corson and lorrains 3rd Ed 1988

Electromagnetic Fields and Waves Including Electric Circuits Third Edition Paul Lorrain Université deMontréal and McGill University Dale R. Corson Cornell University Frangois Lorrain as) W.H.Freeman andCompany New York ‘Coverimage:LinesofEandofHinanopticalwaveguide. Seepage668Backcoverimage:Differentiating circuit.Seepage144. Preface vi : 1 VECTOR OPERATORS —_1 To7:AATUKOGL 2.PHASORS 29etan N N 3 ELECTRIC FIELDS Ia—— Coulomb's LawandGauss'sLaw 4 ELECTRIC FIELDS II 64 TheEquations ofPoisson and Laplace. Charge Conservation. LibraryofCongressCatalogingin-Publication Data ConductorsLorrain, Paul 5 ELECTRIC FIELDS Ill #4 Electromagnetic fields andwaves. Flectric Multipoles Includesindex 6ELECTRIC FIELDS IV101 1.Electromagnetic fields. 2.Electromagneticwaves. Energy,Capacitance, andForces 1.Corson, DaleR.11.Lorrain,Frangois *7BLECTRIC FIELDSV1 ML TiteElectricCireuitsA:RCCircuits QCH6S.EAL67 1957 S30.1°81 8.31808 * SLECTRIC 5ISBN0-716-71823-5 8=ELECTRIC FIELDS VI149 ISBN 0-716-71869-3 (pbk.) Electric Circuits B:Circuit Theorems 9 ELECTRIC FIELDS VII iz Copyright©1988byW.FreemanandCompany DiseareMeeraA:BoundChargesandheEisPex Nopatofhisbookmaybereproducedbyanymechani, 10ELECTRIC FIELDSVIL192photographic, oFelectronic process, orintheform ofa DielectricMaterialsB:RealDielectrics,ContinuityConditionsat phonographic recording, normayitbestored inaretrieval ‘anInterface, andStored Energy system, transmitted, orotherwisecopiedforpublicor SLECTRIC FIE privateusewithoutwrittenpermission fromthepublisher HL ELECTRIC FIELDS IXut Images. Laplace's Equation inRectangular Coordinates Printed inthe United States ofAmerica 234567890 Wo6Sa3210K9 *Asterisks mean that thechapterseanbeskippedwithoutlosingcontinuity wi contents CONTENTS vi 12. ELECTRIC FIELDS Xs 28. PLANE ELECTROMAGNETIC WAVES I 514 Laplace's Equation inSpherical Coordinates. Poisson's Equation Uniform Plane Waves inFreeSpace, Nonconductors, and fork Conductors "130 RELATIVITY | 238 “29. PLANE ELECTROMAGNETIC WAVES II 536 TheLorentz Transformation andSpace-time Waves inGood Conductors andinPlasmas “14 RELATIVITY Il2st 30 PLANE ELECTROMAGNETIC WAVES IIIss TheLorentz Contraction andTimeDilation. TheTransformation Reflection andRefraction A:TheBasic LawsandFresnel’s ofaVelocity Equations “15 RELATIVITY IL 265 31 PLANE ELECTROMAGNETIC WAVES IV 57% Mass, Momentum, Force, andEnergy Reflection andRefraction B:Nonuniform Plane Waves andTotal “16° RELATIVITY IV 284 Refection32. PLANE ELECTROMAGNETIC WAVES V__ 594 maidof0MowingBicoicCompe ReflectionandRefraction C:ReflectionandRefraction atthe “17RELATIVITY V 308 Surface ofaGoodConductor Maxwell's Equations. TheFour-Potential A 33.GUIDED WAVES 16 18MAGNETIC FIELDS 1326 GeneralPrinciples. TheCoaxialandMicrostrip Lines The Magnetic Flux DensityBandtheVectorPotentialA 34GUIDEDWAVESII@s 19MAGNETIC FIELDS sas TheHollowRectangular Waveguide The Vector PotentialA.Ampére’sCircuitalLaw “35GUIDEDWAVESIll66 20 MAGNETIC FIELDS 111360 ThePlanar Optical Waveguide A:TheEigenvalue Equation Magnetic Materials A:TheMagneticFluxDensityBandthe “36GUIDEDWAVESIV«az ‘MagneticFieldStrength H ThePlanar Optical Waveguide B:TheGuided Wave *21 MAGNETIC FIELDS IV 7 37 RADIATIONI «ss Magnetic Materials B:Ferromagnetism andMagnetic Circuits ThePotentials VandAandtheFields EandB 22. MAGNETIC FIELDS V_—387 38 RADIATION Ila TheMagnetic Forces onCharges andCurrents TheElectric Dipole Transmitting Antenna 23. MAGNETIC FIELDS VI az 39 RADIATION Il 12 TheFaraday Induction Law TheHalf-Wave Antenna, Antenna Arrays, andtheMagnetic *24©MAGNETIC FIELDS VII437 DipoleAntennaEleciteCreateC:MutualandSelfEnductonce AppendixASIPREFIXES ANDTHEIRSYMBOLS 730 25 MAGNETIC FIELDS VIII «sz Appendix BCONVERSION TABLE 731 ElectricCircuitsD:InductiveCircuitsandTransformers Appendix CWAVES 732 26 MAGNETIC FIELDS IX amt ANSWERS 739 Magnetic Energy andMacroscopic Magnetic Forces INDEX 4s 27MAXWELL'S EQUATIONS —492 2 3 as t “ | ae Fig.te.AvectorAandithree “ .| 1an sa Uidcareequivatent toTheunit A |theunitvectoré:normaltothe x > vectors #,¥,2pointinthepositive yy, yplanecontaining AandB.The adirectionsofthecoordinateaxes Positivedirectionsfor@and¢ bei-andareofunitmagnitude q follow theright-hand serewrule 4 Thevector product AXBisequal toABsin66,andBXA=“AXMathematically, afieldisafunction thatdescribes aphysicalquantity 8B. atallpoints inspace. Inscalar fields thisquantity isspecified byasingle number foreachpoint. Temperature, density, andelectric potential are ep e examples ofscalar quantities thatcanvaryfromonepoint toanother in AxB=|A, A,A,|=ABsinge=C, (es) space. Invector fields thephysical quantity isavector, specified bybotha BRB. number and adirection. Wind velocity and gravitational force are examples ofsuchvector fields. asinFig.1-2,where Vector quantities willbedesignated byboldface italie type, andunit vectors willcarryacircumflex: £,§,2. A=(Ar+ Aj+AD)? (1-6) Scalar quantities willbedesignated bylightface italic ype. |Weshallfollow theusualcustom ofusingright-hand Cartesian 'sthemagnitude ofA,andsimilarly forB.coordinate systems asinFig.1-1:thepositive z-direction isthedirection Thequantity A+B,whichisread“AdotB,”isthescalar, ordotance atighthard sotewrotated intheservethatturnsthe product ofAandB,whileAXB,read“AcrossB,”istheirvector,or positive x-axisintothepositive y-axisthrough the90°angle crossproduct.1.1.1 Invariance 1.1VECTOR ALGEBRA Thequantities A,B,and@areindependent ofthechoice ofcoordinatesystem, Such quantities aresaid t0beinvariant. Avector, saytheFigureLlshowsavectorAandistheecomponents Ay,Ay,Ay.Ifwe ztavitational forceonabrick,isinvariant,butitscomponents arenot;aaa covectors theydepend onthecoordinate sytemBoth thedotandcross products arefunctions ofonlyA,B,and@and ATAG+AS+AS B=BE4+B +B, sy srethusalsoinvariant The sum andthedifference, A+BandA—B, arethemselves vectors where #,§,2 aretheunitvectors along thex-,y-,and.z-axes sndinvariant respectively, then A+B=(A, +B)+(A,+B)+(A.+BE, (1.2) 1.2THEGRADIENT Vf ARB=(A,=BOEH(A,— BFA BIE (13) ‘scalar pointfuncton isascalar quantity, sytemperstre, thatisA-B=A,B,+A,B, +A,B,=ABcos (14) functionofthecoordinates. Considerascalarpoint-function fthatis 4 (2THE GRADIENT 5 , What direction should one choose for dltomaximize df? That dlirection istheone forwhich cos @=1or@=0, that is,thedirection of vf. beaWtdip ww" ‘Therefore thegradient ofascalarfunction atagivenpointisavectorZ| havingthefollowingproperties: ‘rs ' ' (1) Itscomponents aretherates ofchange ofthefunction along the H ' directions ofthecoordinate axes. : i Fig.13.Ascalar-point function (2)Itsmagnitude isthemaximum rateofchangewithdistance. of—-- 4. chaneesfromftof+dfoverthe (3)Itsdirection isthatofthemaximum rateofchangewithdistance. (4) Itpoints toward larger values ofthefunction. continuous anddifferentiable. Wewish toknow howfchanges over the infinitesimal distance dlinFig.1-3.Thedifferential thegradient isavector point-function that derives from a.scalar point-function, a=Lag tdy+Say a7) Again,wehavetwodefinitions: Vfisavectorwhosemagnitude andax By Fz ulirection arethose ofthemaximum space rateofchange off,anditis alsothevector ofEq.1-9.Itisclear from thefirstdefinition that¥fis isthescalar product ofthetwovectors. ‘ovariant dl=dxt+dyj+dzé 1-8)BPOYSES ad Example|THEELEVATION OFAPOINTONTHE and SURFACE OF THE EARTH oy. 9 ‘Asanexampleofthegradient,considerFig.1-4inwhichE,the Wwzf+2p+22 a9) elevationabovesealevel,isafunctionofthex-andy-coordinatesy?*Be Thesecond vector, whose components aretherates ofchange offwith distance along thecoordinate axes, iscalled thegradient off.The symbol 2,,9.,8 vasc+ — 1-10)8aaytfa: ™ | i)ve isread “del.” { Notethevalueofthemagnitude ofthegradient: wo | 200 =|(ZV+(2LYxy 100. =[(2)+2) cap [ ‘Thus Fig. 1-4. Topographic map ofahill.The numbers shown give the elevation Einmeters.Thegradient ofEistheslopeofthehillat df=Vfdl=|¥/||dl|cos8, (2-12) thepointconsidered, anditpointstowardanincreaseinelevation“The arrow shows PEatonepoint where theelevation is400 where @istheangle between thevectors Pfanddi. meters. 6 VECTOR OPERATORS artox 7 measured onahorizontal plane. Points atagiven elevation define and acontour line. The gradient ofthe elevation Eatagiven point asthefollowing properties: 4 a, Fo thenhasth1properti SpreeaLa,L (417) (1) tis perpendicular tothecontour line atthat point (2)Itsmagnitude isequal tothemaximum rateofchange of Since thisistrue foranydifferentiable f,weknow that elevation with displacement measured inahorizontal plane atthat point. Satutayo t0,,2 (1-18)(3)Ttpointstowardanincreaseinelevation. dx"OrBy*Orgyeas andsimilarly for3/3y’ and3/32". v hn 1.3INVARIANCE OFTHEOPERATOR Thecomponents ofVin$",namely9/8",3/3y’,and3/32’,relateto - . those ofVinS,3/8x, 3/3y, and3/32, inthesame wayasthe WehavejustseenthatVfisinvariant. Istheoperator¥itselfalso UPin ane ayinvariant? Thisrequires careful consideration because thecomponents of. components ofanyvector Ain5‘andin§.Therefore Visinvariant likeVarenotnumber’batoperators anyvector,andittransformsasavector.Weshallusethispropertyofare not . bu a Let§and5”beanytwosetsofCartesian coordinates. Figure 1-5shows inthefollowing sections. twosetshaving acommon origin, forsimplicity. Thenagivenvector A 14FLUXhasthecomponents A,,A,,A.inS,and Ay, Ay, AyinS’,with Ap=duA, +a,A, +A, (1-13) Itisoften necessary tocalculate thefluxofavector quantity through a surface. Bydefinition, the fluxd@ ofBthroughaninfinitesimal surface Ay=A,+4,A,+4,-A-, (1-14) ids ii roughaainfinitesimal Ay=4,,A,+aA,+a,,A, (1-15) d®=B-ds4, (1-19)heacoefficients dependonlyontheorientation of$"withrespecttoS.Teaco Trentepeeonlyonthe wherethevectordefisnormaltothesurface.Thefluxd®istherefore . thecomponent ofthevector normal tothesurface, multiplied bydid,For7 ed (1-16) asurfaceoffiniteareasf,aap By Ge wa ofBeds. (1-20)\|4 Ifthesurfaceisclosed,thevectordfpointsourward,byconvention \ \\ Example |FLUID FLOW \ Consider fluidflow,andletpbethedensity, vthevelocity, and : ds@ anelement ofarea situated inthe fluid. The scalar product pvdefisequaltothemassoffluidthatcrosses SS|dedin1second,inthedirectionofthevectordef.Thenthe afluxofpythrough aclosed surface, ortheintegral of 4 * Fig. 8. AvectorAandtwo pu-defoverthatsurface,isequaltothenetrateatwhich x1selsofcoordinates SandS ‘mass leaves theenclosed volume. Inanincompressible fluidf having acommon origin thisfluxwould beequal tozero, 5 1.6THE DIVERGENCE THEOREM 9 pees Ifwecalculate thenetfluxthrough theotherpairsoffacesinthe enae maa same manner, wefind that thetotal outward flux fortheelement of ” “ t volume duis “ OB,,3B,|OB, pi lceeneeeee do,=(28:42B:,OB:)gy, e A 1 ®,(Sere B)ao (1-24) ' Ne | ‘Suppose nowthatwehavetwoadjoining infinitesimal volume elementsH Peat :$ fi andthatweaddthefluxemerging throughthebounding surfaceofthe| el ' firstvolume tothefluxemerging through thebounding surfaceofthe|fMfiEat wa second.Atthecommonface,thefluxesareequalinmagnitudebutet eeeal” oppositeinsign,andtheycancel.Thesum,then,ofthefluxfromthefirst ‘UAT willbe -- Fig.1-6,Elementofvolume volumeandthatfromthesecondisthefluxemerging throughthe tatLe Fr dxdydzandthevectorBatthe bounding surfaceofthecombined volumes, aa pointP. Toextend thiscalculation toafinitevolume, wesumtheindividual fluxesforeachoftheinfinitesimal volume elements inthefinitevolume,andsothetotal outward fluxis 1.5 THE DIVERGENCE V-B OB,,3B,ab, ‘Theoutwardfluxofavectorthroughaclosedsurfacecanbecalculated Dui(Gage Sn)av. (1-25)either from theabove equation orasfollows. Consider aninfinitesimal v y volumedxdydzandavectorB,asinFig.1-6,whosecomponents B,, wv . . B,,B,arefunctions ofx,y,z.Thevalue ofB,atthecenter ofthe Atanygiven point inthevolume, thequantity righthand facemaybetakentobetheaverage valueoverthatface on‘Throughtheright-hand faceofthevolumeelement,theoutgoingfluxis es+=-2 3B,dx dO_=(8.+S) dydz, (1-21) isthustheoutgoing fluxperunitvolumeandisinvariant. Wecallthisthe slivergence ofBatthepoint since thenormal component ofBattheright-hand faceisthe Thedivergence ofavector point-function isascalarpoint-function. x-component ofBatthatface.Thevolume being infinitesimal, we According totheruleforthescalarproduct, wewritethedivergence ofneglect higher-order derivatives ofthecomponents ofB. Bas 1,theoutgoing fluxis B, Attheleft-handface,theoutgoing vep=2B:,2B,,3B. (1-26)Bde ox"By*oe a,=-(8,-2S)dyaz. (1-22)* thedivergence isinvariant alsobecause both Vand thescalar product There isaminus sign before theparenthesis because B,#points inward at weinvariant. this face and dsf outward. ‘Thus theoutward flux through thetwo facesis 1.6THEDIVERGENCE THEOREM 3B, 3B, d®, +d&g =—dxdydz=a (1-23) Nowthetotaloutward fluxofavectorBis‘equaltothesurface integral orid ofthenormaloutwardcomponent ofB.Thus,ifwedenotebyfthearea where duisthevolume oftheinfinitesimal element, ofthesurface bounding v,thetotal outward fluxis 10 VECTOR OPERATORS " 3B,,2B,3B.))| se =|Beast=|(B+>Se)ae=[vrBdv.(1-27) 1cs 4 ‘Theserelationsapplytoanycontinuously differentiable* vectorfieldB. oe) Thusaa |Beast=|v-Bdv (1-28) ee“ ‘ oo , Fig1-7.Closed, rectangular Thisisthedivergence theorem, alsocalled Green's theorem, orGauss’s ee |Pathinthexy-plane, centered on ;at |thepointP(x,y,0)wherethe theorem. Note that thefirst integral involves only thevalues ofBonthe Oe nrented surface ofareasfwhereasthesecondinvolvesthevaluesofBthroughout i re bythearrow.Theintegration thevolume v. sf ioe around thepathproceeds inthe : direction ofthe arrows, in = accordance withtheright-hand Examples |tnanincompressible fuid,letpbethemassdensityandwthe ? g screwruleappliedtothez-axis,velocity atagiven point. Then ¥-(pv) iseverywhere equal to zero, since theoutward mass fux perunit volume iszero. Within anexplosion, V(pe)ispostive \.8THE CURL 7xB L.7THE LINE INTEGRAL JB-dl. VoranygivenfieldBandforaclosedpathsituated inthexy-plane, CONSERVATIVE FIELDS B-dl=B, dx+B,dy (1-30) The integrals snd . . . $B-dt~$B,dr+.B,dy (31) [a-a. [axa and{fai“ ° Now consider the infinitesimal path inFig. 1-7. There are two evaluated from thepoint atothepoint bover some specified curve, are contributions tothefirstintegral ontheright-hand sideofEq.1-31, one examples oflineintegrals. y= dy/2 andoneaty+dy/2: Inthefirst, which isespecially important, theterm under theintegral signistheproductofanclementoflengthdfonthecurve,multipliedby $8.dea(8.a988)a_(a..9B.®)ae,(1-32) thelocal value ofBaccording totheruleforthescalar product ay2 dy2 Ate feldBisconteraive iheHineintegralofB+dlaroundany thereiaminussgnbeforethesecondtermbecausethepathclementat closed curve 182v+dy/2 points inthe negative x-direction. Therefore, for this in- ‘unitesimalpath, paramo (129) $2.de=—2Baydy (1-33)ey ‘Thecircle ontheintegral signindicates thatthepathofintegration is Similarly, closed. ona $8dy-Deasdy, (134) +A function icontinuously diferenable ititsfrst derivatives arecontinuous and R VECTOR OPERATORS 11STOKES'S THEOREM B Bedt (2B 2Be Wehavearrived atthisresult foranelement ofareadxdyinthe $“a=(2~3)dedy (1-35) \v-plane.Isthisresultgeneral?Doesitapplytoanysmallarea,whatever usorientation withrespect tothecoordinate axes?Itdoesifitis fortheinfinitesimal pathofFig.1-7. invariant. Wehavealready seenthatthescalar product isinvariant. Thus Ifweset theabove lineintegral isinvariant. Wehavealsoseenthattheoperator V sndthevector product areinvariant, Therefore VXBisinvariant. Thisy=Br9B (1-36) meansthatPXBisavectorwhosevalue,definedbyEq.I-41,isoxby independent oftheparticular coordinate axesused,aslongastheyform then \right-handed Cartesian system. Then Eq. 1-41 isindeed invariant; it sloes apply toanyelement ofarea df, and $Bd=,dst, (1-37) 1(¥xB),=lima4Bedi. (1-42) where dof=dxdyistheareaenclosedbytheinfinitesimal path.Note . thatthisiscorrectonlyifthelineintegralrunsinthepositivedirection in thusthecomponent ofthecurlofavectornormaltoasmallsurfaceof thexy-plane, that is,inthedirection inwhich onewould turn a seasfisequaltothelineintegralofthevectoraroundtheperiphery C right-handscrewtomakeitadvanceinthepositivedirectionalongthe clthesurface,dividedbysf,whenthisareaapproacheszero. zaxds Ingeneral, ¥XBisnotnormal toB.SeeProb.1-7. Considernow gsand theother two symmetric quantities asthe Thecurlofagradientisidentically equaltozero: components ofavector Vx(Pf)=0. (1-43) 2B,9B,\.|(3B,2B). (3B,3B, vw(S-Byes(BeBy(BB), (1-38)yoz! ozax} oxay Jvample |FLUID STREAM which maybewritten as Nearthebottom ofafluidstream thevelocity wisproportional to to the distance from the bottom. Set the z-axis parallel tothe ee F direction offlowandthex-axis perpendicular tothestream a 8 bottom. Then vxe- |S22 (1-39)& By Fz .=0, %=0, w=e, (44) BOB, OB, , and thecurl ofthevelocity vector isThisisthecur!ofB.Thequantity gsisitsz-component. Ifwechoose avector dsfthatpoints inthedirection ofadvance ofa oj 2 right-hand screwturned inthedirection chosen forthelineintegral, then aoa a r vxv= |2 2 Sf) ane (1-43) ds=dol (1-40) zy & 0 0 ex and $Bd=(VXB)det (141) 1.9STOKES’S THEOREM ThismeansthatthelineintegralofB-dlaroundtheedgeofthe area df Hquation 1-41 istrue only forapath sosmall that VXBisnearly isequal tothescalar product ofthecurlofBbythiselement of'area, constant overthesurface dsbounded bythepath. What happens when withtheabove signconvention thepath issolarge thatthiscondition isnotmet? Wedivide the 4 1)ORTHOGONAL CURVILINEAR COORDINATES 1s % J\ample|CONSERVATIVE FIELDS c lass. oo“) Tnvabitaryclosedpathequal10zero?FromSikestheorem,Neii WA thelineintegralofB-dfaroundanarbitraryclosedpathiszeroifRY s4) ¥XB=0everywhere. Thisconditionismetifuh 0 tH Hy : av (48)SEE AG Fg18Anarbitrary suraceofarea ort Siy ;oundedbythecurveC.Thesumo - |. SSE Ye otis cureThesumof PxB=0 (49) =BS squaresshownisequaltothelineintegral around C.Thepositive Afield Bthatisthegradient ofsome scalar point-function fis direction forthe vector df follows therefore conservative. ? theright-hand screw rule. 1.10 THE LAPLACIAN OPERATOR V? surface—any finite surfacet bounded bythepathofintegration in thedivergence ofthegradient offistheLaplacian off: question—into elements ofarea ds#,, df, andsoforth, asinFig. 1-8. Foranyoneofthese small areas, of FF averavyohoh,oF (1-50)Ox?ay?az’ $Bedi=(0xB)- dst (1-46) . lcwhere ?istheLaplacian operator. ‘The Laplacian isinvariant because itisthe result oftwo successive Weaddtheleft-hand sidesoftheseequations forallthedsf’sandthen \nvariant operations. alltheright-hand sides. Thesumoftheleft-hand sides isthelineintegral Wehave defined theLaplacian ofascalar point-function f.Itisalso around theexternal boundary, since there arealways twoequal and useful todefine theLaplacian ofavector point-functionB: ‘opposite contributions tothesum along every common side between adjacent ds4’s, Thesumoftheright-hand sidesismerely theintegral of V'B=V°B,, +V°B,, +VB... (LSI) (VX B)- dA over thefinite surface. Thus TheLaplacian ofavector isalso invariant. Equation 1-S1 applies only to faa (VxB)+a, 47) Cartesian coordinates. SeeSec.1.11.6. 1.11 ORTHOGONAL CURVILINEAR COORDINATES where .fisthearea ofanyopen surface bounded bythecurve C. This isStokes's theorem, Itrelates thelineintegral overagivenpathto Itisfrequently inconvenient, becauseofthesymmetries thatexistin asurface integral over anyfinite surface bounded bythatpath. Figure 1-8 certain fields, touseCartesian coordinates. Ofalltheother possible illustrates thesignconvention. coordinate systems, weshall restrict our discussion tocylindrical and spherical polar coordinates, thetwo most commonly used. a We could calculate thegradient, thedivergence, and soon, directly in thbs bothcylindrical andspherical coordinates. However, itiseasierandmore +Thismustbeanorientable surface, thatis,asurface withtwosides.Notalsurfaces i havetwo sides:aMobiusstrip,forinstance,hasonlyoneside.See}.SawinandA generaltointroducefirsttheideaoforthogonal curvilinearcoordinates. ‘Tromba,VectorCalculus,Freeman, NewYork,1976,p.332 Consider theequation 16 11ORTHOGONAL CURVILINEAR COORDINATES ” 4 4 The volume element is VanIa dy=dl,dlydl,=hyhshy(dqydsdq,). (1-56)Nica|ree Wecannowfindtheq's,theh’s,theelementsoflength,andthe f pose dl, Fig.1-9.Element ofvolume in clements ofvolume forcylindrical andspherical coordinates. |ae | curvilinear coordinates. Theunitvectors berefad Tbe aiceiponeThese 1.11.1Cylindrical Coordinatesaanwaft°andovcmtcdinsuchabapareanilar Incylindricalcoordinates, asinFig.1-10,41=p,d=,42.a 4:X4.= 45 AtPthere are three mutually orthogonal directions defined bythe three unitvectors p,@,and2.Theunitvectors pand@donotmaintain fay2)=4q (1-52) thesame directions inspace asthepoint Pmoves about. However, atany uiven point, thethree unit vectors aremutually orthogonal. inwhich qisaconstant. ‘Thisequation defines afamily ofsurfaces in Thevector thatdefines theposition ofPis space, each member characterized byaparticular value oftheparameter q.Anobvious example isx=q, which defines surfaces parallel tothe r=pp+22. (1-57) yz-plane inCartesian coordinates. Consider nowthree equations Note that @does notappear explicitly ontheright-hand side; itis specified bytheorientation ofp. AGW D=G My2)=G Aay.z)=42 (1-53) Ifthecoordinates @andzofthepoint Premain constant while p increases bydp,then Pmoves bydr=dp p.Ifpandzremain constant defining threefamiliesofsurfaces thataremutually orthogonal. The while@increases bydp,[email protected],ifpand@arefixed intersection ofthree ofthese surfaces, oneofeach family, then defines a while zincreases bydz,thendr=dz2.Forarbitraryincrements dp,dp, point inspace, andqy,q2,q3aretheorthogonal curvilinear coordinates dz,thedistance element isthus ofthatpoint, asinFig. 1-9. Calldl,anelement oflength normal tothesurface q;.This isthe dr=dpp+pdpgo+dzé. (1-58) distancebetween thesurfaces q,andq,+dq,intheinfinitesimal region considered. Then —— — dl,=h,dq,, (1-54) fd |whereh,is,ingeneral,afunctionofthecoordinatesq,,q2,qs.Similarly, hee | dl,=h,dqz and dly=hydq. (1-55) a WithCartesian coordinates h,,hz,hyareallunity. ae ° Theunitvectorsgy,G2,4sarenormal,respectively, tothe41,42,qs pefe surfaces andareoriented toward increasing values ofthese coordinates. ae z - Weassign thesubscripts 1,2,3 tothecoordinates inorder that Pe ee 4 Theorientations ofthethree unitvectors vary, ingeneral, with Bj 1.42,Qs.Only inCartesian coordinates dotheunitvectors point infixed —directions. Fig.1-10.Cylindrical coordinates 18 9 Be, ae rants 6Pot N pds *S ee aeOe a 8dio 3, isd ) y ew |\ lsig wa YaD g cylindrical coordinates. spherical coordinates. Figure 1-11 shows thevolume element whose edges aretheelements of length corresponding toinfinitesimal increments inthecoordinates atthe dr=dré+rd00+rsin Odo@. (1-60) point PofFig. 1-10. The infinitesimal volume is Ihe volume element, shown inFig. 1-13, is du=pdpdodz (1-59) dv=Psin@drd0do. (1-61) 1.11.2 Spherical Coordinates Inspherical coordinates theposition ofapoint Phasthecoordinates Toble 1-1shows thecorrespondence between curvilinear, Cartesian,+8,@asinFig.1-12.Again,theunitvectorsF,8,@donotmaintainthe Mindrical, andspherical coordinates.Sameorientations inspace asPmoves about Notethattheangle @inbothcylindrical andspherical coordinates is Thevector rthatdefines theposition ofPisnowsimply r#,the sulefined forpoints onthez-axis, coordinates @and being given bytheorientation of&Also. WithCartesian coordinates, oneusestheoperator Vforthegradient of \scalar point-function and for the divergence and curl ofavector : jount-function, Asingle expression defines V,and we obtain the 4 vuulient, thedivergence, orthecurlbyperforming theappropriate ee ’ H ‘Table1-1 oad | CURVILINEAR CARTESIAN CYLINDRICAL SPHERICAL Pen.9 —_———oe—erree a a x p A vanes, /| a y ® aa hy 1 1 1 Lis vy hy 1 1 sin@ iy a z 2 i BE a i ¢ 6 5 5 2 2Fig.1-12.Spherical coordinates 4 § 20 vectorOPERATOR 21 multiplication. This relatively simple situation ispeculiar totheCartesian hig coordinate system. With other coordinate systems, thedivergence, ae wot gradient, andcurldonotpermit asingle definition forVbutrequire more a a“ H elaborate expressions thatweshall now derive wa-------4 95 ! 1.11.3 TheGradient H Thegradient isthevector rateofchange ofascalar function f: a ' Se,Sf.Ff. ners ' v=LgeFerte, (1-62) af Ba} a an? ed Ley. 1. 1. ge " Fig.1-14. Element ofvolume, =eatSete (1-63) naLE Pe centeredonP(qy.42,4»)where hy3q, "hy3q2©hySas 4 aR. pe thevectorBhasthevalueshown <0 bythe arrow Forcylindrical coordinates, then, ” o..15.. of, iyasimilar argument,w=pritet 7 T=3p)*pag?*az? asp . a- < dayWithspherical coordinates, dOn=Bihahs dasdan+37(Bihzhs) >"daadas (1-69) e..1F,. 1 F. : 4 ' =F,,1% 9,1Ff 5 wortheright-hand face.Thenetfluxthroughthispairoffacesisthen =3,"+7509*raindag (1-65) 3 Onthez-axis,@isundefined andbothpandsin@arezero,sothese dO,~=aq,Bikahs)daydq,dq. (1-70) twoexpressions aremeaningless. ‘a 1.11.4 TheDivergence Ifwerepeat thecalculation fortheother pairsoffacestofindthenet «utward fluxthrough thebounding surface andthendividebythevolume Tofindthedivergence, consider thevolume element ofFig.1-14.The “theelement, weobtain thedivergence:quantity B,istheq,component ofBatthecenter,andfy,hts,hyarethe hvalues atthat point. Since thefaces aremutually orthogonal, the 1 oe 3 3outwardfluxthroughtheleft-handfaceis FB |sp.Gihahny+(Bahn)+3(®nyh.)]. 70) d= ~Byhashy, dq:dqy (1-66) “hd hed sheaIncylindrical coordinates, dq 2day 3day =~(BSA) -2A(hy-2D)gadgs.(1-67) oq,2 Sq,2 3q;2 13 13By,OB.VB="—(pB,) +et (1-72) Remember thatfyandhymaybefunctions ofq,,justasBy.Wemay pep Pop oz neglectdifferentials oforderhigherthanthethird,andthen B,3B,13By,3B.=Be5Be,LBs,Oe (1-73)> 4 pop papoz deb,=~Byhshy dsdqy+=(Byhshs)E dq,dqy.(1-68) 2qy 2 Inspherical coordinates, 2 VECTOR OPERATORS |ORTHOGONAL CURVILINEAR COORDINATES 23 1a2 a a 1[a 3 | v-B= Fage 2Bye 2 2 xB)= |©(Bhs)dsdys~=(Byhs)dqdgs|(1-77 B=regLArsin)+5(Borsin8)+=(Byr)](1-78 =Fatdaadgs|aga(BM9)dasdas—55(Bala)dadgs|(1-77) 2,3B,By 1B, 1_3B, 1[a 8 | a2,+25Be, 19By 1OBy 75) =——|— (Bs)-—(Bh.)|- 1-78)ptat0+50rind36” (1-75) nhs3a,)3a ) (1-78) ‘Thesedivergences arealsomeaningless onthez-anis,wherepand Corresponding expressions fortheothercomponents ofthe cur follow sin6areboth zero. yyrotating thesubscripts. Finally, Igyhagahads LILSTheCurl (foe Weapplythefundamental definition giveninEq.1-42: "XBT an|3aBanOa (79) 1 IB,hpByhsBs (x)= tim$Ba (1-76)0 torcylindrical coordinates, where thepathofintegration Cliesinthesurface q,=constant and > pe where thedirection ofintegration relates tothedirection ofthe unit vector @,bytheright-hand screw rule.Forthepaths labeled a,b,c,din yxp-t|2 2 28), (1-80) Fig.1-15,wehavethefollowing contributions tothelineintegral p\sp 36 a B, B. ~Byhy dq, »phe B, 2B, ah, suulforspherical coordinates By+dgs)(hy+5"ds)das, , . (842)(42a)‘ -6rsinag hs dq>, a +Bahadg yxp-zt J22 (181)OB, Sh; Psind |dr 30 3~(B+5g.)(5g) B,rlysin0 ‘Thesumofthese fourterms, divided bytheelement ofareaisequal to ‘hese definitions arenotvalidonthez-axis. the I-component ofthecurl ofB.Neglecting higher-order differentials, . 111.6 The Laplacian \Wecalculate theLaplacian ofascalar function fincurvilinear coordi- % sates bycombining theexpressions forthedivergence and forthe | svatv=.MF (1-82) 7 a 1 3(hzhy Sf 3(hh, of hay = [2 (tesF),2(har intl (hy3g)tagsheSq) | pig.1-15.Pathofintegration for .nds component1ofthecutincurvilinear +2(22#4)] (1-83) w coordinates. qs\hy8qy 24 VECTOR OFERATORS 2SUMMARY 2s Forcylindrical coordinates, vp=28:,9B,,9B: (1-26)ox) By” Ge yp19 (Ff) 1oF, oFF4~555ap)pag?ae (84) itheoutwardfluxofBperunitvolumeatapoint: Thedivergence theorem statesthat La, af 1 oF aa1FF Mot,oF (1-85)pap"ap"pag?*32" [v-adv=| Bast, (1-28) except onthez-axis. Forspherical coordinates,‘ sere sfisthearea oftheclosed surface bounding thevolume v, ypeO/H), 1a of 1of ThelineintegralP45 (P5p) in030(856) *aeoa—? . 22,7coroar197,1of (87) [aaroror 88 aersin’ 8ag* weraspecified curve isthesumoftheBdltermsforeachelement dl except, again, onthez-axis olthecurvebetween points aandb.ForaclosedcurveCthatbounds an44 oe sentedsurfaceS(seefootnote toSec.1.9),wehaveStokes’ I. WehavealreadyseeninSec.1.10thattheLaplacian ofavectorBin medsurface5(seefootnotetoSec.1.9),wehaveStokes’stheorem Cartesian coordinates isitselfavectorwhosecomponents arethe fa.af(VxB)-dst, (1-47)Laplacians ofB,,B,,B,.Then A a Vx(VXB)=(PB) BR (1-88) oe a g z isanidentity inCartesian coordinates. 3With other coordinates, V°Bis,bydefinition, thevector whose yxe- |2 2 2 (1-39) components arethoseof¥(V-B) —7x(¥XB),andnotthesumofthe oxOyaz Laplacians ofBy,By,By: BB, OB V°B=V(V-B)-Vx(PXB) (1-89) thecurlofthevectorpoint-function B.Theabovesurfaceintegral\yplies toanyoriented surface ofarea ofbounded bythecurve C. VheLaplacian isthedivergence ofthegradient: 1.12 SUMMARY 2 2 vewpa2b2h,ot (1-50) Thegradient Vfisavector whose magnitude anddirection arethose of Ox? dy? az themaximum rateofincrease ofthescalar point-function fwithdistance |heLaplacian ofavector inCartesian coordinates isdefined as atapoint. Theflux®ofavectorBthrough asurface ofareaofisthescalar WB=VBS+VBS+OBE (1-51) Ineylindricalcoordinates (Figs.1-10and1-11), o=|Beast (1-20) le r=pp+28, (1-58) dr=dpp+pdoo+dz2, 1-58) Ifthesurface isclosed, thevectordfpointsoutward, byconvention. wbleuad (158) Thedivergence ofB dv=pdpdo dz. (1-59) 26 VECTOR OPERATORS PROBLEMS 7 ‘Theformulas forthegradient, thedivergence, thecurl,andthe (0)aeoriginisatO,intheplane,andinsideC.ShowthattheequationLaplacian aregivenonthebackofthefrontcover.Thesevector {b)Theoriginisat0",againintheplane,butousideC.Showthattheoperators aremeaningless onthez-axis, wherep=0. equation isstillvaiInspherical coordinates (Figs.1-12and1-13), (c)TheoriginisatO",atsomepointoutside theplane.Showthatthe integral isagainvalid rai, dr=dré+rd00+rsin Odo>, (1-60) I-4.(1.2)ThevectorrpointsfromP’(x’,y’,2’)toPlx,y,2)." (@)ShowthatifPisfixedandP”isallowed tomove, thenF'(I/r) = dv=Psin@drd0do. (1-61) 417,whereFistheunitvectoralongr. (b)Showthat,similarly, ifP”isfixedandPisallowed tomove,then ‘Thevector operators forspherical coordinates alsoappear ontheback of Vir) =Fr thefront cover. They aremeaningless onthez-axis, where sin0=0. Inother thanCartesian coordinates, onedefines VBasfollows: 1S.(1.6) (a)ShowthatP-r=3 . (b)Whatisthefluxofrthroughasphericalsurfaceofradiusa? VeB=V(F-B)—~Vx(PXB) (1-89) . 6.(1.6) Show that Weshall have occasion touseseveral other identities, given onthe pagefacingthefrontcover.t [v9d0= [fase, where istheareaoftheclosed surface bounding thevolume v.Youcan prove thisbymultiplying bothsidesby¢,where eisanyvector independent ofthecoordinates. ThenuseIdentity 3(frominsidethefrontcover) and PROBLEMSthedivergence theorem. 17. (2.8) Sine isnormaltoB,itseems,offhand, tha 1-1.(1.1)ShowthattheanglebetweenA=28+39+2andB=i-69+2is ee ae‘cena(05,Hs9ems,offmnd,thatVxmustbe 130.6 Asacounterexample, showthat(WXB)- B=1ifB=)+2. 1-2.(1.1) (a)Show that(AxB)+Cisthevolume oftheparallelepiped whose edgesareA,B,C,whenthevectorsstartfromthesamepoint. 18.(LILA)(a)Check,byinspection ofFig.1-10, unitvectorsin{Show that(AXC)-B=(AX.B)-C. Observe howthesign Caren soindi cordatsareitdtalons changes when theeyclic order ofthevectors changes. 1-3.(1.1)LetCbeaplaneclosedcurve.Provethattheareasfenclosed byC PmcOspetsingy, $=—singktcospy, 2=2isgivenby (b)Deduce fromthissetofequations that tifrxar F=cospp—singg, F=singptcspg, 2-2 2h’ Youcancheckthissecondsetbyinspection, where thevector rgoes from anarbitrary origin totheelement dlonthe ‘curveandwhere thepositive directions forsfandfordlobeythe 4(1.11.2) (a)Check, byinspection ofFi '2 eck,byinspecti ig.1-12,thattheunitvectorsin right-handscrewrule.Youcanprovethisasfollows. Cartesian andspherical coordinates arerelated asfollows: 1SeeJeanVanBladel, Elecromagnetic Fields, McGraw-Hill, NewYork, 1964, Appen: F=sin Bcos@¥+sinBsin99+005 08, dixes 1and 2,foranextensive collection ofvector identities and theorems.., .- "Sectionnumber 4=cos8cosPE+COSHSINPH-sinBz,—G=-sin Ps+cospy 8 VECTOR OPERATORS (b) Show that JasinOsin97+005BsingO+0099, 2=cosBF-sinod, 1-10.(11.2) AvectorFhasthesamemagnitude andditection atallpointsin PHASORS space. Choose thez-axis parallel toF.Then, inCartesian andincylindrical coordinates, F=Fé Express Finspherical coordinates. ‘COMPLEX NUMBERS ” 1-11. (1.11.2) Show, bydifferentiating theappropriate expressions forr,that |ADDITION AND SUBTRACTION OFCOMPLEX NUMBERS u thevelocity ¢incylindrical coordinates ispp+p96 +22,while inspherical >MULTIPLICATION AND DIVISION OFCOMPLEX NUMBERS ucoordinates itis#+706+rsin896. Exumples 2 1-12. (LIS) Aforce Fisoftheform (K/r)#insphericalcoordinates, where PHASORS * Kisa constant. Isthefield conservative? USING PHASORS “ Example: SOLVING 4SECOND-ORDER LINEAR 1-13, (1.11.6) Show that, incylindrical coordinates, DIFFERENTIAL EQUATION WITH PHASORS, as % )Mom) FoWph)=2, 6)PHUGAI=0, FRODUCTSOFHASONS, * ; 3 ” SUMMARY » 1-14, Inthecoordinate systems that wehave used until now, vectors and the OBLEMS “0 operator ¥allhave three components, However, inrelativity theory (Chaps. 13t017), itisoften more convenient toconsider only wo components, one that isparallel toagiven direction and one that is perpendicular, For example, one writes that r= +, Ifthe chosen direction isthe x-axis, then nest and raypeat ‘hisshort chapter discusses asecond mathematical prerequisite forthe tunly ofelectromagnetic fields, namely, phasors. Also, P=Fy4+¥., with One uses phasors torepresent quantities that aresine orcosine ‘wnetions ofthetime, orofspace coordinates, orofboth. > 38 Thefunctions sincrandcoswtplay amajor role inmodern technol- Wats a5+85, x,mostlybecause oftherelative easewithwhichtheycanbe erated. They arealso relatively easy tomanipulate mathematically. Then \ilother periodic functions, square waves, forexample, aremuch more ‘uiticult togenerate and much more difficult tomanipulate Werveny. svathematically. One often hastosolve linear differential equations involving sineand Show that ‘ovine functions with constant coefficients, Asweshall sec, theuseof piuasors then has the immense advantage oftransforming these VeAR PAPA, PXASB)KALT PLATE KA ‘ntlerential equations intosimple algebraic equations. butfirst wemust review complex numbers. iy Wehave here acomplex number expressed both inCartesian form andin bey polar form. or aes Iaispositive, then @isincither thefirstorthefourth quadrant. Ifais f wexative, @isineither thesecond orthird quadrant. Use theproper , | wizle! Forexample, theargument of-1+is3/4, not—1/4. | Note that of Fig.2-1.Thecomplex number 2+3/plotted x 3x 5 >11. inthecomplex plane. " ewizth expia=-1, exp(-7F)=- exp2z=1. 29) 2.1COMPLEX NUMBERS " z=atbj=rexpi6, (2410) ‘Acomplex number isoftheform thenthecomplex conjugate of2is =a+jb, 2. ; reat 1) zt=a—bj=rexp(-j0) 1) where j=(~1)'? andwhere aandbarerealnumbers, such as2.5,3.of =10.Complex numbers canbeplotted inthecomplex plane, asinFig Toobtain thecomplex conjugate ofacomplex expression, onechanges2-1.Thequantityaissaidtobetherealpart,andbjtheimaginary part, thesignbeforejeverywhere. Forexample, if ofthecomplex number. a+hj a-bj Onecanexpress complex numbers inCartesian form, asabove, orin reg? then (2:12)polar form, asfollows. First, erg ed 2=a+bj=rcos 0+jrsin0=r(cos 0+jsin8), (2-2) 2.1.1 Addition andSubtraction ofComplex Numbers where Withcomplex numbers inCartesian form,onesimplyaddsorsubtractshherealandima; r tel pa(areby? es)the real and imaginary parts separately: isthemodulus ofthecomplex number z,andtheangle @isitsargument. (a+bj)+(+di)=(atc) +(b+dy @13) With the angle 0 sed inraiththeangle©expressed inradians, 1thenumbersareinpolarform,onefirsttransforms themintoCartesiancoso= 1-24 28, 2 ‘ormsl-atg et (24) eee 2.1.2Multiplication andDivision ofComplex Numbers sinO=Oe at (2-5) InCartesian form, oneproceeds asfollows: @je’ 58° expjO=1+j0—s artat Spm (2-6) (a+bj(c+dj)=(ac—bd)+(ad+be)j, (2-14) and a+bj(at ble-d)_(ac+bd)+(ad+b) a45)cos0+jsin@=expj0. 7 etd)(c+dje—dj) Ore Ofcourse,jxj==I,§xjxj=—j, ete Inpolarform, Thus, from Eq.2-2, 2=a+bj=rexpj0. 28) (7,exp0,)(rsexpj82)=nr2exp(0,+82), (2-16) 2 PHASORS 22PHASORS 3 rexpi®, ri 1=1,008 0 (224)eT="expil,~02) 17)semp is Wehaveset«=0forsimplicity. Therefore Remember toexpress theanglesinradians. f=Re(i,cxpjot)= Rel, 25) Examples (+5)+@-3)=642i, 18) wheretheoperatorRemeans“Realpartofwhatfollows.”Thequantity (4+ S97=(16-25)+40)=~9+40) 19 inparentheses isthephasorIofFig.2-2 ee oe oe ee sy onTeGTSE)TTTL0B}, 20) I=expjot=6050+sino. (2-6) (sexp/2)(2ex/2) =1wexp/5. ea SothephasorIisequaltothevariable/plustheparasiticimaginary term2 ® lySiont. Sexp(jx/3) x ThenSpurl ->sexp(17) 222)Zexp(x72) 6 a a w osOSjal, =(joyt=-0', T= (joyt, ete.(227) 2.2PHASORS aIh geI ae Electric currents andvoltages, electric fields, andmagnetic fields are ‘Youcanearily check that often sinusoidal functions ofthetime, Forexample, analternating current a a isoftheform Re(joh= 4, Re(-w= 4ete (228) a de I=Ipcos(wt+a), (2-23) Inother words, where J,isthemaximum value ofthecurrent, w=22fisthecircular, oF Rel=/, (2-29) angular, frequency, andfisthefrequency. Thequantity inparentheses is a althephase, orphase angle, abeing thephase att=0. ReHt Rejot=", (2-30) ‘Thepoint IofFig.2-2rotates onacircle ofradius J,inthecomplex dt a plane atanangular velocity «,Then itsprojection ontherealaxisis a ei tad "I=s5. 3:Refa=Re(jujl=4a, ete (231) al =e Therefore, ifonereplaces thevariable /bythephasor |,then the — Hh operator d/debecomes afactorjw,andadifferential equation involving/| \ timederivatives becomes analgebraic equation! vA Aphasor canbemultiplied byacomplex number: a -|fa (a+Bj)=(a+bj)lyexpjot=rlyexpj(wt+8), (2-32) Fig.22.Pointlinthecomplex [email protected] describesacircleofradius Onecanalsodivideaphasorbyacomplexnumber: | 1,about theorigin O.It | represents thephasor .,expjot tLItsprojectiononthex-axisis nn exp(wr-0) 2. | Ineosen arn7OPK) es) M Puasons 35 Instead ofhaving acosine function ofthetime 1,one might have a q cosine ofacoordinate: | — | HU Ecos : a E=Eqcoske. (24) Lor oe Thecorresponding phasor would thenbe oonhoc 4pm st E=F,, expike (2:35) «) Inawave, one has acosine function ofboth ¢and, say, 2: Fig. 2-3. Mass msubjected toa force F=F,cosot,ta restoring » force ~kx exerted byaspring, and toadamping force ~bdr/dt exerted byadashpot. Atrest,x=0 E=E,,cos(wt~kz), (2-36) wherekisthewavenumber. Thiswavetravelsinthepositive direction of Example |SOLVING ASECOND-ORDER LINEARthez-axis,SeeAppendix C.Inphasorform, DIFFERENTIAL EQUATION WITHPHASORS Oneofthcommontypesofdifferential equationisthe E=Eyexpitwi ke) ean Oneofthemoscommonspeofrental uation fh and 2 mo6Bsx=coset (2-42) Faint, Toe 2:38 a ale Ge 238) Hereallthetermsarereal.Thisequation describes, forexample, themotion ofamass msubjected tothe applied force Fy60swt, Vector quantities canalsobeexpressed inphasor form. Forexample, a toarestoring force —kx, andtoadamping force —bu, ainFig. fore could beaconine function oftime. 2.3:theproduct ofthemass mbytheacceleration d2x/dis equal tothesum oftheapplied forces F=(F, coswt)é 239) Thesteady-state solution isoftheformx,cos(w+8),where 4mistheamplitude ofthemotion. Itisasimple matter tosolve ‘Then thisequation with phasors. Weusethephasor F=(F,expjoni (2-40) x=x,cxpj(ot+0), (2-43) isboth vector andaphasor whoserealpartisthedisplacement x.Wealsoset F=F,expjot. (44) 2.3 USING PHASORS Substitution into thedifferential equation istrivial: Tousephasors, onefirstexpresses thesine orcosine functions inthe " form x,,cos(wt+8),andthenoneusesthephasor Tmo+bjax+lex=F (245) andX=EXPj(OE-+8)=XpC08(Wt+8)+jtSin(wt+8).(2-41) F Onethenperforms thecalculation withthephasors. ‘Theresult almost "ate (2-46) invariably stays inphasor form. However, ifone requires areal function, onesimply rejects theimaginary part. =Thus, expressing thedenominator inpolar form, 6 PHASORS. #ROTATING VECTORS. 37 = -- bo a, A=A,,coswt and B=B,,cos (wt+8). (2-54)"“[komo}+boye 9=~arctang—s. 247) Then thetime-averaged value oftheir product is ‘Theactualdisplacement istherealpartofthephasorx,or A,,(COS@t)B,,cos(cot+8)) =YuCOs(cot+8) (2-48) =(AnBm608(t(cos«wtcosOsinwtsin)) (2-55) 2.4 PRODUCTS OF PHASORS =(ApB(C08? ctcos8—cosatsinatsin8),(2-56) ex ° wherethesigns(---)mean“average valueof.”Nowtheaverage value Oneoftenrequires theaverage valueoftheproduct oftwosinusoidal ‘ofcos?wroveronefullcycleis},aswesawabove, whiletheaveragequantities. Now ifone tries tomultiply phasors, one runs into trouble.. value ofcos wfsinwtiszero. ThenConsider asimple example. Suppose analternating voltage V=V,,coswt isapplied across aresistance R.Then [=(V,,cos wt)/R. Theinstan- (Ap(COSWt)By,C08(Wt+8))=4AByCOSO=Arn,Byms6088.(2-57) taneous power dissipated intheresistor is Ifoneusesthephasors V3,cos?wt Pos=IVROE, (2-49) A=A,expjot andB=B,,expj(wr +0), (2-58) andtheaverage poweris thentheaverage valueoftheproduct oftheirrealpartsisgivencorrectly by Vin_View PuRR” (2-50) }Re(AB*)=}Re(A,,(exp jot)B,,exp[-j(wt+@)]} (2-59) , =}Re[A,.Byexp(—78)]=44,.B,COS0 (2-60) theaverage value ofcos’wtbeing |,Here Visistherootmean square voltage, orthesquare rootofthemean value ofthesquare ofV =ArmsBrms 6088, (261) v, ayabove. Vor=313=0.707V (2-51) 2.5 QUOTIENTS OF PHASORS for asine oracosine function. Ifoneusesthephasors Dividing onephasor byanother ofthesame frequency yields acomplex v number::-mEXPjt V=Vqexpjor and p="SPJOE (2-52) nexpi(wt+a)_n PyR rexplot+a)1iapy, (2-02)rexpj(+B) re then V2,exp2jeot 2.6ROTATING VECTORSe— (2-53) Ifone expresses arotating vector inphasor form, one runs into another whose realpart (V3,/R) cos2wtisneither theinstantaneous northe kindoftrouble. Suppose average power. Sophasors must notbemultiplied inthisway. Suppose onehastwosinusoidal quantities ofthesamefrequency . E=2E,, 00swt+§Eqsinwt. (2-63) 8 2SUMMARY » jok b 2.8SUMMARY SE H Acomplex number zisoftheform a+bj,where j=(—1)" andaandb H are real numbers. Itisthe custom toplot complex numbers inthe G i complex plane asinFig,2-1,andthus rata - —_ z=a+bj=rexp (0). (2-8) 1 The complex conjugate ofacomplex number isitsmirror image with Fig. 2-4. The vectorEhasaconstantmagnitude, butrotatescounterclockwise at respecttotherealaxis: ananguiarvelocityof«radians/second. ItstimederivativeisdE/dt,orjoE,in thedirectionshown.Itssecondtimederivative is~°E. zt=a—bj=rexp(~j0). (2-41) Addition and subtraction ofcomplex numbers are simpler with the ‘Then thevector Erotates inrealspace asinProb. 2-10andFig.2-4,and Cartesian form: dEdtisperpendicular toE. Inphasor form, (a+bj) +(c+dj)=(a+c) +j(b+d). (2413) - However, multiplication anddivision aresimpler withthepolarform: E=E,,(expjur)i+E,,expj(wt~2)5 (2-64); (r:exp0,)(rsexp02)=nrsexp(0,+82),(2-16) and rexpi®,_r UEXP/O11045(0,—03). 2-17)‘ expo,roll ).(2-17) ar10. (2-65) Ifonehastodealwiththetimederivatives ofaquantity oftheform 1=1,,coswt,itisusually advisable tosubstitute thephasor ‘This equation iscorrect. The trouble here isthat itappears tosaythat dE/dtiscollinear with E,which iswrong, asonecanseefrom thefigure 1Ly,expjoo=lyCOS1+jySiOt. (2-26) Thefactorjrotates aphasor by+2/2. Then 2.7 NOTATION A a ”Sajal, s=-0t, <=(joy, (2-27) ‘Wehaveusedboldface sans-serif typeforphasors, andtheusual lightface dt ar de italic type fortheother variables. However, itiscustomary touse lightface italic type forphasors, asforanyother variable, andtoomit the with theunderstanding thatonly therealparts aremeaningful. operator Re.So,inpractice, wewrite Thephasor dx dx, E=E,,expj(wt~kz) (237) xexnexpjot,Tajor, =-o'r, ete, (2466)it ar epresents aplanewavetravelinginthepositivedirection ofthez-axis, where &isthewavenumber. Thequantities EandE,,canbevectors withthetacitunderstanding thattheimaginary partsareparasitic. ThenEisbothaphasorandavector. 0 PHASORS: PROBLEMS, 41 ‘One occasionally requires thetime average oftheproduct oftwo ' "sinusoidalquantitiesofsamefrequencysuchasJandVinanalternating- |current circuit. Thisisgiven by}ReIV*,where IandVarephasors. | ™, Theratiooftwophasors, again ofthesamefrequency, isacomplex at number. PROBLEMS ~ f -_ t 21.(2.1)Complexnumbersinpolarformareoftenwrittenasr2,whereris imthemodulus and @istheargument, expressed inradians Express1+2/inthisway “y12-2.(2.1)(a)Expressthecomplexnumbers1+2j,-1+2j,-1-2j,and | 1~2)inpolarform \_f/ \(b)Simplify thefollowing expressions, leaving theminCartesian coordi- | ,7 5 'nates: (1+ 2/)(1=2/),(1+2),ML+IP,1+M1~3), : |if 2-3.(2.1) What happens toacomplex number inthecomplex plane when itis wo ary - (a)multiplied byj,(b)multiplied byj°,(c)divided byj? Fig,2-5, 2-4.(2.2) Find the real parts ofthe phasors (1+3j)expj(ot+2) and lexpit+2))/(1+3) ifaispositive. Thenthepointa+bjliesineitherthefirstorthefourth 2-5.(2.3)Solvethefollowing differential equation bymeansofphasors: quadrant. If positive, becarefultousetheproperangle(Sec.2.1). 2-10, (2.6) Show that the phasor V=VL expjor +§exp (ox ~/2)] repre- x de sents avector ofconstant magnitude V,,that rotates inthepositive AGE 8ayAE=Scone direction inthexy-plane attheangular velocity 2-6,(24) Find theemsvalues forthewaveforms shown inFig.25. 2-7,(24)Acertainelecttccircuitdrawsacurrentof1.00amperermswhenitisfedat120voltsrms,60hertz.Thecurrentlagsthevoltageby2/4radian(a)Express Vand/intheform ofphasors, and calculate thetime- averaged power dissipation (b) Now calculate thepower Vimfm6088,where6isx/4 2-8. (2.5) Find the value of 5.14exp(wt+3) ITexpilor+5) 29. (2.5) Show that vps b In(a+b/)=$1n(a? +6°)+jarctan” “8 by CHAPTER3 umZn” Fig.31,Charges Q,andQ, |, separated byadistance r ELECTRIC FIELDS I Coulomb's lawgivestheforceFyexerted byQ,onQ,ifQ,is Coulomb’s Law and Gauss’s Law . stationary Fy=2054,, 6) 3 COULOMB'SLAW 2 Ancor ALL THE AMPERE 44 32. THE ELECTRICFIELDSTRENGTHE4 wheretheunitvectorfspointsfromQ,toQn,asinFig.3-1.Thisis 3.3.THEPRINCIPLE OFSUPERPOSITION 45 Coulomb's law.t The force isrepulsive ifthetwocharges have thesame 34 THEELECTRIC POTENTIAL VAND THECURL OF E46 ‘vn,andattractive ifthey have different signs. Thecharges aremeasured34.1THEELECTRIC POTENTIAL VATAPOINTa8 sncoulombs, theforceinnewtons, andthedistanceinmeters.The 35.THEELECTRIC FIELD INSIDE ANDOUTSIDE constant €yisthepermittivity offreespace andhasthefollowing value: MACROSCOPIC BODIES 49 36EQUIPOTENTIAL SURFACES ANDLINESOFB50 ca8854187817x107"farad/meter G2) 37 GAUSSSLAW 50 Example: THEFIELD OFAUNIFORM SPHERICAL. ChakGe DistRipunon ae Substituting thevalue of¢,wefindthat Example: THE AVERAGE POTENTIAL OVER ASPHERICAL SURFACE EARKOHAWS THEOREM kyx9x 10°22 pewtons, os)Example: THE AVERAGE FINSIDE ASPHERICAL VOLUME r CONTAINING APOINT CHARGE Q 56 34SUMMARY 57 where thefactor of9istoolarge byabout onepartinathousand PROBLEMS 38 Weshall notbeable todefine thecoulomb until Chap. 22.Forthe swoment, wemay take thevalue of€9tobegiven, and use this law asa iwovisional definition oftheunit ofcharge1 wi sCoulomb’ “ 0.are InChaps. 3to12westudytheelectric fieldsthatresultfrom FowhatextentdoesCoulomb's lawremain validwhenQ,andQyar accumulations ofelectric charges. Thecharges areusually stationary. notstationary When thecharges domove, weassume that their velocities and ° accelerations aresmall.Thisensuresthattheelectricfieldsarenearlythe 1)IfQ.isstationaryandQ,isnot,thenSeutomb's lawooanesameasifthechargeswerestationary.Wealsodisregardmagneticfields. torceonQp,whateverthevelocityotQs.hisianexperimental fact.ThisfirstchapteronelectricfieldsconcernsCoulomb's andGauss’s Indeed,thetrajectories o}arePare owculatedonthatlaws.Botharefundamental andwellestablished. rectrographs, ‘madfonacoslerators areinvariably calculated cotha 3.1 COULOMB'S LAW, (2) IfQ,isnotstationary, Coulomb's lawisnolonger strictly valid. Experiments show thattheforce exerted byastationary point charge Q, - ‘onastationary pointchargeQ,situated adistance rawayisgivenby Theexponent ofrisknowntobeequalto2withinonepartin10". “ ELECTRIC FHBLDS 1 \THEPRINCIPLE OFSUPERPOSITION 45 Coulomb's lawapplies toapairofcharges situated inavacuum, Italso Fs Qeappliesindielectrics andconductors ifFayisthedirectforcebetween Q, F.= O,"ameart? —Nemtons/coulomb, orvolts/meter, (3-5) andQ,,irrespective oftheforces arising from other charges within the medium. where1voltequals1joule/coulomb. ThefieldofQ,isthesame,Withextended charges, “thedistance between thecharges” hasno whether thetestcharge Qsliesinthefieldornot,evenifQyislarger definite meaning. Moreover, thepresence ofQ,canmodify thecharge than@,distribution within Q,,andviceversa, leading toacomplicated variation “ offorce withdistance 3.3THE PRINCIPLE OFSUPERPOSITION Electric forces innature areenormous when compared togravitational forces, forwhich Itthereareseveral charges, eachoneimposes itsownfield,andthe resultant Eissimply thevector sumofalltheindividual £’s.This isthe F=6,67259 x19-1)Mette (4) principle ofsuperposition. r Foracontinuous distribution ofcharge, asinFig.3-2,theelectric field Forexample, thegravitational forceonaproton atthesurface ofthesun strength at(x,y,2)is (mass=210kilograms, radius=7x10"meters)isequaltotheelec- 1(pe tricforcebetweenoneprotonandonemicrogramofelectrons,separated E=x=!dv’, (36) byadistance equal tothesun’s radius. Ortheelectric repulsion between ied twoprotons(mass=9.1x10°"!kilogram) isabout4.2x10”timeslarger rytof thanthergravitational attraction. = wherepisthevolumechargedensityatthesourcepoint(1x',y’,2’),asP it cepoint Therearetworeasonswhy,fortunately, wearenotnormally conscious 1teae nePeepointingfromneeeOe oftheenormouselectricforces.First,ordinarymatteristrulyneutral,or r(x’,y',2’)tothefiepoueevemedc’dy’ds’.Ifthereext soitseems, Experiments haveshown thatnoatomormolecule carries a twopoints, anddu’istheelemen y 1with chargegreaterthan10-*”timestheelectronic charge.Second,the surfacedistributions ofcharge,thenwemustaddasimilarintegral,wit mobility ofsome oftheelectrons inmatter prevents theaccumulation of anyappreciable quantity ofcharge ofeither sign. | 3.1.1TheAmpere ae Ifcharge flows through, say,awireattherateof1coulomb/second, then ten thecurrent is1ampere. Thisisnotanappropriate definition ofthe |ya -ampere, because itrests ontheabove provisional definition ofthe \ a coulomb. SeeSee.22.3.1. \ ry 3.2THEELECTRIC FIELD STRENGTH E | ee TheforcebetweentwoelectricchargesQ,andQ,resultsfromthe JS interaction ofQ,with thefield ofQ,attheposition ofQs,orvice versa. Wethus define theelectric field strength Eatapoint astheforce exerted onaunittestchargesituated atthatpoint.Thus,atadistance r tig.32.Chargedistribution ofvolumedensity poccupyingavolumev'.The fromchargeQ,, TeatofvolumeatP'(r"9",2°)hasafielddlatPCs,9.2) oy ELECTRIC FIELDS 1 a7 preplaced bythesurface charge density oandvu’bythearea «f’ofthe / y / chargedsurfaces. y¥j3.4 THE ELECTRIC POTENTIAL VAND THE CURL OF E / | Consider atestcharge Q'thatcanmove about inanelectric field. The / energy #required tomove itataconstant velocity from apoint Atoa a Fig.3.3.ThepotentialpointBalongagivenpathis potasagedtothelineapa» SLEatomAtB,whereBi ee, thee! strengthand e=-fEQ’dl G7) LVe v2>famclementofthepathalong,‘hich theintegral runs. The light Because ofthenegative sign,@istheworkdoneagainstthefield.We linesarelinesofE. assume that Q’issosmall that itdoes notdisturb thecharge distributionsappreciably. ;TTthepathisclosed, thetotalwork done onQis Wecannowshow thatthework done inmoving atestcharge ata constant velocity from apoint Atoapoint Bisindependent ofthepath. eefeowal es LetmandnbeanytwopathsleadingfromAtoB.Thenthesetwopaths together form aclosed curve, and thework done ingoing from AtoB alongmandthenfromBbacktoAalongniszero.Thentheworkdone Letusevaluate thisintegral. Wefirstconsider theelectric fieldofa ingoing from AtoBisthesame alongmasitisalongn. single stationary point charge Q.Then Now letuschoose adatum point R(x», Yo,2»),andletusdefine a scalar function VofP(x,y,z)suchthat ° v=E-dl. Gan) > Now theterm under theintegral ontheright issimply dr/r?, or—d(1/r). Butthesumoftheincrements ofI/roveraclosed pathiszero, since r Thisdefinition isunambiguous because theintegral isthesame forall hasthesame value atthebeginning andattheend.Sothelineintegral is paths leading from PtoR.Then, foranypairofpointsAandB,zero,andthenetworkdoneinmovingQ'aroundanyclosedpathinthe os »fieldofQ,whichisfixed,iszero. -{VV-dl=V.—V» -fE-dl, (3-12)Iftheelectric field isthat ofsome fixed charge distribution, then the a “ lineintegrals corresponding toeach individual charge ofthedistributionareallzero.Thus,foranydistribution offixedcharges. vsinFig.53,andtherefore E=-Vvv. (3-13) fe-dl=0. (3-10) The electric potential V(x, y,2)describes the field completely. The negative signmakes Epointtoward adecrease inV. Anelectrostatic fieldistherefore conservative (Example, Sec.1.9).This NotethatVisnotuniquely defined, because point Risarbitrary. In important property follows fromthefactthattheCoulomb forceisa fact,onecanaddtoVanyquantity thatisindependent ofthecoordinates central force: theforceinthefieldofapointcharge isradial without affecting E. 48 ELECTRIC FIELDS1 STHEELECTRIC FIELDINSIDEANDOUTSIDEMACROSCOPIC nODIES 49 From Eq.3-10andfrom Stokes’s theorem (Sec. 1.9), Forsome geometries, thisintegral diverges. Then onecalculates E VxE=0, ow directly, afterwhichoneintegrates tofindV. Thi “ 3.5THEELECTRIC FIELD INSIDE ANDOUTSIDEisisalsoobvious fromthefactthat MACROSCOPIC BODIES VxXE=-Vx PV=0. (3-15) Macroscopic bodiesconsistofpositivelychargednucleiandnegative Remember that wearedealing here with static fields. Ifthere were clectrons. This brings upthree questions time-dependent currents, 7XEwould notnecessarily bezero, and—PV would then describe only part ofE.Weshall investigate these more (1) Canonecalculate thefield ouside anelectrically charged body by complicated phenomena later. issuming thatthecharge distribution inside thebody iscontinuous? Ifso, then one can calculate thefield byintegrating over thecharge distribu- 3.4.1 The Electric Potential VataPoint tion. Otherwise, onemust findsome other form ofcalculation. Equation 3-12shows thatEconcerns onlydifferences between the Iti,infact,usually appropriate totreatthediscrete charges carried by potentials attwopoints. Whenonewishes tospeakofthepotential ata nucleiandbyelectrons within macroscopic bodies asthough theywere givenpoint,onemustarbitrarily define Vinagivenregion ofspacetobe continuous. Eventhelargest nucleihavediameters thatareonlyofthezero.Intheprevioussection,forinstance,wemadeVequaltozeroat orderof10™meter.Nucleiandelectrons aresosmallandsocloselypointR,Whenthecharges extend overonlyafiniteregion, itisusually packed, compared tothedimensions ofordinary macroscopic objects, convenient tochoose thepotential Vatinfinity tobezero. Then, at thatonemayassumeasmoothlyvaryingelectricchargedensitymeasured point P, incoulombs percubic meter orpersquare meter. . (2) Now what about theelectric field inside acharged body? Clearly v=fEdt (3-16) theelectricfieldstrengthintheimmediate neighborhood ofanucleusor. ofanelectron isenormous. Also atagiven fixed point, thiselectric field ‘Theenergy @required tobringacharge Qfromapointwhere Vis changes erratically withtime,sincethecharges arenever perfectly zero,bydefinition, toPisVQ.ThusVis£/Q,andtheunitofVis Mationary. Itisnotuseful forourpurposes tolookattheelectric fieldas |joule/coulomb, of|volt, closely asthat.Weshallbesatisfied tocalculate space- andtime-averaged Ifthe field isthat ofasinglepointcharge,then saluesofEandVinsideachargedbodybyassumingacontinuousulistribution ofcharge. vefQdr__Q- (17) (3)Isit,then,reallypossibletodefinetheelectricfieldatapointP |,Ancor 4n€or inside acontinuously distributed charge? Itappears atfirstsight thatthe IVcontributed bythechargeelement pdu’atPisinfinite, sinceris Thesign ofthisVisthesame asthatofQ. ‘ero. Infact, itisnotinfinite. ‘Theprinciple ofsuperposition applies toVaswellastoE,andforany Consider aspherical shellofthickness drandradius rcentered onP. chargedistribution ofdensity p, Ihechargeinthisshellcontributes atPadVof(427°drp)/(4x€or)= rdrp. Another shell of smaller radius contributes asmaller dV. The yeipd’ (aay) cleetricpotentialVthereforeconverges,andtheintegralisfinite.A axed or umilar argument shows that Ealso converges. One can therefore calculate theelectric fields ofreal charge distribu- with rasinFig.3-2.Thevolume u’encloses allthecharges. Ifthere are ons bytheusual techniques oftheintegral calculus, both inside and surface charges, one adds asurface integral outside thedistributions. 0 ELECTRIC FIELDS1 GaussLAW st 3.6EQUIPOTENTIAL SURFACES AND LINES OFE Tofindtheoutward fluxofE,weintegrate overtheareaof,orovera solid angle of42x. Thus The setofallpoints inspace that are atagiven potential defines an equipotential surface.Forexample, theequipotential surfacesabouta [reat -2 (321) pointcharge areconcentric spheres. SinceEF=—VV(Eq.3-13),Eis le € 7 everywhere normal totheequipotential surfaces (Sec. 1.2). Ifwejoinend-to-end infinitesimal vectors dspointing inthedirection IfQisoutside thesurface atP",theintegral isequal tozero. Thesolid ofE,weobtain alineofEthatiseverywhere normal totheequipotential ingle subtended byanyclosed surface (orsetofclosed surfaces) is42xat surfaces. Weshall return tolines ofEinSec. 6.5.1. ‘point P"inside andzero atapoint P”outside. Ifmore than onecharge resides within v,thefluxes addalgebraically 3.7 GAUSS’S LAW indthetotal fluxofFleaving visequal tothetotal enclosed charge Q divided by€9: Gauss’s lawrelates thefluxofEthrough aclosed surface tothetotal . chargeenclosed withinthatsurface. |F-aa= 2 (3-22)Consider Fig. 3-4, inwhich afinite volume vbounded byasurface of “ €0 encloses acharge Q.Wecancalculate theoutward fluxofEthrough sf fisieGauss's lawinintegral form.+asfollows. ThefluxofFthrough theclement ofareadifis S eIfthecharge occupies afinite volume, then QOrd r - Edt Gs) [e-aa=2[ oa, (3-23) P eo NowF-defistheprojection ofdefonaplanenormal to#.Then whereofistheareaofthesurface bounding thevolume v,andpisthe e clectric chargedensity. Weassumed thattherearenosurface charges on E-dst=77—d9, (3-20) heboundingsurface.° Ifweapply thedivergence theorem totheleft-hand side, wehavethat where dQisthesolidanglesubtended bydfatthepointP’ . 1v-Edv=2 [pdv, (3-24) wlA Since this equation applies toany finite volume v,theintegrands are equal and K » y\ vipa? (3-25) ow ) atevery point inspace. \ Ee +Wehavefollowed theusual custom ofstarting outwithCoulomb's lawandthen — deducing fromitGausslaw.Thisprocedure seemsrationalenough, butthelatterlawis, Fig.3-4.Apoint charge Qlocated inside avolume vbounded bythesurface of infact,more gencral. Indeed, Gauss's lawapplies tomoving charges, whatever betheit area sf.Gauss's lawstates thatthesurface integral ofE-dsfoverfisequal to selocity oracceleration, while Coulomb's law,astated inSec.3.1,isvalidonlyifQ,is Q/eq. Thevector dspoints outward. stationary 32 ELECTRIC FIELDS 1 3 ‘This isGauss’s lawindifferential form. Observe that itrelates thelocal o* charge density tothederivatives ofE,andnottoEitself. * When itisexpressed indifferential form, Gauss’s lawisalocal lawin that itrelates thebehavior ofEintheinfinitesimal neighborhood ofa given point tothevalue ofthecharge density atthat point. However, my A whenitisexpressed inintegral form,asinEq.3-22,Gauss’s lawis bes H nonlocal, because itconcerns afinite region and notaspecific point in I space. | v Many laws ofnature, inparticular thefundamental laws ofelectromag- ons netism, canbeformulated intwosuch equivalent forms, onelocal and . ‘one nonlocal. With thelocal forms ofphysical laws, intheguise of £ differential equations, one views phenomena astheresult ofprocesses occurring intheimmediate neighborhood ofevery point inspace. i + S , Example |THEFIELDOFAUNIFORM SPHERICAL ‘Fig.3-6.Thepotential Vandtheelectric fieldstrength Eas CHARGEDISTRIBUTION functionsoftheradialdistancerfromasphericalcharge Aspherical charge distribution hasaradius Randauniform distribution ofradius Randvolume charge densityp. density p,asinFig. 3.5. Letusfind Eand Vasfunctions ofthedistancerfromthecenterofthesphere.Bysymmetry, bothEandVare independent ofthespherical coordinates @and. Inside thesphere, atP*,thecharge enclosed bytheimaginary Bysymmetry, Eisradial. Itpoints outward ifQispositive sphere isQ(/R)’. Using again Gauss’s law,wefindthat (a)Theelectric fieldstrength E , Outside thecharge distribution, atP*wherer>R,weimagine a gaQU/RY Orpr (28)sphere ofradius randsurface area47°. Theenclosed charge is 4x67" ane” 36 Q=taR’p. (3-26) Figure3-6showsEasafunctionofr.Note that E,= E,atr=R.Thisisinaccordance withGauss’s From Gauss's law, lawbecause aspherical shell ofinfinitesimal thickness justinside oethesurface ofthesphere carries zero charge. --2,-%2 : (b)Theelectric potential V |her Jer 627) ALP, z because the field isthe same asthat ofapointchargeQatO. xTofind thepotential atP",weuse Eq. 3-16: f */yer Vim[bdrm [ede [Bar 330) eB “Thelastintegral issimply thepotential atr=Rofapoint charge . Nwy QsituatedatO,orQ/(4xeoR). Thus ia * 2eSag =f Ore 22 32 :SS v=‘xe.R*GregR”aneR(Gam). 630 Fig.3-5.Spherical charge distribution. SeeFig.3-6. 4 ELRCTRIC FIELDS 1 55 Example THEAVERAGE POTENTIAL OVER A +SPHERICAL SURFACE. EARNSHAW'S THEOREM AQ, sa | Asanother illustration oftheuseofGauss’s law,weshallprove Ba A \thattheaveragepotential(V)overanysphericalsurfaceofradius aake Le)Randcentr Ohasihefolowing twoproperties ifhereano NS Eechargesinside,then(V)isequaltothevalueofVatthecenter J SP ho—Z J O.Ifthere isanetcharge Qinside andnocharge outside, then ae (V) isequal toO/(4x6,R), « ” © (a)Letusfirst think ofaspherical shell ofradius Rcarrying a uniform surface charge density 0andatotal charge Q=41R°0, as inFig.37a.Then, fromGauss law,atsomepoint Poutside, at Pn —— =[244__@fdat a © o ae Fig.3-7. (a)Spherical surface carrying auniform charge where afisthearca ofthe shell distribution 0.(b)Point charge Qoutside animaginary sphericalusingthesetovaletofVgiesapretypeometicelation surfaceofaisR.()tmainary sphericalsurfacei concerningasphereofradiusRandapointPatadistancer>R charge-free region.Theaveragepotentialoverthesurfaceisequal fromitscenter: tothepotential atthecenter. (d)Spherical surface carryinga uniform charge distribution 0.Point Pissituated inside. (e)Point‘chargeQinsideanimaginarysphericalsurfaceofradiusR.(f) oAfoad oo Imaginary sphericalsurfaceenclosingachargeQ.Theaverage Seon4aJAner potentialoverthesurfaceisequaltoQ/(4x€0R). We now shift our attention toFig. 3-7b. We now have an imaginary sphere ofradius Rand acharge Qsituated outside ata (b)Wecan proceed inasimilar fashion tofind theaverage distance r.The left-hand side ofEq,334 isequal tothepotential potential over aspherical surface when thecharges areinside. atthecenter oftheimaginary sphere, andtheright-hand side is Westart again with acharge Qspread uniformly over the theaverage potential (V) onthespherical surface. Sowehave surface, asinFig. 3-7d. Atanypoint Pinside, Eiszero forthe ‘demonstrated that, inFig. 3-7b, theaverage Vonthesphere is following reason, Bysymmetry, E,andE,arezero. TofindE,, just thevalue ofVatthecenter. ‘weapply Gauss’s law toaconcentric spherical surface having a‘Thisresultappliestoanychargedistribution situatedoutside radiussmallerthanRandthusenclosing zerocharge.Wefindthatthe sphere, asinFig. 37c, because ofthe principle of E,=0.Then E=0inside, andtheVatapoint Pinside isequal to ‘superposition (Sec. 3.3). theVatthesurface, namely, Q/(4x¢5R). So,atPinFig.3-7d, Now imagine foramoment that there isapotential maximum at somepointOinaregionwherep=0.Thentheaveragepotential y-2-fadelaafQua 33s; ‘oversomesphere centered onOmustbelowerthanthepotential Trek Jaxer aa Jeaxer OS) atO,which iscontrary totheabove result. Thus there cannever beapotential maximum inacharge-free region. Forthesame Thelastterm isjusttheaverage potential overtheimaginary reason, there can neverbeapotentialminimumether. sphereofFig.37e.Sotheaveragepotentialoveraspherical ‘ThisisEarnshaw’s theorem. surfaceofradiusRcontaining apointchargeQisQ/(4x¢0R), Itisoccasionally desirabletocreateapotentialwellinspaceso TegardlessofthepositionofQinside!Thesameappliestoa astotrapeither ionsorelectrons. Earnshaw’s theorem shows that Charge distribution Qoffinite volume inside thesphere, asinFig. thisisimpossible withelectrostatic felds Be 56 ELECTRIC FIELDS I 7 Example |THEAVERAGE EINSIDE ASPHERICAL 7VOLUME CONTAINING APOINT CHARGE Q \ Figure38showsanimaginary spherical whineofradsR \containingapointchargeQatr,andapointPatr’+r.Wewish N\\tofindtheaverageEinsidethesphere. \\First,atthepointP, —— \g-%, 636) | a are es \ 7 Then, over thevolume vofthesphere, theaverage Eis sy-t [2; -[Badu (3:38)Anco}take Fig.39.Positivecharge@situatedatr’insideanimaginary sphere.Theaveragefieldof@insidethespherepointsinthe ‘ThankstoGauss’slaw,wecanfindthevalueofthislastintegral direction opposite toF’. withoutmuch effort. Suppose that, instead ofhaving asingle point charge Q,wehadauniform charge distribution ofdensity 7 clearfromFiQ/(aR’). Then, fromCoulomb's law,theEatr’would begiven |Thereason forthenegative signshould beclearfromFig.3-9 bytheterm ontheright inEq.3-38, andthisispr'/(3¢,), from theexample onpage 52. \.8SUMMARY So,finally, theaverage Einside aspherical volume containing a Charge @situated atr’is Theforceexerted byastationary point charge Q,onapoint charge Q,, oF citherstationary orinmotion, isgivenby (y=-2, 3-39) a Fay=22286, 9x10722205, +1)eee, 4neor F fo Ny where€)=8.85x10""*farad/meter, risthedistancebetweenthe v4 \ charges, andiistheunitvector pointing from Q,toQ,.This is / . \ Coulomb's law. fk: \ Weconsider theforce toresult from theinteraction between Q,and | \ theelectricfieldE,ofQ,attheposition occupied byQ,,and \ o Sutif 7 yest\ = eo)a B=ei, (5)\ ae 4neor\ Cet Ly \ whey wheretheunitvector#pointsawayfromQ,. NS Seg” Acurrent flowing attherateof1coulomb/second hasamagnitude of Sa cone ee” lampere. Sa According totheprinciple ofsuperposition, twoormore E’sacting at Fig.38.Point charge Qinside aspherical volume ofradius R. thesame point addvectorially. 58 ELECTRIC FIELDS1 pRonLans 9 Anelectrostatic fieldisconservative: droponebyonebetweenapairofphotocell. Ifthecolorinotright, voltageisappliedtoaneedlethatdepositsachargeontheseed.Theseeds fe-a=oom thenfallbetween apairofelectrically charged plates thatdefect the . a 3 undesired ones intoaseparate bin.Onesuch machine cansortpeas attherateof100persecond,orabout2metrictonsper24-hourday. whereCisanyclosed curve. Itfollows that (a)Iftheseedsfallattherateof100persecond, overwhatdistance must they fallfthey must bespaced vertically by20millimeters when they VXE=0 (3-14) pass between thephotocells? Neglect airresistance. (b)Assume thattheseeds acquire acharge of1.5x10-°coulomb, that and that thedeflecting plates areparallel and $0millimeters apart, andthat the potential difference between them is25,000volts. How long should theE=-vv. (6-13) platesextendbelowthechargingneedleifthechargedseedsmustdeflect, 3-13 by40millimeters onleaving theplates? Assume thatthecharging necdle where andthetopofthedeflecting plates areclose tothephotocell 33.(3.4)Rutherford discoversthenucleus yah[ee cos) in1906.inthecourseofahistoriexperiment thatdemonstrated the Greg. or small sizeoftheatomic nucleus, Rutherford observed thatanalpha particle (Q,=2%1.6% 10"coulomb) having akinetic energy of7.68% 10° istheelectricpotentialatapointP.Herepisthevolumechargedensity, electronvolts0.osxwe1oxpeereboundscoal)ina ; ,we . head-oncollisionwithagoldnucleus(Q;=79x1.6x10"coulomt risthedistancebetweenPCs2)andtheelementofvolumedo’at aceclonwit9godputes(01579L610"coulomb (x',y’,2"),andv’enclosesallthecharges.Thisintegralappliesto potentialenergyisequaltotheinitialkineticenergy?Expressyourresultin finitecharge distributions anditassumes thatV=0atinfinity. femtometers (10° mete). Gauss’s lawfollows from Coulomb's law. Inintegral form, (b)What isthemaximum force ofrepulsion? (c)What isthemaximum acceleration ing's? The mass ofthealpha 0 particle isabout 4times that ofaproton, or4x1.7 107” kilogram.[e-aa-2, 21) . ls & 34,(3.4)Electrostatic ionthruster TonthrusterscortecteithertheattitudeoFthetrajectory ofsatellites. where@isthenetchargecontainedinsidetheclosedsurfaceofarea Theforceexertedbyathrusterisequaltom’v,wheremithemassof Indifferential form, propellant ejected persecond andvistheexhaust velocity withrespect to the thruster vee? 25 Figure3-10showsaschematic diagram ofathrusterthatejectsabeamof &. chargedparticles. Thepropellant entersatPandisionizedinS.Electrodes PROBLEMS , ; . a a a 31.(3.)t Coulomb's law relay (i~\ A“Theforceofatraction betweentwochargesof1coulomb andof (WIG (See ‘opposite signs, separated byadistance of1meter, isabout9x10°newtons. r_{ Ha fe ee How large isacube oflead that hasaweightof9x10°newtons?Lead Sic (oot hasadensity of1.13x10"kilograms/meter’ \o\= ar ce 3-2.(3.4)Electrostatic seed-sorting device |QA\\S C7 (Itispossibletoseparatenormalseedsfromdiscolored onesandfrom “SS 3}+foreign objects bymeans ofadevice that operates asfollows. The seeds + se onns a7 Fig.3-10 1Section number. oo ELECTRICFIELDS1 OBLEMS 4 AandBformalensthataccelerates thepositiveions.Abeamofpositive ‘Showthat,atadistancepfromtheaxisintheregionbetweenthetwoionsexits ontheright atavelocity determined bytheaccelerating voltage conductors, E=A/(2neyp), where Aisthecharge perunitlength onthe V.Theionsofmass mcarry charges ne,where ¢isthemagnitude ofthe inner conductor. Thevector Epoints outward if4ispositive electronic charge. Thecurrent is1.Electrons emitted bythefilament F neutralize thebeamsoastoprevent thesatellite fromcharging up. 39.(3.7) |Theforcebetween apointcharge andalinecharge(a)Showthatthethrust isgivenbyF=/[2Vm(ne)}!™ ‘Auniform lineardistribution ofcharge of2coulombs/meter issituated (b)What isthevalue ofFfora0.1-ampere beam ofprotons when stadistancerfrom«poietcharge©ofoppositesign. V=30kilovolts? (a)Calculate theforce ofattraction, (©IfPisthe powerIV spentin acceleratin icles, show (b)Show thattheforce isthesame asifthelinear distribution were) ps Pentinaccelerating theparticles,showthat replacedbyasinglechargeQ’=2Arsituatedatthefootoftheperpendicu- “ lardrawnfromQ. F=Q2Pm'y 22(2)vo \nev, 410. (3.7). Proton beam ‘A1.00-microampere beamofprotonsisaccelerated through adifference ‘Thus,forgivenvaluesofPandm',thethrustisindependent ofthe ofpotentialof10,000volls. es chargetomass ttf theions.OF,foragivenPF&inversely (2)Calculate thechargedensityinthebeam,oncetheprotonshavebeen Proportionaltov.Thelastexpression showsthat,foragivenpower accelerated, assuming thatthecurrentdensityisuniformoveradiameter of expenditureP,itispreferabletouseheavyionscarryingasinglecharge 200millimeters andiszerooutside. (n=1)andtouseaslowanaccelerating voltageVaspossible (b)Calculate theradialEbothinsideandoutsidethebeam. (4)Iftheelectron source isturned offandifthebeamcurrent 1is {c)Draw agraph oftheradial Eforvalues ofrranging from0toampere, howlongwillittakethebody oftherocket toattain avoltage 10.0millimetersequaltotheaccelerating voltage,ifVis50klvolts?sumethatthe (a)Thebeamissituatedontheaxisofagroundedcylindricalconductingrocket isspherical andthatithasaradius ofImeter. Atthatpoint the tubewithaninside radius of10.0millimeters. DrawagraphofVinsidethe thrusterceases tooperate because theionsfollow thesatelite. tobe. 355.