops of M 12=4.901 nH. In the case of partial induc- tances in Fig. 7.14, this coupling between the two loops occurs only throughthe mutual partial inductances between elements of the two different loops:M p15,Mp17,Mp35,Mp37,Mp28,Mp26,Mp48,andMp46. These have magni- tudes that are on the order of 10−8, which is an order of magnitude greater than the loop mutual inductance of Fig. 7.13. How can mutual partial induc-tances that differ by an order of magnitude from the loop mutual inductanceM 12produce the same current in the second loop? The answer to this is that in the partial inductance circuit of Fig. 7.14, the effects of pairs of mutualpartial inductances representing the coupling between the two loops subtract in the production of induced voltages across the segments of the perimeter of 332 “LOOP” INDUCTANCE VS. “PARTIAL” INDUCTANCE loop 2 to produce Vout. For example, the portion of the magnetic ﬂux threading loop 2 due to the current in loop1 on segment 1 of that loop via the mutualpartial inductance between that segment and segments 5 and 7 of loop 2 isψ 2=⎜parenleftbigMp15−Mp17⎜parenrightbigI1. This is sensible since the mutual partial inductance between two segments iandj,Mpij, gives the magnetic ﬂux between segment jand inﬁnity due to the current on segment i(see Fig. 5.7). Hence, the total ﬂux between two parallel segments of loop 2 due to the current on anothersegment of loop 1 is the difference between the two mutual partial inductances(see Fig. 7.5). Hence, we may write the total magnetic ﬂux through loop 2 ( ψ 2 out of the page) due to the current around loop 1, I1, as (use the right-hand rule) ψ2=⎜parenleftbigMp35−Mp37⎜parenrightbigI1−⎜parenleftbigMp15−Mp17⎜parenrightbigI1+⎜parenleftbigMp28−Mp26⎜parenrightbigI1 −⎜parenleftbigMp48−Mp46⎜parenrightbigI1 =⎜parenleftbigMp35−Mp37+Mp17−Mp15+Mp28−Mp26+Mp46−Mp48⎜parenrightbig ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright M12I1 Therefore, the loop mutual inductance between the two loops could be com- puted using partial mutual inductances as M12=Mp35−Mp37+Mp17−Mp15+Mp28−Mp26+Mp46−Mp48 =4.901 nH giving precisely the same value for M12. So the difﬁcult and tedious derivation of the equation for the mutual loop inductance between the two loops, M12, given in (4.110) in Section 4.10.1, could have been derived more easily interms of the prederived mutual partial inductance formulas of Chapter 5. It may appear that since the circuit for the partial inductance method in Fig. 7.14 is more involved than the circuit for the loop inductance method inFig. 7.13, using loop inductances is preferable to using partial inductances.But when examined carefully, this is not the case. Solution of either circuit istrivial using SPICE or the personal computer version, PSPICE. The heart of the solution is the values of the circuit elements of the circuit model! For the loopinductance method one must compute the self inductances for each of the twoloops as well as the mutual inductance between the two loops. The derivationof the equation for the loop self inductance, even a square loop, from theelectromagnetic ﬁeld equations is very involved: (see Section 4.1.1). Next, thederivation of the equation for the mutual inductance between two rectangularloops, even ones that lie in the same plane, from the electromagnetic ﬁeldequations is also extremely involved and tedious: (see Section 4.10.1). Youwill not ﬁnd these equations in handbooks or textbooks and must derive themyourself. Had we not already derived these self and mutual loop inductances FINAL EXAMPLE: USE OF LOOP AND PARTIAL INDUCTANCE 333 for this speciﬁc conﬁguration , you would be required to carry out these detailed derivations from the electromagnetic ﬁeld equations. For every new problemyoumust rederive the formulas for that speciﬁc conﬁguration! On the other hand, calculating the values of the self and mutual partial inductances inthe partial inductance model of Fig. 7.14 is simple! We already derived theformulas for the self and mutual partial inductances of and between segmentsof straight wires: No more derivations need be done for a new conﬁguration.Simply “build” a model of the problem by constructing it with piecewise-linearsegments, compute the self and mutual partial inductances of and between thesegments with the prederived formulas in Chapters 5 and 6, and then simply program PSPICE or any other lumped-circuit analysis tool to perform thecircuit analysis calculations! So by using partial inductances, a circuit designercan build a circuit model without ever having to deal with the complicated electromagnetic ﬁeld equations to derive the values of those elements ! Aside from the very serious requirement in using loop inductances to identify thecomplete current loop , this is the essential beauty in using partial inductances over using loop inductances. APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS Fundamentally, the laws governing the calculation of capacitance and in- ductance are written in terms of vectors of the four electromagnetic ﬁeld vector quantities , which are the electric ﬁeld intensity vector E, the electric ﬂux density vector D, the magnetic ﬁeld intensity vector H, and the magnetic ﬂux density vector B. Therefore, if we are to calculate and understand the no- tions of capacitance and inductance of a physical structure correctly, as wellas use them correctly to construct a lumped-circuit model of that structure, wemust understand some elementary properties of vectors and some elementaryvector calculus ideas. Trying to avoid the use of vector calculus ideas by re-lying on one’s daily “life experiences” to compute and properly interpret themeanings of capacitance and inductance of a structure has caused many ofthe incorrect results and misunderstanding, as well as the numerous erroneousapplications that are seen throughout the literature and in conversations withengineering professionals. We assume that the reader has a rudimentary familiarity with vectors, so this appendix is a review of those important concepts. The reader is re-ferred to other textbooks on electromagnetics listed in the references for moredetails [3–6]. For the computation of inductance, this brief review will besufﬁcient. Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 335 336 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS A.1 VECTORS AND COORDINATE SYSTEMS A vector, as distinguished from a scalar, contains two items of information about a physical quantity: its value and its direction of effect. A vector is shownin the ﬁgures as a line with an arrowhead to show that direction of effect and isdenoted in the text as boldface (e. g. , F). The magnitude or length of a vector is denoted as For asF=|F|. To compute with vectors properly requires a coordinate system. We use primarily the rectangular (Cartesian) coordinatesystem that consists of three axes x,y, and z, as shown in Fig.A.1. These axes are mutually orthogonal. In a rectangular coordinate system, a vector is described as F=Fxax+Fyay+Fzaz (A.1) where the components of Falong (projections of Fonto) the x,y, and zaxes are denoted as Fx,Fy, andFz, respectively, and the unit vectors along the axes are denoted as ax,ay, and az. These unit vectors are of unit length and are directed in the direction of increasing value of the coordinate axis. There are other coordinate systems, such as the cylindrical and spherical coordinate systems described at the end of this appendix. Although a problemcan be solved in any coordinate system, the choice of coordinate system usedto solve the problem will simplify the solution considerably. The rectangular xyz axayazF Fxax FyayFzaz FIGURE A.1. Rectangular coordinate system. VECTORS AND COORDINATE SYSTEMS 337 coordinate system is more suitable for problems whose