(3.7) Possible andimpossible fields (€)Calculate theelectric charge density perunitlength ontheinside of ‘Anelectric field points everywhere inthez-direction thetube (a)What canyouconclude about thevalue ofthe partial derivatives ofE :11,(3.7).Thefieldofanatomicnucleus th tox,,2(i)ifthespacesitypiii)ifpis conta b)Sketchlinesof andforonei= stom =pall—F/e’), sstemption thatPREon onefFoneimporsibefed,onthe 72a,whereps5.010?coulomb/meter anda=3.4femtometes. (a)What isthetotal charge Q? 36.(37) The conduction electron density atthesurface ofelectrically (b)FindEandVoutside thenucleus. What arethevaluesofEandVat chargedcopper thesurface? ‘Acopper atom hasadiameter ofabout 0.3nanometer (6)Find Eand Vinside thenucleus. What isthevalue ofVatthe (a)Calculate (i)theapproximate number ofatoms persquare meter. (ii) center? theapproximate charge density thatwould result ifeach atom gained one (4)Show thatEismaximum atr/a=0.745, freeelectron, and(i)thecorresponding electric fieldstrength (e)Draw graphs showing E/(2pq/15¢,) andV/(2ps/1S¢,) asfunctions (b)Themaximum possible electric field strength inairis3x10*volts/ ofrforr/a=0to5. meter.Howfarapartaretheexcesselectrons atthatvalueofE? 12.(27) VandeGraaff sccelerator 37.(3.7). Theearth's electric charge ‘AVan deGraaff particle accelerator hasahigh-voltage electrode Theelectric fieldstrength intheatmosphere near thesurface oftheearth ‘maintained under pressure ingaseous SF,inametal tank. Itipossible toisabout100volts/meter andpointsdownward. Thepotential increases with maintain muchhighervoltagesinthiswaythaniftheelectrode wereinair.increasing height, uptoabout 300,000 volts. This fieldismaintained by Assume thattheelectrode isspherical andthatitsradius isr,.Itsvoltage thunderstorms, which deposit negative charge ontheearth attheaverage isV.Thetankhasaradius r,andisgrounded. Theelectric fieldstrength is rateofabout 10°amperes. highest atthesurface oftheelectrode, You arerequired tofindvalues ofr, Calculate theelectric charge carried bytheearth and7,thatwillminimize thisE. Foragivenvalueof7,theoptimum valueof1isinfinite, whichis 38.(3.7) Thecoaxial line absurd. Ofcourse, cost,weight, andspace limitr.Soyoumustoptimize r,Figure33-4showsacoaxialline. foragiven1,whichis483millimeters inonespecificase. e muxcraicFIELDS1vmomuens e (a)Show thatEatthesurface ofthehigh-voltage electrode (r=,)hasa Show that,foranyr,=2arpME, where pisthespace charge density ‘minimum value of2V/r, when 7=15/2 El (b)Explain qualitatively whythereisanoptimum radius r, (b)Thedriftvelocity ofthedustparticles isgiven byStokes's law:itis(©)Identifying anoptimum condition isnotsufficient. Youmustalso theforceEQdividedby6xna,where1istheviscosityofthegas. evaluate how critical theconditionis.SoplotE/Vatr=r,forr;=0.483meter Showthattheirdriftvelocityvis2egE*a/n. andforvaluesofr;rangingfrom100to400millimeters. (©)Calculate 1,p,v,andthetimerequiredforadustparticletodrift(2),Watrangofvaluesof,permissible ifcanbe10%agerthan fromthecathode totheanodewhen.d=2x10*meter'/(volt-second),avin’ =5micrometers, and1)=2x10*kilogram/(meter-second).(©)Caleulate2V/r,forV=5x10°voltsandfortheoptimumr, “Thisimplifiedtheoryneglectturbulence,whichsimportantinpractice 3-13.(3.7)Theequilibrium potentialatthesurfaceofastar 3:15.(2.7)Theexpansionoftheuniverse(a)Collatetescapeenergyforapartsofmassmandcare in193LyalctonsoBondsugestedthatheexpansionofthewiversesitwatedatthesurfaceofstarofmassM,chargeQ,andradiusR. couldbeexplainedonthebasisofNewtonian mechanics ifmattercarriesa{b)Cala theequilibrium potential Vofthestar.Assume asphereof tetelectricchargefullyionized atomic hydrogen, withtheelectrons andprotons atthesame izetemperature andzeronetcurrent.Thefractionoftheelectrons, orprotons, sonyeeCfmahemdustyNsoonpotedsmotelnadensue ‘hatpossess enough energy toescape is thattheproton charge e,isequalto(1+y)e,whereeisthemagnitude of , theelectroncharge. exp(-scapeKinetenergy): (a)FindEattheradiusR. AT (b)Show that, fory>10°", theelectric repulsion becomes greater than thegravitational attraction, sothegasexpands.where&isBoltzmann’s constant, 1.37x10*jouleperdegree.Atequi- (©)Showthattheforceofrepulsiononanatomisthenproportional tolibrium, theelectron andproton currents areequal. Itisthisphenomenon itsdistance Rfrom thecenter and that, asaconsequence, theradial that causes thesolar wind velocity ofanatom atRisproportional toR.Assume that thedensity is (©)Show that V=10"volts forthesun. ‘maintained constant bythecontinuous creation ofmatter inspace. This phenomenon does notappear tohave anyappreciable astrophysical (4)Show thatthevelocity visR/T, where Tisthetime required forthe significance. Even giant galaxies have center-to-surface potential differences radial distance Rofagiven atom toincrease byafactor ofe.This time T that areonly oftheorder of1000 volts, like thesun, canbetaken tobetheageoftheuniverse (c)IntheMillikan oil-drop experiment, anelectrically charged droplet FeeeeaePrecision soeliminateGustparticlesfromin ofoilissuspendedintheelectricfieldbetweentwoplanehorizontal dustrial gases,forexample, toeliminate flyashfromthesmokeof electrodes. Itisobserved thatthechargecartiedbythedropletchanges bycoal-fired electricpowerplants.Acoronadischarge ionizesthegos,and integralamounts withinanaccuracy ofabout1partin10" i electrodes, where they collect. Periodically, theelectrodes areshaken, and thedust falls into acontainer. 416, (3.7) The volume average ofEover aspherical volume isequal totheInonetypeofprecipitator, theanodeisagrounded cylinderhavinga ValueofEatthecenter.Analternative proof.radiusRof150millimeters, andthecathode isanaxialwiremaintained ata ‘Weknowthattheforceexerted byauniform spherical chargedistribu-potentialVof~50kilovolts.Thegasionizes,andionsofbothsignsformin tiononanoutsidechargeisthesameasifthesphericalchargewere thecorona discharge near thewire. Thepositive ionsquickly reach the ‘concentrated atitscenter. center wire, while thenegative ionsmove outradially tothecylinder. The (a)Usethisfacttoshow thatthefieldofapoint charge issuch thatits space charge isthusnegative overmostofthevolume ofthecylinder. volume average overasphere isequal toitsvalueatthecenter. Under those conditions, experiments show that Eisapproximately (b)Show thatthesame applies toanyelectrostatic fieldinacharge-free uniform andequal toV/R forallvalues ofr.Ifthedust particles areat region least slightly conducting, they acquire anegative charge Qof12n¢,Ea°,whereaistheirradius.Thechargeissomewhat smalleriftheyarenonconducting. (a)LetIbetheradial electric current permeter and#themobility (speed/E) ofthenegative ions. 41.THE EQUATIONS OFPOISSON AND OFLAPLACE 65 4.1 THE EQUATIONS OF POISSON AND 4 OFLAPLACE CHAPTERLetusreplace Eby—PVinEq. 3-25. Then ELECTRIC FIELDS II vy--2 oy & TheEquations ofPoisson andofLaplace. ThisisPoisson's equation. ItrelatesthespacechargedensitypatagivenCharge conservation. Conductors pointtothesecondspacederivatives ofVintheregionofthatpoint. 4.1 THEEQUATIONS OFPOISSON ANDOFLAPLACE 65 Inaregion where thecharge density piszero, Example THEFIELDOFAUNIFORM SPHERICAL vv<0, (42) CHARGE DISTRIBUTION 65 Example: THE VACUUM DIODE 66 .which isLaplace's equation. OOARSE NYATIONOFFLECTRIC ‘ThegeneralproblemoffindingVinthefieldofagivencharge 43 CONDUCTION 6 distribution amounts tofinding asolution toeither Laplace's orPoisson's 4331 RESISTANCE 70 equation that willsatisfy thegiven boundary conditions. Examples 71 4.3.2CONDUCTION INASTEADY ELECTRIC FIELD 72 Example |THEFIELD OFAUNIFORM SPHERICAL4.3.3THEMOBILITY .OFCONDUCTION ELECTRONS =73 CHARGE DISTRIBUTION 434CONDUCTION INANALTERNATING ELECTRIC FIELD 74 Consider againaspherical chargedistribution ofuniform volume4335THEVOLUME CHARGE DENSITY pINACONDUCTOR 75 ensty pandradiusRasinFig.35.4.36THEJOULE EFFECT 76 ‘Ouiside thesphere, p=0and44 ISOLATED CONDUCTORS INSTATIC FIELDS 77 Example: HOLLOW CONDUCTOR ENCLOSING ACHARGED BODY 77 v°V, =0. 43) 45 SUMMARY 78$ Now,bysymmetry, V,isindependent ofboth@and¢.Therefore,PROBLEMS 79 from Sec. 1.11.6, Poisson’s equationrelatesthelocalvolumechargedensityptothespatial 18/,a% 238¥0)_ 4ratesofchangeofthepotential V.Thisisagainafundamental relation. It FE(rPB)-0 Z(rB)-o. 4follows immediately from Gauss’s law. Laplace’s equation isPoisson's, withpequaltozero.Bothequations servetocalculate electric fields.We MA paA (45)shall return tothem inChaps. 11and12. aor r Charge conservation isanexperimental fact.Whatever thecircumstan- whereAis@constant ofintegration. Thisisinagreement withEq.ces,thenetelectric charge carried byaclosed system isconstant. We 3.27withA=-Q/(4x6,). shallframe thatlawinasimple mathematical formandthenapply ita ‘Insidethesphere, few sections later. We shall return toitonseveral occasions. ‘Themajorpartofthischapter pertains toconductors. Ordinary electric py--2 4s)conductors contain conduction electrons that drift inthedirection €0" ‘opposite totheapplied E.Weshallfind,among otherthings, thatthis 13/,a%)__pdriftvelocity issurprisingly lowandthatthenetvolumechargedensityis ag(rQ- er “normally zero. 6 ELECTRIC FIELDS 11 4.THE FQUATIONS OFPOISSON AND OFLAPLACE o 3(2.9%)__oF 4s clectrons havezeroinitialvelocityandthatthecurrentisnota(-=)nara 8) limitedbythecathodetemperature butcanbeincreased atwillby , increasing V,, peep=+B, (49) SinceVdepends onlyonx,byhypothesis, Poisson's equation or Pe, reduces to or B 2Emap 4-10) v23.7 a7: (12) where Bisanother constant ofintegration. ItisintuitivelyobviousthatE,cannotbecomeinfiniteatr=0; wheretheelectronspacechargedensitypisnegative.Thus soBwneroand @°Vdx’ispositive,butwehavenotyetfoundp. Nowpisequal tothecurrent density divided bytheelectron 7 velocity v,so a(uy * 3c,”bre aio wa (413) asintheexample inSec. 3.7onpage 52 ewe whereJisthemagnitude ofthecurrent density, Example|THEVACUUM DIODE Byconservation ofenergy, Letusfind thepotential distribution between theplates ofa : vacuumdiodewhosecathode andanodeareplane,parallel, and munev, (414)separated byadistance sthat issmall compared totheir linear 2 extent, asinFig. 4-1 ‘Weassume thatthecathode isatzero potential andtheanode ‘where misthemass ofanelectron and~¢isitselectric charge:atapositivepotentialV,.Thehotcathodeemitselectronsthat €=+1.6x10"coulomb.Thenaccelerate inthe direction ofthe anode, We also assume that the av J a~ReVImy* os Fh | Tointegrate,wemultiplytheleft-handsideby2(dV/dx)dxand Fi bie theright-hand sideby2dV.Then f Sag) AV)?_ANmV/26)"" aPity| (By-Seeonea, (4-16) ig gleae| ae re ion ee aR |aeia Fge whereAisaconstantofintegration,ia pi [ee ed WenowfindthevalueofA.Atthecathode, V=0andnyteA FBei A=(dV/dx)’. ButdV/dxiszeroatthecathodeforthefollowing eae a reason.Ifoneappliesavoltagetotheanodewhenthecathode isLe [ee cold,dV/dxispositiveandequaltoV,/s.Ifnowoneheatsthe———P cathode, itemits electrons, there isanegative space charge, and + ov dV[dxdecreases. AslongasdV/dxispositiveatthecathode,the =emitted electrons accelerate toward the anode and cannot return Fig.4-1. Schematic diagram ofaplane-parallel vacuumdiode. tothecathode.Thecurrentisthenlimitedbythethermionic ‘Thecathode,ontheleft,emitselectrons.Anelectronofcharge emissionandnotbyV...Thisiscontrarytowhatweassumedatthe =edrifts inthedirection oftheanode atavelocity v.Wehave beginning. However, ifdV/drwas negative, electrons could never shown widely separated electrodes forclarity. leave thecathode, andthere would bezero space charge, which is 6 eLecTRICMELDS11 \ conpucrion Py absurd, SodV/dx canbeneither positive nornegative. Soitis andforelectrons emitted with zero velocity. However, foranyze10,andAisalsozero.(However, seebelow.)Then geometry,Jisproportional toV3. Figure4-2showsV,E,andpasfunctionsofx/s. av_4(1)7(m)" uw Inanactualdiode,theemissionvelocityoftheelectrons isfinite $72) (gv an andthereis@potential minimum immediately infrontofthe 127m cathode. Only electrons with aninitial velocity larger than avMeL.s(2)(2)+B. (4-18) certainvaluegetpastthisminimum. “\eo/ ee. - \.2THE LAW OFCONSERVATION OF “Theconstantofintegration BiszerobecauseViszeroatx=0.So y * ELECTRIC CHARGE deo) \2e 5 Consider aclosed surface ofareaofenclosing avolume v.Thevolume Whenx=5,V=V,. Therefore <hurge density insideisp.Charges flowinandout,andthecurrent o lensityatagivenpointonthesurfaceisJamperes/meter. vev(2) (420) Itisawell-established experimental factthatthereisneveranynet a-eation ofelectric charge. Then anynetoutflow depletes theenclosed Also, disregarding thesignofE, barge Q:atanygiven instant, 4V,(xy? df dQ=th(¥ 42 -ast=—4[ pdy=—2e=$%(Z)", (21 feat=—5{ pdv=—, (424) JnSeRelVE 2.33510 °UEamperes/meter, wherethevectordifpointsoutward,according totheusualsign. an onvention, lew. Applying nowthedivergence theorem ontheleft,wefindthat San (423)Ws"(x/sy ap[v-sav=-[ Pav. (425) Equation4-22isknownastheChild-Langmuir law.Thislawis hy lyOt valid onlyf "lane-parallel diode with ne ible edge effects waltcalySr@plane-pern caligible edge Wehavetransferred thetimederivative undertheintegral sign,butthen ‘semust useapartial derivative because pcanbeafunction ofx,y,z,as well asoft. 20] ole Nowthevolumevisofanyshapeorsize.Therefore op vere2, (4-26) to} -------------->> e/e, V/V. Equations 4-24 and4-26 are, respectively, theintegral anddifferential “ torms ofthelawofconservation ofelectric charge. Taya 0 a to 1.3 CONDUCTION .4-2.Thespacechargedensityp,theelectricfieldstrengthE,tngu ts Fig42eenaerysevenreymeenmtat x00dconductorssuchascopperoraluminum,eachatompossesses anenena: Thesohecnpnaretortthe ‘neoFtwoconductionelectronsthatarefreetoroamaboutinthe Valueofthe variable attheanode. nate » 13conpucrion n ‘Table 41 VvR=7, (4-28) Conductivity 0, ‘Conductor ‘siemens/meter_ where Visthepotential difference between thetwoelectrodes and/is ‘Aluminium 3.54%10" thecurrent.Brass (65.8 Cu,34.2Zn) 1.59x10"Chromium 3.8107 Examples |Foraeylinderof iy 1acylinderofcrosssectionsf,lengthL,uniformconductivity ConnerPeter @,andwithelectrodesontheendsasinFig.43a, GraphiteTAx 10Iron 1.0107 1=ot)=toe=2, (4.29)Mumetal (75Ni,2Cr, 5Cu, 18Fe) 0.16x107 L Nickel13x10" Seawaters Silver 6.15x10” ATin 0.870 x10" A . Zine 1.86x10” , Ee go og2 Semiconductors maycontaintwotypesofmobilecharges:conduction @eSelectrons andpositive holes. Aholeisavacancy leftbyanelectron 33 liberated from thevalence bond structure inthematerial. Ahole behaves asafreeparticle ofcharge +e,anditmoves through thesemiconductor much asanairbubble rises through water,. Inmost good conductors andsemiconductors, thecurrent density Jis proportional toE: J=0E, (4-27) « where @istheelectric conductivity ofthematerial expressed insiemens permeter, where Isiemens’ is1ampere/volt. ThisisOhm's lawina Legemoregeneral form.Asweshallscelater,anelectric conductivity canbe SFcomplex. WeshallfindastillmoregeneralformofOhm'slawinChap. eg“na23. InnerradiusR, i Table 4-1shows theconductivities ofsome common materials userrasRs ae AaOhm's lawdoesnotalways apply. Forexample, inacertain typeof PO ceramic semiconductor, Jisproportional tothefifthpower ofE.Also Ne some conductors are not isotropic. 4.3.1 Resistance IfOhm's lawapplies, theresistance between twoelectrodes fixed toa ¥ sample ofmaterial is w _ _ Fig.43.(a)Cylinderofweaklyconducting materialwith+AfterErnstWernervonSiemens(1816-1892). Thewordthereforetakesaterminalsia electrodes atbothends.(b)Tubeofweaklyconducting materialthesingular: onesiemens. Thesiemens wasformerty calleda“mho. withelectrodes ontheinnerandoutersurfaces. n ELECTRIC FIELDS 11 \conpucrion B R=L (430) 0millimeters/hour! Thenthedriftvelocityissmallerthanthethermaloat ~ sgitation velocity bynineordersofmagnitude! InEq. 4-34 vgissmall, but Ne isvery |: It f,ThetubeofFig.4-3bhasinnerandouterradiiR,andR,, 4 “ nyUerae,HeOPPS, respectively, alength L,andauniform conductivity 0.There are Ne=8.5 x10x 1.6 10-10" coulombs/meter’. (4-35) ‘copper electrodes ontheinner andouter cylindrical surfaces. The resistance ofacylindrical element ofthickness drisdr/o2arL., ‘Thelowdriftvelocity ofconduction electrons isthesource ofmanyThen paradoxes. Forexample, aradiotransmitting antennaisabout75meters mdr 1 an longandoperates atabout1megahertz. Howcanconduction electrons R=[,DarkIxokRy” (431) fromoneendtotheotherandbackin1microsecond? ‘Theansweris thatthey donot,They drift back andforth byadistance oftheorder of1 4.3.2. Conduction inaSteady Electric Field womic diameter, andthatisenough togenerate therequired current, Forsimplicity, weassumethatthechargecarriersareconduction 1.3.3TheMobility /ofConduction Electrons electrons. ‘Thedetailed motion ofanindividual conduction electron isexceedingly Ihemobility ofconduction electrons: complex because, every nowandthen,itcollides withanatomand lworebounds, Theatoms,ofcourse,vibrateabouttheirequilibrium posi- MrailNe (4-36)tions, because ofthermal agitation, and exchange energy with the conduction electrons. »».bydefinition, apositive quantity.t Itisindependent ofEinlinear However, ontheaverage, each electron hasakinetic energy of3&7, onductors. Thus wherekisBoltzmann's constant andTisthetemperature inkelvins. o=Nedt (37Thus, atroom temperature, thevelocity va,associated with thermal agitation isgiven by where, asusual, wehave taken ¢tobethemagnitude oftheelectronic hharge. mvs_er4.38x10°x300)=6x10?"joule,(4-32) Ifthedrivingelectricfieldisconstant,thenthedriftvelocityis 2sonstant, This means that thetime-averaged netforce onaconduction and lectron iszero, orthattheaverage braking force duetothecollisions wstcancels the—eE force exerted bythefield. -21\ v2 W ”van(Rxi)=10°meters/second. (433) hatisthemagnitudeofthisbrakingforce?Itis9.110 elule—(-eE)=eE="=-“y,, (4-38) ‘Under theaction ofasteady electric field,thecloudofconduction mM Mm electrons driftsataconstant velocity v,suchthat ‘somthedefinition ofthemobility ..Thebraking forceandv,pointinaay »ppositedirections. J=0B=—Nevs, (434) thissituation isanalogous tothatofabodyfallingthrough water;after +while,theviscous forceexactly cancels thegravitational force,andthe wherevgpoints inthedirection opposite toJandtoE,andNisthe snagnitudes ofthetwoopposing forcesareequal number ofconduction electrons percubic meter. ‘Thedrift velocity islow. Incopper, N=8.5x10.Ifacurrentof ampere flows through awirehaving across section of1millimeter’, J=10° andvyworksouttoabout10~‘meter/second, orabout ‘Someauthorsassigntothemobilitythesignofthechargecarrier. on ELECTRIC FIELDS 11 ‘CONDUCTION 15 ‘The quantities N,.l, and oforgood conductors (gc) andfor Theimaginary terminEq.4-42isnegligible forf<7x10".Uptoaboutsemiconductors (sc)arerelatedasfollows: |gigahertz thecloudofconduction electrons movesinphasewithE,and oa=Ned. Ng Mer Oye>>OrcsAgeKM (439) Decreasing thetemperature increases themeanfreepathofthe carriers, whichincreases themobility 4.Atverylowtemperatures the 4.3.4 Conduction inanAlternating Electric Field conductivity ofpure metals iscomplex: Ifwedisregard thermal agitation, there arethus twoforces acting ona Ne conduction electron: thedriving force eEandthebraking force ofEq. O=-i a (4-46) 4-38. Inanalternating electric fieldthese twoforces areunequal, andthe 4.3.5 TheVolume Charge Density pinaConductor equation ofmotionis (1)Assume steady-state conditions andahomogeneous conductor. dv, e then3p/3t=0and,fromSec.4.2,¥-J=0.IfJistheconduction mnt7—CEmexpjot—Var (4-40) currentdensityinahomogeneous conductor thatsatisfiesOhm'slaw J=oE, then where m*istheeffective mass. Thisquantity takes collisions into V-J=¥-oE=o0¥-E=0, V-E=0 (4-47)account. As arule, m® issmaller than the mass ofanisolated electron. Insilicon m*=0.97m, butingallium arsenide (GaAs) theeffective Nutthedivergence ofEisproportional tothevolume charge density p, massisonly0.07m. Electron velocities insolid-state devices areofthe ‘tomSec.3.7.Thus, under steady-state conditions andinhomogeneous order of10°meters/second insilicon andabout 4times larger inGaAs. conductors (0independent ofthecoordinates), piszero. Replacing thetimederivative byjaandsimplifying, wefindthat Asarule, thesurface charge density onaconducting body carrying a 4current isnot zero, =~TSjom*Ale E,,expjut. (4-41) (2)Nowsuppose thatoneinjectschargeintoapieceofcopperby bombarding itwith electrons. What happens tothecharge density? In ButJ=o£=—Nev. Itfollows that thatcase, from Sec.4.2, 2pv-J=-— (4-48)onweste (4-42) 3+iom* Mle Wut,fromSec.3.7, With w=0werevert toEq.4-37. Thisrelation does notapply at v op-J=0V-E=—e (4-46 frequencies oftheorderof1gigahertz (f=10°)orhigher, where atomic ° ee (49) phenomena become prominent. Forcopper atambient temperature, where €,istherelative permittivity ofthematerial (Sec. 9.9). Thus 3p__op ot om*M_o2, papsexp(--%), 4-50) omwma (443) a7eeePeet (~<e,) 5) sy , andpdecreases exponentially withtime.=2afX91NOTX58X10" (4-44) Therelativepermittivity ¢,ofagoodconductor isnotmeasurable8.5.x10"(16x10) because conduction completely overshadows polarization. Onemay =15x10, (4-45) presume that¢,isoftheorder of3,asincommon dielectrics. 76 ELECTRIC FIELDStt 44ISOLATED CONDUCTORS INSTATICFIELDS n ‘The inverse ofthecoefficient of¢intheabove exponent isthe IfEandJaresinusoidal functionsofthetime, relaxation time.emandisth ‘Wehave neglected thefactthat isfrequency-dependent andisthus . Pa,itselfafunctionoftherelaxation time.Relaxation timesingood Poy=Enea=OEras= (4-55) conductors are, infact, short; and pmay besetequal tozero, inpractice. For example, the relaxation time forcopper atroom temperature is about4x10™*second, insteadof~10~"’secondaccording totheabove 4.4ISOLATED CONDUCTORS INSTATIC FIELDS calculation. (3) Ina homogeneous conductor carryinganalternating current,p1s Ifonechargesanisolatedhomogeneous conductor, theconductionzerobecause Eq,4-47applies. electrons moveabout untiltheyhavereached theirequilibrium positionsand then, inside the conductor, there iszero E. (4)Inanonhomogeneous conductor carrying acurrent, pisnotzero. Itfollows that(1)allpointsinsidetheconductor areatthesame Forexample, under steady-state conditions, potential; (2)thevolume charge density iszero, from Eq.4-47; (3)any netstatic charge resides onthesurface oftheconductor; (4)Eisnormal V-J=¥-(0E)=(¥o)-E+o¥-E=0 (451) atthesurface oftheconductor, forotherwise charges would flowalong thesurface;(5)justoutsidethesurface,E=o./€9, whereoa,isthe and surface charge density, from Gauss’s law. Note theparadox: one canexpress Eatthesurface ofaconductor in vege =(POE (4-52) termsoftheJocalsurface charge density alone,inspiteofthefactthatE fo ° depends onthemagnitudes andpositions ofallthecharges, whether they reside onthe conductor orelsewhere. (5)Ifthere aremagnetic forces onthecharge carriers, thenJ=oF What iftheconductor isnothomogeneous? Forexample, onemight doesnotapply andthere canexistavolume charge density. SeeSec. haveacopper wirepressed ontoagold-plated terminal. Then conduction 24.1. electrons drift across theinterface and establish acontact potential, usually ofafraction ofavolt. The magnitude and sign ofthecontact 4.3.6 The Joule Effect potential depend onthenature ofthematerials. Intheabsence ofanelectric field,thecloudofconduction electrons Example HOLLOW CONDUCTOR ENCLOSING A.remains inthermal equilibrium with thelattice ofthehost conductor. CHARGED BODY Upon application ofanelectric field, theelectrons gain kinetic energy betweencollisions,andtheysharethisextraenergywiththelattice.The Figwe44omgwiereginahallowconductorwithmt Conductorthusheatsup.ThisistheJouleeffect. withintheconductor inazeroEenclosesazeronetchan Whatisthekineticenergygainedbytheconduction electrons? becauseofGauss’slaw.Thenthesurfacechargeontheinside Consideracube oftheconductor, with side a.Apply avoltage V surface oftheconductor is-Q. between opposite faces. The current is/.Then thekinetic energy gained Iftheconductor carries azeronetcharge, thenthetotal charge i 5 (ontheoutside surface is isVI,and thepower dissipated asheat percubic meteris ‘thecarlocharge2sitystagivenpolotonthooutide surfaceoftheconductor isindependent ofthedistribution ofQin pave(4) -E) (453) thecavity.Itisthesameasiftheconductor weresolidandcarried a a)\q anet charge Q. p Inversely, thefieldinside thecavity isindependent ofthefield =o8t=~ —wats/meter (4-54) outside theconductor. Theconductor thenactsasanelecrostatic * PROBLEMS ” es Inatime-independent electric field,ge Wepee NX J=0E=—Nev,, (4-34)S LENS, oe= whereNisthenumberofchargecarrierspercubicmeter,~eisthe a° a charge onone ofthem, and u,istheir drift velocity. The mobility iswoe =\ \lefinedbytheequation \+ bal -- velo aBeareca aD a=W (4-36) WwWig yi* E Ne es ees and yo aerageeen’ + o=Neu (437) +o Fig.44,Sectionthroughahollowconductorenclosingabodythatcarriesanet Inanalternating electricfieldtheconductivity oiscomplex:charge Q.The dashed lineisasection through aGaussian surface lying entirely in theconducting material, whereE=0. ___Nedt 7°T¥jom* Ale” 42) where m*istheeffective massofacharge carrier. 4.5SUMMARY Ifthecurrent density isJandtheelectric fieldstrength isE,thenthe\ime-averaged powerdissipated percubicmeterintheformofheatis Poisson's equation follows fromthedifferential formofGauss’s lawand BedPos atedpet mm from therelation E=—VV: FoonPoy=Exes =CEng == watts/meter*. (4-55) o vv=-£. (4-1)jo thisistheJouleeffec. Setting thevolume charge density pequal tozero yields Laplace's Under static conditions there exists anetelectric charge solely attheequation: surfaceofaconductor, andEiszeroinside.Justoutside,Eisnormalto thesurface, anditsmagnitude iso.4/€o, where og,isthesurface chat rv=0 2) semity ° ™ ‘The law ofconservation ofelectric charge states that, whatever the circumstances, the netelectric charge ofaclosed system isconstant Mathematically, PROBLEMS ap 1-1,(43)‘The conduction electron vege 2, 4.26 ).Theconductionelectron densityincopper a (25) Theobjectofthisproblem istoillustrate theenormous magnitude ofthe lectric charge densities inmatter. where Jarepare, respectively, thecurrent andcharge densities. Wetake theexample oftheconduction electrons incopper. Acopper tore obey Ohm's laws: atom contains 29electrons, oneofwhich isaconduction electron. Copper Mostconduct y hasanatomicweightof64andadensityof8.9x10°kilograms/meter.J=0E. (427) Suppose thatyouhavetwocopper spheres, eachonehaving avolume of, Lcentimeter’. Thespheres aredepleted oftheir conduction electrons and whereoistheconductivity, expressed insiemenspermeter. separatedbyadistanceof100millimeters. 80 ELECTRIC FIELDS11 81 Calculate theforce ofrepulsion. Show thatthisforce isequal toabout0.5%oftheforceofattractionbetweenthesunandtheearth.Seethetable ‘| sofphysical constants onthepage facing theback cover, “ “ 42.(4.3)Thedriftvelocityofconduction electrons “ & PR Py Copper hasanatomic weight of64andadensity of8.9x10"kilograms/ =meter’. a(a)Calculate thenumber ofatomspercubicmeterandtheapproximate * diameter ofanatom. Fig. 45(b)Calculate thecharge4carriedbytheconduction electrons in1meter Me45. ofcopper wire 1millimeter indiameter. There isoneconduction electron peratom.(c)Calculate thedriftvelocity oftheconduction electrons inmeters per Showthatifthesubstrate conductivity isuniform andequaltoa,then hour when thewire carries acurrent of1ampere. 43.(43)Refraction oflinesofEattheinterface between mediaofdifferent Yoatconductivities bo”“xoaP‘WeshallseeinSec.10.2.3thatthetangential component ofEis The lesacontinuous attheinterface between twomedia cacti edesateofnitesize.However, youcanperform theShowthat,attheboundary between twomediaofconductivities o,and fatthatBond7eateonthattheyareinfinitelysmall,disregarding theox,alineofE,ofalineofJ,is“refracted” insuchawaythat Pann andJwouldthenbeinfiniteattheirsurfaces.tan6,/0,=tan85/0,where6,and8;aretheanglesformedbyalineofE anarethePrineofsuperpositionafollows,Thecuentinthe intsumof listrit emanating fromC,plusanoths withthenormaltotheinterface. radialdistributionconvergingonC;.Thus,atapointnn “ 4-4.(4.3) ‘The surface charge density attheinterface between media of" different conductivities. oth‘AcurrentofdensityJflowsinthedirection normaltotheinterface Eon Ino between twomedia ofconductivities 0. and 04.2. The current flows from‘ ° medium 1tomedium 2. 17.(43,) Theresistance ofaspherical shellsphericalshellofuniform conductivity h .dShow thatthesurface charge density o4iS€,€0J(1/O.2~ 1/0.) Ry,napectivel tyhasinnerandouterradiiRyandAssume that€,hasthesamevalueonbothsides.Ifthecurrent isnot Ba,respectively. Ithascopperelectrodes platedontheinnerandouter normal totheinterface, thentheabove Jisthenormal component ofthe Showthattheresistance is(1/#,~1/R,)/Ava current density. "18. (43.1) Resistive film 45,(4.3) Conduction inanonhomogeneous medium ‘Asquare filmofNichrome, analloy ofni a coordinates. Showthattheresistance between theelectrc ndsoF nowhere equal tozero, ‘Thissurface resistance isexpressed inohmspersquare. " VeV40V-Pr=0, 19.(43.1) Atheorem ontheresistance ofaplate Arectangular plate ABCD hasathick it sine. ickness 5andaconductivity o.With where+ conducting electrodes onedgesAandCD,theresstance isRy,With 446.(4.3)Geophysical prospecting bytheresistivity method electrodes onBCandDA,theresistance isR,,‘Onecanlocate resistivity anomalies intheground asinFig.4-5.The ShowthatRiR:= 1/(0°s?) current/flowingbetweenelectrodesC,andC,establishesanelectricfield Thisequationalsoappliestoanyregionboundedbyequipotentials and intheground, andonemeasures thevoltage Vbetween apairofelectrodes linesofcurrent flow.Weshallusethistheorem inProb.33.5.ItwasfirstP,andP,maintained atafixedspacing b.With6<a,V/bisequaltoEat Proved byIsukada.t theposition x.Anomalies inground conductivity show upinthecurve ofEasafunctionof2 DJ.Epstein,Proc.IEEE,vol.56,p.198(1968). 28 P e4 — R — if_— 9~~— “Vo * re \.' Fig. 47. — Fig.4-6. 4-13, (4.3.6) The resistojet 4:10.(4.3.1) EandJinsideabattery Figure4-7showstheprinciple ofoperation ofaresistojet usedasa"AbatteryfeedsaresistanceRasinFig.4-6.Thebatteryactsasapump, thrusterforcorrectingthetrajectoryortheattitudeofasatelite.forcingconduction electrons towardthenegative electrode, Thebatteryis Assuming complete conversion oftheelectric energytokineticenergy,cylindrical, oflengthsandcross-sectional areasf,withelectrodes ateach calculate thethrustforapowerinputof3kilowatts andaflowof0.6gram end. Then |PV| =V/sand, inside thebattery, J=o(£,—|¥V|), whereE, ofhydrogen persecond. isthe“pumping field.” Findtheoutput voltage asafunction ofthecurrent. SetR=s/ac asthe ‘output resistance ofthebattery 411. (4.3.4) Mobility andelectron drift ‘Asimple model forthedrift ofaconduction electron isthefollowing Theelectron describes aballistic trajectory forawhile, under theaction of the ambient electric field, and then theelectron suffers animpact. Its velocity justafter theimpact isunrelated toitsvelocity before theimpact, andwesetitequal tozero. Theelectron then starts outonanother ballistic trajectory, andtheprocess repeats itself. Letthemean timebetween the collisions beAtandtheeffective mass (See. 4.3.4) bem* Find themobility interms ofArand m*. 412. (4.3.4) Conduction byholes Wederived Eq.4-40 ontheassumption thatthecharge carriers are electrons. Suppose thecarriers areholes. Then thecharge changes sign, landboth terms ontheright arepositive. Ifthatisso,thecomplementaryfunction,whichoneobtainsondisregarding theforcingterm¢£jit, is v=vmexp(<7) “Then vsincreases exponentially with time, which isabsurd. Show that, ifthecharge carriers aveholes, then ds, £y mtt=sebexpjt~Fe and that ee or gnNel a+Tyjomedtfe ™PIO “TFjom*Me 85 ° P08) 5 / CHAPTER Electric Multipoles +o 5.1 THEELECTRIC DIPOLE 84 wae5.2.THELINEARELECTRICQUADRUPOLE —§7°53 ELECTRIC MULTIPOLES 88 a *54. THE ELECTRIC FIELD OUTSIDEACHARGEDISTRIBUTION. 2 EXPANDEDINTERMSOFMULTIPOLES 89 *54.1THE VALUEOFV.THELEGENDRE POLYNOMIALS 90 -0 *54.2 THE MONOPOLE TERM 93é *54.3 THE DIPOLE TERM 93 154.6THEQUADRUPOLE TERM—94 Fig.5-1,Twocharges+0and—Qseparatedbyadistancesandforminga Example:THEFIELDOFASETOFSIXPOINTCHARGES SET dipole.Thedipolemomentisp.WecalculatethepotentialatpointPbySYMMETRICALLY ABOUT THEORIGIN 95 summing thepotentials ofthetwocharges. 55 SUMMARY 97 PROBLEMS —98 11; v=,2(4-2), (1) Electric multipoles aresets ofpoint charges possessing certain sym- Ame \rsMa metrics. Their interest liesinthefactthatrealcharged objects, suchas whereantennas andatomic nuclei, possess electric fields thatmaybeexpressed e ‘assumsofmultipole fields. a2(3) =r+(8 5. Asidefromthemonopole, whichisasinglepointcharge, themost rear +5)+75cos. 62) useful type ofmultipole isthedipole, which consists oftwo charges of equal magnitudes andopposite signs, some distance apart. Most mole- Wenowdivide both sides byr?andtaketheinverse: cules actlikesmall dipoles. Also many antennas radiate likeoscillating ; teelectricdipoles. re[i+(3)+Scoso| (53) r 2) Tr 5.1 THE ELECTRIC DIPOLE#11s,480)+3(5Sos2 504)=1-3(GatF088)+5(Gatzea) SH ‘The electric dipole isacommon type ofcharge distribution. We return to itlater inthischapter andinChaps. 37and38. 1weneglect terms oforder (s/r)? andhigher, then The electric dipole consists oftwocharges, one positive andone aa:negative,ofthesamemagnitude, andseparated byadistances.Wefind oy-Scosa+ 5Scosa1 (5-5)VandEatapointPsituatedatadistance r>>5,asinFig.5-1.AtP, no wr2 86 FLECTRICFIELDSut a Similarly, r s 8?30s? 8-1altseas0+St (5-6) and Qs 3 v=00s(>) (7) reor \KRAM > i Notethatthepotentialinthefieldofadipolefallsoffas1/r°,whereas WES ZzLY thepotential ofasinglepointchargevariesonlyas1/r.Thiscomesfrom SOVFISECO* thefact that thecharges ofadipoleappearclosetogetherforanobserver eSa@®39 somedistanceaway.s0theitfieldscancelmoreandmoreasthedistance i: SRRX/ rincreases. LY =SIi\\‘Thedipolemomentp=Qsisavectorthatisdirectedfromthenegative /To BT\ tothepositivecharge.Then S_Y pie=25. 5-8) PEVane G8) NZ Wecannow find theelectric field strength E.Inspherical coordinates, xp=-22Pos? (5-9)or4xeor Fig.§2.LinesofE,shownwitharrows,andequipotential linesforéheelectric3 dipole ofFig.S-1.Inthe central region thelines become tooclose toBether tobe=—19V_psind shown.Thecentralarrowisthevectorp. B= --5 8S, (5-10)730” 4xeor 1wv whereQisthenettotalchargeoccupying thevolumev’,and#’defines Fe=~Tangag (11) theposition ofthecenterofcharge, byanalogy withthecente#ofmassin mechanics. Pp _ IfQ=0,then#’—>= andQF’isindeterminate. However, theintegral E=——,(20s67+sin66) (5-12) Gregg(2008OF+sin68), G12) ofr’pdv’stillprovidesthecorrectvalueofp.IfQ=0,thedipole moment isindependent ofthechoiceoftheorigin(Prob.5-19 ThusEfallsoffasthecubeofthedistance r. IfQ#0,thedipole moment ofthedistribution iszerowhe theorigin Figure 5-2shows linesofEandequipotential linesforanelectric ‘satthecenter ofcharge, forthen7’=0. dipole. Rotating equipotential lines about the vertical axis generates equipotential surfaces. 5.2THE LINEAR ELECTRIC QUADRUPOLEZ More generally, thedipole moment ofacharge distribution is Thelinearelectricquadrupole isasetofthreecharges,asinF'ig-5-3.The frpdv' separation5isagainsmallcomparedtothedistancertothepointP. p-{r'pdv'=Q-—— =?'", (5-13) AtP, lyde’ ve(2-72+2)- Q(E+2-2), (5-14) Ma Amey \r, rr) Ameo \te ty“) 88 SATHEELECTRIC FIELDOUTSIDEACHARGEDISTRIBUTION 89 B Pir 89 known asmultipoles. Asingle point charge isamonopole. Adipole is obtained bydisplacing amonopole through asmall distance s,and replacing theoriginal monopole byanother ofthesame magnitude butof opposite sign. Likewise, aquadrupole isobtained bydisplacing adipole , byasmall distance s,and then replacing theoriginal dipole byoneof4 equal magnitude butofopposite sign. Forthelinear quadrupole, s)=s, , The multipole concept can extend indefinitely. For example, the quadrupole can bedisplaced byasmall distance s,,and theoriginal +0 y quadrupole replaced byone inwhich thesigns ofallthecharges have |: heenchanged.Thisgivesanoctupole.A2'-polerequires!displacements )Sis S25 00S Wehaveseenthatthedipolepotential variesas1/r?andthatthe 7 quadrupolepotentialvariesas1/r°,Forthe2/-pole,Vvariesas1/r!*!and |Eas ifr We have calculated thepotential Vinthefield ofadipole and ofa “of quadrupole inthefollowing way. Wefound thesum ofthepotentials of | theindividual charges, expanded thesumasapower series, andthen Fig.5-3.Charges +0,~20, +0forming alinear quadrupole truncated. Thisapproach isstraightforward, andithastheadvantage of showing exactly what approximations areinvolved. Problems 5-5and 5-6 Wecanexpand theratios r/r,andr/r,aspreviously, except thatsnow explore adifferent method thatismore elegant, butthatdoesnotreveal replaces s/2.Thus, ifweneglect terms oforder (s/r)’ andhigher, theexact nature oftheapproximations. ros 5?(3cos?6-1) *5.4THEELECTRIC FIELD OUTSIDE ACHARGEripe eta (15) DISTRIBUTION, EXPANDED INTERMS OF :MULTIPOLES r5 5?(3.cos?6-1)Prar ir 6-16) Achargedistribution ofdensityp(x’,y’,z')occupiesavolumev’and and extends toamaximum distance rj,from anarbitrary origin O,asinFig. 20s? Goose -1) 5-4.Weselect Oeither within thevolume orclosetoit. ya sh. (5-17) Weshallseethatthepotential Vatapoint Poutside thecharge distribution such that >rig isthesame as(1)thepotential V,ofa Thepotential Vofalinear electric quadrupole varies as1/r?,whereas point charge, ormonopole, equal tothenetcharge ofthedistribution, E,calculated asforthedipole, varies as1/r*. Thefields ofthethree plus(2)thepotential V;ofapoint dipole withadipole moment equal to charges cancel almost completely forr>>s. thatofthechargedistribution, plus(3)thepotential Vsofapointquadrupole with aquadrupole moment equal tothat ofthe charge *5.3 ELECTRIC MULTIPOLES distribution, andsoon,themonopole, dipole, quadrupole, etc.,being all located atthearbitrary origin Itispossible toextend theconcept ofdipole andquadrupole tolarger Similarly, theEatpoint PisthesumoftheE'softheabove numbers ofpositive andnegative charges. Suchcharge arrangements are monopole, dipole, quadrupole, etc. IfQ=0, then Vsisindependent ofthe choice oforigin. More —— xenerally, V;isindependent ofthechoiceoforiginifallthemultipole *Starred sections can beomitted without losing continuity, However, theLegendre 1polynomials ofSec.5.4.1arerequired forthestarred Sec.12.1 ‘moments uptothe2!-polearezero. %” S4THEELECTRICFIELDOUTSIDEACHARGEDISTRIBUTION a The factor wisdimensionless. Werequire (1), which is1/r” call ~ 1 Ow) 1(2w) ag. : (iy=t=woy+ (2) 4h (2" o (5. IO =f Fe G wherew(0)=1/r.Tocalculatethederivativesontheright-handside,we \ require thepartial derivative ofJur’—r|withrespect tou: .i ¥}8a > /Flue’—|==[(ue’—29°+(uy=y+(ue’~29") (5-22) De/ 3u3u Lr ~ 1 2 2: 2 95 >Fpgprcg(RE=Dex+2uy?2'y+2uz"?—22'2) (5-23) Fig.5-4.Arbitrary chargedistribution enclosed withinavolumev'.Thepotential awrParen (urlan) (5-24)atP'sthesameasifonehad,attheorigin, amonopole, plusadipole, plusa lu lure °quadrupole, plusanoctupole, etc.However, rmust belarger than themaximumvalueofr'. Thus ow L_(ur'=r-r f= AereeaCals le (5-25) Since theVofamonopole decreases as1/r,thatofadipole as1/r?, ue dur efjue! 4 ur’—FI that ofaquadrupole as1/r°, etc., then along distance away, where and >rieays thefieldofanycharge distribution issimple. Itisthatofapoint \ Dep opaw ake chargeattheoriginandV~V,.Closerin,V~V;+Vs.Stillcloserin,V, (=) aeee (5-26)becomes discernible, then V;,etc.,and thefield becomes more and more Hiwo r Icomple. where#andaareasinFig.5-4.Also, *5.4.1TheValueofV.TheLegendre Polynomials Sw Jerk » ew (ur'—r)er'P re WewishtofindVatsomepointPsuchthatr>rug.FromFig.5-4,this Beuror arae (5-27) is and pdu" 2 vepypthpi 2 =|— =] ew 3r'- rr?r°QBcosa—1 ve]cee 8 a en Wy r where > Ingeneral,rs|r—r'|=[e-x' P+(ye (e-2FP (5-19) 8 ow ad The point P(x,y,z)isfixed.Thusr”isafunctionofx',y’,2’,andwe (3)yoprt! Pale0s a) (5-29) canexpand 1/r”asaTaylor series near theorigin _Let where 1 cosa) =)— Pa@-1)" 5-30=: (5-20) (608@)=7GleogayeCOSa—D) (5-30) 2 3 ‘Table 5-1Legendre polynomials . fn P.(cosa) o 1 1 cose 5 josta-t | J wr 43Scos*a—3000s?a+3rn 5S3eus'a—70cos'a+1Scosa “to s Fig.5-5.ThefirstfourLegendrepolynomialsP,(cosa). isaLegendre polynomial. Table 5-1gives thefirst five, while Fig. $-5 shows thefirstfourasfunctions oftheangle a.Foranya, *5.4.2 The Monopole Term ThefirsttermistheVthatonewould haveatPifthewholecharge was {F.(cosa)|=1. 631) concentrated atthearbitrary origin: ‘ThusfromEq.5-21, ° “ey (5-35) rp"Geosa— : or Fat,PepeGee 4.=SFrpcosa) (5-32)nore mol where Qisthenetcharge inthedistribution. Thisisthemonopole term. Since|P,(cosa)|<1asabove,thisseriesconverges forFac<r. «iszeroiftenetchargeiszero.Itsvaluedepends onthepositionSubstituting inEq.5-18,wefinally havethat chosen fortheorigin. 1 1 *5.4.3 TheDipole Term V~trer[edusner7’cosapdv ‘Thesecondtermvariesas1/r,liketheelectricpotentialofadipole. . From Eq.5-33 12300sa—1 . os[17aSepdv 1 ; areorJ,2 Yeps[reosapd'=—5-[ rod’, (536) .4xeor* J, axeor Jy a[oieesstese #A30084)yy (5-33) axe J, 2 e where theintegral ontherightisthedipole moment ofthecharge 1f.38costa—30e0s@ +39ey distribution: Loe rdKr ri p=[rod s37 ThusEV,+VetVatVatVet oes (5-34) poh=P, (5-38) Aneor” Letusexamine thefirstthreeterms insuccession. asinEq.5-8. 7 suet vie mS4-THE ELECTRIC FIELD OUTSIDE ACHARGE DISTRIBUTION 9s *5.4.4TheQuadrupole Term Pye=Pay-{y'z'pdv' =Qy'e, (5-46) Now consider theterm V,ofEq.5-33. Itinvolves a1/r*factor, liketheVofthelinearquadrupole ofSec.5.2.IfwecalculateV;forthelinear Px=Pu=f2'x'pdv'=OF¥, (5-47)quadrupole with charges Q,20, and Qatz=s, 0,and —s, . respectively, wefindthatitisequal totheVofEq.5-17.Then Vsisthe bo a——sameasifonehadasmallquadrupole attheorigin.Now PoP=[xy'pdv"=OF (6-48) 1 ‘The barindicates, asusual, anaverage value. ‘Thuswea 4°(3cos?a—1)pdu’ 1: Wea (3mnpy.+3nlpec+Simpy=f ipe-ry-r'jp av’ 5-39) ". Trap[ABOeen om) hetlm Set 3m?—1 3n?=1 B=Le+my+n, (+40) ee ee) (49) where /,m,narethedirection cosines of#andwhere Ifthecharge distribution displays circular symmetry about thez-axis, P+m+n?=1 (5-41) Pyz=Pox=Pay=0, Pax=Pyy: (5-50) ‘Thenexpandandgroupterms.Thisyields Itisthenconvenient todefineasinglequantity 1Yonge (3mmy'z'pdv’+nlf2'x'pav’ 0=2p.—Par) (5-51) 4£3lmix'y'pdv’ole{x"pdv’ thatisalsocalledthequadrupole momentofthechargedistribution. I 2h Then, remembering that +m? +n?= 1,wefindthat wet 53n?—1 2 P| yea[27 oav’). (5-42) ye! ant)=2 cos?= a 2 2I =Freep Om =FeeapsGeos0-1) (5-52) These integrals, liketheintegral ofEq.5-37,depend solely onthe atthepointr,8,@inspherical coordinates.distribution ofelectric charge within v’,andnotonthecoordinates Wecan,ofcourse, deduce theelectric fieldstrength from therelation x,y,z ofthe field point P.They specify thenine components ofthe E=-v. quadrupole moment ofthecharge distribution: Example|THEFIELDOFASETOFSIXPOINTCHARGES P.=[e'pdo'=F", (5-43) SETSYMMETRICALLY ABOUTTHEORIGIN. Figure 5-6shows sixcharges. Apoint Pinspace isatadistance 1afromtheorigin.ThepotentialatPisgivenbytheseriesof ee (sas) |‘5347UetretehethecmeFramBg.938 : | y=Be 659 Pes-{2"pdv'=z", (5-45) dre %6 5.5SUMMARY 97 . Itturns out that V,=0, and that a AtP, ne"oSJ VeViaVtVedoe (5-62) qa 60 [31a 1 can=ae(iES +DUmtan) a“ =900?=12)+} (5-63) , a ofunity,with°+m?+n®= 1,thecoefficient ofa"/r’isalsoofthe order ofunity, and atr=10a the series reduces toitsfirst term 2 withanerrorofatmost1%.ThetermsVs,Vs,...become go progressively more prominent asr/adecreases. Fig.5-6.Sixcharges arranged symmetrically about theorigin O. 5.5SUMMARY ‘The electric dipole consists ofapair ofcharges ofequal magnitudes Qbut From Eqs. 5-37 and5-38, ofopposite signs, separated byadistance s.Itsdipole moment pisOs, a vector directed from thenegative tothepositive charge. P=DrO=0, V:=0. (5-54) ‘Atadistance rfrom thedipole, FromEqs.5-43to5-48, vat ys) (5-8)4neqr” =20" =204 =80" 5 , Pox=20°Q, Pry=20°Q, Pee=80'Q, (5-55) ‘Thelinearelectric quadrupole consists offourcharges Q,-20,Q,as inFig.5-3.Then Pos =Pex=Pry=0. (5-56) 208?(3008-1) Finally,fromEq.5-49, =208"Boos’OD ayygs 5.49) nally, Eq.5-49, Vara 3 (P>s') (5-17) 1Pp22a22, YomGyegilOl—1)0"0+Bm—1)e°Q+(an?1)4a°Q](5-57) wherecos=2+ eo, Some distance outside acharge distribution thepotential can be =Fe,piOF+3m?+120-6) (558) writtenasaseriesa 4a" VEVtVyt VtVetVetoes 5-34) =FS0-3)22n-0, (5-59) oH Here V,isthepotential atPdue toasingle charge, called amonopole, and equal tothenetcharge ofthedistribution and situated attheposition of thearbitrary origin. Similarly, V;isthepotential atPdue toadipole @=120°9. (5-60) whosedipolemoment isequaltothatofthedistribution andalsosituated 98 ELECTRIC FIELDS It PROBLEMS: 9 attheorigin, ete. ‘Wefound that, inthefieldofanelectric dipole, Q ve2 (1!w=. (535) irene) . Refer toFig. 5-7.Show that, ifthelength ofthedipole issmall, then pry=fs, (5-38) Opa (i 4 =Bf a (1 or v4me,le()],-< yay (snp. +3nlp,,+3imp., where2’isthepositionofapointonthez-axisandr’=f+9+(=~ are Pie 3P-1 3m? —1 32-1—, ——p,, +——..). }-49)FPFFPwFZPx)(5-49) |pave whereJ,m,narethedirection cosines ofthevectorrdefining the oO y| position ofPandwhere H ¥ Pec=OE",—Py=OV",—pez=OE,(5-43),(5-45) LY 4J ard a Z---- ----” Fig.5-7. Py=Py=QY'Z", Pax=Paz=QEX', Pay=Pyx =O'Y’ 5-6.(5.2) Analternate expression forthepotential inthefieldofanelectric (5-46),(5-48) quadrupole SeeProb. 5-5andrefer toFig.$-8.Show thatthepotential inthe field of PROBLEMS alinear electric quadrupole is ve[2(2)] $-1. (5.1) The dipole moment ofachargedistribution whosenetchargeiszero. negbd\7?No Showthat,ifthenetchargeQiszero,thenthedipolemomentofa 5-7.(5.4)Multipolar expansion ofthefieldofasinglepointcharge chargedistribution isindependent ofthechoiceoforigin. AsinglepointchargeQissituatedatP’(0,0,s)asinFig.5-7. 52.(5.1) Thedipole moment ofparallel linecharges Firstexpand itspotential atpoint Pinterms ofmultipoles. Thevector r ‘Two linecharges +0and ~Qextend, respectively, from (—a, 0,¢)to thatdefines theposition ofPforms anangle @with thez-axis, andr>>s (a,0,c)andfrom (~a,0,~c)to(a,0,~c).Calculate their dipole moment. ‘Thedistance from QtoPisr’.Disregard terms oftheorder of(s/r)* and higher. Then write outthevalues ofV;,V:,Vs 5-3. (S-1) The dipole moment ofaspherical shell ofcharge . Calculate thedipole moment ofaspherical shellofradius Rbearing a 58(S44) |Thepotential coset0adipolesurface ‘demsity o'=aocos8. Calculate Vforadipole exactly, andidentify thequadrupole andcharge ‘octupoleterms.Theoctupoletermvariesas(s/r)*.Youcantherefore5-4.(5.1)Thedipolemoment ofaspherical shellofcharge disregard termsin(s/r)*,(s/r)*,etc.(a)CalculatethedipolemomentofasphericalshellofradiusRwhose 59,(SAA)‘Thefieldof«chargedcabooarscareporensoll+nOnieoftsgisswat Acubeofside2acarriesauniformvolumechargedensityp.Theorigin(c)Whatisthedipole moment ifthecenter ofthesphere isat ofcoordinates isatthecenter. Caleulate Vi,Vs XE+ ¥9+227 ‘5-10.(5.4.4) ThefieldofalinechargeAlinechargeQextendsfromz=—a/2toz=a/2. ‘5-5.(5.1) Analternate expression forthepotential inthefieldofanelectric (a)Calculate themonopole, dipole, andquadrupole terms inthedipole ‘expansion forV. f ELECTRIC FIELDS IV Energy, Capacitance, and Forces 6.1 THE POTENTIAL ENERGY #OFACHARGEDISTRIBUTION EXPRESSED_ INTERMS OFCHARGES AND POTENTIALS 101 6.4.1THEPOTENTIAL ENERGYOFASETOFPOINTCHARGES 101% 6.1.2 THE POTENTIAL ENERGY OFACONTINUOUS CHARGE . DISTRIBUTION 104 6.1.3 TRUE POINT AND LINE CHARGES 104 52. THE POTENTIAL ENERGY @OF ANELECTRIC CHARGE DISTRIBUTION EXPRESSED INTERMS OF E105 Example: THE POTENTIAL ENERGY €OF ACHARGED CONDUCTING | Fig.58. SPHERE 105 6.2.1 THE POTENTIAL ENERGY #OFACHARGE DISTRIBUTION EXPRESSED (b)Forwhatvalueofthedistancertothecenterofthechargeisthe 6aMECAPACITANCE OEANeOLMTEDCONDUCTOR 10x quadrupole term less than 1%ofthemonopole term, if3n?— 1isofthe Tuer ofunity? Example; THECAPACITANCE OFACONDUCTING SPHERE 108, 4THECAPACITANCE BETWEEN TWOCONDUCTORS 109 ‘S-11.(5.44) Thefieldofasetofsixequal pointcharges Example: THEPARALLEL-PLATE CAPACITOR 109 InFig.5-6,letallthecharges beQ.Calculate V,andVs, Example: CAPACITORS INPARALLEL ANDINSERIES 110 65 ELECTRIC FORCES ON CONDUCTORS 110 5.1 ELECTRIC FORCES AND LINESOFE 112 66CALCULATING ELECTRIC FORCESBYTHEMETHODOFVIRTUALWORK 113 Example: THE PARALLEL-PLATE CAPACITOR 114 67 SUMMARY 115 PROBLEMS 116 This chapter concerns the energy stored inanelectric field and the resulting forces exerted oncharged conductors. Capacitors aredevices designed tostore electric energy. 6.1 THE POTENTIAL ENERGY €OF ACHARGE DISTRIBUTION EXPRESSED IN TERMS OF CHARGES AND POTENTIALS 6.1.1 The Potential Energy ofaSetofPoint Charges Imagine asetofNpoint charges distributed inspace asinFig. 6-1. There ‘urenoother charges intheneighborhood. Agiven charge occupies a 02 6.1THEPOTENTIAL ENERGY #OFACHARGE 103 ha &. ‘Thetotalpotential energy oftheoriginal charge distribution isthen \ \ =8,4GtBt by (63) NN Y =2(042422424,2) "\ dnc rahshe hwJ a ; \[Aor ieNo, +B(or0+24Oty...Qu)al~ _\-}-—— Ane, tsho ry ;fu +2(or0+0+ 2++2). : iFig.6-1.Setofpointcharges eo ™ fw Y 01, Os, Qs... separatedby Qn e distancesrirasFay16 +2%(04040404 ---+0). (6-4) ane, point where thepotential duetotheother charges isV.That particular Wenowrewrite thisarray, adding, totheleftofandbelow thediagonal charge therefore possesses apotential energy, which iseither positive or lineofzeros, terms thatareequal totheir counterparts totheright ofand negative. Thesystem asawhole possesses apotential energy @thatwe above thediagonal. Then every term oftheseries appears twice and shall calculate, ‘Assumethatthechargesremaininequilibrium undertheactionof 2-2(0+22422424+=2a) boththeelectric forces andrestraining mechanical forces. Axe, mon he nw ‘Thepotential energy ofthesystem isequal tothework performed by Or /Q. OO Owtheelectricforcesintheprocessofdispersingthechargesouttoinfinity. +gee(Stor Et ee)After dispersal, thecharges areinfinitely remote fromeachother, and oon fae ON thereiszeropotential energy. (W204 Sip 4H) First,letQ;recedetoinfinityslowly,keepingtheelectricandthe axe,(Ftatetw) mechanical forces inequilibrium. There iszeroacceleration andzero Ov(01.0: 0kineticenergy.Theotherchargesremainfixed.Thedecrease inpotential +o(SiSepOryeyneg0). (6-5)energy €,isequaltoQ,multiplied bythepotential V;duetotheother TeoMuttata Te charges attheoriginal position ofQy Ontheright, thefirstlineisQ,V;, thesecond lineisQ,V2, andsoforth, 0,(OsQ Ow whereV;isthepotential intheundisturbed systemduetoallthecharges &=&(2+24. +22) (6-1) exceptQ,atthepointoccupiedbyQ;.Thus Allthecharges except Q,appear intheseries between parentheses. 2€=Q:Vi +O.Vot QWs++++OVer (6-6) With Q,removed, letQ:recede toinfinity, tosome point infinitely . idistant fromQy.Thedecrease inpotential energy isnow andthepotential energy oftheinitialcharge configuration is, Q:(2Qs Qn’ git =Be(Os,Ory... Oe 6-2) €=5> aw, 7 Boole tt tn) o 2122 ad ee eonforallthe ‘Thereasonforthefactorof}followsfromtheabovecalculation. Letane alrari finallytheNthchargecanstayinposition, si allthechargesbepositive.Thenthepotential atthepositionofagivenitliesinazero fiel charge, justbefore itmoves outtoinfinity, isless(except forQ,)than 104 ELECTRIC FIELDS IV (HEPOTENTIAL ENERGY #OFANELECTRIC CHARGE 105 thepotentialatthesamepointintheoriginalchargedistribution. Onthe e-!{(QJLQG-an)Jarrear=—2_ 19) average, thepotential justbefore removal isone-half thepotential inthe 24, \4aR°/3/L4meoR \2-2K? 20me oR” Ichidistributi nhsenergy‘.whichdoesnotincludetheenergyrequiredtoassemble 1.Riszero,then@isinfinite,whichisnonsense. Electrons aretheindividual chargesthemselves, canbepositive, negative, orzero.For yeesumably truepointcharges, anddealingwiththisabsurdresultposesexample, fortwochargesofthesamesign,#ispositive. Forchargesof \utticultproblems whosesolutions arewellbeyondthescopeofthisbook.Withatruelinecharge, @issimilarly infinite. oppositesigns,isnegative.Forasinglecharge,#iszero. fi ‘linech heref lowedinth Butwhatistheenergy required tosimply modify acharge distribution, tuepoint andlinecharges aretherefore notallowed inthepresent ape ontext. Nonetheless, wefollow theusual custom ofspeaking loosely of withoutdispersing ittoinfinity? Thisenergyisclearlyequaltothefinal idli‘hi henthi Jigibl 1.Tfipotential energyminustheinitialpotential energy, whatever method one pentandlineoewhentheradiusisnegligibly small. Truesurfacemaychoose toeffect thechange. rarges cause noproblems. 6.1.2 The Potential Energy ofa 2THEPOTENTIAL ENERGY @OFAN Continuous Charge Distribution ELECTRIC CHARGE DISTRIBUTION EXPRESSED INTERMS OFE Foracontinuous electric charge distribution, wereplace Q,bypduand thesummation byanintegration overanyvolumevthatcontains allthe Wehaveexpressed thepotential energy €ofachargedistribution incharge: iormsofthechargedensitypandthepotential V.NowbothpandVaret=afVpdv. (68) iwlatedtoE.Soitshouldbepossibletoexpress€solelyintermsofE. “he 7 ‘his iswhat weshall dohere. Weshall find that This integral isequal tothework performed bytheelectric forces in en [Ookgoingfromthegivenchargedistribution tothesituation-where p=0 e-[24 (1) everywhere, bydispersing allthecharge toinfinity, orbyletting positiveandnegative charges coalesce, orbybothprocesses combined. sherethevolumevincludes alltheregionswhereEexists.ThuswecanAtfirstsight,thisisanobviousextension ofthepreviousequation. Itis siculate&byassigning tocachpointinspaceanelectricenergydensitynot,because wehavenowincluded theenergies required toassemble the 1CyE7/2. individual macroscopic charges. Infact, asweshall seeinSec. 6.2,the Since theabove ispositive, thatofEq.6-8isalsoalways positive. above integral isalways positive. These twoequivalent expressions for,oneinterms ofpandVand Observe thatthepotential Vunder theintegral signdoes notinclude theother interms of£,areboth important. They emphasize different, thepartthatoriginates intheelement ofchargepduitself.Wesawin hutcomplementary, aspects ofelectrical phenomena. WiththefirstSec. 3.5that theinfinitesimal clement ofcharge atagiven point spression, @isthepotential energy ofasystem ofcharges; withthe contributes nothing toV. cond, @istheenergy stored inafield. Ifthere aresurface charge densities 0,thentheirstored energy is Wefirstapply theabove formula tothefieldofacharged spherical ductor, and then wegive ageneral proof.é=afoVded, (6-9) at tTvample ‘THEPOTENTIAL ENERGY €OFACHARGED where ofincludes allthesurfaces carrying charge. CONDUCTING SPHERE . We find the potential energy #ofaconducting sphereofradiusR 6.1.3TruePointandLineCharges Carrying charge Qinthrediferent ways Suppose wehave aspherical charge Qofuniform volume density and First method radius R.Then, using thevalue ofV,that wecalculated inSec. 3.7.1, Thewhole charge Qisatthepotential Q/4x¢4R. Then 106 ELECTRIC FIELDS 1 THE POTENTIAL ENERGY @OF AN ELECTRIC CHARGE 107 1,2 @ > ‘Thirdmethod 693OCR anak (612) Letf(r)betheareaofanysurface ofradius r,concentric with theconducting sphere andoutside it,asinFig.6-2.Then,from Secondmethod Gauss’slaw(Sec.3.7), Imagine that the radius ofthe charged spherical conductor . increasesslowlyandeventually becomesinfinite.Thetotal €=10v-sfcobdelar. (15) ‘mechanical workperformed bythechargesisequaltotheinitial lawn le potential energy. Sincethetwointegrals areindependent ofeachother, During theexpansion, thefieldoutside remains unaffected . Gauss's law again!). An element ofcharge ode=€,Edelis , the work performed bythis element, when the radius ofthe . . sphereincreasesbydR,is =f(Garegydr=[SEgrdr(6-17)dé=(cEdet)(=)dR=dv, (613) at!wheredvistheclementofvolumesweptbytheelementofarea [Fo (618) def. After the radius hasexpanded toinfinity, thetotal work performed bythecharges is 8spreviously. ef6Bty-[$( 2)sera oy 6.2.1ThePotential Energy&ofaChargeDistributionIe2 In2\4xeor" SreoR Expressed inTerms ofE:General Proof asabove Weexclude unrealistic cases where Vwould bediscontinuous, forthis ae wouldrequireaninfiniteE.WealsosetV=0atinfinity,whichexcludes us~~ chargesofinfiniteextent.ThenVhasafinitemaximum Viyy.andafinite vy~\ minimum Vouig,WithVingx=O,Vipin0. ia\ LetViunbenegativeandimagineaconductor thatoccupiesallpoints / \ where V=Visiq. Theconductor expands outtotheequipotential Visin+ /\ AV.Thisdoesnotaffecttherestofthefield.Eventually, theconductor /f \ reachestheequipotential V=0.Inthecourseoftheexpansion,any i ig © ‘ \ chargeencountered accumulates onthesurfaceoftheconductor.H 3 \ According tothesecond method above, thework performed bythe ! ee 0. | charges isequaltotheintegral of€9£7/2 overthevolume sweptout.\ x a ! NowletVmaxbegreater thanzero,andimagine another conductor\ e ann] ‘occupying theregion where V=Vinx.Itexpands asabove untilitreaches \ te E: ~ theequipotential V=0. Again theworkperformed istheintegral of \ ha Peg / +E?/2 overthevolume swept out.\ Se / Ifthetwoconductors meet,then,immediately before contact, the \ / surface charge densities areequal inmagnitude (same E),opposite in NN A sign,andatthesamepotential. Theycancel. Thecharge density isnow SA 7 zeroeverywhere, cither because thecharges aredispersed toinfinity or Wass ----7 because positive andnegative charges haveneutralized. Fig.6:2.Spherical conductor carrying acharge Q,andacon- Theinitial stored energy isthusgivencorrectly bytheintegral ofcentric imaginary spherical surface ofareas4(r). EP 108 FLECTRIC FIELDS IV. 4-THE CAPACITANCE BETWEEN TWO CONDUCTORS 109 6.3THE CAPACITANCE OFAN CudneR=111%10-"Rfared ISOLATED CONDUCTOR | FiI1R plcofarads (20 Imagine afinite conductor situated alongdistance fromanyother body 6.4THE CAPACITANCE BETWEEN andcarrying acharge Q.IfQchanges, theconductor's potential also TWO CONDUCTORS changes. Asweshall see,theratio Q/V isaconstant. Thecapacitance of theisolated conductor is Wenow have twouncharged isolated conductors. Transferring acharge cue 19) Qfromonetotheotherestablishes apotential difference VbetweenVv them.Bydefinition, thecapacitance between theconductors isQ/V.The Thus thecapacitance ofanisolated conductor isequal totheextra charge capacitance depends solely onthegeometry oftheconductors andon required toincrease itspotential by1volt. The unitofcapacitance isthe their relative positions. farad, orcoulomb pervolt. Pairs ofconductors arranged specifically topossess capacitance are Theenergy stored inthefield ofanisolated conductor is called capacitors. é-&=Ss-¢ (6-20) Vxample|THEPARALLEL-PLATE CAPACITOR Wenowshowthatthecapacitance Cofanisolatedconductor depends “Deralchplatecapacitor(Fig.6-3)consistsoftwoconductingsolelyonitssizeandshape.Theconductorisinair.Thepotentialinthe EhangesQand%WonenectseenanceEine,platescarryregion surrounding theconductor isV(x, y,z). Itobeys Laplace's‘equation, anditiszeroatinfinitybecause theconductor isoffinitesize, e-2e2 Ve& (6-22)byhypothesis. Atthesurfaceoftheconductor thechargedensityois Then weEo od€9E (Sec. 4.4), or€times therate ofchange ofV(x, y,z) inthedirection normal tothesurface. Animmediate consequence isthatthe cat (623) valueoftheconductor potential determines thesurface charge density ‘Also,thestored .o=—€)|¥V| ontheconductor. Therefore theconductor potential also +hestorenenergyiedetermines thetotalcharge Qontheconductor. eoQV_CV_O_ O's (624)Observe nowthattheequation V*V=0islinear, sothatanymultiple 222C2east" “ ofV(x,y,2)isalsoasolution. IfVincreases everywhere bysomefactor on . .a,thisnewVobeysLaplace’s equationoutsidetheconductor, iszeroat é=fEy=2(2) ds=2S (625)infinity, andisequaltoaV,ontheconductor. Furthermore, itistheonly 2 2Neosat) eos” continuous function ofx,y,zthatsatisfies thesethreeconditions. We conclude that iftheconductor's potential increases bythefactor a,then Vincreases everywhere bythesamefactora,andbothoandQlikewise increase bythesame factor. - ThechargeQonanisolated conductor isthusproportional toits 5voltage,anditscapacitanceC=@Q/Vdependssolelyonthesizeand vt LEshapeoftheconductor. 5aExample THE CAPACITANCE OFA —aeCONDUCTINGSPHERE|Ifanisolated conducting sphere ofradius Rcarries acharge Q, thepotential atitssurface isQ/47¢0R and Fig.6-3.Parallel-plate capacitor connected toabattery. 110 m A fi ° aa Vy" Ql . © a Qi |e: oe ys c ry i-O| fo Re ao ms Fig.6-5.Thelocalsurfacechargedensityoonaconductor givestwooppositeldirected electric feds,asshownbythetwoarrows ontheleftallheole Fig.6-4.(a)Two capacitors connected inparallel. (b)Two capac- charges together givethefieldshown ontheright. Thenetresult isafield itors connected inseries. ~trength ofo/¢, outside andzero inside. Theunitvector Apoints outward, as usual Example CAPACITORS INPARALLEL ANDINSERIES Capacitors connected inparallel share thesame voltage. Thus, for thesurface. From Gauss’s law,thisEiso/¢9. Now theforce on0dedis twocapacitors, asinFig.6-4(a), notEodef,because thefieldthatactsonodfisonlythatoftheother CaM O_O O_o ye 629 charges inthesystem,Vv voev us ~ WecanfindthefieldofodsitselffromGauss’s law.ThefluxofE Capacitors connected inseries carry thesame charges, asinFig. emerging from odoisodof/€o, halfofitinward andhalfoutward, asin 6-4(b). Then Fig.6-5. Then odsf provides exactly half thetotal Eatapoint outside, y-2, 2.2 2 closetothesurface, andcancels thefieldoftheothercharges, inside."ota Therefore theEacting onoddis/2€o, andtheforceontheelement ti. CG ofarea dfoftheconductor isfeted, coOS (628) Coa OG> ? Socapacitors connected inparallel addasresistors inseries,and dF=5%odsl=5doh (629) capacitors connected inseries addasresistors inparallel. Thesurface forcedensity is 6.5ELECTRIC FORCES ON CONDUCTORS ©Survace Foes Censty 'S ,_dF_of_ek 2 Anelement ofcharge odfonthesurface ofaconductor experiences the Feta de) MeMtons/meter (6-30)electric field ofalltheother charges andistherefore subjected toan . electric force, Under staticconditions thisforceisperpendicular tothe Theforceperunitareaonaconductor isequal totheenergy density in surface, forotherwise there would beatangential field andatangential thefield. current. Theforce alsoactsontheconductor, towhich odefisbound by Thenetelectrostatic force onaconductor ofareasfis internal electric forces. Tocalculate themagnitude oftheelectric force, consider aconductor eof pscarryingasurfacechargedensityowithanelectricfieldstrengthEnear Fes$edA, (6-31) 12 ELECTRIC FIELDS IV 113 where the vector dsf points outward. The local electric force tends to pulltheconductor into thefield. Inother words, anelectrostatic field vennse ‘exertsanegative pressure onaconductor. Thenetforceonaconductor Yo ~~depends onthewayoandEvaryalongitssurface. Inair,orina i, \ vacuum, electric forces areusually negligible. However, theycanbequite y \ large indielectrics. Y i - F i 6.5.1 Electric Forces and Lines ofE \\ / Figure 6-6shows linesofEforfourpairsoflinecharges. SD 4 InFig.6-6(a) and(c),theforce isattractive andthelines ofEare Sneee7 clearly “under tension.” Indeed, thetensile force per square meter is €oE7/2, aswecaninfer from theprevious section. © InFig. 6-6(b) and (d), ontheother hand, wecan see lines ofE “repelling”eachotherlaterally.Therepulsivesurfaceforcedensityinthe \/ . * Vig.6-6.LinesofEbetween pairsoflinecharges. Itisusefultovisualize @ wiractive electric forces asbeing caused byatension inthelinesofE,asin(a) ‘od(€).Similarly, repulsive electric forces may bethought ofasbeing caused by“lateralrepulsionbetweenlinesofE,asin(b)and(d).In(c)and(4)thecharge obapefal : »theleftistwiceaslargeastheother. ORT A segion where thelinesofEareparallelisalso€9E2/2.SeeProb.6-8. POST peso Later weshall seethat magnetic fields behave similarly. SAR TESS 6CALCULATING ELECTRIC FORCES BYTHE SOGPLTTTT ISS METHOD OFVIRTUAL WORK PL AVVAX Wecanalsocalculate electric forces bythemethod ofvirtual work.This method consists inpostulating aninfinitesimal displacement and then ® \nplying theprinciple ofconservation ofenergy. Wefirstdefine asystem, 4 ELECTRIC FIELDS 1 SUMMARY us and then wecalculate theenergy fed into itinthe course ofthe Here displacement. This energy isequal totheincrease intheinternal energy aol a ofthesystem. bp=V(VdC)= Vd=VietF=(eyMobds).(6-34) The method ofvirtual work isageneral andreliable method for . calculating forces, butontwoconditions: (1)onemustbeperfectly clear Ifdsispositive, dCisnegative andenergy flowsfromthecapacitor intothebattery aboutexactly whatsystem oneistalking about, and(2)onemustbe Prsaly nceBeVin particularly careful tousetheproper signs yaince BV cE? (eos) lV2/_ ds) eu®Example |THEPARALLEL-PLATE CAPACITOR fend(Sas)=d(H) =AE(-S)=- Saas ‘Aparallel-plate capacitor isconnected toabattery supplying @ (635) fixed voltage V.SeeFig.67. Weassume thatthedistance s Note thenegative sign! Theenergy density €eE%/2V decreases virtual work. Wearegoing tocalculate energies related tothe Thus capacitor. Thus thenetenergy fedintothecapacitor inthecourse netofthedisplacement dsofthetopplatewillbeequaltothe Fds~ealds=~otds, (6-36) increase intheelectric energy stored inthecapacitor. 2 Let@>bethemechanical energy fedintothecapacitor bythe eel! force F,let%ybetheelectric energy fedinbythebattery, andlet Fa", (637) 4,betheincrease intheelectric energy ofthecapacitor. Then 2 by+hyn (63) antheforeperuiareasequalftheneenitvolume withHalftheenergy supplied bythesource becomes mechanical &=Fds, t= Vd0 (633) energy, and the other half becomes electric energy. This isa where dQistheextracharge fedintothecapacitor bythebattery general rule 7 SUMMARY F ‘hepotential energy ofacharge distribution isgiven byeither one oftwo ~of ntegrals: (| 2 \ @=3[Vode (6-8) i a“ = ! “h |e » a” oe : “si&we—_ogLo| e=a]EE?dv. (611) aD ) ; EE Iuthefirst integral thevolume vcontains allthecharges, while inthe second itincludes allthe field. The assignment ofanelectric energy density. €oE?/2 toevery point inspace leads tothecorrect potential Fig.6-7.Parallel-plate capacitor connected toabattery supplying energy ofacharge distribution. afixed voltage V.Wecancalculate theforce ofattraction between Ifthepotential ofanisolated conductor isVwhen itscharge isQ,then theplatesbyimagininganequalbutPeoreFed uscapacitance CisQ/Vfarads.ThisquantitydependssolelyontheCurveremindsusthatweapplytheprineipleoftheconservation of vcometry oftheconductor. Thecapacitance between twoconductors isenergy tothecapacitor alone. uystin Q/V, where Qisnow thecharge transferred from one totheother 116 ELECTRIC FIELDS 1V PROBLEMS 47 andVisthepotential between them. Inboth cases thestored energy is According tothereciprocity theorem forelectrostatics, 92¥_cV?_ 0 won QV)+OV5+O5VS+==O1Vi+OV+OW,+o>,ee 26 (6-20) or Dov'=Dorv. Theforceperunitareaexerted onaconductor situated inanelectric Weshallfindanalogous reciprocity theorems inChaps. 8and27. fieldisequal totheelectric energy density atthesurface: ‘Youcanprovethistheorem bycalculating theenergy required tochange thecharges from QtoQ’andequating thisenergy to,Q’V'/2~-¥, OV/2.r-oF?_ Todothis,setthechargeandvoltage onconductor 1equalto(1—x)Q: +"26," (6-30) xQ}and(1—x)V,+xVj. Thenyoucangofromonestatetotheotherby letting xgofrom zero tounity whereaisthesurfacechargedensity 65.(6.4)Cylindrical capacitorItisusefultovisualize electricforcesasbeingcausedbylinesofEthat (a)Showthatthecapacitance perunitlengthofacylindrical capacitor isareundertension andthatrepeleachotherlaterally '&axeafin(Ro/R:), whereRyandRearetheinnerandouterraiWecanalsocalculate theelectric forceonaconductor bytheprinciple (©)Calculate thecapactance permeterwhenRa/R,=¢=2.718 ofvirtual work, which issimply theprinciple ofconservation ofenergy 6-6.(64) Spherical capacitor applied toaninfinitesimal disturbance ofasystem. 4)Showtatthecapacitance ofaspherical capacitor ofinnerandouter caSreeRe PROBLEMS RR” 6-1.(6.1) Thedipole andthequadrupole (b)Calculate thecapacitance when R,=100 millimeters and R:= Calculate thepotential energies ofanelectric dipole andofanelectric 200millimeters. quadrupole, 6-7.(64)Connectingchargedcapacitorsinparallel 62.(6.1) Thepotential energy ofasphere ofcharge ‘Twocapacitors C,andC,arecharged tovoltages V,andVs,respectively, (a)Calculate theelectric potential energy ofasphere ofradius R andthenconnected inparallel, positive terminal topositive terminal and carrying atotal charge Quniformly distributed throughout itsvolume. negative tonegative (b)Calculate thegravitational potential energy ofasphere ofradius R' (a)What isthefinal voltage? andtotal mass M. (b)What happens tothestored energy? (©)Calculate thegravitational potential energy ofthemoon. SeeTheOl ceatthee° 68.(65.1)ElectricforcesandlinesofETableofPhysicalConstantsattheendofthebook. (@)UsethemethodofSee.6.5.1tocalculatetheforceofattractions youcanassemble asphereofprotons withadensity between twochargesQand~Oseparated byadistance 2D. equal tothat ofwater. What wouldbetheradiusofthissphereifitselectric (b)Repeatthecalculation fortwochargesofequalsign. potentialenergyweresufficienttoblowupthemoon? Pe nBesoFquanSED. (©)What isthevoltage atthesurface ofthesphere ofprotons? 69,(6.5) High-voltage generator . Imagine thefollowing mechanism forgenerating highvoltages. Oneplate 63.(6.2) Theenergyinthefieldofasphereofcharge ‘ofaparallel-plate capacitorisfixedandconnected toground.Theother Asphere ofradius Rcontains acharge Q,uniformly distributed lateismovable. When theplatesareclose together atadistance s,athroughout itsvolume,Calculate(a)theenergy,(b)theenergystoredin g , fe : t ve " contact closes andthemovable plate charges tothevoltage V.Then the thefieldinsidethesphere.and(c)theenergystoredinthefieldoutsidethe themovableplat roadist iitsvoltsphere,Thereisfivetimesmoreenergyoutsidethaninside deeceedintecardine cddeeclfecte.Atthisnointanothercontact v ey increases tonV,disregarding edgeeffects. Atthispoint another contact 6-4. (6.4) The reciprocity theorem forclectrostaties closes, and the movable plate discharges toground through aload Consider asetofnconductors ofarbitrary sizes, shapes, andpositions. resistance R.Conductors 1,2,3,...carrychargesQ;,Qs,Qs,...,andtheirvoltages (a)Verifythatthereisconservation ofenergy.areVi,VorVs... Without disturbing theconductors, youchange the (b)Canyousuggestamoreconvenientgeometryforsuchahigh-voltage charges toQi,5,Qi generator? 610. (6.5) Thesurface fore ona ballon carrying aneletic charge itissuggested thataballoon made ofight conducting material could bekeptapproximately sphericalbyconnetingitoahigh-voltage supply.The 7 balloon has adiameter of100 millimeters, and the maximum breakdown «HAPTER field inairis3megavolts/meter. (a)What isthemaximum permissible voltage? (b) What gas pressure, inatmospheres, inside the balloon would have * ant, a8 : ELECTRIC FIELDS V (c)How large could thesurface mass density oftheballoon be? Electric Circuits A: RC Circuits 6-11, (66). Stored energy Four charges +0, -Q, +Q, ~Qoccupy thecorners ofasquare ofside 4,withthe poste charges‘on onediagonal andthe negative charges on 1)SYMBOLS AND DEFINTTIONS 20 theother. 