boundaries ﬁt a rectan- gular shape. The cylindrical coordinate system is more suitable for problemswhose boundaries ﬁt a cylindrical shape, whereas the spherical coordinatesystem is more suitable for problems whose boundaries ﬁt a spherical shape.The unit vectors of a rectangular coordinate system are mutually perpendic-ular at a point. Hence, the rectangular coordinate system is said to be anorthogonal coordinate system . The cylindrical and spherical coordinate sys- tems discussed at the end of this appendix are similarly orthogonal coordinatesystems. Vectors in any orthogonal coordinate system are added or subtractedby adding or subtracting their corresponding components: A±B=(Ax±Bx)ax+⎜parenleftbigAy±By⎜parenrightbigay+(Az±Bz)az (A.2) There are two ways of performing the multiplication of two vectors: the dot product and the cross product. The dot product of two vectors gives the result as a scalar and is deﬁned by [3] A·B=ABcosθAB =AxBx+AyBy+AzBz (A.3) where θABis the angle between the two vectors as illustrated in Fig. A.2(a). In plain terms this gives (1) the product of the length of Aand the projection of Bonto A, or (2) the product of the length of Band the projection of A onto B. The result for the dot product in terms of the vector components in a rectangular coordinate system given in (A.3) is easy to remember: It is thesum of the products of the corresponding components of the two vectors. Thiswill also be the case for the cylindrical and spherical coordinate systems. Twovectors are perpendicular ifA·B=0. Also, the dot product of a vector with itself is its magnitude squared: A·A=|A| 2. B AB AABθ ABθA×Bna AB AB θcos = •B A (a) (b)n AB AB a B A θsin =×AB B θcos FIGURE A.2. Dot and cross product of two vectors. 338 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS Thecross product of two vectors gives a vector and is deﬁned by [3] A×B=ABsinθABan =⎜parenleftbigAyBz−AzBy⎜parenrightbigax+(AzBx−AxBz)ay+⎜parenleftbigAxBy−AyBx⎜parenrightbigaz (A.4) where θABis the angle between the two vectors, as illustrated in Fig. A.2(b). The result gives a vector that is perpendicular to the plane containing Aand B. The unit vector perpendicular to (normal to) this plane containing AandB is denoted as an. Since there are two sides to this plane, which contains Aand B, the direction of the unit normal is determined by the right-hand rule ; that is, if the ﬁngers of our right hand curl from AtoB, the direction of the normal to this plane for A×B will be given by the thumb of our right hand. The reader should practice this since it is used throughout this book. The axes of therectangular coordinate system are assumed to be ordered cyclically accordingto the convention of x→y→z→x→y→z→··· . In other words, if we cross the xaxis into the yaxis, we get the zaxis: a x×ay=az. Note that, for example, ay×ax=− az. The vector result for the cross product in a rectangular coordinate system in terms of the vector components in (A.4) iseasily remembered. Each component is of the form⎜parenleftbigAβBγ−AγBβ⎜parenrightbigaαin the orderα→β→γ→α→β→··· according to the cyclic ordering of the axes. This rule for determining the cross product is the same in the cylindricaland spherical coordinate systems. Two vectors are parallel ifA×B=0. Note thatA·B=B·A and the order in the dot product does not matter. However, the order in the cross product does matter: A×B=−B×A. EXAMPLE Two vectors lying in the yzplane are deﬁned, as shown in Fig.A.3, as A=3ay B=2ay+az The lengths of the two vectors are A=3 andB=⎜radicalBig (2)2+(1)2=√ 5. The dot product is A·B=⎜parenleftbig0ax+3ay+0az⎜parenrightbig·⎜parenleftbig0ax+2ay+az⎜parenrightbig =3×2+0×1 =6 VECTORS AND COORDINATE SYSTEMS 339 z yAB ABθ x FIGURE A.3 From the dot product in (A.3), cos(θAB)=A·B AB =6 3·√ 5 =0.894 Hence, the angle between the two vectors is θAB=cos−1(0.894)=26.57◦. For these simple vectors, we can obtain this angle directly by trigonometry: θAB=tan−11 2 =26.57◦ so we again obtain A·B=AB cosθAB =3×⎜radicalbig 22+12×cos(26.57◦) =6 The cross product is A×B=⎜parenleftbigAyBz−AzBy⎜parenrightbigax+(AzBx−AxBz)ay+⎜parenleftbigAxBy−AyBx⎜parenrightbigaz =(3−0)ax+(0−0)ay+(0−0)az =3ax 340 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS Directly, we obtain the same result: A×B=AB sinθABan =3×⎜radicalbig 22+12×sin(26.57◦)an =3an Since both vectors lie in the yzplane, the unit normal perpendicular to the plane containing AandBis in the ±xdirection. Using the right-hand rule and crossing AtoBgives the unit normal in the positive xdirection: an=ax. A.2 LINE INTEGRAL The fundamental equations governing the electromagnetic ﬁeld vectors (re- ferred to collectively as Maxwell’s equations) involve two basic integrals: theline integral and the surface integral . Hence, it is important that we understand what these mean and how to evaluate them. The vectors in the electromagneticﬁeld equations are functions of the coordinate system variables x,y, and z, which is denoted by F(x, y, z )and hence are said to constitute a ﬁeld. There are two possible types of ﬁelds: a scalar ﬁeld and a vector ﬁeld. An example of ascalar ﬁeld is a plot of the temperature distribution in a room. Lines of constant temperature (a scalar) show the distribution of that ﬁeld in the room.An example of a vector ﬁeld would be the plot of ﬂow rates and directions of the water ﬂow in a river. The directions of these vectors show the direction ofthe water ﬂow at that point, and the lengths of these vectors are proportionalto the rates of ﬂow at that point. Theline integral of a vector ﬁeld is denoted as ⎜integraldisplayb aF(x, y, z )·dl=⎜integraldisplayb aF(x, y, z )cosθd l (A.5) The line integral means that we take the products of the projection of the vector Fonto the path, Fcosθ(alternatively, the component of Ftangent to the path), and the differential lengths, dl, along the path and sum them with an integral from the starting point ato the endpoint b, as illustrated in Fig. A.4. An example of a line integral is the computation of the work required to push an object from one point to another when the force Fis exerted on the object at an angle to the path as shown in Fig.A.5. The work doneisW=⎜integraltextFcosθd x=⎜integraltextF·dl. The line integral is a very sensible result. LINE INTEGRAL 341 ab cdl c Fdl θ FIGURE A.4. Line integral. There are two components of F: One component is parallel to the path and the other component is perpendicular to the path. Only the component parallel to the path should contribute to the result. The actual computation of the line integral in a rectangular coordinate system is very simple. In a rectangular coordinate system a vector differentialpath length is dl=dxa x+dyay+dzaz (A.6) Hence F·dl=Fxdx+Fydy+Fzdz (A.7) and the line integral becomes ⎜integraldisplayb aF·dl=⎜integraldisplayb aFcosθd l =⎜integraldisplayxb xaFxdx+⎜integraldisplayyb yaFydy+⎜integraldisplayzb zaFzdz (A.8) where the path extends from (xa,ya,za)to(xb,yb,zb)and each component ofFis a function of x,y, and z:Fx(x, y, z ),Fy(x, y, z ), andFz(x, y, z ).I f massθF Fcosθ x FIGURE A.5. Line integral in computing work. 342 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS the integral is taken around a closed path, it is denoted with a circle on the integral sign as⎜contintegraltext cF·dlandcrepresents the contour of that closed path. EXAMPLE A vector ﬁeld in the yzplane is given as F(x, y, z )=zay as shown in Fig. A.6. Determine the line integral of Falong a straight-line path between the two points in the yzplane from point aat (0,1,3) topoint b at (0,2,4). Observe that at all points in the yzplane the vector is directed in theydirection. However, its magnitude depends on z: for positive, increasing values of z, its magnitude (length) increases. For znegative, it is pointing