2KIRCHHOFF’S LAWSFORDIRECT CURRENTS —121(a)Calculate thestoredenergy @,andsketchacurveof#asafunction 2.1)THEMESHMETHOD FORCALCULATING BRANCH CURRENTS 122 ofa Example; SIMPLE TWO-MESH RESISTIVE CIRCUIT WITHAVOLTAGE(b)Amechanism constrains thechargestostayatthecornersofa SOURCE 122"aCCateteforcesonthechargerthemethodofvirtualwork Example:SIMPLETWO-MESHRESISTIVECIRCUITWITHACURRENT (a)Compare withthevalues deduced from Coulomb's law. SOURCE 123Evonple:SIMPLETWO-MESH RCCIRCUITWITHAVOLTAGESOURCE. 6-12.(6.6) Theforces ontheplates ofaparallel-plate capacitor SOLVING ANORDINARY DIFFERENTIAL EQUATION WITH CONSTANT Showthatthe foreofatraction between theplates of paralle-pate SOLVINGANORDINA capacitorthatisnotconnected toabattery is€,£*sf/2, asinSec.6.6. 22THENODE METHOD FORCALCULATING NODE VOLTAGES 126 6-13. (6.6) Halfthebattery energy becomes mechanical work, andtheother Example: SIMPLE TWO-MESH RESISTIVECIRCUIT WITHAVOLTAGE halfstored inthe electric Rel Example: SIMPL Rewrite Eq.6-32intheform =~ @,,andshow that, ifdsis \ALTERNATING CURRENTS 127negativeonehloftheenergysuppliedbythebaerybecomescletie UU-THREE.WIRE SINGLE-PHASE ALTERNATING CURRENTS 12873.2THREE-PHASE ALTERNATING CURRENT =128 1° ALTERNATING CURRENTSINRESISTORS 130SALTERNATING CURRENTSINCAPACITORS 132 THEINPEDANCE 2AND THE ADMITTANCE Y 134EuompleSIMPLERCCIRCUTT 15 >POWERINACCIRCUTTS 137 KIRCHHOFFS LAWSFORALTERNATING-CURRENTCIRCUTTS 138rample:SIMPLERCCIRCUIT SOLVEDBYBOTHTHEMESHANDNODEMETHODS 138 o SUMMARY 139 PROBLEMS 4 this isthefirst offour chapters onelectric circuits. For themoment, we limit ourselves toresistive and capacitive components. The next chapter , sleals with circuit theorems and isgeneral, even though itappears to +coapten and atentrequted forwhat flows, However, Chap. saprerequisite 120 ELECTRIC FIELDS V rat suffer from thesame limitation. Wehave towaituntil Chaps. 24and25 tostudyinductive circuitsandtransformers. oFUnless specified otherwise, weassume that allcircuit components are linear and time-independent Ifyouarenotfamiliar withphasors, youshouldreadChap.2before Fig.73.Typicalcurveofoutputworking onthematerial onalternating currents thatstarts inSec.7.3. current asafunctionofoutput voltage for acommercial current- 7.1 SYMBOLS AND DEFINITIONS =F stabilized powersupply. Figure 7-1shows thecircuit-related standard symbols thatweuseinthis book. Anideal voltage source supplies avoltage thatisindependent of thecurrent drawn. Agood commercial voltage-stabilized power supply is close toideal, uptoaspecified current, afterwhich thevoltage drops off, ~s a asinFig.7-2.Similarly, anideal current source supplies acurrent thatis independent oftheoutput voltage. Commercial current-stabilized power supplies arenearly ideal,uptoaspecified voltage, beyond whichthe _ —currentdrops,asinFig.7-3, /—\ /)Figure7-4showspartofanelectriccircuit.PointssuchasA,B,Care a) _ (.callednodes;connections betweennodessuchasAB,orBC,are Na /)aoebranches; andaclosedcircuitsuchasABCisamesh ty EomsaQuinLOEO--O- —~ Ane i (4) NWA AE OSTTTIN IOITI -0o } \ paccaret Fig.7-4.Sectionofanelectriccircuit.Thismeshhasthreebranchesandthreessi nodes. Fig.7-1.Symbols forcircuitcomponents Anactivecircuitcomprises sources, whileapassive circuitdoesnot.A circuit component issaid tobelinear ifthecurrent passing through itis proportional totheapplied voltage. 4 Byconvention, current flows from plus tominus, outside asource, in {-—————— thedirection opposite totheelectron drift. | ‘ 7.2 KIRCHHOFF’S LAWS FOR DIRECT CURRENTS Fig.7-2.Typicalcurveofoutput Kirchhoff’s twolawsareself-evident andappear trivial.Theyare,infact,Yotage28aFunctionofoutput thefundamental lawsofcircuittheory,andtheyarecompletely general: — 1 Stabilized power supply. they arevalid forboth linear and nonlinear, and forboth passive and I12 ELECTRICFIELDSV| 123active circuits. They remain valid even ifthecomponents aretime- |) ‘ .dependent oriftheydisplay hysteresis.+ t 4ost‘Theyalsoapply,whatever bethetimedependence ofthevoltagesor| currents orwhatever betheinitialconditions. Allcircuittheorems derive, | _ 4 — insome way, from these two laws. Kirchhoff’s laws serve tocalculate }} —— \ branch currents andnodevoltages incircuits comprising knownsources +L/ \ \g,andknown components. nnWedisregard alternating currentsuntilSec.7.3. \ J \Z The Kirchhoff current law(KCL) states that thealgebraic sum ofthe currents entering anodeisequaltozero.Thisstands toreason because, » otherwise, charge would accumulate atthenodes. Anode, byitself, has zero capacitance toground. Ifagiven connection Apossesses an 1 appreciable capacitance toground, then thatcapacitance must appear on 7 thecircuit asaseparate branch between Aandground. t Fig.7-5.Two-mesh resistive circuit fedbyasource supplying a‘TheKirchhoff voltage law(KVL) isequally obvious: thesumofthe! steady vohage V voltage dropsround ameshisequaltozero. { 5 ‘ x ‘ 7.2.1 The Mesh Method forCalculating Branch Currents } With themesh method oneassigns mesh currents lL,ly,ke, ---,a8inFig. ~~ — 7-4.Thiscurrentdistribution satisfiestheKCLautomatically. Themeshes Oo f,\maybechosen arbitrarily, aslongasnobranch isleftout.ThentheKVL g ‘ *\ )s* provides oneequation foreachmesh. Solving yields themeshcurrents, \ Z \ Z thebranch currents, andthen thenode voltages, asweshall seeinthe ‘examples below. Ifthecircuit comprises onlysources andresistors, thentheequations o arealgebraic. However, ifthere arealsocapacitors, then theequations + involve time derivatives. = Fig.7-6.Two-mesh resistive circuit fedbyasource supplying @ Example |SIMPLE TWO-MESH RESISTIVE CIRCUIT WITHA steadycurrent[, VOLTAGE SOURCEL20.6 heo2Vv (73)Refer toFig.7-5.Choosing mesh currents asinthefigure, we R R apply theKVLtothetwomeshes, stating atBinbothcases. For “Thedownward current inbranch ABis0.4V/R amperes, andthus mesha, thevoltage atAis0.4Vvolts. V-LR-(,-1)R=0. 7-1) Formeshb, Example |SIMPLE TWO-MESH RESISTIVE CIRCUIT WITHA ~(h-W)R~24R=0. 2 CURRENT SOURCE ‘ We now have the circuit ofFig. 7-6, with acurrent source Known, butthevoltage Vacross thesource isunknown, Equations on i 7-1and7-2stillapply. Solving, wefind that +Hysteresis isaproperty ofcertain devices oFmedia whose parameters depend ontheir 1 previous history. Ferroelectric mediaarehysteretic (Sec.10.1.5. V=SLR hao (7-4) 124 ELECTRIC FIELDS V ).2KIRCHHOFF'S LAWS FOR DIRECT CURRENTS 125 Example |SIMPLE TWO-MESH RCCIRCUIT WITHA ‘Asweshallsee,thereexistaninfinitenumberofsolutionsQ(t)VOLTAGE SOURCE. SOLVING AN ORDINARY thatsatisfy thisequation. However, itisasimple matter toselect DIFFERENTIAL EQUATION WITH CONSTANT thecorrect one,inaparticular situation.COEFFICIENTS WefirstfindoneobvioussolutionthatwecallQ,, cv ‘ThecircuitofFig.7issimilartothatofFig.7-5,exceptthatwe => (7-10) haveadded aswitch and substituted acapacitor forone ofthe resistors. Theswitch closes at1=0. Themesh equations arenow ‘Youcaneasily check thatthisisasolution,bysubstitution. Thisis afollows for1=0: theparticular solution. We found itbyconserving only thelast termontheleft. V~LR~(L.—h)R=0,(7-5) NowletQbesomeothersolution. Bydefinition, italsosatisfies Hh=I)Rbe20 6) Eq,7-9.Thenthefunction2-0-0, ny Rewriting,withJ,=dQ/deif I,andQarechosen asinthefigure, satisfies thehomogeneous equation 40 dUR-RE2=v, a) AQ,4Qe z dt 3kdt+26 0. (7-12) LR-2R 2-226, Thefuncti thle L 2. 78) function Q,isthecomplementary solution.ac Inotherwords,thegeneralsolution QisthesumofQ,andQ.: Eliminating nowJ,yields 0-0,+0. (7-13) sn 2422-y os) Nowthesolution ofEq.7-12issimple. Rewriting, weget :© de2 (7-14) ‘This isanordinary differential equation withconstant “ae aCe oy coefficients. The term ordinary refers tothefactthat theequation and comprises ordinary, andnotpartial, derivatives. The equation is . 2nomhomogencousFeeuse irighthand sideinotsfenton of 2.=Aes0(~ 3): os ©.Itislinear because itcomprises only thefirstpowers ofthevariable Qandofitsderivatives andnoproducts suchasQor where Aisaconstant ofintegration. Q(dO/dt), ‘Thus ra a ee 0-+sew(-24) 16) = Y= Since Aisany constant, there exists aninfinite number of solutions. For given conditions, thevalue ofAfollows from, say, — - thevalueofQat1=0.Inthisexample,Q=0at1=0,and (~~4 a~| A=~-CV/2. Thus + \ / \ ve{« )OR ne ~. -&[~exp(-2} z 7|ysl ) o-F[1-e9(-3%) 7)\ ra Figure7-8(a)shows@asafunctionoftforCV=2andRC=2. For1>3RC/2, Q~1. Then /,=0 andthevoltage across the Ocapacitor isequal tothevoltage between Aand ground. Figure 7-8(b)shows1,=dQ/dtasafunctionof¢. =We have here acircuit that can exist intwo states, first with S Fig.7-7.RCcircuit fedbyavoltage source. Switch Sremains open and then with closed. With $open and until 1=0, ‘open until ¢=0. L=V/2R, 1,=0, Q=0.With$closedand1>>3RC/2, thesame 126 3ALTERNATING CURRENTS vd hoy Example SIMPLE TWO-MESH RESISTIVE CIRCUIT WITH A VOLTAGE SOURCE Refer again toFig. 7-5. According totheKCL, f,=f+Jy,with v-v y, v, os aYoM aM als 0. 4rR bR bhR (7-18) ‘Therefore V~Va_ Va Va \ n 1 arentagMe 7-19) ' Solving, wefind that “ V,=0.4V, (7-20) asinthefirst example inSec. 7.2.1. The values ofthebranch 03 currents {,,f,fsfollow immediately. Observe that with thenode method wehad asingle equation to solve, namely Eq. 7-19, versus two with themesh method that we oe used forthesame circuit inthefirst example inSec. 7.2.1. With h thisparticular circuit, thenode method issimpler. " 7.3 ALTERNATING CURRENTS o 1 4, Most electric andmagnetic devices operate witheither fluctuating or : alternating currents. There aremany reasons forthis, butthetwomajor wo ones have todowith power technology and with thetransmission and storotFig.27 processingofinformation. fon ofthetinesferCV22,RC=D(O)Thecurreninmesh First,withalternatingcurrents,theelectricpowersuppliedbyasource basafunctionofthetime. atagivenvoltagecanbemadeavailable atalmostanyotherconvenientvoltage bymeans oftransformers. This makes electric power adaptable to valuesofI,andfyapply,butQ=CV/2.Intheinterval between «broadvariety ofuses.(Thepower supplied byadirect-current source1=0and£>>3RC/2, thecircuit adapts itselftothenewsituation. canalsobechanged fromonevoltage toanother. Thisisdone byfirst Phenomena thatoccur during such periods ofadaptation are switching thecurrent periodically, toobtain analternating current, next called transients. tveding thistoatransformer, and then rectifying andfiltering theoutput. - Theoperation isrelatively costly.) 7.2.2TheNode Method forCalculating Node Voltages ‘Thesecond reason forusingalternating, orfluctuating, currents isthat Thenodemethod isanalternative tothemeshmethod. Asarule,oneof theycantransmit information. Forexample, amicrophone transforms thetwo methods requires asmaller number ofsimultaneous equations theinformation contained inaspoken word intoacomplex fluctuating. and istherefore preferable. current, Weassume that one ofthenodes isatground potential (V=0)and We assume that thevoltages, currents, and charges areallcosine thattheother node voltages V,,V,,V., areunknown. Application of functions ofthetime, with appropriate phases. This isnotalways the theKCL ateach node insuccession yields thenode voltages andthus the case, Thecurrent through amicrophone isnotnormally sinusoidal. Or branch currents. ‘onemight have only thepositive part ofthecosine function oracosine Weillustrate thenodemethod withasingle example. function whose amplitude isafunction ofthetime. However, any 128 PLECTRIC FIELDS V us periodic function isthesumofaninfinite series ofsineandcosine terms ° called aFourier series (Sec. 11.3, example). a : A R, 7.3.1Three-Wire Single-Phase Alternating Currents ©) InNorth America electric utilities supply electric power toindividual houses andapartments atboth 120and240volts rms, 60hertz, asinFig. 7-9.Thisisthethree-wire single-phase alternating-current system.‘Thus _ & V4=120x1.414c0s(2x601)=170.cos(22x601),(7-21) 1 Oya Vo==120x1.414c0s(27X601)=~170¢0s(27x601),(7-22) F < V4—Ve=240x1.414cos(221x60t)=340cos(27x608). (7-23) o Low-power devicesoperate at120voltsrmsacrossADorCD,while high-power devices such aswater heaters operate at240volts rmsacross ‘ Aand C. () R 4,4> f “OL a” R ©La NeaS v= ’ 8 Rwy Fig.7-10. (a)Three identical sources ofalternating current with phase differences of120°, feeding three resistors R,,R.,Ry.(b)Three-phase supply. If ~ theresistances areequal,theconnection DEisunnecessary. a) reflect their relative phases. For example, ifthe phase atAiszero at 1=0, then ¢ 4h La=Vin mM, (7-24) Fig.7-9.Three-wiresingle-phaseacsystem.Theplusandminussignsmeanthat v,cos> (724) 1two sourceshaveoppositephases.IfRy=RsionBDis a thetwo ourceshaveopposephases.If,=Ri,theconectionBD p=Vgcos(or+2), 728) . 3 4a Veo=Vo00s(an+52) (7-26) 7.3.2 Three-Phase Alternating Current Figure 7-10(a) shows three sources ofalternating current feeding resistors Ifthethree resistances areequal, thesumofthecurrents flowing inthe Ry,Ra,Rs.Wehaveoriented thesources 120°apart onthefigure soasto three grounded wires is | Vow 2a 4x 1Ta|coswr+0s(on+24)+cos(or+“2)]=0=72) ‘Then thewire DEinFig. 7-10(b) canbedispensed with. Iftheresistances areunequal, thecurrents donotcompletely cancel inDE.Wenowhave ‘ bd fourwires doing thework ofsix,with low/°Rlosses inthewire DE. Asetofthree sources, star-connected and phased asinFig. 7-10(b), supplies three-phase alternating current. The main advantage ofthree- phase current isthatitcangenerate therevolving magnetic fields oflarge co) electric motors. See Prob. 18-6. ‘ Electric power stations usually generate three-phase alternating cur- rent. That iswhy ordinary high-voltage transmission lines have either three orsixwires, plus one ortwo light wires. Three-wire single-phase power isobtained from atransformer whose primary isconnected \ v, Vat,emanbetween onephase ofathree-phase lineandground. eR \ a Except forafewproblems attheendofthischapter, weshall be \ concerned henceforth solely with single-phase currents. 7.4ALTERNATING CURRENTS INRESISTORS " }+ —t Figure 7-11(a) shows analternating voltage source connected toa' resistor. Let . V=V,,00sot, (728) ‘ where Vj,isthe maximum, orpeak, voltage; w=2zf isthe circular frequency, expressed inradians persecond; andfisthefrequency \ expressed inhertz. Inthefigure, w Vp,€08wt Fig,7-11.(a)ResistorRconnected toasourcesupplyingavoltageVandapay coset, Iga (7-29) current.(b)VandIplotted asfunctions offoraresistor. R R ‘Thecurrent isinphasewiththeapplied voltage, asinFig.7-11(b). Pyy=VenVin Hn| (7.31) ‘Atanygiveninstant thepower dissipated intheresistor is 22R 2 Vim 2Pigs=VI=(VoyCOSO1)(Lp COSWt)=VingCOS?ot =Vemalems =po=Fras (7-32) v2= cos?wt=PRcos?wt (7-30) asinFig,7-12. R Recall that thesubscript “rms” stands forroot mean square, orthe : squarerootofthemeanvalueofthesquare(Sec.2.4).Inpractice, the Since theaverage value ofcos?wtis4,theaverage power dissipation is subscripts “av” and“rms” prove tobeunnecessary, andwesimply write 132 133 Pras ®Ue ut 1 2 % ARAAS1» v < ‘a ° + 7 * rs ’ 1wcrc (ur47. ene . VaVcomae (+3) Fig. 7-12. Instantaneous and average power dissipated inaresistor. Vv v2 \ =v payraVep : \ aswith direct currents. We shall nonetheless use thesubscript “rms” an 4 wherever itapplies. ° ° Unless stated explicitly otherwise, allvoltmeters and ammeters show rms values. ra Ifthevoltage orcurrent isanalternating butnotasinusoidal function ofthetime, then theratio Vies/Vm, depends onthenature ofthefunction. Forexample, ifavoltage simply alternates between +VqandVy,the i” » ims value isVp. Fig.7-13. (a)Capacitor Cconnected toasource supplying avoltage V.The circledVandJrepresent avoltmeter andanammeter. (b)Vand/asfunctions of 7.5ALTERNATING CURRENTS INCAPACITORS thetimeforacapacitor.ThecurrentleadsthevoltagebyO°andw=0.5,here Figure 7-13(a) shows asource ofalternating current connected toa itor. Thicapacitor. then whereJ,Q,Vareallphasors, Atagiveninstant theenergy storedinthe O=Cv, conan, 3) capacitor is _4Q_ a ca yCV?_CV;,cos? wt 1=B=—CVsinwt=CV005(oF+3). (735) bag=F=Lar (737) ‘Thecurrent leads thevoltage by2/2radians, asinFig,7-13(b). andtheaverage stored energy is Inphasor notation (Chap. 2), CV3/2_ CV2ag1=jwQ =jwCV, (7-36) nar (7-38) | 7.6 THE IMPEDANCE ZAND THE ADMITTANCE Y Byanalogy with Ohm’s lawfordecircuits, wehave Ohm’s lawforac,circuits: k Vv=v . wex|= 2 - 1=5. (7-39) %, 6, c where both [and Varephasors and Zistheimpedance ofthecircuit ‘The impedance ofaresistance isR. ° ° Foracapacitor, from Eq. 7-36, j o ©arma (7-40) Fig.7-14,(a)Anypassiveandlinearcircuitcomprising onlyresistorsandcapacitors. Theimpedance attheinput terminals isR+jX.(b)Equivalent series ; ; circuit. (c)Equivalent parallel circuit. Impedances inseries and inparallel operate like resistances inseries andinparallel Impedances areingeneral complex andthusoftheform whereGistheconductance andBisthesusceptance. Theadmittance Y Z=R+ix, (7-41) ofacapacitor isjwC, anditssusceptance BiswC. Any two-terminal passive and linear RCcircuit isequivalent toa whereXisthereactance. Given atwo-terminal passive andlinear circuit conductance inparallel with asusceptance asinFig.7-14(c). You can that comprises only resistors and capacitors, however complex, the easily show that impedance between itsterminals isacomplex number R+jX. ‘Thisreactance Xisnegativeforthefollowingreason.Thereactance X o-—R_-R peXX (745) ofasinglecapacitance Cisnegative: R+X? Z" R4+X? Zz GG B B 11 R-—2-8 =--8_._8 =peeX=-6 (7-2) GrpyyXOaye yO Nowthecurrent inacapacitor leadstheapplied voltage, andintuitively, ‘Thequantity Gispositive, likeR,while Biseither positive ornegative, youcanguessthat,whatever thearrangement ofresistors andcapacitors, likeX. thecurrent attheinput terminals willleadthevoltage. Thus the Ifagiven circuit comprises onlyresistors andcapacitors, thenXisreactance forthecomplete circuitwillbenegative, orcapacitive. negativeandBispositive.Also, ThusthecircuitofFig.7-14(a) isequivalent tothatofFig.7-14(b) with 1_R4X \ RaGaa (7-47) R=R G=-—y (7-43) ox B x V=0C, ¢,=2=-—*__. (7-48) Theinverse ofanimpedance isanadmittance: oR X) 14 Example |SIMPLE RCCIRCUITY=5=5-—=G+iB, 7-44)ZRix ?*! ow InthecircuitofFig.7-15(a)thesourcefeedsanimpedance e e ; components ofsuch equivalent circuits arealmost invariably Z+ frequency-dependent. Thecircuit ofFig.7-15(a) isalsoequivalent tothatofFig. e 7-15(c), withG’andB’givenbyEq.7-45: ++ rR _aR yeLX gg, v & cay = Cape z8=-ggat jpOH) ~ c | wherejX"=1/(j@C’) andX"=—1/(wC’). Aftersimplification, we find that B 4+9R'0*C2AFOROCps 55} | z+Rwe® 55) = | Then 21+3R*w’C oCry » pe21HROC gC -| CRROC FTIR 8) 1 | InFig.7-15(c), theresistor hasaconductance G’andhencearesistance 1 red 57) + | Rae (757) =v ee c i [Astothecapacitance, 7 [B'=joC, C=z (7-58) LetR=1000ohms,@=500radians/second, C=1microfarad eThen, from Eq. 7-50, -7-15.(a)SimpleRCcircuit.(b)Equivalentseriesc'rcuit.(c)Equivalent =10,99248%10"xS007x10°)—10"500%10% =1750—250) (7-60) gareRIREUGOC)_ p2+3Kj0C a '% 7-0)2R+IGaC) 1+2RjoC =1768exp[arctan(-254)=RETSRIOCI(I =2RjoC)_p(2+6R*w*C*) —RiwC =1768exp(—0.142/) ohms. (761) ~ Parc 14+4R °C tration (7-50) Angles areexpressed inradians. IfVis10volts rms, Ineffect, thismeans thatthecircuit ofFig,7-15(a) hasthesame 1=oexp(0.142))=5.66exp(0.142) milliamperes rms. impedanceasthatofFig.7-15(b),wheretherealpartofZ, 1768°°?(0-142/)P(0.142) pe;boplt3RwrC? mst (7-42) R=RR (7-51) Thecurrentleadsthevoltageby0.142radian,orby8.13degrees. ‘Theinstantaneous value ofthecurrent is isfrequency-dependent. Thevalue ofC’isgiven bytheimaginary 1=2!9x 5.66008 (500 +0.142) PartofZ: 7 =8.00.cos (500r+0.142) milliamperes. (7-63)aoc (7-52)Jal 1+AR@°C 7.7POWER INACCIRCUITS1+4R*w°C? 1 ve Ces 7-53) C=Rare“RaretC53) SupposearesistanceRisinserieswithacapacitorC.Oneappliesa The value ofC’isalso frequency-dependent. Indeed, the voltage V,,expjeattothecombination. Then { 138 ELECTRICFIELDSV| 139 , k A k Vo.expje 1aPOhyexpj(wt-8),8) where: _ zorin(R+te) expia, o=arctan(—2"-), (7-65)| +°joC we) SxPse Roc)” °\ © 1 k 1 “ Vin ~ dn= I a+ 7-66) TR?+wy? (66) ‘The average power dissipation is a Pa,=Waly 608.8 =Viryslinn 6088, (7-67) } = ' Fig.7-16. Simple RC circuitfromSec.2.4.Thetermcos@=R/IZ| isthepowerfactorofthe ® pleBCceca. impedance Z. 1 1 vor 10Sec 4,wecanalsowrite =v)—+——__ =——__ BsAccordingtoSee.2makow evigtarmal |3Re2joo) Py=4ReVI", (7-68) i AtA, wheretheoperatorRemeans“realpartof”andwhere/*isthecomplex Vj=V-LR= vittie’ 7-7)conjugate of/(Sec. 2.1).Oneoccasionally writes that yo ‘The ratio V/l, should beequal totheimpedance Zthat we IVI =P+jun (7-69) calculated intheexample inSec.7.6.Letuscheck: ¥_IUGR+4/GoC)] _2RGRWC+4) _p2+IRWC ny where thetermontheleftisthecomplex power, P.,isthetime-averaged 1OR+4GoC)+2R~ BRjwC+4 ~*i+2Rjoc? 774) power, andQ,.isthetime-averaged reactive power. aspreviously 7 _ (b)Withthenodemethod, weapplytheKCLatnodeA, 7.8 KIRCHHOFF’S LAWS FOR ;where wehave, aswith adecircuit, ALTERNATING-CURRENT CIRCUITS hehth hel, heh, belly 025) with BoththeKirchhoff current lawandtheKirchhoff voltage lawapplyto yeVoMs po Ms ne. 6alternating-current circuits ifoneusesphasors andimpedances. MR BER TiGacy R79) Thus Example SIMPLE RCCIRCUIT SOLVED BYBOTHTHE VaVa_Vaacy (7-77)MESH AND NODE METHODS R R+1/(jac)” R* sie «usemesh curre v 14RjwC (a)Referring toFig.7-16,weusemeshcurrents andapplythe Veep RI ag KVL asifwehaddirect currents: TFROC+Rod)2+3Kjoc” —7) V=LR=(,-1)R =O. 7) 4 7.9SUMMARY ==) =4R-=0. 7) ;joC Kirchhoff's current law (KCL) states that thealgebraic sum ofthe Solving yields currents entering anode isequal tozero. According toKirchhoff's voltage law(KVL), thesumofthevoltage drops around amesh isequal Po.=Vinal6088, 07-67) to zero, ‘These two laws serve tocalculate voltages and currents atvarious where cos@isthepowerfactorR/\Z|ofZ.Also points inacircuit, given thesources andtheimpedances. With themesh] , method weassign mesh currents J,,[y,f,... toeach mesh. Wethen Po=4Re VI". (7-68) apply theKVL toeach mesh insuccession tofind themesh currents. | With thenodemethod weassign node voltages V;,Vs,V.,... toeach Kirchhoff’s current andvoltage laws apply toalternating-current node.Oneofthenodes issetatV=0.ThenweapplytheKCLtoeach |“ifcuits. node insuccession toobtain thenode voltages.Ifthecircuitispurelyresistive,theneithermethodgivesasetof|PROBLEMSsimultaneous algebraic equations. Ifthecircuitcomprises bothresistors | andcapacitors, weobtain asetofsimultaneous differential equations. 7-1.(7.2) Thepotentiometer‘Thealternating current inatrueresistor isinphasewiththeapplied | Figure7-17shows»potentiometer circuit,Showthat,whenJ=0, voltage.Thetime-averagedissipatedpoweris | veRy,ORR, alaR op Thisisacommon typeofcircuit. Itserves tomeasure avoltage, inthisPS =Fel (739),7-32) caseVe,without drawing crrent. Incurveplotes thecurren, after amplification, actuates amotor that displaces thepenandsimultaneously where Jnisthemaximum value ofIandlipistherootmean square moves thetapinthedirection thatdecreases ,Theresistances RyandRy‘ots“ay”andl“ema” actasapotentialdivider. value ofI.Inpractice, thesubscripts “av” and “rms” areunnecessary. ‘Thealternating current inacapacitor leads theapplied voltage by90°: 7-2.(7.2) Theoperational amplifier Figure 7-18(a) shows acommon type ofamplifier. Thetriangular figure isanoperational amplifier whose gainis—A.Such t=jucv. (736) aptshavegainsoftheorderef101010anadrareliamount ofcurrent attheirinputterminals, Theaccuracy ofthegainofthis ‘Thetime-averaged stored energy isCVm,/2. circuit islimited onlybythestability oftheratioRy/R,. ThedriftinRy/R, More generally, duetoaging, temperature changes, andsoforth isnormally smaller than thedrift inAbyorders ofmagnitude. v Asafirstapproximation, (a)theoperational amplifier draws zero I=Z (7-39) current, andthesamecurrentJflows inR,and inR;,(b)Aisinfinite, and thepotential atthejunction between Ryand Ryistherefore zero, Then 1V,/R,~-V,/Rz, andthegainisabout—R,/R,. whereZistheimpedance ofthedevice across which thevoltage isV.For (a)Findamoreaccurate expression forthegainV,/V,, Youcantake aresistor, Z=R. Foracapacitor, Z=1/(jwC). Impedances areusually intoaccount thefactthatthegainisnotinfinite bysetting V,=—AVz, in complex: Z=R+iX, (7-41) whereXisthereactance ofthecircuit.Theinverseofanimpedance isanadmittance:. R 1 " a yey (7-44) “OF.; Thetime-averaged power dissipated inacircuit is Fig.7-17. 7 “ » x » | HI |8 2 8 = ‘| Fig.7-20. | V=0means0andV=V'means1,whereV’isaprecisely regulated Fig.7-18. \ positivevoltage. 7 eapproximation ~ va 7 RefertoFig.7-20.SayV’=1volt.Thevoltmeter connected betweenA (b)WhatistheminimamvalueofAifR,=1000ohms,R=2000ohms, terminals B,CD,andtheanalognumberappearsatA.Forexample, andthecircuitgainmustbeequalto2within0.1%: {i ‘UsetheresultofProb.7-3tofindthevalueofV,when ‘7-3.(7.2) TheR/2R ladder network (a)Vy=1,Ve=0,andVp=0; Figure 7-19 shows aso-called R/2R ladder network thatserves forboth (b)V,=0, Ve=1, andV;,=0; digital-to-analog (D/A) and analog-to-digital (A/D) conversion. See (©)Ve=0, Ve=0, andV,= 1 me =a-”calles1input ‘7-5.(7.2)Chargingacapacitorthrougharesistor tat teesas bree4andgroundledtenpr| Aeee eeaextorAtoawolageV ‘. Calculate theenergysupplied bythesource,thatdissipated bytheresistor, 7-4.(7.2) Digital-to-analog conversion andthatstored inthecapacitor, after aninfinite time. ‘TheR/2R ladder network ofProb. 7-3canconvert abinaryt number to ‘Youshould findthattheresistor dissipates halftheenergy andthatthe kn —) Kn, —* Kn Figure 7-21shows anRCdifferentiating circuit. ‘Theloadresistance connected atV,islarge compared toR. a ee ee i epiavengred ta moar » . x awV,=RCa“ (b)The input isasquare wave. Sketch V,(1). 9 a j| 1Qy-@ four=100,five=101,etc. ‘Fig. 7-21. 144 | 145» R ¢ ' Tr A:¢ y \ oy Qy-@ tR | (a)ooo mene ® 0\)-0c tH gw) r—9 !Me i | 7-7. (7.2) Differentiating circuit usinganoperationalamplifier ' Fig.7.24 “ThecircuitofFig.7-21(a)issimpleandinexpensive, butV.<V,Figure » 7.24, 7-22(a)showsamuchsuperior, butmorecomplex, differentiating circuit. {| > ies ‘Thetriangle represents anoperational amplifier asinProb. 7-2.Figure 7-9.(7.2) Integrating circuit using anoperational amplifier Th rangle representsanoperationalampli i) ‘TheimeyatingcreatshownaFig7-280)ptformsintegrationswithoutthelimitation V,«V,thatappliestothecircuitofFig.7-23.The ret: triangle represents anoperational amplifier, asin Prob. 7-2.Show that V,=-RC-a if aslong as A>1and|V,l/RC>>|dV,/dtl/A. NotethatRCcanbemuch v=~ae[Yau unity,sothatV,neednorbemuchsmallerthanV, largerthanunity,20thatV_neednotbearch ifA>1andif(V.I(RC)<A[dVdel.UsetheequivalentcircuitofFig, 7-8.(7.2) RCintegrating circuit 7-24(b), andset Figure7-23showsanRCintegrating circuit.Thecurrentthroughtheload 4y,-4(v,- 2)connected atV,isnegligible compared todQ/dt inC. aa Me (@)Showthataslongasthevoltage acrossCissmallcompared tothat {|-19,(7.2)Motion transduceracross R, Figure 7-25shows howthedifferentiating circuit ofProb. 7-7canserve 1 tomeasure adisplacementvag[va ‘Showthat,withanalternating voltageattheinput,theoutputvoltageis . proportional tothespacingsoftheparallel-plate capacitor, neglecting edge AsinProb.7-6,V,<V,,WeassumethatV,=0at¢=0. ettects.a ve tionoftimeifV, ©)Stetch‘curveoftheoutputvoltageasafunctionoftimeifV,isa|111.(7)Compensatedpotentialdivider sa7 ‘Thepotential divider ofProb. 7-1isnotuseful assuch athigh frequencies forthefollowing reason. Thete arestray capacitances duetothewiring in k parallel with RyandRs.Ifthefrequency ishigh enough, these stray capacitances carry anappreciable current and V/V, isafunction ofthe frequency ‘Showthat,withthecircuit ofFig.7-26,therelation ofProb.7-1applies if ¥« 4 RyC\=RsC.. Iftheadded capacitances arelarge compared tothestray capacitances, thepotential divider issaidtobecompensated. Fig. 7-23. |> 5 )© 5 o | c +f R impedance ofcomponents. These bridges havenowbeenlargely super- |} ©ingens ofcommons Thee ridges avnoweearyie ms ments oncomplex circuits. Figure 7-27(a) shows suchanimpedance bridge. Ry meIfZ,/Z,=Z/Zy,thenV=0.Theimpedances mustsatisfytwoindependent —||‘equationstosatisfyboththerealandtheimaginarypartsofthisequation. i one sets R,=R,/2, Ry=Ry, Cy=Cy. Find thecondition forbalance. i ‘TheWien bridge isusedintuned amplifiers andinoscillators aswellas} formeasuring ormonitoring frequency. Tomeasure afrequency, one Fig.7-27. 7-13.(7.7) Phase shifter (b)Draw agraph ofthephase ofV,withrespect toV;intherange Itisoften necessary toshift thephase ofasignal. Figure 7-28 showsa RoC=0.1to10.Usealogarithmic scaleforRwC. The resistances areadjustable, butequal. Usethepolarities shown. They From many points ofview, alternating current ismuch preferable to mean that ¥,isthevoltage ofthetopterminal with respect tothebottom direct current forpower distribution. However, linelosses arelower with left-hand one ‘Onahigh-voltage overhead transmission line, themaximum operating(a)ShowthatV,/V,=exp(2)arctan[1/(RoC)]} voltagedependsonseveralfactors,suchascoronalosses(currentlosses 7 hy Vr 148 ELECTRICFIELDSV_| never exceed acertain value, say V,.Otherwise, the downtime and thecostofmaintenance becomeexcessive. Thecostofalineincreasesrapidlywith CHAPTER ‘the curent nthe linecanbemade neatly 3large asone likes without damaging it,sincetheconductors arewellcooled bytheambient ir *ELECTRIC FIELDS VI lowerthecatrent thebeter " Electric Circuits B:Circuit Theorems With direct current, twoconductors operate at+V, and —V, with respect toground. The power delivered totheload is2V,/y.. With single-phase coe, (SP)alternating current, there aretwowires at+V,coswtand—V,cos wt. 8.1THEPRINCIPLE OFSUPERPOSITION “ (a)Showthatforthesamepowerattheload,thermscurrent /ypis 82THESUBSTITUTION THEOREM ==1502!14..TheIRlossesinthelinewithsingle-phase alternating current are 8.3.THEVENIN'S THEOREM —150twicealargeasthonwithdirectcurrent { Puample THEZINC-CARBON BATTERY St {oy With threephase (TP) siterating curtent, wehave thee wires at | 84. NORTONSTHEOREM 152Vjc0sat,Vgeos(r+24/3),andVgcos(t+42/3).Weassumethatthe{|$3MILLMAN'S THEOREM 153 ‘Then thecurrent intheground wire iszero. Show thatforthesame total} 56 TELLEGEN'S THEOREM 154 power dsivered 10thetheeloudrestnces, thermsCensor fe} NyAMORE GENERAL FORMOPTELLEGENS THEOREM 16 NVitre-phaseakernating cutent,thermscurtentsarethusaboutthe 87THERECIPROCITY THEOREMS 157 instead oftwo,sothatthelosses are50%larger thanwithdirect current. A. Example: SEARCHING FORANOMALIES INGROUND three-wire lineisalsomore expensive thanatwo-wire one. CONDUCTIVITY 160 | 88 THEPOWER-TRANSFER THEOREM 16 89DELIASTARTRANSFORMATIONS 163 | Example 165 |gto SUMMARY 166 t pRosLeNs ter | Generally speaking, circuit theorems serve tosimplify calculations. There | exists aseemingly infinite number ofcircuit theorems, buttheeight that follow areprobably themost useful. The delta-star transformations of Sec, 8.9 are even more useful. ' Allthematerial inthischapter isgeneral andisnotrestricted toRC circuits. We, however, assume that allcomponents are linear and time-independent, asinChap. 7.Alinear component isone that follows Ohm's law, J=V/Z. '8.1 THE PRINCIPLE OF SUPERPOSITION The principle ofsuperposition ofSec. 3.3applies tolinear circuits: each source acts independently ofalltheothers. Inother words, thecurrent “Caples and8are0reuired forwha follows, However, Chap. 7 prteqise for Cu. that flows through one branch isequal tothe sum ofthe currents pooo-----3 ascribable toeach individual source. . | y H I 1 I 8.2 THE SUBSTITUTION THEOREMhom lt !i” 4 1 Thesubstitution theorem isanobvious consequence oftheKirchhoff | 1voltage law(KVL): ifthevoltage drop across apassive component ina i 1 circuit isV,thenonecanreplace thatcomponent byanideal voltage = aaa source supplying thesame voltage, without disturbing any ofthebranch currents. Clearly, this does not disturb the voltage drops around the meshes. Figure 8-1showsatrivialexample. 'p= &« | — a ) (vw th(©) (ov 1, ‘Ye (as) Swe) oe (se) * a .-\ ~ \ Fig.8-2.Thévenin’stheorem.(a)Anactive circuitAfeedscurrenttoaresistance tR.(b)CircuitAisequivalent toanidealvoltagesourceV,inserieswithan{output impedance Z...Asarule, Z,isaresistance R,..(c) Onecanmeasure R,by + 7 | plotting V=V,~R.7asafunction of1.Decreasing Rincreases1anddecreases .» V.Theslopeofthecurveis~R., Fig.81. The substitution theorem. (a)The voltage drop across theresistance on jtherightisV/3.(b)Replacing thatresistance byabatteryV/3ofthecorrectpolarity leavesthemeshcurrents unchanged. |where ¥,isthevoltage thatwould appear between thenodes attheends |ofabranch6ifthebranchwereremoved,andZr,istheimpedance 8.3THEVENIN’S THEOREM calculated atthecut,including theimpedance ofthebranch, withallthe sourcesreplacedbytheiroutputimpedances. Inotherwords, Thévenin’s theorem states that any active, linear, two-terminal circuit is equivalent toanidealvoltage source inserieswithanimpedance Z,, | 4-— 82calledtheoutput impedance ofthecircuit, asinFig.8-2.Asarule,this OOZ4+Zy" (62) impedance isresistive, and thesource issaid tohave anoutput resistance R,,.Agood voltage source hasalowoutput resistance. where Z,istheoutput impedance ofthecircuit feeding thebranch. ‘Thévenin’s theorem applies tosimple sources such asflashlight batteries, butitalsoappliestocomplexcircuitssuchasaudioamplifiers, Example |THEZINC-CARBON BATTERYittors,etc.When appliedtosimp.Thévenin's heenPlantgenerators Stalfact,ateastforaInnitedrangeatoutput Azine-carbon battery,liketheonesusedinflashlights,performs jeoremisanexperimental fact,atleastforalimitedrangeofoutput asifitcomprised anidealvoltagesourcesetatV,=1.5101.6currents, Granted thatthetheorem applies tosimple sources, thenitalso volts, depending ontheactual materials used, inseries witha applies tocomplex circuits (Prob. 8-3). resistance R,.Forafresh size-D cell, R,=0.27ohm. The value of ‘Thévenin’s theorem isuseful forcalculating branch currents. Wewrite R,increases gradually with useandwith time andeventually | growsbymanyordersofmagnitude.Withtime,¥,fallstozero. v.‘Ahigh-resistance voltmeter connected across. the terminals nae, (1) measures V,.Tomeasure R,onecaneither proceed asinFig.8-2, Zn ‘ormeasure both V,andthe“flash” (peak short-circuit) current 152 ELECTRIC FIELDSVE ASMILEMAN'S THEOREM 153 8.4 NORTON’S THEOREM where J,andY,aretheparameters oftheNorton source and¥,isthe admittance ofthe branch. Norton's theorem isthedual ofThévenin’s. Itstates thatanyactive, Tocalculate J,andY,,weagain remove thebranch. Then /,isthe linear, two-terminal circuit isequivalent toanideal current source J,in current through azero-resistance ammeter connected toitsterminals, parallel withanoutput admittance Y,,which isusually aconductance. + and¥,istheadmittance calculated atthose terminals, with allthe SeeFig.8-3,Agoodcurrent source hasalowY, i sources replaced bytheiroutput admittances. Itiseasy toshow that, foragiven device, . 8.5|MILLMAN’S THEOREM ZY,=1, (83) Millman’s theorem serves tocalculate thepotential ataspecific node ina atleast when Z,andY,arereal. From Fig.8-2,R,,isequal tominus the circuit, andhence theincoming branch currents. Itisasimple application slope oftheVversus /curve. But, from Fig.8-3,G,isminus theslope of, ofKirchhoff’ current law(KCL). Itisbestexplained byanexample theIversus Vcurve. SoG,=1/R, andR,G, =1 InFig.84, +h+=0. Then Norton’s theorem isuseful forcalculating branch voltages. We again consider thecomplete circuit minus thebranch asasource, and the (Y= VM+(Va-V2+(VyV)¥S=0, (85) branch asaload. Thus yMktara < Veo(8-4) Athth “° PTY, +Ys" This isMillman’s theorem. Ofcourse, | h=(Y-V)%, —ete. (8-7) oO | ia J=IO ZR 1 \ 1 2%\ oF i 6, i ° 6 6 v % 4 ' fo) 9 9 Fig. 8-3. Norton's theorem. (a)Anactive circuitAfeedsaresistanceR. i = (b)CircuitAisequivalent toanidealcurrentsourceinparallelwithanoutput 'admittance ¥,.ASarule, Y,isreal and isthusaconductance G,.(¢)Onecan Fig.84,Millman’stheoremrelatesthevoltageVatanodetotheY"sandV's, measureG,byplotting"=I,~G,VasafunctionofV.IncreasingRincreasesV connectedtoit.Thesquaresrepresentunspecifiedcircuitsthatmaybeeither and decreases /.Theslope ofthecurve is~G,. passive oractive. 154 155 G y- P Ww k= aa /f.~ = =: Fig. 86. The sign convention forTellegen’s theorem. o Fig.85. (a)Simple circuit fortesting Millman’s theorem. (b)Thesame circuit, redrawn toresemble thatofFig.8-4. where the summation runs over allthe branches ofthe circuit atthat instant. This issimply astatement ofthelawofconservation ofenergy. You will beable toshow inProb. 8-7 that Tellegen’s theorem follows ‘There canbeanynumber ofnodes connected tothespecific node in from boththeKCL andtheKVL.question. Nowsupposethatwehavealternating currentsinacircuitcomprising_ only sources and linear passive components. Then thecomplex powerExample iFig,85,fn‘clearlyequaltoV./3.Applying Millman’s (Sec.7.7)flowingintobranch6isV,/3/2,andthesumofthecomplex orem we Bing that ipowers iszero, VIR_Vs v=sq7R)~3" aa DvV,lt=0, (8-10) 8.6TELLEGEN’S THEOREM |forthefollowing reason.Substitute foreachstarredbranchcurrentthe | difference between theneighboring starred.mesh currents. Then, reat- Tellegen’stheorem,initsoriginalform,islikeKirchhoff's laws:itis\oleaenesumwehavethefststarredmeshcurrentmultipliedPytheself-evident andappears trivial. Again, thisisafalseimpression; ot tocadeymounditeteEach hihein Tellegen’stheoremisrelatedtoatleast150othercircuittheorems!+ |bythevoleKULthewaetc.Eachtermofthissumbeingzero, Supposewehaveonlydirectvoltagesandcurrentsinacircuit.SetV, accordingtodfhitlokoveequationiscorrect equaltothevoltageacrossbranchband/,equaltothecurrentflowing |Se ee putportthatthrough it.Also, choose thesigns ofV,andJ,insuch awaythatVil,is; 7 SPs ‘ Salle m thepowerflowinginfothatbranchatthatinstant,asinFig.8-6.Some weeneSercoconnectsehatentwriheinputpoemerle, branches comprise sources, oth it.Then, ding toTellegen’s | aetheorem, p “FeONE BOE NOTEENS circuitsalsocomprise oneormoreoutputports,asexemplified againby: theWheatstone bridge. Tellegen’s theorem, asstated above, applies to SVs =0. 9) thecomplete circuit, including sources andload.7 Itisuseful torewrite Tellegen’s theorem inaform that shows explicitly thepower attheinput and output ports. With thesign convention ofFig. |8:8fortheports, thetheorem becomes ++Paul Penfield, Je., Robert Spence, and Simon Duinker, Tellegen’s Theorem and .EteciricalNetworks, ResearchMonograph 58,M.LT.Press,Cambridge, Mass.,1970 DVols—DVol=O, (8-11) 156 S9-THERECIPROCITY THEOREMS 137 4 around thefirst mesh oftheunprimed circuit, thesecond...ete.Each term ofthis new series iszero, because ofthe KVL, and JVeli=0. (8-13) vt a » Bysymmetry, DVib=0. (8-14) 8.7 THE RECIPROCITY THEOREMS G The reciprocity theorems aremost useful. Weusetheabove form of Tellegen’s theorem toprove oneofthem. Fig.8-7.Wheatstone bridge.TheinputportisAB,and is *Be:neImaport theoutputportisCD. Figure8-9(a)and(b)showsatwo-port circuitPconnected toexternal elements xintwo different ways. Circuit Pispassive and linear. The x's can beeither sources orpassive linear elements. 1 i Wenow rewrite Eqs. 8-13 and 8-14 intheform ofEq. 8-12, with the ‘ power flowing outofthedevices connected totheports ontheleft-hand + side oftheequations and thepower flowing into Pontheright-hand side: v a - . Vili+Vals =DVoli. (815) ' Viht Vib=D Vibe (8-16) Fig. 88. The sign convention forTellegen's theorem,asstatedinEq.811 Wecaneasilyshowthatthesumsontherightareequal.Saythe wheretheprimed summation overthebranches excludes theports,and resistance ofBranch bofcircuitAisRp.Then where thepower with subscripts pisthepower flowing outofthedevices A , ,.Vol=(UnRols=by(Rols)=LVS 8-17) connected totheports.Thus tb~RM =(Rel) @17) , Itfollows that,inFig.8-9, DLVals=DVoll (8-12) Vali +Val Vih +Vile (8-18) 8.6.1 AMore General Form ofTellegen’s Theorem Weagainlimitourselves todirectcurrents. Suppose thatwehavetwo OFo HO OFO HO circuits thatareunrelated, except thattheyshare thesamegeometry. The ct] © OL 6 LJ branch voltages andcurrents areV,andJ,foronecircuit, andVj,and[;, 00° rome: fortheother. What isthevalue of3!Vili? Thisisadmittedly aweird om w expression, butweshalluseitinthenextsection, Identical i P(edtoexternalelementFirstwesubstitute for£,thedifference betweentheneighboring mesh ToediferentwaysAeusualacheledVrepresentsaniahmerestance currentsintheprimedcircuit.Rearranging thesum,wehavethefirst Voltmeter,andacircled[azero-resistance ammeter.Thesignsandthearrows meshcurrentoftheprimedcircuit,multipliedbythevoltagedropsil‘showwhichdirectionsaretakentobepositive. 158 159 oO ° oO oO a to= LA, = o Fig.8-10. (a)Circuit Pwith anideal voltage source ontheleftandanammeter contheright. (b)Thesame circuit P,withavoltage source ontherightandan oO oO jmmeteton thelet.Themostcommonly usedreciprocity theorem atesthatthe . RS ratios V/Tareequal i Q aS Si NowconnectPasinFig.8-10(a),whichmakesV;=0.Reconnect Pas oe_0 °° inFig.8-10(b) soastomake V;=0. Then, from Eq.8-18, ‘ Mah=Wi “ Vili=Vib (8-19) 5 oO Seti =" *7 4S Q SettingV,=V3makesJ=hy me) 1Po©)©rh SE Inother words, interchanging thevoltage source and theammeter does y o notaffecttheratioV/I.Wehaveassumed steadydirectcurrents and ' o_o °°linear elements, buttheresult isgeneral. A Vow, = ® Thisisanimportant result. Itmeans thatgivenanylinearandpassive | circuitwithoneinputportandoneoutputport,theshort-circuit output \ oO oO current, foragiven inputvoltage, isunaffected iftherolesoftheinput a y <andoutputportsareinterchanged. Inotherwords,theratioV/Jremains nS) 1Pe ©“@)thesame. Theoutput impedance ofthesource neednotbezero, butthe oo ©0 impedance oftheammeter mustbezero.Thisisthemostcommonly used = we reciprocity theorem. j . IHNIMS ti‘Thisresultisunexpected, becausecircuitPneednotbesymmetric at Fig.811.Thefourreciprocity theorems. InallcasesPisanypassiveandlinear it. (a),(b)Interchanging the positionsoftheidealvoltagesourceandofthe all.Thepowersuppliedbythesourceisnorthesameinthetwo eeeOO aeteieteatinVV.(oh(@)Enterchanging sheidealcurrentconfigurations. sourceandthevoltmeterdoesnotaliertheratioV/I.(e),(f)and(g).(h)TwoFigure 8-11illustrates thefourreciprocity theorems. | Gtherreciprocity theorems. Example |Letuscheckthereciprocity theoremillustrated inFig.8-10(a)and | R(b)byapplying ittothecircuit ofFig.8-12(a). Wefirstapply V n= yy. (8-21)volts atport Iand connect anammeter atport 2,asinFig. | RERAF RRIF RAR: 812(b). Werequire thecurrent f,whenport2isshort-circuited by Withthenodemethod, wesetthesumofthecurrents flowing theammeter. Letuscalculate J,inthreedifferent ways:bythe intonodeAequalto2er0:meshmethod ofSec.7.2.1,bythenodemethodofSec.7.2.2,and byMillman’s theorem ofSec.8.5. V~Vs_Va Va (8-22)With themesh method, weapply theKVL tomeshes @and6in Ry Ry R; succession: Solving,wefindthesamevaluefor/,:V-LR,~(L ~LR.=0,(LR ~1,R,= 0.(8-20)wm | ve RRs v. (23) Solving, 4RRs +RR, +RRs j 160: 451H: POWER-TRANSFER THEOREM 161 R & According tothereciprocity theoremillustrated inFig.8-11(c) ‘i, Ny ow : o and(d),theratioV/Jwouldbeunaffected if,instead,thecurrent ‘wasinjected atP,P, andthevoltage measured atC.C,. ‘ *_©) © 8.8THEPOWER-TRANSFER THEOREM ‘Thepower-transfer theorem states thatthecondition formaximum power °° transfer toaload isthat : ZZ, (8-26) a & where Z,istheload impedance andZ}isthecomplex conjugate ofthe + ‘source output impedance. “ ‘ ‘The proof issimple. According toThévenin’s theorem (Sec. 8.3), a 7 real source isequivalent toanideal voltage source inseries with an impedance Z,,asinFig.8-13. Let " Zo=R AIX Zr=RetiXe (827) Fig.8-12. (a)Simple circuit fortesting thereciprocity theorem ofFig.8-10(a) and (b). (b)We have addedavoltagesourceontheleftandanammeteronthe !‘Thenthepowerdissipatedintheloadis right. (c)Wehave now interchanged thesource andtheammeter ve \=PR,=-——* ___,k, (8-28) eM RW oy P=PRORR+RtRD PRS”RRRR,+RR, -° i setX= —X,. Then WecanapplyMillman’stheorembyreferringagaintoFig|‘TomaximizeFy,se 8-12(0) R.a VIR, i ree ta (8-29) =IR __ ' R,+R Thisyields thesamevalueforVyandthesameforf, r | Now,tocheckthereciprocity theorem, weinterchange the | ! 2 |voltage souee andtheammeter ofFig81210)toobtan Pig | i8:12(c). Werequire 1,Thatiseasy!Circuit(A)isidentical to|| f o circuit (b),except forthefactthatR,andR,areinterchanged. So ' wecanwrite down the/,ofcircuit (d)byinterchanging R,andR, ! intheaboveexpression forthe/,ofcircuit(b).Thetwocurrents ! aarethesame, \~Problem 6-4concerns areciprocity theorem forelectrostatic |fields Example |SEARCHING FORANOMALIES IN | iGROUND CONDUCTIVITYi \ dt Geophysicists usemany methods forlocating potential orebodies. sutput i isZ,,connectedtoaloadimpedance Figure 4-7shows one method. The ground actsascircuitA,port1|Fig.&13.Source,whoseoutputimpedance isZ,,connectedt ypedane isthepairofelectrodes C,C,, andport2isthepairP,P, " i 162 ELECTRIC FIELDS Vt AYDELTA'STAR TRANSFORMATIONS 163 Foragiven source, with agiven value ofR,,,wesetdP,/dR,=0andfind Sothecondition formaximum powertransferfromasourcetoaload thatRy=R,.SoP;ismaximum when requires thatone-half ofthetotal power bedissipated inthesource. In . _ _os otherwords, theefficiency atmaximum powertransfer is50%. 2,=Ri+ikiRo~iXo=Zi. (8-30) Figure 8-14showsthatthecondition formaximum power transfer is asabove. notcritical. Underthoseconditions, thepowerexpended inthesourceis $9DELTA-STAR TRANSFORMATIONS ve =PR,=PR,=“ r=PRo=VRu= aR (831) {Figure 8-15(a) shows threenodes A,B,Cforming partofsome { unspecified circuit. The nodes aredelta-connected in(a)andstar- m ' connected in(b). Asweshall see, wecansubstitute thestar forthe i equivalent delta, orinversely, without disturbing therestofthecircuit. ' Indeed,ifwehadtwoboxes,onecontainingthedeltaandtheotherthe & | equivalent star,withonlytheterminals A,B,Cshowing, therewouldbe im 1 noway oftelling which box was which. | Itisuseful tobeabletotransform adeltaintoanequivalent star,or 1 inversely, onpaper. This often simplifies thecalculation ofmesh currents - | ‘undnodevoltages.I | WecouldfindZ,,Zp,ZcintermsofZ,,Zp,Z.,and,inversely, by i assuming thesame mesh currents [,,Iy,Jcinthetwocircuits, andthen! |making V,~Vp, Va-Ve,Ve—Vq inthedeltaequaltothecorrespond-otf |ingvoltages inthestar.SeeProb.8-11.Hereisanother approach thatis 1 somewhat lessconvincing, butshorter.| J|. H | oe ea | | ; | Ya z(, hn 1 i ! | 1 % % or 7 00 c 2 ct ® a j « » Fig.8-14.TheratioP,/P;a.andtheefficiency #asfunctions oftheratioR,/R, Fig.8-15.ThethreenodesA,B,Carepartofalargercircuit.(a)Thethreeforthecircuit ofFig.8-14, with X,=—X,. The power P,dissipated intheload is nodes aredelta-connected. (b)The nodes arestar-connected. Under certainmaximum whentheloadresistance R,isequaltotheoutputresistance R,ofthe conditions thedeltaiscompletely equivalent tothestar,butonlyatagivensource. Butthenonlyhalfthepower goestotheload; theother halfislostinthe } frequency. IfthesixZ'sareallresistances, thentheequivalence isvalid atall source. frequencies. 164 uremicros i 165 Suppose wehaveadelta andtheequivalent star.Theimpedance 1 lwo ' : : between AandBisthesame inboth. Then ‘s 7 hea a 8 ” oar 45510, s Z(Z,+2) Zan =Ot) 27,4 Zp, 5.077,47, Atee (8-32) Similarly, Vat tat ni ZZ,+Z) Zu =SO =Zp+Ze, 3 297 BeZ,+Z,+Z, +Z (8-33) Tt rBAZ42)_ “~ a LeaTaga7,tee (8-34) bo » js Fig.816.(a)Delta-connected circuit,(b)Itsequivalentstarat1kilohertz.Note Observe thatZ,isassociated withZ,Z,,ZnwithZ.Z,,andZ.with thenegative resistance " Z,2,.Therefore vy. +ererere (8-42 2,2% Yu+YatYe (42) = =a (635) yy, Ld ree are 643)20-757 Z (8-36) Ifastaroradeltacomprisesonlyresistances,thentheequivalent Ze= (637) circuit isalsopurely resistive. However, iftheoriginal circuit comprisesZ,4+2,+Z, ' capacitors (orinductors, orboth), then thevalues oftheresistances and capacitances (andinductances) intheequivalent circuitarethemselves Tofindtheinverserelationship, weusetheadmittance between node frequency-dependent., Aandnodes BandCshort-circuited together. Thisgives Forexample, refer toFig.8-16(a) andcallthecomponents RandC. Then theresistance inthelower branch oftheequivalent star ontheright y,+y=SatatYo) (6-38) turnsouttobe—R/(4+R’w°C?).Thismeanstosaythat,forgiven Xa +Yat Ve ° components, theequivalence isvalid atonly onefrequency. Moreover, theresistances ofequivalent circuitscanbenegative, asinthisinstance. Similarly, Soarealcircuit canbeequivalent, onpaper, toanimpossible one. Yo(Yo+ Ya) ¥+¥,=So (8-39) Example |‘ThedeltaandthestarcircuitsofFig,8-16areequivalent, asyouYu+¥a+ X (639) cancheckbysetting =YAYat Yo) | =2=s— ohms, =1000hms, (8-44) Lehoe (8-40) nha yohmsZeal (B44) Again byinspection, ate ohms, 5)pes Za=TnM5145 oidxIoy |ONMS(BH) enna (8-41) — OVA +YatYe" Ze218+ oxaixoy ON (6-46) Cea re. ~*~ ) . s ., |Vvc c Vva(1») k; J a hd ns Fig.620. “eo(+) & '8-9.(&7)Thereciprocitytheoremappliestotransients Show that theratio 1/V forthe circuit ofFig 8-20(a) isthe same asfor athecreuit ofFig. 820(b). Find Q(0) fist 8.18,® 8-10.(8.8)Audioamplifier ‘Anaudio power amplifier hasanoutput resistance of8ohms and feeds a$5.(84)Norton'stheorem resistiveloadof6.4ohms.Calulatetheefficiency. pili thewaagarsherate RofFig.18byNorton's $11.(9)Dettstar transformation,Xeoutput conductance ofthesource isinfinite, ‘| Findtheequations foreither thedelta-star orthestar-delta transforma 8-6,(8.5) Millman’s theorem { tionbyassuming mesh currents asinFig.8-15andmaking thevoltagesInthecircuitofFig.819theswitch5closesat=0,Findthevoltage Ve{| Va—Vn. VoVe»Ve~Vuinpart(a)thesameasthoseinpat(). aes‘as.afunctionoftheimebymeansofMillman’stheorem.Set, Findancavationaneom«inC.MeBA.Mem0.Sinceis = Lohm, Re=2ohms, Ry=3ohms, C= 1microfarad. V=100volts equation onst bevalid whatever thevalues ofthemesh currents, the , theses mustallbeidentically equal tozero.Thiswillgiveyouoneof 8-7.(8.6)Tellegen’s theorem Parent yeq(ashowherTels theorem isaconsequence oftheKCL. theequations ofonese;theothertwoequation fllowbysymiety,(b)ShowthatTellegen’s theoremisalsoaconsequence oftheKVL. |812.(89) Delta-star transformation ; 8.(87)Thereciprocity theorems I FindtheresistanceofthecicultshowninFig.8:21oC tsfspoiyhorsofFig6byspyingthom|)SNVanPusene tthesimpleercuitofFig.8120 isdificulttomeasuretheconductivityofsmallsamplesofsemiconduc- |tor.First, they arebrittle andthusdifficult tomachine. Second, contacts to ‘| thematerial areresistive. WithVanderPauw’stheorem, however, itis : t possible tomeasure theconductivity ofasample intheformofathinplate ‘|‘ofarbitraryshapewithfourcontactsaroundtheperipherywithout fane fs| a 40 v‘; PeFig.8-19,i Fig.8-21. |170 |PROBLEMS im© © equation applies totheupperhalf,ifisnowthecurrent atAflowing into[~] : theupperhalf.Sowenowhaveasemi-infiniteplateasinFig.8-22(b). 7 ' (©)Now suppose thatacurrent Jcomes outofBasinFig.8-22(). 4 CalculateVo—Voagain;thensuperpose cases(b)and(¢)toobtainFig. > oN 8-22(d).Showthat,forcase(d), a ae i Ve-Vo_ 1,@+b+ob ; T “x0 "(a+b\b +e) This ratio has the dimensions ofaresistance; call itRyp.co- The contact “ co resistances atAand Bateunimportant because only thecurrent Ibetween ‘AandBissignificant.ThecontactresistancesatCandDarealso |unimportant iftheir sum ismuch smaller than the resistance ofthe voltmeter that measures Vo~Vp 4 (@)Showthat,withcurrents asinFig.8-22(e), aoafeIo Gao | Rrcon=eneTOO+e) H| (e)YoucannowderiveVanderPauw’stheorem: 1 ! 1 ; exp(105R4n.co)+XP(OSRge.pa) =1- . } ‘The only unknown iso. This result, infact, applies toalamella ofany © | shape,withcontactsA,B,C,Daroundtheperiphery ! 1 Fig. 8:22. interference from the contact resistances. We deduce this theorem for the case ofasemi-infinite plate (a)Imagine aninfinite thin plate ofthickness sand conductivity 0.A current 21lows into point AinFig. 8-22(a). ‘Show that, intheplate, E=I/-tors, where ristheradial distance toA. (b)Now consider three points B,C,DasinFig. 8-22(a), onalinegoing throughA.Showthat Ve-Vp=insete, xo" a+b Ifwecuttheplatealongthelineandremovethelowerhalf,theabove| 2THE ELECTRIC POLARIZATION P 13 |Somemoleculespossessapermanentdipolemomentandaresaidtobe CHAPTER9 |polar.Othermoleculesarenonpolar.Agiven substance canbeadielectric under normal circumstances and | secome aconductor under appropriate conditions. For example, ELECTRIC FIELDS VII |shotoconductors arenormally nonconducting, butbecome conducting . -whenexposed tolight Dielectric Materials A: Inthisfirstchapter ondielectric media weareconcerned withthebasic Bound Charges andtheElectric Flux Density D concepts ofpolarization, susceptibility, andrelative permittivity, mostlyinstatic fields. 9.1 THE THREE BASIC POLARIZATION PROCESSE:92THEELECTRICPOLARIZATION TS ”.1THETHREE BASIC POLARIZATION 9.3 FREE AND BOUND CHARGES 174 PROCESSES 9.3.1 THE BOUND SURFACE CHARGE DENSITY 0, 174 9.3.2THEBOUND VOLUME CHARGE DENSITY p,175 1)Under theaction ofanapplied electric field,thecenter ofcharget 9.3.3 THE POLARIZATION CURRENT DENSITYJ,176 oftheelectroncloudinamoleculemovesslightlywithrespecttothe 94 THE ELECTRIC FIELD OFAPOLARIZED DIELECTRIC 176 centerofchargeofthenuclei.Thisiselectronicpolarization. The ‘Example:THEFIELDOFANINFINITE-SHEET ELECTRET 177 displacement isminute, evenontheatomic scale,typically 10-*timesthe 88GAUSS'SLAW 178 Jiameter ofanatom. 9.6 POISSON'S AND LAPLACE'S EQUATIONS FOR VIN DIELECTRICS 178 9.7THEELECTRIC FLUXDENSITY D.THEDIVERGENCE OF D179 2)Polarmolecules alignthemselves andbecome further polarized inExample: THEINFINITE-SHEET ELECTRET —180 inapplied electric field.Thisisorientational polarization. However, Example: THEBARELECTRET 180 collisions arising from thermal agitation partly disrupt thealignment 9.8 ‘THEELECTRIC SUSCEPTIBILITY x, 181 9.9THERELATIVE PERMITTIVITY «, 181 (3)The third basic process isatomic polarization, inwhich ions of Example: THEFREE AND BOUND VOLUME CHARGE DENSITIES 182 Jifferent signs inasolid such asNaCl move indifferent directions when Example: THEFREEANDBOUNDSURFACE CHARGE DENSITIES ATTHE subjectedtoanelectricficld,Theferroelectric dielectrics ofSec.10.1.4 INTERFACEBETWEEN ADIELECTRIC AND ACONDUCTOR 183 exhibit atomic polarization. CAPRCTOR Te ATEDPARALLEL PLATE {Apolarized dielectric possesses itsownfield,whichaddstothatofthe Example: THEFIELDOFFREECHARGES EMBEDDED INA othercharges. Thetwofieldscanbecomparable inmagnitude. DIELECTRIC 186 9.10THEDISPLACEMENT CURRENT DENSITY30/31186 9.2.THEELECTRIC POLARIZATION P9.11SUMMARY —187 PROBLEMS 189 If,inthe neighborhood ofagiven point, the average vector dipole moment per molecule inagiven direction isp,and ifthere areN Dielectrics differfromconductors inthattheypossess nocarriers offree molecules pereubiemeter,then charge thatcandriftabout under thecontrol ofanexternally applied; P=Np (9-1) electric field. Inatruedielectric thecharges areallbound totheir atoms | ‘ormolecules, andtheycanbeforced tomove byonlyminute distances, | 8theelectric polarization atthatpoint, SoPisthedipole moment per positive charges going onewayandnegative charges theother way.A unitvolume atagiven point, Recall fromSec.5.4thatthedipole moment dielectric inwhich thisdisplacement hastaken place issaidtobe ~ polarized. +Thecenterofchargeisanalogous tothecenterofmassinmechanics. 175 174ELECTRIC FIELDS vit ofacharge distribution isindependent ofthechoice oforigin when the — netcharge ofthedistribution iszero, asitisforanormal, neutral. molecule. 5 \ 7 ¢ a 9.3 FREE AND BOUND CHARGES a OO Polarization causeschargestoaccumulate, eitherwithinthedielectricor\ >atitssurface.Werefertosuchchargesasbound.Otherchargesaresaid o— = - —————— tobefree. Examples offreecharges aretheconduction electrons ingood -. conductors, thecharge carriers insemiconductors, andelectrons injected > —0" intoadielectric bymeansofahigh-energy electron beam.’ -¢——______»»: 9.3.1TheBound Surface Charge Density o, |Fig.9-1.Element ofareadfsituatedinsideadielectric. Thevectord.sfis ImagineanelementofareadodinsideadielectricasinFig.9-1.Saythe|—_dipoleswithinthedielectric.UndertheactionofthisF,themoleculeseither. dielectric isnonpolar. When thedielectric ispolarized, thecenter of | stretch,orrotateandstretch,andanetchargeNQsdf crossestheelementof positive charge +Qofamolecule liesatadistance sfrom thecenter of area, negative charge —Q. This sisthesame forallthemolecules over an infinitesimal region. | on=fe=Pea, (9-4)Upon application ofanelectric field, m,positive charges cross the dab element ofareabymoving inthedirection ofs,andn_negative charges cross itbymoving intheopposite direction. Thenetcharge thatcrosses where fiistheunitvector normal tothesurface andpointing outward.dodinthedirection ofsistherefore Thuso,isequalinmagnitude tothenormalcomponent ofP,pointing outward. dQ=n,Q~n(-Q)=(n, +n_)Q. (9-2) it 9.3.2 The Bound Volume Charge Density p; Now1,+n_issimply thenumber ofmolecules within theimaginary Wenowdemonstrate thatinsideadielectric thebound volume chargeparallelepiped ofFig.9-1,whosevolume iss-ds.Then density p,isequalto~V~P.Thenetboundcharge thatflowsoutofa |\otume1acrossanelementdefofitssurfaceisP+ds,aswefound dQ=NOQs-ded=Np-dod=Pds (9-3) above. Thenetbound charge thatflowsoutoftheclosedsurface ofarea | delimiting avolume ventirely situated withinthedielectric (soasto whereQsisthedipolemoment pofasinglemolecule. |excludesurfacecharges) isthus Ifdsfliesonthesurfaceofadielectric,thendQaccumulates thereand thebound surface charge density is ie cae ens Onu=|Past, os) "The concepts offreeandbound charges, andeventheconcept ofpolarization, arenot, however, asclear-cut asonemightwish.Thedistinction between freeandboundcharges andthenetcharge thatremains within vmustbe~Quoe- Ifppisthefests ontheassumption thatthedielectric consist ofelectrically neutral molecules, Ithere "remaining within v,then ‘arenowell-defined molecules asinacrystalofNaCl,forexample, thevaluesofthefree volume density ofthecharge remaining ' andbound surface charge densities arearbitrary. Weconfine most ofourdiscussion here to dielectrics composed ofmolecules SeeEdvard.Purcell,ElectricityandMagneism, BerkeleyPhysicsCourse,vol.2 [psdv=-Qsu=—[ Pedst =~[vpaw (96)McGraw-Hill,NewYork,1965,p.44 i ; a 176 PLECTRIC FIELDS VIL {ATHF ELECTRIC FIELD OFAPOLARIZED DIELECTRIC 7 ‘Wehave used thedivergence theorem totransform thesecond integral 1 PF yy oF agintothethird.Sincethisequationappliestoanyvolumechosenasabove, B=el,pwtaeIa, outheintegrands areequal atevery point inthedielectric andthebound volume charge density is where #points from P’toPandwhere pand@aretotal charge densities, _freeplusbound. Pen—W-P. on Itdoesnorfollow thatonecancalculate Efrompandobecause both 9.3.3 ThePolarization Current Density Jj charge distributions depend onE: Themotion ofbound charges under theaction ofatime-dependent Example ‘THEFIELD OFANINFINITE-SHEET ELECTRET electric field generates apolarization current. Consider asmall surface difsituated insideadielectric asinFig.9-1,butnormal toP.Asthe Anleethelesaequivalentof»permanentmagnetIn polarization increases from zero toP,anetcharge dQ,=Pdetcrosses femoval oftheelectric field. butsome dielectrics retain theit definthedirection ofP.More generally, ifPincreases bydPinatime polarization forlongperiods. Forexample, certain polymers have interval dtand ifdsf isnormal todP, then acurrent ‘extrapolated lifetimes ofseveral thousand years atroom temperature dQ, \aP|det ‘Asarule, electrets have theform ofshects, with the ar nr a (9-8) polarization normal tothesurface. Onewayofpoling anelectret‘ . istoplacethematerial inanelectric fieldofabout 10°volts/meter atabout100°C. Abound surface charge buildsupasthemolecules flowsthrough dsinthedirection ofP. orient themselves, andthesample isthencooled down toroom ‘Thus if,atagiven point inspace, Pisafunction oftime, themotion of temperature without removing theelectric field. bound charge results inapolarization current density ‘One commonly used material ispolyvinylidene fluoride (PVF;).. This isapolymer composed ofachain ofCH,—CF, units. Its oP remanent polarization istypically 50to70millicoulombs/meter” Woe (9-9) ‘Thismaterial isusedinvarious typesoftransducers, microphones" forexample, because italsohastheproperty ofgoing intoa metastable polarized statewhenstretched, 9.4THE ELECTRIC FIELD OF A Forasheet electret, neglecting edgeeffects, thepolarization PPOLARIZED DIELECTRIC isuniformandequalt00,asinFig.9-2.Thenpiszero,fromEq, 9-7. Inside the sheet, Wehaveseenthatpolarization causescharges toaccumulate, eitherat Ea%, (9-12)thesurface ofadiclectric orinside, with charge densities 0,and ps, €0 respectively. Now Coulomb’s law applies toany netcharge density, regardless ofanymatter that may bepresent. The potential Vascribable toapolarized dielectric thatoccupies avolume v’bounded byasurface ofarea sf’istherefore thesame asifthebound charges were located ina vacuum: 1 Pr»du’ 1 o,dA’ = =vagh[mdetfsoda ' peapeo AmeodyorAneyla or (9-10) where risthedistance between theelement ofbound charge at ‘ P"(x',y',2")andthepointP(x,y,z)whereonecalculatesV. ” Iftherearealsofreecharges present, thenoneaddssimilar integrals E=0d=0 Ceayer VecueeBindPoot forthefreecharges.Therefore al > inoppositedireetions. 178 ELECTRICFIELDSvit 17THELECTRICFLUXDENSITYBANDITSDIVERGENCE 179 9.5 GAUSS’S LAW 9.7 THE ELECTRIC FLUX DENSITY D AND ITS DIVERGENCE Say agiven volume vcontains various dielectrics, some ofwhich may be partly inside andpartly outside. The total freeandbound charge within v According toEq.9-15, isQ,+Q,=@. There arenosurface charges onthesurface ofv.Then Gauss’s law relates theoutward flux ofEthrough thesurface ofarea sf p+ Potothenetenclosed charge Q: Eee @-18) o| ButwefoundinSec.9.3.2thatpy=—¥+P.Therefore[e-aa-2. | (9-13)“ s| F(co+P)=Pp (19) ThisisGauss’s lawinitsmore general form. Weconclude thatthevectorIfthevolumevliesentirely insideadielectric, therearenosurface charges and D=6E+P (9-20) 1 1 [,5-it=— [(0,+Ps)du==[eodv, 14) issuchthatitsdivergence isequaltothefreevolumechargedensity.This. quantity iscalled theelectric fluxdensity. Thus where p=, +ppisthetotal charge density. Applying thedivergence theorem tothesurface integral ofEgives thevolume integral ofV+E. V-D= py. (9-21) Equating theintegrands then yields Itfollows that, foranyvolume vthat liesentirely inside adielectric and p thatencloses afreechargeQ,withnosurfacecharges, V-E=Pau (9-15) ; [v-pdv=f p-dst=|p,av=0, (0-2) which isagainGauss’s law,expressed indifferential form. be a Ie This isone of Maxwell's four fundamental equations ofelectromagnetism. ThisconcurswiththerelationD=oF+P: 9.6POISSON’S ANDLAPLACE’S EQUATIONS FOR f f f { .= “dA -dd= v-Pdv VINDIELECTRICS |,D-dst=co) E-ds+) P-da&-Q+) V-Pdv(923) SinceE=—PV,fromSec.3.4,itfollowsthat =Q-[P.dv=O~Qy=Q, (9-24) VV=—pleo. (9-16) ‘Thefactthatthedivergence ofDdepends solelyonthefreecharge ThisisPoisson's equation forVindielectrics. densitydoesnotmeanthatDitselfdepends onlyonp,.TofindD,we Laplace's equation applies toregions where thetotal electric charge mustintegrate Eq.9-21subject towhatever boundary conditions apply. densityiszero: Inusingthedivergences ofE,P,andD,weassume implicitly thevv=0, (9-17) ‘existence ofthespace derivatives. Ifwehave todealwith theinterface between two media, where these derivatives donot exist, then wemust asinSec.4.1. ausetheintegralformofGauss’slaw,whichistherefore moregeneral. 184 eLPCTHICLDSvit 18s a,+0 enn (9-34) wT4 || =| +4 ObservethatGauss’slawforE,asstatedinEq.9-13,leadstothe GRE - [__] samevalueofE. \t [}—SinceD=eeE, aoea1 i —_ a=D=el0,+0), o%=-(1-2)o, (028) aS — =o,+0,=2 (0-36) + 4 — h ++ —_ ComparewithEqs.931and9-32. on Example |DIELECTRIC-INSULATED or +-4PARALLEL-PLATE CAPACITOR 4a a | haveshownairspacesoneithersideofthedielectric sheetsoasto itt—4 t— incontactwiththedielectric. \+-—t IWeassumestaticchargesandfields.Also,almostinvariably, psee 1,=0iinthedielectricandthuspy=0and¥=P=0,fromthefirst — i thesurface charge densities o,anda,areuniform. Similarly, the vectorsP,E,andDareuniformwithinthedielectricanduniform beopetes Withintheaiespaces. pan tu mo ‘The Gaussian volume Gstraddles the air-conductor interface. Within theconductor, E=0andD=0. Therefore, intheair Fig.9-6.Pictorial representation ofthemanner inwhichE’space, variesinsideaparallel-plate capacitor containing asheetof,% dielectric with€,=2.Ofcourse,individual linesofEdonotexist. Bn%, Demo (9-37) Here||=loy|/2. Similarly, fortheGaussian volume G', ‘Thevoltage Vacross thecapacitor isthus to 4 % p22 2Ente, Dimoy (9-38) vaLigthsLanZ(g+h+2), (0-41) But where 0,=Q/sf, Qbeing themagnitude ofthecharge andsfthe area ofone plate. bywPen (039) ‘Thecapacitance isoes cH2-9 eet _ (9-42) Itfollows that VV “gthtsle, a=-(1-bo, (9-40) Withoutairspacesoneithersideofthedielectric,g+h=0and . y--2.5, (9-43) Ofcourse, |o,|=P,fromSec.9.3.1. fob 190 ELECTRICFIELDSvi pRowtens 191 Inoneparticular instance, a0.1-microampere beambombarded anarea 9-10,(9.9)‘Theresistance andthecapacitance between twoelectrodes of25centimeter’ ofLucite (¢,=3.2)for1second, andessentially allthe Whenthespacebetween theplatesofaparallel-plate capacitor isfilledelectrons weretrapped about 6millimeters below thesurface inaregion withadielectric ¢,ithasacapacitance ofCfarads. Ifthedielectric is about 2millimeters thick.Theblockwas12millimeters thick. replaced byamaterial whose resistivity p=1/0ismuchsmaller thanthatInthefollowing calculations, neglect edgeeffects andassume auniform Oftheelectrodes, theresistance between theelectrodes isRohms. density forthetrapped electrons. Assume alsothatbothfacesoftheLucite (a)Show thatRC=pe,neglecting edgeeffects. areincontact withgrounded conducting plates. (b)Show thatthisresult alsoapplies tocylindrical andspherical(a)Whatisthebound charge density inthecharged region? capacitors.(b)What isthebound charge density atthesurface oftheLucite? (©)Show thatthisresult applies toanypairofelectrodes submerged ina (©)Sketch graphs ofD,E,Vasfunctions ofposition inside the ‘medium whose resistivity pismuch larger thanthatoftheelectrodes.dielectric. Youshouldbeabletoshowthatthefieldisunaffected bythe(a)Showthatthepotential atthecenter ofthesheetofcharge isabout4 conductivity, withtheabove restrictionkilovolts. Oneimportant application ofthisfactistheelectrolytic plottingtank,(e)What istheenergy stored intheblock? Could theblockexplode? which isusedforplotting electric fields intwo,andinsomecases three, limensions. 9-7.(9.9)Capacitance withandwithoutadielectric otemedium‘occupies onlytheregionbetweentheelectrodes, thenRCAconducting bodyA.ofarbitrary shape, isgrounded. Another isagainequaltope,except thatCdoesnotinclude thefringing fieldandis‘conducting bodyB,ofarbitrary shapeandposition, ismaintained ata therefore smaller thanthetruecapacitance.potential V.Inair,thecapacitance isCy.Showthatthecapacitance is€,Cy (a)Thecapacitance perunitlength between twoparallel wiresofwhenthebodiesaresubmerged inalargebodyofdielectric €,. diameter dandseparated byadistance DisC’=xe/cosh"" (D/d). Find ice betw wires 10millimeters indiameter sepa- 9.8.(9.9) Thecoaxial line theconductance between parallel wires iE Figure33-4showsasectionofcoaxialline.Adielectric¢,fillsthespace fatedby100millimeters andsubmerged inseawater(0=5) betweenthetwoconductors. FromProbs.6-5and9-7,thecapacitance per 9-11,(9.9)Thecurrentbetween twoelectrodes inaconducting medium. unitlengthC"is22r¢,¢u/In(Rz/R,).Theouterconductorisgrounded,and SeeProb.9-10.Theelectrodes ofpart(c)areinitiallycharged,and theinner conductor isatthepotential V. discharge through themedium. Show thatJ,+9D/3t =0. (a)Calculate thecharge perunitlength 4ontheinner conductor (b)Show that thebound charge perunitlength ontheinner surface of thedielectric is~A(1 =1/¢,). ‘Thus thenetcharge perunitlength attheradius R,is4/e,. Thebound ‘charge perunitlength ontheouter surface ofthedielectric issimilarly +A(1~ I/e,), andthenetcharge perunitlength atRyis~A/e,. (©)Show thatthevolume density ofbound charge iszero. (a)Draw graphs ofD,E,andVasfunctions oftheradius rfrom r=Ry tor=Rz, forV=100volts, R,= 1.00 millimeter, R;= 10.0 millimeters,=3.00. 9-9. (9.9) Aparallel-plate capacitor with anonuniform dielectric ‘Thedielectric ofaparallel-plate capacitor hasapermittivity thatvaries as0+ax,wherexisthedistancefromoneplate.Theareaofaplateissf,andtheir spacing iss (a) Find thecapacitance. (b)Show that, if¢,varies from éot02,0, then Cis1.44 times largerthanifawerezero.(©)Find Pfrom thevalues ofDand Eforthat case. (d)Deduce thevalue ofps (©)Now calculate p,from therelation given inProb. 9-2 (f)Draw curves ofE,p,, and Pasfunctions ofxfor é9=3.00, 4@=€0/s, 5=1.00millimeter when theapplied voltage is1.00volt 196 ELECTRIC FIELDS Vil 102 THE CONTINUITY CONDITIONS AT AN INTERFACE 197 components, andifonechooses thecoordinate axes properly with respect eventually disappears above theCurie temperature. This temperature isa tothecrystal, these sixcomponents reduce tothree. characteristic ofthematerial Ifthevarious components ofthesusceptibility arenorfunctions ofE, ‘Themain advantage oftheferroelectric dielectrics incommon usein then Pisalinear function ofthecomponents ofEandthedielectric is ceramic capacitors isthat they have large relative permittivities, ranging linear. upto10,000.Asarule,thesedielectrics arecompounds oftitanium.Inanisotropic dielectrics, itisstilltrue that Some ferroelectric dielectrics possess definite temperature coefficients that areuseful incertain applications. D=6E+P, VeD=V-(6E+P)=pp(10-8) *10.1.5Hysteresis butthethree vectors areusually notparallel. . 7 |Ferroelectric materials also exhibit aproperty called hysteresis, illustrated *10.1.4 Ferroelectricity inFig.10-3.Then theratioD/Edepends ontheprevious history ofthe material. Such materials are thus said tobenonlinear because Disnot Ferroelectric dielectrics havethepeculiar property ofexhibiting spon- proportional toE. taneous polarization overmicroscopic crystalline regions called domains. Therelative permittivity ¢,=D/eoE ofagiven ferroelectric sample Thewordferroelectric originates fromthefactthattheirbehavior isin thushasnodefinite valueandcanevenbenegative. When onequotes several respects similar tothatofferromagnetic substances (Sec.21.1). relative permittivities forferroelectric materials, aswehavedoneabove When placed inanelectric field,domains thathappen tobecorrectly andinTable 9-1,onerefers totheorder ofmagnitude ofD/egE, some polarizedgrowattheexpenseofneighboring domainsandeventually distancefromtheoriginE=0,D=0,inthefirstandthirdquadrants.coalesce. Ifthetemperature ofaferroelectric substance increases slowly, the 10.2 THE CONTINUITY CONDITIONS AT AN spontaneous polarization varies, usually insome complex fashion, and INTERFACE. 10.2.1 The Potential V ower. Thepotential Viscontinuous across theboundary between twomedia. ne Otherwise, adiscontinuity wouldimplyaninfinitelylargeE,whichis 0.06. 1physically impossible. > 08 10.2.2 The Normal Component ofD hor Consider ashort imaginary cylinder spanning theinterface, and ofcross 44 section sfasinFig.10-4,Thetopandbottom facesofthecylinder =12 -08/ -04 “| Foe 12«0 areparallel totheboundary andclose toit.The interface carries afree a0 & surface charge density 0. According toGauss’s law(Sec. 9.5), thenetfluxofDcoming outof ~00y thecylinder isequal totheenclosed freecharge. Now theonly fluxofD 0K isthat through thetopand bottom faces because theheight ofthe cylinder issmall. Ifnowtheareasfisnottoolarge, Disapproximately “ 008 uniform over it,and then 103. veforatypicalferroelectricmaterial.Startingwithan -= =D)-a= r Unbolaized sampleand&'=atpointaanincreaseinincreasesDitopointb PanPdt=O78,(DaDi)B=Op,(198) TendecreasingFtoerlanofburesoontrepreviousstonyctthe wherefistheunitvectornormaltotheinterfaceandpointingfromspecimen. imedium2tomedium 1 19 198 yo Rs y S if Nig Eg 72 af” Fe i, tegrationspanningtheinterface Diy fee I} & f between media1and2.The (E tangential components ofEare Ssequal: Ey,=Ey, —§ \ Li \ Le \ with @defined asabove. The tangential component ofEiscontinuous -NL acrossanyinterface Fig. 10-4. Imaginarycylinderstraddlingtheinterfacebetweenmedia|and2and 10.2.4BendingofLinesofEatanInterface Seliinganares,ThedfernceDusDibetweenthenormalcomponents TheFandDvectorschangedirectionattheinterfacebetweentwodifferent linear andisotropic dielectrics. InFig. 10-6, ifthere iszero free surface charge density, thecontinuity ofthenormal component ofD Asatule,theboundary between twodielectrics doesnotcarry free requires that charges, and then thenormal component ofDiscontinuous across the interface. Thusthenormalcomponent of£isdiscontinuous. D,cos8,=D3cos82 (10-12)On the other hand, ifone medium isaconductor and the other a dielectric, andifDisnotafunction ofthetime, then D=0 inthe orthat conductor andD,=0;inthedielectric, IfDisafunction ofthetime, Eq 10.9stillapplies, butDisnotzero intheconductor. En€ok 6088)=€n€oE2 C0882 (10-13) The1 7, Also,frthe tinuity ofthetangential itofE, 10.2.3 The Tangential Component ofE SOeTronthecontinulty Bentialcomponent o Consider nowthepathshown inFig.10-5,withtwosidesoflength L \ 1paralleltotheboundary andclosetoit.Theothertwosidesare |\|infinitesimal. IfLisshort, Edoesnotvarysignificantly overthat ;\ |distance,andintegratingoverthepathyields nl| fo\w $Bdt=BL Eyl (10-10) i} ‘ | ga Now,from Sec,3.4thislineintegral iszero, andthus . or Ey=Ext, or(E,~E:)xa=0, (10-11) OFsec feeff BK Fig.10-6.LinesofBcrossing . Z\ __theinterface between media1 1ThelineintegralofBdiszeroinelectrostaticfields.Moregenerally,it6equalto Lf \an 2Thepermiiiyof ‘minusthetimederivativeofthemagneticluxlinkingthepathofintegration,asweshall aNGGhee otium2largerthenthatofisgeneral '200 eLecrnicFIELDSvit|103THEPOTENTIALENERGYOFACHARGEDISTRIBUTION 201 Eysin0,=Esin3. (10-14) asinSec.9.4.Sinceallthedielectrics inthefieldarelinear,by ; ; ; . hypothesis, p,isproportional top,and9,too,(Sec.9.9,examples).Dividingthethirdequationbythesecondgives Westartwithazerochargedensityeverywhere andgraduallypullintan,_tan: {charges frominfinity.Wedisregard surfacechargesforthemoment. Let“aea" (10-15) thefinalpotential andthefinalvolume freechargedensityatapointbe,° |respectively, Vandp,Atagivenmoment, thefreevolume charge Thelargeranglefromthenormalisinthemedium withthelarger densityisap,,whereaincreases gradually fromzerotounity.permittivity. ThelinesofD“prefer” themedium withthehigher Since permittivity. Thisruledoesnotapply toelectrets. SeeProb. 10-10.Since thenormal component ofEisdiscontinuous, thereexists a VD=V-(€,€0E) =~€0¥+(€,VV) =pp, (10-18) bound surface charge density. From Gauss’s law, ;Visproportional top,.Thus thepotential isaVwhen thecharge density 04=E26088;~€oE,0080,=€oEsc0s 0.(1-£2).(1016) |'sapy -& j Suppose thataincreases toa+da.Then,atagivenpoint, Forexample, ifmedium 1isairandmedium 2isadielectric, thenthe} dp,=p, -boundsurfacechargedensityisnegative ifEpointsintothesurfaceof ly=pyda, dV=Vda. (10-19) thedielectric. Theenergy required topull intheextra charge is 10.3 THE POTENTIAL ENERGY OF ACHARGE da DISTRIBUTION INTHE PRESENCE OF ae=(p,dav(a+%) dv,(10-20) DIELECTRICS. ; where thevolume vincludes allthecharges and where V(a +da/2) is InChap. 6wecalculated theenergy stored inanelectric field when the theaverage potential during theoperation. Disregarding the(da)* term, charges reside either onconductors orinfreespace. wefindthat ‘Theaddition ofdielectrics tothefield constitutes amajor complication, buttheproblem becomes tractable with thefollowing assumptions.‘Ourfirsttwoassumptions arerealistic.(1)Weassumethatboththe d=[(pVdv)ada, (10-21) freeandbound charge distributions areoffinite extent. This makes the \potentialVequaltozeroatinfinity.(2)Weassumethatthedielectrics é-iiadaipVdv=3[pVdv. (10-22) arelinear(Disproportional toE),butnotnecessarily homogeneous or 0I, I, isotropic.‘Wealsoassume thatthedielectrics arefixedinposition andrigid.In Thepotential energy stored inthefieldisthusequaltoone-half ofthe ‘otherwords, thedielectrics donotdistort undertheelectric forces. This volume integral ofpyV,exactly asinSec.6.1.2. However, Vnowassumption isnotrealistic, butwecandonobetterinthepresent depends onthenature, shape, size,andposition ofthedielectricscontext situatedinthefield10.3.1ThePotential EnergyExpressed inTermsofp,andV Example |THEDIELECTRIC-INSULATED APARALLEL-PLATE CAPACITOR 1any point, 1 ret 1 pote,Asheet ofdielectric separates the plates ofaparallel-plate veralPetPoay+xz,FOag,(10-17) capacitor,oneofwhichisgrounded.Thentheenergyrequiredto ane), or 4neose 7 charge theother plate tothepotential VisQV/2, where Q=CV. 202 ELECTRICFIELDSVit 103THEPOTENTIAL ENERGYOFACHARGEDISTRIMUTION 203 6 E Ths .dé=(odsa)(5)dt, (10-25) 6-2,(10.23) otathefrstexample taBes,64 thefieldactingon0dstbeingE/2,fromSec.6.5.So ForagivenchargeQ.thepresenceofthedielectricdecreases V cope byafactorof¢,,andhencedecreases&bythesamefactor. pact =a aaaEanHowever, Ciproportional to€and,foragivenpotentialV,the dé=Dodd>dl=detd= dv. (10-26) stored energy isproportional to€,. 10.3.2 TheEnergy Density Expressed inTerms ofEandD Theconducting surface continues toexpand untilitreaches the ; * -|equipotential V=Vix.Intheend,ithassweptthrough allspace,the Ourthought experiment herewillbesimilar tothatofthestarred Sec. freecharges havebeeneither removed toinfinity orcanceled bythead-6.2.1, |ditionofopposite charges ontheconducting surface, andEiszero ‘Assume thatthere isoneandonly oneregion where Visminimum, everywhere. With Ving=O.Similarly, there isoneandonly oneregion where Thusthetotalwork donebythefieldis V=Vinax =0. The dielectrics arelinear Now imagine aconducting surface ofarea along theequipotentialV=Vauq.Thereiszerofieldinside.Theconductor expandsslowly,from t={JE+Ddv, (10-27) coneequipotential tothenext, sweeping through thedielectrics and . picking upthefreecharges thatstand initsway. . .‘Theconductor carries afreesurface charge density 0.Atthesurface, andtheenergydensity is Eisnormal anditsmagnitude isequaltoo/€,, asinFig.10-7.Ifthe g=i8-p.medium isnotisotropic, Disnotparallel toE.However, =u . (10-28) odd=D-det, (10-24) Ifthedielectrics arelinearandisotropic, D=¢,¢yEand from Gauss’s law,Dbeing zeroinside theconductor. GeeSaytheelementofareadfmovesoutbyadistancedlasinthefigure, | t=[“ya. (10-29) dlbeingnormaltothesurface.Thentheworkdonebythefieldonthe| “charge 0dfis ‘Theenergy density isthen€,€o£*/2. Thisintegral extends overallthe | volume occupied bythefield. |Example|THEDIELECTRIC-INSULATED| PARALLEL-PLATE CAPACITOR oly Returning tothecapacitor oftheexample inSec.10.3.1, wefind v that, foragiven E,itsstored energy isproportional to€,,as eo < . previously. Asusual, wehaveneglected edgeeffects * 10.3.3 The Energy Density Associated with Polarization Fig. 10-7. Thought experiment forcalculating theenergy stored inanelectric sfieldinthepresence ofdielectrics, Theshaded volume ofareasfisaconductor Sioce thatexpands slowly, fromoneequipotential surface tothenext.Thesurface vot >chargedensityis0.Theelementofareadsmovesoutbyadistancedl. @=1BD=3E -(6oE+P)=heoE* +4E-P, (10-30) 204 ELECTRIC FIELDSVi PROBLEMS 205 theenergy density associated withthepolarization alone isE-P/2. Thepotential energy stored inacharge distribution, when dielectrics Inisotropic dielectrics, arepresent, isgiven either by deck =H 2 8=decok?=11+Z.)e0E* (10-31) e-ifpvde, (02) Inmost common dielectrics, theelectric susceptibility x,ranges about. from 1to4,andmost of@ispolarization energy. where visanyvolume containing allthefreecharges ofthesystem, orby 10.4ELECTRIC FORCES INTHEPRESENCE OF @=[18-Ddu, (1027)DIELECTRICS where vnow includes alltheregions where theEofthecharge Electric forces between conductors inthepresence ofdielectrics arebest | -—_distribution exists. Forlinear andisotropic dielectrics, calculated bythemethod ofvirtual work, asinSection 6.6 When the conductors are immersed inadielectric, the forces are - 2 smaller than those inairbyafactor of¢,ifthecharges arethesame. e=ifSe,€oB?dv. (10-29) ‘They arelarger bythesame factor iftheelectric fields, and hence the voltages, arethesame. Onemaytherefore ascribe toanelectric fieldanenergy density E»D/2 Thecaseofsoliddielectrics isbestillustrated byProbs, 10-13to10-15, |or,forlinear andisotropic dielectrics, €,¢o£°/2. Theenergy density associated with thepolarization itself inadielectric isE+P/2orx.€9E*/2 |ifthedielectricisisotropicandlinear. 10.5SUMMARY ]PROBLEMSAlthoughsomedielectricsareclosetoideal,mostexhibitamore|complexbehavior. Iftheconductivity oofadielectricisnotnegligible, 10-1.cas1.1)Paralletpate capacitorwithaconducting dielectric thenitscomplex relativepermittivity is confucivity ou,Problems wecallasurfacechargedensityomand= 5 Thedielectric ofaparallel-plate capacitor hasarelative permittivity €, ene-je, fa (10-4) andaconductivity 0...Theconductivity ofthedielectric ismuch lessthan eo thatoftheplates, which makes Euniform between theplates. ' (a)Assume surface charge densities o.,ontheplates, anduseGauss's Generally speaking, c;decreases with frequency and €7increases. lawtorelateEtoogy.Thenshowthattheimpedance ofthecapacitoristhe ‘Temperature effects canbelarge. same asthatofaresistor andacapacitor inparallel. Some dielectrics arenonisotropic. Theneachcomponent ofPdepends (b)Showthatwhenthecapacitor isdisconnected, thecharge ontheonthethreecomponents ofE,andthesusceptibility isatensor. Gerotteapecnns @factorof¢in€/0.5seconds.ThisistherelaxationFerroelectric dielectrics exhibit large permittivities and hysteresis: the | valueofDdepends ontheprevious valuesofE. ae eae aPutecapacitor withtwoconducting dielectrics‘Attheinterface between anytwomedia,bothVandthetangential Figg” Parallatecapacitor ismadeupoftwoparts,wsin component ofEarecontinuous. The normal component ofDcanbe (a)Find theimpedance Z. discontinuous, however, thediscontinuity being equal tothefree surface (b)Call aconductivity 0.andasurface charge density oy,Calculate the charge density ontheinterface. surface charge density ontheinterface. Lines ofEbend ataninterface, thelarger angle from thenormal being 10-3. (10.1.1) Parallel-plate capacitor with anonuniform dielectric inthedielectricwiththelargerpermittivity. i;‘Aparallel-platecapacitorhasplatesofareasfseparatedbyadistances. 206 207 He «. «| 1) ° ° ¥®IT aad To, oe a ee || ° ° ! | 10-10.u Fig.1048, Fie | 10.5. (10.1.4) The hysteresis curve ofaferroelectric material Figure 10-10 shows aschematic diagram ofaninstrument that hasbeen .usedtoplotthehysteresis curves offerroelectric materials. Anoscillator Isicteric hasaconductivity oa+br,wherexthedance 10one vopiientaacensng voltageVWoteeteqeciney tatote,thepencePlate,andauniformrelativePeatthecopachor. platecapacitorC,containing thematerialandanormalcapacitorC>>C,. (a)Calva theresistanceRofthecapaci {ThevoltageacrossCgoestotheYinputofanXYrecorder,whilethe (b)Showthatwith3seedyhipreappli " voltageacrossR»<R,goestotheXinput.Theferroelectric sampleisaaumiform volumedensityoffreeCharge. fewmillimeters thik,andV10Kilovolts,f~10 "here©waeNoesOtforstageaevoestheelectro . ExplainwhythevoltageacrossCisproportional totheDinthesample (@)Withanalternating voltage rodes, containedinC,,whilethatacrossR;isproportional toE,TherecorderoD draws essentially zero current atitsX’and Yterminals,1=a(oe+22) =, ; 3 10-6. (10.1.5) Thestored energy density inanonlinear dielectric Aparallel-plate capacitor whose dielectric isnonlinear isconnected toa Showthat¥-J.=31/31=.Theniindependentos powersupp,ThevoltageVineasssightbydVananextacharge (©)ShowhaE114BF artoZisntthethat Qflowsintothecapacitor.Showthatthedensityofstoredeneraywefoundabove. However, if«=0,wereverttoR,asexpected, SeeProb. increases byEdD. 101 10-7. (10.1.5) Hysteresis inferroelectric materials ()Find p,. Show thatthearea ofthehysteresis loop foraferroelectric material isoa(un stor equaltotheenergydissipatedpercubicmeterandpercycle."One particular ceramic capacitor iscylindrical, withthreeelectrodes asin 10-8.(10.2) Theboundary conditions attheinterface between twoconductors Fig.10-9. Theceramic disks each have adiameter of21millimeters anda ‘Show that, attheinterface between twoconductors, thickness of0.5millimeter. Thenominal capacitance is0.05microfarad .withinarangeof~20%to+80%.Whatistheapproximate valueof«,? (E-Byxa=0, Uy-ayxa-“2, | where fisaunit vector that isnormal totheinterface andpoints away from conductor 1,andwhere avis thesurface charge density ontheinterface. 10.9, (10.3.1) Capacitive energy storage forasmall vehicle Investigate thepossiblity ofpropelling asmallvehiclewithanelectric <= motor fedbyacharged capacitor. Consider only theproblem ofenergy storage. \ (a)Show that themaximum energy density inthedielectric of2 o parallel-plate capacitor is¢a°/2,whereaisthedielectricstrengthoftheFig.10.9, insulator, orthemaximum Ebefore breakdown. Agood dielectric touse 208 209 +o i 3, |nd te » = + Lt 1| |@ Fig.10-12. hk Fig. 10-11. p=2x10™ coulomb-meter andfor700stages. That accelerator hasa would beMylar, which hasadielectric strength of1.6% 10°volts/meter length of10meters. when intheformofthinsheets andarelative permittivity of3.2,The 10-13. (104) Electrostatic d: icclamps si times lesswit lytic capacitors. ' rorimate machined, forholding silicon wafers during electron beam microfabrica- thecapacitor thatyouwould need tooperateaI-kilowattmotorfor1hour tion,etc.Theycompriseaninsulatedconductingplatemaintainedata 10-10, (10.2.4) ‘Thebarelectret Potential ofseveral thousand volts andcovered withathininsulating sheetFigure9-4showstheEandDfields ofabarelectret. Theworkpieceorthewaferrestsonthesheetandisgrounded.Itis (a)Show thatthelines ofEdonotbend atthecylindrical surface, but advisable toapply afilmofoiltothesheet toprevent sparking. that the lines ofDdobend. ‘Oneparticulartypeoperatesat3000voltsandhasholdingpowerof2 (b)Show thattheinverse istrueattheendfaces. atmospheres (2x10°pascals). Iftheinsulator isMylar (€,=3.2), what is 10-11. (10.3.2) The potential energyofadipolesituatedinanelectricfield itsthickness (a)Adipole offixed dipole moment porients itself in@uniform electric 10-14. (10.4) Self-clamping capacitor fieldE.Show thatitspotential energy is~pE, assuming thatthepotential Acertain capacitor consists oftwopolished circular aluminum plates, 237 energy iszerowhen thedipole axisisperpendicular tothefield millimeters indiameter, separated byashect ofplastic 0.762 millimeter (b)Anonpolar molecule acquires adipole moment inauniform electric thick, with arelative permittivity of3.0.Thin films ofairsubsist between field thatgradually increases from zero toE.Show thatitspotential energy theelectrodes andtheplastic. This reduces thecapacitance below therated is+pE/2. value, and there isnoway ofclamping theplates mechanically with sufficient force 10-12. (10.3.2) Anaccelerator forneutral molecules‘There existaccelerators forneutral molecules thatoperate asfollows. (a)Someone suggests thattheelectric forcealone might besufficient to Figure10-11showsapairofspheresthatcarrycharges+@and~Q,anda clamphepate,attheoperating wotageof0klovos, Whatiyourmoleculeofdipolemomentp.Themoleculeaccelerates towardthespheres inion ,untilitreachestheirmidpoint. Atthatinstantthespheresaredischarged, 0)ShowAnat,ithecomplete capacitor issubmerged inanoilwithandthemolecule continues [email protected] €.%3,theforceis3dimesless.Seethenextproblem.accelerators serve tostudy theprocesses that occur during molecular 10-15, (10.4) Theclamping force, with andwithout anairfilmcollisions InProb.10-14wefoundthattheelectricclampingforceonacapacitoris(a)Find thekinetic energy acquired bythemolecule ifthedistance xin larger byafactor ofe,when there areairfilms between theelectrodes and thefigure isinitially much larger thanthedistance Dbetween thepairof thedielectric, foragiven applied voltage, oragiven E.Thisisparadoxical electrodes. Consider thespheres aspoint charges, andapply theprinciple Foragiven E,theenergy density is¢,times larger inadielectric than in‘ofconservation ofenergy.Assumethatpisconstant,andrefertoProb. air.Thentheforceshouldbe¢,timeslargerwhentherearenoairfilms.10-11. Thedipole moment infactincreases asthemolecule approaches, so ‘Youcanexplain thisparadox byconsidering thethree capacitors ofFig.thatwehaveunderestimated theenergy. 10-12.In(2)theairfilmismuchthinnerthanthedielectric. Theelectrode(b)Inoneparticular accelerator theelectrode voltagesare£40kilovolts, spacingis$inallthreecapacitors. Foreachcasefindthesurfacechargetheir radius is0.25 millimeter, and D~=1.00 millimeter. Calculate the densities 0ontheelectrodes aswell asD,E,and#intheairandinthe approximatevalueofthekineticenergyofamoleculeinelectronvolts fori dielectric 20 ELECTRIC FIELDS vit 10-16. (10.4) Electric forces, with andwithout adielectric Electric forces onconductors immersed inliquid dielectrics arelargerthaninaitbythefactor¢,ifthevollagesarethesame.Theyaresmaller 11thaninairbythesamefactor ¢,ithecharges arethesame. Canyoujustify CHAPTER these general statements? This isnotincontradiction with Probs. 10-14 and10-15, where wehada soliddielectric withthinfilmsofairoroilnexttotheelectrodes. ELECTRIC FIELDS IX Images. Laplace's Equation in Rectangular Coordinates 11,1 THE UNIQUENESS THEOREM FOR ELECTROSTATICFIELDS211 "1.2 IMAGES 212 Example: POINT CHARGE NEAR AGROUNDED CONDUCTING PLATE 212 Example: POINT CHARGE NEAR ADIELECTRIC 213 “113 SOLVING LAPLACE’S EQUATION INRECTANGULAR COORDINATES 215 Example:THEFIELDBETWEEN TWOGROUNDED SEMLINFINITEPARALLEL ELECTRODES TERMINATED BY APLANE ELECTRODE MAINTAINED ATA FIXED POTENTIAL; FOURIER SERIES 218 U4 SUMMARY 221 PROBLEMS 223, |Uptothispointwehavelimitedourdiscussion ofelectricfieldsto general considerations and tosimple charge distributions. We now address thecalculation ofmore complex fields, both inthischapter and in thenext one. That wil!complete ourstudy ofpurely electric fields. 11.1 THE UNIQUENESS THEOREM FOR ELECTROSTATIC FIELDS According totheuniqueness theorem forelectrostatic fields, there cannot exist more than one potential V(x,y,z)thatsatisfiesbothPoisson's equation and agiven setofboundary conditions. This theorem is important because itleaves usfree touseanymethod, even intuition, to , find V.Ifwecan, somehow, discover afunction V(x, y,z)that meets these two requirements, then itistheonly possible potential function. Several proofs ofthistheorem areknown, each one based onitsown particular setofassumptions. We simply assume that the uniqueness theorem always applies. 2a 1113SOLVINGLAPLACE'S EQUATION 2s reqh ~(;21)0D|— E.=(1ime (1-6)ee) attey 2 oD > - 7 =Fe+Ladner ay rk ODvine,ky, Since€,Ey=En,thenormalcomponentofDiscontinuous -T4\ acrosstheinterface.Thiswastobeexpectedbecausethefree i,‘one } surfacechargedensityiszero. *a{ ‘Thankstotheuniqueness theoremandtoimages,ourproblem |p isnearly solved. | E 4 i ‘TheFieldontheLeftoftheInterface (a) The boundary condition isstated inEqs. 11-4 and 11-5. ( From Eq. 11-4, wecanremove thedielectric and replace itbyan a) image charge Q',asinFig.11-4(a), without disturbing E,..Then, from theuniqueness theorem, thefield everywhere ontheleftof Fig.11-3. Point charge Qnear alarge block ofdielectric, The theinterface issimply thefield ofQplusthatofQ’. arrows forthefields areoriented ontheassumption that both Q (b) Now, according toEq. 11-5, E,,isalso unaffected ifoneand9,arepositive.IfQispositive,ois,infact,negative removesthedielectricandreplaces@byacharge FromGauss’slaw,0,providesequalnormalfieldstrengths o=2e,0. cas)10,/2¢y oneither side ofthe interface, oriented asinthefigure. We ealshallfindthatthesignof0,isoppositethatofQ. we gedistribution, andhencethe“Therefore,justinsidethedielectric,thenormalcomponentofE aldathevogodwoageinterstedinBothacororteoetelentPointing totherightis fieldisthatofFig.11-4(a. - OD,»Furer*de, arty ‘TheFieldontheRightoftheInterface Now, from Eqs. 9-4and9-28, (a) From Eq.11-6, Eyisthesame asifonehad, instead of - 5 andthedielectric, thechargeQandacharge~Q"atadistance D Op=PomPym —E€, —Ew (ba) totherightoftheinterface. Thisisofnointerest because this is a changesthefieldontheright serethewieconAfnoretodheitefcnandpoief Ta. 117,oeidomthesighofthsDetectsistssame asifthere were nodielectric andacharge Qwere substituted o.~ f=DOD as) forQ,asinFig.11-4(b). oa(e +Dr (¢)Figure11-4(c)showsanotherpossibility: thedielectricextends onbothsidesoftheinterface, andQ"replaces Q. Asexpected, Qand0,have opposite signs. ‘igure 11-5shows: ind equipotentials forthisfield.Ifwetaketherighthand direction aspositive, thenormal Figure11-5showslinesofDandequipotentatsfor components ofHoweandinsidethedelet, respectively, “11.3SOLVING LAPLACE’S EQUATION IN are as follows: - * 1) opt RECTANGULAR COORDINATES B=(14St)OP ay“ eer Solutions ofLaplace’s equation V?V=0areknown asharmonic func- 26 OD (ans) tions. These functions possess anumber ofgeneral properties, ofwhich©+Laxey” ‘weshall usethefollowing one. Ifonecanfindsolutions V,,V>,V3... of 218 LECERIC FIELDS 1x 219 Observe how astute this is:wehave transformed apartial differential , equation inallthree variables x,y,z into three simple ordinary differential equations inx,andy,andz Sometimes, asinthe example below, one has tosum aninfinite number ofsuch solutions, each multiplied byasuitable coefficient, tofita aegivenboundarycondition. [Aora*Example |THEFIELDBETWEEN TWOGROUNDED 22 ieSEMI-INFINITE PARALLEL ELECTRODES. G ae a TERMINATED BYAPLANE ELECTRODE nol eeree pt MAINTAINED ATAFIXED POTENTIAL; a a FOURIER SERIES Figure 11-6 shows theelectrodes. Wewish tofind V(x,y)between cL theplates. Byhypothesis, thefield isindependent of2and C,=0. = Then@Xpay€Y_ps Fig.11-6.Grounded, plane,parallel,andsemi-infinite electrodesBeneS Gacy. (ui-t4) terminated byaplaneelectrode atthepotential V, Wehavesubstituted k?forCyand~KforCstoavoid square whe ‘are constants ofintegration. Wecanagain verootsinthesolution.Asweshallseeimmediately. C,mustbe |tesSolsuonoysusttonsinceemiswotrowta negative.infinity, G=0.Thus We solve the ¥equation bysetting =x¥=Csin™ exp(2 2 Y=Asinky+Bcosky, (ais) Vax¥=Csin™ exp(-"). (1-20) whereAandBareconstants. Thissolutioniseasilyverifiedby whereCisanotherconstantsubstitution, ‘Butthisisnotright!AlthoughthisVsatisfiesconditions(1)and venvmysayvoglowingrarigpnivions@ (3),itclearly‘doesnotsatisfycondition(2).Wecanalsosatisfy EOatxnO,y=;2)V=Yoatx=0;(3)V0asxm, conditi addinganinfinitenumberofsuchsolutions SinceV=0aty=0,thenB=0. ondition(2)byaddinganinfinitenumberofsuchsolut ‘Also,thecondition V=0aty=brequites that = any nmna v=>Cysin™™exp("2 an Ayan, k="(21,23...) (1H16) =’(7) So ~‘Then weusecondition (2)toevaluate thecoefficients C,,bysetting Y=Asin™™ (n=1,2,3,...) (117) *=0 So b Se int!Vo=DCsin (11-22) Weignore thevalue n=0because itcorresponds tozerofield 4 b ‘Thedifferential equation forXisnow forallybetween 0andb @X_ (nay? ‘Aninfinite series oftheform a-(F)x (118) /i. andD(Gin +,cos). noxnaw a b > X=Ges"+Hexp(-™*), (419)il whereC,andD,areconstants,isaFourierseriesandformsa 20 FLECTRIC FIELDS 1X 2 complete set:givenareasonably well-behaved’ function V(y) . “ hoegdefinedintheintervaly=0toy=b,thereexistsaFourierseries °f\ /\\ thatsequaltoVy)inthisinerva Lr» > LNWecanfindthevaluesoftheC,coefficients byaningenious PARA FARA Ytechnique devised byFourier. Fist,wemultiply bothslesofEq i11-22 bysin(pzy/b), where pisaninteger, andthenweintegrate onfromy=01t0y=b: \| [rosin ay=S[casin2sinPay,(11-28) | Ontheleft-handside, / [rosin ay=prKPO (1-24) wl. 0 itpiseven . aE Wa mn v0 ‘The terms ofaFourier series arethus said tobeorthogonal. On . theright,ry 0ifpen Fig.11-7.TheconditionV=V,atx=0,assatisfiedbyaFourierseriesbytakingfsin" sin?ay<4eb (1125) (a)onlythefirstterm,(b)thefirst3terms,(c)thefirst10terms,and(d)thefirstna Soa) SS itpen * 100terms. Thefitimproves asthenumberoftermsincreases, butthespikesat 2y=Oand y=, where Vis discontinuous, remain. This istheGibbs Itfollows that ‘phenomenon. 4v condng itmisodd (129 |thatV(y/b)rapidlybecomesasinecurve,withincreasing x. ”i. yure11- theequipotentials 0ifmiseven. Figure11-8showstheeqpipotcatiats. Finally, wed 11.4SUMMARYfy na nox vant Ssinexp(-“*) (1127) . Tn b b ‘The uniqueness theorem states that, foragiven setofboundary ‘Thesuccessive termsintheseriesbecomeprogressively less conditions, theeisonlyoneposible fl.Thistheorem isofenormousimportant, because ofthefactor1/n,butmostly because ofthe Practical importance. If,somehow, onecanfindaV(x,y,2)thatsatisfies ‘exponential function, Figure 11-7shows thedegree ofapproxima- both Laplace's (orPoisson's) equation andthegiven boundary condi- tionachieved with 1,3,10,and100terms oftheseries. tions, then itisthecorrect potential function. Note howfasttheexponential function decreases withx.For ‘Themethod ofimages canoften simplify thecalculation ofelectricn= Landx=bitisdown(oexp(~),or49%.Roughlyspeaking, fieldsthatinvolveinterfaces between different media.Itconsists in thefieldofthechargedplatedoesnotpenetrateinthexdirection diffe fieldontheothersideofthefi beyond adistance equal tothespacing bbetween theplates. setting up,onpaper, adifferent omtheother sideoftheinterface, Notice alsohowfasttheexponential decreases withn.Forn=1 with fictitious, orimage, charges.andx=bitis0.04,butform=2andx=bitis0.002.Thismeans Solutions ofLaplace's equation VV=0aretermedharmonic func- tions.Itisoftenpossible toreduceLaplace's equation toasetofthree "Bya“reasonably wellbehaed anetion, wemeanonethiicontinuous, oatleast independent, ordinary differential equations, oneforeachcoordinate. Piecewise continuous, and that does not become infinite atany point. Most boundaryconditions encountered inpractice possessthesecharacteristics. * 1el ealcataee“Exceptpossiblyattheendpointsy=0andy=b,butthisdiscrepancy betweenVantheFourseriesusualyealmportantinpeat me"i V(xyz)=X@)YO)Z(), ais) ad PROBLEMS 23 1 (a)Calculate thesurface charge density induced ontheplate asa | functionoftheradiusrfromthefootoftheperpendicular drawnfromthecharge &(b)Show that thetotal induced charge is-Q- \ 11-3,(11.2) Linechargenearaconducting plate— Alinecharge ofAcoulombs/meter isparallel toaflatconducting plate, atadistancea,asinFig.11-9.FindBatapoint(x,y).Thesurfacecharge Ndensityontheplateisgivenby€,Eat(x,0). REX‘N 11-4,(11.2)Thefieldofachargeinsideahollowconductingsphere Sd RAK ‘®hollowconductingsphereofradiusacontainsapointchargeQatthe SA radiusbasinFig,11-10. SSSA (a)Showthatthefieldinsidethesphereisthesameasiftherewasno SSS QV sphereand,instead, achargeQ'=—(a/b)Q atD=(a/b)a.YoucanproveSSS <<, thisbyshowingthattheVofQplusQ’isuniformoverthesurfaceoftheSEEE> sphere.Ss (b)Calculatetheforceofattraction. ()Calculatethesurfacechargedensityontheinsidesurfaceofthe NN‘conducting sphere. . 11-5. (1.2) The field ofapoint charge near ablock ofdielectricFig.11-8.Three-dimensional plotofthepotentialVforthefieldofFig,11-5.The Showthatthepotentialatthesurfaceofthedielectric, opposite@inFigU-shaped curves areequipotentials; theothers show theintersections ofthe 11:3, isthesame, whether one calculates thefield inair,asinFig. 11-3(a), Potential surfacewithplanesparalle!tothe1V-plane. orinthedielectric, asinFig.11-3(c). Set€,=3.11-6,(11.3)SolutionsofLaplace'sequationcanbeoftheformX(x)+¥(y)+ js . Z@) whereXisafunctionofxalone,etc.Thisoperation iscalledseparating ‘ShowthatthereexistsolutionsofLaplace'sequationthatareoftheform thevariables. X(x)+¥(y)+Z(2). Forcertain fields onecanfittheboundary conditions onlybysumming 11-7.(11.3) ‘Thefunction 1/randitsderivatives aresolutions ofLaplace's aninfinite number ofharmonic functions. Weused aFourier series equation(a)ShowthatF°(1/r)=0. <y (b)Usethisfacttoshowthat Y(c.sin™+,cos 3B(Gan"P+0.00%") a1#1wtBr ber Bear where C,andD,areconstants, arealsosolutions ofLaplace's equation PROBLEMS M-1,(IL1)Theuniqueness theorem‘According tothistheorem, thePoisson equation ¥°V =~p,/¢, canhave ‘only onesolution ifthepotential Visdefined attheboundaries ofthe field. Show that two solutions can differ atmost byaconstant ifthenormal “ component ofPVisdefined attheboundaries. 11-2, (11.2) Point charge above aconducting plate . [email protected] Fig.11-9. 226 BLECTRIC FIELDS X 12.1SOLVING LAPLACE’S BQUATION 227 2/2), 137 av Rewriting the©equation yields(55)*saan(859)"° (ay _—- = ——)+ +1)Osin 6=0. 12-10) ToseparatethevariablesasinChap.11,weset a(snGp)+mC+NOsin (21) V=R()O(A), (122) Nowset U=0s8, (2-11) whereRisafunction ofronly and©afunction of8only. Then, bysubstitution, remembering that,foranyfunctionf(s), 3(29R), R 9/3008(PB)+oS(sin02)0 (123) dfdfdu df df aor.sin630 30 == —= S=—(1- yp)?=. 12-12)" "407dudo 8ayOO ay (i212) Dividing byRO, weget ; ‘Then the©equation becomes Legendre's equation: ld/,dR), 1d. dopa o) teana aa(inaa)=0. 12-4) Rarx)Sundae(a) (24) J[a12]+nr+no=0 (12:13)lu a ‘Wehave now written total instead ofpartial derivatives because Rand@areeachfunctions ofasinglevariable. When»isaninteger, itssolutions aretheLegendre polynomials ofSec. The second term isindependent ofr.Then thefirst term isalso S41 independent ofrandisequaltoaconstant: e=28) (i214) id(P4)=k (a2) Weshallusethefollowing propertyoftheLegendre equation. SinceRdrdr ‘Then nin+l)=n'(n'+1),ifn’=-(n+1), (2-15) L_d(. doSapp (Sin) =-k, (12-6) thenBanoo (Ge) P(neis(c08 8)=P,(cos8). (12-16) since thesumofthetwoconstants must equal zero. . .ve 's fe Letusexamine theRequation first.Multiplying bothsidesbyRand Itfollows that,foreverysolution ofLaplace's equation oftheform differentiating thetermenclosed inparentheses, wehavethat V=Ar'P{cos8), (2.17) od+ok=kR=0. (12-7) thereexistsanotheroneoftheform par =Br-"*P_.,,(c05 0)=Br-"*P,(cos 8). (1218 ‘Thesolutionofthisequationisoftheform VmBOOP anorfers8)=Brfeos6).(1218) , A Observe that thisresult isinagreement with Eq. 12-8. nA eBiren : ne anh . REA+Ber”, (12:8) SothegeneralsolutionofLaplace's equationforfieldspossessing axial with symmetry is n(n+l)=k, (12-9) .V=D(Aur+Bur)P,(c08 8). (12:19) asyoucancheckbysubstitution. f 2 eaKeos0) =124SOLVING LAPLACE'S FOUATION 29 Table12-1SolutionsofLaplace'sequationinsphericalpolarcoordinates forfields (1)Vie=0, Q)Vee=—Eo2=Euros@.(12-21)possessing axial symmetry—_AEEee Fromboundarycondition 1andfromEq.12-19, —_ 0=3Asa’P,(c0s @)+3BuaP,(c0s 6).(12-22) 1 reos8 Peost Sem?=1 Gx?0-1) Wenowevaluate theAandBcoefficients inamanner 2pGeord-d riGaore-y analogous tothatoftheexample inSec.11.3.Wemultiply both : 2 sides oftheequation byP,(cos 8)andintegrate from cos6=~13pcot=30080) Geos!O=3e080) owe=+i 4(3Sc0s"030608°+3) 5(35.cos*#—30cos*+3 wa Seee voGhoseaee » 0=YfAsa"Py(cos 8)Pa(c0s8)d{cos#)5pleas! Wcos!+150s) |gGBeos*#— 70cos"H+15cos#) an ?. +3itB,a-®"P,(cos 8)Pa(cos8)d(cos8).(12-23) Table 12-1 shows thefirst sixterms According toEq.12-20, theonly nonvanishing terms arethose for This series isanalogous toaFourier series (example, Sec. 11.3) inthe which m=n.Then eachsummation reduces toasingle term: followingways.First,theexpressions underthesummation signforma fo nerf”p4oon6)dleoscomplete set:theseriescansatisfyanyreasonably well-behaved bound- O=Ana’f,Prleos6)d(cos0)+Buaf,Pa(eos8)d(cos6) arycondition exhibiting axial symmetry. Second, (1229 a 0ifmen -aeohe (12-25)P,,(cos@)P,(cos 8)d(cos@)=42 12. mee f(cos8)P,(cos)d(cos@) itmen, (220) soens B,=~A,a". (12:26) Substituting intoEq.12-19yields Legendre polynomials atethusorthogonal. Third,wecanusethis igintoEq.12-19ye orthogonality tocalculate thevalues oftheA,,andB,coefficients. vesAde~a**'r-")P, (cos8). (227) Example |UNCHARGED CONDUCTING SPHERE INA iNowboundary condition (2)concerns thevalueofVatinfinity PREVIOUSLYUNIFORM ELECTRIC FIELD whore inversepowersoftadfoBre.Ths,are, Aninsulated and uncharged conducting sphere issituated ina ~ previously uniform electric fieldEy.Thisapplied fieldoriginates in V=—Eor cos@=—EyrP,(cos 8)=5Agr"P,(cos 8) (12-28) remotely situated charge distributions that areunaffected bythe ~~ presence ofthesphere. forall8.Byinspection, theonlynonzero termontherightisthat‘Atanypoint inspace, either inside oroutside thesphere, forwhich n=1.SeeTable 12-1, Then E=E,+E,, where E,isthefield ofthecharges induced onthe sphere, The induced charges arrange themselves soastorender Ai=-Eo (1229) thenetfieldinside equal tozero. . . Wecalculate thefieldoutside thesphere bysolving Laplace's andalltheother A'sarezero. Also, from Eq.12-26, alltheB's equation intwodifferent ways. arezeroexcept By: (a)Wefirstusespherical polarcoordinates, withtheorigin at Ag =Esa? 1230)thecenterofthesphereandthepolaraxisalongEy.The BeaAaa=Fut (30)boundary conditions arethenasfollows: Finally, atanypoint outside thesphere, 20 SurcTmic FIELDS x 1211SOLVING LAPLACE'S EQUATION 2 Fuacos ° conditionatr=a,whereVmustbeindependent of6.Wemust V=—Eyrcos04°8=(1)Farcos0,(12-31) thereforeaddanotherfunctionthatalsoincludesthecos@factor s0that the coeficient ofcos will bezero atr=a. Then, from ov2a" Eq.12-19, p=Ba(1472)E,cos, 2:Beng (ieess om V=-Eyrcos0+28 (1235) 13V a Ey75g7(1)Eosin8(223) WefinallysetB=F,a’tomakeV=0atr=a,Oursolution 30Ip satisies bothLaplace's equation andtheboundary conditions; itis : ; therefore thecorrect solution, according 10theuniqueness Thesurface charge density ontheconducting sphere is Perce ee TETSeoFigDt 0=CEr(rau)=36k96088. (1234) Example |DIELECTRIC SPHERE INAPREVIOUSLYUNIFORM ELECTRIC FIELD Intheabove expression forV,observe that thefirst term comes from Ey,while thesecond istheVofapoint dipole (Sec. 5.1) Wenow have apreviously uniform field Ey,asinthepreceding situated attheorigin andoriented along the2-axis, ofmoment section, andthefollowing boundary conditions, where aisthe6 iB 14¢0°Ey radius ofthesphere: (b) We can also find Vbyamuch less formal method, as follows. (1)Viscontinuous atr=a.‘Werequiretheterm—Eyrcos@soastofittheconditionat «normalcompon «continuous atr-=ainfinity. Nootherfunction withapositive power ofris @)Th component ofDiscontinuousat permissible. This oneterm, however, isinadequate tofitthe 8) Viw=Evrcos@. Instead ofgoing through aformal solution, aswedidinpart (a) oftheprevious example, weproceed asin(b)and devise a combination ofspherical harmonics that satisfies allthree conditions Pr Seeeeee There nowexists afieldinside thesphere. Sowerequire twoEEE EEeee solutions,onethatsvalidinsideandonethatisvalidoutside.The COCCeeeeeecee fieldoutsidemustsatisfyboundary condition (3).SowerequireaSoca eeseeeee term—[email protected] HEErowers ofrqualify. Inside thesphere, there must benonegative Coy Oc powersofr,because Vcannotbecome infiniteatr=0.CCCcccoo ATCC LetV,andV,bethepotentials outsideandinsidethesphere, HA GRA ET respectively. ThenECE SE V,=Forcos+SBur""P,(cos 8),(12-36) Corre yy) Pr - Coc v=5CyrP(cos 0) (237 Peery er = CORT er Boundary conditions (1)and(2)requirethat,atr=a,aT Fig.12-1. Lines of£(arrows) andequipotentials foran . Mv . uncharged conducting sphere situatedinapreviouslyuniform VeaVeorGr (12:38) electric field. The lines ofEarenormaltothesurface,andthereis zero field inside, Observe thatthefield ishardly disturbed at ‘Then wehave thefollowing twoequations: distances large than oneradius from thesurface ofthesphere. : ;‘Theonginofthespherical polarCoordinates usedforthe Facos0+BigBie 8,BARLCOS A), calculationiatthecenterofthesphere ae a 22 BEECTRIC FIELDSX 12.2POISSON'S EQUATION FOREINELECTROSTATIC FIELDS, 233 =Co+Cyacos@+C,a°P,(cos 8)+ (12-39) e-land B= Cy=0, B= Baa (12.43) Eycos6+B42£088,38:Px(cos9), 5ara a CT ypn BG =O (n>). (12-44) =~€[C, cos@+2CsaP(cos 8)+--+].(12-40) Finally, “ ‘Thesetwoequations applyforallvaluesof0.Itfollowsthatthe v=-(1-S42) ur0080, (12-45) coefficient ofagivenfunction of8ontheleftmustbeequaltothe eter coefficient ofthesame function ontheright. From thefirst 3 .equation, Ver enEngrcos0. (12-46) By Be Bs Ofcourse, rcos ==zSa, -BatS=Cja, =a’, 12-41 a o a ‘) Notice that,if€,>>1,thenV,isapproximately thesameasthe From thesecond Voftheconducting sphere oftheprevious section,fomthesecont ‘Also, theEinside thesphere isuniform: By 2B By,a BoteeR GERRELa BaWe bt (12.47)(1242) a“ ‘Combining theseequations gives SeeFig,12-2. 12.2 POISSON'S EQUATION FOR EIN ELECTROSTATIC FIELDS COCOA Wediscussed Poisson's equation forVinSec.4.1.There alsoexists a COC Poisson equation forE.From Sec.1.11.6, PPP . . .PE VX(VXE)=-VE+ (VE), (12-48)PE PP rr ae Now, inelectrostatic fields, thecurlofEiszero, sothat Pe ereCOCR SEA _ voCOCCEOE ETtee VEWB) =, (12-49)COC "HEREEEEEE HH} fromSec.9.5,wherepisthetotalchargedensity,freeplusbound.We can solve this equation by first separating itinto itsthree Fig. 12-2. LinesofD(arrows)andequipotentials foradielectric components, thex-component being sphere (€,=3)situated inapreviously uniform electric field. The lines ofDcrowd into thesphere, andDislargerinsidethan apOP outside.Theequipotentials spreadoutinside,oEissmaller VERSe (12-50) insidethan outside. There existsaboundsurfacechargedensity, oxeo andEisdiscontinuous atthesurface, Asfortheconducting sphere (Fig.12-1),thefieldishardlydisturbed atdistances larger Thisscalar equation issimilar tothePoisson equation forV(Eq.4-1),thanoneradiusfromthesurface.Thefieldinsideisuniform, :whosesolutionistheVofEq.3-18.Thus 235 2M ELECTRIC FIELDS x PROBLEMS @R., aR 1paplax' 20R,4,dR_ R=0 12-7),(129) B=LfPaw’, (1251) Pea et G2), (29) 4xeole 7 andsimilarly fortheothertwocomponents. Asusual, theprimes referto andLegendre's equation thecharge distribution, andristhedistance between thefieldpoint d deP(x,y,z)andthesourcepointP’(x’,y’,2).ThesolutionofLaplace's x{a-«)8]+n(n+10=0, (12-13)‘equation forEisthus ” 1 (vp+ where y=cos8.Thefirstequationdefinesafunctionofthetype E--[eww (12-52)Seo) 7 B RAMton, (128) This equation relates Etothegradient ofthetotal charge density. Note thenegative signandthefirstpower ofrinthedenominator, This whiletheLegendre polynomials ofSec.5.4.1aresolutions ofLegendre'sequation isvalidwhatever thenatureofthemediathatarepresent inthe cautionfield,aslongasthegradient isdefinable. "Thegeneral solution ofLaplace's equation inspherical coordinates, for ‘Themoreusualexpression fortheEofavolumedistribution ofcharge axialsymmetryisthus isaconsequence ofCoulomb's lawanditistheonethatwefound in ” ‘ Sec.3.3: S V=>(Agr +Bur PP, (cos8). (12-19) B=y [Saw" (12-53)Beier Poisson's equation forEis Although thetwointegrals forEareequal, theintegrands are vpobviously unequal. Oneintegral applies onlywhere Vpisnotzero,while VE a (12-49)theother extends over thecomplete charge distribution ° Example |THEFIELD OFASPHERE OFCHARGE WHOSE, anditssolution is DENSITY ISAFUNCTION OFTHERADIUS 1p¥pE«-7[Tea (12-52) Imagine asphere ofcharge ofuniform density, except near the dae, he7 periphery, where thedensity gradually falls tozero. With Eq, 12-52, itisonlytheregion nearthesurface thatcontributes tothe Forelectrostatic fields, wecancalculate Eeither asabove orthrough integral, since thegradient ofthecharge density iszero Coulomb's leweverywhere else. Nonetheless, thisequation leads 10thesame ‘oulomb’s law: resultasEq.12-53.SeeProb.12-6. 1fotrel (12-53) Areyhor 12.3 SUMMARY Inspherical coordinates, ifVisindependent ofthe@coordinate, weset PROBLEMS V(r, 8)=R(r)O(8). (12-2) 12-1, (12.1) Grounded cylindrical conductor inauniform ‘Agrounded, infinite, cicular cylindrical conductor ofradius aliesina ‘Then Laplace's equation becomes apair ofordinary differential previously uniform electric field, withitsaxisperpendicular toEp,asinFig.i 12:3. Show that V=~E,(1 =a°/p")p cos9. equations L 13.1 REFERENCE FRAMES AND OBSERVERS 239 relate electric fields tomagnetic fields and then establish relations that we shall later rediscover, starting with Chap. 18, without referring to * 13.1REFERENCE FRAMES AND OBSERVERSRELATIVITY I Considerthetworigidreference frames§and5”ofFig.13-1,whereS” The Lorentz Transformation and Space-Time moves atsomeconstant velocity ¥withrespect toS. Both frames areinertial, byassumption. Wedefine aninertial frame as ‘one that does not accelerate and that does not rotate. Inother words, an inertialframeisonewithrespecttowhichtherearenoinertialforces.* Each reference frame carties an observer, which iseither ahuman 13. REFERENCE FRAMES ANDOBSERVERS 239 being equipped withinstruments, orsome device thatcantakereadings 132THEGALILEAN TRANSFORMATION 240 orphotographs, either automatically orunder remote control. Observer 133. THEPRINCIPLE OFRELATIVITY 240 otsituated inS,ando'in5’. 134THELORENTZ TRANSFORMATION 241 Thespecialtheoryofrelativity concerns measurements madebytheseURENTeTeARoetON ee UNDER twoobservers. Asarule,werefertothetwospecificframesofFig.13-2.Example: THEPHOTON 248 Wearemotconcerned withthegeneral theory ofrelativity, which deals 13.5 SPACE-TIME DIAGRAMS AND WORLD LINES 244 with accelerated frames andgravitation.135.1THEMINKOWSKI DIAGRAM 235 135.2 CAUSALITY AND MAXIMUMSIGNAL VELOCITY247 . eo 136 SUMMARY 247 -PROBLEMS —248 y Wehavenowstudied electric fieldsatquitesomelength. Atthispoint e p - either youcanskiptoChap. 18andgoontomagnetic fields, oryoucan it Dy study Chaps. 13to17,which deal with relativity. a fs ‘There aretworeasons forthisdigression. First, relativity reveals the ° fundamental aspects ofelectromagnetism. Second, there aremany Fig.13-1,Reference frames Sand5’,withS°moving atsomearbitrary constant phenomena, mostly related tohigh velocities, thatarebatfling without velocity¥withrespecttoS,withoutrotation.Observersoand0°perform relativity.Thedaggeredproblems, fromChap.18on,areexamplesof variousexperiments, eachoneiahisorherownframe. these. ‘The longer path ismore interesting, asalways, butitmay notbethe better one. The fivechapters onrelativity donotreplace themore conventional approach thatcomes afterward, Selecting onepathorthe Foramoredetiled introduction 10relat, seeEdwinF.TaylorandJohaA.other isamatter oftimeandpersonal taste. Wheclet,Spuceime Physics, W.H.Freeman, NewYork,196 You cantherefore godirectly toChap. 18without losing continuity. "The forces thatonefeelswhile stationary inside anaccelerating vehicle areinertial Chapters 13,14,and15setforth thefundamentals ofrelativity, with forces. Theinertial forceonaperson ofmassmis~ma,whereastheaccelerationofthe littlereference toelectromagnetism. Then, inChaps. 16and17,wefirst vehicle 20 134.THE LORENTZ. TRANSFORMATION 241 P This principle was formulated byGalileo in1638. Itisquite clear in‘ itself,butitissofundamental thatwestateitinanother way,to firt emphasizeitsmeaning.Theprinciplemeansthatanyexperimentleadsto ie4 = precisely thesameresult,whether itisperformed inSorinS’,Theresult Ree r AN isthesame, whether theexperiment takes place inastationary orina eas) er| moving vehicle, aslongasthevehicle movesinastraight lineata ‘ = i P " Forexample, observer 0’inreference frame S”canmeasure theperiod jo. Jo P ofapendulumsuspended atsomepointinS’,orthecollisionofbilliardee ia 7 balls onatable atrest inS',etc. Inallinstances thephenomenon aiaaeane: B observed by0’isprecisely thesameasifsheperformed herexperiment Be Som inreference frameSerect. ey Thisprinciple isfirmly established. Inparticular, itisanexperimental factthatthespeedoflightisthesameforallobservers traveling ata Fig.13-2,Thereference frame5’movesatavelocity V8withrespect toS. constant velocity. 13.2THE GALILEAN TRANSFORMATION Toillustrate, consider thelaw a According toclassical physics, observers inreference frames §and5”of Fajlow), (32) Fig. 13-2 may select thesame time scale. Ifthey set ¢=0attheinstant when their frames overlap, thenthecoordinates ofagiven point inspace, asobserved inreference frame S.According totheprinciple ofrelativity, with respect tothetwoframes, satisfy theequations there exists anidentical law xextL yay 22, ree (3-1) Fem’) (133) ThisistheGalilean transformation. “ Inparticular, suppose observer o’finds thatthevelocity ofacertain thatapplies in$’.Physical lawsarethussaidtobeinvariant. light pulse relative toS”is¢,inthepositive direction ofthecommon Itfollows that there exist mathematical relations that transform x-axis. Then, according totheGalilean transformation andaccording to unprimed quantities toprimed quantities, andinversely. Weshall elementary mechanics, observer oinreference frame Sshould findthat discover several such transformations inthese fivechapters onrelativity thevelocityofthatparticular lightpulseis¢+Y. WiththeGalilean transformation, F=F’, m=m’,t=t',v=v'+Y,‘TheGalilean transformation proves tobeinerror forhighvelocities. andtheprinciple ofrelativity applies. However, nature isnotthatsimple; Experiments show thatthevelocity ofthelight pulse isthesame forthe these fourequations are,infactapproximations, Relativistic transforma- two observers. This fact goes against common sense, ofcourse. The tions show thatF#F', m#m’, c#0', andv4"+V1 Galilean transformation also proves tobeincompatible with electromag: netism because, under that transformation, Maxwell's equations arenot 13.4 THE LORENTZ TRANSFORMATION invariant TheLorentz transformation relates thespaceandtimecoordinates of 13.3THE PRINCIPLE OFRELATIVITY reference frame Stothose ofS’,andinversely. Fortheframes ofFig. 13-2, ‘Theprinciple ofrelativity states that itisphysically impossible todetect whether aninertial frame isatrestorinmotion fromobservations made *SeeWolfgang K.H.Panofsky andMetbaPhillips, Classica Elecriciy andMagneto,entirelywithinthatframe. fpRitoees:ReadingMas1,chap.15 a RELATIVITY 1 13.4 THE LORENTZ TRANSFORMATION m3 vere ,x-V (1)Therelationbetweenquantities inoneinertialreference frameand a7 OO (13-4) thecorresponding quantities inanotherinertialframecanalways.be , expressed bycither oneoftwoequations, orsetsofequations, that areyay, yay (3-5) equivalent.zez', z'ez, (13-6) (2)Oneobtains theinverse relation byaddingprimestoumprimed . jantities, deleting primes on primed quantities, and changing the signpete, tw as») oe BP Prmee=r arya isthe customtoset wherecisthespeedoflightinavacuum. Itisthecustom tose ‘Thissetofeightequations forms thebasisofspecial relativity.’ We ry 1 spendtherestofthischapter,andthenextfour,discussing someofits B=> andy “apy? (13-8) strange consequences. For the moment, we note sixfairly obvious fe 5 atures Note that 6<1and y=1. Then (1) Theright-hand column isidentical totheleft-hand column, except sae . .- y =y@-V0), . thattheprimed andunprimed quantities areinterchanged, andthat—7 says Ve), x=e V0), (13-9) replaces ¥.The reason issimply thatitisimmaterial whether S’moves at yay yy (13-10) thevelocity 7withrespecttoS,orwhether§movesatthevelocity r=2', vez, «aan=VEwith respect toS’. , . vx Ve (2)Thereareonlyfourindependent equations because theright-hand ter(ve23), vr=r(t-S) (13-12)column follows from theleft-hand one, andinversely. You caneasily © « check this. . . Itisoftenmoreconvenient toexpress thetransformation invector (3) The origins OandO'coincide at¢=«' =0because, ifx=y=z= form,asinTable13-1,wherethesubscripts ||and1refer,respectively, £0,thenx"=y"=2'=1 =0,andinversely. tothecomponents thatareeither parallel orperpendicular tothemotion (4) TheLorentz transformation reduces totheGalilean transformation OfS"withrespecttoS. (Eq. 13-1) ifwesetthespeed oflight ¢equal toinfinity. (5)Therelative velocity Vofthetwoframes cannot exceed c,for Table15-1Lorentz transformation otherwise eitherxand¢orx’and¢’become imaginary. revit VW)+r Pedn~V4r, (6)Thefactthat¢#¢’ means that, ifobserver omeasures atime vir) . Yrinterval Tbetween twoevents, theno'will,ingeneral, measure a r(te,) f=(-2) different time interval 7’between thesame pairofevents. Inparticular, es iftwo events aresimultaneous for0,then they arenotnecessarily ‘simultaneous for0’. Example |THEINVARIANCE OFx?+y?+2?-c?? UNDER ALORENTZ TRANSFORMATION Points1and2illustrate twogeneral rules.‘ “ ‘Aquantity issaid tobeinvariant ifitsnumerical value isthesame "The Lorentz transformation follows from theprinciple oftelavty applied tothe inallinertial frames. For example, under aGalilean Wheeler, Spacetime Physics, W.H.Freeman, New York, 1966, p43. same whether oneperforms themeasurement in$orinS" uM RELATIVITY | 245 Note thattheword invariant isnotsynonymous with theword “ iY constant, Indeed, ifpoint aisfixed inS,andbisfixed inS",then ruis function ofthetime but istll invariant under aGalilean transformation, “ With the Lorentz transformation, rgi8mot invariant, but 1°—c't’ is.Youcaneasily check that oo Pty trae Pant bytester, \ A. or P-eP ara (1313) a a Weshallusetheinvariance ofr?~c°r*below.Weshallalsofind WN 4 several otheranalogous invariant quantities. SN a Example |THEPHOTON SAKs Imagine aflashoflightemitted atOorO°attheinstant when the “ - . twoorigins coincide. According totheprinciple ofrelativity, the light propagates inalldirections atthesame velocity ¢inboth-13-3.(a)Worldlineofastationary electron at(x9,0,0).(b)Worldlineof referenceframes.Thereforethepositionofaphotonsatisfiestheeecaron‘whosevelocitydx/dtisfirstnegativeandthenpositivealongthe equationx-axis. The shaded area isinaccessible tothiselectron because |dx/dt| cannot ceyaet peepee exceed¢. 13.5.1 The Minkowski Diagram inagreement withtheprevious equations. Wecanpicture theLorentz transformation bysuperposing the(x,ct)and the(x’,ct’) planes inamanner devised byMinkowski asinFig. 13-4. 13.5 SPACE-TIME DIAGRAMS AND WORLD LINES First, t= —Vi). 13-15) Aspace-time diagram isagraphofctasafunction ofxforagivenevent, ware V0, (315) asinFig.13-3,Suchdiagrams helpusvisualize some oftheimplications Setting x’=0defines aline oftheLorentz transformation forthereference frames ofFig.13-2 YTheworldlineofanobjectoreventisitscurveofctasafunctionofx. x= (")ct (13-16) Sayanelectron isstationary atx=x9,cf=0. Itsworld lineisthestraight lineaofFig.13-3. Iftheelectron moves insome wayalong thex-axis, inthe(x,ct)plane, Thisline,along which x’=0,isthect’-axis, ainthe thenitsworld lineisacurve suchasb.However, bisnotanyarbitrary figure. Also, curve because thespeed ofany object isalways lessthan c.This makes themagnitude oftheslopeeverywhere largerthanunity,andtheelectron rare 4). (3-17)hasnoaccess totheshaded region. « Iftheelectron moves inaplane normal tothex-axis, then y=’, a 7z=2",fromtheLorentz transformation. Setting ¢°=0defines afine Nowsayaflashoflightoccursatx=0,cf=0.Photons travelalongthe _(e (sts) linesct=x.Ifwedraway-axisperpendicular tothex-andct-axes,the a=(©)e "photons travelalongaconewhoseaxisisthect-axis. Infour-dimensional | space-time,theyfollowalightcone. |inthe(x,ct)plane.Thisline,alongwhichct’=0,isthex'-axis.Lines i 246 13.6SUMMARY 247 4 s those events clearly occur atdifferent values of¢and aretherefore not a simultaneous inS.Innumerable paradoxes resultfromthisfact. aL- A IftwoeventsFandGoccuratthesame¢andatthesamex(butnot _— : x necessarily thesameyandthesamez)inS,thentheyarealso ‘ isTST J ;simultaneous inS’,andinversely.“FPtH|Yofe/ 13.5.2Causality andMaximum SignalVelocityyr // TheorderinwhichtwoeventsFandGoccurcanbedifferentindifferentcr / frames because 4 /// to-th=1[lo-t)~(4)leo-*n), as) / |=avs |a andthesignsof1%—‘pandfg~fycanbedifferent. ThisisdisturbingwzVAJf / because,accordingtotheprincipleofcausality,acausenecessarily *pT / recedes itseffect._Ki ys | Forexample,imaginethatobserverothrowsaballinthedirectionofof { { thex-axis.Afteraflightofafewseconds, theballbreaksawindowpane. ® ' > 9 3 ‘The Lorentz transformation surely cannot mean thatthese events could ‘ occur backward, forcertain observers. Fig.13-4.The(x,1)planeofreference frame Sand,superimposed,x’-and SayeventFoccursatoriginsOandO'attheinstantwhenthey SROnewayeiorsTeoneercOrebandesOree coincide.EventFisthecauseofeventG,whichoccursatxgatalater Coincide at@Event Eoccursatx2,3.oratx’=0.973,ef’=2.48,The timefcinSandatrzinS’.EventGcannotoccurbeforeeventFinS’.curves areparabolas x?—c'=4°.ThisisaMinkowski diagram. ‘Then1{,mustnotbenegative. Now x’=constant areparalleltothect’-axis,whilelinesct’=constant are . Y Yparalleltothex-axis. to=y[to- (2)re]=ye(2)a] (13-20) Notice thefollowing points. vw=ro(1-). (13.21) (1) Thescales arenotthesame. Forexample, alength corresponding to é 3metersislongeronthex'-axisthanonthex-axis. wherevisthespeedatwhicha“signal” propagates fromFtoG.Forthe (2)Theangle between thect-andct’-axes isthesame asthatbetween above example, visthehorizontal velocity oftheball. Therefore thex-andx'-axes. This reveals acertain symmetry between space and Vvsc. Since VScfrom Sec. 13.4, vSc.A“signal” cannot therefore time. However, there arethree space dimensions andonly onetime propagate ataspeed thatislarger than c. dimension, sothesymmetry isonly partial. (3). The lines x=ctcoincide with thelinesx”=ct’becauseavelocityj13.6SUMMARY along the x-axis isinvariant. H Aninertial reference frame does notaccelerate and does notrotate. (4).Points onthex'-axis correspond toevents thataresimultaneous for 4) “Thespecial theory ofrelativity concerns observations andmeasure-observer 0”inreference frame S’since 1’=0allalong thisline.However, ‘ments onagiven phenomenon made byobservers situated intwoinertial q 248 RELaTivy 1 PROBLEMS 249 frames,oneofwhichmovesataconstantvelocitywithrespecttothe 133.(144)Sigalingproblemswithafsttrain leitywith other.Asarule,thetw ‘: ip134 weepersonsA,0",andB,rideonatrainmovingatavelocity¥,witTh McaaneioFramesreferredtoarethoseofFig.132. Ainfront,O”inthemiddle,andBintherear.Afourthperson,O,standsieprinciple ofrelativity statesthatitisphysically impossible todetect besidetherails,AtthemomentO'passesO,lightsignalsfromAandBwhether aninertial reference frame isatrestorinuniform motion from reach bothOandO'.Persons OandO'areasked whoemitted herlight observations made entirely within thatframe. signal first. What dotheyanswer? Thusphysical lawsareinvariant: thelawdescribing agiven phenome- 13-4,(13.4) Transformation ofanangle nonismathematically thesame, inwhatever inertial frame thephenome- ‘Astraight linepassing through theorigin O"ofS”forms anangle awith non occurs. thex-axis. (a)Find arelation between aand a’ ‘TheLorentz transformation relates x,y,2,¢in Stox',y',2', 0’inS* (b)What isthe value ofawhen tends to€? (Fig. 13-2): 13-5, (13.4) Things thatmove faster thanaphoton ‘TheLorentztransformation impliesthattherelativevelocity ¥°oftwo xaYe'4+ Ve), x= ya- V0, (13-9) frames ofreference cannot exceed thespeed oflightc.Wehave alsoshown : , that asignal cannot exceed thespeed oflight. Discuss thefollowing cases.yay", yy, (13-10) {a}Avlong,straightrodformeasmallangle6withanotherrod.whichis zee wer (341 horizontal andstationary. Thefirstrodmoves downward atavelocity v, : +11) Whatisthespeedofthepointofintersection oftheloweredgeofthe Ve Ye moving rodwiththefixedrod?Canthisspeedbegreater thanc?Canthe =r("+4), vsr(t-*), (13-12) pointofintersection beusedtotransmitasignal?e (b)Theupperrodisinitiallyatrestwiththepointofintersection atthe = rey" origin, Therodisstruck adownward blow attheorigin withahammer. wherey=(I~Vile) Canthemotionofthepointofintersection beusedtotransmit asignalat Table13-1showstheseequationsinvectorform.Thesubscripts ||and ‘aspeedgreaterthanthespeedoflight? Lrefer, respectively, tocomponents thatareparallel orperpendicular to (c)Apowerful laser rotates rapidly about anaxisperpendicular toits thevelocity of5’with respect toS. length. Aquantity issaid tobeinvariant ifitsnumerical value isthesame inall Cantheazimuthal speed ofthebeam exceed thespeed oflight? Canthe inertial reference frames, beam transmit asignal between twopoints ataspeed greater thanc? ; (a)Themanufacturers ofsome oscilloscopes claim writing speeds in Aspace-time diagramshowsctasafunctionofxforagivenobjector cexcensofthespeedoflight.Isthispossible?event.Theworldlineofanobjectoreventisitscurveofcfasafunction 135)Space-time dofx.Onecanvisualize theLorentz transformation bysuperposing the 126,(145)|Spaoc-timediagrams space-time diagrams for§andfor5’ weAsignalcannotpropagateataspeedlargerthan 1 ates(54)(x+00) " j Substituting ~cforcgives PROBLEMS’{ rdgia} vna'=(1*8)"Q en), 13-1.(13.4) TheLorentz transformation t 1-6) Forwhat value of Bisthevalue ofyequal to1.01? | 13-7, (43.5) cistheultimate speed 13-2.(13.4) TheLorentz transformation ; Imagine aseries ofreference frames $,S'$$", ete,with$?moving at Calculate y,8,andvforaconduction electron whose energy is10 ' avelocity ¥¥withrespecttoS,S*movingatthesamevelocitywithrespect electronvolts.Therestenergy ofanelectron is5.11X10electronvolts. to$",etc.According totheGalilean transformation, aparticleatrestinS ;moves atavelocity n¥/é with respect toS,where n¥”isarbitrarily lage. Show that, according torelativity, the velocity ofthat particle with — — j respect toSisalways lessthan c. "Several oftheproblems onrelativity areadapted, with permission, from Edwin F Taylor andJohnA.Wheeler, Spacetime Physics, W.H.Freeman, NewYork, 1966. 13-8. (13.5) Three-dimensional space-time x,y,ct 250 *RELATIVITY I ¥ The Lorentz Contraction and Time Dilation. - —— —____— The Transformation ofaVelocity Fig.13-5. 14.1TRANSFORMATION OFALENGTH. THELORENTZ CONTRACTION 251 Figure 13-5 shows the2y-plane, thect-axis, andtheplanes x=ctand 14.1.1 TRANSFORMATION OFANELEMENT OFAREA def 253 x=nct Example: THE APPARENT SHAPE OFARAPIDLY MOVING (a)ShowthataLorentztransformation contracts thisspacebythefactor OBJECT 254 [C+B)/(.~ p)]'* inthedirection oftheplane x=ct,anddilates itbythe 14.2 TRANSFORMATION OFATIMEINTERVAL.TIMEDILATION254 samefactoritthedirection ofx=et Example: THETIMEREADONARAPIDLY MOVING CLOCK 2551gSew8POI,3ctansormstoanotherpontontesame Example:THERELATIVISTIC DOPPLEREFFECTFORELECTROMAGNETIC WAVES 257 13.9. (13.5.1) TheMinkowski diagram 143. ‘THEINCREMENT OFPROPER TIME FOR ANACCELERATED (a)Draw aMinkowski diagram similar tothat ofFig. 13-4 with PARTICLE 258 Example: TRANSFORMING THESPEED OF LIGHT 260, 14S SUMMARY 260 PROBLEMS —261 This second chapter onrelativity concerns afew immediate consequences ofthe Lorentz transformation. The Lorentz contraction makes anobject appear shortened inthedirection ofitsmotion, while time dilation makes atime interval onamoving object appear longer for astationary observer. However, amoving clock can also appear torun fast, aswe shall see. We end this chapter with thetransformation ofavelocity. 14.1 TRANSFORMATION OF A LENGTH. THE LORENTZ CONTRACTION Imagine that observer 0’onreference frame S"fixes aruler oflength fy onthex’-axis ofS’sothat itsextremities are atx’=0 and x'=ly, The quantity [yisthelength oftheruler, asmeasured initsown reference frame, andiscalled itsproper length. According to0,theruler sweeps theshaded area inFig. 14-1 ‘What isthelength ofthesame ruler forobserver 0on$?That observer performs hismeasurement bynoting thepositions ofthetwo extremities al 282 253 ‘ *ear cee Hit “a He,pee n° ee | x es ied xe oe Se a oo y e | a, { ea[ { o —L 0mC 5 1 ad 2 ¥ 4 hy. “7 Fig.14-2. Theruler ofproper length /,isnowfixed inreference frame S.It Fig.14-1, Arulerofproper length /,,fixedinreference frame S’,sweeps the ‘sweeps theshaded region and,foranobserver onframeS",itslengthislo/y. ‘shadedregionastimegoesby.InreferenceframeSitsapparentlengthislu/y. Note that thescale onthex’-axis isdifferent from that onthex-axis. The curves: areparabolas x?—"=a, Thus osays that themeters ofo”aretooshort, and 0”maintains that those ofoaretooshort. This isnotabsurd because thecomparisons are oftheruler atthesame time, say ¢=0. These areevents Qand (Iy/y, 0) quite complex; they involve eight separate measurements. inFig. 14-1, Clearly, thelength /o/y isshorter than [.How much Iftheruler moves relative tooinadirection perpendicular toits shorter? Atthe right-hand end x=/ and ¢=0. From the Lorentz length, then youcanshow thatthere isnoLorentz contraction: itslength, transformation (Sec. 13.4), measured by0,isequal toitsproper length /,.Alength /therefore ex’=ya-M)=Yh, in, asn transformsasfollows: k [aMeh, 1 ‘Thusarulermoving inthedirection ofitslengthatavelocity 1” Y 42) relative toanobserver appears tobeshortened bythe factor 1/y. Remember that y=1,fromSec.13.4.ThisLorentz contraction is wherefo,isthecomponentoffythatisparalleltothemotionandfy,is independent ofthesignofV’ theorthogonal component. Ofcourse, the Lorentz contraction applies ifthe ruler isanywhere else .alongthex-axis, mr » {14.1.1 Transformation ofanElementofAreads Iftheruler liesonthex-axis inS,then o’finds itshortened bythe i Say theelement ofarea isarigid parallelogram ofsides dly and dlay of ‘same factor 1/y, asinFig. 14-2. arbitrary magnitudes andorientations initsown reference frame. Then 2s keLaniviny ass dy=dyXdl, (14-3) 4 4 Thisvector isnormal totheelement ofarea. / 4 Y NowifSisaninertialreference framewithrespecttowhichthe — Yelement ofareamoves atavelocity Y,itturnsout(Prob. 14-3)that,inS, ik Hj theelementofareasuffersaLorentzcontraction inthedirectionof¥' 3 / L J/ J, df=dl,xdl,=dsAy,+Sins (14-4) / YL)/ Example|THEAPPARENT SHAPEOFARAPIDLY ,0@4VA/MOVINGOBJECT ‘yYj/ Supposeonelookedthroughatelescopeatafarawaycube |/th /‘movingatavelocity =perpendicular tothelineofsight.Then | —_—thefacenormal tothelineofsight would appear tobe . WA / Tv foreshortened inthedirection ofmotion bythefactor 1/y.Also, é {/— conewould seethetrailing faceforthefollowing reason. Atagiven /Y Tiinstant, theeye senses thephotons that arrive atthat instant. - : Photons originating fromdistant partsoftheobject haveleft Tf | earlier than the others, and the object has moved inthe — ' meantime. The neteffect isthat thecube would appear tobe 0 : > 7 rrotatedthroughananglearctan(1'/e).Ifthecubewerenotfar :away, then itwould appear distorted inpeculiar ways, depending Fig.14-3. Theproper timeinterval between eventsQandQ"thatoccuratx’=0 ‘onitsdistance andvelocity inreference frame S"isT;,with cT,™ 2.Foranobserver onS,thistime intervalThiseffect,whichhasneverbeenobserved, wasdiscovered by isytimeslonger.Thescaleonthe¢r'-axisisnotthesameasonthect-axisJames Terrell in1959, 54years after thepublication ofEinstein'sfirstpaperonrelativity.” pap uivity, From Fig.14-3,theinterval T,asmeasured inS,islonger thanthe proper time interval T;.Indeed, according totheLorentz transformation, 14.2 TRANSFORMATION OF ATIME INTERVAL. theevent Q’forwhich x’=0and¢’=T,occurs att=yT,foroinS.So TIME DILATION T=yTo (145) Observer o’measures theduration ofacertain phenomenon that occurs atx’=0inS".Thephenomenon starts at’ =0,namely [email protected], Foro,thetimeinterval islonger thanTj.Thisisthephenomenon called anditends at«’=Tp,oratQ".Thetime Tymeasured inthereference timedilation. Inother words, amoving clock appears torunslowbythe frameofthephenomenon isthepropertime.Observer 0’hasasingle factoryifonemeasures itsrateasabove."clock located atx’=0. Observer0measuresthesametimeintervalwithtwoidenticaland Example Movine chock.ONARAPIDLYsynchronized clocks, one atx=0,where thephenomenon starts, andthe other attheposition ofO’attheend ofthetime interval. Wehave seen that, if0uses twoidentical clocks attwodifferent x'sonStomeasure agiven time interval that o’measures ata Vs i ics "Sce‘the filmentitled “Time Dilation: AnExperiment onMu-Mesons” byF.Friedman, SeeV.F.Weisskopf,PhysicsToday,September 1960,p.24 D.Frisch,andJ.Smith,producedbytheEducational Development Center,Newton,Mass a 256 142TRANSFORMATION OFATIMEINTERVAL TIMEDILATION 237 _! isalsonegative, since B=1. Thus, when oreads @time¢’onan y =) r approaching clock,hisowntimeis‘ —<— re(58) 49) —— | vs 1 ' Does theapproaching clock appear torunfast, orslow, with respect tothestationary clock? Sayobserver 0reads atime 4 a :{=I second ontheapproaching clock, and theabove square root equals 0.9.Then =~0.9. When themoving clock reaches (0,both 1”and¢"willbezero. So,intheinterval, themoving clock willhave advanced by1second, andthefixed clock by0.9second. 7 Theapproaching clock appears torunfast Fig. 14-4, Observer0usesadoublemirrorMtophotogea Example |THERELATIVISTIC DOPPLER EFFECTPOR ‘The relation ¢~ytholds, butthelight from themoving clockdoesnotreachthecamera untila latertime > “TheDoppler effectisthefrequency shiftobserved whenasource cofwaves moves with respect toadetector. This phenomenon is fixedpointonS'withasingleclock,thenofindsatime interval ‘elka inthefinoffacomsiics.thatislongerthanthatofo”bythefactory. Imagine asource ofelectromagnetic wavesofproper frequency‘Whatif0usesasingleclockand/ooksatthemovingclockofo” fsituatedat"and=detectorstO.Whatistheapparent asinFig.14-4Lettheprimed clockbeat0”,andletbothYand frequency at{thepositive. Then0istotherightofOandmovesaway. Thisproblem isidentical totheclockproblem thatwejustImagine thatohasasetofidentical andsynchronized clocksall solved, because thesourcebeatsperiods I/finstead ofsooundsalonghisx-axis.Astheprimedclockgoesbyeachone,the Therefore, witharecedingsource,theperiodmeasuredbyafixed relation =yt’holds. Butostands attheoriginOofS,andthe observer atOis 2 lightfromtheprimedclockatO”takessometimetoreachO. r-(28+8", (4-10) Suppose observer oreads atime t'onthemoving clock. What 1-p) time isitonhisown clock? Call this time ¢.Then 1sthe above ¢, and plusthetimerequiredforlighttotravelthedistanceV1: we(2B (De anrartte(ienasar=a+in, 046) Foranapproachingsource, withB=7/c,asusual. Thus, when oreads ¢°onaclock thatis 1p)"movingaweherownimei 1-(28)"eo8 aan atthe =(18)"r 147) a-By! “\iz_p) (14-7) Notethatitisonlytherelative velocity thatcounts. SeeFig.14-5, . .If,atagiven instant, therelative velocity forms aright angle With ¥andfbothpositive,¢> withthelinejoiningthesourcetothedetector,thereisstilla Therefore, ifone looks ataclock that ismoving away, the Doppler effect becauseoftimedilation.Thisisapurelyrelativistic moving clock appears torun even slower than with the effect. The frequency measured atthereceiver iseither larger or measurements ofSec. 142. Smaller than fo,depending onthereference frame inwhich the ‘What ifthemoving clock isapproaching? Then theorigin O'is angle is9°(Prob. 14-7): totheleftofO.WithY’positive, both¢and”arenegative and f vive =yh (source), orf= (detector). (14 cortaped pywss) Fo(source), orFay(Getecion). (O41) 258 so (/ >\\ [/ \\\ 1+ap-———-——4 fe(oe 1((jm)44 pf é ey I .¢ . |4a w ° the 5 Fig.14-5. TheDoppler effect. (a)Source Smoves tothe rightwithrespect to Fig.14-6. Theworld lineofaparticle moving along thex-axis. Inthe fixed stationary receivers A,B,C.(b)The source isstationary, andthereceivers all reference frame S,thetime interval between events Aand Bisdt.Intheframemovetotheleft.Thefrequencyshiftsarethesameinthetwofigures.The 5"ofconstantvelocity¥occupiedmomentarily bytheparticle,thepropertimemeasuredfrequencyislowerthantheproperfrequencyatA,andhigheratB.At intervalisdig,whichisequaltodt/.G,themeasured frequency iseither lower orhigher thantheproper frequency, as inProb. 14-7. of proper time repeatedly inthefollowin; 14.3THEINCREMENT OFPROPER TIME chenoenean OfPrope Peary ® FOR AN ACCELERATED PARTICLE m ° ie it 14.4TRANSFORMATION OFAVELOCITY Consider aparticle moving atsome arbitrary time-dependent velocity. Saytheparticle movesinastraight linealongthex-axis.Thenitsworld Observer 0’onreference frame$”notesthatanobjectmovesatsomepineomecurveasinFig1R6.Anobservernotesthepositionofthe velocityv'.Thevelocityneednotbeconstant.Whatisthevelocityofthis parejeattimes¢and¢+di.ThisdefinestwopointsAandB,asinthe Sameobjectforobserver oonreference frameS? urthevectorformofthe Lorentz transformation (Table 13-1), ‘ThepropertimeintervaldiybetweenAandBisdefinedasthetime FromthevectorformoftheLor c »interval between these twoevents, asmeasured intheunaccelerated dr_y(dr,+ Vat’) +dr’, 7 i y =Ht a)4ae (14-15) frame $'occupied momentarily bytheparticle: adyd+V-dr'/e) a a\12 vi , t +givesdy= Hoa0-4)" eu) Dividingnumerator anddenominator byydt’give: sey ayt, yey (14-16) asinEq.14-5,whereuistheaveragevelocityoftheparticleduringthat T+ujVle time interval. , ‘Wehave simply applied therule fortime dilation that wefound inSec, The Galilean relation v=v' +¥istherefore valid only ify~1 and if 14.2forinertial frames. This seems illogical because areference frame v'¥<«c* Itfollows from Sec. 13.4that attached totheparticle does accelerate. This peculiar procedure follows fromtheexperimental factthatthehalf-life ofafastparticle ofagiven yetwy (14-17) velocity isthesame whether thetrajectory isastraight lineoracircle. 1-u,V/e* a co RELATIVITY PROBLEMS 21 Example |TRANSFORMING THESPEEDOFLIGHT Aparticledescribes somearbitrary motioninspace-time. Ifthetime |Aphoton trevels atthespeed insomesbitrary direction in interval between twosuccessive positions isdi,thenthecorrespondingreference frame S*.Then proper time interval is ve+v2ec (14-18) -# and to=F (14-14) ayy a(t”) (wily veei- (i) +4 14.19)G+uVTe)+65+0Ta) 19) Avelocitytransforms asfollows: Expanding andusing Eq.14-18 twice, wefindthatthespeed of eresthephotoninSisakoc.whateverthevalueof petittelly (1416) ThisisinagreementwiththefistexampleinSec.13.4,where T+u,Vle wediscussedtheinvariance ofx?+y?+2*~ct! aticTeele asin 14.5 SUMMARY =uVe Anobject hasaproper length /,oriented insome arbitrary direction. ‘Then, inanother reference frame moving atsome constant velocity 7 PROBLEMS withrespecttotheobject, ni 14-1,(14.1)Lorentzcontractionforalengthf=4h, (14-2) Aone-meter rulermovesataspeedc/2.Initsownreference frameitY forms anangle of45°withitsvelocity.heltothemoni ‘Whatisitslength,asmeasuredbyafixedobserver? Lengthsparalleltothemotionareshorterbythefactory.Thisisthe 142.(14.1)“Twosuccessive eventsatagivenpola Lorentzcontraction. Lengthsorthogonal tothemotionareunaffected. ‘Twoeventsoccuratthesameplaceinthelaboratory atanintervalof3 Aprocess lastsapropertime7,initsownreferenceframe.Inanother seconds. frame asabove, thetime interval isytimes larger: ‘What isthespatial distance between these twoevents inamoving frame with respect towhich theevents occur Sseconds apart, and what istheT=7Tp (14-5) relativespeedofthemovingandlaboratory frames? |14-3,(14.1.1) Transformation ofanelement ofarca Thisistime dilation. Asmall rigid parallelogram ofsides dyanddla», initsownreferenceIfasourceofelectromagnetic wavesofproperfrequency f,moves frame,hasanarea awayatavelocity Y=Bcfromanobserver, theapparent frequency is ds,=dl,xdly, 1-y? Show that,withrespect toanother reference frame, theelement ofareaisr=(55) Lo<ho- (14-11) givenby , ; dst. Foranapproaching source, thesignbefore changes, at=dX l= dates 14By" 14-4,(14.2)Theredshift f=(4) fo>ho (14-12) ‘Theradiogalaxy3C295hasaredshiftof46%.Astronomers meanbyB thisthattheobserved wavelength is1.46timesthewavelength ofthesame ; ; radiation originating inthelaboratory. ‘ThisistheDoppler effect. Only therelative velocity between source and (a)Calculate theradial velocity ofthegalaxy detector matters. (b)Some quasars haveredshifts of200%. What istheirradial velocity? i ws RELATIVITY PRoMLEMS 208 145.