in the –y direction. Performing the line integral gives ⎜integraldisplayb aF·dl=⎜integraldisplay0 x=0Fx⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0dx+⎜integraldisplay2 y=1Fy⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright zdy+⎜integraldisplay4 z=3Fz⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0dz =⎜integraldisplay2 y=1zd y z yxy za F= ab (0,1,3)(0,2,4) c FIGURE A.6 SURFACE INTEGRAL 343 =⎜integraldisplay2 y=1(y+2)dy =7 2 and we have substituted the equation of the path, z=y+2. A.3 SURFACE INTEGRAL Thesurface integral is ⎜integraldisplay sF(x, y, z )·ds=⎜integraldisplay sF(x, y, z )·ands =⎜integraldisplay sF(x, y, z )cosθd s (A.9) The surface integral gives the integral of the products of the components of Fthat are perpendicular to the surface s and the differential surface elements dsas shown in Fig. A.7. The unit normal perpendicular to the surface is denoted as an, and the differential surface area is ds=dsan. The surface integral gives the ﬂux of the vector ﬁeld Fthrough the surface s . This is like shining a light through an opening. There are two components of F: One component is parallel to the surface and the other component is perpendicular Fθs dsna FIGURE A.7. Surface integral. 344 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS to the surface. Only the component of the light ﬂux that is perpendicular to the opening contributes to the net light ﬂux passing through that opening. Ifthe surface sis a closed surface, the surface integral is denoted with a circle on the integral sign:⎜contintegraltext sF·ds. Hence, the surface integral in (A.9) is said to give the net ﬂuxof the vector ﬁeld through the surface s. Observe that there is a major difference between the line integral and the sur- face integral. The line integral involves the components of Fthat are parallel to (tangent to) the path, whereas the surface integral involves the componentsofFthat are perpendicular to the surface. The evaluation of the surface integral in a rectangular coordinate system is very simple. The vector differential surface is ds=dy dz a x+dx dz ay+dx dy az (A.10) Note that the components of this are the differential surface areas whose unit normals are perpendicular to them (e. g., dy dz ax). Hence, the surface integral simpliﬁes, in a rectangular coordinate system, to ⎜integraldisplay sF(x, y, z )·ds=⎜integraldisplay sxFxdy dz+⎜integraldisplay syFydx dz+⎜integraldisplay szFzdx dy (A.11) EXAMPLE A wedge-shaped surface lies in the yzplane as shown in Fig. A.8. Determine the ﬂux of the vector ﬁeld F=(x+2)ax z yxs ) 1 , 3 , 0 ( ) 1 , 1 , 0 ((0,1,3) ()x x a F2+ =4+ − =z y FIGURE A.8 DIVERGENCE 345 through the surface. The surface integral becomes ⎜integraldisplay sF(x, y, z )·ds=⎜integraldisplay sxFxdy dz =⎜integraldisplay3 z=1⎜integraldisplayy=−z+4 y=1⎛ ⎝x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0+2⎞⎠ dy dz =⎜integraldisplay3 z=1⎜integraldisplayy=−z+4 y=12dy dz =⎜integraldisplay3 z=1(−2z+6)dz =4 We have substituted x=0 over the surface into Fx=x+2 and the equation of the top part of the wedge, y=−z+4, in the limit of one of the integrals. A.4 DIVERGENCE The line and surface integrals apply over regions of space. The following vector calculus results, the divergence and the curl, are the point forms of these integrals which apply to points in space and are differential relationsthat give the relationships between the ﬁeld vectors at points in space. Thedivergence of a vector ﬁeld gives the net outﬂow orﬂuxof a vector ﬁeld from a point, hence the name divergence , and is deﬁned by ∇·F(x, y, z )=lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1v→0⎜contintegraltext sF·ds /Delta1v(A.12) This is illustrated in Fig. A.9. If we surround a point by a closed surface s that contains a differential volume /Delta1v, compute the net ﬂux of Fout of the closed surface per unit of volume enclosed by s, and then let the surface and enclosed volume shrink to zero, the limit of that is the divergence ofFat that point. Essentially, this gives an indication of any sources of Fthat are located at the point. If the divergence of Fis negative at the point, we say that a sink exists at that point. So the divergence indicates whether there is a net outﬂow ofFat that point. If we puncture an inﬂated ballon, we get a divergence of the air contained in that ballon. 346 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS Fnads s Δvpoint FIGURE A.9. Divergence of a vector ﬁeld. The “del operator,” ∇, is somewhat equivalent to a derivative in scalar calculus and is an “operator” deﬁned by [3] ∇= ax∂ ∂x+ay∂ ∂y+az∂ ∂z(A.13) Using the del operator, we obtain the divergence of a vector ﬁeld in a rectan- gular coordinate system as ∇·F(x, y, z )=∂Fx ∂x+∂Fy ∂y+∂Fz ∂z(A.14) It is very important to observe that the divergence of a vector ﬁeld gives a scalar quantity as the result. EXAMPLE A vector ﬁeld is described by F=xax+yay+zaz as plotted in Fig. A.10. Determine the divergence of the ﬁeld. The divergence of this ﬁeld is ∇·F(x, y, z )=∂Fx ∂x+∂Fy ∂y+∂Fz ∂z =1+1+1=3 Since this result is independent of x,y, and z, there is a net outﬂow of the vector at every point in the space. This is a sensible result since the ﬁeld isconstant over any sphere of radius r=⎜radicalbig x2+y2+z2centered at the origin of the coordinate system and is directed normal to the surface of that sphere.Hence, from the basic deﬁnition of the divergence given in (A.12) we can DIVERGENCE 347 z yx FIGURE A.10 calculate directly ∇·F(x, y, z )=lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1v→0⎜contintegraltext sF·ds /Delta1v =r×4πr2 4/3πr3 =3 A.4.1 Divergence Theorem We can interchange certain surface and volume integrals with the divergence theorem [3]: ⎜contintegraldisplay sF·ds=⎜integraldisplay v(∇·F)dv (A.15) This result provides that if we integrate the divergence of Fthroughout some volume v, we can obtain the same result by performing the surface integral of Fover the closed surface sthat contains the volume v. This is a very sensible 348 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS result if we think about what these quantities mean. According to (A.12), the divergence ∇·Fgives the net outﬂow or ﬂux of Fthroughout the volume /Delta1v per unit of that volume. Rewriting (A.12) gives ⎜contintegraldisplay sF·ds=lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1v→0⎜bracketleftbig∇·F(x, y, z )/Delta1v⎜bracketrightbig =⎜integraldisplay v(∇·F)dv (A.12) Hence, it makes sense that we can obtain the net ﬂux out of the closed surface sthat encloses that volume,⎜contintegraltext sF·ds, by performing the volume integral of ∇·Fthroughout that volume. EXAMPLE Verify the divergence theorem for the vector ﬁeld F=xax+yay+zaz for the square volume whose corners are at (0,0,0), (0,0,1), (0,1,0),(0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1) as illustrated in Fig. A.11. The surfaceintegral over the closed surface sis ⎜contintegraldisplay sF·ds=⎜integraldisplay1 z=0⎜integraldisplay1 y=0Fxdy dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright front−⎜integraldisplay1 z=0⎜integraldisplay1 y=0Fxdy dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright back−⎜integraldisplay1 z=0⎜integraldisplay1 x=0Fydx dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright left +⎜integraldisplay1 z=0⎜integraldisplay1 x=0Fydx dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright right−⎜integraldisplay1 y=0⎜integraldisplay1 x=0Fzdx dy ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright bottom+⎜integraldisplay1 y=0⎜integraldisplay1 x=0Fzdx dy ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright top =⎜integraldisplay1 z=0⎜integraldisplay1 y=0x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 1dy dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright front−⎜integraldisplay1 z=0⎜integraldisplay1 y=0x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0dy dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright back−⎜integraldisplay1 z=0⎜integraldisplay1 x=0y⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0dx dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright left +⎜integraldisplay1 z=0⎜integraldisplay1 x=0y⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 1dx dz ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright right−⎜integraldisplay1 y=0⎜integraldisplay1 x=0z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0dx dy ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright bottom+⎜integraldisplay1 y=0⎜integraldisplay1 x=0z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 1dx dy ⎜bracehtipupleft ⎜bracehtipdownright⎜bracehtipdownleft ⎜bracehtipupright top =1−0−0+1−0+1 =3 DIVERGENCE 349 xyz bottomtop leftback right front z y xz y xa a a F+ + =x =1y = 11=z Fz = z = 1 Fz = z = 0Fx = x = 1Fx = x = 0 Fy = y = 0Fy = y = 1 FIGURE A.11 Notice that the surface integral determines the net ﬂux leaving the closed sur- face. A vector component points into one side and out of the other side.Hence, half the integrals are positive and half the integrals are negative.Observe also that each integrand is 0 or 1 over a surface and the dimen-sions of each side are 1. Therefore, the integral over a side is either 0 or1.Since ∇·F(x, y, z )=∂F x ∂x+∂Fy ∂y+∂Fz ∂z =1+1+1=3 the right-hand side of the divergence theorem in (A.15) also gives the same result: ⎜integraldisplay v(∇·F)dv=⎜integraldisplay1 x=0⎜integraldisplay1 y=0⎜integraldisplay1 z=03dx dy dz =3 350 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS F(x, y, z) paddlewheel () z y x, ,F×∇ FIGURE A.12. Curl (circulation) of a vector ﬁeld. A.5 CURL While the divergence gives the net outﬂow or ﬂux of a vector ﬁeld from a point, the curl of a vector ﬁeld gives the net circulation or rotation of the ﬁeld about a point . For example, consider the vector ﬁeld shown in Fig. A.12. This ﬁeld might represent the ﬂow of the water in a river. If we insert a smallpaddlewheel as shown, the ﬂow pattern will cause the paddlewheel to rotatein the clockwise direction. If we turned the paddlewheel such that its axis wasparallel to the ﬁeld lines, it would not rotate. Figure A.13 shows how we might deﬁne the circulation of a vector ﬁeld in one plane. Deﬁne a ﬂat surface sin that plane and the associated contour cenclosing it. Deﬁne the unit normal to that plane as a n, with its direction c anΔ sF(x,y,z) FIGURE A.13. Deﬁning the curl of a vector ﬁeld. CURL 351 according to the right-hand rule with respect to the direction of caround that surface perimeter. The net circulation at the point per unit of the enclosedsurface area in this plane would be circulation per unit area =a n⎛ ⎝lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1s→0⎜contintegraltext cF·dl /Delta1s⎞ ⎠ (A.16) By performing the line integral of Faround the contour cenclosing the surface /Delta1sand dividing by that surface, we get a measure of the circulation (in this case in the counterclockwise direction). A direction is given to that circulationby the unit vector a nnormal to the surface. The direction of the unit normal is obtained in accordance with the right-hand rule. Since the result is circulationor rotation of the ﬁeld, we should obtain the total circulation or rotation inthree orthogonal planes. The result gives the curl of the vector ﬁeld as ∇×F(x, y, z )=ax⎛ ⎜⎝lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1syz→0⎜contintegraltext cyzF·dl /Delta1syz⎞ ⎟⎠+ay⎛ ⎝lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1sxz→0⎜contintegraltext cxzF·dl /Delta1sxz⎞ ⎠ +az⎛ ⎜⎝lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1sxy→0⎜contintegraltext cxyF·dl /Delta1sxy⎞ ⎟⎠ (A.17) where /Delta1sxy, for example, is a ﬂat surface in the xy plane which is perpendicular toaz, andcxydenotes the contour around the enclosed surface /Delta1sxy. Applying the del operator that is deﬁned in (A.13) gives a mechanical way of determining the curl in a rectangular coordinate system [3]: ∇×F(x, y, z )=⎜parenleftbigg∂Fz ∂y−∂Fy ∂z⎜parenrightbigg ax+⎜parenleftbigg∂Fx ∂z−∂Fz ∂x⎜parenrightbigg ay +⎜parenleftbigg∂Fy ∂x−∂Fx ∂y⎜parenrightbigg az (A.18) Observe that each of these components can be remembered easily using the cyclic rule for the cross product, the cyclic ordering of the three axes, andthe deﬁnition of the del operator given in (A.13). For example, each compo-nent of the curl is of the form ⎜parenleftbig∂Fγ/∂β−∂Fβ/∂γ⎜parenrightbigaα, where the ordering is α→β→γ→α···. 352 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS EXAMPLE Determine the curl of the vector ﬁeld F=zay that is illustrated in Fig. A.14. First we see clearly that there will be circulation and the rotation will be clockwise with the unit normal being in the negativexdirection. Substituting into (A.18) yields ∇×F(x, y, z )=⎛ ⎜⎜⎜⎝∂Fz ∂y⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0−∂Fy ∂z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 1⎞ ⎟⎟⎟⎠ax+⎛ ⎜⎜⎜⎝∂Fx ∂z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0−∂Fz ∂x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0⎞ ⎟⎟⎟⎠ay +⎛ ⎜⎜⎜⎝∂Fy ∂x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0−∂Fx ∂y⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0⎞ ⎟⎟⎟⎠az =−ax as expected. z yxyza F= FIGURE A.14 CURL 353 A.5.1 Stokes’s Theorem Similar to the divergence theorem, Stokes’s theorem allows us to interchange a surface integral and a line integral [3]: ⎜contintegraldisplay cF·dl=⎜integraldisplay s(∇×F)·ds (A.19) Stokes’s theorem provides that the surface integral of the curl of Fover an open surface swill give the same result as performing the line integral of F around the contour cthat encloses that open surface. As was the case for the divergence theorem, Stokes’s theorem is a very sensible result. According to(A.17), the curl of a vector ﬁeld, ∇×F, gives the net circulation or rotation of a ﬁeld around a contour that encloses a differential surface per unit of thatenclosed surface. Rewriting the xcomponent of (A.17) gives ⎜contintegraldisplay cyzF·dl=lim⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright /Delta1syz→0⎜braceleftbig⎜bracketleftbig∇×F(x, y, z )⎜bracketrightbig x/Delta1syx⎜bracerightbig =⎜integraldisplay syz(∇×F)·ds Hence, it makes sense that by integrating the curl over the surface with a surface integral we will obtain the same result as the line integral around thecontour enclosing that surface would give. EXAMPLE Verify Stokes’s theorem for the vector ﬁeld F=zay and the closed contour cand its enclosed surface sshown in Fig. A.15. The curl of Fis ∇×F=⎛ ⎜⎜⎜⎝∂Fz ∂y⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0−∂Fy ∂z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 1⎞ ⎟⎟⎟⎠ax+⎛ ⎜⎜⎜⎝∂Fx ∂z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0−∂Fz ∂x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0⎞ ⎟⎟⎟⎠ay+⎛ ⎜⎜⎜⎝∂Fy ∂x⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0−∂Fx ∂y⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 0⎞ ⎟⎟⎟⎠az =−ax 354 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS z yxs ) 1 , 3 , 0 ( ) 1 , 1 , 0 ((0,1,3) y za F=4+ − =z y 1=z1=y 1c2c 3c FIGURE A.15 Hence, the right-hand side of Stokes’s theorem is ⎜integraldisplay s(∇×F)·ds=⎜integraldisplay3 z=1⎜integraldisplayy=−z+4 y=1(−1)ax⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ∇×F·(axdy dz )⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright ds =⎜integraldisplay3 z=1⎜integraldisplayy=−z+4 y=1(−1)dy dz =−2 Since F·dl=Fydy=zd y, the left-hand side of Stokes’s theorem is ⎜contintegraldisplay cF·dl=⎜integraldisplay3 y=1Fydy ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright c1+⎜integraldisplay1 y=3Fydy ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright c2+⎜integraldisplay1 y=1Fydy ⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright c3 =⎜integraldisplay3 y=1z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright 1dy+⎜integraldisplay1 y=3z⎜bracehtipupleft⎜bracehtipdownright⎜bracehtipdownleft⎜bracehtipupright −y+4dy+⎜integraldisplay1 y=1zd y =2−4+0 =−2 which is the same. A.6 GRADIENT OF A SCALAR FIELD Perhaps one of the best illustrations of the use of the gradient is a topographi- cal map. Contours of constant elevation (above sea level) are shown as closed GRADIENT OF A SCALAR FIELD 355 contours. We might denote this as the scalar ﬁeld EL (x, y, z ). Think of this scalar function as depicting a three-dimensional map with the xandycoor- dinates giving the horizontal position over the Earth’s surface, and the zaxis giving the elevation of each point above sea level. The closer the contours ofconstant elevation are to each other, the steeper the slope (i. e., the greaterthe change in elevation with a change in horizontal distance). If we wanted tochart a course for hiking that would avoid the steep slopes, we would choosea path between points on adjacent contours of constant elevation with thosecontours being as widely separated as possible. In doing so, we would makethe vertical distance we move as long a horizontal distance as possible. Also,to make the trip as expeditious as possible we would choose a route that isperpendicular to those contours. Denote some general scalar ﬁeld as f(x, y, z ). A differential change in the function (the scalar ﬁeld) as we move between contours of constant value offis df=∂f(x, y, z ) ∂xdx+∂f(x, y, z ) ∂ydy+∂f(x, y, z ) ∂zdz (A.20) Using the del operator in (A.13): ∇= ax∂ ∂x+ay∂ ∂y+az∂ ∂z(A.13) we deﬁne the gradient offas ∇f=∂f(x, y, z ) ∂xax+∂f(x, y, z ) ∂yay+∂f(x, y, z ) ∂zaz (A.21) Note that the gradient of a scalar ﬁeld f(x, y, z ),∇f(x, y, z ), gives a vector as the result. Recalling the vector differential path length in (A.6), dl=dxax+dyay+dzaz (A.6) 356 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS we can write (A.20) in terms of the gradient as df=∇f·dl (A.22) which you should verify. Now we interpret the meaning of the gradient. The differential change in (A.22) is df=∇f·dl =|∇f|dlcosθ (A.22) where θis the angle between the gradient vector, ∇f, and the differential path length vector, dl. The rate of change of the scalar ﬁeld along this path is df dl=|∇f|cosθ (A.23) If we want to move in the direction of the maximum rate of change of the scalar ﬁeld (i. e., perpendicular to the contours of constant f), the path taken must be perpendicular to the gradient vector (i. e., θ=90◦): df dl⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle max=|∇f| (A.24) Therefore, the gradient vector gives both the direction and the magnitude of the maximum space rate of change of the scalar ﬁeld. EXAMPLE Show that the gradient of the scalar ﬁeld f(x, y, z )=x+yis normal to the lines of constant f. The scalar ﬁeld is plotted in Fig. A.16. The gradient is ∇f=∂f ∂xax+∂f ∂yay =ax+ay which is plotted in Fig. A.16. Obviously, the gradient is perpendicular to the lines of constant f, and it also points in the direction of the maximum rate of change of f. IMPORTANT VECTOR IDENTITIES 357 xy y = 1y = 2 x = 1 x = 21=f2=f 0=fy x f a a+ = ∇ FIGURE A.16 A.7 IMPORTANT VECTOR IDENTITIES An important vector identity that will prove very useful in deﬁning the concept of partial inductance is ∇·(∇×F)=0 (A.25) Note that it would make no sense to write ∇×(∇·F)because the divergence ∇·Fgives a scalar and we cannot take the curl of a scalar. With our under- standing of the meaning of curl and divergence, this identity is sensible. Thecurl of a vector ﬁeld, ∇×F, gives the net circulation orrotation of the ﬁeld, whereas the divergence of a ﬁeld, ∇·F, gives the net outﬂow orﬂuxof the ﬁeld from a point. We have two situations to consider: (1) If the vector ﬁeldhas circulation at a point, ∇×F/=0, it can have no divergence (net outﬂow of the ﬁeld) at that point and (A.25) is satisﬁed; (2) on the other hand, if thevector ﬁeld has no circulation at a point, ∇×F=0, the divergence of this is zero. A simple way to prove this important identity is to carry out the operation in a rectangular coordinate system using symbols. For example, ∇×F= ⎜parenleftbigg∂Fz ∂y−∂Fy ∂z⎜parenrightbigg ax+⎜parenleftbigg∂Fx ∂z−∂Fz ∂x⎜parenrightbigg ay+⎜parenleftbigg∂Fy ∂x−∂Fx ∂y⎜parenrightbigg az 358 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS If we next take the divergence of this result, we obtain ∇·(∇×F)=∂ ∂x⎜parenleftbigg∂Fz ∂y−∂Fy ∂z⎜parenrightbigg +∂ ∂y⎜parenleftbigg∂Fx ∂z−∂Fz ∂x⎜parenrightbigg +∂ ∂z⎜parenleftbigg∂Fy ∂x−∂Fx ∂y⎜parenrightbigg =∂2Fz ∂x∂y−∂2Fy ∂x∂z+∂2Fx ∂y∂z−∂2Fz ∂y∂x+∂2Fy ∂z∂x−∂2Fx ∂z∂y =0 Another useful vector identity is that the curl of the gradient of a scalar ﬁeld is zero: ∇×∇f(x, y, z )=0 (A.26) Integrating this over some open surface sand using Stokes’s theorem on the result gives ⎜integraldisplay s⎜bracketleftbig∇×∇f⎜bracketrightbig·ds=⎜contintegraldisplay c(∇f)·dl =⎜contintegraldisplay c∂f ∂xdx+∂f ∂ydy+∂f ∂zdz =⎜contintegraldisplay cdf =0 The result is due to integrating dfaround a closed path . This identity can be directly proven by carrying out the operations in (A.26) symbolically in arectangular coordinate system: ∇×∇f=∇ ×⎜parenleftbigg∂f ∂xax+∂f ∂yay+∂f ∂zaz⎜parenrightbigg =⎜parenleftbigg∂ ∂y∂f ∂z−∂ ∂z∂f ∂y⎜parenrightbigg ax+⎜parenleftbigg∂ ∂z∂f ∂x−∂ ∂x∂f ∂z⎜parenrightbigg ay +⎜parenleftbigg∂ ∂x∂f ∂y−∂ ∂y∂f ∂x⎜parenrightbigg az =0 A.8 CYLINDRICAL COORDINATE SYSTEM A point in a cylindrical coordinate system is deﬁned by the three variables r, φ, andz, as illustrated in Fig. A.17. The coordinate ris the radial distance of the CYLINDRICAL COORDINATE SYSTEM 359 xyz φr z φa raza FIGURE A.17. Cylindrical coordinate system. point from the zaxis (parallel to the xyplane), the coordinate φis the angular displacement (in radians with 0 ≤φ≤360◦) of the projection of the point on the xyplane measured counterclockwise from the positive xaxis, and the coordinate zis the distance of the projection of the point along the zaxis. The corresponding three unit vectors ar,aφ,andazare directed in the direction of increasingvalue of the variable and are mutually perpendicular. Hence, thecylindrical coordinate system, like the rectangular coordinate system, is anorthogonal coordinate system . A vector in cylindrical coordinates is again described in terms of its unit vectors as A=Arar+Aφaφ+Azaz (A.27) Two vectors are again added or subtracted by adding or subtracting their corresponding components: A±B=(Ar±Br)ar+⎜parenleftbigAφ±Bφ⎜parenrightbigaφ+(Az±Bz)az(A.28) 360 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS The dot product of two vectors is, again, the sum of the products of the corresponding components: A·B=ABcosθAB =ArBr+AφBφ+AzBz (A.29) The cross product of two vectors is, again, A×B=ABsinθABan =⎜parenleftbigAφBz−AzBφ⎜parenrightbigar+(AzBr−ArBz)aφ+⎜parenleftbigArBφ−AφBr⎜parenrightbigaz (A.30) Note that the coordinates are ordered r→φ→z→r→φ→z→r→ ···such that ar×aφ=az. Note that aφ×ar=−azandar×az=−aφ. The vector result for the cross product in a cylindrical coordinate system in termsof the vector components is, again, easily remembered. Each component isof the form ⎜parenleftbigAβBγ−AγBβ⎜parenrightbigaαin the order α→β→γ→α→β→··· according to the cyclic ordering of the coordinates r→φ→z→r→φ→ z→r→··· . The algebra results above are the same as for the rectangular coordinate system. However, the vector calculus results will be different from those fora rectangular coordinate system since one of the variables of the cylindricalcoordinate system, φ, does not have the dimensions of distance. Differential changes in the coordinates give differential arc lengths dr, r dφ, anddz,a s illustrated in Fig. A.18. Note that the φvariable is the only one of the three whose units are not a length. (The units of φareradians . )For a differential change in φ,dφ, the corresponding change in arc length for a radius of ris rsindφ∼=rd φ using the small-angle approximation for the sine. Hence, a vector differential arc length is dl=dra r+rd φaφ+dzaz (A.31) and the line integral is ⎜integraldisplayb aF(r, φ, z )·dl=⎜integraldisplayrb raFrdr+⎜integraldisplayφb φaFφrd φ+⎜integraldisplayzb zaFzdz (A.32) A vector differential surface is ds=(rd φd z )ar+(dr dz )aφ+(d rrd φ )az (A.33) Each of these components is formed by the products of the two sides of each differential surface in Fig. A.18 that is perpendicular to the unit vector for CYLINDRICAL COORDINATE SYSTEM 361 xyz φz rdz drφdφd r φa raza FIGURE A.18. Differential elements in a cylindrical coordinate system. that side. For example, the side perpendicular to arhas sides of length dzand rd φ, while the side perpendicular to aφhas sides of length dzanddr. Hence, the surface integral is ⎜integraldisplay sF(r, φ , z )·ds=⎜integraldisplay srFrrd φd z +⎜integraldisplay sφFφdrdz+⎜integraldisplay szFzd rrd φ (A.34) The divergence and the curl are a bit more complicated than for the rectan- gular coordinate system. The derivations of these are given in reference [3,6]and become ∇·F(r, φ, z )=1 r∂(rFr) ∂r+1 r∂Fφ ∂φ+∂Fz ∂z(A.35) ∇×F(r, φ, z )=⎜parenleftbigg1 r∂Fz ∂φ−∂Fφ ∂z⎜parenrightbigg ar+⎜parenleftbigg∂Fr ∂z−∂Fz ∂r⎜parenrightbigg aφ +⎜bracketleftBigg 1 r∂⎜parenleftbigrFφ⎜parenrightbig ∂r−1 r∂Fr ∂φ⎜bracketrightBigg az(A.36) 362 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS A.9 SPHERICAL COORDINATE SYSTEM A point in a spherical coordinate system is deﬁned by the three variables r, θ, andφ, as illustrated in Fig. A.19. The coordinate ris the radial distance of the point from the origin of the coordinate system , the coordinate θ(inradians with 0 ≤θ≤180◦) is the angular displacement from the positive zaxis, and the coordinate φis the angular displacement (in radians with 0 ≤φ≤360◦) of the projection of the point on the xyplane measured counterclockwise from the positive xaxis. Note that the rin a spherical coordinate system is different from the rin a cylindrical coordinate system. The corresponding three unit vectors ar,aθ,andaφare directed in the direction of increasing value of the variable and are mutually perpendicular. Hence, the sphericalcoordinate system, like the rectangular and cylindrical coordinate systems, isanorthogonal coordinate system . Some textbooks denote the radius dimension in a cylindrical coordinate system as ρinstead of rto distinguish it from the radius rin a spherical coordinate system. It is usually rather simple to distinguish between the two.Ifris the distance perpendicular to the zaxis and parallel to the xyplane, this is the cylindrical coordinate system. If ris the distance from the origin of the coordinate system, this is the spherical coordinate system. xyz φrθar aθaφ FIGURE A.19. Spherical coordinate system. SPHERICAL COORDINATE SYSTEM 363 A vector in spherical coordinates is again described in terms of its unit vectors as A=Arar+Aθaθ+Aφaφ (A.37) Two vectors are again added or subtracted by adding or subtracting their corresponding components: A±B=(Ar±Br)ar+(Aθ±Bθ)aθ+⎜parenleftbigAφ±Bφ⎜parenrightbigaφ(A.38) The dot product of two vectors is, again, the sum of the products of the corresponding components: A·B=ABcosθAB =ArBr+AθBθ+AφBφ (A.39) The cross product of two vectors is, again, A×B=ABsinθABan =⎜parenleftbigAθBφ−AφBθ⎜parenrightbigar+⎜parenleftbigAφBr−ArBφ⎜parenrightbigaθ+(ArBθ−AθBr)aφ (A.40) Note that the coordinates are ordered r→θ→φ→r→θ→φ→r→ ···such that ar×aθ=aφ. Note that aθ×ar=−aφandar×aφ=−aθ. The vector result for the cross product in a spherical coordinate system in termsof the vector components is, again, easily remembered. Each component isof the form ⎜parenleftbigAβBγ−AγBβ⎜parenrightbigaαin the order α→β→γ→α→β→··· , according to the cyclic ordering of the coordinates r→θ→φ→r→θ→ φ→r→··· . The algebra results above are the same as for the rectangular and the cylin- drical coordinate systems. However, the vector calculus results will be differ-ent from those for a rectangular coordinate system since two of the variablesof the spherical coordinate system, θandφ, do not have the dimensions of distance. Differential changes in the coordinates give differential arc lengthsdr, r dθ, andrsinθd φ, as illustrated in Fig.A.20. Hence, a vector differential arc length is dl=dra r+rd θaθ+rsinθd φ aφ (A.41) 364 APPENDIX FUNDAMENTAL CONCEPTS OF VECTORS xyz φr θar aθaφ dφdθdr r dφ rsin θ dφ r dφ FIGURE A.20. Differential elements in a spherical coordinate system. and the line integral is ⎜integraldisplayb aF(r, θ, φ )·dl=⎜integraldisplayrb raFrdr+⎜integraldisplayθb θaFθrd θ+⎜integraldisplayφb φaFφrsinθd φ (A.42) A vector differential surface is ds=(rd θ rsin θd φ)ar+(dr rsinθd φ)aθ+(d rrd θ )aφ(A.43) Each of these components is formed by the products of the two sides of each differential surface in Fig. A.20 that is perpendicular to the unit vector forthat side. For example, the side perpendicular to a rhas sides of length rd θ andrsinθd φ, while the side perpendicular to aθhas sides of length drand rsinθd φ. Hence the surface integral is ⎜integraldisplay sF(r, θ, φ )·ds=⎜integraldisplay srFrrd θrsin θd φ +⎜integraldisplay sθFθdr rsinθd φ+⎜integraldisplay sφFφdr rdθ (A.44) SPHERICAL COORDINATE SYSTEM 365 The divergence and the curl are again a bit more complicated. The deriva- tions of these are given in references [3,6] and become ∇·F(r, θ, φ )=1 r2∂⎜parenleftbigr2Fr⎜parenrightbig ∂r+1 rsinθ∂(sinθFθ) ∂θ+1 rsinθ∂Fφ ∂φ(A.45) ∇× F(r, θ, φ )=1 rsinθ⎜bracketleftBigg ∂⎜parenleftbigFφsinθ⎜parenrightbig ∂θ−∂Fθ ∂φ⎜bracketrightBigg ar +1 r⎜bracketleftBigg 1 sinθ∂Fr ∂φ−∂⎜parenleftbigrFφ⎜parenrightbig ∂r⎜bracketrightBigg aθ+1 r⎜bracketleftbigg∂(rFθ) ∂r−∂Fr ∂θ⎜bracketrightbigg aφ (A.46) TABLE OF IDENTITIES, DERIV ATIVES, AND INTEGRALS USED IN THIS BOOK Identities (1)lna+√ x2+a2 −b+√ x2+b2=sinh−1a x−sinh−1−b x =sinh−1a x+sinh−1b x(5.24) (2) ln⎜parenleftbig x+√ x2+1⎜parenrightbig =− ln⎜parenleftbig −x+√ x2+1⎜parenrightbig (3)sinh−1x≡ln⎜parenleftbig x+√ x2+1⎜parenrightbig =− sinh−1(−x) (D700.1) (4) K(k)=⎜integraldisplayπ/2 ζ=0dζ⎜radicalbig 1−k2sin2ζ(D773.1) (5) E(k)=⎜integraldisplayπ/2 ζ=0⎜radicalbig 1−k2sin2ζd ζ (D774.1) (6) tan−1θ1±tan−1θ2=tan−1θ1±θ2 1∓θ1θ2θ1,θ2≥0 (7) tan−1(x+y)+tan−1(x−y)=tan−12x 1−x2+y2 (8) tan−1(x+y)−tan−1(x−y)=tan−12y 1+x2−y2 Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 367 368 TABLE OF IDENTITIES, DERIV ATIVES, AND INTEGRALS USED IN THIS BOOK (9)lnb a=ln⎜parenleftBigw a+1⎜parenrightBig ∼=w aw/lessmucha (D601) (10)⎜integraldisplayπ θ=0⎜radicalbig 1−k2cos2θd θ=2⎜integraldisplayπ/2 θ=0⎜radicalbig 1−k2cos2θd θ =2⎜integraldisplayπ/2 θ=0⎜radicalbig 1−k2sin2θd θ (11)⎜integraldisplayπ θ=01√ 1−k2cosθdθ=2⎜integraldisplayπ/2 θ=01√ 1−k2cosθdθ =2⎜integraldisplayπ/2 θ=01√ 1−k2sinθdθ (12) (1−x)−1/2∼=1+1 2x+··· (D1) (13)sinh−1x=ln⎜parenleftbig x+√ x2+1⎜parenrightbig =− sinh−1(−x) =− ln⎜parenleftbig −x+√ x2+1⎜parenrightbig(D700.1) (14) sinh−1x a=− sinh−1⎜parenleftBig −x a⎜parenrightBig =ln⎜bracketleftBigg x a+⎜radicalbigg⎜parenleftBigx a⎜parenrightBig2 +1⎜bracketrightBigg =ln⎜parenleftbig x+√ x2+a2⎜parenrightbig −lna(D700.1) (15)ln⎡ ⎣l d+⎜radicalBigg⎜parenleftbigg l d⎜parenrightbigg2 +1⎤ ⎦=ln2l d+1 4⎜parenleftbigg d l⎜parenrightbigg2 −3 32⎜parenleftbigg d l⎜parenrightbigg4 +···l d>1 =l d−1 6⎜parenleftbigg l d⎜parenrightbigg3 +3 40⎜parenleftbigg l d⎜parenrightbigg5 −···l d<1(D602.1) TABLE OF IDENTITIES, DERIV ATIVES, AND INTEGRALS USED IN THIS BOOK 369 (16)⎜radicalBigg 1+⎜parenleftbigg d l⎜parenrightbigg2 =1+1 2⎜parenleftbigg d