(142)Theeaseofthespeedingphysicist ‘ThefigureshowsS”attwosuccessive beatsofthesource,separated bya Aphysicist isarrested forgoing through aredtraffic light. Incourt she time pleads thatsheapproached atsuch aspeed thattheredlight appeared oped green. Thejudge, agraduate ofaphysics class,changes thecharge to nae =To speeding and fines thedefendant $1forevery kilometer perhour she exceeded thespeed limit of50kilometers/hour. itVyh<«r. What isthefine(ygey ™5.3X10”meter, Aygy6.5 x10-7meter)? (a)Show thattheperiod measured atOisT=y(1~fcos8)T,andthat 146,(142),Thetwinpardon | F=felly(1~B00s8))=f'/Ly(1~Bcos8)]. Jnhistwenty-first birthday, Peter leaves histwin Paul behind onthe ,us,if@=2/2,thenf=fly,f=f"/7. earth and goesoffinastraight linefor7yearsofhistimeata speedof Tus0.56Petestheareneea eaecesipias (b)Compare thisresultwiththeDopplereffectcalculated inSec.14.2.1 (a)WhataretheagesofPeterandPaulatthemoment ofreunion? ag eee(b)PeterandPaul,expecting astrange result, perform thefollowing © that‘experiment duringPeter's trip.Theybothobserve adistant variable star awhose lightalternates fromdimtobright atafrequency fwhen observed (1+ Bcos6") from theearth. Thevariable starisin adirection perpendicular toPeter's . yf’i ‘Thus,if6"=2/2,thenf=f/yandf=yf"trsjecory. They,ofcourse,bothcountthesumenumberofpulsations (a)From(a)and(c),¥*(1~fcos0\(1+Bcos8)=1.Usetheexpression fortheDopplrshittoverifythediference inage ChecktheaotOsean eed ee andbetween PeterandPaulattheendofthetrip.Seethenextproblem. at0=/2,RefertoProb,110ontheheadlight effect 14-8. (14.2) Transforming visible light tohigh-energy radiation 14-7.(14 fora aty=constant ¢Powe17acites‘ourceofelctiomagnctic radistionofproper Visiblelightcanbetransformedintohigh-energyradiationbyreflectinga frequencyf,situatedattheoriginO”ofreferenceframeS”movingatthe laserbeambackwardon2high-energyclectronbeam.Saytheinitial velocity1withrespecttoS.Asusual,O'isatx=0at1=0,1'=0.We arettcuems eetromolss theelectronenergy18 wish to asmeas osituated a ronvolect snHoamenrymemeaneredbycbeerverosnuatedatthe (a)Calculatethephotonenergyhy’inthereferenceframeS’oftheelectrons. (b)Now calculate the energy Av" ofthe reflected photons inthe laboratory frame Asafirst approximation, youcanneglect therecoil oftheelectrons, but thisgives toolarge avalue forhv’. sys os 1 14.9,(14.4)Thespeedoflightinamovingmedium 1Light moves more slowly through amaterial medium than through a ve ! vacuum, itsphasevelocity vbeingc/n,wheremistheindexofrefraction of I themedium.“ a Ifnowthemediumitselfmovesatavelocity7”<«ewithrespecttothe= 7 laboratory, show that thephase velocity ofthelight with respect tothe ri “1 laboratory isapproximately c/n+V(1— I/n*).SinceVee | 14-10.(14.4)Theheadlight effectyooie 1 Asourceoflightmovesatavelocity12.Consideraraythatformsan /a 1 angle6”withrespect tothex-axis.(74 H (a)Showthatinthereference frameS,tan@=sin8'/Ly(cos6"+f)] LA ol H Thentan6"=sin6/[y(cos 6~B)],fromSec.13.4.7 (b)Plot@asafunctionof6”forB=0,0.5,0.9,0.9999. "i i . Observe that, forlarge values ofB,8ismuch smaller than 0",except for 97>Fig167 values of6"nearx.Ifthesource radiates isotropically initsownreference a 2ot ) CHAPTER15 “feK *RELATIVITY II “ Mass, Momentum, Force, and Energy Fig. 148, 15.1THEFOUR-VECTOR® —26515.2 FOUR-VECTORS 267 frame, then, forastationary observer, thesource radiates mostly inthe 18.2.1 THE SCALAR PRODUCT OFTWO FOUR-VECTORS 268 forward direction. Thisistheheadlight effect. 15.22 THENORM OFTHEFOUR-VECTOR® 268 (©)Anisotropic source oflightmoves ataspeed c/3withrespect toan 15.2.3 THENORMOF AFOUR-VECTOR® 260 observer. Calculate thesolidangle defined byaconethatpoints forward \nk THERELATIVIsTICMASS mw andthatcontains 25%ofthetotallightflux. tlTucmeattvisticwoweserumy 290 14-11, (14.4) Thecollimator paradox ISS THE RELATIVISTIC FORCE F 270 Figure 14-8shows asource oflightandacollimator Cfixedinareference Example 271 frameS’thatmovesatavelocity Vtwithrespect toafixedframeS.The 15.6THEFOUR-MOMENTUMp =271detector Dmeasures thelightthatgoesthrough thecollimator. 7 =meAccording toProb.14-10,theangle@formedbythebeamoflightisless 17TERELATINISTICENERGY €= 7sthan6”becauseoftheheadlight effect.However, theLorentzcontraction Examples 'ott eteonthecollimator makes itsangle @larger than0".Butthisisabsurd! If Example: THERELATION @=mic+p'c? 275 lightreaches DinS’,thenitdoessoinS. 15.8KINETICENERGY =276 ‘You cansolve thisparadox byusing theLorentz transformation. 15.9. THE LAW OFCONSERVATION OFFOUR-MOMENTUM p 276 14-12, (14.4) Three reference frames 15.10 TRANSFORMATION OFAFORCE 276 Wehave three reference frames A,B,C. Frames BandCmove, Examples 277 respectively, atvelocities 8/2 andVwithrespect toA.Usesubscripts to 15.11 THEPHOTON 277 identify thevelocities: tp=1/2, Yoa= V. 15.12 SUMMARY 278 Showthatucy(velocityofCwithrespecttoB)islargerthan'/2.Thus, PROBLEMS —280with respect toB,Cmoves away faster than A, ‘There are five chapters onrelativity; thefirst three provide thebasic ideas that areprerequisite fortheother two. This third chapter concerns mechanics. Asyou will see, relativity makes havoc ofmechanics. Allits basic concepts, such asmass, momentum, force, and energy, crumble, andclassical mechanics sinks totherank ofanapproximate theory. Itisa curious fact that classical electromagnetism, bycontrast, iscompletely compatible with relativity andisthus spared. 15.1 THE FOUR-VECTOR r Anevent E,such asthecollision between twoparticles ortheemission of 7 aphoton, occurs atagivenpoint(x,y,z)andatagiventime¢,Theevent son 152FOUR-VECTORS 267 , , PHO OSHC Zee (15-2) where x,y,z,ctandx’,y’,x’,et’satisfy theLorentz transformation. This four-vector istheposition vector inspace-time. This isnot anew situation. For example, saypoint Oinordinary three-dimensional space istheorigin, andpoint Ais1meter above O. The vector rhas amagnitude ofunity and points upward. But the components ofrdepend ontheorientation ofthecoordinate axes. —b- _ Inthree dimensions, avector Fcanserve torelate apoint P,toapoint -fi] P,,insteadoftotheorigin: —_{| fa=(2—m)8+02yO+(22~208 (183)+ _——_— | Similarly,thefour-vector o—Ho feahonlin=Adele0)]) (154) Fig. 15-1. Eventoccursat(x,0,0,ct)inS,orat(x',0,0,et’)inS*.The four-vector rdefines thecoordinates ofEwithrespect totheevent &thatoccurs relates event 2toevent 1 at(0,0,0,0). 15.2 FOUR-VECTORS isthussaidtooccur atthespace-time coordinates (x,y,2,ct).These are Inthree dimensions, thevector r(Z,y,z)denotes theposition ofthe thecoordinates ofevent £withrespect tothedatum event @thatoccurs point (x,y,z)withrespect totheorigin ofcoordinates attheorigin (0,0, 0,0). Other vector quantities inthree dimensions, such asvelocities, Wecanthus imagine afour-vector tpointing from QtoEin accelerations, forces, etc., transform inthesame wayastheposition space-time. Inparticular, ify=0 andz=0, wecandraw FasinFig vector r.Indeed, anythree quantities thattransform asthecomponents15-1,Wedenotefour-vectors bymeansofboldfacesans-serif type:F.Itis ofthepositionvectorarethecomponents ofathree-dimensional vector,thecustom towrite outthecomponents ofrinoneoftwoforms: bydefinition. Similarly, infour dimensions, any four quantities a,,@2,as,a4that r=(x,y,2,ct)=(r,et). (15-1) transform asthecomponents oftheposition vectorinspace-time arethe components ofafour-vector ‘The four-vector ftransforms asinTable 15-1, because ofthe Lorentz transformation (Sec. 13.4). B=(44,da,as,da)=(@,ay) (15-5) serve that the four-v. 0Observe thatthefour-vector Frelates oneeventtoanother. Itthus Table 15-2states therulesfortransforming thecomponents ofapossesses itsownidentitity, irrespective ofthechoice ofreference frame: four-vector. ‘Table 15-1 Transformation ofthefour-vector =(r,ct) Table 15-2 Transformation ofafour-vector @=(a,a.) j eeu. raysVier, ayn-Vo+r, i} arraievneas«——04Ve Yn a=(a+ ai=y(a,-7 amr(er+Zti) ater) (eie™) (0-7) Ws RELATIVIFY 269 15.2.1 The Scalar Product ofTwo Four-Vectors joo “ Given the two four-vectors 8=(01,03,05,0.) b=(by,bs,by,by), (15-6) airi theirscalarproduct is 5 a+b=a,b, +azb2+abs~aaby. (15-7) a : H 7 Notethenegative signbefore thelasttermontheright.Otherwise, this i tproductissimilartothescalarproductoftwovectors. w rs 7 —, 7 Forexample, if o wo) .oe Fig.15-2.(a)EventFoccursatye=0,zp=0,andx¢>cte.Itoccursatr=0in r=(x,y, 2c)=(Rot) (158) acertainframe§”(b)EventEoccursatyp=0,2g~0,andX¢<ety,Itoccurs - atx’=Oinsomeotherframe R=(X, Y,Z, =(R, 5s(cT)=(R,cT), as9) Letusset,successively,x¢><cte,x¢<cte,andxp=cte.First,we thenchoose anevent Esuch that x>ctr, asinFig. 15-2(a). Then wecan Zi 7 frameS’inwhichEoccurs at eeRaxX +y¥422-chP =rRAAT. 10) imagineaxesx’andct’ofareference y=r R= (45-10) t=O. In’,[te]=x.Thismeansthatifx;>cte,thenthenormofreisThescalarproduct oftwofour-vectors isaninvariant. Thisisa equaltothespatialdistancebetweeneventsQandEinareference frame ‘consequence oftheLorentz transformation. withrespect towhich QandEaresimultaneous. Thisresultapplies even ifye#0, zp40. 15.2.2 The Norm oftheFour-Vector r ‘Now suppose thatx»<tp,asinFig. 15-2(b). Wenow select axes x" “Thequantity andct’sothatEoccursatx’=0,Then,inS’,eventEoccursatthetime Ils[ere=pety?+2?—cr]!=[crY2(15-11) tote (15-13) isthe norm ofthe four-vector r,byanalogy with ordinary vectors. i itObserveagaintheminussign.Thisisnotasdcanpomenry!Then ‘Therefore, ifxp<cty,thenormoftrdividedbycisequaltothetimeobserve theabsolute valuebare;bydefinition, thenormofafour-vector interval between QandEinaframeS’wherebotheventsoccuratthe isreal and positive.. °same place. This istheproper time interval between Qand E. ‘Thenormofafour-vector isaninvariant, likeascalarproduct. Remember thatwehavesetyg=Oandzp=0. 0.Letuspauseabittoreflect onthemeaning ofthenormofthe Ifnowxp=ctr, then|r|=0,despite thefactthatt,#0. Forfourevector rythatdefines theposition ofanevent inspace-time with example, suppose aphoton travelsalongfeinreference frameS.Then@respect to8givenreference frameS.Wesetye20,2p, soantobe andErepresent twopointsontheworldline(Sec.13.5)ofaphoton withvoletoshowtyomapace timediagra eeita * x,=tp,andtheredoesnotexistareference framewithrespect towhich© pace-time diagram asinFig, 15-2(a)and(b).Then QandEtakeplaceeitheratthesametimeoratthesamex. rel=bee C23! -Weel=behehh (as-12) 15.2.3TheNormofaFour-Vector a Clearly, thenorm ofrgisnotitslength. ‘Thenorm ofanarbitrary four-vector aisdefined likethatofF: Ab lal=la-al!?=aj+03+a3—a3|!* (15-14) F-%, (as-17) Since all four-vectors transform asr,itfollows that the norm ofa four-vector isinvariant. where pisnow therelativistic momentum. Weshall findthetransforma- tion equations forFlater, inSec. 15.10. 15.3 THE RELATIVISTIC MASS mExample |Ahigh-energy particleofrestmassmo,velocityv,andchargeQ crosses aregion where theelectric fieldstrength isE.Theelectric Initsown reference frame, anobject hasarestmass my.Iftheobject foes movesatavelocityvwithrespecttoanobserver,then,forthatobserver, pLqu)=[0] 08.(1518 m a dt ala-vey" :m=ayaaYMm0>Mo (15-15) 15.6THEFOUR-MOMENTUM p Wetake thisfairly well-known result forgranted.’ The quantity misthe relativistic mass. Allmass measurements areinagreement with theabove Consider aparticle ofrest mass myand velocity vat equation. For example, the mass ofahigh-velocity electron inan _ . accelerator isytimes thatofaslowelectron. r=(yy, 2,et)=(x,ct) (15-19) Therelativistic masstends toinfinity asvapproaches c.Then what withrespect toareference frame S,asinFig,15-3 about thephoton? Ithasarelativistic mass m,buta(presumably) zero Test mass mp. See Sec. 15.11. a Weshall seehow totransform arelativistic mass inSec. 15.6 (Table 1533) poof i} iv 15.4THE RELATIVISTIC MOMENTUM p ; ot\ Therelativistic momentum ofamassmmoving atavelocity visdefined a | asinclassical mechanics, except thatmistherelativistic mass: | =my== ym, (15-16) |peme= Tayajeya moe r+at-——44 4——Ne Itisthis quantity that isconserved incollisions, and not themomentum H movofclassical physics. ! Weshall find how totransform arelativistic momentum inSec. 15.6 H (Table 15-3). 15.5 THE RELATIVISTIC FORCE F f 7 : The relativistic force isalso defined asinclassical mechanics: "Seeforexample,DavidBohm,TheSpecialTheoryofRelativi,W.A.Benjamin,New Fig.18.3,Worldlineofparticlemovinginthe(x,f)plane.Thefour- Yorkiosnt ” of Rea ‘momentum pisalongdr,andisthustangent mn m nenativery i 187THE RELATIVISTIC ENERGY =mc? Over thetime interval dt,asmeasured inS, Table 18-3Transformation ofafour-momentum P=(p,mc) dr=(dr,edt) (15-20) p=vpitm'V) +P.p=(Pp)—mV)+P. =y(m'e+228 te=7me—U0" ‘Thisincrementdrofristangenttotheworldlineoftheparticle.Now m={mma*B) mes{mec) Vey Vuy Ir\2)"2 =ym'(1488 im’=ym(1—"8 (dr=dr?cdr=ea1-4(4)|(18-21) maym'(1+ 72) ym(1-"3') o\de ——————————————————— =cdti-|""note can, (as.22) Sothenormofthefour-momentum ofagivenobjectisindependent ofe Y thevelocity! Itisaninvariant. of ticle thatisstationary inspaceisnotstationary in fromSec.14.2,wherediyistheincrement ofpropertimeintheinertial spece-time:ittravelsparalleltothectaxisatthespeedoflight. frameoccupied momentarily bytheparticle. Thevelocity vneednotbe ‘AStothephoton, seeSec.15.11constant. . ; ;Thefour-t itthasanother peculiar property. Since p-pis Foragivendy,say1second,thecorresponding dtdependsonthe invanant. momentum *" ¥ppreference frame oftheobserver, butdiyisthesame forallobservers, and ° hence dfyisaninvariant d(p-p)=2p-dp-0. (15-28) Thefour-momentum oftheparticleisdefinedasfollows: ‘Thismeansthatdpis“orthogonal” top.Butpis“parallel” todr,from dr Eq, 15-23. Thus Ppem-. (15-23)diy dr-dp=0, (15-29) ‘Thisisafour-vector because drisafour-vector, while bothmyanddiy anddpis“orthogonal” todr,inagreement withthefactthatpis areinvariants, Thefour-momentum istangent totheworld lineofthe invariant. particle. Tosummarize, thefour-momentum pofaparticleisafour-vector (1)Tofindthecomponents ofp,wewriteoutthecomponents ofFand thatistangent totheworld line,(2)whose space component ispanddifferentiate: whosetimecomponent ismc,(3)whosenormismoc,and(4)whose dro drt de differential dpis“orthogonal” tobothpanddr. P=mogemmo(ret=(mite omoer). (15-24) Sincepisafour-vector, ittransforms likeF,asinTable15-3.. . ° . Ifv?,v,¥?areallnegligible compared toc*,thenwerevert to But, from Sec. 14.2, classical mechanics: a j p=pitm'Y, m=m’ (15-30) dg awe (15-25) .° 15.7 THE RELATIVISTIC ENERGY €=mc* Thus ‘Wehave justseen thatdr-dp=0.Thus P=(moyv, moyc) =(mv, mc)=m(v, c)=(p, mc). (15-26) dr-dp=(dr,edt)(dp,cdm)=dr-dp—c2dtdm=0, (15-31) The norm ofphasapeculiar value: or wy? dp IPI(P=pl!?=mjv?~ c=me1-5)=moc,(15-27) arBeedm, (15-32) 1 keLaTiviry 1 275 Now weareconcerned with anobject ofrelativistic mass m,velocity v, oO andrelativistic momentum p=mv.Sodp/dtistheforceFappliedtom, p=mv.Sodpepp Fig.15-4.Apositiveanda andthetermontheleftistheenergyexpended byFonmoverthe negativeelectronannihilate todistance dr.Anincrement ofmass dmaccompanies thisincrement of form twophotons. Ifthekineticenergy. energiesoftheelectronsarelow,Moregenerally,inanyphysicalprocess,anincreaseofenergyd& oe eeneheangnereyiy results inanincrease ofmass dmsuch that mw | electron, 5.110"le electronvolts. dé=c' dm, (15-33) and €=me? (15-34) unchanged. Strictly, oneshould, ofcourse, think interms ofthe potential energy ofthe brick-earth system and take into account istherelativistic energy ofanobject ofmass m. theupward motion oftheearth asthebrick falls. Ifaforce Factsonamass mmoving atavelocity v, When apositron meets anelectron, bothparticles disappear to form apair ofphotons, asinFig. 15-4. Charge, relativistic mass, dime) dé energy, andmomentum areallconserved. irir conservation ofcharge,buttheconservation ofmass,energy,and > momentum requiresthepresenceofanotherparticle Aparticleatrestpossessesarestenergymyc Arelativistic energytransforms likeamass(Table15-3). Example |THERELATION @=mic*+pc? Anobjectofrestmassmaliesatrestinareference frameS’.With Examples |Sayamixtureofhydrogen andoxygenexplodes. Whathappens to i a athemass? Before thereaction, themolecules maybeassumed to respect toanother reference frame S,itsmassismandits beatrest.Lettheirtotalrestmassbemy,Afterthereaction, the momentum isp.Sincethenormofthefour-momentum isrelativistic mass isunaltered because noexternal energy hasbeen invariant, fedintothegas.Then pra mic =mie (as-37) My=May=Ma,+KE, (15-36) ‘Therefore where KEisthekinetic energy ofthehigh-temperature steam ame =mitt pe, (1538) ‘The new rest mass issmaller than the initial rest mass. After awhile, thesteam cools andbecomes water atroom asinFig.15-5. vs :temperature, KEtends tozero, andm,decreases tomy,.Inthe ‘Thetermpic*isnegligible when y"f"<1, orwhen 261. process thekinetic energy hasspread outtoneighboring bodies, ‘Then€~moc*. Ontheotherhand, ify*p"> 1orif~1 and thereby increasing theirmasses. Themassofthewater islessthan v=, then thatoftheoriginal mixture, buttherelativistic mass ofthe €=me~pe (1539) universe isunchanged. Fission and fusion reactions arequalitatively similar toexother- ‘mic chemical reactions: there isaloss ofrest mass and arelease of thermal energy. What happens when youdrop abrick? The restmass ofabrick depends onitsposition, because ofitspotential energy: thehigher . itis,thelarger isitsrestmass. Asthebrickfalls,itsrestmass Fig. 18-8.Therelativistic eneray #decreases anditskinetic energy increases. When thebrick hitsthe ietclated tothe restenergy mart ground, itskinetic energy becomes thermal energy. Itloses rest andtotherelativistic momentump ‘mass, but, again, the relativistic mass ofthe universe is 7 asinthisright-angled triangle 1% neLarivery ao 15.8 KINETIC ENERGY ‘Table 15-4 Transformation ofaforce Bydefinition,thekineticenergyofapointmassmmovingatavelocityv| F=F;amor erFakeaewithrespect toagivenreference frame isequaltotheenergy expended vl Vile Lace diedinincreasing itsvelocity fromzerotovinthatframe. Thisis | vey 22*Be }wherew’isthevelocityofthepointofapplication ofF’inreference (15-40) | frame S". mov?3myv" | Nowdé"/dt’istherateatwhichtherelativistic energybuildsupinS’ arar een (15-41)|undertheactionofF’.Then ‘Thefirsttermontherightisthekineticenergyofclassicalmechanics. It aSpeeFlul+Fey’ isequaltotherelativistic kinetic energy ifv?<c* 1 aePw Pt Piow (15-45) 15.9 THE LAW OF CONSERVATION OF Substituting into Eq.15-44 andsimplifying leads tothetransformation FOUR-MOMENTUM p equations ofTable 15-4. The transformation does notinvolve the coordinates ofthepoint ofapplication oftheforce. Saytwoparticles interact, yielding twoorthree other particles. There is Note that F.#F,. Since afour-vector hasthesame perpendicular conservation ofthetotal four-momentum component inallinertial frames, therelativistic force isnorthespatial component ofafour-vector. yp=(Yp.cd 15-4;ze(=pdm) (15-42) Examples|1fFisparallelto¥,thenF=F’. Two forces that areequal and opposite inone frame arenot forthesimple reason that both Epand Ym areconserved. Sothe rnceessarilysoinanotherframe.Theyremainequalandopposite conservation offour-momentum groups theconservation ofmomentum, onlyiftheir points ofapplication have equal velocities. ‘ofmass, and ofenergy! Theconservation andtheinvariance ofthefour-momentum implythat 15.11 THE PHOTON ifpand mareconserved inone inertial reference frame, then they are conserved inanyother suchframe Thephoton haspresumably azerorestmass. Itsspeed’ isc,andits energy is 15.10TRANSFORMATION OFAFORCE #=hy, (15-46) Aforce (Sec. 15.5) transforms asfollows. From Tables 15-3 and 13-1, where hisPlanck’s constant, 6.626x10”joule-second,andvisthe yfthe wave. 15.7, atld-tVACIC) ase) frequencyoftheassociated wave.FromSec.15.7,itsmassis at y+ Vdrile)" g mas, (as-47) Dividing above and below byydt’yields paFit(Vie)déide’+Fy (15-44) "Photonsalwaystravelatthespeedc,eveninsidematter.SeeSec.37.4.Forthe140, le moment, wecanthink ofphotons traveling inavacuum. 178 RELATIVITY tH 15.12 SUMMARY 29 anditsmomentum is r=(x,y,z,et)=(r, ef). (1S-1) € hvih ‘The four-vector pemeni = e4, (15-48)coc & fa=ho-n=[n-n),cb-n)) (1s-4) where Aisthewavelength oftheassociated wave. The relation between p and &foraphoton hasbeen verified bymeasuring both theradiation relates anevent Eswith respect toanother event Ey. Afour-vector pressure andtheenergy fluxofalight beam. transforms asinTables 15-1 and 15-2. Any setoffour quantities that The four-momentum ofaphoton is transform asthecomponents ofafour-vector arethecomponents ofa four-vector. P=(p,me)=(me,me). (15-49) ‘Thescalarproduct oftwofour-vectors aandbis Itsspatialcomponent isme,wherethevector¢pointsinthedirectionof a-b=ayb, +asb;+asby—asbs. (as-7)propagation. Also, =|e—mej"?= ‘Thisproductisinvariant. =|e? =me?|'? =0. 15-50) inl\ (sso) Thenormofafour-vector isalsoinvariant: According toEq.15-27, thisimplies that thephoton hasazero restmass. ae oneAlso,fromSec.15.6, (|=[a-al!?=lay+a}+a}—ail!” (as-14) _del Themass ofanobject depends onitsvelocity with respect totheIpl=m =O, (ssi) observer: mo and|dr|=0.Theincrement ofproper time(Sec.14.3)overdriszero: m= rey Vme7mo (15-15) dry=S20, (15-52) Thisquantity isalsocalledtherelativistic mass. « Therelativistic momentum ofapoint mass is ‘Aphoton’s clock always reads thesame time! mv =‘Thekinetic energy ofaphoton is(m—mg)c?=mc?,andnotme?/2, as Puan =ymee, (15-16) wewould expect from classical mechanics. Although agiven photon hasa vistispeedcforallobservers, itsenergy @anditsmomentum arenot andtherelativistic forceis invariant, because oftheDoppler effect. Forexample, ifS’moves inthe dp samedirection asthephoton, thenv’<v,andthus r-2. (15-17) hv’<hy, e<é, n< 15-53)psp (553) ‘Thefour-momentum ofaparticle isthefour-vector 15.12 SUMMARY dr Afour-vector hasfourcomponents, thefirstthreeofwhicharethespace Pama (p,me) (15-23),(15-26) components ofathree-dimensional vector. One specifies thelocation of anevent inspace-time with respect tothedatum event 2(0,0,0,0) by whose norm ismoc. means ofthefour-vector Four-momenta and masses transform asinTable 15-3. = RELATIVITY tt PROBLEMS Therelativistic energyis Yr de’ dm=y(1t+uis)dm', (b)dr=——— >, “me (15-34) EY ccaBEN Th 2 Ky ©p=r(1+4t)o p=r(1-“F)e. ferestenergyismoc’.Kineticenergyisrelativistic energyminusrest e °energy, mc?—moc’ 15-4,(5.5) Therelativistic forceInanyinteraction there isconservation ofthetotalfour-momentum p, Inclassical mechanics, F=maifthemassisconstant andhence conservation ofmomentum, mass, andenergy. Showthatwithrelativity, Forces transform asinTable 15-4. my mo 4 Forthephoton, therestmassispresumably zeroand FGreye *Tavepe esmanmae é where a)anda,arethecomponents oftheacceleration thatare,m=s, 15-47 respectively, parallelandperpendicular tothevelocityvofthepointof 2 ) per «application oftheforce wh 'Aforcethatisperpendicular tovchanges thedirection ofw,butnotthepame=—= "=" (15-48) mass,andhencenotthespeed,aswecouldexpectbecausetheforcedoeseek nowork. Then, ifFisperpendicular tov,F=maapplies! . | However, aforceparalleltochangesthemagnitude ofvandhencethe where fhisPlanck’s constant, 6.626 10°“ joule-second, visthe massalso.Theresistance toacceleration islarger because ofthe7*term,frequency, and4thewavelength oftheassociated wave. “Thequantity missometimes calledthetransverse inertialmass,andy'm | thelongitudinal inertial mass | ,PROBLEMS 15.5.(15.7) ‘Thegravitational redshift| ‘Aphoton ofenergy fv,leaves thesurface ofastarofradius Randmass M. 15-1, (15.3) Burning hydrogen (a)Show that after thephoton hasescaped toinfinity, Av/v9 isequal to Youignite amixture ofhydrogen andoxygen inside aclosed vessel, and GM/(Rc’), where Gisthegravitational constant. Thischange issosmall thenallow thewater vapor tocool. | thatyoucansetv=¥inyourcalculationofthechangeinpotential Sketch graphs ofmc®andofmc* asfunctions oftime. energy. What isthesign ofAv? This change offrequency isthe 15-2.(15.3)Relativistic effectswith40-GeVelectrons weer redwuforth fortheearth,Se facingthe‘Alinearaccelerator accelerates electrons uptoenergies of40gigaelec- (©)Calculate v/vforthesunandfortheearth.Seethepagefacingt tronvolts (40x10”electronvolts).° Bienes back cover. vt ; (a)Calculate themassofanelectron thathasthefull energy.How (c)Calculate Av/vforaphotonthattravels fromthesurfaceofthesun energy. How doesthismasscomparewiththatofaprotonatrest?Seetheporefacl tothesurfaceoftheearth,takingintoaccountbothgravitational fieldsbackcover. peme pagefacingthe (a)Siriusandasmallerstarrevolvearoundeachother.Themassofthe(b)Whatisthelengthoftheaccelerator inthereferenceframeofan smallerstarisaboutequaltothatofthesun,butisighthas@Av/yofflectronthathasthefullenergy?Thelengthoftheaccelerator, as 7107.Whatisitsaveragedensity?measured ontheground, is3000meters (€)Theperiodofrotation ofthesunis24.7days.WhatistheDoppler(©)Howmuchtimewould suchanelectron taketogofrom shiftfor500-nanometer lightemitted fromtheedgeofthesun’sdisk,atits electron’s frame ofreference? (f)Thesunejects ionized hydrogen. Howdoesthemassofaproton vary itflies away from the sun? 15-3, (15.6) Transformation ofamass densityas 7om te ‘Weusethesymbol rforavolume inthisproblem. 15-6,(15.8) Themassofahigh-energy proton ;Asmallelement inanobjecthasapropermassdims,apropervolume ‘Aprotonhasakineticenergyof500millionelectronvolts. Finditsmass dro, andaproper mass density py=dmig/dry.WithrespecttoSandS’the andvelocity. massdensities arep=dm/dt andp'=dm'/dr’, respectively, andthe 15-7.(15.9)Theconservation lawsforcolliding particlesvelocities oftheelement arevandv" Inthecourse ofacollision between twoparticles thereisconservation of‘Showthat,if¥isthespeedofS’withrespect toS,then | energy andconservation ofmomentum. (282 RELATIVITY Ht PROBLEMS: 283 (@)Show that, ifthese conservation laws apply inoneinertial reference (4)Now thishydrogen must fistbebrought uptospeed. This slows the frame, then theyapply inanyother inertial frame ship. Show that thenetgin ispositive up )=0.707 andnegative (b)Show that if,inagiven reaction, relativistic mass isconserved inall afterward. inertial frames, thenpisalsoconserved,andinversely, 1s12(1511)“TheDopplreectagain 15-8. (15.10) Transformation ofaforce Refer toFig. 14-7. Anobserver attheorigin Oofthereference frame S theorigin O'of$7Inrob, 1-9 wefound that f=(1+ cos) FartAtVREXY)ux(exv)| ‘Checkthisequationbytransforming thefour-momentum ofaphoton.m é 15-13. (15.11) ‘TheMéssbauer effect , sorceand pow Anexcited nucleus ofFeformed bytheradioactive decay of"Coemits 1ingbonBaaIT,showthat agammarayof1.44%10"eletronvolts. Intheproces,thereisconserva-tionofenergyandmyc?=m,c?+hv,wheremgistheinitialmassofthe Pus alsoconservation ofmomentum, /v/c=m,u,where uistherecoil velocity ‘ofthe iron nucleus. Let my bethe rest mass ofthe nucleus after the 15-10.(15.11)Theultimatespaceship eeethenwearereleasedbythereaction‘Eo(meemae4‘Thethrustofaspaceship engineisequalfotheproductm'v,wherem'is Oaeeeeeeee eee aeyoustaaline themasofpropellant ejectedpersecondandistheexhaustvelowith Square.Thensquarethesecondequation andsubstitute. Youshouldfind respect totheship. Theultimate spaceship would transform all.its that Elim+Mo) épropellant intoradiation andejectphotons backward atthespeedoflight, ty= (Ima) ‘Themass ofthepropellant would then beminimum. a “ (a)Show thatthepower-to-thrust ratio P/Fforaphoton engine isc. Sohv<@:partof€goestothephoton, andtheother partsupplies kinetic Since P/FanddM/dt areindependent ofthefrequency, thesource of energy totherecoiling nucleus. . radiation need not bemonochromatic. (b) Set_ my=57x1.7x10",andshowthat€/(2mgc*)=1.3x10”. (b)Thenaphotonshipburning|gramofmatterpersecondwouldhave Thusthefractionoftheavalabieenergy&thatappearsaxrecoilissmall athrust of3x10°newtons. Thedifficulty istotransform anappreciable (©)Méssbauer discovered in1958 that, with solid iron, asignificant fraction ofthepropellant massintoradiation, asthefollowing example will fraction oftheatoms recoil asiftheywere locked rigidly totherestofthe show solid This isthe Mossbauer effect. Ihesample hasamass of1gram, by ‘Anordinary flashlight hasacapacity ofabout 2ampere-hours atabout 2 what fraction isthegamma rayenergy shifted intherecoil process? volts. Show thatitsterminal velocity isoftheorder of10)*meter/second. (d)Asample ofnormal “Feabsorbs gamma raysof14.4kiloelectron- IS-L1, (15.11) Isinterstellar travel possible? gamma raysofanynearby energy. Theexcited nuclei thusformed remit (a)First,timeshould bedilated by,say,afactor of10.Then y=10. 14.4-kiloelectronvolt radiation inrandom directions sometimelater.Thisis Calculate u/c. | resonantscattering.(b)Imagineaspaceship equipped withaphotonmotor.SeeProb.15-10, Ifasampleofactivated “Femovesinthedirection ofasampleofnormal Yowanfindhetonofhealmanhatemaiathssp Fe,whatmustbethevalueofthevelocity thatwilshiftthefrequency thasattained theproper B,from theconservation ofenergy andthe ofthegamma rays,asseenbythenormal nuclei, by3partsin10!*?Thisis conservation ofmomentum. Take intoaccount theenergy andmomentum ‘onelinewidth.oftheradiation. Youshouldfindthatf=0.05. (c)ADopplershiftinthegammarayresultsinamuchlowerabsorptionThespaceship must then brake toastop. Thisrequires 95% ofthe byanucleus iftheshiftisoftheorder ofonelinewidth ormore. Whatremaining mass.Attheendofthereturntripweareleftwith(0.05)*= happenstothecountingrateofagamma-ray detectorplacedbehindthe 6.25x10*oftheinitialmass.Ifthemassoftheshipanditspayloadis1 sampleofnormal“’Fewhenthesourceofactivated“Femoves(i)toward ton,thenthepropellant hasamassofabout200,000tons. thenormal“Fe,(ii)awayfromit? (c)Inprinciple, thespaceship could collect andannihilate interstellar (f)Ifa14.4-kiloclectronvolt gamma raytravels 22.5meters vertically mater. There itabout oneatom ofhydrogen percub centimeter upward, bywhatfraction wiltsenergy decrease? Calculate themassofhydrogen collected during 1yeariftheshipsweeps (g)Anormal “Feabsorber located atthisheight mustmove inwhat ‘outavolume 1000square meters incross-section atthespeed oflight. direction andatwhatspeed inorder forresonant scattering tooccur? 162. THE FOUR-CURRENT DENSITY J 285 charge-to-mass ratio e/m foraparticle moving atanincreasing velocity v 16 variesasfollows: CHAPTER e_e/,_vy?moml!-@) as *RELATIVITY IV ‘Theelementary charge ¢therefore remains invariant, whilethemassm TheField ofaMoving Electric Charge varieswithvelocity asinSec.15.3.Thisrelation applies uptothehighestenergies attained todate. ‘Another proof oftheinvariance ofelectric charge isthefactthat a 16.1 INVARIANCEOFELECTRICCHARGE 284 metalobjectdoesnotacquireanelectricchargewhenitiseitherheated Example 285 orcooled (excluding thermionic emission), despite thefact that the 16.2 THE FOUR-CURRENT DENSITYJ285 averagekineticenergyofitsconduction electronsismuchlessaffected Example:THECONDUCTION CURRENT INAWIRE 287 than thatofitsatoms.’ Itisbecause theenormous positive andnegative 163THEFOUR-DIMENSIONAL OPERATORO 286 charges inapiece ofmatter (Sec. 3.1)cancel perfectly atalltemperatures 14THECONSERVATION OFCHARGE 290 thatordinary matter remains neutral when itstemperature changes.16.5 THE FIELD OFAPOINT CHARGE QMOVING ATACONSTANT vELonyat im enkilogra 10%atomsand10°x 1651 THEFORCEFi,22 anne|Teei016x10"colombsofconductionelectrons the 16.5.2THEFORCE Fy,292 positivechargeincreased ordecreased byonly1partin10"upon 165.3THELORENTZFORCE 233 heating, thecopper would acquire anetcharge of1.6% 1016554THEELECTRIC ANDMAGNETIC FIELDS 204 | coulomb. A10-klogram coppersphere,witharadiusofabout65 Example: THE FIELD OFA10-GIGAELECTRONVOLT ELECTRON 295\ millimeters,wouldthenacquire@potentialofabout2kilovolts, Example: THE MAGNETIC FIELD NEAR ASTRAIGHT WIRE CARRYING Such aneffect has never been observed. ASTEADY CURRENT 298 16.5.5THEFORCEFg300 |16.2THEFOUR-CURRENT DENSITY J166 TRANSFORMATION OF EAND B 301 Example:THEPARALLEL-PLATE CAPACITOR —303 '‘WenowshowthatJandcparethecomponents ofafour-vector J,called 167SUMMARY 304 thefour-current density. PROBLEMS 305 fl Imagine amacroscopic chargeQmovingatsomearbitrary velocityv|withrespecttoareferenceframeSattime1.Intheinertialreference Inthischapter wefirstexamine twofundamental properties ofelectric frame S,occupied momentarily byQ,thevolume ofthecharge isVo.Incharge,namely,invarianceandconservation.However,ourmainobjec-|S,thechargeisalsoQ,butthevolumeisshorterinthedirectionofvand tive here istocalculate the electric and magnetic fields ofacharge movingataconstantvelocity.Wedo,ofcourse,availourselves ofthe vevi(18)" (162)mathematical apparatus thatwedeveloped inthelastthree chapters. N e The simplest field ofallisthat ofthestationary point charge. The next a ‘one,inorderofcomplexity, istheonetowhichweaddress ourselves |InSjthevolume chargedensity ishere. Q| Pome (163) 16.1 INVARIANCE OF ELECTRIC CHARGE ° Electric charge isinvariant: abody carries thesame electric charge forall "See, forexample, Charles Kittel, Introduction toSolid State Physics, Sthed.,John observers. Itisanexperimental factthat, inanaccelerator, the ‘Wiley, NewYork,1976,p.166. 26 ceLariviry 1 287 while inS Table 16-1 Transformation ofafour-current density 5=(.ep) Q Po OePVGavia YPo>Po- (16-4)| Taye Ve)+T Isy=Vp)+d. VG-vie)| Me : i. { p=("+73 p=(0-*3) NowconsidertwonearbypointsAandBonthewordline(Sec.13.5) | oS ofQ,separated bythefour-vector dr=(de,cd ‘Theanalogy between thefour-current density Jandthefour-6 » 65) momentum pisstriking. YouwillrecallfromSec.15.6that inS,with dr=wdt. AswesawinSec. 15.6, theproper time interval - =betweenAandBis P=(p.me), —Ipl=mee, (16-12) ” andthatthefour-momentum istangenttotheworldlineofthemassm, dty=a (16-6) pointinginthedirection ofmotion.Similarly, thefour-current densityis Also. J=.cp), (16-13) its norm is dt 1 iardig vey (16-7) i=0SE=poe, (16-14) Thus itistangenttotheworldfineofthechargeQ,anditpointsinthea direction ofmotion ifthecharge ispositive. P= 1P.= Po» 16-8)dty ass) iExample |THECONDUCTION CURRENT INAWIRE ‘Thecurrent density inSjiszerobut,inS, Awirethatisstationaryinreference frame $cares current densityJ.ThenetvolumechargedensityinSiszero: dr)dr_ ddr dd P=, +p.=0. (16-15) I=pv=P>=Po==po =pu=r. 16-9) ne a adigdtPag (6s) | ‘Therearesurfacecharges, butwecandisregard thembecause they justsuperpose another electric field over theonethatweare Comparing nowEqs. 16-9and16-8,weseethatJandcpstand outas interested inhere. Thisextra electric fielddepends onthethecomponents ofthefour-vector fevandgeometryofthewire,8wellstonthecaret ing through it a ar ‘Thewireisparalleltothex-axis,andthecurrentflowsinthe I=, cP)=por, ct)=pos (16-10) negative direction, asinFig.16-1.Thentheconduction electronsdo diy flowinthepositivedirection ofthex-axis.Foranobserver onS"moving totheright atavelocity Y,the called thefour-current density. Here both poanddfyareinvariants. Lorentz contraction forpislessthanforp,andthewireappears ‘Therefore Jandptransform asinTable 16-1,andthenorm ofJis to‘bepositively charged. From thetransformations oftheinvariant: | four-vectorJ(Table16-1),vr, Y 2pitta yegyre fenysyaysu, Say 16-16) =Ucp=U?p22(16-11)i premvad=yall»(1s16) 288 i163THEFOURDIMENSIONAL OPERATOR & 289 Yj y tcalledquad.Notethenegative signbeforethefourthcomponent.Remember thatthefourth component ofris+ct(Sec. 15.1). ¥ ‘Thespace andtime derivatives transform asfollows: axxo vo a VaS2S +S -r(S-a5). 16-213xGxSx’*x3se257) asa) QD = 223, 2-2. (16-22) Fig.16-1.Conducting wirecarrying acurrentdensityJ.The eya =eeconductionelectrons,ofvolumechargedensityp,,flowtothe 282 2222) ae) Tightatthevelocityv.ReferenceframeS”hassomevelocity7 GLH8we (2_¥2), paralleltotheire. ad 3raeeae NG” Invector form, Multiplying both equations bythecross section ofthe wire, whichisthesamein§andS’,wefindthattheprimedlinearcharge _48 .densityis v=%-2S)+0, (16-24) yMays U'=yt 16-17) aa ;¢ . on 2=x(2-rim). 1625) Another way ofexplaining theexistence ofapositive charge is thefollowing. Imagine that, foranobserver inthereference frame , .ofthewire,thecurrentstartstofloweverywhere alongthewireat Theoperator Otransforms asafour-vector, asinTable16-2,andacts t=0.Then,fortheobserver onS’,thecurrent flowstartsat likeafour-vector.Y IfFisascalar function ofx,y,z,f,thenitsfour-gradient is ta-yds, (16-18)“ OF\_(3F,, 8F,,OF,_oF) accordingtotheLorentztransformation (Sec.13.4).Thatis,the oF=(FF,-Z)-(GerFoe-Z)(16-26) electrons start flowing tothe right atthe right-hand end ofthe wire and themotion propagates totheleft. This hastheeffect of : tetk‘depleting thepopulation ofconduction electrons inthewire | Thefour-gradient isafour-vector ifFisinvariant. Weshallnotrequire| thefour-gradient.16.3THEFOUR-DIMENSIONAL OPERATOR 0 ‘Thefour-divergence ofafour-vector isascalar: Inthree-dimensional space,thethreeoperators 3/3x,9/y,3/32arethe D-a=(¥.-2)(a,ca)=Va (16-27)components ofthedeloperator: Set ot 39 ,,9,,9 >waste (iss) ratte162Transformation of3=(¥.-2) AswesawinSec. 1.3,thisoperator transforms asavector. Va OL ve‘Thereexistsacorresponding four-dimensional operator ver(ri-Bg)trs War(mtag)+h a2 a_ aa,,8,,8 2a a 2=(Z-vivi1) Z=(S+rimi) =(#2452422, -2)-(¢,-2 an” Ge ar Nae O°(54 ae)"(R-zy) 629 ee 2” “RELATIVITY1|165APOINTCHARGEQMOVINGATACONSTANTVELOCITY 21a Ie yf ationofchargecanalsobewrittenas+J=0: DegaSte4Say5Bae,Bay cis2s) | ‘Thelawofconserv: reOxdyazot 3 ai d=(¥,-35)-(nep)=P-s+ Pao, (16-33) Remember, from Sec. 15.2.1, that thefourth term inthescalar product ' et ot oftwofour-vectors isminus theproduct ofthefourth components ofthe four-vectors. Now(-Jisinvariant, likethescalar Product oftwofour-vectors. Thefour-divergence isinvariant because itisthescalar product oftwo Then theconservation ofcharge ingiven circumstances applies toany four-vectors. Then inertial reference frame. eaSeg24Os5Ot_Se,It,Ser,Be1G96,16.5THEFIELD OFAPOINT CHARGE QaxByBe ax"*ayaztHOP) MOVING ATACONSTANT VELOCITY Weshallnotrequire thefour-dimensional curl,which isasecond-order Weconsider twopoint charges, asinFig.16-2.Thecharge Qisatthe tensor . originO’oftheinertialreference frameS’andthusmovesattheThescalar product ofOwithitselfisthed°Alembertian: constant velocity 1’withrespect tothefixedframe S.Charge qmoves at 2 7 op 2 2 some unspecified velocity vwith respect toS,That velocity need notbeFf FF F 1Ff 218 ae0-O0=0?=~44+54+5-55=P-555. (16-30) constant. There arenoother electric charges orcurrents inthevicinity. ax?By®827car or ( Forexample, ifFiseither ascalar oravector function ofx,y,z,¢,then : ’ 22,13°F .OF =PF -5a (16-31) y©ar 1 sl Thewaveequation forFisO°F=0(App. C). | x The d’Alembertian isinvariant, like the square ofthe norm ofa J s four-vector. i( oo fo16.4THECONSERVATION OFCHARGE emAswesawinSec.4.2,thelawofconservation ofchargestatesthat NS io") iyRL 1\Ron op - toveg=-2. s. iox (16-32) Se By Thus,inanyframeofreference, thereisnevercreationorannihilation of . Hl\ thenetcharge ofaclosed system. en ed Donotconfuse thislawwithcharge invariance. Charge invariance “Se > means that theelectric charge carried byanobject isindependent of thevelocity oftheobject with respect totheobserver. Inother words, ] aiis ii } .,velocit \echatattheoriginO'ofS'isV¥withrespect to thechargeisthesuemalin ramesofreferenceMassisconerved |_——_‘i 462Tevlsyofhecharsattheorigin”ois7wthee inthecourseofaninteraction(Sec.15.9),butitisnotinvariant. \variablesshownaremeasuredwithrespecttoS. m De | 165APOINT CHAKOF. @MOVING ATACONSTANT VELOCITY 293 The force Fo,exerted byQonq,asmeasured inS,willgive usthe Ogr+(Vic?)r sv.~(Ve? )ugr.fieldof@atthepositionofq,in§.Wecalculate thisforce,firstinS’and Foo"ane, yl—Bisin Oy? (16-41) then inS. where, fromIdentity 2onthepageinsidethefrontcover, 16.5.1 The Force Fi, InS',chargeQisstationary. Nowitisawell-established experimental | Very)—Vor=Vwen)—neV) (16-42) factthattheforce exerted byastationary charge onamoving charge is =uXx(VXr)=ux(V Xr). (16-43) independent ofthelatter’s velocity (Sec. 3.1). We may therefore calculate Fg,from Coulomb's law: Finally, 5 . Qq r+vx(VXA/c*»Og?__Qar' FougrePPL=Brain?0) (16-44) - - drePPOBsi8) PoeGreg? dnegr™ (16-34) eyY'r'(1—B*sin?8) . 16.5.3 TheLorentz Force wherer’and#aredefined asinFig. 16-2. | ‘ i\ TheforceFe,comprises twoterms.Thefirsttermisindependent ofv 16.5.2 TheForce Foy I andistheelectric force.Thesecond termdoesdepend onvandisthe Referring nowtoTable15-4,wecandeduceFp,fromFy: i magnetic force.SowehavetheLorentz force Fo,=WE+¥xBo), (16-45) r,-—24[xc+MMICwt| 45)' Ogre oeuvre) (16-35) wherevisthevelocityofqandwhere ‘Westillhavetotransform theprimedquantities ontheright.FromTable | Eo>oe 2oat (16-46) 13-1andfrom Fig.16-2, 4neoy'r'(1 —Bsin’8)? H=R,-V)=",, r=Ro=r, (16-36) |istheelectricfieldstrengthinthefieldofQatthepositionofq,B=V/c, 2 2 a, id r= P(R,-Vo?+Ri=UR,-V0)?+R2(1-6).(1637) ian .| Bo=—20xe ase Now 4xeo7(1—BFsin’8)? P=(Ry~ VIP+RE (16-38) i issimilarly themagnetic fluxdensityinthefieldofQattheposition ofq. ‘Theunitofmagneticfluxisthetesla. ‘ThusInthese three equations, allquantities concern thesame reference 2 2_g2p? 22sin? frameS.r=7(P—BR?)=77°(1—B*sin”8), (16-39) Moregenerally, ifachargeqmovesatavelocity vinafieldE,B,the where0ismeasuredinreferenceframeS.Also,fromSec.14-4, LorentzforceisP=q(E+uXB), (16-48) oy 7 ve"ure Yaw, VTA) (16-40) ‘Thevelocityvneednorbeconstant.Note that, foranobserver inS,thecharge Qexerts onqboth an ‘Substituting andsimplifying, wefindthat electric force gEandamagnetic force quxB.However, foranobserver 204 sweLaTiviry tv ns inthereference frameS’ofQ,thecharge Qisstationary andtheforceexerted byQonqispurelyelectric. ‘The factthat themagnetic force exerted onanelectric charge is L C7proportionaltothevectorproductofitsvelocityvbythelocalmagnetic NN 4 BY fluxdensityBhasfourobviousconsequences. Inagivenreferenceframe. ™ae Ccacethemagnetic force onapoint charge (1)exists onlyifthecharge moves AN : \-withrespect tothatframe,(2)isindependent ofthecomponent ofvthat 7 NN 9 isparallel toB,(3)isperpendicular toboth wandB,(4)does notaffect a ~ = thekinetic energy ofg. _ Then, if@istherelativistic energy mc? ofaparticle ofmass mand charge Qmoving atavelocity vinafield E,B, ow ty dé Fig.16-4,(a)LinesofEand(b)linesofBforapostivechargeQmovingataTPURME+eXB)-v =gE. (16-49) fconstantvelocity¥,asseenbyastationaryobserver. Only anelectric fieldcanaffect thekinetic energy ofacharged particle; i OF amagnetic fieldcandeflect suchaparticle, butcannot change itsspeed. E-yPd-P amOy (1651) 16.5.4 TheElectric andMagnetic Fields j MoV sing _ MoV XF 1652| BPrdaRsim aay—pintaye (O52) Itisconvenient toset 2 1 : Sce Figs. 16-4 to16-7 and Prob. 16-3. eet (1650) Observe thatEisradial, asiftheinformation concerning theposition ofthecharge traveled ataninfinite velocity! Actually, itisonlywhenthe ‘Thisisthepermeability offreespace. Bydefinition, to=4x107”. velocity ofthecharge isconstant thattheelectric fieldisradial. Ifthe Wenowrewrite thefields andBatapointPasinFig.16-3neara charge accelerates, thelinesarenotradial. SeeFig.38-1. charge Qmoving ataconstant velocity V: ThelinesofBarecircles centered onthetrajectory ofQ. Atagiven distance r,both Eand Baremaximum at@=1/2 and HA minimum at6=0,6=z. oh Example THEFIELD OFA10-GIGAELECTRONVOLT aN \ ELECTRON Zan | ‘Whatarethemaximum valuesofEandofB10millimeters from , yo fl thepathofasingle10-gigaelectronvolt electron? Onegigaclec-f BUN \ tronvolt is10”electronvolts.foe, il BothEandBaremaximum at@=90°,orwhenthelinejoiningWN NN Hl theparticletothepointofobservation isperpendicular totheoa SU! { trajectoryinthereferenceframeSofthelaboratory.Then~ sin@=1, Fig.16:3.TheEandBvectorsatapointPinthefieldofachargeQmovingata —— (1653)constant velocity ¥withrespect toareference frameS.Thedistance randthe aren) angle6arebothmeasured inS. orytimeslargerthaniftheelectronwerestationary, 296 ww po / : YN\ s : : En 3 : : — —= f / % “ a 0 ae 780° A / } ' EN : d Fig.166, TheEofamoving point charge asafunction ofthe polar angle 0of b 3 Figs.16-3and16-5,forsevenvaluesofB=V'/c.Theobserver isstationary and ‘jseesthechargemovingattheuniformspeedY.ForB=0thefieldisisotropic.It % i vA ; ishardly disturbed at=0.25.Asthespeedincreases, thefieldincreases nearPSE y / 8=90°anddecreases bothahead ofthecharge (near 8=0)andbehind it(near K —S . 2 0=180"). Atextremely highvelocities, most oftheelectric fieldconcentrates AS = . - near 6=90°.These curves explain qualitatively thevalidity ofGauss’s lawfor — pnors 4-o00 moving charges: asthespeed increases, thefluxofEshifts from theregion where . 00 and 0 181 to =A, and thetotal flux ofEremains constant. (Then é < . whyaretheareas under thecurves notequal?) Note thattheelectric field is &é é symmetric about 90°.Thus there isnowayoftelling fromtheshape ofthefield A 4 whether thecharge ismoving totherightortotheleftf y. ’f f , Eon=e, (1654)f / f y treeX A ‘ i intheradial direction and ae 7Gas é j Ba=Hero (16-55)SES aurNeg NEE J intheazimuthaldirection. <li S > Since therelativistic kinetic energy is10gigaelectronvolts, 7 8-090 (m—my)=mely—1)c*=10"electronvolts, (16-56) Fig.16-5. Lines of£forachargeQmovingalongthediameterofanimaginary withmoc?=5.11x10°electronvolts. Thus stationary sphere, Thedotsshowwherethelinesemerge fromthesphereatthe “e instant when thecharge goesthrough itscenter. Thedensity ofthedotsisa 10"‘measureofthemagnitude ofE.Thetotalnumberofdotsisthesameinallsix Y=Sarxipt220%10" (1657) figures,tosatisfyGauss'slaw(Sec.3.7).Notehowthefieldshiftstoward8=90° asthevelocity increases. Forf=1thefield isallconcentrated at6=90°. and ¥~c. Then 298 299 09 oon 10 > yoews AZ on A 5 D Fos = 168. Apositive charge moves atavelocity vparallel toaSoa EiBigncwirecarryingacurrent [Themagnetic forceQvXBisin theditecion shown because, frthecharge Q,thewite is / \ positively charged. IfQispostive, theforce isrepulsive Ao» ~~ wherepreferstothelatticeofpositivechargesandntothe<i Conductionelectrons.Theconductionelectronsdriftatthe olBEI i Seren tics nearchargedemsy,m theirOwnreference "“ “ “ ne “ - frame S’,isA,Because oftheLorentz contraction, ’ neni ne(i-8)™ ase Fig.16-7. The Bofamovingpointchargeasafunctionofthepolarangle@for da=YA—Y=(I » seven values of =V/e. For P=Othere's nomagnetic eld. Asincreases, Bfirstincreasesatallangles.ThenBcontinuestoincreasenear@=90",while In5,thepositivechargesarestationary, andexertonQadecreasing bothahead ofthecharge andbehind it.Atextremely highspeeds, repulsive forceintheydirectionmost ofthemagnetic eld concentrates near theplane #=9". The magnetic field, liketheelectric field, issymmetric about 90°.Themaximum ordinate on a, 2)anycurveis. F-08=0(5-5) (16-62) (9x102«109(1.6%10-") Similarly, theconduction electrons exertintheirownframean Ena= attractive force =0.29volt/meter, (16558) pnOiOe as) n "2me0y Yn2TEay’ = ONEx1091.6x10°")x10") a10 where 2;isnegative andy’=y. This force also points inthey =9.610-™testa (16-59) direction. FromTable 15-4, 1(4uate).Oke(_Yat Example|THEMAGNETIC FIELDNEARASTRAIGHT Fy=Fay(1-58)032(1-782). 664WIRE CARRYING ASTEADY CURRENT ThisforcedoesnotquitecancelF, Wereturntothecurrent-carrying wireoftheexample inSec. "Thenetforceexerted bythe‘wireonthechargeQ,inthe 16.2.Figure 16-8 shows apositive charge Qmoving atavelocity reference frame Softhewire, is toparallel toastationary wirecarrying acurrent [atadistance y wowLetuscalculate theforceexerted bythewireonQ. Fak,+b=Gir4Oke Oar 6.45)InthereferenceframeSofthewire,thenetlinearcharge Anew*Bae2newe density inthewireiszero: NowAvsisequaltothecurrent thatflowstotheright,andathe=0, (16-60) wiv,©fasinthefigure.Sosubstituting wyfor1/ée, 300 *RELATIVITY IV| 301 F= Ovo.fat (16-66) . . “= : MFetd (16-67) fi pt inthedirection shown inthefigure. Weshallrediscover thislawin / fd Thismagnetic force isinfinitesimal compared totheforce / pe fF between Qandtheconduction electrons inthewire andtothe Ae Joe force between Qandthelattice ofpositive ions. i> e4Saywehaveacopper wirewithacross-sectional areaof1 LA “eemillimeter andcarrying acurrent of1ampere. Thensuppose that 1 @jsanelectron traveling atthedrift velocity ofconduction ra electrons inthewire. Wecancalculate these forces asfollows. o Fe We found inSec. 4.3.2 that thedrift velocity isabout 10*meter/second orabout40centimeters/hour. Then7isequalto Fig.16.9.(a)TheforceFa,exertedbyQong.(b)TheforceFoexertedbyqon unity within 1partin10", Q.These twoforces arenotequal andopposite, asonewould expect from Copper contains 10°atoms percubic meter andoneconduction classical mechanics electron per atom. Then |meter ofthe wire contains 1.6x10" . , coulombs ofconduction electrons. Ifyis1centimeter, thenthe Fig.16-9; (2)interchanging ¥andw;(3)replacing @by6’;and(4)replacing force ofattraction between theelectron andthepositive lattice is B=Viebyp’=vlc andyby7’. poega16x10 _1gy(16.68) ms OrtVXxnl“Tne 2mX8.85x10 =x10°” " Fo=- 22ttVwXie 16-71| 90Axe)y7P(A—BPsin?OY? aa =4.610 newton (16-69) Clearly, Fj#—Fo,. The electric forces areinopposite directions, but Thisforce, ifacting alone, would impart totheelectron an their magnitudes arenotthesame because ofthey',A’,6”terms. The acceleration of5x10°"meters/second"! Theforceofrepulsion magnetic forces aretotally different. Theforces Fy,andF,areequal ifbetween Qandtheconduction electrons inthewireisslightly Vrece VW Kelarger «ec?andifVuKe ‘The netforce onanelectron ofvelocity 10°“meter/secondisthe ‘ThereisadifferencebetweenFgandFo,thatisnoteworthy. The magnetic force ofEq.16-66: expression forFo,isvalid ifthevelocity ¥ofQisconstant, while that 6x10" x10 7 forFygapplies ifvisconstant. ;PHFax107x1Qnx10)22%10™newton.(16-70) ThefactthatRotForisnotpeculiartocleaphenomena. Itisativistic effectthatoccurswithanytypeofforce. Sothemagnetic force,inthisparticular instance, issmaller that purely retauyists ytype theelectrostatic forces by25orders ofmagnitude! This isapurely relativistic effect that takes place for c= 10-*/(3 10")=3x 16.6TRANSFORMATION OFEANDB 10”. Ofcourse, theforce exerted onasecond wireisappreciable, for We have just seen that apurely electric field inone reference frame thesimple reason thatitalsocontains anenormous number of becomes both anelectric andamagnetic field inanother frame. Wenow conduction electrons, deduce thegeneral rulefortransforming electric andmagnetic fields by 16.5.5 TheForce Fo 7 again transforming theforce onamoving charge ‘Apaintcharge Qmovesatavelocity vinaregionwherethereexists Tocalculate theforce exerted byqonQinFig.16-2, weinterchange the anEandaB,allmeasured with respect tothesame inertial reference roles ofqandQinEq.16-44. This means: (1)replacing #by—7,asin frame S.Then theLorentz force onQis 302 “RELATIVITY IV 303 F=O(E+uxB). (16-72) Table 16-3 Transformation ofEandB ‘Thissame force, when itismeasured inadifferent inertial frame S’ E=E\+y(EL-V XB’) E'=E)+(E,+VxB) movingatavelocity ¥with respect toS,is " VXE . * PBaBi+y(Bi+2")pawsr(a,-EX") F'=Q(E'+v' XB’), (16-73) SincethechargeQisinvariant, ithasthesamenumerical valueinboth frames,fromSec.16.1. F-O(E;+re40xLe’), (16-80) Wecanfindexpressions forFandforBintermsofFB’andYin | ¢ the following way. Starting with the above F', wefind Fand then transform v'tov.This gives Fasafunction ofv,E',B’,and ¥, } Comparing with Eq.16-72, Comparing thisFwiththeoneabovegivesEandBasfunctions ofE", yB’,and¥. E=E,+yE., B=LVxe’ (16-81) Since thecalculation isstraightforward andfairly similar tothatofSec. « 16.5,werunthrough onlypartofit.WesetB’=0. Then ‘Thepurely electric fieldE’inS’becomes bothanEandaBfieldin F'=QE'=Q(E, +E!) (16-74) reference frame S. Table 163shows thetransformation equations thatapplytoany and,fromTable 15-4, electromagnetic field.|.CVLev",OF",+OF! Itisfairlysimpletoshowthat F=QE, +20 OF,FOF (16-75) vtuiVe) B Ps an Ce (16-82) Now, from Table 14-1, e ce _y¥ and thata (16-76)1-v,¥le yd—v,V/e) E-B=E'-B’. (16-83) anditisasimple matter toshow that Therefore both B*— E?/c? andE-Bareinvariants ee (16-77) Example |THEPARALLEL-PLATE CAPACITOR °VU=vyVlery Initsownreference frameacharged parallel-plate capacitor i possesses only anelectric field, Butsuppose thecapacitor moves Substituting and simplifying, wefind that atavelocity ¥parallel toitsplates. What field does afixed x wy Goose aesonFig1610,andell8thereferenceframeof F=0(Bit+ yeu,i+eyee: 16-78) ChooseaxesasinFig.|,andcal1¢reference frame OFtraeRierelygre). 067) encgucioradStata anesThe Grouping thesecond andfourth terms ontheright yields E00, EwE' Ein0, BimB,=Bi=0. (16-6) In§,from Table 16-3, CLA ree“BL~yiBL=Sex(vxe) (16-79) E,=0, E,=YE",E,=0, (16-85) e ; B=0, B=0, Ba=yLer (1686) So,finally, e 304 16.7SUMMARY 305 voy . ‘The four-current density isafour-vector that relates thecurrent densityrr _ . tothecharge density. ov b 3 -m ~~+ UAT drane Wa ake J=(I, cp)=pos, (16-10) ae Lk aA‘Sa "Gaza where Jistheusual current density, pistheelectric charge density fora :: t fixedobserver, poisthecharge density inthereference frame ofthe - + ” Febly moving charges, andristhefour-vector defining theposition ofthe co co charges. The four-current density transforms asinTable 16-1 Fig.16-10. (a)Parallel-plate capacitor, asseen initsownrefrence frame. The The four-dimensional operator Ocorresponds tothedeloperator FV: electric field strength is£',andB’=0. (b)The same capacitor moves totherightatthevelocityV.HereE=yE",andB=y(V/c°)E’ inthedirection ofthe a.8.e@ a aZ-axis.TheoriginsofSand3’aveonthelowerplate.Wehavedisregarded edge -(e2s92422,-2)-(r.-2). aoa effects. Daath gyteapber Bet (16-20) Thisisnotdifficult toexplain. First, theelectric field. InS’the Ittransforms asinTable 16-2. Thefour-divergence ofafour-vector isa electric charge densities are0"=sey". InSthecharges arethe scaler:same, the plates are shorter bythe factor 1/y, and 0=yo’. So E=yE". (We shall seeinSec. 17.1thatGauss’s lawapplies to 2a, moving charges.) Qra=Veatot (16-27) InS"thepotential difference between theplates isVi=E's. In S,thepotential difference V,isEs= V6, The spacing sisthe .sameinbothframes because thatlength isperpendicular tothe Thed’Alembertian operator is velocity ¥.Wereturntothisquestion ofvoltages inSec.17.8 LeNowthemagnetic field, Anobserver inthe“fixed” reference coer 16-30)frameSseesacurrent flowing totherightinthelowerplateand aos or i) anequal current flowing totheleftintheupper plate, From Table 163, Thelawofconservation ofcharge canbewritten as v - B.=GE!=HyeoE")Y =tooV=tol, (16-87) ap vel=-— (16-32) where /isthecurrent perunitwidth thatresults from themotion ot ofthecharged plates Finally,letuschecktheinvariance ofE*—c*B*andofE«B: InthefieldofachargeQmoving ataconstant velocity ¥,theelectric vy indthe magnetic flux density are given, respectively, bye-eBt=ye*—er(2) a (1688) fieldstrengthare yaregispectively, by a1) peepteeeapt r-—, 2, 16-46) ar(i-Ber- e889 0689) Tre —BainOy?” (iso) Also,E+B=E'+B'=0. aeOVxAye? (1647)©4e97?'(1 —Bsin?0)?” 16.7 SUMMARY asinFig. 16-3. Anelectric charge isinvariant: itsnumerical value isthesame forall The transformation equations for Eand B,inany field, arethose of observers. Table16-3. 6 “RELATIVITY IV PROBLEMS 37 PROBLEMS moving sidebysidealong parallel paths 1millimeter apart iftheyeach have akinetic energy of(a)1electronvolt and(b)1megaclectronvolt. UseTable 16-1.(16.1) Theinvariance ofelectric charge 154. Imagine that electric charge isnot invariant and that Q=Q,{1~ 16-6. (16.6) E-Bis invariant (v'/e*)'®. (Remember thatcharge is,infact, invariant, according toall ‘Show thatE-Bisinvariant under aLorentz transformation, experiments performed todate). The charge Qyisthat measured byan 2 pe-rovingwit chemge, on isthechat =msoberver 16-7.(16.6) B°~E’/c’isinvariantcoringanveieinyheTete eeareformmcher ShowthatB°~E*/c?isinvariantunderaLorentztransformation, Iftheelectrons inagiven sample have anaverage kinetic energy of100 16-8. (16.6) The angle between EandBisnotinvariant electronvolts, what percentage increase intheir charge must weexpect if Show that theangle between EandBisnotinvariant. theirvelocity increases by1%6? 16-9.(16.6) Transformation ofarelativepermittivity €, 16-2. (16.2) Conduction andconvection currents inamoving ring Adielectric-filled parallel-plate capacitor moves atavelocity Vwith its ‘Asquare conducting ringcarries acurrent 1initsown reference frame, plates (a)parallel tothexzplane (Fig. 16-10) and(b)parallel totheyz asinFig. 16-11, Itscross section is1’ plane. Show that €,=€;inboth cases. (a)Theringmoves atavelocity ¥inthedirection normal toitsplane tet0. \+-10.(16.6) Thetransformation ofP’ fred thecurtent andthecharge density withrespect toafixedreference Adielectric situated inframe "contains N"atoms percubicmeter, (b)Theringmoves totherightatavelocity 2. ‘cachatompossessing adipole momentp’=Qs’.SoP’=N'Qs' Findthecurrents andthecharge densities inafixedreference frame. Showthat,withrespect toframe S,P=P,+yPi. (©)The motion ofthespace charge atavelocity V#gives aconvection 16-11.(16.6) Time-independent magnetic field, current, Calculate theconvection andconduction currents inthefour sides. Inreference frame S$wehave aconstant magnetic field and noelectric expresai field. 163.(16.5.3)|Alternateexpressions forEandB Showthat£’='¥XB"inS’,Notetheprimeontheright-handside.So ° E’isperpendicular toboth'YandB per pte Xr ©aze(ye tye rye dattyeeye "Thetransformation forMismoredificulttoprove,SeePaulPenfieldandHermannA 16-4, (16.5.3) Thefieldofa10-megaelectronvolt proton Haus, Electrodynamics ofMoving Media, NAT, Press, Cambridge, Mass, 1967,p.209 PlotEandBasfunctions ofthetimeatapoint Ponecentimeter away ‘Thetransformation isthesameasforP. from thepath ofa10megaelectronvolt proton. SetPat(0,0.01,0), with thecharge at(V7, 0,0). 16-5. (16.5) The force between electrons moving side byside Calculate theforce, asobserved inthelaboratory, between twoelectrons Le Fig.16-11, 171. THE DIVERGENCE OF # 309 17.1 THE DIVERGENCE OF E charter 17 Indiscussingelectrostatic fieldsinChap.9wearrivedattheequation *RELATIVITY V |ved,| any)Maxwell's Equations. The Four-Potential A — where pisthetotal charge density, free plus bound, py+p,.This is Gauss's law, and itisone ofMaxwell's equations. Inintegral form, 17.1THEDIVERGENCE OFE309 [e-aa- 2 (17.2) 172. THE DIVERGENCEOF#310 a©" 17.3 THECURLOFE 311 14 THECURLOFB 312 where ofisthe area ofaclosed surface enclosing the total charge 175 MAXWELL'S EQUATIONS 313 0=0, +0, 12.6 THEVECTORPOTENTIALA 314 ; Weshallcheckthevalidity ofGauss’s lawformoving charges inthe 17.6.THEVECTOR POTENTIAL INTHEFIELD OFACHARGE MOVING ATA following special case. Acharge Qmoves ataconstant velocity V.AtcoANTETT ne theinstantitgoesthroughtheorigin,asinFig.17-1,thefluxofEoveran Example:THEVECTORPOTENTIAL NEARTHETRAJECTORY OFA ee neneeatteaul.vemeted attheorigand10.GIGAELECTRONVOLT ELECTRON 316 imaginary sphereofradius7andarea,centeredattheorigh 1.7 THESCALAR POTENTIAL V.THEELECTRICFIELD STRENGTH E stationary inthereference frame Sofanobserver, should beequal to EXPRESSED INTERMS OFVANDA 316 Olé. Example: THESCALAR POTENTIAL INTHEFIELD OFACHARGE Atthatinstant, ¢=0,and, from Eq.16-46, MOVING AT ACONSTANT VELOCITY 317 Example: THE VALUES OFV,PV, AND 3A/3¢NEAR THE TRAJECTORY OFA10-GIGAELECTRONVOLT ELECTRON 319 4 Example: THE PARALLEL-PLATE CAPACITOR 321 Coad 1719 THE LORENTZ.CONDITION 321 ‘ 17.10 SUMMARY 322 os,PROBLEMS 323 Ce7? This isourlastchapter onrelativity. Wefirst check thevalidity oftwo of Maxwell's equations forthefield ofapoint charge moving ataconstant Thesimpl yftranst these tionsth velocity, “Thesimple process oftransforming thesetwoequations then Fig.17-1.Gauss's lawappliestoamovingcharge:thesurfaceintegralofthe yieldstheothertwo.WeshallnotdiscussMaxwell’s equations any FekinatComponent oftheEeldofamovingchargeO,evaluatedoverafined furtherforthemoment;thatwillcomeinChap.27. ohoreisSquatD/enasithechargewerestadonary,Theencenfedneat Thelatterpartofthischapter concerns thefour-potential Aandits @=x/2justcompensates theweakfieldnear@=0and@=x.SeeFigs.16-5andcomponents, thevector andscalar potentials AandV. 16-6. 310 smanvevyfi pa tmcunors su [E-at-—2 f2arsin0d0 (73) magneticfieldofanysteadycurrentdistribution, Thisequation,infact, a 4x7 byPUL=Bsin?0)? applies toanymagnetic field, excluding thefields ofmagnetic monopoles. O(*sinoao ThisisGauss's lawformagnetic fields,another oneofMaxwell’s-;[Fearn (17-4) equations.Applyingthedivergencetheorem(Sec.1.6)yieldsthesamecoy"Wy(Bcos!@+Uy lawindifferential form Tointegrate ontheright wesetcos@=u.Then : iv-n-0. (7-7) [e-ae=52[ane? Ca| - PZevButy? €o |Thesetwoequationsapplyinanyinertialreferenceframe. Gauss’s law therefore applies tothefields ofcharges moving at 5constant velocities, WehavealreadyutilizedthisfactinFigs165and {|17-3THECURL OFE 16-6.Thelawapplies evenifthecharges accelerate. WehavejustseenthatVB=0inanyinertial frame, sayS.Then, ina 17.2THEDIVERGENCE OFB frame$”movingataconstant velocity ¥withrespecttoS, Wedisregard thefields ofmagnetic monopoles (Sec. 18.1), andwe |!v'-B'=0. (17-8) consider only thefields ofelectric currents. Such amagnetic field isthe Lap 5sumofthemagneticfieldsoftheindividualmovingcharges. «Fransionming nn,ee Tables16-2and16-3,wefindanWefurtherrestrictourdiscussion tocurrentsinconductors. Conduc- ‘a nyframe3: tionelectrons (1)move inalldirections because ofthermal agitation, (2) Ys reehavekineticenergiesthatvaryrandomly,(3)driftinthedirection [r(rie%S)or.]-[a +9(B,-738)|-0. (17-9)opposite toEthrough alatticeoffixedpositive charges, and(4) eon Gerateastheygoaroundbendsinthewire.Tietime-averaged Bthatresultsfromthermalagitation iero,Sowe OFcours,thesalarproductofaparallelcomponent andaperpendicu-needthinkofonlythedriftvelocity. tarcomponent iser.Tisevesonlyfourtersonthelt.ExpandingWecandispose ofthecentripetal acceleration ofthedrifting cloudof andthendividing byyleadsto lectrons atbends. Wehave notdiscussed thefields ofaccelerated roe vee charges, andallthatwecansayisthattheseaccelerations giveriseto ee aek (17-10)magnetic fieldsthatareoftheorderofu/c,typically about3x10-", ©ot © times theonethatwecalculated inthesecond example inSec. 16.5.4. . - aConsider Fig.16-4.ItshowsthatthelinesofBforagivenchargeare ‘ThetunofthefirstandthirdtermsisVB,whichiszero.Multiplyingcircles centered on,andorthogonal to,thetrajectory, withBuniform all nowbycy around agivencircle. Thenimagine avolume ofarbitrary shape situated 3Binthisfield.ClearlythenetfluxofBemerging throughitssurfaceiszero. Vay7VA(VRE) 7-11) Inotherwords,foranyclosedsurface ofareasf, Nowtheorientation of¥isarbitrary and [a-aa-0 (17-6) opa 1Bev Vxb)=BUX) V-(PXE) =-V(PRE). forthemagnetic field ofone individual charge, and hence forthe (712) 312 snevaniviry v 175 MAXWELL'S EQUATIONS 313 Wehave used Identity 7inside thefront cover, aswellasthefactthat7 Wenowexpand theleft-hand side, asinthepreceding section, andthen isindependent ofthecoordinates. Thus divide by7.Then VE L(V, OBV+ (yx B)=—-%) 7:19) (vx),=-(Z) (17-13) HET OXD=(0B ars) ‘The first terms oneither side cancel, and But, again, the direction ofY,towhich the subscript ||refers, is arbitrary. It follows that v v-% "YEyyy py=Vd (1720) = oo coe" vxe=- 2. (17-14) ‘Theorientation of¥beingarbitrary, wecanagaindisregard thescalar or multiplication byV.Also, since €oic?=1,fromSec.16.5.4,wehave —_ thelastofMaxwell's equations, ‘Thisisstillanother oneofMaxwell's equations 1 or Applying Stokes’s theorem, wefind that vxpe ty LE, (721)eel teor a doE-di=-“{ peaa=-2, Z t i[aaa=—4 (17-15) or a where Cisaclosed curve, sfistheareaofanyopen surface bounded by |vxB— ato =pol. (1722)C,and@isthemagnetic fluxlinking C. " 174THE CURL OFB Inthese equations Jisthefotalcurrent density atapoint, including . polarization currents indielectrics (Sec. 9.3.3) and equivalent currents in WecanproceedinasimilarmannertorelatethecurlofBtotheother magneticmaveralser20.3)leadstotheintegralformoftheabfeldquantities plying Stokes's theorem leadstotheintegral formoftheabove ‘Wesawabovethat equation orp $Bedl=wo) (1+60%)ast, (17-23) veeae.(17-16) I be 3 where fis thearea ofany open surface bounded byC. This equation isvalid inany inertial frame S,foreither stationary or moving charges. Then, inanother frame S’moving atthevelocity Vwith 17.5 MAXWELL’S EQUATIONS. respect toS, et ‘Table 17-1groups Maxwell's equations. These arethefourfundamental vee (7-17) ‘equations ofelectromagnetism. Weshallusethemconstantly throughout theremaining chapters. Wehave already used thefirst pair repeatedly in ‘Transforming again alltheprimed terms according toTables 16-1, 16-2 Chaps. 3to12.These equations apply inanyinertial reference frame. Of and 16-3, wefind that course, x,y,2,t,p,J,E,andBallrefer tothesame reference frame ‘These equations are invariant: ifwetransform allthe variables to 2.) ¥(,Vs primedquantities, thenweobtainidentical equations thatarevalidin + Vi)(Ey+Ey =—(p-—+ iS . (veeEeSptMa)MEre,+1B]&(eS). a718) anotherinertialreferenceframeS aid 176 THE VECTOR POTENTIAL A 315 Table17-1Maxwell's equations Ano (1725)TO 4ar(1— Bsin’@)'?" DIFFERENTIAL FORM INTEGRAL FORM A Q with Y,r,and@defined asinFig.17-2. Thecurlofthisquantity should First recall from Sec. 16.5.2 that V-B=0 [B-a0=0 . r=yr(1—Bsin? 0)"=[ye —Ve?+y?+27]! (17-26) veer Bo bean fnae ar k a) a Thus Loe 3B vxp4Eas fardtmwof(946%)ase 4-H mayQY ar: dar 4ax[Pe— Vor ty +e) Thefourpartial derivatives of1/r’areasfollows: Theequation O-J=0 (Sec. 16.4) fortheconservation ofcharge follows from theequation forthecurlofB,simply bytaking the al Lio Ya-V)divergence ofbothsides. Ser 73sBY ~VO)=-— a (17-28) Wereturn toMaxwell's equations inChap. 27 Observe that theequations forV+Band forVXEarecloselyrelated a (1729) Indeed, transforming one gives theother, Wededuced thesecond from ayn br r orn thefirst,buttheinverse operation isequally effective. These two 31 Li , PV(x— V1)equations aresometimes calledthefirstpair. B=ARV VOY) = (17-30) ‘Theequations for-Eand¥XBaresimilarly related. They form the second pair. Then 17.6THEVECTOR POTENTIAL A | You will remember from Chap. 3that, inelectrostatic fields, E=—PV. . ‘ThemagneticfluxdensityBcanbeexpressedinananalogousfashionas |ol Is B=VXxA, (17-24) MN. | where Aisthevector potential, expressed intesla-meters. Itcanbe Sin shown mathematically that ifV-B=0, then there exists afunction A . Nn 2 that satisfies theabove equation ‘Thevector potential Aisnotuniquely defined byB:foragiven a * B(x, y,z)there exists aninfinite number ofpossible A(x,y,2)’s. mS, we 17.6.1TheVector Potential intheField ofaCharge ae Moving ataConstant Velocity Fig.17-2. Thecharge Qmoves totherightatavelocity V.Wecalculate AandV Wenowverifythat,atapointPinthefieldofapointchargeQmoving atthepointP.ThedistanceOPisrinthefixedreferenceframeS,anditsr'in ataconstant velocity V, s 316 *ReLATIVITY v 317 2)a9A,_ 9A,_mo52hvt wh 7 VxXA=0=4,_-A,-% 3 z _— / (a=DiFeBor 34%) (731) = _MayOVXrwoVF) D nm 4x? day Bsin? (17-32) which istheBthat wefound inSec. 16.5.4. Fig.17.3. Eisthevector sumof~PVand~2A/3t. Example |THEVECTOR POTENTIAL NEARTHE 5TRAJECTORY OFA10-GIGAELECTRONVOLT Example |THESCALAR POTENTIAL INTHEFIELDELECTRON OFACHARGE MOVING ATACONSTANTVELOCITY Asinthefirst example ofSee. 165.4, y=2010" and ¥~c. ‘Thevector potential ismaximum at@=90°and, at10millimeters Wefrstwrite down theexpression forthescalar potential Vinthe from thepath, fieldofasingle point charge Qmoving ataconstant velocity Y,asinFig.17-2,andthenwechecktheaboveequation aaa etOY_ 4107725 1081.6.10-9IF Inthisfield, ee4a 4xx107 ve 2(1733) ~Sexe =Bsin® 8) =9.6x10"tesla-meter. (07-34) _0| vO 7“free inedyaVotyey1739) 17.7THE SCALAR POTENTIAL V.THE ELECTRIC Observe theanalogy withtheexpression forAgiveninEq.17-25. FIELDSTRENGTH EEXPRESSED INTERMS Figure174showscquiotennls surroundings movingcharge, OFVANDA asseenbyafixedobserver, forsixvaluesofB=V/c.These ‘equipotentials are spheres that are elongated byafactor of We have seen inSec. 17.3 that 0XE=~9B/3t.Nowthecurlofa y=(1—°)"** inthedirectionsperpendicular tothetrajectory gradient iszero. Itfollows thattheequation E=~PVofChap. 3cannot (Prob. 17-1). begeneral. ‘Wenowcheckthevalidity oftheaboveequation byproceeding Electric fieldsresultfromtwophenomena thatwemayconsider, for asinSee.17.6.1 themoment, tobedistinct, First,accumulations ofcharge giveriseto pag NSA (7-37) bothanEandaV,withE=—PV. However, ifsome ofthecharges Ox” By" zt move,thereisalsoaBandanA.IfAistime-dependent, thenthere --22[-85 WH.ys-it]—MoVVV(x=V0) exists afurther electric field E=~3A/81. Sothegeneral expression for *axe, re ae 4xVr™ theelectric field strength is (17:38) oA Recalling thate,44oc? =1,wefindthat g=-w-S, (17:35) 0 ,e-72, [«-vor(i- ery+e8]0739asinFig,17-3. Inthiscontext, Vistermed thescalar potential. Thisisan " important equation, Weshallreturn toitinSec.23.5, =efix-vetyp+z2]=2,17-40) ‘Asweshall seeinSec.17-8, therelative values ofthetwoterms onthe Aner’ Ancor right-hand sideoftheabove equation depend onthereference frame of . oF omantheobserver. So,inthatsense, thetwophenomena thatwehave referred axe —Bain Oy toabove arenotdistinct. asinSec. 16.5.4. a8 178THEFOURPOTENTIAL ATRANSFORMING ¥ANDA 319 Example |THEVALUES OFV,VV,AND34/31NEARTHETRAJECTORY OF A10-GIGAELECTRONVOLT ELECTRON From thefirst example inSec, 16.5.4, y=2.0x 10".Atafixed Vou=Fe (1742) Srey and isytimes larger than iftheelectron were stationary. At10 boo millimeters from thetrajectory, boos 2x10'x1.6%10" ipaors VousJoeRRSETO?2)milivolts. (17-43) ,Tocalculate PV, weuseEq. 17-36 and polar coordinates: ov,18V, |wyro (07-44) - Q 54 Bisin8.058] “nar Fae? (PTpao 8)ars =10 =2saree (°=3) (746 =-0.29% —volt/meter (r=10mm). (7-47) ‘This isalso ytimes larger than ifthecharge were stationary. ‘Thevectorpotentialisparalleltothevelocitybutpointsinthe iN ‘opposite direction because electrons arenegative. At@=9",the7 1|\) distance risminimum andtheangle@ismaximum, soAis\y) maximum,3A/2¢iszero,andY E=-PV=0.29% volt/meter (r=10mm). (17-48) 17.8 THE FOUR-POTENTIAL A.TRANSFORMING V AND A B= 080b=ons ‘Wenowdemonstrate thatAandV/care,respectively, thespace and snowtime components ofafour-vector A. Weareinterested inthepotentials atapoint Pinspace-time defined bythefour-vector r=(x,y,z,c1) (17-49) Fig.17-4. Equipotentials forapoint charge moving either totheright ortothe : 5 left. The equipotentials nearQaretooclosetogethertobeshown.Remember inareferenceframeS.ThesourceofthefieldisachargeQmovingata thatEequals —VV—4/31, not—VV. velocity V.Wefound above that 320 sxeLariviry v sa wYQV ‘Table 17-2 Transformation ofthefour-potential A=(A,V/c)AaatQY vy. (17-50) aEeeeee dar’ Aner’ ¥ ¥sethedi Aayaittv)tai Avar(Ay-SV) +A, where r’isthedistance from QtoPinthereference frame S’ofQ. « “ Foragivenfour-vector ¢thedistancer’isaspecificdistanceina v_(e+Al) v(t) specific frame, Then wemay rewrite Aasfollows: eo Ne*e e Nee =MorQdro_HoOdre An dardt”ar’dig’ crs) ExampleTHEPARALLEL-PLATE CAPACITOR whererydefinesthepositionofQatthetimef,andfoisthetimeinthe in impleinSec.16.6wetransform: ofa referenceframe5”ofthechargeQ.WesawinSec.14.3thatdr=ydo Wedlcwe“altosTeteA,(ransformes,meFe Similarly,frameS’ofthecapacitor,wemaysetA’=0,sincethereiszero yodt ‘magnetic field vn20,Swe 72) Terthepotentialofthelowerplatebezeroandthatofhetopincor’ dar’ diy platebeVj,asinFig.16-10. Then ‘Wehave used therelation €toc?=1. Ley . WecannowseethatAandV/carethecomponents ofthe vey te sn Four-potential FromTable 17-2, vya=(4.") (17-53) Any A,=0, A,=0, V=—yyE'. (17-58) ald ep)=Mote (17-54) se4rd nr’ dy 3A__ OV, 3A av ga—wy a Wy Ae Nea ye, (17-9 Equation 17-53 isgeneral, while Eqs. 17-54 apply toasingle charge Q. . Notetheanalogy withthefour-momentum ofanobject ofproper mass pavxa=45-Az- 42," ee 176 img(See. 15.6) 327 By ay Ve ca dr asintheexample inSec.16.6 P=(p.*)=m7- (17-55) Youcaneasilycheckthat|A|=|A'|, orthatA?-V7/c?=dy anova. andwith thefour-current density resulting from themotion ofacharge ofproper density py(Sec.16.2) 17.9THE LORENTZ CONDITION I=,00)=po (17-56) ‘Theexpressions forAandVofSecs.17.6and17.7aresocloselyrelateddt thatonesuspects theexistence ofsome general relation linking them. We have therefore shown that, inthe field ofapointchargemovingat Suchrelationdocsexist(Prob.17-8),anditiscalledtheLorentz ‘aconstant velocity, AandV/carethecomponents ofafour-vector. This condition: istrue foranyelectromagnetic field. O-A=0, (17-61) SinceAisafour-vector, ittransforms liker.ThusAandVtransform asinTable 17-2. where (is defined inSec. 16.3 and AinSec. 17.8. Thus 322 “RELATIVITY V PROBLEMS: 333 V:A+Catewe“ 72 ‘Thevectorpotential Aisdefinedby B=VXA (17-24) which isthe more usual form ofthe Lorentz condition. Since the divergence ofafour-vector isinvariant, the Lorentz Thegeneral expression fortheelectric field strength is condition applies inany inertial reference frame. Again,thisresultisvalidforanyelectromagnetic field:onecanalways g--yv—24 (17.35)select VandAsoastosatisfy theLorentz condition. Observe that the ot Lorentz condition ismathematically similar tothelawofconservation of charge (Sec. 16.4) ‘The firstterm ontheright results from charge accumulations, andthe Weshall return totheLorentz condition onvarious occasions. second from changing magnetic fields. Thefour-potential Agroupstogether thescalarpotential Vandthe 17.10 SUMMARY vector potential A Gauss'slawforelectricfieldsstatesthat a=(a,"). (17-53) c |v-E=2, (7) TheLorentzconditionrelatesAtoV: O-A=0, (17-61) where pisthetotal charge density, freeplusbound, orpy+py.Thislaw or applies eventomoving charges. ev Gauss's lawformagnetic fieldsis V-A+ culos =O. (17-62) [rao] PROBLEMSv-B=0. (17-7) a 17-1, (17.7) The equipotentials ofamoving point charge areforeshortened We also found that spheres Show that theequipotentils ofamoving point charge areforeshortened 7 spheres, asinFig. 17-4, for astationary observer, Set V=1 and B yO/Arey= | VXE=-s. (17-14) 17-2,(17.8)TheintegralforA Verifythat ante Lo SE 17-3, (17.8) Transl hefieldof lel-pl XBcquyee= 7 3.(17.8)Transforming theficldofaparallel-plate capacitor XBeolly5=Hol (17-22) ‘Achargedparallelsplate capacitormovesatavelocityVinthedirection normal toitsplates, The capacitor plates have anarea sfandareseparated .byadistance s.ThevectorE”pointsinthepositive direction ofthex-axis, ‘Thesearethefour equations ofMaxwell. They apply inanyinertial andthepositive plateisatx’=0 reference frame. Find V,A,E,andBwithrespect toastationary reference frame, 324 *RELATIVITY v 205 17-4. (17.8) Transformation ofthefield ofasolenoid moving inadirection y Perpendicular toitsaxis, i, Inside along solenoid, B= )N'TifN’isthenumberofturnspermeter andthecurrent isJ.The solenoid moves atavelocity ¥inadirection perpendicular toitslength. (a)Calculate £and B,both inside and outside thesolenoid, asmeasured byastationary observer. The axis ofthesolenoid isthe z’-axis, and VaVE, (b) You can also calculate this field bytransforming thepotentials. First _ show that, intheframe ofthesolenoid, thevector potential B ByBx, | anSa Ey gives thecorrect B’. Note that there exists aninfinite number ofpossible (©)SetV"=0. Now calculate A,V,E,andBinside thesolenoid. Both | ° Vand Adepend onthe expression that we chose arbitrarily for A’ 7 Nonetheless, therelations E=~PV~aA/3t and B=VX.A always apply ‘Atpoints outside thesolenoid, initsownframe, A#0,asweshallseein T Fig.17-5. theexample inSec. 19.1 17-5.(17.8) Transformation ofthefieldofasolenoid moving parallel toitsaxis Show that Thesolenoid ofProb. 17-4moves atavelocity Yinthedirection ofits aar axis.FindEandBinside andoutside thesolenoid, asmeasured bya Bea stationary observer. 17-6. (17.8) Theparadox oftheperpendicular capacitors Youwillhavetoshow that¥(3/3x") =3/31"forthisparticular field. Figure 17-5shows twoidentical capacitors setatright angle, oneparallel 17-8. (17.9) The Lorentz condition tothevelocity ¥and theother perpendicular. Inthereference frame of Show that theLorentz condition thecapacitors 5=53,Vi,=Vi, and E=Ei Forastationary observer, E,=YE;ands,=s1.$0Vy=E,s,=YE‘S\= O-A=0, or WA+ eu’=o, Wie However, Ex= Eb. sn=Sil7. and Vin=E,8y=Biss=Vil= or Vol¥“Thisisabsurdbecause thecapacitors areinparallel andVaymustequal applestothefieldofapointchargemovingatavelocityYwithrespect10 Vas!Youcansolve thisparadox ifyoutransform thepotentials andthe theobserver fields carefully 17-7. (17.8) How the magnetic force QuxB becomes an electric force Q(-PV" =3A"/3r") ‘Acharge Qmoves atavelocity vinaconstant, but not necessarilyuniform,magnetic fieldB.Themagnetic forceisQvxB.Allthreevariables refer toastationary frame S.There isnoelectric field. “The charge accelerates. However, atagiven instant, itoccupies an inertial frame S”that travels attheinstantaneous velocity voftheparticle With respect (05S’,Qisatrest and F’= QE’. The charge has the same value inboth frames. From Prob. 16-11 B= VXB'=VX(V KAY), 182 THE MAGNETIC FLUX DENSITY 327 InChaps. 3to12westudied theEfields ofcharges that arefixed in positionorthatmoveslowly.Fixedchargeshavenomagneticfield;a 18 magneticfieldexistsonlyiftherearemovingcharges. (CHAPTER: Inthis chapter westudy the magnetic fields ofconstant electric currents. Weassume that theelectric charge density pisalso constant. Thus3p/3t=0,andhence,fromSec.4.2,V+J=0.Wealsoassumethat MAGNETIC FIELDS I therearenomagnetic materials, andnomoving materials, inthefield, TheMagnetic FluxDensity B *18.1MAGNETIC MONOPOLESand the Vector Potential A Weassume here that magnetic fields arise solely from themotion of electric charges. "18.1 MAGNETIC MONOPOLES 327 However, Dirac postulated in1931 that magnetic fields canalso arise 182. THE MAGNETIC FLUX DENSITYB.THEBIOT-SAVARTLAW 327 frommagnetic“charges,”calledmagneticmonopoles. Suchparticleshave 18.2.1THEPRINCIPLE OFSUPERPOSITION —329. notbeenobserved todate(1987). Thetheoretical valueoftheelemen- Example: ALONG STRAIGHT WIRE 330 tarymagnetic charge is Example: THE CIRCULAR LOOP 331 Example:THESOLENOID 331 h4356692x10-"weber,” (18-2) 183 THE DIVERGENCE OF #333 e 184 THE VECTOR POTENTIALA 333 Example:AANDBNEARALONG,STRAIGHT WIRE335 wherehisPlanck'sconstantandeisthechargeoftheelectron. SeetheExample:PAIROFLONGPARALLEL CURRENTS 336 tableinsidethebackcover. Example: ANTHEFIELDOFAMAGNETIC DIPOLE 337 Atadistance rfromastationary magnetic monopole of“charge” Q*, Example: BINTHEFIELDOFAMAGNETIC DIPOLE M0 wowouldhavethat 18.5.THEMAGNETIC DIPOLE MOMENT OFANARBITRARY CURRENT oO.DISTRIBUTION 340 BaTat (18:3) 186 SUMMARY 341 PROBLEMS 343 Also,theforceofattraction orrepulsion between twomonopoles Q}and Q}would be 030%. r=2:0, (18-4) Imagine asetofcharges moving around inspace.’ Atanypointrinspace tHe andatanytime1thereexistsanelectric fieldstrength E(r,1)anda Amagnetic fieldwouldexertaforceQ*B/sy onamonopole infreemagnetic fluxdensity B(r,1)thataredefined asfollows. Ifacharge Q space. moves atvelocity vat(r,£)inthisfield, then itsuffers aLorentz force 18.2THE MAGNETIC FLUX DENSITY B. F=QE+exB). et) THEBIOT-SAVART LAW ‘TheelectricforceQEisproportional toQbutindependent ofv,while Intheneighborhood ofanelectriccircuitCcarrying asteadycurrent1themagneticforceQu*Bisorthogonal tobothvandB. thereexistsamagnetic fieldand,atapointPinspace,asinFig.18-1, "IfyouhavenotstudiedChaps.131017onrelativity,simplydisregardreferences10 "ThistheDiraccharge:theSchwingerchargeistwiceaslarge them from here “ D >| (_ E J D | +] ae 1 | |Fig. 18-2. Atagiven point inavolume distribution ofcurrent, thecurrentdensityisJ.Thevectordi”specifiesthemagnitudeandorientationofthe |shaded area. Shifting thiselement ofarea totheright bythedistance dl’alongJ- sweeps outavolumeditdi’=dv’ a fl Fig.18-1. Circuit Ccarrying acurrent Iandapoint Pinitsfield. AtPthe magneticuxdensityisB Mo[Ix? aed eee (18-8) pall¢axt (18-5) inwhichv’isanyvolumeenclosingallthecurrentsandristhedistancedtde re between theelement ofvolume du’andthepointP. Thecurrent density Jencompasses moving freecharges, polarization ‘Asusual, theunitvector #points fromthesource tothepointof currents indielectrics (Sec.9.3.3), andequivalent currents inmagneticobservation P.ThisistheBiot-Savart law.Theintegration canbecarried materials (Sec.20.3) outanalytically onlyforthesimplest geometries. Seebelow forthe Canthisintegral serve tocalculate Batapoint inside acurrent definition of49. distribution? Theintegral appears todiverge because rgoestozerowhen Thisintegral applies tothefieldsofalternating currents, aslongasthe f dv’isatP.Theintegral doesnot,infact,diverge: itdoesapplyeveniftimer/c,where cisthespeedoflight,isasmallfraction ofoneperiod thepointPliesinside theconducting body.Weencountered thesame(Sec.37.4). | problem whenwecalculated thevalueofEinsideachargedistribution inThe unitofmagnetic fluxdensity isthetesla. Wecanfindthe ges dimensions oftheteslaasfollows. Aswesawintheintroduction tothis LinesofBpointeverywhere inthedirection ofB.Theyprovetobechapter, vBhasthedimensions ofE.Then | justasusefulaslinesofE.Thedensity oflinesofBisproportional tothe magnitudeofB. Tesla=Xoltsecond_weber (18-6) Aswithelectricfieldsagain,agreatdealofconvenience attendsthemetermeter—meter” useoftheconceptofflux.Themagneticfluxthroughasurfaceofareasf is ‘One volt-second isdefined as1weber. Bydefinition, j =fB-dswebers. (18-9) do=421x107weber/ampere-meter. (18-7) | ;The surface isusually open; ifitisclosed, then &=0, asweshall see Thisisthepermeability offreespace. | below. Wehave assumed acurrent /flowing through athinwire. Ifthecurrent a “aflowsoverafinitevolume,wesubstitute Jdef’forI,Jbeingthecurrent |18.2.1ThePrinciple ofSuperposition density inamperes persquare meter atapoint anddif’anelement of Theabove integrals forBimply thatthenetmagnetic fluxdensity ata area, asinFig. 18-2. Then Jdef’dl’isJdv’ and, atapoint P, point isthesumoftheB’softheelements ofcurrent Idi’, orJdu’. The 330, MAGNETIC FIELDS1 331 principle ofsuperposition applies tomagnetic fields aswell astoelectric \ fields (Sec. 3.3): ifthere exist several current distributions, then thenetB isthevector sum oftheindividual B's “ NG Example |ALONGSTRAIGHT WIRE Anelementdl’ofalong,straightwirecarryingacurrentJ,asin ("" Fig.18-3,gives, atthepoint P(r,8,),amagnetic fuxdensity 2 > H y=HoldS085_wotdl'cose 5ag. 0 4aa ‘ > ly z ‘The relative orientations ofJand Bsatisfy theright-hand screw i rule. Here Teptana, a=24eFda asin) aS cosa p ‘Thus pp 1 Fig.18-4,CoilofwireofradiusRcarryingacurrent/,thefieldatel"cosadad=He6(sa 448thatoriginatesintheelementId,andthetotalfieldB. aap ban 2xp Lines ofBarecircleslyinginaplaneperpendicular tothewire Example|THECIRCULAR LOOP andcentered onit.Themagnitude ofBfallsoffas1/p. Tocalculate thevalueofBontheaxisofacircular loopofradius 4,refer toFig. 18-4, The figure shows the dBofanelement ofAt current/dl’,Bysymmetry,thetotalBpoints along theaxis and +Cee a,=210050, (18-13)Ae pe apot AE i bo2xal bolaWV =Heelcos9=Holt 18-14) _ : B=fe 00sOS (18:14) | / } Alongtheaxis,B=yol/2a atz=0andfallsoffas1/z°for ' | |L/ jExample |THESOLENOID mae ' ‘Theabove result canserve tocalculate Bontheaxisofthe | \ solenoid ofFig.18°5bysumming thedBofthe individual turns 7V’ ‘Thesolenoidisclose-wound, oflength L,with N’turns permeter, and itsradius isR.Atthe center, Lg (1RINIde _: | eed as) Fig183Long,scrughtwiecarrying acurrent1.AtthepointP a L theelement/dlcontributes adBinthedirectionshown.Aline =BN WNTsinOy18-16 ofBisacirclecenteredonthewire. i 2RE (ste) 332 184 -THE VECTOR POTENTIAL A 333 | See Fig. 18-5 forthedefinitions of@,,and @,.Atone end, again ‘on the axis, Saez aUL QQ . "Tsin8,@<TEOO\ patest (s.r)SAR \WWig ‘ThemagneticfuxdensityislargeratthecenterthanattheendsRA} veciusethelinesofBfareoutatheends,anFig18-6,Inside PAN}\ys alongsolenoid,atpointsremotefromtheends,B=juoN'T. ‘WICalculating Batapointofftheaxiswouldbemuchmore ) siielt Fig.185.Soteacid 18.3THE DIVERGENCE OFB Assuming that magnetic monopoles donot exist (Sec. 18.1), oratleast that the netmagnetic charge density iseverywhere zero, allmagnetic fields result from electric currents, and the lines ofBforeach element of current are circles, asinFig. 18-3. Thus the net outward flux ofB through anyclosed surface iszero: [a-as-0 (18-18) Applying thedivergence theorem, itfollows that v-B=0. (18-19) Ss‘These arealternate forms ofoneofMaxwell's equations. Observe that Eq. 18-19 establishes arelation between thespace derivatives ofBata given point. Equation 18-18, onthecontrary, concerns themagnetic flux overaclosedsurface. OA RAS 18.4THEVECTORPOTENTIALA We have just seen that V-B=0.Itisconvenient toset B=VXA, (18-20) where Aisthe vector potential, asopposed toV,which isthe scalar potential. The divergence ofBisthenautomatically equaltozerobecause thedivergence ofacurliszero. Itisimmediately apparent that, foragiven B,there exist aninfinite number ofpossible A’s. Indeed, one can add toAany quantity whose Fig.1846.LincsofBforasolenoidwhoselengthisequaltotwiceitsdiameter curliszero,forexample258,withoutaffecting B.Themagnetic flux 34 MAGNETIC FIELDS1 M44THEVECTORPOTENTIAL A 335 densityisameasurable quantity, butAisknownonlywithinanadditive Aad[a (18-26) term." ante r” Notetheanalogy withtherelation where theelement di’ofcircuit CisatP’(x’,y’,2"),andristhedistance between PandP’. =-V 8-2 e(s21) Thesetwointegrals applytothefieldsofalternating currents ifthe ofelectrostatics. time delay r/cisasmall fraction ofoneperiod. The vector potential isanimportant quantity; weshall useitasoftenasV. Example |AANDBNEARALONG, STRAIGHT WIRE NoticealsothatBisafunctionofthespacederivativesofA,justasE WefirstcalculateAandthendeduceBinthefieldofthelong, isafunction ofthespace derivatives ofV.Thus, todeduce thevalue ofB straight,current-carrying wireofFig.18-7.Weshouldfindthe fromAatagiven point P,onemust know thevalue ofAintheregion same value ofBasinthefirstexample inSec. 18.2.1. around P. Ata distance p Wenowdeduce theintegral forA,starting fromtheBiot-Savart lawof dantelat 0827) Sec. 18.2: awe ; ‘Thevector Aisparallel tothewire andpoints inthedirection of xto oak stao’=e[(wt)xsav’, (18-22) thecurrent.an) 4xhr Foraninfinitely longconductor, dAisproportional todl'/Ifor largevalues ofrwhere r~LThenAtendstoinfinity fromIdentity 16inside theback cover. Applying now Identity 11,wefind logarithmically. However, thefactthatafunction isinfinite does that ‘not mean that itsderivatives areinfinite; that is,Bcan befinite 1 Jyvxs eventhoughAisinfinite. (v2)x= x7 2X7, (18-23) Wecancircumvent thisinfinite value ofAbyfirstcalculating A rror andBforawire offinite length Landthen setting L>>p atthe ‘endofthecalculation, Referring toFig.18-7,weseethat where thesecond term ontheright iszero because Jisafunction of x',y', 2",while Vinvolves derivatives with respect tox, y,2.Thus jar to wfJ a=28|(rat)avr=ox(ef“av'), (18-24) dale aa der and | Ho[Jay Ante|nai 2 red(18-25) | ‘This expression forAhasadefinite value foragiven current distribution. ‘This integral, like that forB,appears todiverge inside acurrent- carrying conductor, because oftherinthedenominator. Actually, itis wellbehaved, liketheintegral forVinside acharge distribution aIfacurrent/flowsinacircuitCthatisnotnecessarilyclosed,then,at apoint P(x,y,2)inspace, ° v "See Richard P.Feynman, Robert B.Leighton, and Matthew Sands, Lectures onPhysics,Vol.2,See.15-5,Addison-Wesley, Reading,Mass.1964,foradiscussionofsome Fig.18-7.TheelementofvectorpotentialdAduetotheelement‘quantummechanical aspectsofA Tdi’.ThevectorBisazimuthal, asinFig.18-3. 336 MAGNETIC FIELDS 1 18.4 THE VECTOR POTENTIAL A 337 col[dll (PEE calculateA,weusetheaboveresultfortheAofasinglewireand _bol, (LM 447/097)fol,LeeL aalelInty)=MelgMel(D9) 18:31 a1nyaie Ly aiea) (|PePy)2pyAMEY? ase) wtlig® apres (18-29) ‘ThevectorApointsinthedirection ofthecurrentthatiscloserto 2H p it iszero intheplane p,=py.Then Inthislastexpression wehave neglected aterm inIn(L/2) and _2A__wol (D=y, y) » wrehave arbitrarysetA~Oattheradius&.SealsoPro.1813, Bmoy"anoeta ass) Tocalculate B=VXA, weusecylindrical coordinates, keeping inmindthatAispaletothef-ansandindependent ofboth n=Atl), x33)and2.From theexpression for7XAontheback ofthefront ax 2a\phps cover, B.=0. (18:34) pavxantls Uplt). (830) Alongthinemidwaybetwoenthewowits, 2uol asinSecs. 16.5.4 and18.2.1 B= B=0, BHO. (18-35) Example |PAIROFLONGPARALLEL CURRENTS Figure 18-8shows twolongparallel wires separated byadistance mn 5DandcarryingequalcurrentsIinoppositedirections. To Example |AINTHEFIELDOFAMAGNETIC DIPOLE Amagnetic dipole isaloopofwirecarrying acurrent [,asinFig. 218.9. We calculate Ainthis section and Binthe next. This will | lead ustoaninteresting relationship between theBfield ofa magnetic dipole and theEfield ofanelectric dipole. ' Wecalculated thefieldontheaxisofacircular loopinthe /second example inSec.18.2.1.Wenowcalculate AandBatany |4 remotepointinspace,atdistancesrthataremuchlargerthanthe Varadius aoftheloop. Figure 18-10(b) shows thefield close toand inside theloop. The field inthat region, away from theaxis, is PRES difficult tocalculate—a . ‘AtthepointPinFig.18-9, eo > bol{a 4I antl fae : <v/| ahr (18-36) 7 /M vA Bysymmetry, Aisazimuthal: foranyvalue ofr’wehave two / symmetricdiswhosey-components addandwhosex-components b/ cancel. Then weneedcalculate onlythey-component oftheAin “ thefigure, and olfad@ cosa ante [ates (0837) Fig.18-8.Pairoflongparallelwirescarryingcurrentsofthesame Wecanexpressr’[email protected] inoppositedirections, ThevectorAiszerointhe Refertothefigure.First, vertical plane that passes through thedashed line, and itpoints ‘upwardontheleftanddownward ontheright. rteP4a—2arcosy (1838) 338 339 | i\ \\\ wi’\\ " I \ H \6 eta ~©<A g Fig. 169, Magnetic dipole. The vector Aisazimuthal " Now Fig.18-10. Thefcids (a)ofan electric dipole and(b)ofa magnetic dipole, intheimmediate vicinity ofthedipoles. X0OSH =FCOy (18-39) and 1_1f,_lfa_ja@ 3fa*_ a? @raP44°~2axcosg, (18-40) pF(-3[F-220]1B-GO1 4 . « ax - (18-44)varie [S-2 (Ecos)|}=retneasa ’ Discardingnowalltermscontainingthethirdandhigherpowers Observe that x/rcos@=1 ofalr a P Hu a 3a) weexpand1/r'asaninfiniteseriesanddisregardtermsinvolving Fat48 [h3c] 18-45)migherpomersofaiThan peebto-[-30]8}, css ait 1 3, Finally, substituting into yields LMtaped ase) Finally,substitutingintoEq,18-37yield Settin telaa4cong)—(13%cos9)“)cos . . ante[[etFeoe)-(gheee)5]cosoue (Feo)=), (8-43) (18-46) wefindthat Only thesecond term between thebrackets survives, and MO MAGNETIC FIELDS I Mi Hola®x_jiola®sin0— ‘ 7Ae >a) (18-47) D lefinition, A) By definition,A) manati (a8) des A)Woe isthemagnetic dipolemoment oftheloop.IfthereareNturns, We Hae) SinceAisazimuthal, NECEEE EA AaeOXt We’ ro) Dp y ExampleBIN THE FIELD OF AMAGNETIC DIPOLE al 1valueofB=9XAfollowsimmediately -II.LoopCcarryingacurrentf.Theloopisnotplane.Wehavedivided ‘ThevaleofB=#24followsimmediately tritiarysurfaceboundedbyCiniaitesinal planeces,eachcaryngaB=Ho(cos64+sin08) (18-50) current ‘The analogy with thefield oftheelectric dipole ofSec. 5.1is each one carrying acurrent /around itsperiphery. Adjoining currents obvious. Theanalogy, however, applies solely atdistances rthat cancel, andthemagnetic dipole moment ofCisthevector sumofthe arelargecompared tothesizesofthedipole. Figure 18-10shows magnetic dipole moments oftheindividual cells.Thusthe near fields. 18.5THEMAGNETIC DIPOLE MOMENT OFAN m=Dilp rxat'=sibexdr’ (18-53)ARBITRARY CURRENT DISTRIBUTION “* « andEq.18-52applies toanyclosed circuit. Assumefirstaplaneloopofarbitrary shapecarryingacurrent[.Thenwe Anarbitrary currentdistribution possesses amagnetic dipolemoment can set m=Ali, (18-51) m=3{rxado (18-54) where ofistheareaoftheloop,andwhere theunitvector 2isnormal to . cap var ¥theloopandsatisfiestheright-hand screwrule.Theaboveexpressions Weheereplaced1a"byJdst’di’anddst’dl"bydu’.TheoriginofrforAandforBapplyaslongasa°<r°, whereaisnowthelongest canbeanywhere. dimension oftheloop. According toProb. 1-3,wecanalsowritethat 18.6SUMMARY <1 . ‘Acharge Qmoving atavelocity vinthefield ofanarbitrary distributionm=upexa’, (18-52) ofchargesandcurrentsissubjectedtoaLorentzforce where thelineintegral runsinthedirection ofthecurrent J.Theorigin of F=Q(E+v xB), (18-1) can beanywhere. What ifloop Cdoes notlieinaplane? Imagine anarbitrary surface where QEistheelectric force, QuXBisthemagnetic force, andBisthe bounded byC,asinFig. 18-11, divided into infinitesimal plane cells. magnetic fluxdensity, expressed inteslas. 342 MAGNETIC FIELDS 1 PROBLEMS 33 Inthe field ofacurrentJflowingthroughacircuitC, loop,timesthenumber ofturns,timesthecurrent. Thevectormisnormal totheplane oftheloop, inthedirection defined bytheright-hand palo! falxe screw rule,Refer toFig.18-9.Also,“ah co)lr | 5om ; 5Inthefieldofavolumecurrentdistribution, | B=7p(200sOF+sin68). (18-50) paloiIXiy: (188) ‘Themagnetic dipolemomentofanarbitrary circuitCcarryinga dade PF current Lis LinesofBpointeverywhere inthedirection ofB. m=grxar. (1853)‘Themagnetic fluxthrough anareasfis « Foranarbitrarycurrentdistributionoccupyingavolumev, o-fa+d webers. (18-9) 7 m=ifrxtdo’, (18-54) Theprinciple ofsuperposition applies tomagnetic fields. “ Thenetmagnetic fluxthroughaclosedsurfaceiszero: whereJisthecurrentdensityatapointandristheposition vectorfor thatpoint.Theoriginofrisarbitrary. [a-ast~o (18-18) Hence v-B=0. (1819) PROBLEMS ‘These are,respectively, theintegral andthedifferential forms ofoneof 18-1.(18.1) Theforceonamagnetic monopole situated inamagnetic fieldMaxwell’s equations, Showthattheequation F=Q7Q/(4xyor*) isdimensionally correct.‘Theequation ‘ThismeansthatF=Q*B/j1o, andnotQ*B,asstatedbysomeauthors. - 18:2,(18.2.1)Thefieldoftwoparallelwires Baxa(18-20) Twoparallel witesofradius Randseparated byadistance 2Dcarrya defines the vector . ; current Jinopposite directions vector potential A.Foravolumecurrentdistribution, (a)FindBalongaperpendicular linepassingthroughthewires. MofJF (b)PlotBforR=1.00millimeter, D=10.0millimeters, and/= A=Aldu’ (18-25) Lampere.18-3, (18.2.1) Saddle coilswhile,foracurrent/flowingthroughacircuitC, Figure18-12(a)showstwoviewsofapairofsaddlecoils.In(a)wehave .shown justoneturn ineach coil, and(b)shows across section C.More bol[dl’ generally,wecouldhavethecurrentdistribution ofFig.18-12(b),where A=ral= (18-26) thetwopartscarryequalcurrentdensities.Thereiszerocurrentinthe Atler centralregion.Wecouldalsohavethecurrentdistribution ofFig.18-12(c) ‘Asweshallsee,themagnetic fieldsinthecavities areuniform. Inthefield ofamagnetic dipole, (a)Show thatB=4,JXr/2 inside aconductor ofcircular cross section. 3 ‘Theorigin ofrisatthecenter ofthecross section. anomxe asso) Wo)InFig.8:2(a). Bisthesameasfeachconductor occupied fullax Pl circle, with opposite currents inthecentral region. Find Binthecentral region. where themagnitude ofthemagnetic dipole moment mistheareaofthe (c)FindBinthecavities ofFigs. 18-12(b) and(c). a PROBLEMS, 3S LY CalculatethevalueoftheBohrmagnetonws,whichisthemagnetictf\ j momentofanelectronorbitforwhichn=.ThenumberofBohr co! i i i magnetons peratom orpermolecule isoftheorder ofafewandis,infact,ct)\ @ ae) meeneaPea a 18-6. (18.2.1) Rotating magnetic field \ a 2m 4x\lt pal B.=Bncosot, B=Bacos(or+™), —B,=B,cos(or+S),. andpoint asinFig. 18-13(b). (a)Show thatthe resulting field has«magnitude of1.58, androtates at anangular velocity .This isthe method used togenerate rotating magnetic fields inlarge electric motors : (b)Does thefieldrotate clockwise oranticlockwise? @ a) 18-7.(18.2.1) TheFabryequation forsolenoids 4 Asolenoid hasaninner radius R,, anouter radius R,, and alength 2L. . The current is |a (a)Show that atthecenterff 2apy k—p—>} keps a+(a?+B) B= yonll In———— ae ow woa 10+) Fig.18-12. where nisthenumber ofturns persquare meter (~1/cross section ofthe wire), «= Ri/Ry, and B= L/R, (b) Show that thelength ofthe wire is 1=nV =2an(a? —1)BR}, 18-4, (18.2.1) Themagnetic uxdensity atthe center ofasunspot ‘TheZeeman effect observed inthespectra ofsunspots reveals. the where Vithevolume ofthewindingexistenceofmagneticfieldsaslargeas0.4tesla.Thesefieldsareassociatedwith pancake-shaped current distributions intheplasma near thesurface “ a Ineffect, onehasadisk ofelectrons, with aradius of,say, 10”meters, rotating atanangular velocity oftheorder of3x10?radian/second. The thickness ofthedisk issmall compared toitsradius. (a)Calculate thesurface density ofelectrons required toachieve aBof b O.dtesla atthe center (b)Calculate thecurrent, & 18-5. (182.1) The Bohr magneton ‘According totheold Bohr model oftheatom, electrons describe orbits around thenucleus, Atomic and molecular magnetic moments areex- . pressed inBohr magneton. (a)Find the magnetic moment ofanelectron onacircular orbit of radius r. a (b)According totheBohr postulate, theangular momentum isquan- ro ry tized: mur =nfi= nh2=nx1.0546 x10-™, where misaninteger anda quantum number. Fig. 18-13. 346 MAGNETIC FIELDS1 347 (6)Check theFabry equation, which states thatatthecenter ofany ,solenoid toe Pio\'* 0.040:s=a(F) A ‘VA>:ZA HereGdependsonthegeometry,Pisthedissipatedpower,=nar?isthe Yr \ filling factor, ofthefraction ofthecoil cross section occupied bythe 18-8, (18.2.1) Ashort, thick solenoid q Figure 18-14 shows thecross section ofacoil. The dimensions shown are inmillimeters. The wire hasasquare cross section of2millimeters’ and a resistance of8.93ohms/kilometer. The current islampere. See the \ preceding problem. A y(a)Calculate Batthecenter. Usetheformulas given inProb. 18-7. 4 (b)Calculate thepower andtheapplied voltage. 0.20 0 P+020 (c)PlotBasafunction ofzalongtheaxis,fromz=~0.3 to (a 7 oy) ° = 0,3 meter 18-9.(18.2.1) Helmholtz coilsprovide auniform field Fig.18-15. ‘TheHelmholtz coils ofFig.18-15(a) provide asimple means ofobtaining uniform magnetic field over agiven volume. Roughly speaking, B.is uniform within 10% inside asphere ofradius 0.1a.(a)FindBasafunctionof»alongtheaxis. 18-15(a)butwithaspacingof2a,insteadofa,andwithcurrentsflowingin Ifyouhavethepatience toexpand thisexpression about z=0,youwill Opposite directions. ThisisaMaxwell pair. YouwillfindthatdB/dz is findthat surprisingly linear between about z=—0.7a and z=0.7a. _ 1442" 18-11,(184) Thevectorpotential ABea(tTsai*) Inagivenregion,B=Bg,SuggestpossibleA’sandacharacteristic ofthe corresponding current distribution This means thatthefirst, second, andthird derivatives ofBwith respect to zarezer0 at7=0.Sothecurve ofB(z) isexceptionally flatnear the 18-12. (18.4) Intwo-dimensional magnetic fields alineofconstant Aisaline center. ofB(b)Plot B/(u NI/a) asafunction ofz/afrom z/a=—0.5 toz/a=0.5. Acertain magnetic fieldhasazeroz-component. Figure 18-16(b) shows B.(2) forvalues ofrranging upto0.16 (a)Show thatA=AZisonepossible value ofA. 18-10.(18.2.1) AMaxwell pairprovides auniform gradient ofB 1)ShothatthinspolnathefelofsoniaPlotB/(uqNI/2a) asafunction ofz/aforapairofcoilslikethoseofFig. (c)Showthatthisappliestothefieldofastraight current-carrying wire. 18-13. (18.5) ‘The magnetic field ofaspinning electrically charged sphere = =k ‘Aconducting sphere ofradius Rischarged toapotential Vand spun aboutadiameteratanangularvelocity«. ee(a)Show that the surface current density isa=€,oV sin@=Msin8, ¥ where MiseV. | | (b)Findthatthemagnetic fluxdensity B,atthecenter. oo ato (©) What isthenumerical value oftheByforasphere 100millimeters in | radius, charged to10.0kilovolts, andspinning at10,000 turns perminute? + (d)Show thatthedipole moment is{R*M2, where 2isaunitvectorSs alongtheaxis,relatedtothedirectionofrotationbytheright-handscrew =< tule.oe ae0ay (c)What isthedipole moment oftheabove rotating sphere? —k———20—__» Fig.18-14. (QWhat current flowing through aloop100millimeters indiameterwould have thesame dipole moment? 49 CHAPTER19 eS 4ES — The Vector Potential A. Fig.19-1. (a)Asimple closed circuitC.(b)AnN-turncoil.Theturnsareall Ampére’s Circuital Law closetogether. (c)Amorecomplex closed circuit where fisthearea ofanysurface bounded byC.Wehave used Stokes’s theorem.19.1THELINEINTEGRAL OFA-dlAROUNDACLOSEDCURVE 348 5ee eeeeeeneeeeeActeonatons NowsupposethecoilhasNturnswoundclosetogether,asinFig. ee we 19-1(b). Overanycrosssection ofthecoil,sayatP,thevarious turnsare 02IIELAPLACIANOFA SI allexposed toapproximately thesame A.Then 19.3 THE DIVERGENCE OFA31 194THECURLOFB 382 fa-d=N| B-dt=No~A, (19-2) 195 AMPERE'S CIRCUITALLAW 352 © “ Example: LONG CYLINDRICAL CONDUCTOR 353 sane ss ofaeeeeeeentsoIe ee where“isthefluxlinkageandfistheareaofanysurfacebounded by Example: THE REFRACTION OFLINES OFBATACURRENT «coi Sumer 3% ‘Theunitoffluxlinkage istheweber turn. 196THE LAPLACIANOF B 357 What ifonehasacircuit such asthatofFig. 19-1(c)? Then 19.7 SUMMARY 357 PROBLEMS 358 faa=|Bed=A, (19.3) Inthis second chapter onmagnetic fields, wefirst derive adirect except thatnow itisdifficult todevise asurface bounded byC.Luckily consequence ofthedefinition ofthevector potential A:thelineintegral enough, thissurface isofnointerest because thefluxlinkage Aiseasily ofA-dloveraclosedcurveisequaltotheencircledmagneticflux.This measurable (Sec.24.2) result isgeneral. However, therest ofthechapter applies only tostatic .fields.Theexpressions thatweshallfindherefor¥°A,V-A,0XB,and Example |THEVECTOR POTENTIAL AINTHEFIELD OFV°Barealltruncated: theyalllacktime-dependent terms. Itisonlyin ALONG SOLENOID Chap. 27thatweshallfindthefull-fledged expressions. Letusfirstsee,qualitatively, howAvaries withposition, both inside and outside along solenoid. Remember theintegral forA 19.1 THE LINE INTEGRAL OF A-dl that wefound inSec. 18.4: AROUND A CLOSED CURVE .bolfat a=ege (19-4) Consider first asimple closed curve, asinFig. 19-1(a). The line integral“ of A-dlaround Cisequal tothemagnetic fluxlinking C: Atapoint ontheaxisofthesolenoid, thedAofanelement {1dt" situated somewhere onthe solenoid cancels the dA ofthe element/dl’situateddiametrically oppositethefirstone.Onthe facdt=|(vxA)-dst=|Bedet=o, (9-1) axisofalongsolenoid,Aisthereforezero. 1 AtapointPinsidethesolenoid, butofftheaxis,theelements | 1’Ag=ARUN’ =MoNIR®ZI’closesttoPwillcontribute most.SoAisazimuthal, asinFig. dard aR'wN'L Aoe Ge as) 19-2, and itincreases with the radius r. , ‘AtapointP'outside thesolenoid, Biszero,asweshallseein Thinkhowlaborious itwouldbetocalculate Abyintegratingthesecondexample inSec.19.5.ButAisclearlynotzerobecause 1dl/roverthewinding!theelements Idi’ closest toP’contribute most Avector potential Acantherefore exist inaregion where Bis 19.2 THE LAPLACIAN OF A zero. This simply means that VXA=0 forA+0, which is perfectly sensible. Forexample, ifA=ki,where kisindependent ; ofthe coordinates, then VXA=0.Wearealreadyfamiliarwitha ‘YouwillrecallfromSecs.3.4.1and4.1that similarsituationinelectrostatics: Vcantakeanyuniformvaluein| 1 4regionwhereE=~VV=O. vai[fa,wv=-2. (9-7) ‘Outsidealongsolenoid,thevectorAisagainazimuthal,but Anehur & nowitdecreaseswithrasinthefigure.NowletuscalculateA. ThefirstequationrelatesthepotentialVatthepointP(x,y,z)tothe etconnietheae heslcnoidAtapointremote completechargedistribution,pbeingthetotalvolumechargedensityat UN'L, where N’isthenumber ofturnspermeter(Sec.18.24). P'(x',y’,2’)andrthedistancePP’.Thesecondequationexpressesthe ‘Then,fromSec.19.1andataradiusrasinthefigure, relation between thespacederivatives ofVatanypointtothevolume. charge density patthatpoint. daramarun ante (a9) ‘Thereexistsananalogous pairofequations forthevectorpotential A. ‘WehavealreadyfoundtheintegralforAinSec.18.4: OutsideasolenoidofradiusR,attheradiusr’, J \ anolTau’,(19-8) “a> where v’isanyvolume enclosing allthecurrents. Thexcomponent of v \ thisequation is/v4 A-f/Lay’, (19-9) af/ 7 \ 4adyr +\\ // \Then,byanalogywithEq.19-7, ||(|D| VA,=—Hole. (19-10) \\\ \ Ofcourse,similarequationsapplytothey-andz-components,and\ / | VPA=—tiol. (19-11)XS CAWA Thisequationappliesonlytostaticfields.————— 4- i] ee\19.3THEDIVERGENCE OFA Fig. 19-2. Long solenoid seen endwise and lines ofAinside and Siulside,ThemagnitudeofAsproportional toviassat |Wecanprovethat,forstaticfieldsandforcurrentsoffiniteextent,theinverselyproportional toroutside divergence ofAiszero.First, 352. MAGNETIC FIELDSIt 353 of J 0 Z =al*veanvete fay=Hfv.(2)au, (19-12) fa —1|}— aea 1/ a where thedeloperator actsontheunprimed coordinates (x,y,z)ofthe / / ’ fieldpoint, while Jisafunction ofthesource point (x',y',z'). The 4—-—--_f integral operates ontheprimed coordinates. Asusual,risthedistance o. i \between thesetwopoints, andtheintegration covers anyvolume SP o£PD yedty yenclosing allthecurrents. eeeara TOWN WsWenow usesuccessively Identities 15,16,and 6from theback ofthe (- ) front cover: ed otal . 1 w » vant|(pt)-zdv'=—22(vt).sav'(19-13) ath ath Fig.19-3.(a)Closed pathofintegration Clinkedbyacurrent/.Ampére'sboven cireuital lawstates thatthe lineintegraofB=dloverCisequaltoynl.(b)Here at(pte PS)a cos) thelineintegralofBdoverthedashedcarveisequaltoGul az), ry Inatime-independent field, 5p/3¢=0 and, from the conservation ofcharge(Sec.4.2),¥’«J=0.Therefore ¢Bedl=[(rxa)-asa=oftvdsf=ol.(19-18) veA=-Hfvelays-ffJoga’=0,(19-15) InthissetofequationswefirstusedStokes’stheorem,ofbeingthearea 4)7 ater ofanysurfaceboundedbyC.Thenweusedtherelation7XB=jigthathereal”isth . , wefoundabove.Finally,/isthenetcurrentthatcrossesanysurfacewae eetheareaofthesurfaceenclosing thevolume u’.Wehave bounded bytheclosedcurveC.Theright-hand screwruleappliestotheusedthedivergence theorem totransform thefirstintegral intothe direction ofJandtothedirection ofintegration around C,asinFig. second. Thesecond integral iszerobecause, over.f,Jiseitherzeroor 19-3(a)tangential ThisisAmpdre’s circuitallaw:thelineintegralofB+dlarounda closed curve Cisequal touotimes thecurrent linking C.Thisresult is 19.4 THE CURL OF Bagain valid only forconstant fields. FromDefinitions § sack Sometimes thesamecurrent crosses thesurface bounded byCseveral‘rom Definitions 5,10,and15onthebackofthefrontcover | times. Forexample, withasolenoid, theclosed curveCcouldfollow the VX B=9x(PXA)=P(F-A)— PPA (19.16) axisandreturn outside thesolenoid, asinFig.19-3(b). Thetotalcurrent linking Cisthen thecurrent inone turn, multiplied bythenumber of ‘Thus, from Secs. 19.2and19.3, turns, orthenumber ofampere-turns'‘ThecircuitallawcanbeusedtocalculateB,whenBisuniformalong VXB=pol. (19-17) thepathofintegration. Thislawisanalogous toGauss'slaw,whichweused tocalculate anEthat isuniform over asurface, This equation isvalid only forstatic fields. Example|LONGCYLINDRICAL CONDUCTOR 19.5 AMPERE’S CIRCUITAL LAW ore’ f LetusapplyAmpére’s circuital lawtocalculate Binsideand; outside thelong, straight cylindrical conductor ofFig.19-4 The line integral ofBdlaroundaclosedcurveCisimportant Sanying1creatJuniformlydstrbuedovertscrosssection ‘st 19.5AMPERE'SCIRCUITAL LAW ass _— r. Example |THELONGSOLENOID i ae ] ‘Wereturn tothelongsolenoid andrecalculate Binside, ina A | regionremote fromtheends.Thisrenders endeffectsnegligibleOK fg | SeeFig.19-6.Weassume thatthepitchofthewinding small‘ RS) Weagainusecylindricalcoordinates. Thefigureshowsasolenoid <3ofcircular cross section. However, our main conclusions will be > validforanycross section. PP / i FirstnotethatBpossesses thefollowing generalcharacteristics: r 7 (1) Bysymmetry, Biseverywhere independent ofzandof@. e(2)Imagineanaxialcylinderasinthefigure.Itsradiusiseither Cineaeenecasetoeofsiralarcrosssection smallerorlargerthanthatofthesolenoid.Theintegralsofcalculating B. rePathsofintegration for B-dsfovertheendfacescancel.ThentheintegralofB-dstover thecylindrical surface iszero, from Gauss's lawforB(Sec. 18.3). Then, both inside and outside, B,=0 We usecylindrical coordinates. wi axis alony jlresaclay naricalcoordinates withthez-axisalong.the (3)ThecurlofBisze10everywhere exceptinsidethewire,whereJ#0.Then,fromtheexpressionforthecurlincylindrical spirale theconductor, isazimuthal andindependent of¢. Coordinates, outsiethewite,9B,/ap=0.Bysymmetry,3B,/39 isalso zero. Then B,isuniform inside thesolenoid and itis also ol uniform outside, neglecting endeffectsBa (19.19) Inside theconductor, foracircuital path ofradius p, UMGR*)|p"_Hole | B=yyLMaRze"|wale : i ee ora ao (19.20) i SeeFig.19-5. ' (ONE 7)) t a, y :1 i LA sh 1 7 Hy ' \ qi i Sa en g1x ' Woy Fs H . |Ys‘ WW 1 — GF ' SSH fe —=—N1 oA = YA 5 + + TH 0 —— plier) SS = Fig.19-6Longsolenoidcarrying Fig.19-8. Basafunction ofpforawire1millimeter inradius ‘current /,withpaths ofintegra- carryingacurrentof1ampere | tiona,b,¢ 357 356MAGNETIC FIELDS 11 1927 SUMMARY ~BybL= a= By 19-21) Nowconsider thefieldoutsidethesolenoid, ByL=ByL=poatl. By=By~Wom. C AlineofBtherefore deflects intheclockwise direction foran (1)WecanshowthatB,=0byconsidering pathainthefigure. observer lookinginthedirection ofa ‘Thenetcurrent linkingthispathiszero,andthelineintegral of Wecanalsoarriveatthisresultinanother way.Themagnetic B--dlaround itistherefore alsozero.Now,sincethelineintegrals fluxdensity Bresults fromtheexistence ofacurrent inthesheetalongsides1and2arezero(B,=0),thelineintegrals alongsides andtocurtents flowing elsewhere. According toAmpére's3and4cancel, Butsides3and4cancachbesituated atany circuital law,themagnetic fieldofthesheet,justbelowthesheet,distance fromthesolenoid, soB,iseitherzero,ornonzero and istoa/2andpointsleft.Justabove,thefieldisagaintoar/2,butit independentofp.Nowthefluxoutsideisequaltothefinitefux pointsright.Adding tisfieldtothatoftheothercurrents leadsto inside.Therefore, ouside,B,tends10zero. Tangential components thatdiferasabove (2) Apath such asbislinked once bythecurrent. Thus, outside thesolenoid, B,=uol/2p. Thisfuxisusually negligible 19.6THE LAPLACIAN OFB Now letuslook inside the solenoid, Wecandeduce thevalue oftheLaplacian ofBfrom thatoftheLaplacian (1) There iszero fluxinthe@-direction inside because theline ofA(Sec. 19.2). Since . ox integralofB,dloveracircleofradiusp,saythetopedgeofthe VPA=—Hod, (19-22) smallcylinder shown inthefigure, is2xpB,; andthisiszero, then according toSec.19.5because thepathencloses zerocurrent. PX(2A) =—1oXJ. (19-23)(2)Considernowpathcinthefigure.Remembering thatB,=0 bothinsideandoutside, andthatB,=0outside, weseethat,if . a Laplacian ofacurlandthus thereareN’turns|meter, B,S=UoN'/Is andB,=jN'T. NowthecurlofaLaplacian isequal tothe Laplacian 24-- 7 19-24) Example|THEREFRACTION OFLINESOFBATA 1 VAUXA)=—Ho¥XJ. ‘ CURRENT SHEET Finally,VB=—oPXJ, (19:25) Athinconducting sheet carries asurface current density of@ amperes/meter, asinFig.19-7.Inpassing through thesheet, the againforstaticfields.lines ofBbend asfollows. Since FB =0, thenormal component ofBisthesame onthetwosides: By,=Boa. 19.7 SUMMARY Applying Ampére’s circutal lawtothe pathoflength J.shown -intefigure m ThelineintegralofA-dlaroundaclosedcurveCisequaltothe magnetic fluxlinking C: fava|B-dst=A, (193) e le i“ a whereofistheareaofasurface bounded byC.c| e _ For static fields, J S _ VPA=—1od, (19-11) ly x V-A=0, (19-15)a PXB=tod, (19-17)Fig.19-7.Refraction oflineof 2 . /. BYsoningncuronsec FBHVS. (1925) 358, MAGNETIC FIELDS It PROBLEMS 359 Ampére’s circuital lawstates that (b)How isBoriented withrespect to@?{@)Aconductingbodycaresahightrequeneycurtentthaticonfined whereJisthenetcurrent thatcrosses anyopensurface bounded bythe 19-5,(19.5) Themagnetic fieldneartheaxisofacircularloop curveC,inthedirection given bytheright-hand screw rule. ‘Acircular loop carries acurrent J.Choose theaxisofsymmetry asthe z-axis, andcalculate B,and B.near theaxis. 19-6. (19.5) ‘Theaverage Bover asphere isequal toBatthecenter PROBLEMS Refer toProb. 3-16 concerning theaverage Eover aspherical volume. 194,(19.1) ‘Thevestor potential inside acurrent-carrying conductor spherical volume isequaltotheBathecentrShow tha, inside straight current-carrying cond sR,owtha sentcurenvcarrying ctorofradius 19-7.(19.5) Thefieldofashortthicksolenoid, compared tothatofalong mle? f solenoidani (ine) "ThevalueofBatthecenterofashort,thicksolenoid giveninProb.18-7 " f canbewritenasB=iN'lg, wherewoN'Tithefieldofalongsolenoid. ifAissetequal tozero atp= R. Find g. 19-2. (19.5) Van deGraaff high-voltage generator l charge tothe high-voltage electrode | (a)Calculate thecurrent cared byaSO0-mlimetr-wide beltdive by 2.100-millimeterdameter ple thatrotates at0revolutons/seconds if) =2x 10volts/meterat thesurface ofthe Belt (b)Catala theBlose tothebel } Find (a)theszimthal field along paths «and, (b)Binside thetoroid, and (che lineintegral ofBslong path | 194. (195) Bnear conducting shet ‘conducting sheet cares acurrent density of@amperesimeter. There | (a)What ithevalue ofB,closet thesheet? OW eee | (op | |= . Se | ¢ eg ae . Fig. 198, electric moment toamacroscopic body. Magnetic materials areanalo- gous inthat their atoms canactasmagnetic dipoles that can also beCHAPTER 20 ‘oriented.Thebodyisthensaidtobemagnetized. Magneticeffectsare weak inallbutferromagnetic substances, and those aregrossly nonlinear. This first chapter onmagnetic materials sets forth afew fundamental MAGNETIC FIELDS III ideas. Then, intheend,itprovides aglimpse ofthecomplexities offerromagnetism. Magnetic Materials A: Chapter 21concerns ferromagnetism andsome ofthemethods thatare The Magnetic Flux Density B available fordesigning devices incorporating ferromagnetic materials. andtheMagnetic FieldStrength H 20.1TYPES OFMAGNETIC MATERIAL 20.1 TYPESOFMAGNETICMATERIAL 361 There existthree main types ofmagnetic material. 20.2THEMAGNETIZATIONM —36120.3 THE MAGNETIC FIELD OFAMAGNETIZED BODY 362 (1) Allmaterials arediamagnetic, This magnetism originates from the Example: THEEQUIVALENT CURRENTS INAUNIFORMLY, factthattheapplication ofanexternal magnetic fieldinduces momentsMAGNETIZED ROD 364 according totheFaraday induction law(Sec. 23.4). This effect isusually 204 THE DIVERGENCE OFBINTHEPRESENCE OFMAGNETIC imperceptible, anditdisappears uponremoval oftheexternal field. dosMMEMAGNETIC FIELDSIRENGTHAL THECURLOFH xs (2)Inmostatomsthemagneticmomentsresultingfromtheorbitaland903.1DIELECTRIC ANDMAGNETIC MATERIALS COMPARED 346 spinning motions oftheelectrons cancel. Ifthecancellation isnot206AMPERE’S CIRCUITAL LAWINTHEPRESENCE OFMAGNETIC complete, thematerial isparamagnetic. Thermal agitation causesthe MATERIAL 367 individual moments toberandomly oriented, but theapplication ofa Example: SOLENOID WOUND ONAMAGNETIC CORE 367 magnetic fieldbringsaboutapartialorientation, 20.7 THE MAGNETICSUSCEPTIBILITY ,,AND THE RELATIVE ,PERMEABILITY »,368 (3)Inferromagnetic materials suchasiron,themagnetization canbe s20.71 THEMAGNETIZATION CURVE OFAFERROMAGNETIC orders ofmagnitude larger than ineither diamagnetic orparamagnetic MATERIAL 369 substances. This effect comes from electron spin, together with group *20.7.2 FOUR DEFINITIONS OFTHERELATIVE PERMEABILITY u,9) phenomena thatalignthemoments throughout asmallregioncalleda 20.8 BOUNDARY CONDITIONS 370 domain. 209 SUMMARY 372 PROBLEMS 578 20.2 THE MAGNETIZATION M ; -Themagnetization Misthemagnetic moment perunitvolume of Thusfarwehavestudiedmagnetic fieldsassociated withmovingcharges, magnetized materialatapoint.IfthereareN‘atomsperunitvolume, although wedidallude tothemagnetic materials thatweshallstudyin eachpossessing amagnetic dipole moment moriented inagiven thischapter andthenext. Section. then Allatoms contain spinning electrons that give risetomagnetic fields. It isourpurpose inthesetwochapters toexpress these fieldsinmacroscopic M=Nm. (20-1) terms Th izationM ia hepola Indielectric materials, individual atomsormolecules canpossess jemagnetization Minmagnetic mediacorresponds tothepolariza- . tionPindielectrics. Theunitofmagnetization istheampere permeter. electric dipole moments which, when properly oriented, confer anet 302 MAGNETIC FIELDSi 203THEMAGNETIC FIELDOFAMAGNETIZED BODY 363 20.3 THE MAGNETIC FIELD OF A Hom XF xMAGNETIZED BODY Ava PS (2) Tofindthemagnetic field ofamagnetized body, weshall integrate the ‘Theunitvector#pointsinthedirectionofP,fromthecenterofthe loop, expression forthevector potential dAofamagnetic dipole Mdvoverthe andrislarge compared tothelargest dimension oftheloop. Then, fora volume ofthe material. This calculation will show that the field isthe volume u'ofmagnetized material, same asifwehad anequivalent volume current density 7XM, plus an equivalentsurfacecurrentdensityMfi,situatedinavacuum. A=fefMX?4,Mo[Mxv(2)dv’ (203) ‘You will recall that wearrived atasimilar situation when wediscussed 4a he 4x), r dielectrics: theelectric field ofapolarized dielectric isthesame asifwe hadvolume andsurface charge distributions p,and0,situated ina WehaveusedIdentity 15fromthebackofthefrontcover. Thevolume vacuum, ofmagnetized materialisv',anditssurfacehasanareaf’.Thisresult isinteresting, butitisofnouseforcalculating magnetic Then, fromIdentity 11andfromStokes’s theorem, fields because Misitself afunction ofB,and unknown, , Inpractice,onecalculatesBeitherbyrathercrude,semiempirical A=-ral(vrx*)a’+efTXgy(20-4) methods suchasthose ofChap. 21orbymeans ofelaborate computer datJw r aale or codes. ofMXa ofOXWeshallfindthattheequivalent volumecurrentdensityJ,isequalto -#[wna! +e[PxMay, (205) VXM.Thus VJ, iszero and charge cannot accumulate atapoint by virtueofJ..Furthermore, theequivalent currents donotdissipate energy fibeingtheunitvectornormaltothesurfaceofareasfofthemagnetized because they donotinvolve electron drift andscattering processes like material andpointing outward, Wemay omit theprime onthedelthat conduction currents. operates onM,since thatdelclearly operates onthecoordinates x’,y’, Letuscalculate Batapoint outside amagnetized body, asinFig.20-1. 2’ofthepoint where themagnetization isM. From thethird example inSec. 18.4, thevector potential atapoint P ‘These expressions forAareallequivalent, butthelastonelendsitself locatedatadistance rfrom acurrent loop ofmagnetic moment mis toasimple physical interpretation. Itisclear thatthevector potential in theneighborhood ofapiece ofmagnetized material isthesame asifone had, instead, volume and surface densities “ Se=VXM and @,.=MXfi (20-6) 4 j Insideamagnetized material noneoftheaboveintegrals diverge, and assi thefieldisagainthesameasifthemagnetized material werereplaced bypote itsequivalent currents. OFte Moregenerally,”\ ’ . under the integra sign. Then one would replace thepolarization current density bythe displacement current density 3/21 (Sec. 9.10). The term €,2E/3t does notbelong here Fig.20: Element ofvolume inside because magnetic fields arisesolely from themotion ofcharge. Thisisconfirmed bythefactamagnetizedbodyandanexternal thatonecancalculatethefieldof&transmittingantennainfreespacefromthecutrents Msc) point P. flowing init, disregarding displacement currents infreespace. SeealsoSec.37.1. 364 MAGNENIC FIELDS1 205THEMAGNETIC FIELDSTRENGTH #1,THECURLOF# 365 Ho oP . that wefound inSec. 18.3 applies even inthepresence ofmagneticAaa" (42a 0x) dv -al(4orM)av’ (07) materials. ThisisoneofMaxwell's equations. where Jjisthe current density offree charges and 9P/3t isthe polarization current density ofSec. 9.3.3. 20.5 THE MAGNETIC FIELD STRENGTH H. ‘Thus amore general form oftheBiot-Savart law(Sec. 18.2) is THE CURL OF H a-%fGASPIA TEMXEgy (208) InSec.19.4wefoundthat,forstaticfieldsintheabsenceofmagneticly materials, Example |THEEQUIVALENT CURRENTS INA PXB=pod. (20-10)UNIFORMLY MAGNETIZED ROD ‘ Suppose themagnetization Misuniform andparallel totheaxis. Henceforth weshall useJj,instead oftheunadorned J,forthecurrent This isanidealized situation because theelementary dipoles tend density related tothemotion offree charges. toorient themselves along B,which isonlyapproximately axial Imthepresence ofmagnetized materials,SinceMisuniform, VXM-=0,andtherearenoequivalent |Volume currents. Also, since Misparallel totheaxis, thecurrent density onthecylindrical surface isM,inthedirection shown in VXB= wold +5.) @o11) Fig. 20-2, and there are nocurrents onthe end faces. The rod therefore acts asa solenoid with N'7=M,whereN’isthenumber Thisequation,ofcourse,appliesonlyinregionswherethespace ofturns permeter, Observe that,inside, BandMpoint inthe derivatives exist, thatis,inside magnetized materials, butnotattheir same general direction, surtaces. Then 20.4 THE DIVERGENCE OFBINTHE PRESENCE VXB=uJ, +XM), (20-12) OF MAGNETIC MATERIAL Bvx(2—M)=4, (20-13) Magnetic fields originate either inthemacroscopic motion ofcharge orin Ho equivalent currents. Therelation Thevectorwithintheparentheses, whosecurlequalsthefreecurrent (20.9) density, isthemagnetic field strength: i n-®—M. (20-14) Ho °NYM, BothHandMareexpressedinamperes/meter.Thus3 | B=u(H+M) (20-15) 5 | and,eveninside magnetized materials, Re- | gSFig.20-2.Auniformlymag- € PP netizedrodactsasasolenoid { VxH=J, (20-16)fe carryingasurfacecurrentdensity | a,=M. forstatic fields 366 MAGNETIC FIELDSut 206AMPERES CIRCUITAL LAW 367 20.5.1 Dielectric andMagnetic Materials Compared 20.6AMPERE’S CIRCUITAL LAW INTHE Compare theabove equation forBwiththecorresponding Eq.9-20, PRESENCE OFMAGNETIC MATERIAL y 20- asurfaceofareasfboundedbya e-1~-P) (017 Letusintegrate Eq.20-16overanopen sur rea y © curve C: Notethedifference insign:minusPinsteadofplusM. [(xH)-dt =fJd, (20-18)Figure 20-3 illustrates thedifference. InFig.20-3(a) thecapacitor “ “ plates carry freecharges thatarenotaffected bythepresence ofthe . ,dielectric, neglecting edge effects. SoD=isfixed. Alllthree vectors or,using Stokes's theorem ontheleft-hand side, point totheright. Without thedielectric, Ewould beequal toD/é>, Withthedielectric, Eissmallerbecausethefieldoftheboundcharges tH-dl=|,, (20-19)‘opposes thatofthefree charges. InFig.20-3(b), thecoilapplies agivenH.Allthreevectors pointto whereJisthecurrent offreecharges linking C.Theright-hand screwtherightagain.Without themagnetic core,B=uoH.Withthecore,Bis ruleapplies tothedirection ofintegration andtothedirection ofz.Notelargerbecause thefieldoftheequivalent currents aidsthatofthefree thatJ,doesnotinclude theequivalent currents. Thetermontheleftiscurrents. themagnetomotance.Remember thatweareconcerned heresolelywiththespace-and Thisisamoregeneral formofAmpere's circuital lawofSec.19.5,intime-averaged fieldsinside matter. Remember alsothatthesimilarity thatitcanservetocalculate Hleveninthepresence ofmagneticbetween thefieldsofelectric andmagnetic dipoles existsonlyatpoints materials. Itisrigorously valid,however, onlyforsteadycurrents.remote from thedipoles. Closer in,thefields aretotally different, as ‘showninFig.18-10. Example |SOLENOID WOUND ONAMAGNETIC CORE Imagine along solenoid wound onamagnetic core, asinFig 20-4.Atpointsremote fromtheends,Hisparallel totheaxis epeeheeestale |inside thesolenoid, andessentially zero outside. Then, from pS aoe Ss ; oie 1 z—7} ee r* e=!o-n B=wi ' “ ®( \ Fig.20-3.(a)Plane-parallel capacitor. Theplatescarryfixedsurfacecharge \ edge effects. Note theorientation ofthesmall dipole anditsfield. (b)Solenoidcarryingafixedcurrent/.Introducing themagneticcoreincreasesBby2factorofw,,neglecting endeffects. Noteagaintheorientation ofthesmalldipole andof Fig.20-4.Longsolenoid wound onamagnetic core.According 10itsfel. Ampere’scitcultlaw,H=NTinsidethecore 368 MAONETIC FIBLDSI 207THEMAGNETIC SUSCEPTIBILITY 24 369 Ampére’s cicuital lawapplied tothepathshown, ifthenumber of *20.7.1 The Magnetization Curve ofa turnspermeter isN’andthecurrent is/, Ferromagnetic Material Hl=N'l, H=N'T (20-20) Ifanunmagnetized ferromagnetic material issubjected toagradually inside thecore, whatever itismade of. increasing H,then Bincreases along aroughly S-shaped curve, asinFig. 20-5. After awhile, saturation sets inand Mincreases nofurther. For 20.7 THE MAGNETIC SUSCEPTIBILITY y,,AND. most ferromagnetic materials saturation occurs atabout 2teslas. THE RELATIVE PERMEABILIT by . . 7 Yo *20.7.2 FourDefinitions oftheRelative Permeability 1, Itisconvenient todefine amagnetic susceptibility x»,suchthat’ Thebehavior offerromagnetic materials issocomplex thatseveral 7 different typesofrelative permeability havebeendefined. Wemention M=nH. (20-21) onlyfour.Alldefinitionsrelatetothemagnetizationcurve.Relative Then permeability isoften loosely called permeability. This iswhat wedohere. B=(H+ M)=Holl+Ym)H=Hol=WH, (20-22) (1)Thewordpermeability, leftunqualified, simplymeanstheorderof where magnitude oftheratioB/joH onthemagnetization curve,possibly over y=1+Xn (20-23) somespecifiedrangeofBorofH. istherelative permeability and (2). Themeaning oftheterm maximum permeability isobvious. =mty (20224) (3)Theinitial permeability istheratioB/1uoH atveryweak fields. isthepermeability ofamaterial. Bothz,,and1,arepurenumbers. 2 — m Ee B 7 | M=yn (20-25) : #‘Themagneticsusceptibilityofpurelydiamagneticmaterialsisnegative |+|4]andoftheorderof10~*.Inparamagnetic materials, z,,variesfromabout _' <7r_| 10°*to10-*.Thesusceptibility offerromagnetic substancescanbeas 4 rt7 ZAtai lai =avae‘Theequation B=4o(H +M)isgeneral, butequations involving either o J 7 1M,OF%massume that thematerial isboth isotropic andlinear. Inother words, they assume that Misproportional toHand inthesame direction. os|Inferromagnetic materials,BandHdonotalwayspointinthesame Vw_| || direction, andwhen they do,u,canvary byorders ofmagnitude, depending onthevalueofHTandontheprevious history ofthematerial eeee| (Sec. 21.2). Inapermanent magnet, Band Hpoint inroughly opposite aay 0 w iw we wedirections. 1Angersusie) Fig.20-5, Magnetization curves forvarious materials: S,Supermendur; Fe, - annealed pure iron; D,Deltamax; P,Permalloy; PI,powdered iron; C,gray cast *Compare withP=ze(Sec.9.8). iron;M,Monel 370 MAONENE FIELDS1 208BOUNDARY CONDITIONS an (4)Thedifferential permeability istheratio dB/1y dH,ortheslope of Bin=Bay (20-26) themagnetization curvedividedbylo,ata onahysteresioop. SuENEANIECEAY Hoe4givenpolonahysteresis ‘Thenormalcomponent ofBistherefore continuous acrossaninterface.Consider now Fig. 20-6(b). The small rectangular path pierces the 20.8 BOUNDARY CONDITIONS interface. From thecircuital lawofSec.20.6, thelineintegral ofH-dl around thepath isequal tothecurrent /linking thepath. With thetwo BothBandHobeyboundary conditions attheinterface between two longsidesofthepathinfinitely closetotheinterface, /iszeroandthe media, Weproceed asinSec.10.2, tangential component ofHiscontinuous across theinterface: Figure 20-6(a) shows ashortGaussian volume ataninterface. From Hy=H. (20-27)Gauss’s law,thefluxleaving through thetopequals thatentering the wee bottom and These twoequations aregeneral Setting B=uHforboth media, the permeabilities being those that un correspond totheactual fields, and assuming that thematerials are rr——- f isotropic, thentheabovetwoequations implythat p~(4 A tan6 / . \AYN ranOy_Has aie \ ant=Bt 0. hK- Be \ tata (20-28) Ii SY \ ifYOFal(eeNa ;{ Wethereforehavethefollowingruleforlinearandisotropicmedia: \ s q “At linesofBliefarther awayfromthenormal inthemedium possessing the \Ofer |or) largerpermeability. Inotherwords, thelines“prefer” topassthrough a \ J themorepermeable medium, asinFig.20-7.YouwillrecallfromSec. t — 10.2.4 that we had asimilar situation with dielectrics. MA a ti Sages. 288 oe \ cf a {? ) \ a a | GN. | A \ hin i\ iUF}| A i oy Fig. 20-7. Line ofBcrossing theinterface between linear andisotropic media 1 and 2.The permeability ofmedium1islargerthanthatofmedium2.Pointsa Fig.20-6.(a)Gaussian surfacestraddling theinterface between media1and2. ‘andbareatequaldistances fromtheinterface. Thepathinhigher-permeability Thenormal components oftheB’sareequal. (b)Closed pathpiercing the ‘medium 1islonger thanthepathin2:thelineofB“prefers”tobeinthe interface. Thetangential components oftheH'sareequal higher-permeabvlity material an MAGNETIC FIELDS A PROBLEMS mm 20.9 SUMMARY PROBLEMS ‘Themagnetization Misthemagnetic dipolemoment perunitvolumein 20-1.(20.2)ThenumberofBohrmagnetons peratominironmagnetized material. Themagnetic field, bothoutside andinside, isthe ‘Themagnetization Minironcancontribute asmuch as2teslas toB.If same asifthematerial were replaced byitsequivalent volume andsurface oneelectron contributes one Bohr magneton, how many electrons per current densities: ‘atom, onaverage, cancontribute toM?SeeProb. 18-5. 20-2. (20.2) The field ofadisk magnetL=VXM and@=MXa (206) 'Adiskofironofradius«andthickness5imagnetized paralleltoitsaxis. Calculate Bon the axis, outside the iron ‘Thedivergence ofBiszero even inthepresence ofmagnetic materials: 20-3. (20.2) Thefieldinside atubular magnet Youareaskedtodesignapermanentmagnetthatwouldsupplya 209) we Sees8ret sions hog el28 millimeters indiameter. Someone suggests atubular magnet, magnetizedislength,thatwouldsurroundthisvolume,Whatdoyouthinkof ‘ThemagneticfieldstrengthHhasthesamedimensions asM,and Siteeaer {hisvolume, Whatdoyouthi B 20-4.(20.5)ThedivergenceofHisnotalwayszeroH=B Mu, B=udit+M), (20-14), (20-15) Show thatF=f isnotzeroinanonhomogeneous magnetic materia a 20-5.(20.5) Thefieldinacavityinsidemagnetic materialVxH=J, (20-16) FindBandHinsideathin,disk-shaped cavitywhoseaxisisparalleltoB,inside magnetic material whereJ;isthecurrentdensityattributable tofreecharges.Thislast 20-6,(20.7)Thefreeandequivalentvolumecurrentdensitiesequationappliesonlytostaticfields.Itfollowsthat,forstaticfields, Showta,inahomogeneous, isotropic,andlinearmagneticmaterial, J.=(u,- -dl= 20-7. (20.7) Current-carrying wirealong theaxisofanirontubefeat, eo) hicay entPaton thatsofholorn ovlinder. whereCisaclosedcurvelinkedbythecurrent JjThisisAmpére's 5)FindHB,andMinheineein,inheon,andinteote rewieaei8moregeneralform.defi wi ©Findtheequivalentcurrents.Asiciclecr,itoersi ineamagneticsusceptibility an.(20.7)Themagnetization intermsofandB %mandarelative permeability 1,asfollows ‘Show thatinalinear andisotropic magnetic medium, M=tall, (2021) etfall+x=) B= pat, 4,=14%—q (20-22), (20-23) ‘Then j1=4, isthepermeability. Inferromagnetic materials thevalue of ,can vary byorders ofmagnitude, and even insign, depending onthe value ofHand ontheprevious magnetic history ofthesample ‘Attheinterface between two media thenormal component ofBand thetangential component ofHarecontinuous. There isrefraction ofthe lines ofBataninterface, thelarger angle with thenormal being inthe medium with thelarger permeability -212 HYSTERESIS 375 direction, even inthe absence ofanexternal field. Intheunmagnetized statethespinsofthevariousdomainsarerandomlyorientedandthenet CHAPTERmacroscopic field iszero. Upon application ofamagnetic field, those domains thathappen tobe orientedinaboutthecorrectdirection growattheexpenseoftheir MAGNETIC FIELDS IV* neighbors bythemigration ofthedomain walls. Eventually, near . 5 saturation, themagnetization rotates totheimposed directionMagnetic Materials B. Ferromagnetic substances areusuallyanisotropic.Ferromagnetism and Magnetic Circuits *21.2 HYSTERESIS "21.1FERROMAGNETIC MATERIALS —374 ‘Onecanmeasure Basafunction ofHwithaRowland ring,asinFig. 1212 HYSTERESIS 375 21-1." Winding ahasN,turns andbears acurrent J.From Sec. 20-6, it °21.2.1 ENERGY DISSIPATED INAHYSTERESIS CYCLE 377 applies anazimuthal magnetic field strengthExample:HYSTERESIS LOSSESINATRANSFORMER IRON377 °21.3MAGNETIC FIELDCALCULATIONS 377 Hane (21-1) *21.3.1THEBARMAGNET —377 2ar Example: THE BAR E ACommie, ARMAGNET ANDTHEBARELECTRET whereristhemajorradius.Wehaveassumed thattheminorradiusof did MAONETICCIRCUITS 381 thetoroid ismuch smaller than r. Example; MAGNETIC CIRCUIT WITH AN AIR GAP 382 "215 SUMMARY 384 fr st oy1 P PROBLEMS. 3 VA . 5 Inthischapter wefirstdescribe briefly theproperties offerromagnetic Gmaterials.Thesematerialspossesshighpermeabilities, buttheyare ~~) 4Wygrosslynonlinear andlossy ma amesMagnetic fieldsinvolving ferromagnetic materials donottherefore lend f“i “aethemselves torigorous mathematical analyses. Elaborate computer codes Geareavailable forperforming numerical calculations anddrawingfield AV" AGlines.Otherwise,oneusesapproximatemethodsliketheonesthatwe a <<a surveybrieflyhere TiCyyyil *21.1 FERROMAGNETIC MATERIALS a Fig.21-1. Rowland ring formeasuring Basa funetion ofHinaferromagneticmaterial. WindingaappliesauniformH.ThevoltageY'acrosswindingbgivesB Ferromagneticandferroelectric (Sec.10.1.4)materialsare,toacertain seinehateaensec3382 aad extent, similar. Ferromagnetic materials arepartitioned intomicroscopic domains within which thespins areallspontancously aligned inagiven — "The Rowland ring geometry has long been abandoned for obvious reasons: itis ‘elatively dificult tomachine ating, especially ifthematerial ibrit, and winding two Coils oneach sample takes time. Instead, one paces acylindrical sample ofthematerial in *Thischapter maybeomitted without losing continuity, except thattheconcept of theairgapofanelectromagnet. However, thecalculation ofHinthesample isthen reluctance (Sec. 21.4) isaprerequisite forSec.25-4, ° ‘relatively dificult,TheRowlandringremainsaninstructive exercise. to,--------4 Soft magnetic materials possess ahigh permeability and anarrow & hysteresis loop. They serve inelectromagnets, transformers, motors, etc. & Hard materials arecharacterized bybroad hysteresis loops. They are used mainly forpermanent magnets. jk / ‘Asarule, onerecords ahysteresis loop withHandBeither parallel or antiparallel. Inmost applications however, Hisonly approximately Biteslas) collinear with B,because ferromagnetic materials areusually anisotropic. y ‘Onecandemagnetize anobject byexposing ittothemagnetic fieldofa coil carrying analternating current and then gradually removing the object from thefield, soastotake thematerial round smaller andsmaller hysteresis loops. wo teefo tei "21.2.1 Energy Dissipated inaHysteresis Cycle fampereures) Aferromagnetic materialdissipates energyingoingaroundahysteresis aeter loop, asweshall seeinthesecond example inSec, 23.4.2. The energy dissipated percubic meter ofmaterial and percycle isequal tothearea ofthe loop, measured intesla-ampere-turns per meter, orinweber- ampere-turns percubic meter, orinjoules percubic meter. a Example HYSTERESIS LOSSES INATRANSFORMER IRON y ‘TheloopofFig.21-2encloses anareaof150joules/meter’ cycle,7 Fig,21-2. Magnetization curve ab or1.1watts/kilogram at60hertz. This ironissuitable forpower andhysteresis loopbedefgb for transformers, butthere existlesslossy alloys. Hysteresis losses in “ “o ‘onetypeoftransformer iron, some amorphous alloys called metalic glasses arelower byan order ofmagnitude The magnetic flux density Bisb/s, where ®isthemagnetic flux and fisthecross-sectional areaofthecore. Onemeasures ®byamethod *21.3 MAGNETIC FIELD CALCULATIONS described inthefirst example inSec. 23.4.2. . Ifwestartwithanunmagnetized sample offerromagnetic material and Asnotedabove, magnetic fieldsinvolving ferromagnetic materials donot increase thecurrent incoila,thenBincreases alongacurvesuchasabof lendthemselves torigorous mathematical analyses. Fig.21-2.Thisisthemagnetization curveofSee.20.7.1 Simple-minded calculations, suchastheoneintheexample that Reducing now thecurrent inwindingatozero,Bdecreases alongbe. follows,areuseful.However, theyyieldnomorethanthegeneralThemagnitude ofBatcistheremanence, ortheretentivity. Ifthe features ofafield.current thenreverses indirection andincreases, Bdecreases tozeroatd. Theconcept ofmagnetic circuit (Sec. 21.4), when applicable, usually Themagnitude ofHatthispoint isthecoercive force. Onfurther offersasomewhatmorerealisticapproach. increasingthecurrentinthesamedirection, apointe,symmetric topoint *21.3.1 TheBarMagnet b,isreached. Ifthe current now decreases, then reverses and increases, point bisagain reached. Theclosed curve bcdefgb isahysteresis loop. ‘Aswesawintheexample inSec.20.3, theBficld ofanidealized bar Although hysteresis loops arealways ofthesame general shape, they magnet thatismagnetized uniformly parallel toitslength isthesame as takemany forms. They canbenarrow asinthefigure, orbroad, oreven thatofasolenoid ofidentical dimensions andbearing acurrent density rectangular withnearly vertical andhorizontal sides. N’/equal tothemagnetization M.Figure 21-3shows thegeneral features 378, 379 {/ \| 7"\ y~ |YS\iLES |/7 (3 {\ 7\\ A JEL \ i\ \ ringtoobtainabarmagnet.Insidethebar,BandHpointinoppositedirections. '\ / Lines ofBemerge fromtheNorthpole. Fig.213. Lines ofH oli) fra uniformly magnetized cylinder, with M identical oFig94 that points inthe direction opposite toB.The operating point ofa permanent magnet therefore liesinthesecond quadrant ofthehysteresis loop. The operating point ofalong slender magnet isclose to¢,andthat ofa ofthis field. Observe how the lines ofBbreakatthecylindricalsurface. stubbymagnetapproaches d,onthehysteresisloop. However, thelines ofHpierce thecylindrical surface undisturbed. The The field ofarealbarmagnet isnotthat simple. Since themagnetic normal component ofBand thetangential component ofH’are moments oftheindividual atoms tend toalign themselves with theB continuous atthecylindrical surface andattheends, asinSec. 20.8. field, themagnetization M,andhence theequivalent current density on Outside, H=B/1o,whileinsideH=B/{to—M. thecylindrical surface, areweaker neartheends.Moreover, sinceMis Observe that inside themagnet Band Hpoint inapproximately notuniform, there areequivalent currents inside themagnet. Theend ‘opposite directions. faces alsocarry equivalent currents since MX@atthefaces iszero only Where isthepoint (H,B)situated onthehysteresis loop? Suppose one ontheaxis. The netresult isthatthere are“poles” near theends ofthe hasaringofthematerial with windings asinFig.21-1. One varies the magnet from which lines ofB,outside themagnet, appear toradiate in current inwinding asoastogo,onthehysteresis loop ofFig.21-2, from alldirections. The poles aremost conspicuous ifthebarmagnet islong ato6andthentoc.AtthatpointH=0andB=1M.Thisisthe andthin. situation near thecenter ofalong barmagnet. ‘One now cuts outasection ofthering, asinFig. 21-4, toform abar magnet. FromAmpére’s circuital law,thelineintegral ofH«dlaround Example THEBAR MAGNET AND path Ciszero, because there arenofreecurrents linking C.Outside, H THE BAR ELECTRET COMPARED andBpoint inroughly opposite directions. Inother words, upon cylinder ofmagnetic material, Fig.21-5(a), withthatofits removing asection ofthering asinFig. 21-4, there appears inside anH electrical equivalent, theuniformly polarized cylinder ofdielectric 380, 21.4MAGNETIC CIRCUITS: 381 ») 21.4MAGNETICCIRCUITS DKte) ap Imagineaferromagnetic ringcarryingashortN-turncoil.Thecurrentis ae~7 1.Wewish tocalculate themagnetic flux@through across section ofthe a Sm core Weexpect theBinside thecore tobemuch larger near thewinding than ontheopposite side. Onthecontrary, Bis,infact, ofthesame order ofmagnitude atallpoints around thering \ ; This isnotdifficult tounderstand. The magnetic field ofthecurrent / ei magnetizes thecoreintheregionnearthecoil,andthismagnetization6\\) givesequivalent currents thatbothincrease Bandextend italongthe CQ Wy 5 core.Thisfurther increases andextends themagnetization, andhence B. Si t Some ofthelines ofBescapeintotheairandthenreturntothecoreto =a passagain through thecoil.Thisconstitutes theleakage fluxthatmay, or Fig.21-5. (a)Barmagnet. (b)Barelectret, (c)Solenoid whose B maynot,benegligible. Forexample, iftheringisalong, thinironwire,fieldisthesameasthatofthebarmagnet.(A)Pairofelectrically mostofthefluxleaksacrossfromonesideoftheringtotheother. chargedplateswhoseEfieldisthesameasthatofthebarelectret. Ifthewindingextendsallaroundthering, ofFig.21-5(b). Wediscussed thefieldofthebarelectret inthe -M -secondexample ofSec.9.7. Hoo (21-10) ‘TheBfieldofthemagnet isthatofanequivalent solenoid, Fig. asi a 21-S(c), thatcarries acurrent density N'=M.Then,tofindthe asinSec.21.2,and Hfield,weusetherelation patN ru)a - B=y(H+M),12)ae ‘TheEfieldoftheelectret isthatofthebound charges onthe Thefluxisthesame asifthecoilhadnocoreandu,times more endfaces, asinFig.21-5(d), orofapairofparallel! andoppositely ampere-turns. Inother words, foreach ampere-turn inthecoil,there are charged disks carrying uniform charge densities +Pand~P.The u,—1ampere-turns ofequivalent currents inthecore. Theamplification Dfieldthenfollows from tnbeashighas10°. 1 IfRyistheradius where Bhasitsaverage value andifRjistheminor E=—D-P) (21-3) radius ofthering,then iNT NI Mathematically, thefieldsobeysimilarequations: =aRi- (21-12)‘4 2aR2ARy/(uaR3) ElectretMagnet1 This relation isreminiscent ofOhm’s law. Here NJisthe E-L@-P) @r4) B=u(H+M) (15), magnetomotance, while v-D=0 016, F-B=0 en, ant=2th (21-13): aR}VxXE=0 (21-8), VxH=0 (21-9). ane Indeed, theHfieldofthemagnet isthesame asifonehad isthereluctance ofthemagnetic circuit, where Lisitslength and4isits magnetic pole densities +M and ~M ontheendfaces cross section. Thus ws MAGNETIC FIELDS 1V 383 r Magnetic lux=Magnctomotance (21-14)reluctance Palye Theinverseofareluctance isapermeance. C€my |Thecorresponding quantities inelectricandmagnetic circuitsareas <x © 7alfollows: - 4 AK Current1 Magneticflux® Ki\a)| CurrentdensityJ Magnetic fluxdensityB N\A :\Conductivity 0 Permeability iy if{}oyApplied voltage V Magnetomotance NI i) 4 etElectricfieldstrengthE MagneticfieldstrengthH WS eo YZ,]B| Resistance R Reluctance % aaeaaee ConductanceG=1/R Permeance1/% ||ta ‘Thereisoneimportant difference betweenelectricandmagnetic NE b~ circuits: themagnetic fluxcannot bemade tofollow amagnetic circuit as » anelectric current follows aconducting path. Amagnetic circuit behaves Th muchasanelectric circuit would ifitweresubmerged insaltwater: part Fig.21-6.Electromagnet, Theartisthasremovedpartofthecil ofthecurrent would flow through thecomponents, andtherestwould torexpose theleftthand sideoftheyoke flow through thewater. ‘The leakage flux caneasily beone order ofmagnitude larger than theiei =Ba,= bee (21-16) usefulfluxifthecircuitisnotdesignedcarefully. OmBat=Bes @ri6) whereafisacrosssection ExampleMAGNETIC CIRCUIT WITH AN AIR GAP Combining these twoequations, Figure 21-6 shows anelectromagnet with asoftiton yoke. Each Lyk) lywinding provides NI/2ampere-tums. Weshallseethatthe Bet(Tatae) erin magnetic flux intheairgap isequal tothemagnetomotance NI divided bythesum ofthereluctances oftheironyoke andofthe and airgap,assuming thattheleakage fuxisnegligible. wr Niet,‘Thisisageneral rule:reluctances andpermeances inamagnetic ©=Bt,=—_—___.. Mo a.isy circuit addinthesame wayasresistances andconductances inan Lil(ust) +Lgloste) Leg electric circuit From Ampére's circuital law, ‘ontheassumption thatthereluctance oftheyokeisnegligible‘compared tothat ofthegap. You cannow show that HL, <HL, Ni=HL, +HL, (21-15) ifthis assumption iscorrect. Since wehave neglected theleakage flux, thisequation canonly where thesubscript /refers totheiron and gtotheairgap. The serve toprovide anupper limit for@andforB,. quantity L,canbemeasured along thecenter ofthecross section ‘There exist empirical formulas forcalculating leakage fluxes.” allaround the yoke a We assume that the flux ofBisthesameoveranycrosssection ‘SeeMalcolmMcCuaig.PermanentMagnetsinTheoryandinPractice,Wiley,New ofthecircuit: York, 1977,Chap. 6 LetN=10,000,1=1.00ampere,.f,=100centimeters*, sf,= 2Ssitimcen—>} fe—28mii»—>| 50.0centimeters’, uz,=1000, L,=900millimeters, L,=10.0milli- 9 5 =1.710°ampere-turns/weber, (21-20) See lot ieee =get O00x10weber, (21-21) _ Setaaa _ $.9x107_ |5 5) B=ygahteslas. (21-22) _ With thisparticular desig theleakage Muxiapproximately 70% Fig.217. ofthefay inthegap. Inother words, onecanexpect theBin the gaptobeonly 1.2/1.7 =0.7 tesla. 