l⎜parenrightbigg2 −1 8⎜parenleftbigg d l⎜parenrightbigg4 +1 16⎜parenleftbigg d l⎜parenrightbigg6 −···d l≤1 =d l⎜radicalBigg⎜parenleftbigg l d⎜parenrightbigg2 +1 =d l+1 2⎜parenleftbigg l d⎜parenrightbigg −1 8⎜parenleftbigg l d⎜parenrightbigg3 +1 16⎜parenleftbigg l d⎜parenrightbigg5 −···l d≤1(D5.3) (17) ln⎜parenleftbig x+√ x2+a2⎜parenrightbig −ln⎜parenleftBig −y+⎜radicalbig y2+a2⎜parenrightBig =ln⎜parenleftBig y+⎜radicalbig y2+a2⎜parenrightBig −ln⎜parenleftbig −x+√ x2+a2⎜parenrightbig (18)lna+√ x2+a2 −b+√ x2+b2=sinh−1a x−sinh−1⎜parenleftbigg −b x⎜parenrightbigg =sinh−1a x+sinh−1b x (19) tanh−1x=1 2ln1+x 1−xx2<1 (D702) Derivatives (1)d drln⎜bracketleftBigg a r+⎜radicalbigg⎜parenleftBiga r⎜parenrightBig2 +1⎜bracketrightBigg =−a r√ a2+r2 (2) d dxsinh−1a x=d dxcsc h−1x a =−a |x|√ x2+a2 (D728.8) (3)d drln⎜parenleftBig a+⎜radicalbig a2+r2⎜parenrightBig =−a r√ a2+r2+1 r (4)∂ ∂utan−1u=1 1+u2(D512.4) (5)d⎜parenleftBigu v⎜parenrightBig dx=vdu dx−udv dx v2(D65) 370 TABLE OF IDENTITIES, DERIV ATIVES, AND INTEGRALS USED IN THIS BOOK Integrals (1)∂ ∂y⎜integraldisplayb af(x, y)dx=⎜integraldisplayb a∂f(x, y) ∂ydx (D69.3) (2)⎜integraldisplay1⎜parenleftbig a2+x2⎜parenrightbig3/2dx=x a2√ a2+x2(D200.3) (3)⎜integraldisplay1 a2+x2dx=1 atan−1x a(D120.1) (4)⎜integraldisplaydx⎜parenleftbig ax2+b⎜parenrightbig⎜radicalbig fx2+g=1√ b√ag−bftan−1x√ag−bf√ b⎜radicalbig fx2+g(D387) (5)⎜integraldisplayπ 0(a−bcosx)dx a2+b2−2abcosx=⎜braceleftBiggπ aa>b>0 0 b>a>0(D859.124) (6)⎜integraldisplay1√ x2+a2dx=ln⎜parenleftBig x+⎜radicalbig x2+a2⎜parenrightBig (D200.01) (7)⎜integraldisplay ln⎜parenleftbig x2+a2⎜parenrightbig dx=xln⎜parenleftbig x2+a2⎜parenrightbig −2x+2atan−1x a(D623) (8)⎜integraldisplayx a2+x2dx=1 2ln⎜parenleftbig a2+x2⎜parenrightbig (D121.1) (9)⎜integraldisplaydx x√ x2+a2=−1 aln⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsinglea+√ x2+a2 x⎜vextendsingle⎜vextendsingle⎜vextendsingle⎜vextendsingle(D221.01) (10)⎜integraldisplay sinh−1x adx=xsinh−1x a−⎜radicalbig x2+a2a>0 (D730) (11)⎜integraldisplaydx⎜parenleftbig ax2+bx+c⎜parenrightbig3/2=4ax+2b⎜parenleftbig 4ac−b2⎜parenrightbig⎜parenleftbig ax2+bx+c⎜parenrightbig1/2(D380.003) (12)⎜integraldisplayxd x⎜parenleftbig ax2+bx+c⎜parenrightbig3/2=−2bx+4c⎜parenleftbig 4ac−b2⎜parenrightbig⎜parenleftbig ax2+bx+c⎜parenrightbig1/2(D380.013) (13)⎜integraldisplay√ x2+a2 xdx=⎜radicalbig x2+a2−alna+√ x2+a2 x(D241.01) (14)⎜integraldisplayx√ x2+a2dx=⎜radicalbig x2+a2 (D201.01) (15)⎜integraldisplay lnax dx=xlnax−x (D610.01) TABLE OF IDENTITIES, DERIV ATIVES, AND INTEGRALS USED IN THIS BOOK 371 (16) ln⎡ ⎣l rw+⎜radicalBigg⎜parenleftbigg l rw⎜parenrightbigg2 +1⎤ ⎦=ln2l rw+1 4⎜parenleftBigrw l⎜parenrightBig2 −3 32⎜parenleftBigrw l⎜parenrightBig4 +···l rw/greatermuch1 (D602.1) (17)⎜integraldisplaydx√ x2+bx+c=ln⎜parenleftBig 2⎜radicalbig x2+bx+c+2x+b⎜parenrightBig (D380.001) (18)⎜integraldisplay ln⎜parenleftBig a+⎜radicalbig a2+x2⎜parenrightBig dx=xln⎜parenleftBig a+⎜radicalbig a2+x2⎜parenrightBig −x+aln⎜parenleftbig x+√ a2+x2⎜parenrightbig (D740) (19)⎜integraldisplay⎜radicalbig a2+x2dx=a2 2ln⎜parenleftBig x+⎜radicalbig a2+x2⎜parenrightBig +x 2⎜radicalbig a2+x2 (D230.01) (20)⎜integraldisplay xln⎜parenleftBig a+⎜radicalbig a2+x2⎜parenrightBig dx=−x2 4+a 2⎜radicalbig a2+x2 +x2 2ln⎜parenleftBig a+⎜radicalbig a2+x2⎜parenrightBig (D740.1) (21)⎜integraldisplay ln⎜parenleftBig x+⎜radicalbig a2+x2⎜parenrightBig dx=xln⎜parenleftBig x+⎜radicalbig a2+x2⎜parenrightBig −⎜radicalbig a2+x2 (D625) (22)⎜integraldisplay x⎜radicalbig a2+x2dx=1 3⎜parenleftbig a2+x2⎜parenrightbig3/2(D231.01) (23)⎜integraldisplay2π x=0ln⎜parenleftbig 1+b2−2bcosx⎜parenrightbig dx=⎜braceleftbigg 4πlnbb > 1 0 b<1(D865.73) (24)⎜integraldisplay xlnxd x=x2 2lnx−x2 4(D610.1) (25)⎜integraldisplay xln⎜parenleftbig x2+a2⎜parenrightbig dx=1 2⎜parenleftbig x2+a2⎜parenrightbig ln⎜parenleftbig x2+a2⎜parenrightbig −1 2x2(D623.1) (26)⎜integraldisplay tan−1x adx=xtan−1x a−a 2ln⎜parenleftbig a2+x2⎜parenrightbig (D525) (27)⎜integraldisplay x2ln⎜parenleftbig a2+x2⎜parenrightbig dx=x3 3ln⎜parenleftbig a2+x2⎜parenrightbig −2 9x3+2 3xa2−2 3a3tan−1x a(D623.2) (28)⎜integraldisplay xtan−1a xdx=ax 2+x2+a2 2tan−1a x(D528.1) 372 TABLE OF IDENTITIES, DERIV ATIVES, AND INTEGRALS USED IN THIS BOOK (29)⎜integraldisplay x3ln⎜parenleftbig x2+a2⎜parenrightbig dx=x4−a4 4ln⎜parenleftbig x2+a2⎜parenrightbig −x4 8+x2a2 4(D623.3) (30)⎜integraldisplay tan−1a xdx=xtan−1a x+a 2ln⎜parenleftbig x2+a2⎜parenrightbig (D528) (31)⎜integraldisplay x2tan−1a xdx=x3 3tan−1a x+ax2 6−a3 6ln⎜parenleftbig x2+a2⎜parenrightbig (D528.2) (32)⎜integraldisplay lnxd x=xlnx−x (D610) REFERENCES AND FURTHER READINGS [1] C.R. Paul, Analysis of Linear Circuits, McGraw-Hill, New York, 1989. [2] C.R. Paul, Fundamentals of Electric Circuit Analysis, Wiley, New York, 2001. [3] C.R. Paul and S.A. Nasar, Introduction to Electromagnetic Fields , McGraw-Hill, New York, second edition, 1987, and third edition, 1998. [4] C.R. Paul, Electromagnetics for Engineers: With Applications to Digital Systems and Electromagnetic Compatibility, Wiley, Hoboken, NJ, 2004. [5] C.R. Paul, Introduction to Electromagnetic Compatibility , second edition, Wiley- Interscience, Hoboken, NJ, 2006. [6] C.T.A. Johnk, Engineering Electromagnetic Fields and Waves , second edition, Wiley, New York, 1988. [7] H.B. Dwight, Tables of Integrals and Other Mathematical Data , fourth edition, Macmillan, New York, 1961. [8] C.R. Paul, Analysis of Multiconductor Transmission Lines , second edition, Wiley-Interscience, Hoboken, NJ, 2008. [9] W.B. Boast, Vector Fields, Warren B. Boast, Ames, Iowa, 1964. [10] W.R. Smythe, Static and Dynamic Electricity, third edition, revised printing, Hemisphere Publishing Company, New York, 1989. [11] E. Weber, Electromagnetic Fields, Theory and Applications: V ol. I, Mapping of Fields, Wiley, New York, 1950. [12] W. Kaplan, Advanced Calculus, Addison-Wesley, Reading, MA, 1952. Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 373 374 REFERENCES AND FURTHER READINGS [13] R.M. Fano, L.J. Chu, and R.B. Adler, Electromagnetic Fields, Energy, and Forces, Wiley, New York, 1960, second printing 1963. [14] F.W. Grover, Inductance Calculations, Dover Publications (Instrument Society of America), New York, 1973, (ﬁrst published in 1946). [15] A.E. Ruehli, “Inductance Calculations in a Complex Integrated Circuit Envi- ronment,” IBM Journal of Research and Development , pp. 470–481, September 1972. [16] C. Hoer and C. Love, “Exact Inductance Equations for Rectangular Conductors with Applications to More Complicated Geometries,” Journal of Research of the National Bureau of Standards: C. Engineering and Instrumentation , vol. 69C, no. 2, pp. 127–137, April–June 1965. [17] G.A. Campbell, “Mutual Inductance of Circuits Composed of Straight Wires,” Physical Review, vol. 5, pp. 452–458, June 1915. [18] C.L. Holloway and E.F. Kuester, “Dc Internal Inductance for a Conductor of Rectangular Cross Section,” IEEE Transactions on Electromagnetic Compati- bility, vol. 51, no. 2, pp. 338–344, May 2009. [19] G. Antonini, A. Orlandi, and C.R. Paul, “Internal Impedance of Conductors of Rectangular Cross Section,” IEEE Transactions on Microwave Theory and Techniques, vol. 47, no. 7, pp. 979–985, July 1999. [20] E.B. Rosa and F.W. Grover, “Formulas and Tables for the Calculation of Mutual and Self Inductances (Revised),” Bulletin of the National Bureau of Standards , vol. 8, no. 1, 169, pp. 1–231, 1912. [21] P.L. Kalantarov and L.A. Tseitlin, Raschet Induktivnostie, Energoatomizdat, Leningrad, Russia, 1986. [22] P. Silvester, Modern Electromagnetic Fields , Prentice-Hall, Englewood Cliffs, NJ, 1968. [23] J.C. Maxwell, A Treatise on Electricity and Magnetism , V ol. II, second edition, Clarendon Press, Oxford, 1881. [24] W.D. Stevenson, Jr., Elements of Power System Analysis , second edition, McGraw-Hill, New York, 1962. [25] E.B. Rosa, “Calculation of the Self-Inductance of Single-Layer Coils,” Bul- letin of the National Bureau of Standards , vol. 2, no. 2, 31, pp. 161–187, 1906. [26] E.B. Rosa and L. Cohen, “On the Self-Inductance of Circles,” Bulletin of the National Bureau of Standards, vol. 4, no. 1, 75, pp. 149–159, 1907–1908. [27] E.B. Rosa, “The Self and Mutual Inductance of Linear Conductors,” Bulletin of the National Bureau of Standards, vol. 4, no. 2, 80, pp. 301–344, 1907–1908. [28] E.B. Rosa and L. Cohen, “Formulae and Tables for the Calculation of Mu- tual and Self-Inductance,” Bulletin of the National Bureau of Standards , vol. 5, no. 1, 93, pp. 1–132, 1908–1909. [29] A. Gray, The Theory and Practice of Absolute Measurements in Electric- ity and Magnetism, V ol. II, Part I, Macmillan, London and New York,1893. REFERENCES AND FURTHER READINGS 375 [30] T.J. Higgins, “Formulas for the Geometric Mean Distances of Rectangular Areas and of Line Segments,” Journal of Applied Physics , vol. 14, pp. 188–195, April 1943. [31] C.L. Holloway and E.F. Kuester, “Net and Partial Inductance of a Microstrip Ground Plane,” IEEE Transactions on Electromagnetic Compatibility , vol. 40, no. 1, pp. 33–45, February 1998. [32] C.R. Paul and T.S. Smith, “Effect of Grid Spacing on the Inductance of Ground Grids,” 1991 IEEE International Symposium on Electromagnetic Compatibility , Cherry Hill, NJ, August 1991. [33] C.R. Paul, “Modeling of Electromagnetic Interference Properties of Printed Circuit Boards,” IBM Journal of Research and Development , vol. 33, no. 1, pp. 33–50, January 1989. [34] C.R. Paul, “What Do We Mean by ‘Inductance’? Part I: Loop Inductance,” IEEE EMC Society Magazine , Fall 2007, pp. 95–101, and “What Do We Mean by ‘Inductance’? Part II: Partial Inductance,” IEEE EMC Society Magazine, Winter 2008, pp. 72–79. INDEX Ampere’s law, 34 for time-varying currents, 85, 98point form, 101 Biot-Savart law, 19capacitance, 4 deﬁnition, 6energy stored in, 6generalized, 12 circular loop by vector magnetic potential, 144inductance of, 126self inductance of by the Neumann integral, 153 coaxial cable inductance of, 130magnetic ﬁelds of, 43 common-mode currents, 301conducting loop inductance of, 113 conduction current, 85conductors of rectangular cross section partial inductances of, 246 connector pins modeling with partial inductances, 320 conservation of charge, 83conservation of energy, 111 Coulomb choice of gauge, 111cross product rectangular coordinate system, 338 curl cylindrical coordinate system, 361example, 352general deﬁnition, 350rectangular coordinate system, 351spherical coordinate system, 365 current loop magnetic ﬁelds of, 31, 71vector magnetic potential of, 58 current return path, 309, 311current sheet vector magnetic potential of, 63 cylindrical coordinate system, 359 del operator, 18 differential-mode currents, 301displacement current, 85divergence, 18 cylindrical coordinate system, 361example, 346general deﬁnition, 345 Inductance: Loop and Partial, By Clayton R. Paul Copyright © 2010 John Wiley & Sons, Inc. 377 378 INDEX divergence (Continued) rectangular coordinate system, 346spherical coordinate system, 365 divergence theorem, 347 example, 348 dot convention, 169, 243dot product rectangular coordinate system, 337 electrical dimensions, 3, 102 electrically short, 3electrically small dimensions, 90electromotive force (emf), 88, 118elliptic integrals, 61, 149, 129 Faraday’s law, 88, 118 point form, 97 ferromagnetic materials, 6, 134,137ﬂux linkages method of, 119 Gauss’s law, 132, 174 electric ﬁeld, 16magnetic ﬁeld, 16 geometric mean distance (GMD) between a shape and itself, 273between two circular shapes, 287between two lines, 287between two rectangles, 289deﬁnition of, 266of a circular shape, 274of a line, 278of a rectangle, 280partial inductances from, 268 gradient example, 356general deﬁnition, 355rectangular coordinate system, 355 ground bounce, 11, 312ground grid, 312ground plane, 81, 310 Helmholtz coil, 33 high-frequency partial inductances numerical methods, 291 hysteresis curve, 14 images method of, 80 inductance deﬁnition of loop, 6energy stored in, 7of a coaxial cable, 165of a two-wire line, 165Inﬁnite length of current magnetic ﬁelds of, 23, 35 intentional vs nonintentional inductances, 307internal inductance of a coaxial cable, 157of a wire, 164of wires, 155with partial inductances, 239 Leibnitz’s rule, 67 Lenz’s law, 89, 91line integral cylindrical coordinate system, 360example, 342general deﬁnition, 340rectangular coordinate system, 341 spherical coordinate system, 364 loop inductance, 8 by vector magnetic potential, 139concept of, 117energy method, 163methods for computing inductance, 182of coupled coils, 167 loop inductance vs partial inductance an example, 328 loops of various shapes inductance of, 324 Lorentz choice of gauge, 109lumped circuit analysis, 2, 102 magnetic dipole moment, 32, 60 magnetic ﬁeld intensity, H, 13magnetic ﬂux density, B, 13 ﬁnite-length currents, 21, 23 magnetism history of, 1 Maxwell’s equations, 105 iterative solution, 106 method of ﬂux linkages for multiturn loops, 133 microstrip line inductance of, 181 multiconductor transmission lines inductance of, 171n wires above a ground plane, 177n wires within an overall shield, 178n+1 wires, 175 mutual inductance between two circular loops, 147loop, 120 mutual partial inductance between two parallel, aligned wires, 209between two parallel, offset wires, 213 INDEX 379 between two skewed, offset wires, 236 between two wires at an angle to each other, 224 by the Neumann integral, 225of wires, 202, 209 neighboring conductor currents modeling effect with partial inductances, 318 Neumann integral for computing loop inductances, 145 partial inductance, 8, 11 by the Neumann integral, 205, 212concept of, 195from vector magnetic potential, 197, 201general meaning of, 196plots of, 240 passive sign convention, 170PCB inductance of, 182 perfect conductor, 80permeability, 13 incremental, 15initial, 15 power density, 111power rail collapse, 11, 312Poynting’s theorem, 111printed circuit board lands inductance of, 179internal inductance of, 251mutual partial inductance of, 262partial inductances of, 246, 252self partial inductance of, 254 proximity effect, 122 current redistribution, 158 rectangular coordinate system, 336 rectangular loop by vector magnetic potential, 141inductance of, 121magnetic ﬁelds of, 25self inductance of by the Neumann integral, 150 relative permeability, 13relaxation time, 80retardation, 110right-hand rule, 23, 90 self partial inductance of wires, 201, 205 sheet of current magnetic ﬁelds of, 27, 29,46 signal integrity, 312skin depth, 155, 298solenoid inductance of, 134 spherical coordinate system, 362Stokes’s theorem, 353 example, 353 stored energy electric ﬁeld, 79, 113magnetic ﬁeld, 79, 113, 249 stripline inductance of, 180 superposition of magnetic ﬁelds, 25, 86 surface current, 27surface integral, 16 cylindrical coordinate system, 361example, 344general deﬁnition, 343rectangular coordinate system, 344spherical coordinate system, 364 time delay, 103 toroid inductance of, 137 transmission line one-wire above ground, inductance of, 161two-wire, inductance of, approximate, 160 two-wire, inductance of, exact, 160 transmission lines loop inductance from partial inductance, 314 two rectangular loops mutual inductance between, 184 uniform plane wave, 98, 101 vector identities, 357, 358 vector magnetic potential, A, 47 ﬁnite-length currents, 54, 57for time-varying currents, 107inﬁnite currents, 53line currents, 50surface currents, 50 vectors electromagnetic ﬁeld, 4 vias modeling with partial inductances, 318 voltmeter leads effect of, 92 wavelength, 3, 102,104 wire internal and external magnetic ﬁelds of, 37 wires in series and in parallel net partial inductance of, 321