25.0 millimeters long, itsdiameter is20.0millimeters, and itoperates aits *21.5 SUMMARY ‘optimum Hof4x10*ampere-turns/meter.Calculate Binthegap. The reluctance oftheyoke isnegligible. The Ferromagnetic materialsspontaneously magnetize overmicroscopic magnetomotance ofaera aliketngof0302s Thedomains thatcangrow,oneattheexpenseofanother, uponapplication 5 Pe ~"ofamagnetic field. 21-2,(21.4) Magneti creweee neaffHfe ‘i samp! ‘Amagneticcircuitcomprisesanairgap,asinFig.21-8,withRyRy« oneplotsBasafunctionofforaninitiallyunmagnetized sample, a,Findanapproximate expressionforthereluctanceofthegap. Bfirstincreases withHalongthemagnetization curve.Overacomplete 13.(24) I thathinairpupcycle ofHoneobtainsaclosedcurvecalledahysteresisloop.Theareaof Anironringcarriesa300-turncoil.Theringhasameandiameterof theloop isequal totheenergy lostpercubic meter ofmaterial andper 400millimeters, across section of1000millimeters*, andarelative per- cycle. meability of500.The concept ofmagnetic circuit isuseful when the magnetic flux is (a)Calculate Bwhen thecurrent inthecoilis1ampere. mostly circumscribed tomagnetic material. The magnetic equivalent of Ohm's law isthen Magnetic flux=T2snctomotance | (21-14) reluctance ,, withcorresponding quantities asinSec.21.4. Themagnetomotance ofa re, coil isNI, where Nisthe total number ofturns. PROBLEMS 2I-1.(21.4) Apermanent-magnet loudspeaker JSD po ‘The topplate hasadiameter of25.0millimeters, andthegapis ee 2.50millimeters. wide. The magnet ismade of“Alnico V,itis ¥ Fig.21-8. 386 MES CHAPTER22Carey 4iy MAGNETIC FIELDS VCS ikege TheMagnetic ForcesonChargesandCurrents eeaansere>2 aes 22.1 THELORENTZFORCE 387ed: Example:THEHALLEFFECT 388 Fe.219. Example: THEMAGNETOHYDRODYNAMIC GENERATOR 389+219. "22.11THEMAGNETIC FORCEINSIDEFERROMAGNETIC MATERIALS 39222.2 THE MAGNETIC FORCEONACURRENT-CARRYING WIRE393 (b)Calculate Bforthesamecurrent whentheringhasagapof Example: THEFLOATING-WIRE HODOSCOPE 396 1.00millimeter 23. THEMAGNETIC FORCE BETWEEN TWOCLOSED CIRCUITS 3% 21-4, (21.4) Plotting amagnetic field with anelectrolytic tank. Example: THEFORCE BETWEEN TWO PARALLEL CURRENTS 397 Figure 21-9 shows oneexample oftheuseofanelectrolytic tank for 22.3.1 THE DEFINITIONSOFyy,THEAMPERE, THECOULOMB, plottingamagneticfieldinaregionwheretherearenocurrentsandno ANDeG,398 magnetic materials. Here theelectrodes areshaped likethepolepieces of 224 THEMAGNETIC FORCE ONAVOLUME DISTRIBUTION OFanelectromagnet. Themodeisstyvalidiftherelativepermitof CURRENT. 98 1¢polepiecesisinfin ‘ThelinesofEforthe modelaeidentical totheinesofBforthe Exenple: THEHOMOPOLAR MOTOR 9 clectromagnet. Letusseewhy, Example: THEHOMOPOLAR GENERATOR 403 (a)Which three differential equations does Bsatisfy infreespace? The 25 SUMMARY 404 ‘equation B=VXAisnotusefulhere.Whichthreedifferential equations PROBLEMS =405 does Esatisfy inthe model? (b)Show that Bisderivable from apotential: B=~Pu.The function isthescalar magnetic potential Weshall study magnetic forces intwochapters, thepresent oneand (c)Show thatasurface ofconstant wisorthogonal tothelinesofB. Chap. 26.Thisisbecause werequire theforceQvxBnow,butwemust TAG)SaataPon eTesPonds (0anequipotent deferamoregeneraldiscussion ofmagnetic forcestoalaterstage. {e)Onthemodel, J.E-dl=V, where Cisanycurve thatgoesfrom ‘TheQuxBforce manifests itself most clearly onelectron andion foneelectrode totheother andVistheapplied voltage. What isthe beams, sayintelevision setsand certain ionaccelerators. Itisalso the corresponding equation forB? force that drives electric motors. (f)Onthemodel, /=foF-dsf, where Iisthecurrent between the electrodes, 0theconductivity oftheelectrolyte, anddfanelementofarea ‘onanelectrode. Whatisthecorresponding equation forthemagnetic flux? 22.1THE LORENTZ FORCE ‘One could also deduce the reluctance from the resistance V/I Experiments show that the force exerted onaparticle ofcharge Q moving inavacuum ataninstantaneous velocity vinaregion where there exist both anelectric and amagnetic field is F=Q(E+v xB) (224) 388 MAGNETIC FIELDS¥ 22.THELORENTZFORCE 389 ‘This istheLorentz force. This equation isvalid, even ifvapproaches the Ifavoltmeter connected between theupper and lower speed oflight. The variables E,B,and vcan bespace- and time- electrodes drawsanegligiblecurrent,thentheplateschargeuntil dependent, butthey allconcern thesame reference frame. their transverse electric field halts thetransverse drift. ThisThetermQuXBisthemagneticforce.Observethatthemagnetic transversefilistermedtheHalfl.Thenettransverseforceforceisperpendicular tov.Itcan therefore change thedirection ofv,but Itisasimplemattertocalculate thevoltageVifwedisregard itcannot alter itsmagnitude, norcanitalter thekinetic energy ofthe tendeffects andassume thattheexternally applied Bisboth particle. Itcannonetheless douseful work, asweshall see. uniform andmuch larger than thatofthecurrent Jofthefigure. Once the transverse drift has subsided, Example |THEHALLEFFECT ovoe,-2"- ove, (222) Inabarofconducting material subjected toelectric andmagnetic DfieldsasinFig.22-(a)or(b)thechargecarriersdrift wtisthesample thicknessasinwre,andvisthe force. ASaresult, there appears apotential difference between theupper andlower electrodes. This tendency ofcharge carriers V=vBb. (223) todrift sideways inatransverse magnetic field iscalled theHalleffect. IfthemobilityofthecarriersisM(Sec.4.3.3)andthelongitudinal electric field strength isEsoy, then V=EvagBb. (224) ~ > ‘The voltage isproportional tothemobility + | = cj Notethatinthefiguresthechargecarrierstendtodrift »ane x N downward, whethertheyarepositiveornegative. Thusthe ac‘ polarityofVdependsonthesignofthecarriers. pee Ne Seaeeen “TheHalleffecservestomeasureB,butitalsoservesseveral leTK 1SrTK : otherpurposes.SeeProb.22-8vb Boos a ‘The mobilities ofsemiconductors being larger than those of goodconductors byorders ofmagnitude, Halldevices invariably ”oo make useofsemiconductors. Some devices aremicroscopic and form part ofintegrated circuits. L » _——. Example |THE MAGNETOHYDRODYNAMIC GENERATOR 2 4 a ‘Themagnetohydrodynamic (MHD)generatorisalarge-scale f‘ (o«< y « application oftheHalleffect.Itconvertspartofthekineticenergy ONSTe. |(PEAS 7 ofahotgasdirectly intoelectricenergy. Figure22-2showsits 3ie 1aN principleofoperation. Ahotgasentersontheleftatavelocityof |:| oe a theorderof1000meters/second. Itcontains asaltsuchasK,CO,LClore ee ea thationizesreadilyathightemperature, formingpositiveionsandv \ PA 2 electrons. Thetemperature approaches 3000kelvins andthe NL Ne conductivity isabout 100siemens/meter. (Theconductivity of °” copper is5.8x10”siemens/meter.) Thegasremains neutral Fig.22-1. TheHalleffect insemiconductors. (a)Inp-type material thechar Positive ionscurve downward, andelectrons upward. ‘Thecarriersareholes,andtheHallvoltageasshown.(b)Inn-typematerialthe, resultingcurrent1flowsthroughtheloadresistanceR.This carriersareelectrons, andtheHallvoltagehastheopposite polarity, (e)Inp-type establishes anelectricfieldFasinthefigure. materialthetransverse forcesQE,andQuXBareequalandopposite. (d)The Oneobvious advantage oftheMHDgenerator isthatittransverse forces inn-type material comprises nomoving parts, except forthegas.Another isthatit 390 391 ~§ or te ~ — e \ ore Fig.22-2. Schematic diagram ofamagnetohydrodynamic (MHD) .generator. Partofthekineticenergyofaveryhotgasinjectedon EEtheleftatavelocity visconverted directly intoelectric energy. *‘ThemagneticfieldBisthatofapairofsuperconducting coils w »situatedoutsidethechamberandnotshown.Themovingionsdeftect either upordown, according totheitsigns. Fig.22-3. (a)Theelectric force QEandthemagnetic force OvXBacting onapositively chargedionintheMHDgenerator ofFig.22-2.ThesumFofthosetwoforces points downward andtotheleft.(b)Theforce Fhastwocomponents: canoperate atsuch@highinputtemperature thattheoverall atangential braking force6andanormal centripetal forcethermodynamic efficiency Ton=Towe => (225) kinetic energy ofthegasserves togenerate electric energy. Also, " some ofthekinetic energy associated with thetransverse motion exceeds 50% ifthehotoutput gasfeeds aconventional turbine ofthecharged particles onlyincreases therandom thermal energy generator. Thefueliseither coaloroil.Thethermodynamic ofthegas. . efficiencies ofconventional thermal plants range from about 30% Wesuppose thatE,B,andtheparticle velocity vareuniform to3% and mutually perpendicular inside thegenerator, asinthefigure. One transportable MHD generator serves togenerate current ‘These arecrude assumptions indeed pulses of10,000 amperes forgeophysical exploration. Thelargest TheLorentz force Q(E+XB) actsonacharge Qasifthe MHD generator atthis time isunder construction andwill electri fieldstrength wereB+vXB.So,foragasofconductivity produce 500 megawatts ofelectricity ® Partofthekineticenergyassociated withthebulkmotionofthe Yi=lo+vxB)=o(u8~F)=o(vB-F), 26)gasbecomes electric energy inthefollowing way. Themagnetic 5 force onacharged particle isnormal tothevelocity andhasno IReffectonthekineticenergy. Thefunction ofthemagnetic forcesis I=<i)=sto(vB-5). (22-7) tocompel thepositive particles togotothepositive electrode and thenegativeparticlestogotothenegativeelectrode. Sotheions whereIandRareasinthefigureandwheresfistheareaofone oftheforces shown inFig. 22-3, andslow down. Since thegas * pressure isabove atmospheric, themeanfreepathbetween p= veh acollisions isinfinitesimal andthecharged particles areembedded. “body +R inthegas.Thus, slowing theparticles slows thegasandruns down itsbulk kinetic energy. Observe that vBb istheopen-circuit output voltage andthatb/ost Ideally, thecharged particles should arrive attheelectrodes at istheresistance ofthegasinthetransverse direction. Wecould,zerovelocity.Theyarriveinfactatafinitevelocity,theelectrodes therefore,havearrivedatthisresultdirectlyfromThévenin’s heatup,and only part ofthekinetic energy oftheparticles theorem (Sec. 8.3) becomes electric energy. Aneven smaller fraction ofthebulk The output voltage is[R.After eross multiplication, 302 MAGNETIC FIELDS V 222 THE MAGNETIC FORCE ON ACURRENT-CARRYING WIRE 393 a Ib 5 WeshouldreallywriteQuXByforthemagnetic force,butwerefrain V=IR= vB -* (229)od from doing so,toavoid complicating thenotation. Theimportant point is “Thevoltage Vdecreases Kneatty with thattheeffective Bforeitherconduction electrons orholesisoH,and Nowletuslookintotheefficiency. Wearenotprepared to notHHH. calculate theefficiency with which the bulk kinetic energy ofthe ‘gasbecomes electric energy. However, wecancompare theJoule losses intheload resistance Rtothose inside thegas. Soletus 22.2 THE MAGNETIC FORCE ON A define theefficiency as CURRENT-CARRYING WIRE fe Joulelosses inR (22-10) i iToulelossesinR+Joulelosses inthegas.2 Astationary wireofcrosssectionfcarriesacurrent/inaregionwhere PR R 7 thereexistsamagneticfieldBoriginatingelsewhere.Thewirecontains PREPoNod“RFbNOD’ (22-11) Nchargecarrierspercubicmeterdriftingatavelocityv,eachoneofcharge Q. aswewouldexpectfromThévenin’s theorem. Anelement oflengthdlofthewirecontains sfNdlchargecarriers.Fromtheaboveexpressions forIandfor/R, Thenthemagneticforceondlis eed 22412) dF=iNdlQuXB=(stNQv) dlXB=1dIXB, (22-13) ‘The efficiency istherefore equal tounity when /=0, orwhen since /isequal tothecharge contained inalength vofthewire. R=. Itisequal tozerowhen /=osfuB, orwhen R=0,V=0. ‘Themagnetic force perunitlength onawire bearing acurrent /is - therefore IXB. *22.1.1TheMagnetic Force Inside Ferromagnetic Materials Nowwehavecalculated theforceonthecharge carriers. Howisthis Imagine acharged particle, forexample, aconduction electron ora loree transmitted tothe wire? The charge carriers move sidewise, high-energy cosmic rayproton, moving through magnetized iron. The slightly, asinFig. 22-4, which sets upaHall field, and theresultant particle “sees” anexceedingly inhomogeneous magnetic field because electric force onthestationary positive lattice pulls thewire sideways.” ‘each individual electron inthematerial acts asasmall coil. The magnetic The total magnetic force onaclosed circuit Ccarrying acurrent Jand force QuXBthus varies erratically with time, both inmagnitude and in lying inamagnetic field is direction Itwould bepointless toattempt adetailed description ofthemotion. All that matters isthat, onamacroscopic scale, the magnetic force . corresponds tosome effective Bthatisnotnecessarily thesame asthe me macroscopic Bthatwehavebeenthinking ofuntilnow. Cc Forslow particles, such asconduction electrons inferromagnetic Et € Fig.22-4.Section through awirematerials, themagnetic forceisQuXoH,whereHisthemacroscopic ara bearingacurrent/inamagnetic fieldmagneticfieldstrengthandvisthedriftvelocityoftheparticle is A, Mehavegrosexaggeratedthe However, Rasetti showed, many yearsago’thatwithhigh-energy ca ‘ontheelectrons pushesthemtothe particles the magnetic force isQuXB.Sotheeffectivefluxdensityin FE € left,leavinganexcessofpositive relationtothemagneticforceinferromagnetic bodiesdependsonthe c" chargefntheright.theresultantHall velocity oftheparticle." N ofpositive charges totheleft. ",Rasetti,PhysicalReview,vol.65,p.1(1944) See W.R.MeKinnon. S.P.MeAlister. and CM. Hurd, American Journal ofPhysics, 'G.H. Waniet, Physical Review, vol.72,p.4(1947) Vol,p.493(1981). fora mote rigorous discussionofthisforce 94 MAGNENE FIELDSv 205 1 ra1faxe 22-14 ? o. ~~ a ? IfBisuniform, thenetmagnetic forceFiszero. £8 Rye Example |THEFLOATING-WIRE HODOSCOPE ' The floating-wire hodoscope’ isadevice that simulates the “ ” trajectory ofachargedparticleinamagneticfield.Sayacharged particle ofmass m,charge Q,and velocity vfollows acertain Fig. 22-6. (a)Apositive charge Qmoving atavelocity vina trajectory ingoing from apoint atoapoint binamagnetic field magnetic fieldB.Theradius ofcurvature ofthe trajectory isR,. Then alight wirecarrying acurrent /,fixed at@andgoing over @ (b)Light wirecarrying acurrent /intheopposite direction inthe pulleyat6,inthesamemagneticfield,asinFig.22-5,willfollow bepadmagneticfoldThetensioninthewiresT,andtheradiusBratgajctory if feurvature sRTheworaiareequalfmvQsequalto adz (22.15) ‘obvious, butitiseasy toprove, asweshall see, The advantage of where Tisthetension inthewire. This statement isbynomeans thefloating-wire hodoscope isthat itismuch easier toexperiment with awire than with anion beam. Suppose thebeam: isnormal toBasinFig. 22-(a). Then, ifR, ape istheradius ofcurvature ofthetrajectory, t) mo mv a rd 22-16) y7%) v=, ROE 22.16) | NY Now suppose thewire isalso normal toB,asinFig. 22-6(b).—. WU(/ ‘Theelementdl,witharadiusofcurvatureR..,isinequilibriumif Aad)theoutward magnetic force BI! just compensates theinward y ‘component ofthetension forces T: Z| rent) oe)» Z i .r 2¢. ya d Fig. 22-8. Floating-wire hodoscope. Alight wire bearingacurrent mat 1fixed atagoesoverapulley atb.Asmall weight provides a . tension T.This device simulates thetrajectory ofanionbeam travelling through thesame magnetic field, With themagnetic fields shown thewire adopts anS-shaped posture * la, "Thetermhodoscope alsodesignates various devices thatrecord thetrajectory followed “i byahigh-energy particle. The principles involved inthose devices bear norelation tothe ‘material inthissection Fig. 22-7. Two closed circuits «and b 396 MAGNETIC FIELDS V 2.3THEMAGNETIC FORCE BETWEEN TWO CLOSED CIRCUITS 397 4o_dl Tr Thesecondintegralontherightiszeroforthefollowing reason.Itisthe Bldl=2T sin>=TE, R= (22-17) 2my TR, iB ordinary integral ofdr/r’, withidentical upper andlower limits, because ‘Thetworadiiareequalif circuit bisclosed, byhypothesis. Sothedouble integral ontheleftis rzero, and mee (22-18) .Qu Fo=—Bhs{fide (22-23) Notice thattheparticle deflects downward ifthemagnetic force *" mi isdownward, butthewire curves downward ifthe force isupward. at F.=~ use #pol . ,Coat oetectkebeamienotperpendicular toB, ItfollowsthatFi,=—F.because#pointstowardthecircuitonwhich thenthewiredoesnotalways follow thetrajectory. Forexample, theforceacts. ‘4magnetic fieldcanservetobothfocusanddeflect anionbeam, Theabove double lineintegral isusually difficult tocalculate analyti- ‘Then focusing forces onthebeam canbecome defocusing forces cally, unless youhave access toacomputer thatcanperform symbolic ‘onthewire, which then moves away from thetrajectory, calculations, Weshall find more useful expressions inChap. 26. 22.3 THE MAGNETIC FORCE BETWEEN 7 ,TWOCLOSED CIRCUITS Example |THEFORCEBETWEEN TWOPARALLELCURRENTS ‘Wesawabove thatthemagnetic force exerted onastationary circuit Wecancalculate theforce perunitlength between twolong carrying acurrent /is/timesthelineintegral ofdlxB.Then,applying parallelwiresbearingcurrents asinFig.22-8,without havingtotheBiot-Savart lawofSec.18.2,themagnetic forceexerted byacurrent carryouttheintegration. Attheposition of1,B,isHol,/2*D inLon» an Fig.22-7.isgivenby thedirectionshowninthefigure.SeethefirstexampleinSec J,onacurrent J,,asinFig.22-7, isgiven by 18.4, The force onaunitlength ofwire bisthus loHl,XF Fa=hfdyxMe,part, (22-19) Ip 4n°J,or 0 dl, x?i | =iffllxSZ, (22-20) . , aa 1S P }'. 1 where risthedistance between dl,anddl,and#points from dl,todl), i ‘Thefactthatdl,anddl,donotplaysymmetric roles inthisintegral is disturbing. Theasymmetry appears toindicate thatFy,#F.., which iscontrary toNewton's law.’Thatimpression isfalse.Wecantransform ~~ a thedouble lineintegral toasymmetric oneandshow thatFyy=—Fyo as ‘ follows. First,weexpand thetriplevector product: “ dl,x(dl,x#)=dl,(dl,«#)~#(dl,-dy). (22-21) i |Then,rearranging terms, ir D ] (dl,+? dl,-# | ffA $a,§St (22.22) \ | Lh oF an al \ |v t — - Fig.22-8, Two long parallel wires bearing currents f,and.The *Action isequal toreaction atspeeds much lessthanthespeed oflight, force isattractive when thecurrents flowinthesame direction. 98 MAGNETIC FIELDS v 399 omBp,=olle 5. 44Mag P=B= ay | fff, ‘Theforceisattractive ifthecurrents flowinthesamedirection ae andrepulsive otherwise. Theforceisnormally negligible — (£7 L! 22.3.1 The Definitions ofjo,theAmpere,theCoulomb, | _TrtJ. and€ —fad ——i LDA)Asstatedpreviously inSec.18.2, Pai >: / L-T-F fo410-7weber/ampere-meter. (22-25) AK |/ If1,=J,=amperes, thentheforceperunitlengthis a||/ 2x10Fi=——)—— _newtons/meter. (22-26) Fig.22-9.Elementofvolumedu=dldainacurrentdistribution J. This equation defines the ampere.‘Theefaitionofthecoulomdfollows:itisthechargecartiedbya canfindthemagneticforcedensitywithoutmucheffort.Consider asmall current of1ampere during 1second, clement ofvolume oflength dfparallel toJandofcrosssectiondf,asin Sothedefinitions ofuo,,andQarearbitrary andrelated. Fig,22-9.Themagnetic forceexerted ontheelement is ButCoulomb's lawrelates theforceofattraction between twoelectric dP=(Idol) dxB=3x Baw (22:29)charges totheir magnitudes, and force isdefined inmechanics. Coulomb's lawmusttherefore involve aconstant ofproportionality andtheforceperunitvolume iswhose value must bemeasured, Wecould, inprinciple, deduce thevalue of¢9from themeasurement F'=IxB. (22-30) ofF,the Q's, and rinCoulomb's law. However, this would not make much sense because none ofthose measurements canbevery accurate. Thetotal magnetic force onagiven distribution ofconduction currents Instead, weusethefactthat occupying avolume visthus eek, (2227) F=[axBav. (2231) wherec,thespeedoflight,isdefined to9significant figures.Thus Example |THEHOMOPOLAR MOTOR : Ahomopolar motor comprises acopper disk that rotates inamore y=8.854187817 x10"? farad/meter, (22-28) ‘oflessuniform axialmagnetic field, withcontacts ontheaxisand ‘ontheperiphery, asinFig. 22-10. The magnetic force onthe 22.4 THE MAGNETIC FORCE ON AVOLUME current flowing radially through thedisk provides thedriving torque. DISTRIBUTIONOFCURRENT Homopolar motors areinherently low-voltage, high-current devices. Theyusually operate withdirect current. Theycan Inthischapterwehaveassumedthattheconduction currentsflowinthin providehightorquesatlowvcloctesandaresaitabc,forwires. What if,instead, wehave avolume distribution ofcurrent? We example, forship propulsion. Thecurrent isthen supplied bya ‘ oN —* or // £ \\ QueNe|7 v/@|\\ofYo |>\©01. ME=ort oe_— Commas’ \®SJ Nor ‘%> wa5\ A ‘ ype- UV Fig.22-11.(a)Therotorofahomopolar motor.Theelectricfieldisradial,while4 themagnetic field isaxial. Acharge carrier Qhasanazimuthal velocity or (b)ThevelocitycomponentsofQ(straightarrows)andthevariousforcesexerted o<(on@(wavyarrows). i” Foradiskofthickness s,thetorque perunitvolume exerted on thecharge carriers isrxF",andthetotal torque is Fig. 22-10. Homopolar motor. (a)Principle ofoperation. (b)Cross section *showingthecoilC,thediskD,thebrushcontactsBR,andtheterminalsV.The T=[1[2ars0(E ~orB)B)dr=reso[PE=rB)dr(22-33)torque ontherotating disk Disequal andopposite tothat onthefixed disk l l Moreover, thefield ofthecurrent through DandD'isazimuthal. Thus there is 10force oncoilC.This isimportant because Cissuperconducting andenclosed inthecounterclockwise direction. Here aandbare,respectively,inacryostat. theinnerandouterradiioftheregionofthediskwherethecurrent isradial. ‘We must still find Easafunction ofr.That iseasy because the radial current isindependent ofr: diesel motor-generator through astep-down transformer anda 1=2arsJ =2arso(E ~eB). (2-34) rectifying circuit. Superconducting coilsprovide theaxialB. Then ‘Weassume thattheelectric fieldinthediskispurely radial borg at a‘This requires contacts allaround theperiphery, which istheusual 7 lane‘configuration. FinallyRefertoFig.22-11. (1)Thecharge carrier Qhasanazimuthal reai{ rarBO) 22-36velocitywr.(2)Itistherefore subjected totheradialmagnetic =BI)dr e369)forceQwrB.(3)ThechargeQisalsosubjectedtotheradialforceQE. (4)Itsradial velocity istherefore .M(E —wrB), where Mis This isthecounterclockwise torque exerted onthecharge carriers.themobility(Sec.4.3.3).Themobilityofachargecarrierisequal Itisalsothetorqueexertedonthediskbecausethecarrierstotheforce divided byQu. This radial velocity issmaller than «or continually collide with theatoms ofthecrystal lattice. bymany order ofmagnitude. (5)This small radial velocity gives There arethree interesting points tonote here. anazimuthal force Q[-M(E rB))B. (1)The magnetic force does useful work here because it ‘Theazimuthal force perunitvolume attheradius risthus possesses anazimuthal component. Thefactthatitdoes notaffect, thekineticenergyofthechargecarriersisirrelevant. F’=NOQ[M(E —orB)]B=o(E-crB)B, (22-32) (2)Equation22-35showsthat£isapeculiarfunctionoftheadi fromSec.4.3.3,Nbeingthenumber ofcharge carriers perunit mas volume. ThisforcedensityissimplyJB].Itpointsinthe E=(wByr+—_! (237)counterclockwise direction inthefigure. 2xasr — ~ Equation 22-42 shows that, foragiven V,thecurrent is Y% Ar 1 st maximum when@=0,FromEq.22-36,Tismaximum whenJ// i \\ ?ovgunt) |w.derh ismaximum,andthuswhenw=0.Thenthemechanicalpoweris ¢/sB |F i zero. Conversely, from Eq.22-40, /and Tareboth zero when {® \ <~-(6) - ~og-2Y { 09. } igitenth HN) Tar i ©=OnnBa (22-43) ery / age? : This isthemaximum angular velocity. The mechanical power is \ we then alsozero. Beyond thisangular velocity, thedevice actsasa - , generator, with theaxisaspositive terminal, feeding current into o 7 thesource V. Under what condition isthemechanical power maximum? Set Fig.22-12. This figure issimilar toFig. 22-11, except that itapplies toahomopolar generator. (a)With«andBinthesamedirections asforthemotor, 4ory=4ivi PR)=0. (22-44) theaxle ispositive andEisalsointhesamedirection.(b)TheproductwBis aai larger than B,and there isnow abraking force (5) Then v r=, vi=2rR, (22-45) “The volume charge density is andthemechanical power isequal totheJoule power loss , ‘The mechanical power ismaximum when half the energy peer Exe! 2¢8)=2608. 238) | supplied bythesource disipates asheat. Then theefficiency isrr | 30%, and, from Prob. 22-13, ‘Thischargedensityisuniform.ItipostivewithwandBasin | 7 ve o6Fig.22-12(a). Itresults fromthevXBfield.SeeSec.4.3.5and muaowesBibai)” (240) Prob. 22-9, | (3)Thewltageappliedbetweentheaxisofradius@andthe| OFWpaa/2.periphery atbis | . * 14|Example|THEHOMOPOLAR GENERATOR vefecd=[ (wore 1)d aay I,I Basar ‘The homopolar generator, like the homopolar motor, isa vee 1b low-voltage, high-current device. Oneapplication isthegener-=opee nt (22-40) ationofthelargecurrentsrequiredforpurifyingmetalsbynaa electrolysis onanindustrial scale. ‘Thecoefficient ofFistheresistance ofthediskbetween radii@and Figure 22-11 applies, except thatnowoBilarger thanE,asin b(Sec.4.3.2). Also,thecoefficient ofwisthetorque thatwe Fig.2-12. Theabove calculation alsoapplies, withthe’same found above, divided by/.So ProvisoThe force density F'isnow clockwise. The magnetic torque vee ai) exertedonthechargecarriersBIG =a)! 7=Bee) (2-47) Wecouldhaveexpected this:multiplying bothsidesbygives 2 , | isalsoclockwise andtherefore opposes themotion. VizoT+FR (22-42) Conservation ofenergynowrequiresthat ThisequationsimplysaysthatthepowerVIsuppliedbythe OT=VI+PR= PRs+R), (22-48)source isequal tothemechanical power wTplusthethermal | power lossinthedisk/°R.Wehave disregarded other losses where wTisthemechanical power fedintothegenerator, VIis 404 MAGNETICFIELDSV\PROBLEMS 405 theelectric power fedtotheload resistance, and[°Risthe PROBLEMS thermal power dissipation. We have again disregarded other losses. Substituting thevalue ofTandsimplifying, 22.1. (22:1) Electrons intheCrab nebula’ oB(b-2’) ‘IntheCrabnebulathereisamagneticfieldofabout2x10°*teslaand=e (2249) electrons whoseenergy isabout2x10"electronvoltsPRs +R) (a)Findtheradius ofgyration. Compare thisradius withthatofthe h'sorbit. (Seethepageinside thebackcover). ‘Thehomopolar generator therefore actsasanidealvoltage source car >9B(6?~a°)/2,withanoutput resistance equaltotheresistance R (b)Howlongdoesanelectron taketocomplete oneturn,indaysofthedisk. 22-2,(22.1)Thepincheffect‘Abeam ofcharged particles ofcharge Q,mass m,and velocity vhas a 22.5 SUMMARY radius r.Thebeam current is/.Assume thatthecharge density isuniform. (This isapoor approximation; thecurrent density asafunction ofthe ; .radiusfollows, infact,aGaussian curve.) Theforce onacharge Qmoving atavelocity vinafieldE,Bis Findtheoutward force onanionsituated attheperiphery ofthebeam. ‘You willfind that there isanoutward electric force andaninward magnetic F=Q(E +xB) (22-1) force. The magnetic force tends to“pinch” thebeam, ortoconcentrate it along theaxis. This istheLorentz force. Theterm QuXBisthemagnetic force. Ifyoucancel theelectric force byadding ionsoftheopposite sign, then Charge carriers flowing alongaconductor situated inamagnetic field themagnetis forosatssonsaadthsbeescontr. Thsphosomesen byi,i whee easytoobservewithpositiveionaccelerators. Residualgasinthepathof tendtodriftsidewaysbecauseofthemagneticforce.ThisistheHall thebeamionizesbyimpact,andtheresultinglow-energyelectronsremain effect.;inthebeam,whilethepositive ionsdriftaway,Ifthepressure increases Themagnetic force petunitlength onacurrent-carrying wire is1XB. somewhat, thefocusing improves. Thisisgasfocusing The total netforce onaclosed circuit isthus ‘Athigher gaspressures thelow-energy ions remain mostly inside the ‘beam because their mean free path between collisions isshorter. The beam thenbecomesunstablebecauseofphenomena thatarenotwellunderstood F-1paxe (22-14) atthistime 22-3, (22.1) ‘Theacceleration ofanelectron inafield ,B" Themagnetic forceexerted byaclosed circuit aonaclosed circuit bis (a)Show thattheequation ofmotion foraparticle ofrestmassmo,charge Q,andvelocity vin afield E,Bis = dx ®_ole+uxe-Su- FaMhffarxa (22-20) ymt=O(B+uxB-Su-k) =Hepp$4jie (22 IfEiszeroandBis static, ard or WayPeoxe,, am Bydefinition, jig=410”weber/ampere-meter. Thentheforceper andtheelectrondescribesacircleattheangularvelocity@,=QB/m, meter between parallel wires carrying thesame current Jandseparated which iscalled thecyclotron frequency byadistance Dis2x10-’/'/D. The force isattractive ifthecurrents (b)A12.0-MeV (million clectronvolt) electron moves inthepositive flow inthesame direction. direction ofthe2-axis inafieldE=1.00x10°, B=1.008. “The magnetic force exertedonavolumedistribution ofcurrentis Calculateitsacceleration. Therestmassofanelectronis5.11x10° electronvolts P=[yxade (2231). "This problem requires aknowledge ofrelativity 408 MAGNETIC FIELDSV 409 (a)ShowthatF’=oB*(u~v), wherew=EXB/B?andvisthefluid 4velocity, whichisperpendicular toB.Thismeansthatthemagnetic force an triestomake wequal tou. ——— (b)Calculate theefficiency, ontheassumption thatapermanent magnet ro isupplies themagnetic field. Neglect edge andend effects. Istheeflcienehigh,orlow? * 4 >j_ 228.(22.1)TheHalleffect (ees Letusinvestigate theHall effect more closely. Weassume again thatthe eX Fig.22-15. charge carriers areelectrons ofcharge e.Their effective mass ism*. The effective mass takes intoaccount theperiodic forces exerted onthe 22.15, The element then hasonly twoterminals andiscalled a electrons astheytravel through thecrystal lattice. Asarule,theeffective ‘magnetoresistor. Magnetoresistors areuseful formeasuring magnetic fluxmass issmaller thanthemass ofanisolated electron. densities. The force onanelectron is F=~c(E+v XB), whore Ehastwo components, theapplied fieldE,andtheHallfieldE,.Theaverage drift 22.9.(22.1) Theelectromagnetic flowmetervelocity is The eleciromagneti flowmeter isthe inverse oftheelectromagnetic pump . aF (Prob, 22-7). Itoperates asfollows. SeeFig.22-16. Aconducting fluidflows v=__aEsux), inanonconducting tubebetweenthepolesofamagnet.Electrodes on .either side ofthetube and incontact with thefluid measure thevXBfield,andthusthequantityoffuidthatflowsthroughthetubepersecond.Thisis where Misthemobility (Sec.4.3.3). ThelawF=m*a applies onl indthusthequantityo T between collisions withthecrystallattice. ¥ ‘Halleffect,exceptthathereionsofbothsignsmovewiththefluidinthe (@)Show that same directionFaraday attempted tomeasure thevelocity oftheThames River inthis =-M(E,+1,B), y=ME, 1.8), 0,20 wayin1832.Themagnetic fieldwas,ofcourse, thatofthe earth Tntheabsence ofturbulence, thefii velocity inatube ofradius Risof (6)Show that theform v=u,(1 r/R). The vXB field inthefuid istherefore not uniform. This gives risetocirculating currents withJ=o(—PV +vXB). janeu MED yyyEotHE rane eeraioniechesthatsecumbitsontheletoses eae?eB (a)Sketch across section ofthe tube, showing qualitatively, bymeans of Thus, ifJ,=0, arrows ofvarious lengths, themagnitude anddirection ofvX’B. b (b)Sketchanothercrosssection,showingthelinesofcurrentflow.The E,=-ME,B or V,=—MV.B. current drawn bytheelectrodes isnegligible. a (©)Neglect endeffects bysetting 3/42 =0.Usethefactthat +J=0 to NotethattheHallvoltage isproportional totheproduct oftheapplied showthat voltage V,andB.TheHalleffectisthususefulformultiplying onevariable vv=BB sing.byanother. ay or‘Whenitisconnectedinthisway,theHallelementhasfourterminalsa ien annecedinthway.theHalelementasorteinsand sincethisLaplaciansequal©pe,thewlumechargedensitypisero (©)Calculate V,forB=1millimeter, a5 millimeters, 4¢=7 meters!/ volt-second (indium antimonide), V,=1volt, B=10°‘ tesla. +8 ZF(@)Showthat,ifE,=0,then ‘| _AA ARcp, ozd Ry N/a where Ryistheresistance oftheprobe inthex-direction when B=0,and \ TheHallfield£,canbemadeequaltozerobymakingcsmall,sayafew \pe .micrometers, andplatingconducting stripsparalleltothey-axis,asinFig. ‘s -6 Fig.22-16. ontheaxis,where 3u/ar=0 andat@=0,andatx.Seepage402. 22-13. (22.4) ‘Thepinch effect onaconductor py? .nm Iftheforcedensity isenormous, thenanysolidcanbetreated asafluid,€vrLav!Vv"gd andifpisthepressure,Vp=JxB. dPrdsPdr (a)Aconducting wireofradiusRcarriesacurrentJ.Atthesurface, = yw that liusr,p=[Mol?/(4x°R*)\(1-1°/R*). The (6)Youcansolvethisdiferential equation asfollows:(1)expressthe Papeteonetamprentsthewie,PelIGeROISPIR).The thevalueoftheotherbyremembering thatJ-=0at#20,j=Ritthe Yaporios,Suchlargecurrentsareobiainedbydischarging larvoltmeter draws zerocurrent. (c)Showthat,withatubular conductor ofinnerradius R,andouter Notethattheoutput voltage isindependent oftheconductivity ofthe radiusRz,thepressure inthecavityisgivenby Findtheoutputvoltageasafunction ofthevolume offuidthatows ol?1—(R/RG¥IL+2 InRR] inonesecond, aR OL-(R/R TE andemerges frome themagnetic Held. Sketch linesofcurrent flowforthese Discegard theskineffect. Transient pressures approaching 10*atmos: 22-10, (22.2) Improving (2)electric motors ‘Oneypeofxray source implodes thin, aluminized plastic wbes by (a)Someone suggests that,iftherotorsofelectricmotorswerewound discharging Imegajoule inthem.Theresultingplasmagenerates a with ironwire instead ofcopper wire, thetorque could increase bya 150-kilojoule pulse ofradiation. factor of1000. 22-14, (22.4) Thehomopolar motor Show thatthetorque would indeed increase, butbyanegligible (a)Findthemechanical power asafunction ofwforahomopolar afactor ofmaybe3or4. (c)Showthatthemechanical powerismaximum attheangular Youcanshowthatthisisincorrect ifyouarecarefultodistinguish Velocity@masimumponer=V/[B(b* —4°) between themagnetic ficld ofthestator and that oftheiron oftherotor. - 22-15. (22.4) Thehomopolar motor 22-11,(22.3) Theforcebetween twoparallel currents. ‘Show thatthecentrifugal forceontheconduction electrons intherotor Twolong,straight, parallel wiresoflength2.separated byadistance ofahomopolar motoriscompletely negligible. SetB=1,o=2n.The Dcarry equal currents Jflowing inthesamedirection, ratioofthemagnetic force tothecentrifugal force isenormous because 22:12, (223) The magnetic force law does notapply totheforces between single particles.” p=U2BY0.0, ‘4n€,s" Now calculate thissame force from Coulomb's lawand from themagnetic force lw substituting Qo ford, and Qo forhls 282 MOTIONAL ELECTROMOTANCE 413 usual, and that VX£=—9B/3t, The Faraday induction law neatly 23 groupsbothphenomena. CHAPTER 23.1 THE vXB FIELD INSIDE ANONCONDUCTOR MAGNETIC FIELDS VI (1)Suppose thatanonmagnetic nonconductor moves insomearbitrary TheFaraday Induction Law fashion inaconstant magnetic field.Thenacharge Qcarried alonginsidethebody atavelocity w,inaregion where themagnetic flux density isB, experiences amagnetic force QuxB. 231THEwXBFIELD INSIDEANONCONDUCTOR 413, AswenotedinSec.22.1,thevelocityvcanbeanyfunctionofthe23.2MOTIONAL ELECTROMOTANCE. THEFARADAY INDUCTION LAW coordinates andofthetime.Itcanbeuniform throughout thebody,oritconeaSIMPLEmeEDGENERATOR 414 canvaryfromonepointtoanotherandwithtime.‘SIMPLE: 4 "B ates, ee ERMATING:CURRENTGENERIOR 48 __Ofcourse,Hcanitselfbeanyfunctionofthecoordinates. However, 23 LENZSLAW 419 isconstant, byhypothesis; weshallcome totime-dependent magnetic 234FARADAY’S INDUCTION LAWFORTIME-DEPENDENT B's.THECURL eldsindoetime. ore am Now XBhasthedimensions ofE,because QuxBisaforce, 234.1 LENZ'SLAWAGAIN 21 Indeed, vXBadds toanyEthatmaybepresent. Thepolarization is 2342 FLUXLINKAGE 21 therefore given by Example: MEASURING BIN AROWLAND RING 422 _Example: THEENERGY DISSIPATED INDESCRIBING AHYSTERESIS P=cax(E +xB) (3-1) LOOP 423 sts cauD5_TiheELECTRIC FIELDSTRENGTH EEXPRESSED INTERMSOFTHE (2)Ifthenonconductor ismagnetic, itsequivalent currents, ofcourse,POTENTIALS VANDA oe followthemovingmedium, buttheycanbetime-dependent iftheExample: EDDY CURRENTS 225 ambient Bisnonuniform. Sothesituation canbecomplex. Example: THE INDUCED ELECTROMOTANCE INARIGID CIRCUIT 426 .23.2 MOTIONAL ELECTROMOTANCE. 236 THEE, ~FV, -34/31, ANDwXBFIELDS 427 THE FARADAY INDUCTION LAW "23.7 RELATING THETWOFORMSOFTHEFARADAYINDUCTION FORvXBFIELD! LAW 228v S 123.1 TRANSFORMATION OFAMAGNETIC FLUX 429 . j ©7372 TRANSFORMATION OFANELECTROMOTANCE 429 Consider aclosed circuit Cthatmoves asawhole anddistorts insome BA SIXKEYEQUATIONS <0 arbitrary wayinaconstant magnetic field, asinFig.23-1. Then, by 239 SUMMARY 431 definition, theinduced, ormotional, electromotance is PROBLEMS 432 v-$vxB)vat=—$B-yx 5. Intheprevious chapter westudied themagnetic forcesexerted on pv xB) p.B(wxdl) (232) moving charged particles and oncurrents situated inmagnetic fields. ; These forces wereoftheformQuxB. ‘Thenegative signcomes fromthefactthatwehavealtered thecyclic Wenowinvestigate theeffect ofavB ficldinside amoving ‘orderofthetermsunder theintegral sign. macroscopic body.Weshallseethatinconducting bodies v Bactslike NowvXdiistheareaswept bytheelement diin1second. ThusE.Weshallalsoseethatatime-dependent magnetic fieldgivesrisetoan B-(vXdl)istherateatwhichthemagnetic fluxlinkingthecircuitelectric fieldofstrength —2A/3t, where Aisthevector potential, as increases because ofthemotion oftheelement di.Integrating overthe 414 41s .\ na Se @ =~ © a —— x xX a oye4 a +t - \ \ Zt A . — +Or <= \ J \ \ exe X: > \ \ b \. w \ \ | Fig,23-2.(a)Conducting rodmoving atavelocity vinamagnetic fieldB,The ‘vXBfield drives conduction electrons upward, Current flows until theelectric Fig.23-1. Closed circuit Cthatmoves anddistorts insome arbitrary wayina fieldEexactlycancelsthew*Bfield.(b)DiskrotatinginamagneticfieldB. ‘constant magnetic fieldB.Theelement dfmoves atavelocity vandsweeps an HerethevXBfieldcauses conduction electrons tomoveoutward. Aradialarea wXdlinIsecond, ‘current flows until theelectric fieldcancels thew*Bfield completecircuit,wefindthattheinducedelectromotance isproportional hTheseesendmostimpractical) ‘ypeofgeneratorisiiaof totheti ink : ig.23-3(a).Thelinkslidestotherightataspeedvsuchthat (0thetimerateofchangeofthemagnetic fluxlinkingthecircuit: tice,where¢isthespeedoflight,inauniformBthatis de normal tothepaper. Theresistance attheleft-hand endoftheline v=- (23:3) isR,andthatofthelinkisR).Thehorizontal wireshavezeroa resistance. |The electromotance is The positive directions for¥°andfor®satisfy theright-hand screw rule. a Thecurrentisthesameasifthecircuitcomprised abatteryofvoltage VoxGy=Bde. (23-4) This‘istheci‘araday induction lawforvXBfields. Thislawis Wehavedisregarded themagneticfluxresultingfromthecurrentJ important.Asfarasourdemonstration goes,itappliesonlytoconstant itselfInotherwords,theresistanceRislarge.Then Bis,butitis,infact,general, asweseeinSec.23.4. Quite often ®is bbw difficult todefine; then wecanintegrate vXBaround thecircuit to IRR (23:5)obtain ¥, " IfCisopen,asinFig.23-2,thencurrentflowsuntiltheelectricfield asondoston,seconinsidetelekfotomslong,atthevelocityw,andthevXBfieldprodsitdownward.Itsvelocityw, seeetioefromthesccomeletions ofcharge exactly cancels the»xB relative tothelinkisstrictly vertical, understeady-state condi- onl tions, because otherwise negative charges would accumulate indefinitely along one edge ofthelink, leading toaninfiniteExample |ASIMPLE-MINDED GENERATOR potential.SeeFig.23-3(b)Inafixed reference frame 5,the force onaconduction electron ‘Anelectric generator transforms mechanical energy toelectric ofcharge Qinside thelinkisQ(E +vXB). Thus, inthelink, energy, usually by moving conducting wires inadirectionperpendicular toamagnetic field J=oF+0XB)=o(-PV +vxB). 236) 416 a7 Ceeeeeeeee ee oot++++¢+++++ _ ime ‘ | | & .veal Hh Qo |¥ , L5 5 | we ° 7 o> —. J i Fig.23-4, (a)Voltmeter Vconnected across thelinkinthegenerator ofFig.23-3 Wd (b)Equivalent circuit, SupposeweconnectavoltmeteracrossthelinkasinFig. f3 23-4(a).CalltheresistanceR,,withR,>R,.Thishardlyaffectsthecurrent /.What will bethereading onthevoltmeter? Ifthe.23:3. (a)Simple-mindedel cae Q Pekcrator,Shaieg helatothet currentthroughthevoltmeteris/.,thenitwillreadavoltage atavelocity vgenerates acurrent in 1.Ry, withthepolarity shown inthefigure. thedirection shown. Themagnetic Nowrefer toFig.23-4(b). Clearly, /,R,=/R,andthevoltmeter field isconstant anduniform. The reads thevoltage drop /,R). resistanceRistheload.(b)Section ‘Chapter 17,onrelativity, isaprerequisite fortherestofthiswooeB through thelink: visthevelocityofthe ‘example, wire,v,thevelocityofaconduction AllweknowaboutthemagneticfieldisthatB=—B2;the eldTheHaleldexactadheHall currentdistributionthatgeneratesBisunspecified.Letusset ov,XBfield. A,=nBy, A,=(n+1)Bx, (23-10) Letuscalculate VinS. wherenmisapurenumber. Itisasimple matter tocheckthatAtbinFig.23-3(a), V,=JR.Ineitherhorizontal wire,J=oE B=VXA. isfinite.Sinceo->=,byhypothesis, thenE=0,VV=0,and Ifn=0,thenthecurrents supplying Bareallvertical. Ifn==1, they are allhorizontal. With asolenoid whose axis Vi=V.=0, V.=Vy=IR. (23-7) coincides withthez-axis, n=~|.Therefore, insideRandR, InsideRandR,,withthey-axisasinthefigure, E=-wv-A-_w (3-1) 7 y__vBRV=IR-=—— _y. (23-8) vBR = 2BR 2312) D RR aR! @3.12) ThevoltageV,acrossR,isIR Letusnowseewhathappensinsidethelink,initsown« We ve,whiwakesy~1. V.=IR=I(R+R)~IR,=vBD~IR,(23.9) ferenceframe9”Weassumethatvcs,whichmakesy ‘This means thatthemotion generates avoltage vBD inthelink, . wv wvBR whileitscurrent causesavoltage dropIR, AinA,—Senby- Seg =nBy 13) ais MAGNETIC FIELDS vt 419 AL=A, =(n+ 1)Bx =(n+ 1)Bur, (23-14) * vBR R 1Vavd,=OR_ynpy =(—* 3. - VinV0A,=Piey~onBy(aeeBy.(23-15) “ R viz (R= n 23.16 Note that thevalues ofA’andofV"depend onthevalue ofn.In bet eb ‘other words, they depend ontheparticular geometry ofthecoils selected forgenerating H.Observe also the appearance ofa DNS . 9A/9t termin 7 ™ Now a F ~ een (3.17) y where a, 8 v v Fig.23-5. Loop rotating ina wate Sy, ver Sear-Se=e (318) constantanduniformmagnetic eded ©© fieldB.Theslipringsprovide Thus contacts between the voltmeter av,3a’av".aay landtheloop. greOV, GAN OV BAG og Bay? or a a” G19) Alongtheright-handsideof» =[-(qegonetn+nealy 2320) (2)AlongtheHiehehandsideoftheloop, R R bvx=2Bbsina=PBsinore 2324) =(8 1va5-2ony, 23.2 2 (gag) -Fe Re = oe Along theleft-hand sidewehavethesameinduced electromot- Ingeneral, —2A"/3¢' isnotequaltovXB ance,butdirected downward asinthefigure. Alongtheupperand‘Thequantity hasdisappeared! Wecould haveexpected this lower sides, vxisperpendicular tothewire,Thiscrowds the because, clearly, £”mustbeindependent oftheconfiguration of conduction electrons sideways, thereby increasing theresistance thecoilsthatgenerate thegiven magnetic ld imperceptibly, butvXHcontributes nothing totheelectromot-Wecould alsohave found £”directly, bysimply transforming ame. So E,withv?«ce? V=abBeosinwt. (23-25)=Byt(E,+uX BE+uxB (23-22) ® , 7 . Noticethatthereiszeroelectromotance whenwt=na,wherenis VBR Sg) Rg > awhole number. Then vandBareparallel, andvXBiszero.(Ren) RRM 2H) (b)Thetimerateofchangeofthemagnetic faxgivesthesame result asabove, Thisshows that, inthemoving reference frame ofthe wo 4 link,BisequaltoBplusvXB. Va-O.4ppconor=abBusinot.(2326)The current 1"isequal toJ,and thevoltmeter reading isthe same for an observer inS°asfor an observer inS : 23.3 LENZ’S LAW Example ANALTERNATING-CURRENT GENERATOR “TheloopofFig.23:5rotatesatanangularvelocity «inauniform, NowletusreturntoFig.23-3(). Observe thatslidingthebartotherightGonstaan B.Wecalculate theintheod sletirorwtence ¥.feet increases thelinking flux, buttheinduced current tends todecrease it.So through wxBandthen through d@/de theinduced electromotance drives acurrent whose field opposes achange 420 MAGNETIC FIELDS VI 23.4 FARADAY’S INDUCTION LAW FOR TIME-DEPENDENT B'S 421 inthenetmagnetic flux linking thecircuit. This isLenz's law. Ifthe Assuming thecorrectness oftheabove result, theelectromotance circuit was superconducting, theenclosed flux would remain constant. induced inarigid and stationary circuit Clying inatime-varying Wereturn toLenz’s lawlater. magnetic field is 7 ” de a 23.4FARADAY’S INDUCTION LAWFOR vagBedt=|(rxe)-dt =~2-[Boast, 329TIME-DEPENDENT 8's. THE CURL OF E “ “ “ Wehave used Stokes’s theorem ingoing from thefirst tothesecond Imagine nowtwoclosed andrigid circuits asinFig.23-6. Theactive integral, sfbeing anarbitary surface bounded byC.Also, wehave a circuitaisstationary, whilethepassivecircuit6movesinsomearbitrary partialderivative underthelastintegralsign,totakeintoaccountthefact,way,sayinthedirection of@asinthefigure. Thecurrent /,isconstant. thatthemagnetic fieldcanbeafunction ofthecoordinates aswellasof From Sec. 23.2, theelectromotance induced incircuit is thetime. The right-hand screw rule applies. The path ofintegration need notlieinconducting material do Observe that theabove equation involves only theintegral ofE-dl.It y=fxB)-ar= 2% : : pv XB)di= —"7, (2327) doesnotgiveEasafunction ofthecoordinates, except forsimple geometries, and only after integration where®isthemagnetic fluxlinking b.Thisseemstrivial, butitisnot, Sincethesurface ofareaofchosen forthesurface integrals isarbitrary, because d¢/dt could bethesame ifbothcircuits were stationary andif1, theequality ofthethird andlastterms above means that changed appropriately. This means thattheFaraday induction law OB do \yxe=-—. (23-30) v=- (23-28) iL ieswhether th , . Thisisyetanother ofMaxwell's equations. Thisequation, liketheotherapplies whether therearemoving conductors inaconstant Borstationary two(Eqs. 9-15and18-19), isvalidonthecondition thatallthevariablesconductors inatime-varying B.However, ourargument isnomore than relate tothesame reference frame. plausible. Aproper demonstration follows attheendofthischapter. It requires relativity. 23.4.1 Lenz’s Law Again The negative sign inEqs. 23-28 and23-30 isimportant. If®points into the paper and increases, then d/dr points into thepaper. Then, according totheright-hand screw rule, thenegative sign means that the induced electromotance iscounterclockwise. Observe that theinduced electromotance tends togenerate amagnetic field thatcounters theimposed change influx. SoLenz's lawofSec. 23.3 applies here also. 23.4.2 Flux Linkage « Ifthe closed circuit comprises Nturns, each intercepting the same L magnetic flux, then theelectromotances addandthenetelectromotance isNtimeslarger.Thenthequantity N@®istermedthefluxlinkage: Fig.23-6. Circuit aisactive and fixedinposition. Circuit issive andmove y daFig,2346.Circuitaisativeandfixedinposition,Ciruitbispa1dmoves A=No andy=— (23331) 422 MAGNETIC FIELDS VI 23.4FARADAY'S INDUCTION LAWFORTIME-DEPENDENT 8'S 423 Ofcourse, thegeometry ofthecircuit andtheconfiguration ofthefield LfydB Nascanbequitecomplex.Thenthisequationstillappliesandthegeometric Va~Re[veoio (23:35)meaning ofAbecomes obscure, but ¥,and hence A,aremeasurable quantities. WehavesetV=0at1=0,Sothevoltage attheoutput oftheintegrating circuit isequal to Example |MEASURING BINAROWLAND RING aknownconstanttinesB. Wearenowinaposition tounderstand howtomeasure Binthe Example THEENERGY DISSIPATED INDESCRIBING ARowland ring ofSec. 21.2 toobserve thehysteresis loop ofa HYSTERESIS LOOP sample offerromagnetic material. Letthelow-frequency alternating current through winding ain ‘Wecannowalsocalculate theenergy dissipated indescribing theFig.23-7be hysteresisloopofFig.21-2.WerefertotheRowlandringofSec.=, 2) 21.2.Whenthecurrent J,increases, theflux®inthecoreIshooax. G32) increases, andtheclectromotance —N,d@/drtinduced inwinding ‘Then,fromSec.20.6,Hisasinusoidal function ofthetimeand,if 4opposes thisincrease, according toLenz’slaw.Theextrapowerwinding ahasN,turns, spent bythesource isthen Nala6080 dt_)(yd a8 HaNeeser 23-33) 48_1(n,4%) <n? Dar 333) (Ne Ge)=UN 336) ‘Weassume thattheminor radius oftheringissmall compared ifofisthecross-sectional areaofthesample. Also, toitsmajor radius. Then risthemean radius. Ifthemagneticfxdensity,averagedoverthecrosssection,is FENeyrsHy, (337) B,then themagnetic fluxisBefand, disregarding signs, the dt2ar at aelectromotance inducedintheN,turnsofwindingbis ‘tromotance inducedintheN,turnsofwindingb wherev=2xrafisthevolumeofthesampleand py tO. dBv=Ma=Neta (23-34) t=ofadB (23-38) Tithewotngsattheterminalsofwindingbifthemeasuring istheenergysuppliedbythesourceingoingfromgto6onthe ‘Toobtain B,weconnect anintegrating circuit (Prob. 7-9)tothe loopofFig.21-2.Thisintegral isequaltotheareadefined bytheterminalsofwinding},asinFig.23-7.Then,withanoperational pointsagbhandtotheenergysuppliedbythesourceperunit. _ . volume ofcore. amplifier ofgain A>1,andifA@RC>1, ‘Whenthecurrentisinthesamedirectionbutdecreasing, the polarity oftheinduced electromotance reverses, andingoing from « toc,theenergy YS) (> n- enofnas 2339) @MA AY ;C(x) 3 returnstothesourceel Proceedinginthiswayallaroundtheloop,wefindthat,inthe d course ofonecomplete cycle, thesource supplies anenergy Fig.23-7.TheRowlandring.Thisisbasicallyatransformer withaprimaryaand ev asecondary b.Thesecondary isconnected toanintegrating circuit. Aisan S=vpHas, (23-40) ‘operational amplifier. Theoutput alternating voltage Visproportional tothe ‘magnetic fluxdensity Binthecore. whichisutimestheareaenclosed bythehysteresis loop. 44 MAGNETIC FIELDS Vt 235THE FLECTRIC FIELD STRENGTH 25 23.5 THE ELECTRIC FIELD STRENGTH E The Faraday induction law, indifferential form (Eq. 23-30), relates EXPRESSED IN TERMS OF THE POTENTIALS space derivatives ofEtothetimederivative ofBatagivenpoint. VANDA Observe that PVisafunction ofV,which depends onthepositions of thecharges. However, 3A/3t isafunction ofthetime derivative ofthe An arbitrary, rigid, and stationary closed circuit Clies inatime- current densityJ,henceoftheacceleration ofthecharges. dependent B.Then, from Sec. 23.4, The relations d e=-w-4 andB=VXA (23-46) f.e-a=~5[Beast, (23-41) " 'cdha arealways validinanygiven inertial reference frame.” wherefistheareaofanyopensurface bounded byC. Inatime-dependent B,theelectromotance induced inacircuitCis Now,fromSec.19.1,wecanreplacethesurfaceintegralontheright yee[Aa (23-47) bythelineintegral ofthevector potential Aaround C: ar da 3A Exal cl S fe-d=-fh4-a=-§Aa (23-42) Example|EDDYCURRENTS «© « Imagine asheet ofcopper lying inside asolenoid, inaplaneendicular 10theaxis ‘analternatin Thereisnoobjectiontoinsertingthetimederivativeundertheintegral perpendtHartotheaxis.Thesolenoidcarriesernating sign, butthen itbecomes apartial derivative because Aisnormally a ‘According toLenz’s law,the-2A/dt electric field induces function ofthe coordinates aswell asofthe time, currentsinthecopperthattendtocancelthechangesinthenetB. Thus These currents are azimuthal because Aand 3A/3t are azimuthal (example inSec. 19.1) 3A Currents induced inbulk conductors bychanging magnetic fieldsp(e+Z)-a-o, (23-43) aetermededdycurrentEddycurrentscanbeuseful.Forexample,theydissipateenergy ; . invarious damping mechanisms. whereCisaclosed curve, asstated above. Then, fromSec.1.9.1,the ‘Theyareharmful intransformers because theyciuseJoule expression enclosed inparentheses isequal tothegradient ofsome losses inthecore.Transformer cores areusually assembled from function: thinsheetsoftransformer iron,calledlaminations, afraction ofa aA nilimeter thickinsmallunits,insulated fromeachotherbyathin Ets=-Wv, (23-44) layerofoxide.Withasolidcoretheeddycurrents wouldlargely "cancel changes inmagnetic flux. Also, theJoule losses would be aA excessive E=-vv-a (23-45) Inaudiotransformers theironalloyissometimes intheformofa ‘4powder molded inaninsulating binder. Thetransformer isthen saidtohaveapowdered ironcore. whereVis,ofcourse, theelectric potential. Ferrites serve ataudio frequencies andabove. These are SoEis thesum oftwo terms, —PVthat results from accumulations of ceramic-type materials thataremolded from oxides ofironandof charge and —3A/3t whenever there aretime-dependent fields inthe various other metals. Their main advantage isthattheir electric given reference frame. conductivity islow,oftheorderof1siemens/meter, or10-*times thatoffransformer iron.Eddy-current losses inferrites arethus Thisisanimportant equation; weshalluseitrepeatedly. Observe that Olymaneeableitexpresses Eitself,notitsderivatives oritsintegral,atagivenpointin terms ofthederivatives ofVandofAintheregion ofthatpoint. Its Unlessstatedotherwise, allthereference framesthatwerefertoareinertial: theydo magnetic equivalent is B=VXA(Sec.18.4) sotaccelerate,andtheynotrotate 26 MAGNETIC FIELDS VI 236 THE B,PV, ~24/01, AND wa FIELDS 427 Example THEINDUCED ELECTROMOTANCE IN Insidetheresistance R,,A~0byhypothesis and ARIGID CIRCUIT Ir Intherigidcircuit ofFig.23-8,weassume that(1)thehorizontal E,=-We=~T 9, VomdR (23:51) wires have zero resistance, (2)theresistance R,isalong distance away from 1',and (3)both R,and R,arelarge enough torender with E,and PV, pointing asinthefigure themagnetic field ofFnegligible compared tothat of1’.Thus B The potential ateiszero forthefollowing reason. Inside the andAareessentially thoseof/’andpointinthedirections shown. horizontal wires,E=J/o,whereJhassomefinitevalueandoisAccording toLenz's law(Sec. 23.4.1), anincrease inI’induces an infinite, byhypothesis. So electromotance andacurrent /inthecounterclockwise direction. BV.8A, ‘One may ascribe the induced electromotance cither tothe E,=0, -2t-Ang, (23-52) changing magnetic flux®ortotheelectric fieldstrength —3A/2r a oF Ry, with: m" ButA,iseverywhere zerobecause thecurrents /’haveno o=fava~ap (23-48) sx-component.SoWoo,veo. (23-53) ‘Thelineintegral runsclockwise because oftheright-hand screw ox rule. The induced electromotance isthus ‘Similarly, V.=V,=IR. (2354)v=-2.2p (23-49)" " ‘Thispotential onthecdwireresults from theaccumulation of Thisiscounterclockwise if®points intothepaperandincreases, positive surface charges ontheupper halfofthecircuit, andTie counterclocswis negative charges onthelower half,asinthefigure. Trside Ry, therefore, (8A,/3)D— 3-5 TR. IRR, (23-50) vy,5 (2355)D inthedirection shown, againif"increases. pointsupward, ad 3A,_IRyPoaaaeaeee) B= p (356) he aa) + alsopointsupward.Ofcourse,Epointsinthesamedirection asJ.. | ReRythenn + | :| e\=iE, |4|=219%1 (357) ti||e.@ : * || eS de'| ! ' 23.6THEE,-VV,—9A/3t,ANDvXBFIELDSwn ' t ' lp: es ;Inany given inertial reference frame, sayS,theequation 7 Ex Fig.23-8.Rigidcircuitgcdeterminated byresistances R,andR,andsituated E=-¥v- a (23-58) near apairofwirescarryingacurrent/'.Thelengthsofthearrowsshowthe relativemagnitudes ofE,PV,and3A/3t, with R,=Ra. always applies. 28 MAGNETIC FIELDS v1 217 TWO FORMS OFTHE FARADAY INDUCTION LAW 29 Ifacharge Qmoves atavelocity vwith respect toS,then foran from Sec.13.4, Thus, atanygiven point in$’,x’isfixed andobserveron$theforceis 3A d=yd’, (23-64) F=QE+vxB)=0(-vv-Ssuxa) (23-59) , a Proving theother twoequations takes abitlonger. Allthevariables aremeasured with respect tothesame reference 23.7.1 Transformation ofaMagnetic Flux frameS.Theseequations arevalidevenifvapproaches thespeedof ‘ light‘The magnetic fluxlinking agiven closed circuit bounding anarea ofis For anobserver onthemoving body, sayinreference frame $’,thebodyisatrestand o=[adh (23-65) F'=QE'= o(-rv -a) (23-60) inareference frameS$and *23.7RELATING THETWOFORMS OFTHE °-[2at 3-66) FARADAY INDUCTION LAW' ins’ Ofcourse, thesurface ofarea sfinframe Shasadifferent shape in Wefound above that ifarigid circuit moves inaconstant B,then the frame S',because oftheLorentz contraction, andadifferent area of’.A induced clectromotance follows Faraday's induction law.Thenwe givenelement ofarea(sayitispaintedred)carriesafluxBdfinframe concluded that thesame lawapplies toastationary circuit lying ina S,and B’def" in$'.Thus time-dependent B. Sedat! Passing fromoneformofthelawtotheotherrequires relativity. do"Bidet’ (23-67) Intherestofthischapter weshallcallanelectromotance Vm,andthe do” Bede velocity of$with respect t0$Yee. aEquation 23-28referstotheinduced electromotance, asmeasured in But,fromSecs.14.1.1and16.6,andsetting E=0, thefixed reference frame S,From ourexperience with relativistic ad,calculations, itisbynomeansevidentthat,intheframe$”ofthemoving BY-dst’=(B,+yB,)-(dst, +“) (23-68)circuit, . do’ =B,-dd,+B, +d, =B-ds, (23-69) n= -F (23.61) and@'=.Themagnetic fluxlinkingarigidclosedcurveisinvariant. ‘That is,infact, true because, asweshall see, y . dt *23.7.2 Transformation ofanElectromotanceVim=Vem, 0=O, dt’==, (23-62)Y Refer again toFig. 23-6 and call S"thereference frame ofthemoving ‘Wecanimmediately accept thethird equation forthefollowing reason. circuit. InS’, For any point inS’,Vox’ VinaGE-dl (23-70) t=re+), (23-63) Sinceweareonlyinterestedinmotionalelectromotance forthe --moment, wemay assume that Bisconstant andEiszero inS.Then, Relativity isaprerequisite forSec. 23.7, from Sec. 16-6, 430 MAGNETIC FIELDS VI 23.9SUMMARY, 431 Vin=$1VeaXB) (23-71) 23.9SUMMARY ‘Abodymovesinsomearbitrary fashion inaconstant, butnotnecessarily Nowthevector product isperpendicular toV;q).Itistherefore onlythe uniform, magnetic field. Atapoint fixedtothemoving body, acharge Q Perpendicular component ofdl’thatmatters, dl’,=dl,and experiences aforceQuxB.Themotional electromotance is 7 v= v . (23-2)Vann1§GoaXB)edl=Vu (23-72) [exsa 2) Here visthevelocity ofachargefixedinthebody,andBisthemagnetic Wehavetherefore proved allthreeequations 23-62. Asaconsequence, fluxdensity atthatpointinspace wehaveshown that,under anycircumstance, butinasingle reference When thecurve Cisclosed, themotional electromotance isalsogiven frame, theelectromotance induced inaclosed circuit associated witha by changing magnetic fluxisgiven by do V=-—, (23-3). dd dt (23-3) Vom=“Gr (23-73) where®istheenclosedflux.ThisistheFaradayinductionlaw.The ‘Thepositivedirections chosenforV.,,andfor@followtheright-hand Fighthand serewrulesapplies.Thislawalsoapplies(0afixedcircuitscrew rule. situated inatime-varying magnetic field ‘Atanypoint inspace, inagiven reference frame, 23.8SIXKEY EQUATIONS a VXE=—s (23-30) Itisuseful atthisstage togroup thefollowing sixequations: L ms) oA ThisisoneofMaxwell's equations. (G)E=-vv —o° (23-46) Lenz's lawstates thattheelectromotance induced inaclosed circuit tends tooppose changes inthemagnetic flux linking thecircuit. OB ‘itha . )fea-[at, (2329) Withamultiturnclosedcircuit« _ dA OB v--4, (23-31) @ YxE=->, (23-30) . where Aistheflux linkage (G) B=VXA, (Sec. 18.4) and (23-46) Inanygiven inertial reference frame, dl . 5 2A $aatasy{adst,(Sec.19.5) pe-py-A angBax (23-46) VXB=jod. (Sec.19.4) ‘AchargeQmoving atavelocity vinsuperposed electricandmagnetic Thefourequations preceded by(G)aregeneral, whiletheothertwo fieldsissubjected toaforce apply only toslowing varying fields (Sec. 27.1). Ineach equation allthe 3Atermsconcern thesamereference frame. F=Q +uxB)-0(-PV-S+uxe). 23.59) 432 MAGNETIC FIELDS v1 433 PROBLEMS ss 23:1,(23.2)Thethoughtexperiment ofFig.23-3 / % Show thatthere isconservation ofenergy inthethought experiment of ia Fig.23-3(a. <R 23-2. (23.2) Tides andthemagnetic fieldofthecarth e Discuss how tides affect the magnetic field ofthecarth byconsidering AN thecase ofariver flowing into theseaintheeast-to-west direction inthe ' northernhemisphere.Rememberthatthemagneticpolesituatedatthe x{!north geographic pole isasouth magnetic pole. The vector Bpoints Li downward inthenorthern hemisphere. i 23-3. (23.2) The magnetic braking force onasatellite ‘Anatural satellite whose diameter is10'meters moves atavelocity of 1kilometer/second inthedirection normal tothemagnetic field ofaplanet inaregion where B=10 "tesla. The satellite has anappreciable Qconductivity. Fig.23-9, (a)The satellite moves inaperfect vacuum. What happens? (b)The ambient gashasadensity oftheorderof10"particlespercubic e)Calculat ” i m eter,theparticlesbeingeitherelectronsofsinglychargedions.Eachhalf sol,CmenlatetepowerIVdisputedinthepate,Thisshouldbeequ ofthesatellite collects particles ofthecorrect signinsweeping through (f)Estimate thevalue ofBrequired ineddy-current diskbrakes foraspace.Caleulatetheorderofmagnitude ofthecurrent. smallbus.‘Theconductivity ofcopperis5.8x10”siemens/meter. Estimate(©)Calculate theorder ofmagnitude ofthebraking force. thepower dissipated ateachwheel. Why does avehicle equipped with (4)Someone suggests thatthiscurrent could provide power foran eddy-current brakes sillneedconventional brakes? artificial satellite traveling inthesamefieldatthesamevelocity. Inversely, Tnmountainous regions some buses andtrucks areequipped with acurrent intheopposite direction could servetopropel thesatellite. What ddynamos thatbrake bygenerating electric power thatisdissipated inalarge isyouropinion? resistance ontheroof Artificial satellite velocities range from about 4to8kilometers/second, and vXB intheionosphere and magnetosphere ranges from about 23-5. (23.4) Detecting flaws inmetal tubing 100microvolts/meter to320millivolts/meter. Figure 23-10 shows theprinciple ofoperation ofadevice fordetecting flawsinmetal tubing, orrod.Thecoilsaprovide alargegradient of 23-4,(23.3) Eddy-current damping magnetic fieldalong theaxis,asinProb. 18-10. Coilbisconnected toa Figure259showsonecommon typeofeddy-current damper. Motionof monitor. ThetubingTmovesataconstant velocity walongtheaxisofthecopperplateinthefieldofthepermanent magnetinducescurrentsthat , volta aryscseascollBwien2flawpassesthroty tendtoopposethemotion,accordingtoLenz'slaw.Joulelossesinthe pana oltageapeSScollbwhenaHawpassesthrough plate dissipate itskinetic energy. Dampers ofthis general type are used mostly, but not exclusively, in low-power devices suchaswatt-hour metersandbalances. Asyouwillsee, athebraking force isproportional tothevelocity, asinaviscous fluid. y G (a) Explain qualitatively, but ingreater detail, theorigin ofthebraking force (b)Howcouldyoudesignanautomobile speedometer thatuseseddy (2currents? i(c)SayBisuniform over thepole face. The path followed bythe \ current iscomplex; setR~3a/obs. This quantity isoftheorder 3.The 3 g plate hasathickness sand aconductivity a.Calculate thecurrent. (4) Caleulate the braking force F.This isproportional tothe conduc- tivity. Sothe plate should beeither copper oraluminum. Aneven better solution istouseanironplate faced with copper. 7 Fig.23-10. 434 MAGNETIC FIELDSVt 435 23-6.(23.4)Theflux-gatemagnetometer Solenoid "Amagnetometer measures B.Onecommon typeistheflux-gate €&‘magnetometer, whichputstousethehysteresis curve.Thereexistmany So retingsinforms, oneofwhich isshown inFig.23-11(a). Thetworodsaremade ofa i 4ferromagnetic materialsuchasaferrite,whosehysteresiscurveisshownin i {_-PrtupcoiFig.23-11(b). Thetwincoilsareinseries andarewound asinthefigure so f h-T ‘astomagnetize therodsinopposite directions. Thecurrent through these f re |coilsissufficient tocarry thematerial through acomplete hysteresis loop. 1byIntheabsence ofanexternal fieldH.,themagnetic fluxesthrough the WY |rodscancel,andV=0. fp<a (a)Sketch(¢)andV(t)foreachrodwhenHy,=0. I | (b)Sketch thesame quantities forH.,#0. You will notice that ifthe }csslatroperatesasGequenyfthefundamentalequeeyofVb3 Ne{ Fig. 23-12. 8 A a ff cancelstheambientBandthecurrentinthesolenoidisthenameasure (CaofB. ) y Thepeaking striphasaratherlimitedrangeofapplications. (1)The g+) / solenoidhastobeatleastabout10centimeters longbecauseitmustbeat ay)on Li. _ leastafewtimeslongerthanthestrip,toavoidexcessive endeffects.But nl(0ee Jif =F thelength ofthestrip must bemuch larger than itsdiameter, again to ‘x y | signal proportionately. (2)Theambient Bcannot belarger thanafewo— TP) /\ hundredths ofatesla,forotherwise thepowerdissipated inthesolenoidCy JS) becomes excessive, (3)Ifonemeasures Bintheneighborhood ofa U pole-piece, thefieldofthesolenoid alters thepermeability oftheiron locally bal » Calculate thepeakvoltage induced inthepickup coilunder thefollowing Fig.23-11. conditions: stripdiameter, 25micrometers; number ofturns inthepickup coil, 1000; maximum value of1,75,000; frequency, 60hertz; amplitude of ‘Thisfacilitates themeasurement because thedetector canbemade toreject thealternating H,7ampere-urns/meter. thefrequency f. “<8.(23.4) Measuringaresistivitywithoutcontacts, Flux-gate magnetometers canmeasure fields down toafewnanoteslas. Itisuseful tobeable tomeasure theresistivity ofasample without 27. (2 7 having tocement contacts toit.One method involves placingadiskofthe 7bayeratprrnaha tomeasureB.Itconsistsof@finewireof materialinsideasolenoidcarryinganalternatingcurrent,withthetwoaxes permalloy (seebelow) oriented inthedirection ofBwithasmall pickup parallel, andmeasuring thepower absorbed bythedisk.Thediskhasa Coilofafewthousand turns near thecenter, ontheaxisofawolenoid, te radius a,athickness s,andaconductivity 0.Themagnetic fieldisuniform, Fig.23-12 andB=B,,cosa. Weneglect themagnetic fieldoftheinduced currents “Tomeasure theambient B,thesolenoid carries adirectcurrent thatjust Wetherefore restit ourselves tolow-conductvity materialscancels B,plusasmal alternating current. Then theHontheaxisofthe Findtherelation between @andtheaverage dissipated power P. solenoid isthatofthealternating current, andthestrip goes through a \¥,(23.5) Theinduction linear accelerator hysteresisloopateverycycle. Figure23-13showsaschematicdiagramofaninductionlinearac- Withmolybdenum permalloy theloopisapproximately rectangular, and celerator. Itconsistsofaseriesofferritetoroidslinkedbytheionbeamand thevoltageinducedinthesmallcoilhastwosharppeaks,onepositiveand byone-turnloopsthatcarrylargepulsedcurrents. tonenegative, whichcanbeobservedonanoscilloscope. ‘Onesuchaccelerator comprises 200toroidsandaccelerates a10-When theoscilloscope sweep issynchronized withthealternating current kiloampere pulsed electron beam to50million electronvolts. Itstotallength inthesolenoid, thetwopeaks aresymmetric ifthetime-averaged Honthe issimeters, andthepulses are70nanometers wide axisofthesolenoid iszero. Then thesteady fieldofthesolenoid exactly Explain itsoperation qualitatively. 436, 24 oo.a (CHAPTER *o_o © MAGNETIC FIELDS VII Electric Circuits C.Mutual and Self-Inductance 241 MUTUALINDUCTANCE M437 = , 2411 THENEUMANN EQUATION 437 Example: THE MUTUAL INDUCTANCE BETWEEN TWO COAXIAL SOLENOIDS 499 M12 THESIGNOFM 440 a Fig.2313. Example: THESELF INDUCTANCE OFALONGSOLENOID 442 Example:THE SELFINDUCTANCE OF ATOROIDALCOIL442 23-10, (23.5) Amagnetometer thatuseseddy currents 321 TUE MPEDANCE OFANINDUCTOR ats Figure 23-14 shows theprinciple ofoperation ofamagnetometer thatcan eaeveecacten us measuremagnetic fieldsassmallas10-*tesla andupto10-*tesla. The 242 *aluminum platePturnsontheaxisAAintheambient fieldBythatwewish 0ee ee encMrR stetomeasure. Thefluctuating eddycurrentsinducedinPproducea 243THECOUPLING COEFFICIENT k446 fluctuating magnetic fluxthrough thefixedcoilC,which hasNturns, and Example: COAXIAL SOLENOIDS 447thevoltage Visameasure ofBy. 244SUMMARY =448 ‘Theplate is10millimeters square andiscemented inside theplastic rotor PROBLEMS 448 ofasmall airturbine that operates at1000 revolutions/second. The only metallic parts aretheplate and thecil ‘Anexact calculation ofVasafunction ofgeometry, ofw,andofBy ‘This chapter concerns theelectromotance induced inacircuit when its wouldbedificult: Butthisisunnecessary because mecanclrat the magnetic fluxlinkage changes. Thechange influxcanoccureitherinatrament withHelmhoite coils(Prob. 18-9) because ofachange inthecurrent flowing through thecircuit itselforin(a)HowdoesVvarywithB,andwithwSetwt=0whentheplateles : intheplane ofC currentsflowingelsewhereorbecauseofachangeinthegeometryufthe (b)What isthefrequency ofV? circuits. A) 24.1MUTUAL INDUCTANCE M ZA . 24.1.1 TheNeumann Equation cf InFig.24-1theactivecircuit@carriesacurrentJ,.Themagnitude flux ’| ©,thatoriginatesinaandlinksbis -a oly{dl \a Oy=$Asndly=p(Hb) a, (24-1)—_ I, ,\4aJor l=- Fig.23:14. whereristhedistancebetweentheelementsdl,anddl,.Thus 438 11MUTUAL INDUCTANCE 39 Ifcurrent /,issinusoidal, then a oe Vy=-joMipl. (24-6)oe pee »Mi ars| ‘Adevicecomprising twocircuitsdesignedtopossessmutualinductance ~~:{|WAufi istermedamutualinductor,oratransformer.ela | pan ae ‘TheNeumann equation isseldomusefulbecause thedoubleintegralises} £1 Vi difficult toevaluate, evenforsimple geometries. Thisisnotamatter for; Ht\ie"] LI concern, because mutual inductances areeasilymeasured withimpedance el Te7| 7 boridges.OnecanalsocalculateMyyfromtheratioVjtodl,/dt. - Se ‘TheNeumannequationisnonethelessinteresting.Itshowsthatmutual 7]a 7 inductance depends solelyonthegeometry ofthesystem. Wehada similar situation with respect tocapacitance. Also,wecaninterchange thesubscripts intheNeumannequation Fig.24-1.Theactivecircuitabearsacurrent[,thatincreases.Partofits withoutalteringthemutualinductance. Therefore ‘magneticfluxlinksthepassive circuitb.Theelectromotance 1,induced inbisin thedirection shown. Mus=Mog=M. 47) This issurprising because the circuits can have different shapes and oywialeffdladle «lifferentnumbersofturns.ThisisastrikingexampleofthereciprocityaPPMaan (24-2) theoremofSec.8.7.Thatis, where dl, dl,ifYya-M then v=—-M (248)M,,=efMerde Ie aOa TS (24-3)Example |THEMUTUAL INDUCTANCE BETWEEN TWO isthemutual inductance between thetwocircuits. Mutual inductance is COAXIAL SOLENOIDS expressed inwebers perampere, orinhenrys. ThisistheNeumann Inthethirdexample inSec.18.2.1wefoundthat,insidealongequation. solenoidwithN’turnspermeterandbearingacurrent/(ignoringIfthegeometry ismore complex thanthatofFig.24-1, theabove endeffects), reasoning stillapplies, except thattheflux,,becomes thefluxlinkage " 2:Aap(Sec.23.4.2),and 7 B=WN'l (249)Aas=Mal (4s) Weaasecondwindingoverthesolenoid,ainFig,242,and weassume thatbothwindings are long compared totheir common“Theelectromotance inducedinbbyach: diameter,inordertorenderendeffectsnegligible. yachange inJ,is First,weassume thatsolenoid a,ofradius Randnumber of dA, dl,dh turnsN,,bearsacurrent[,.Themagnetic fluxofathatlinksY=-“aMew=-1tes (24-5) solenoidb,ofthesameradiusandwithN,turns,isthen,fromthea thirdexample inSec,18.2.1, ‘The mutual inductance isusually constant. Then thelast N: cthy =Ry ' Tenteanates termonthe b= ARwyL, (24.10) ‘Therefore themutual inductance between twocircuits isIhenry ifa a , current changing attherateof1ampere/second inonecircuit induces an MagNie HeRRNN: (+11)electromotance of|voltintheother. 4 « 40 412SELEINDUCTANCE 1 441 rc oe ThesignofMisdefinedasfollows.Firstweselectarbitarypositive — directions forthecurrents inthe(wocircuits. Then Mispositive ifa nyNN wy positivecurrentinagivesinbafluxlinkageofthesamesignasthati 4) be)|— resultingfromapositivecurrentinb.SeeFig.24-3, \ \ § 24.2SELF-INDUCTANCE L Fig.242.Twocoaxial solenoids. Wehaveshown different radiiforlarity, but Asimple circuit carrying acurrent Jis,ofcourse, linked byitsown collbiswounddirectlyovera magneticflux,asinFig.24-4.Theratio Alternatively, weassume acurrent finsolenoid 6,Then L-4 (24-14) 7 Oy,=ARyh, (412) istermedtheself-inductance ofthecircuit,Asformutualinductance, Thisfluxlinksonly(6/a)N, turnsofcoila,sinceBfallsrapidly to selfinductance depends solely onthegeometry ofthecircuit andis zerobeyond theendofalongsolenoid. Thus measured inhenrys. Self-inductance isalways positive Ifthecurrent /inaself-inductance Lchanges, thenavoltage DNs_MoRENAN, My,=2MePae_HoNaNe (2413) ‘ I @ an q y=-B_-1E (2415) andMis=My,asexpected a dt Itisparadoxical that avarying current intheinner solenoid shouldInduceanelesromotance intheouterone,singeB~0 appears between itsterminals. Thenegative signmeansthatthisvoltageavalfomimsthetiederivativeofAnovBrandAcsot ‘posesthechangeincurrent.SeeFig.24-4. vanishoutside alongsolenoid, eventhough ¥XAiszero. ‘Therefore theself-inductance ofazero-resistance circuit is1henry ifthecurrent increases attherate of1ampere/second when thedifference 24.1.2 The Sign ofM ‘ofpotential applied between theterminals is1volt. Asarule,thesignofMisimmaterial. Thereareoccasions, however, "bothZandI:aretime-dependent, thenwhen phases matter, and then one requires thesign ofM.Also, the directions offorces andtorques between circuits depend onthesignof M,asweshall seeinChap. 26. t AA “NIT N - ‘ Fig.24-3.Convention forshowingthesignofMonacircuitdiagram.Ifthe -Positive directions chosen forthecurrents inthetwowindings aresuchthat Fig.24-4, Circuit linked byitsownflux.Ifincreases, thevoltage across L current enters through thewirecarrying adot,then Mispostive, aitishere opposes f 442 “MAGNETIC FIELDSvit 483 dA__d di_jdb oa. Wire Ve~The Da ba (24-16) FZ As arule, Lisconstant, é Team (||! ‘One can calculate aself-inductance, atleast inprinciple, from thes \ >=n Neumann equation (Sec. 24.1.1), with both lineintegrals running over Zz a7) << : thesamecircuit.Iftheconductor crosssectionisinfinitely small,thenthe | s PUAN)linkingfluxandtheself-inductance turnouttobeinfinite.Thisarises oy <—e UIfrom thefact that Btends toinfinity intheimmediate neighborhood of oe UD thewire. Theregion where Btends toinfinity isitself infinitely small, but 7 theflux tends toinfinity logarithmically. With currents distributed over % aninfinitely thin surface, asinthenext example, Bremains finite andL isalso finite: Fig.245. Toroidal coilofsquare cross section Self-inductance iseasily measured. Acircuit designed topossess self-inductance iscalled aninductor. 24.2.1 The Impedance ofanInductor Example|THESELF-INDUCTANCE OFALONGSOLENOID thevoltageVacrossanidealzero-resistance inductordetermines therate ofchange ofthe current flowing through it.Disregarding the 111sthelengthoftongsolenoid,WhenumeroftsandR negativesignofEq.24-15, itsradius, then A_N@_Nak*p(N/I_ woN'AR® v-ie (2421)1SeNONAR(NIDwoNARE 4.17) at 17 7 T ‘Theself-inductance ofashort solenoid issmaller byafactor 11Tisasinusoidal function oftime, then that isafunction ofthe ratio R/l Ahigh-permeability nonconducting magnetic core increases V=jolt (24-22) B, and hence the self inductance, by afactor ofabout 1.3(length/diameter)'” andtheimpedance ofaninductor is Example |THESELF-INDUCTANCE OFATOROIDAL COIL Z=joL (2423) TiemagneticOarentee ee Realinductorsarenotthatsimple.Unlessaninductorissuperconduct-sng, One must usually take itsresistance into account and write toNI 5Bon e418) Z=R+jol. (2424) Thus WNT (6 NIb a+b Increasing thefrequency increases theratiowL./R, andthislessens theCal Oedeleravee telativeimportance ofR.However,thisbringsinanotherphenomenon. NO N° 2a+b lecause ofthevoltage difference between turns, theinductor alsoactsaseeredeed (2420) 4capacitor, anditsequivalent circuitisthatofFig.24-6,whereCisits2 vray capacitance. Also, Rincreases with frequency because thecurrent Iftherelative permeability ofthecoreism,,thenthe flows closer andcloser tothesurface ofthewireasthefrequency self-inductance is1,times larger. increases. Thisistheskineffect. 48 44s L k ae .z\ \ | i} c v6 | Fig. 2446, The equivalent circuitofarealinductor.TheparametersR,L,andC ° depend onthegeometry andonthenature ofthematerials‘ Fig.24.8. Impedance Zinthecomplex plane. Thepower factor isequal tocos@ ‘The impedance ofareal inductor isthus ;For =0, Z=R, asonemust expect. Forlarge values of«,only its gaRti@obLGol)_Rtjob (2425 higherpowersremainandZ=1/(j@C): allthecurrentthenflowsR+joL+1/GoC) (R+joLl)jaC +1 +25) throughthestraycapacitance C,andtheinductoriscapacitive. Aste idstoinfinity, Ztendstozero.SeeFig.24-7 R+jol. R+ joL)(1 -@°LC -fen = ee Bie)(24-26) InductorsaresodesignedthatjoL,oratworstR+L,isagoodLC+Rio @*LCY’+(Racy approximation forZoveragivenfrequency range._RtjolL—C(R?+0°L?)| ,,Tele oe RX (2427) 24.2.2 ThePower Factor4 ‘The ratio R/|Z| isthepower factor ofanimpedance. This quantity is usually denoted byAandexpressed asapercentage. With Zplotted in ox0" thecomplex plane asinFig.24-8, A=cosg.If@is45°,then 2is70.7%. 95 x10 4 Example_| MINIMIZING LINE LOSSES Figure 24.9 shows aload 4 Z=Ris+JX=|Zlexpid, (24-28) Pax w | 00«1" , ° = ee . - yams x0 ° Fig.24-7. Locus oftheimpedance Z=R’+jXofacertain inductor asafunction offrequency. The parameters R,L,andCofFig.246are,respectively, Fig.249.AloadimpedanceZ=Rise+X,oFanadmittance 1000ohms,1.00millinenry, and100picofarads.Wehavedisregarded thefactthat Y=Goss+/B,connectedtoasourcethroughalineofresistanceRincreases withfrequency. Re.Thelineresistance isequaltotwicetheresistance ofone wire 446 swanene Hees vin 243THECOUPLING COEFFICIENT & “7 14 andthemutual inductance between thetwoloops is Y=s= pew (-i0) =G+jB (24-30) zZ\2Z\ -abbpPoaMu=teakewe=kuoLy. (2439) connected to asource Lk Likewise, V=Vpexpjot (2431) Myo =KroLy (24-40) through alineofresistance Rige |Z).Wewish tominimize the Now My=Myg=M,fromSec.24.1.1.Thus losses intheline, fora given line agiven V,,andagiven power dissipation intheload. M=kaskslalen (24-41) Defining V’asinthefigure, thecurrent is . oM= £|keokoe|'® (Lols)'? =k(Laly)', (24-42) 1aV'¥~V¥= VG+18), (432) (KL=Theakeel!?, (24-43) Thenthepowerintheloadis wherekisthecoupling coefficient forthetwoloops.Superposing thetwo Pos }ReVI"=|Re[VV"(G —/B)]=1V26=V3,G, (2433) loops makes kequal tounity. ThesignofkisthatofM. For coils ofarbitrary shapes and sizes, one replaces thefluxes bythe whilethepowerlostinthelineés fluxlinkages Mew MewWie=|Re(V~V')I*} =Re[URja)" (2434) ar eewe (24-44)=RueVE(G?+BY)=RwVinsG?+B’) (24-35) “ROW, V2.8) uw andEq,2440applic.Hower, ofcanwbelagerthan ‘Thepowerlostinthelineisminimum whenB=0.Then¢iszero ‘Themaximum possible magnitude of&isunity.Wecanprovethisand thepower factor cos@is100% statementeasilyfortwosolenoidsofthesamelengthandnearlythesame Statedotherwise, thecomponent ofthecurrent that isin radius, one inside theother. Then ka»andky,areboth maximum. Say quadrature (phase leadorlagof90°)withtheapplied voltage solenoid ahasN,turns, andsolenoid BhasNeturns. Then Yields zero useful power intheload, butnonetheless gives tisetoa power lossandtoavoltage drop intheline N, N, Electric motors areinductive loads. Therefore, inlarge kav=ay hoa=H (24-45)installations, oneconnects capacitor banks across theline, close to “ ° themotors, tomakeBequaltozero. Eitheroneoftheseratioscanbelargerthanunity,buttheirproduct is equaltounityandk=1. 24.3THE COUPLING COEFFICIENT k ‘InSec.24.2wesawthattheself-inductance ofacircuit tendstoinfinity ,asthewirediameter tendstozero.However, mutual inductance remains ConsideraloopofwireabearingacurrentJ,andlinkedbyitsownflux wellbehavedevenifbothcircuitsarefilamentary. Thereasonisthatthe ®...Another loopofwire6,nearby, intercepts afraction ofthisflux fluxclosetoathinwiredoesnotlinkanother circuit somedistance away. o.- Asthediameter ofthewiretendstozero,Ltendstoinfinity, ktendsto0=esPons (437) zero,andMremainsfinite where[kao]<1.Theself-inductance ofloopais Example|COAXIAL SOLENOIDS =< “Thecoanial solenoids oftheexample inSec.24.1.1 areofdifferentLas (24-38) lengths.Then,fromEq.24-42, 448 “MAGNETIC FIELDSvi 449 =—M__(b)"* 2 / / |coe () (24-46) / sar with MasinEq, 24-110 and theL’sasinEq. 24-17, on en ee 24.4 SUMMARY / / vig.2410, Themutualinductance Mbetween twocircuitsaandbisgivenbythe (a)Calculatethemutualinductance M.Neumann equation (b)Does theradius oftheshort coilaffect M? bo{[dl,-dly Notehowmuchmoredifficultitwouldbetocalculatethefluxlinkingthe My =F pp, (243) solenoid foragiven current intheshort coil4h, 242. (24.1) ‘The mutual inductance between atoroid andanaxial wire wherethelineintegrals runaround eachcircuit andristhedistance Alongstraight wireliesalongtheaxisofatoroidofNturns,majorbetween theclements dl,anddl,Because ofthes radiusa,andsquarecrosssectionofsideb,witha>>b.integral, ofthesymmetry ofthe Calculatethemutualinductance(a)assumingacurrent/inthewireand .M. (b)assumingacurrent/inthetoroid.l=Mie=Me ean 243.(24.1) ‘Themutual inductance between astraight wireandaloop IfAgsisthefluxthatoriginates inaandlinksb,then ‘AloopofwireofradiusRiscentered atadistance 2Rfromalongstraight wire. The wire isin theplane oftheloop.Mao=Mave (24-4) Calculatethemutualinductance ‘Theelectromotance induced in6bythecurrent inais 24-4.(24.1) Azero-mutual-inductance magnetic dipole pairAcertain device forgeophysical exploration comprises twoshort coils in ~—aN» yydle,dM thepositionshowninFig.24-10.Themanufacturer statesthatthemutualW= =-M—*-1,— (24-5) Le dt dt dt inductance iszero. Isthattrue? Theself-inductance ofacircuitis 24-5.(24.1)CurrenttransformerFigure 24-11 shows aside-look current transformer formeasuring large L-“ eats) currentpulses.Showthatforasingle-turn coil v=itin(248) where Aisthefluxlinkage when thecurrent is/.Theimpedance ofan x "\p=al di ideal inductor isjo. The coupling coefficient between twocircuits is M 1 k=" 7aa (24-46) ‘This coefficient takes thesign ofM,and itsmagnitude isatmost unity. The power factor ofanimpedance isR/|Z|. PROBLEMS 24-1. (241) ‘The mutual inductance betweenasolenoidandashortcoaxialcoil > ylang solenoid ofradiusRandNts permetercariesashortcilof Fig.2611. turns near its center. 450 *MAONENE FIELDSvu 451 Onecanobtain /(¢)withanintegrating circuit (Prob. 7-9). Q 24-6.(24.2) Aconducting shield forfluctuating magnetic fields aIts often necessary toshield instruments from stray magnetic fields. If theonly disturbing field isthat oftheearth, then onecansetupapair of Helmholtz coils (Prob. 18-9) tooppose theearth’s field. Ifthefield isstatic butnotuniform, then onemust useashield made ofhigh-permeabilty ¢ material. Multiple shields, oneinside theother, arebetter thanasingle & thick shield. Couldaconducting enclosure beagoodshieldagainstfluctuating q () »magnetic fields? The answer isyes, asweshall see, butonly atquite high Imagine asimple situation where theexternal magnetic fieldByis Euniform, with B..=Be..m¢xpjwt.Theshieldisalongtube,paralleltothe lines ofB,afewtimes longer than itsdiameter 2a,andafewtimes longer € than theshielded region Y Weassume that thecurrent induced intheshield isuniformly distributed throughout itsthickness 6.Inother words, wedisregard theskineffect Fig.24-12.(Sec. 29.1). Ifthisassumption isnotvalid, then theshielding isbetter than ‘ourcalculation would indicate ource now?(2)CalculatetheresistanceR’ofthetube,perunitlength,inthe Oia2anopaleeieredpatel2 azimuthal direction. Calculate L (b) Let B.,=B...xpjotbethevalueofBinsidethetube,awayfrom 2411,(242.1)Power-factor correctionwithfluorescentlamps theends. Find theratio B./B... 'Afluorescent lamp consists ofanevacuated glass tube containing mercury(©)Showthat,iftheskineffectisnegligible, thisratiocannotbesmaller vaporandcoatedontheinsidewithafluorescent mixture.Adischargethan0.5. Aconducting enclosure therefore actsasashield onlythrough the ‘occurs between electrodes situated ateach end. Thedischarge emits most skin effect. ofitsenergy at253.7nanometers, intheultraviolet. The fluorescent coating absorbs thisradiation andreemits visible light. 24-7.(24.2) Inductance andreluctance . ‘Thedischarge operates correctly onlywhenitisconnected inserieswith‘AnN-turn coillinks magnetic circuit. ShowthatL=N°. animpedance. Aresistor would dissipate energy. soitisthecustom touse 24-8.(24.2)TheMaxwellbridge aseriesinductor. Severaltypesofcircuitareinuse,SeeProb. 7-12. Figure 24-12 showsaMaxwellbridge.Thiscircuitserves Oneparticularfluorescent fixtureoperatesat120-voltsanddissipates tomeasuretheinductance Landtheresistance Rofaninductor.One 80watts.Itspowerfactoris50%adjusts thevalues ofR,,R.,Ra,andCuntil Vequals zero, (a)Find thereactive current.FindLandRin terms oftheother components. (b)What isthesize ofthecapacitor connected inparallel with the discharge tubeanditsinductor thatwillmakethepowerfactorequalto 24-9, (24.2) Electromagnet operating onalternating current 100%? ‘Anelectromagnet with avariable gap length operates onalternating current. How does therms value ofthemagnetic flux depend onthegap length, foragiven applied voltage, andneglecting leakage flux? 24-10, (242) Power-factor correction Aloadisinductive, hasapowerfactorof65%,anddrawsacurrent of 100amperes at600voits. (a)Calculate the magnitude ofZ,itsphase angle, and itsreal and imaginary parts. (b)Calculate thein-phase and quadrature components ofthecurrent. (©)What size capacitor should beplaced inparallel with the load to cancel the reactive current at60hertz? 25.1 CIRCUFTS COMPRISING SELFINDUCTANCES 453 25.1 CIRCUITS COMPRISING SELF-INDUCTANCES .25 Ifacircuitcomprisesself-inductance—we excludemutualinductance for(CHAPTER themoment—then boththeKirchhoff currentlaw(KCL)andtheKirchhoff voltage law(KVL) (Sec. 7.8) apply aspreviously, thevoltage * dropacrossaninductorbeingLdl/dt.MAGNETIC FIELDS VIII Forsinusoidal currents, theimpedance ofaninductor isjo,andthe Electric Circuits D: Inductive Circuits and delta-star transformations ofSec.8.9apply. Ohm’s lawbecomes /=V/Z. Transformers Example |SERIES RESONANCE. InFig. 25-1, asource ofalternating voltage feeds aseries LRC circuit. Then I= V/Z, where25.1CIRCUITS COMPRISING SELF-NDUCTANCES 453 : :Example: SERIES RESONANCE 453 - 1Empl:PARALLELRESONANCE. 4S 2-Rei(al-Ze)=R+ial(\-Fre)- OD 25.2CIRCUITS COMPRISING SELF-INDUCTANCE ANDMUTUAL ‘AtresonanceINDUCTANCE 457 , Example: SIMPLE CIRCUIT COMPRISING ATRANSFORMER 457 1 » 25.3. TRANSFORMATION OFAMUTUALINDUCTANCE 458 ab=Ze oFwhC=h oe) Example: SOLVING THE TRANSFORMER CIRCUIT IN ANOTHER WAY 459 ‘Theimpedances oftheinductance andofthecapacitance then25.4MAGNETIC-CORE TRANSFORMERS 460 cancel,ZisrealandequaltoR,and[=V/R.Theresonance 25.4.1THEIDEALTRANSFORMER 461 sircularfrequency is25.4.2THERATIOV/V; 461 o-— (25-3)254.3 THERATIO L/L, 462 To ee RAMONE wsZo408 Itisconvenient tousethetwodimensionless numbers Example:THEAUTOTRANSFORMER 463 o (LI)!_ 25.4.6 POWER TRANSFER FROM SOURCETOLOADTHROUGHA aaaa (54) TRANSFORMER 464 255SUMMARY —465PROBLEMS 466 Thisisthelastchapter onelectric circuits. Westillhavetodiscuss howto : apply Kirchhoff’s laws tocircuits comprising self- and mutual induc- tances. Self-inductance iseasy. Mutual inductance islesssimple because , itconsists ofaninteraction between branches. However, with either one oftwosimple transformations, theapplication ofKirchhoff’s laws « becomes straightforward. Acareful discussion ofmagnetic-core transformers iswell beyond the c scopeofthisbook.Wetherefore limitourselves toacrude,butuseful, T° Fis:25:1,Series-resonant circuitconnected approximation called theideal transformer. - ‘0aSoureeofalernating vonage 454 “MACHETIC FEEDS VI 25.1CIRCUITS COMPRISING SELF-INDUCTANCES 455 Thesecond oneisameasure ofthe“quality” ofacircuit: the Figure 25-2(a) showscurvesofR/\Z)=R|Y}asafunctionofo" highertheQ,thelowerthedissipation.’ Then forvarious valuesofQ.Bydefinition, thewidthoftheresonance peakisthedifference Afbetweenthefrequencies forwhich Zz 1 zat+i0(o-3) (25-5) R41° p= 51a =0.7071 Stizi72? 1. (25-6) 0, and LapaL 25.7)YG cc |ont Thehigher theQ,thesharper theresonance peak.a Thephase angleofZis Kos 3 ¢=arcan[o(o’-)| (25-8) andvaries from 2/2 to2/2, asinFig.25-2(b) ook est TLL SS | Example |PARALLEL RESONANCE o Figure 25-3showsaparallel-resonant circuit fedbyasource V. ‘There isaresistance Rinseries with L,butnone inseries with C because realinductors arelossy, whereas realcapacitors arenearly lossless. Here je 1, R=job sonYnjoC +gyjop=i+paes (25.9) andafter afairamount ofalgebra, wefindthat _fw?14a RI-[ee(25-10) _| ; Fig. 25-2. (a)R/|Z| asa function ofw'=«w/c, forthecircuit ofFig.25-1 andfor various values ofQ.(b)ThephaseofZasafunctionof«'forthesamevalues ofQO. | “ — a oe *Exil1.Greenhasrecounted“TheStoryofQinTheAmericanScientist,vol.43,p 584(1955). Fig.25-3.Parallel-resonant circuit. so 252CIRCUITSCOMPRISING SELFINDUCTANCE ANDMUTUALINDUCTANCE 457 ow 25.2 CIRCUITS COMPRISING SELF-INDUCTANCE AND MUTUAL INDUCTANCE vw Thereexistsamutualinductance between branches aandbinthecircuitofFig. 25-5(a). Then, toapply Kirchhoff’s laws, one adds avoltage , source —jwMI, inbranch a,asinFig. 25-5(b), and avoltage source 2 > —jwMI, inbranch b.One chooses thesign ofMasinSec. 24.1.2. If > branches @andbhave acommon terminal, then onecanalsoproceed as inSec. 25.3. i ‘ Weusesigns andarrows oncircuit diagrams, aswith direct currents. Then weapply Kirchoff’s laws aswewould with direct currents. beT) ' Ci Example |SIMPLE CIRCUIT COMPRISING A “ TRANSFORMER Figure 25-6(a) shows atransformer fedbyasource Vandfeeding aload resistance. Figure 25-6(b) shows acircuit that isequivalent oe butwithout mutual inductance. The Kirchhoff voltage lawyields (7Vd twoequations, onefortheprimarymeshandoneforthe secondary: ‘ / Vi-joMl-Z,h=0, -jwMl,~Z:h=0. (25-11) e-1/ Here2,istheimpedanceR,+jwL,oftheprimarywinding,and \ <j 7 Z,=R,+jel,istheimpedance ofthecompletesecondary circuit. -~“%_ =eM 5.19) ["“7eeMTZ ""~Zesem 2) ” Fig.25-4. (a)|Z|/R asafunction of«forthecircuit ofFig.25-3 \ ‘ ‘fotvariousvaluesofQ.(b)Thephase¢ofZasafunctionof«forthesamevaluesofQ. ” =!w=o3)\h 2 [2 y \ Figure25-4(a)showstheinverseofthisquantity,namely|Z|/R, ‘| \, att, (Yat asafunctionof«'forvariousvaluesofQ.Themaximum occurs A = f+ slightly below «'=1.Figure 25-4(b) shows thephase angle @of Bydefinition, thewidthoftheresonance peakiaginthe , 7 iweenthetwofrequencieswhere|Z|/Risequalto 2"or0071sandAf=.Thisrelationnowapproxmate Fig.258.(a)Partofactuinwhchthereexistsamutual Teasvalidwithin 1%for Q'=1.7. inductance Mbetween twobranches. (b)Equivalent circuit 458 25.) TRANSFORMATION OF AMUTUAL INDUCTANCE 459 h P 4 b Since thetwo circuits are equivalent, they carry thesame mesh — — currents p,q,r.Weassume thatthecurrents aresinusoidal, andwethus | usephasors. A * . 22 2 InFig.25-7(a), ¥ 2G) wu z = Go .z Va=(p~4)Z, +joM(r—q). (25-14) ‘ 5 . ww. ww ‘Thesecond termontherightisthevoltage induced inbranch AC.We mi have assumed thatclockwise currents arepositive. ForthesignofM,sec Sec. 24.1.2. Also, “ " Vy=(q-1)Z:—joM(p~4). (25-15) Fig. 25-6. (a)Transformer feedingaresistance.(b)Equivalentcircuit InFig.25.70, Va=(P~Q)Z4+(P—riZe, (25-16) Thecurrent intheprimary isthesame asifoneremoved the Vp=(q-")Zy+(p—NZc (25-17) secondary and added thereflected impedance w°M"/Z, inthe primary. Sotheinputimpedance is After equating thetwovalues ofV4,thenthetwovalues ofVp,and simplifying, wefindthat Z.-Maz eM (25-13) 2-2-9 7 “hZz P(Z,~Z,—Ze)+q(Za—Z,—joM)+(Ze+joM)=0, (25-18) P(Zc+ j@M) +q(Zy —Za~jwM) +r(Z—Zy—Zc)=0. (25-19) 25.3 TRANSFORMATION OF A MUTUAL INDUCTANCE These equations arevalid foranysetofarbitrary values ofp,q,r.Then thesixparenthetical expressions arezero, and Figure 25-7(a) shows @mutual inductance Mwithitstwocoilsconnected bez wi yee; yeatC.Wecantransform thismutual inductance totheequivalent starof Z,=Z,+joM, Zy=Z,+jwM, Ze=—jaM. (25-20) Fig.25-7(b) inthefollowing way.Assume thatCisatground potential IfMispositive, thenthereactance —jwMiscapacitive. IfMisnegative, JoM isthereactance ofapure self-inductance |M|. A"8 12aNteryn 4 Example |SOLVING THETRANSFORMER CIRCUIT INANOTHER WAY 4 ‘Thecircuit ofFig.25-8(a) isthesame asthatofFig.25-6(a). We > >a nowfind4,h,andZ,bytransforming themutual inductance.d a" “ , NotethatZ:isthesecondary impedance plusR,..Thisleadstothe equivalent circuit ofFig. 25-8(b) ‘Applying the Kirchhoff voltage law tothe two meshes in < . succession, fay (by V,~(2,+j@M)h, ~(—jaM)(1, —L)=0,(25-21) Fig.25-7.Transformation ofamutualinductance. (a)Theactualcircuit.Observe (~j@M\(I; ~h)~(Z.+joM k=0.(25-22)that, forthis transformation, thetwo inductors must have one common terminal (b)Intheequivalent circuitZ,=Z,+jwM,Zp=Z2+jwM,Ze=~joM. ‘Theseequations yieldEqs.25-12. ‘60 25.4MAGNETIC.CORE TRANSFORMERS 461 2+yu =Ry)tw turns. (2)Themagnetic fluxthrough thesecondary isnearly equal tothat u / OOO OOO through theprimary because themagnetic flux follows thecore. Thus the mo coupling coefficient isclose tounity andtwosuchtransformers can \ ‘operate close together with little interaction. “®) 230wQ 2 = Bjan|a ‘Thecoredesignminimizes theeffectsofeddycurrents. (SeethegS 3 example ofSec. 23.5.) wh The analysis ofmagnetic-core transformers isdifficult forseveral reasons. First, the relationship between Band Hinferromagnetic ome materials isnotlinear (Sec. 21.2). Forexample, ifthefluxlinkage ina w » circuit isAwhen itcarries acurrent J,then theself-inductance LisA/1, . asinSec.24.2.Iftherearenoferromagnetic materialsinthefield,Ais F2580 aneypuveontces 2matofthe proportional toJandLdependssolelyonthegeometryofthecircuit.However, inthepresence ofmagnetic materials, Aisnotproportional to 1,andthevalue oftheself-inductance L=A/Icanonly beapproximate. 25.4 MAGNETIC-CORE TRANSFORMERS Moreover, thelosses inamagnetic-core transformer arecomplex: there areeddy-current losses (Sec. 23.5), both intheiron core andinthe Magnetic-core transformers serve atfrequencies ranging from afewhertz copper windings, hysteresis losses intheironcore(Sec. 21.2), andJoule toabout Imegahertz. Figure 25-9shows onecommon typethatisused at losses resulting from thecurrents flowing inthewindings. Allthese losses60hertz. canbeexpressed asan/°Rlossintheprimarybut,foragiven ‘Amagnetic-core transformer possesses twoessential features. (1)Fora transformer, Rdepends onthevoltage applied totheprimary, onthe given cross section, themagnetic fluxperampere-turn intheprimary is current drawn from thesecondary, andonthefrequency. larger than with anair-core transformer byafeworders ofmagnitude. This permits thedesign oftransformers thataresmaller andhave fewer 25.4.1 The Ideal Transformer Theideal transformer isacoarse, butuseful, approximation. Wemake 19 thefollowing assumptions. (1)There arenoJoule oreddy-current losses. ene (2)Thehysteresis loop forthecore isastraight linethrough theorigin. y/=il ThenBisproportional toH,andtherearenohysteresis losseseither.(3) a Allthemagnetic fluxisconfined tothecore.ThenthecouplingWY] coefficientisequaltounity,andthefluxthroughtheprimaryisequaltopi g thatthroughthesecondary.M Asaconsequence ofthefirst two assumptions, thetransformer isNi lossless.Theefficiencyis,infact,closeto100%forlargetransformers, an butonlyoftheorderof75%forsmallpowertransformers supplying tens ofwatts, dd ‘The assumption that Bisproportional toH,and hence toJ,isnot @ tb) realistic, but itisdifficult toavoid. smontypeofmagnetic-core tr foruseat60hertz surroundthecenterleg:Thecorecomprises¢{wotypesofamination, oneshaped Asatule,theloadconnected tothesecondary terminals isaresistance, 462 “MAGNETIC FIELDSvit 25.4MAGNENCCORETRANSFORMERS pn With anideal transformer thevoltage V,applied totheprimary is 25.4.4 TheInput Impedance Z, proportional tothemagnetic fluxinthecore:- From theexample inSec.25.2, withM?= LL, R,~0, R:~ Ry adae Vi=N,=jon. (25-23) PLL:dt oq=jk+OEe joy(1So 25-2 Revjets tal R,jal) (25-29) ‘The voltage across thesecondary is=Riots big (Re, els)=n,42 - “R,+Jol,L,®wl (25-30) i=, =Mj =bRr, (25-24) Then MywhereN,andNyarethenumbers ofturnsintheprimary andinthe Zu=(E)R (RiKol,), (25-31) secondary windings, respectively, andfisthecurrent inthesecondary. "‘Thus voN andtheinputimpedance isreal.voN (25-25) V, Np 25.4.5 The Ratio h/l, Inanideal transformer andforagiven V,,V;isindependent ofthe Thevoltage induced inthesecondary isequal tojwMI,. Therefore load current. Also V,iseither inphase with V,or=radians outofphase.Ofcourse, onecanchange thephaseofthevoltage onR,byradians by JoMl, =(Ry+jola)hs, (25-32)interchanging theconnections tothesecondary winding. bkjoM=M_(Ly)Ny,NotethatV;isequaltoN,j«@®.Thus,foragivenappliedvoltage,the 1R,+jol, La(2)"he (25-33) magnetic fluxisindependent ofthecurrent drawn from thesecondary.. “ * ‘Also, ®isBtimes thecross section ofthecore. Since themaximum Thus thetransformer islossless, asweassumed atthebeginning: possible value forBdepends onthecorematerial, Eq.25-67 shows that, WM=bV foragiven V,,anincrease infrequency permits theuseofasmaller , andhence ofacoreofsmaller cross section. Fxample_| THE AUTOTRANSFORMER . Figure25-10(a) showsaschematic diagram ofanautotransform. 25.4.3TheRatio Li/L> Itssinglewinding servesasbothprimary andsecondary, "The Let®,bethemagnetic fluxwhen /;iszero,and,similarly, let®,bethe winding isoftenwound overatoroidal core. magnetic fluxwhen f,iszero. Thereluctance (Sec. 21.4) ofthecore is2. Then DMR _Ni 2ee 5-26) ° Sut CS eta a (25.26) 3. : Similarly, © Ils nT) “s L.=of (25-27) -* .and ©: ©¥ Li_(M)?ix) (25-28) o o Theprinciple ofsuperposition applies because, byhypothesis, B/H is «constant. So®=,+, i,25-10(a)Schematic diagram ofanautotransformer. (b)Theequivalent 464 *MAGNETIC FIELDSVit 465 Letuscalculate theratio V;/V,. Weassume acoupling h 4 coefficient ofunity, and wedisregard the resistance ofthe 3 winding. These assumptions arereasonable inpractice. Letthe S 4 numbers ofturns oneither sideofthetapbeNyandNz,with k =corresponding inductances L,andLsasinFig.25-10(b). Then, if . $ R, Alisaconstant ofproportionality, 3 R CS (%)n L)=AN3, L:=ANS, M=(LyL2)'7=ANNs. (25-34) E M2Wese "&) 3 7@) Vs_VaV! 5 =Para (25-35) r where = oe543 (25.36) s °Vi ZjoM Fig,25-11. (a)Transformer Tinserted between asource ofinternal resistance R, and andaloadresistanceR,..(b)Equivalentcircuit (Z=joM)jo(Ls+M) ViZ=jwM+jo(L.+M) (53 ; ;— Vi jamesM)ioMy. replacedbytheresistance (N;/Nz)°R,, asinFig.25-11(b). RememberZ~jwM+jo(L,+M)+ja(L,+h thatwithourapproximation thetransformer hasanefficiency of100%.Then, with atransformer asinFig. 25-16, thepower transfer is ‘After multiplying these tworatios, substituting thevalues of optimum when Ly,Ls,M,andsimplifying, weareleftwith yiM (25-38) ow\w) N,\R,)* (25:39) VON TN Thetransformer isthen saidtobeused forimpedance matching. When SoVzissimplyproportional tothatfractionofthewindingthat thiscondition applies, thepowerdissipation intheloadismaximum, but isspanned bythesecondary. Theoutput voltage iseven theefficiency issillonly50% 5independent oftheload impedance Z!Recall that wehave disregarded theresistance ofthewinding 25.4.6 Power Transfer fromSource toLoad 25.5SUMMARY ThroughaTransformer Kirchhoff’svoltageandcurrentlawsapplytocircuitscomprisingself- AswesawinSec.8.8,thepower dissipated inaloadismaximum when inductances. Thevoltage drop across aninductance isLdl/dt, orjwLI itsresistance R,,isequal totheoutput resistance ofthesource R,,and withsinusoidal currents. when X,=—X,.Asarule,thereactances arezero. Ifacircuitcomprises amutualinductance Mbetweentwobranches, Ifitisimpossible tovaryeither R,orR,,thenonecanstillachieve thenonecanapply Kirchhoff’s lawsbydisregarding Mandadding to optimum power transfer byinserting atransformer between source and cach branch avoltage source —jwM times thecurrent intheother load, asinFig. 25-11(a). Letusseehowthiscomesabout. branch,Ifthetwobrancheshaveonecommonterminal,onecan ‘Assumethatthetransformer hasamagneticcore,thattheapproxima- transformthemintoaYcircuit,asinFig.25-7. tions ofSec.25.4 aresatisfactory, andthatX,=X, =0.Then, from Sec. Inanair-core transformer, theinput impedance is 25.4.4, thetransformer hasaninput impedance of(N,/N2)*Rz ohms. In otherwords, thecurrent andthepower supplied bythesource are 2-240 osprecisely thesameasifthetransformer anditsloadresistance Ry,were Gnat (25-13) 466 *MAGNETIC FIELDSvit 467 where Z,istheimpedance oftheprimary winding and Z,isthe 4 impedance ofthecomplete secondary. Also, LA kh=-7 (25-12) k andV;is4times theload impedance Z, & ‘Theidealtransformer approximates, inacrude way,thebehavior ofa \ magnetic-core transformer.Wefoundthat T ‘Vi=joN,®, (25-23) h_Vs_™:ToynnN,(Rela) 2525),253) = Fig.25-13. DAY »20=(M) Ri,(Riots), (2531) 5-2,(25.1)ThemagneticenergystoredinaninductorNo) Avoltage source Visconnected through aswitchtoaninductor of .inductance Land resistance R.The switch closes at¢=0. Onecanachieve maximum power transfer between asource andaload ‘Show thatatanytimeTtheenergy thathasbeensupplied bythesource, byinterposing atransformer with aturns ratio N,/N:=(R,/Ry)". minustheenergydissipated intheresistance, isequaltothemagneticenergy PL/2. PROBLEMS 253. (25.1) RLcircuit Find thecurrent inthe inductance Lofthecircuit ofFig. 25-13. The switchclosesat1=0.Ifyouhavestudied Chap.8,usethesubstitution 25-1. (25.1) Impedance veore! ‘en Millmatrs theore (a)Calculate theimpedanceZofthecircuitshowninFig.25-12.Whatis ‘theoremandthenMillman’stheorem. thevalue ofZwhen (i)f=0,(ii)f>*? 254,(25.1) Thestar-delta transformation withaselfinductance (b)Calculate themagnitude andthephase angle ofZat1kilohertz. Section 8.9isaprerequisite forthisproblem. Show thatthestarandthe (c)Calculate theamplitude andthephase angle ofY=1/Zatthat deltaofFig.25-14 areequivalent at1kilohert frequency. di mil 25:5.Thecoefficient ofcoupling@tea thepowerdissipation whenthecurrentis100milliamperes, Thecoefficient ofcoupling kbetween twosingle-turn coilswasdefinedin*#(e)Cantheralpartofheimpedance becomenegative? Sec,24.3Ingeneral hyhySHOWthatKalk,=LL (8)Forwhat frequency ranges isthe circuit equivalent to(i)aresistor in ‘ . series with aninductor, (i)aresistor inseries with acapacitor? (g)Atwhat frequency isthecircuit equivalent toapure resistance? wt van ue wa nH)0 10011 sk SxWF wa smitHaaN ‘ ‘a © 2000.0 nn ’ Suk Fig.25-12. Fig.25-14, 468 PROBLEMS 469 2 25-11, (25.4.4) The reflected impedance ‘Show that apositive (inductive) reactance inthe secondary ofa transformer isequivalent toanegative (capacitive) reactance inthe primary, andinversely. 25-12, (25.4.4) Electromagnetic crack detectors and metal detectors Itispossible todetect cracks inmetallic objects asfollows. Ifthepart tobeexamined isplaced inthevicinity ofacoil fedwith alternating % Fig. 25-15. current, the inductance measured atthe coil terminals islowest when there are no cracks. Such instruments can detect cracks only 25-6, (25.2) Measurement ofthecoefficient ofcoupling k 10micrometers deep. Thecoilforms part ofaresonant circuit.’ Metal ‘Atransformer hasaprimary inductance L,,asecondary inductance L2, detectors operate similarly andamutual inductance M.The winding resistances arenegligible. Consider thefollowing simpler situation. Asingle-layer close-wound Show thatZ/Z.=1—K?, where Zand Z.aretheimpedances measured solenoid hasalength 1,aradius a,andNturns. Letuscalculate how its attheterminals oftheprimary, when thesecondary isshort-circuited and impedance changes when oneintroduces into thesolenoid athin brass when itisopen-circuited. tube ofwall thickness 6. When analternating current flowsinthesolenoid, thechanging 25-7, (28.3) Impedances inparallel, with mutual inductanceCaleulatetheimpedanceofthectcuitshowninFig.25-15, magneteuxinduoes2curentinthetube,whichthusacta aeconay 25-8, (25.4) Improving (?)iron-core transformers d@/dt, andhence jo, andhence .Thepresence ofthetube thus Inairon-core transformer, thewindings areoutside thecore, where Bis reduces theinductance atthesolenoid terminals. Theeffective inductance orders ofmagnitude smaller thaninside, Why notputthem inside? Ofthesolenoid decreases when theresistance ofthetubedecreases 25.9, (25.4) Eddy-current losses intransformer laminations Wedisregard theskineffect (Sec. 29.1) inthetubeandthestray Eddy-current losses inmagnetic cores areminimized byassembling them capacitance ofthecoil.Also, weset[>>asoastodisregardendeffects.from laminations. Consider acoreofrectangular cross section asinFig. ‘Thecoefficient ofcoupling isnearly equal tounity. 25-16. Theeddy-current lossisproportional to¥°/R, where Vis the (a)Calculate theresistance R,ofthewindingofthesolenoid.Setthe‘lectromotance induced around atypical current path such astheone conductivity ofcopper equal to0. shown byadashed curve. Theresistance isalsodifficult todefine, butitis (b)Calculate theimpedance Z,ofthesolenoid without thebrasstube. ofthe orderoftwicetheresistanceoftheupperhalf,or2a/{o(b/a)L] (©)CalculatetheresistanceR;ofthebrasstubeintheazimuthal (a)Show thatsplitting thecoreintonlaminations reduces eddy-current direction. Setitsradius equal toa,andcallitsconductivity o,. loses by.afactor of (4)Calculate itsinductance L;andimpedance Z:. {(b)Showthattheselossesincrease asthesquareofthefrequency. {6)Nowcalcsate theimpedance atthesolenoid terminals withthetubeinplace. 25-10,(25.4) Hysteresis losses (O)Calculate impedances, without andwiththebrass tube, whenHysteresis losses areproportional totheoperating frequency f,while N=1000, /=200millimeters, a=20.0millimeters, 6=0.5millimeter, tcy-current losesincrease asf°,a5wesawabove.Youaregiven& f=1000hert2, 9,=5.8%10"siemens/meter, 0,=1.6%10”siemens) umberoftransformerlaminations.Canyoudeviseanexperimentthat meter.NotehowthepresenceofthetubeincreasesR(moredissipation) wilpermityutoevalntetherelativeimportanceofthetwotypes dnddecreasesZ(lessfun), 13. (25.4.5) Soldering gun _ ‘Asoldering gun consists ofastep-down transformer that feeds alarge currentthroughalengthofcopperwire.Onetypedissipates100wattsin 4apiece ofcopper wire (o=5.8x10”siemens/meter) havingacross >S section of4millimeters? andalength of100millimeters.t Ce (a)FindVand/inthesecondary. 4 “ (b)Findthecurrent intheprimary ifitisfedat120volts, assuming an efficiencyof100%. bh ‘ceProb.1713inElectromagnetism: PrinciplesandApplicationsbythesameauthors Fig. 25-16. sndthesame publisher. 470 L t sn26 a tamoter c k - — MAGNETIC FIELDS IX Fig. 25-17.* Magnetic Energy andMacroscopic ae aneeeeaimedanceseenbythetransiter unENERGYSTORAGEINANINDUCTIVECIRCUIT.THEMAGNETIC canal toRe+ike ENERGY ¢,,EXPRESSED INTERMS OFLAND 472(a)UnderwhatconditionisX=0? Example 474 ,(jb)Then what isthevalue ofR/R,? 22.THEMAGNETIC ENERGY DENSITY €,EXPRESSED INTERMS OFJ (6)Calculate the values ofCand Lfor R=S0ohms and f= ANDA 474 14megahertz ifR/R,,must equal 12.5. {3THE MAGNETIC ENERGY DENSITY €),EXPRESSED INTERMS OF (a)Now plot R,,andX;asfunctions ofthefrequency between 13.5, ANDB 475 and 14.5megahertz Example: THE LONG SOLENOID 476 ‘This LCcircuit isinexpensive, compared toatransformer, butthe su THE SELF-INDUCTANCEOFAVOLUMEDISTRIBUTION OF impedance match applies onlyatthedesign frequency. CURRENT 477 +Example: THE SELF-INDUCTANCE OFACOAXIAL LINE 477 “445. THE FORCE BETWEEN TWO CURRENT-CARRYING CIRCUITS EXPRESSED INTERMS OFM,J,,ANDI, 478 “06 THE FORCE BETWEEN TWO CURRENT-CARRYING CIRCUITS EXPRESSED INTERMS OFTHE MAGNETIC ENERGY &, 481 Example: THE FORCE BETWEEN TWO LONG COAXIAL SOLENOIDS 481 "17 MAGNETIC FORCES AND LINES OFB482 “8MAGNETIC PRESSURE 483 Example: THE MAGNETIC PRESSURE INSIDE ALONG SOLENOID 484 “29 MAGNETIC FORCES WITHIN AN ISOLATED CIRCUIT 485 “14.10 MAGNETIC TORQUE 485 Example: THE MAGNETIC TORQUE EXERTED ONA CURRENT LOOP 485 ‘hal SUMMARY 486 PROBLEMS 487 this isthe last chapter dealing specifically with magnetic fields. It concerns energy and macroscopic forces. We shall find several expres- sions forthemagnetic energy stored inafield, andthen deduce theforces exerted onacurrent-carrying body situated inamagnetic field that originates elsewhere. Asinmost ofthematerial that wehave studied until now, werestrict ourselves tolow frequencies and tocurrents that result from themotion offree charges. an MAGNETIC FIELDS 1X 2h.) ENERGY STORAGE INAN INDUCTIVE CIRCUIT 413 26.1 ENERGY STORAGE IN AN INDUCTIVE vou CIRCUIT. THE MAGNETIC ENERGY @, UIDV| IE== gaIR=Vo (26-2) EXPRESSED IN TERMS OF LAND / Part ofthepower supplied bythesource serves toestablish themagnetic ‘The circuit ofFig.26-1 illustrates thestorage ofmagnetic energy in field, andtherestdissipates asheat. inductive circuits. The wire isofuniform cross section ofand uniform (2)Now setthesource voltage Voatzero andincrease itslowly. Then conductivity 0.Allmaterials arenonferromagnetic, andweassume that €,=1forthewire.Thecircuitisrigid E=-yvy 24 (26-3)(1)Thecurrent isconstant. Inside thewirethevolume charge density a iszero, atleast intheabsence ofavXBfield, and thewire nowliesinitsown 84/8t field J Therelation J=o£stillapplies, andbothJandEareuniform B=|PVi=5 (26-1) hroughout thewire,aspreviously. Thismeans thatthesurface chargesdistribute themselves soastomaintain auniform Einside,despitethe Iftherearenosharpbends, bothEandJareuniform. ThelinesofE, presence ofthe3A/3r term. inside, follow thewire,parallel toitsaxis. Inside thewiresthatgofromthesource tothecoil,Aisweakand ‘Atthesurface ofthewire, Ehasbothanormal andatangential E~=—DV. Inside thecoilwire,wehavethesameE,butitcomes partly component. Ifthewiteissituated inavacuum, thefreesurface charge tom=VV,whichpoints inthedirection ofJ,andpartlyfrom—3A/3t, density is€9E;,,where E,isthenormal component ofEjustoutside. Ifit whichpoints intheopposite direction becauseApointsinthedirectionof liesinadielectric, thenthefreesurface charge density isD,=€,€0E Jandincreases. ; ‘At-anypointinspace Eisproportional toV).Ofcourse, Ealso NowwehavejustseenthatJE=1R.Integrating Einsidethewirefromdepends onthegeometry ofthecircuit andontheneighboring objects. thepositive tothenegative terminal ofthesource, Inprinciple, onecould calculate VandEeverywhere from thesurface - aachargedensityallaroundthecircuit,sourceincluded, andonthe J(-rv-) -a=ir, (26-4)neighboring bodies.However, oisitselfafunctionofE. ta a IfListhelength ofthewire, sfitscross section, andRitsresistance - -3A1/(o24),then -fvyeat{iSpa=i. (26-5) L Sincethefirstintegral equalsVo, _— Z8ql / ~3A ap da/ Vy=IR+| “.dt=in+— .dl=IR+o4_\ _ fSa alA-dl=IR+S. (266) - ~~ \w_/ <—ext\gel WehaveusedthefactthatthelineintegralofAisequaltotheflux 7— snkage(Sec.19.1) vaA RON Atanyinstantthepowersuppliedbythesourceis / ¢fT =\ \ \ opath/ f-#-Y h\ Mo=FR+ID (26-7) eae Thefirst term ontheright isthepower dissipated asheat, and thesecond Fig.26-1. Coilfedbyabattery. 1stherateofincrease ofthemagnetic energy. Thus, if>,isthestored magnetic energy atagiven instant, Now consider theidentity dé,_,dA_d(Ll)_,ar ral [oA{2om2pAOy (26-8) {fyade= [sy SFAdv. 26- rel aaa ea (26-8) all Adw=]s a+|SAdv (26-16) ‘Theinductance Lisaconstant because wehave assumed thatthecircuit Thefirstintegral ontheright isequal to/dA/dt, aswesawabove isrigid andthatthere arenoferromagnetic materials inthevicinity. Similarly, thesecond integral isequal toAdi/dt, which isthesame as Clearly, €,,=0when/=0.Then 1!dA/dt iftheinductance isconstant. Therefore LP_IA dé, 1dbmaSaS: (26-9) Set [+Adv : 272 aad 2Adv (26-17) Ifwehavetwocircuitsaandb,then and 1én=3[s-Ade, 26-1 Gut Eps=Mae +InAs) (26-10) aya Cots) =UL(Lole+Mp)+In(Lnly+MI.)| (26-11) where,again,visthevolumeoftheconductor. Thisequation applies Uy 4+Ly +Mh, (26-12) onlyifthesource isoffinite size. Ifthere areseveral circuits, then theintegral runs over allofthem and Example |‘Themagneticenergystoredinalongsolenoidfollowsfromthe thevectorpotential inonecircuitisthesumoftheA’sofallthecircuits.value oftheselfinduetance thatwefound inthefirstexample in OnecanaddtoAanyquantity whose curliszero without affecting this Sec. 24.2 integral (Prob. 26-4). aEE_toRE 2613) ‘Themagneticenergydensityatapointcanthereforebetakentobe in uw“ Em=A. (26-19) 26.2 THE MAGNETIC ENERGY DENSITY €,EXPRESSED INTERMS OFJAND A 26.3 THE MAGNETIC ENERGY DENSITY &;, EXPRESSED INTERMS OF HAND B We can rewrite thetime derivative of&,asfollows. Since A= LI, loexpressthemagneticenergyintermsofHandB,weuseEq.26-9 HntA Zl)=1eif aAay andapplyittotheloopofFig.26-2.Theloopliesinahomogeneous, dt 2\ da dt dt JoOt 3A = . dl 26-14)[2at$Sr (26-14) eon hi“ Taf at =[xBa, (26-15) y SS \ar Ee } where fisthecross-sectional areaofthewire,Cisthecurvedefined by G pee y thewire, andvisthevolume ofthewire. Thevolume visfinite. > Nod Noteinpassing thatJ-(9A/8t) dvistheworkdoneinmoving the vig.262.Sine ‘oopofconduction charges situated intheelement ofvolume dvagainst the tvpicallineofH.Theopensurface,ofareasf,isboundedbyC.andits electric field —9A/8t during 1second. everywhere orthogonal toH. 6 MAGNETICFIELDS1x “84THESELEINDUCTANCE OFAVOLUMEDISTRIBUTION OFCURRENT 477 isotropic, linear, andstationary (HILS) magnetic medium. Thisexcludes "26.4 THE SELF-INDUCTANCE OFAVOLUME ferromagnetic media. From Ampere’s circuital law, DISTRIBUTION OF CURRENT I=¢Hdl, (26-20) WesawinSec.24.2thattheself-inductance ofacircuitcomprisinginfinitely thin wires isinfinite. Arealcircuit comprises conductors of where C’isanylineofH. finite cross section anditsself-inductance L,bydefinition, isproportional Also, letsfbetheareaofanyopen surface bounded bytheloopCand tothestored energy: orthogonal tothelines ofHandofB.Then 1WP=6,={av. (26-26) Aso=fpeas e621) Thus “ le 12 and \ Len[paw (2627)e.=1Aa=34 nalBast. (26-22)© *Example |THESELF-INDUCTANCE OFACOAXIAL LINE NowthelinesofHfandthesetofopensurfacesdefineacoordinate Assumetha istowenoughtosysteminwhichdi-dafisanelementofvolumewithdfanddsfboth panera esanruyAaleypliadipersaridparallel toH.Also, foreachelement dialong thechosen lineofH,one ‘conductors, andneglect endeffects. integrates over allthecorresponding surface. Since thefield extends to Wecalculate successively themagnetic energies perunitlength infinity, thisdouble integral isthevolume integral ofH-Boverallspace, ofthelineinregions 1,2,3,4asinFig.26-3,andthenwesetthe and sumequaltoL'/°/2, where L'istheinductance permeter. Lines :ofBare circles centered on the axis. <1fu“Bdv. (26-23) (1)Region1.FromAmpére’s circuitallaw,Secs.19.5and20.6, 2) attheradius p, ‘Themagnetic energy density innonferromagnetic media isthus 2xpB,=ol2 (26-28) H-B_B?_uH®gt BFe (26.24) andthemagnetic energyperunitlengthis “Themagnetic energy density varies asB°.Thus, aftersuperposing severalfields, thetotalfieldenergy isnotequaltothesumoftheindividual oo lyenergies. SeeEq.26-12 2 \ea ‘Compare thissection withSec.6.2. Z< , Example|THELONGSOLENOID |VeNeglectingendeffects,wefindthatthemagneticenergystoredin ig\\thefieldofalongsolenoidinairis \K»\sellton uninnnZRHME 62s) rex. |uo 2 a.ey asintheexample inSec.26.1. ~ Fig.26-3.Coaxial line. 478 MAGNETIC FIFLDS1X 479 2 f - tax[(feb)zapdp=r (26-29) 2poiNara a aa Wwecegion2.Here “S (2)Region2.Hes Yaar Tox,peeeee (26-30) 8 + K /a oe a : y i Sh — (3)Region 3.Werequire thenetcurrent thatflows within a _— \ : -circularpathofradiusp.Thisis/minusthatpartofthecurrentin Ue eS Lae .Ss theouterconductorthatflowsbetweenradiibandp.Thus (ai)1 altol(,p=bt)_molf=pt26-31 eT Bas(Ga)depoaorOOD wofcc 30-8| ray=Hol|in- 26-32) > toe ap eso) . ‘rom Ampére's circuital law, there iszero field in- wisreson 4FromAmpire’scrcl “ Hg,264.Twoparallelloopsbearingcurrents1,andswithtypicalinsof ‘onginatinginloo iementoffor componi Finally,theselF-inductance perunitlengthofthecoaxiallineis iatingioopa,TheelementofforcedFpossessesacomponentinthe Litblip etye38a8? Luge tgelnats,[Says wea ; nt2x a*Ial(@-bF 6Keb). 1move toward each other. They arefixed inposition byopposing(26-33) mechanical forces.Allmaterials arenonmagnetic.. Now assume asmall virtual translation (Sec. 6.6) ofone circuit,e rmbetweenthebraces from@}andis u .normalythemostimportantthebracescomesfrom#5a withoutanyrotation.Sincethereisconservation ofenergy,theenergy expended bythesources isequal totheincrease inmagnetic energy plus themechanical work done. The displacement takes place slowly soasto *26.5 THE FORCE BETWEEN avoid taking kinetic energy intoaccount. TWO CURRENT-CARRYING CIRCUITS ‘Tosimplify thecalculation, weassume thatthecurrents areconstant. EXPRESSED INTERMS OF M,I.,AND I, Thisassumption willnotaffect ourresult. Wehadasimilar situation in clectrostatics. See Prob. 6-12. Wealready found anintegral fortheforce between twocurrent-carrying Loop bmoves adistance drtoward loopa.Only Mchanges and,from circuits inSec. 22.3. However, aswenoted atthetime, theintegral is iq,26-12, themagnetic energy increases by difficult toevaluate. Hereweexpress theforceinterms ofthemutual inductance between dEq=IndydM=[,dAsa=IpdAyy, (26-34) thetwocircuits. Mutual inductance isjustasdifficult tocalculate asthe | : force,butitiseasytomeasure, muchmoresothantheforceitself.Inthe nwbeingthefluxoriginating inbandlinking a,andsimilarly forAy.process, weshallfindthatwhenever onechanges thegeometry ofa SinceMispositive (Sec.24.1.2)andincreases, dé,,ispositive.circuitorofapairofcircuits, precisely one-half oftheenergy furnished Nowconsider theextraenergy supplied bythesources. Inloopb,thebythesource. exclusive ofJouleandotherlosses, becomes magnetic linkingfluxincreases andtheinduced electromotance tendstogenerate aenergy. andtheotherhalfbecomes mechanical work magnetic fieldthatopposes thisincrease. Therefore theelectromotanceWeconsider twocircuits carrying currents [,andJinthesame induced inbtendstooppose [,.Tokeepthatcurrent constant, itssource direction, asinFig.26-4.Themagnetic forceissuchthattheloopstend supplies theextravoltage dA,,/dt andtheextraenergy dhe *26.6 THE FORCE BETWEEN TWO dy, =2dt=hydp=IolaAM. (26-35) a. — iy=dydt (26-35) CURRENT-CARRYING CIRCUITS EXPRESSED INTERMS OF THE MAGNETIC ENERGY %q Bysymmetry, dé,,isthesame, andtheextra energy supplied bythe twosources is Since theterm ontheright inEq. 21-37 istheincrease inmagneticdé,=21,1, dM, (26-36) energy, wecould alsowritethat which isexactly twicetheincrease inmagnetic energy. Theremainder has Fy-dr=dé, (26-42) gone intomechanical work. Inother words, theenergy supplied bythe sources divides equally between mechanical energy andmagnetic energy. remembering that theforce pulls inthedirection that increases the SeeSec. 6.6.1 magnetic energy. Also, IfEF,istheforcethatcoilaexerts oncoilb,thenthemechanical work 3%, done is Fane = (26-43) Ey+dr=1,1,dM. (26-37) 7 Example |THEFORCE BETWEEN Since thequantity ontheright ispositive, F,,points toward coila,like dr,which iscorrect. TWO LONG COAXIAL SOLENOIDS ‘Thexcomponent oftheforce is Figure 26-5shows twocoaxial solenoids, oneofwhich extends a aM distance/insidetheother.Themutualinductance ispositive.The Fe.=2, (26:38) netforceisaxial,anditisattractive,ascanbeseenfromFig.ox 26-6. Remember that theforce between twoparallel currents flowing inthesame direction isattractive where dxisthexcomponent ofdr.This equation applies toanypair of circuits. n N, Thisisanalternate expression fortheforcethatwefound inSec.22.3. > WN) NtWecanshowthatthetwoexpressionsaeequalasfollowsLetcetb gy)moveasawhole, without rotating, parallel tothex-axis. Then Ly By Ha MeL if)| oolWe G alo dl,-dl,\ ty 1 ox \4rJ,J, or whereristhedistance fromdl,todl,.Thederivative withrespect tox ig:2665.Twocoaxialsolenoids ofapproximately equaldiameters. TheforceFactsonlyonthe1/rtermundertheintegral signbecause thevectors dl thoeaneenanamoetSeeectcrn,Thecleaare arenotaffected byatranslation ofcircuit b.Thus Eye=~Hinggee (26-40) PTTONM SCHOOOE and, more generally, . Mo dl,-dly__ tho jiledly 96.4y ™tagO°i Fy—Hele|[ee—Tela|LF, 26-41) ©eo0% ODO000 , Fig.2646,Section through partofthesolenoids ofFig.26-5.Thereisaforceof asinSec, 22.3. attraction attheends ofthesolenoids. 482 MAGNETIC FIELDS1 83 ‘Wecannot perform arigorous calculation oftheforcebecause it : ;clearly depends ontheveryendeffect thatwehavedisregarded . ' ‘until now. However, wecanfindanapproximate expression by ' applying theformulas thatwefound above. '‘Weassume thatthefieldofasolenoid stopsabruptly attheend. / \ 'Then { q \ '| (CO KS) h 1 \ \ Foracross-sectional areast, ' Fh =yh2alhly- (26-46) Fig.26-7.(a)TheBfieldoftwoparallelcurrents flowinginthesamedirection, ley ‘The magnetic flax betweentwosuccessivelinesofBisconstant.Wehavenot shown thelines near thewires because they aretooclose together. (b)The B wsattractive bec increases with/ fieldoftwoparallel currents lowing inopposite directions THgeri aselaate thefocefom3/3: TheseinesofBareentaltotheeqpotentils inFig.67because,in Nowletus calculate the a two-dimensional field,alineofBisalineofconstantA(Prob.18-12),andboth Vand Avary as1over thedistance from thewire task[wav=E(wiL—9 +B+ B+Bll (26-47) Thisisageneral rule:linesofBare“under tension” and“repel 1 laterally,” justlikelines ofE.Seethenext section.=5g,(Bile+Bile+2BBul)st, (26-48) Ho "26.8 MAGNETIC PRESSURE where B,=oN'l, originates insolenoid a,andsimilarly for6. Then Ifthecurrentflowsthroughaconducting sheet,thenitisappropriate topattnBB ons (2649) thinkintermsofmagneticpressure. Imagineaconducting sheet,inair,“dy carrying @amperes/meter andsituated inauniform tangential magnetic field B/2 originating incurrents flowing elsewhere, asinFig. 26-8(a), Observe thattheforcewould bezeroif&,wereproportional toB. with@normal toB.Theforceperunitarcaisalf/2. Nowincreaseauntilitsfieldcancelstheambientfieldononesideand ‘ = .26-8(b). Then, from Prob, 19-3, * FORCES AND LINES OFB Joubles itontheother, asinFig.2 rc 26.7MAGNETIC a”=B/jo,andtheforceperunitarea,orthepressure, is Lines ofBareuseful forvisualizing magnetic forces between current- B carrying wires.Figure 26-7(a) showslinesofBfortwowirescarrying Pm (26-50) equalcurrents flowing inthesamedirection. ThelinesofBthatcrossthe midplane are“under tension,” andtheforceofattraction perunitareain Thisapplies toanycurrent sheetwithzeroBononeside. themidplane isB/(2uo). Figure 26-7(b) shows linesofBforcurrents Thepressure isequal totheenergy density, aswithelectric fields(Sec. flowing inopposite directions. NowthelinesofB“repel laterally,” and (6.5),Thispressure pushes thecurrent sheetawayfromthefield.Onthe theforceofrepulsion perunitareainthemidplane isB*/(2s) fieldside,thelinesofBareparallel tothesheetandrepellaterally. 484 (0MAGNETIC TORQUE 485 z S Thisisequaltothemechanical workperformed bythemagnetic aa— Pressurepoveranarea2xRIdR,and be4 B 2 | = 52)Gog: [s fi « arn (2552) en]| yy)y?in~ }if Ve ‘26.9MAGNETIC FORCES WITHIN‘ homed ANISOLATED CIRCUIT ‘| SS J a Within asingle isolated circuit onestillhasmagnetic forces, because the current inone part flows inthemagnetic field oftherest ofthecircuit. ‘ lorexample, ifthecircuit isasimple loop, then themagnetic force on thewire tends toexpand theloop. 26.10 MAGNETIC TORQUE Fig. 26-8. (2)Aconducting sheet carries acurrent density of«amperes/meter of ithandesnauniformmagneticeldB/2originatingelsewhere:(0)The ityanalogywithSec.26.5,acircuitathatformsanangle@withanother‘was,whilethetotalBonthefarsideiszero. circuitbexertsonbatorque 38M_36,T=Lh=-= (26-53) Example |THEMAGNETIC PRESSURE INSIDE 3030 (2653) ALONG SOLENOID Thetorquetendstoincrease boththemutualinductance Mandthe Figure26-9showsanendviewofasolenoid. Weusethemethod magnetic energy By. ofvirtual work (Sec. 6.6) toshow that themagnetic pressure is B?/(2). Imagine thatthecurrent remains constant, while the *magneticpressureincreasestheradiusfromRtoR+dR.Then, ‘-xample |THEMAGNETIC TORQUE EXERTED ONAforasolenoid oflength /,themagnetic energy increases by CURRENT LOOP B Arectangular loopofwirecarrying acurrent /liesinauniform B at=57DakdR (26-51) inair,asinFig.26-10.Wecalculatethetorqueattheangle@.° (1)Thesimplest procedure hereistocalculate thetorque from themagnetic force /dlxBontheelement dlofthewire,asin gotttitite, Sec. 22.2 Vs+BSMe T=2asin0Bib=Bidsin8,(26-54)fit wheresfistheareaoftheloop.Thetorqueisinthedirectionof iPsiilpeeythecurvedarrowshowninthefigure iiget aE ‘=>| (2)NowletususeEq.26-53.LetJ,betheunknown current west thatprovides thefieldB.Then % oe noe F Fig. 26-9. End view ofasolenoid. The dots\Aeens))featnesofB,gonatotheponer yaBose (2655)er a¥XYyy ‘Thearrowsshowhowthemagnetic Ia Se pressure pushes thewinding away from theSsaccsz Feld. WerequireanegativesignherebecausethefluxBstc0s@links 486 PROBLEMS: 487 whereuvisthevolumeoccupiedbythecurrent SN‘ ‘Thus theenergy density canbetaken tobe D = BNLA eH B= > 26-24oy } mu (26-24)\o _ a \ Wedefine theself-inductance ofarealcircuit comprising currents ~~ distributed over afinite volume interms ofthemagnetic energy stored in 2Yenc! i thefield:Fee 1DA L=3[Bde (26-27) JA749 Bae wlJ.2 BSps ,Z; ~ Thex-component oftheforce exerted onacircuit bsituated inthefield | ~ ofanothercircuit@is aM_3t,, Fig.26-10.Loopcarryingacurrent/inaconstantB.The ner re (268),(26-43)‘magnetic torquetendstoturntheloopinthedirection ofthearrow. ‘Theforcepullsinthedirection thatincreases bothMandthemagnetic theloop inthedirection opposite tothat ofthefluxofacurrent / energy. flowing intheloopasinthefigure. Then Similarly, thetorque isgiven by =14M14 = T=hoe=Son (26-53) T=LhSe=-15,(Bet00s8)=Blctsin8. (26-56) ly39=36 (26-53) Thepositive signmeans thatthetorque isinthedirection shown, Ifthecurrent flowing through aconducting sheet situated inamagnetic aswefound above. Thetorque tends toincrease @andhence to field issuch thatthemagnetic fluxdensity isBononesideandzero on increase M. theother, thenthemagnetic pressure onthesheet isB°/(2j4o). The pressure tendstopushthesheetawayfromthefield. 26.11 SUMMARY PROBLEMS Whenacurrent /flowsinacircuitofself-inductance L,themagnetic ao.(26:1)Th storedinamagneticcet 4 (2 eenergy it energystored inthefieldcanbeexpressed invarious ways: ‘Showthattheenergy storedinamagnetic circuitis®°R/2, where®is themagnetic flux and2isthereluctance. =P =A) [H-Bdv, (26-9),(26-23) 26-2.(26.1)Theenergytheoremforlinearpassivecircuits2), (a)Thistheorem follows fromTellegen’s theorem (Sec.8.6).Suppose cone hasapassive circuit comprising resistances, self-inductances, and where Aisthemagnetic fluxlinkage andwhere theintegration runs over capacitances. Oneapplies analternating voltage V,toaninput port. allspace.Also,forafinitecurrentdistribution, Showthat Volp=P+2e0(€mage~cians &=)[rad (26-18) wheretheleft-handsideistheinputcomplexpower,Pisthepower2H dissipated inthecircuit, Eup. istheaverage magnetic stored energy, and fastheaverageelectstoredenergy.Thisisteenergytheorem (3)Calculatethenumberofampereturns requiredineachcoilto(@)Itisshown inProb. 25-4thatthestarandthedeltaofFig.25-19 are} support amassofImetric tonwhen R=1meter equivalent. Now, ifoneapplies analternating voltage V(rms) between j (b)DrawasketchshowingthetwocoilsandlinesofB.Canyouexplain terminals BandCofthestar,then V/*isrealandequal toV*/2000. theforce ofrepulsion qualitatively’ According totheenergy theorem, theaverage energy stored inthe 26-7. (26.7) Theforce between twoparallel busbarsoffinite cross section. capacitors ofthedeltamustbeequal totheaverage energy stored inthe ‘Twoparallel busbarshaveequal circular crosssections andcarryequal inductor, atanyfrequency. Show thatthisiscorrect. currents J.Thecurrents areequally distributed overthecross sections. 263, (26.1) Theaverage stored energies incapacitors andininductors Show, without anycalculation, thattheforce isthesame asifthebus ‘Theaverage stored energies incapacitors andininductors are CV?/2and barswerethinwires. LF/2, respectively, where Vand/arermsvalues. 26-8. (26.7) Asuperconducting power transmission line ‘Show that Asuperconducting depower transmission linehasbeen proposed thatae enoe wouldcarry100gigawattsofpowerat200kilovolts over1000kilometers.ew=F Cow=FOaL ‘Theconductors wouldhaveadiameterof25millimetersandbeseparated byacenter-to-center distance of50millimeters. 26-4. (26.2) The magnetic energy interms ofJandA (a)Calculatethemagneticforcepermeter.Seethepreviousproblem.It Wesawthat isclearly preferable touseacoaxial line. eal (b)Calculate thestored energy inkilowatthours. Theself-inductance perbmn[F-Aav, meteris[4o/(42)][1 +4In(D/R)]. where vis anyvolume thatencloses alltheconductors 26-9. (26.8) Large-scale energy storage ininductors andincapacitorshereanywustatenoleconcn Muchworkhasbeendoneonthelarge-scalestorageofenergyininductors, forpublicutilities. Oneauthor proposes ahuge,underground, 26-5.(26.4) Theinductance ofacoaxial lineisslightly frequency-dependent cryogenized inductor thatwould operate atafieldof14teslas. High-frequency currents donot penetrate aconductor asdolow- (a)Calculate theenergy density inkilowatthours/meter: frequency currents. Thisistheskineffect (Sec. 29.1). Does theself- (b)Calculate themagnetic pressure inatmospheres.inductance ofacoaxiallineincreaseordecreasewithfrequency? (©)Itseemsmorereasonabletostoreenergyinacapacitor,becausea 26-6.(26.5) Theelectromagnetic levitation ofhigh-speed tracked vehicles capacitor neednotbecryogenized andbecause theforcepoints inward, not Thesuspension andthepropulsion oftracked vehicles become major outward asinaninductor. Calculate theenergy density inkilowatthours/ problems atspeeds ofseveral hundred kilometers perhour. Wheels are meter’ if¢,=3 andthedielectric strength is1.5x10*volts/meter. then impractical because vehicle vibration, track damage, andpower loss Gasoline canstore over 100kilowatthours/meter’, andflywheels over become excessive. The tractive force also deteriorates with increasing 200, speed. 26-10.(26.9) Themechanical workperformed bymechanical forcesonan Anaircushion provides asatisfactory suspension athighspeeds, butit isolated, active, anddeformable circuit consumes alarge amount ofpower. Propulsion thenrequires either a Weshowed inSec.26.5that,ifoneactive circuit moves withrespect to propeller oralinear electric motor, withthestatorinthetrack. another, themechanical workperformed bythesources isequaltotheItisalsopossible tosupportavehiclebymeansofmagneticforces,and increaseinmagneticenergyifthecurrentsaremaintained constant. severalmethods havebeendeveloped. Inoneofthese,superconducting Hencetheforcebetween twoactivecircuitsisgivenbytherateof coilsinthevehicle generate amagnetic fieldthatextends down intothe increase ofmagnetic energy. track, which isasheetofaluminum. Atrestandatlowspeeds, thevehicle ‘Show that, similarly, ifthegeometry ofanisolated active circuit useswheels. Asthespeedincreases, theeddycurrents induced inthetrack changes, theenergy supplied bythesources divides inthesameway.bythetravelingmagneticfieldexertaforceofrepulsiononthecurrentsin ‘Assumeagainthatthecurrentisconstant.Itfollowsthat,onthis thevehicle coils,andthevehicle fliesabout 10centimeters above thetrack. assumption, theforceonanelement ofanactive circuit isequaltothe‘There are,ofcourse, problems ofstability. Also, thesuspension isnot rateofincrease ofmagnetic energy.TosecounthereareJouleyossesithetrack‘Apairofparalleland 26-11.(26.9)Theaxialcompression forceonasolenoid. coanial colsofradius Rand Nturns areseparated byadistance D.The {@)Show qualitatively, intwodifferent ways, thattheturns ofa lower coilsimulates thetrack. ForD~0.1R, themutual inductance is solenoid tend tosqueeze together. Gren byN(2154.— 1204{(D/R) —O.I)R micohenns, (b)Calculate theaxialcompression force onalongsolenoid. 26-12. (26.9) Magnetic shutterMagneticfieldscanperformmechanicaltasksthatrequireahighpowerj _—levelforaveryshort time. Forexample, magnetic pressure cancrush a z ~ lightaluminum tubethatactsasashutter toturnoffabeamofightorof — —> softx-rays. The tube isplaced inside acoil, parallel totheaxis. When 7 =e thecoilissuddenly connected toalarge capacitor, thechange influx: induces alargecurrent inthetube,whichcollapses underthemagnetic — waepressure _ i Letuscalculate thepressure. IfthecurrentJinthesolenoid increases a themagnetic pressure isnegligible. Letusassume thatdi/dtinthecoilis “ Fig.26-11. solarge thattheinduced current inthetube maintains zero magnetic field inside it.Then there isamagnetic field Bonly intheannular region between thesolenoid andtheconducting tube. field. Themagnetic particles cling tothesteel wires where thefield (a)Calculate thepressure onthetubeinatmospheres at1tesla gradient islarge. Arrays offinesteelwiresnormal toBarealsoused. (b)What would bethepressure iftheconducting tubewereparallel to With @fieldoftheorder ofseveral teslas supplied bysuperconducting theaxisbutofftheaxis? coils, theseparation occurs even with materials that areonly slightly magnetic. The method isalso applicable inairforremoving magnetic26-13,(26.9)|Fluxcompression particles,sayinpulverizedcoal. sme Fluxcompression isonemethod ofobtaining large magnetic fields. For Letusseehowasmall magnetic dipole behaves inanonuniform B. example, onecaninsert alightconducting tube inthefieldBofa Thedipole firstorients itself. Then, asweshallsee,ittends tomove in solenoid andthenimplode thetubebymeans ofanannular explosive thedirection inwhich theapplied Bincreasescharge situated between thetubeandthesolenoid. Currents flowinthe Figure 2611showsasmallcurrent loopofradiusRthatisalready tube, andthemagnetic pressure builds upuntilitisequal totheexternal oriented inafieldBthatincreases. symmetrically about thepositive gaspressure. Thesolenoid isfedbyaconstant-current source Guection ofthes-axis (2)Show that,ifthe radius ofthetubeshrinks veryrapidly, the.B (a)Show, without anycalculation, thatthemagnetic force points toinsideisaboutB,(Ri/R*) attheinstantwhentheradiusisequaltoR.For theright.Notethatthisforcetendstoincreasethelinkingflux. example,ifBois1OteslasandifRe/R=10,thenB=10tesas (b)ShowthatF=2xRIB,, whereB,isthecomponent ofBthatis (b)Calculate thesurface current density inthetubeinamperes/meter. noreal totheaude » » ()Caleulate thechange inmagnetic energy, theenergy absorbed by (©)Nowconsider asmall volume ofthickness Az,asinthefigure. Use theconstantcurent sourcefesding thesolenoid, andtheexplosive thefactthatthenetoutward fluxofBiszerotofindB,andF.energyrequiredtocompress thefield.Assumethatthetubeis ul rateofi magneticener200millimeters long, R=50millimeters, andneglectendeffects (9)CatealatethefreeFromtherateofincreaseofmagneticenersy 26-14. (26.10) Thetorque onacurrent-carryingcoil (a)Show thatacurrent-carrying coiltends toorient itself inamagnetic field insuch away that the total magnetic fux linking the coil is ‘maximum. (b)Show thatthetorque onthecoilismXB, where misthemagnetic moment ofthecoilandBisthemagnetic fluxdensity when thecurrent in the coil iszero. 26-15. (26.10) The torque onacylindrical permanent magnet Show that thetorque exerted onasmall, cylindrical permanent magnet cofdipole moment msituated in@magnetic field ismXB. See the preceding problem. 26-16. (26.10) High-gradient magnetic separation Itispossible toseparate magnetic particles insuspension inafluid by passing themixture through steel wool subjected toastrong magnetic 27.4 MAXWELL'S EQUATIONS INDIFFERENTIAL FORM 493 The above equations are general inthat the media can be nonhomogeneous, nonlinear,andnonisotropic. However,(1)theyapply CHAPTER‘only tomedia that arestationary with respect tothecoordinate axes," and (2)thecoordinate axes must notaccelerate and must notrotate. fThesearethefourfundamental equations ofelectromagnetism. They MAXWELL’S EQUATIONS H formasetofsimultaneous partial differential equations relating certain time and space derivatives atapoint tothecharge and current densities atthat point. They apply, whatever bethenumber ordiversity ofthe 27.1MAXWELL’S EQUATIONS INDIFFERENTIAL FORM 492 sources: 27.2MAXWELL'S EQUATIONS ININTEGRAL FORM 495 Wehave followed theusual custom ofwriting thefield terms onthe 27.3FURTHER COMMENTS ONMAXWELL’S EQUATIONS 498 leftandthesource terms ontheright. However, thisissomewhat illusory 27.4THELAWOFCONSERVATION OFCHARGE S00. because pandJarethemselves functions ofEandB.Asusual, 27.5 MAXWELLS EQUATIONSAREREDUNDANT 500 Eistheelectricfieldstrength, involts/meter; 276 DUALITY SOL ; ite i 5Example: THEFIELDS OFELECTRIC ANDMAGNETICDIPOLES 501 p=p,+pyis thetotal electric charge density, incoulombs/meter’; 27.7LORENTZ’S RECIPROCITY THEOREM 2 pyisthefreecharge density; Examples 503 27.8THEWAVEEQUATIONS FOREANDFORB=S04 6,=—¥-Pistheboundchargedensity; 27.9SUMMARY S05 Pistheelectric polarization, incoulombs/meter’; PROBLEMS S07 Bisthemagnetic fluxdensity, inteslas; J=J,+9P/3t+VXMisthetotalcurrentdensity,inamperes/meter?;* Thischapterconcernstheowefundamentalequationsofelectromagnet Jjisthecurrentdensityresultingfromthemotionoffreecharge; ismthat bearthenameofJamesClerkMaxwell(1831-1879). Hewasthe ji a a firsttostatethemclearlyandtorecognize theirimportance, butitwas AP12athepolarization current densityinadielectric; Oliver Heaviside (1850-1925) whofirstexpressed them intheform that VXMistheequivalent currentdensityinmagnetized matter; weknowtoday.Theseequationsaresofundamental thatweshallexpress ‘Misthemagnetization, inamperes/meter; the vi wi is ic reminvarious waysandthendiscuss theirphysical meaning. cisthespeedoflight,about300megameters persecond; 27.1 MAXWELL’S EQUATIONS IN qsthepermittivity offreespace, about 8.85x10~"?farad/meter. DIFFERENTIAL FORM Inisotropic, linear, andstationary media, LetusgroupMaxwell's fourequations; wediscussthematlengthbelow. T=0B, P=eoxE, M=Xml, (27-5) Wefound them successively inSecs. 9.5,23.4, 20.4, and17.4: ~— nn SOwhere@istheconductivity, x,istheelectricsusceptibility, andz,,isthe |3B magneticsusceptibility. Also, |v-E=2, (27-1) vxE+=o, (272)° ‘ GLxeHe=LtHm (27-6)v-B=0, (273) vxp-t2B_ yy | s Peat(27-3) Byres —QT) "SeePaulPenel,Jt.andHermanA.Haus,Elecrodynamics ofMovingMedi,a a] ResearchMonograph40,M.I-T.Press,Cambridge,Mass,1967 ‘ *Until ow wewere concerned solely with freecurrent densities, andweusedJinstead IfyouhavenotstudiedChap.17,youwillhavetotakethisequationforgrantedatthis stage.YouwilfdprotinSe.28.6. =? AENEatonfrgamed th ofJ10spy notion 494, MAXWELL’S EQUATIONS 27.2MAXWELL’S EQUATIONS ININTEGRAL FORM 495 where ¢,istherelative permittivity andy,istherelative permeability Thisisageneral rulefortransforming anequation interms of€0,Uo,PvF Inside asource, suchasabattery oraVandeGraaff generator, electric toanother oneinterms of€,MyPy»Ip charges are“pumped” bythelocally generated electric fieldE,against TheMinkowski formulation ofMaxwell’s equations isoften useful. It theelectric fieldEofother sources, andJ=o(E+E,). expresses thesamerelations, butinterms ofthefourvectors E,D,B,H: Writing outpandJinfull,Maxwell's equations become on . VD=p, (27-20) vxe+ Bao, Q721) veeiP 7-7) & &2D OB V-B=0, (27-22) VxXH- 34 (27-23)VxE+S=0, (27-8) Inthefollowingchaptersweshallbemostlyconcerned withelectric V-B=0, (279) andmagnetic fields that aresinusoidal functions ofthetime. Then, for Ler op isotropic, linear,andstationary media,notnecessarily homogeneous, PxBSS=mld++Pxm). (27-10)ox 2 VicE=p, (27-24) VXE+joul=0, (27-25) ‘ThisAmperian formulation expresses thefieldinterms ofthefourvectors. V-uH=0, (27-26) VXH~jwcE=J, (27-27) E,B,P,and M. ; ‘With homogeneous, isotropic, linear, andstationary (HILS) media, Itisworthwhile todiscuss Maxwell's equations further, butfirstletus rewrite them inintegral form. _eae (Sec.9.9) 7-11) 27.2MAXWELL’S EQUATIONS IN RAL FORM P=(c,-leoE —(Sec.9.9) (27-12) INTEG! _(= 1) Integrating Eq. 27-1 over afinite volume vand then applying theMaNi B(Sec.20.7) (27-13) divergence theorem, wefindtheintegral formofGauss’s law(Sec.9.5) and 1sn [e-aa=4f pa=2, 27-28) vee=",(27-14) PxE+S=0, (715) “ ow ° «" wherefistheareaofthesurfacebounding thevolumevandQisthe oE total charge enclosed within v.SeeFig.27-1. “B=27-1 epee - veB=0,G16) VxB—ew samy, (7-17) Similarly, Eq.27-3saysthatthenetoutward fluxofBthrough any closedsurface iszero,asinFig.27-2: Recall that €=€,€) and =4,o, €,isfrequency-dependent, and y,is hardlydefinable inferromagnetic materials. Theexpressions forPand fB-dst=0. (27-29)forMarenotsymmetrical, butP,E,and Dpoint inthesame direction, as HikeMand, intropic andlneatmedia Equation 27-2isthedifferential formoftheFaradayinduction lawforith"hefolkatthebstitutions: equations followsfromEqs.27-1to27-4 time-dependent magneticfields.Integrating overanopensurfaceofareawith thefollowing substitutions: sfbounded byacurve Cgivestheintegral form, asinSec.23.4: 6 Mo (7-18) 4 an E-di=—<| B-ast=-—, 27-30) Ppp, Ind. (27-19) ¢Evdi=— 7[ dt (27-30) 496 | a7 ' | ve\>\ Fig. 27-3, Ifthe magnetic flux linking Cincreases, itinduces anelectromotance aroundCinthedirectionofthearrow.Theelectromotance pointsinthesame Fig,27-1,LinesofEemergingfomavolumecontininganechargeQ.The directionifBpointsupwardanddecreases 3E Bedl=wo)(1+eo)dot 27-31 whereAisthelinkingflux.SeeFig.27-3.Theelectromotanceinduced f.Lvres Crs!) aroundaclosedcurveCisequaltominusthetimederivative oftheflux Wefound two! he linkage. Thepositive directions forAandaround Csatisfy theright-hand '¢foundtwolessgeneral formsofthislawinSecs.19.5and20.6.Thewet convention, closed curve Cbounds asurface ofareasfthrough which flowsacurrent Finally, Eq.27-4isAmpére’s circuital lawinintegral form ofdensityJ+€98B/3t. SeeFig.27-4 IN ; le},| c° mJ‘ie B Ite, HF Fig.27-4. The lineintegral ofB-dlaroundCispostiveiftheintepratonrunsin Fig.27-2. Lines ofBpassingthroughaclosedsurface.ThenetoutwardfluxofB thedirectionofthearrowswithacurrentdensityJ+€,3E/21pointing, seeqaaito zero. downward 498 MAXWELL'S EQUATIONS 173FURTHERCOMMENTS ONMAXWELU'S EQUATIONS 499 27.3 FURTHER COMMENTS ON (4)Wefound Faraday’s law MAXWELL’S EQUATIONS OB dvxE=-8 fE-dl=-5fBedst,(27-34) (1)Maxwell’s equations 27-1to27-4arelinear.Thismeansthattheydo a ‘c dhe fotcontain products oftwoormore ofthevariables oroftheir inSec.17.3,andwediscussed itinSec.23.4,These equations areequallyderivatives. ItfollowsthatifthefieldE,BysatisfiesMaxwell’s equations teneral. HeretheclosedcurveCmustbefixedinthecoordinate system whenp=p;andJ=J;,andifthefieldE2,B2,corresponds similarly to withrespect towhich bothEandBaremeasured. However, theopen p=p; andJ=J,, then thetotal fieldisE,+Es, By+B: when curface bounded byCmaymove p=p+psandJ=J, +4;.Thisistheprinciple ofsuperposition (Secs. (5)InSec.20.5wefound that,forstaticfields,3.3and 18.2.1). Inother words, each source ofpandJacts independ- ently ofalltheothers. VxH=J, (27-38) Inlinear media, Pand Mareproportional, respectively, toEand B, and Eqs. 27-7 to27-10 arealso linear. The principle ofsuperposition This equation isnot general, for the following reason. Take the applies. divergence ofbothsides.Thenthedivergence oftheleft-hand sideiszeroHowever, innonlinear media, Pand Maremore orlesscomplicated because thedivergence ofacurl isidentically equal tozero. However, functions ofEandB.Equations 27-7 to27-10 arethen nonlinear andthe thedivergence oftheright-hand side isnotnecessarily zero: principle ofsuperposition does notapply top,andJj.The principle of superposition continues toapply topandJ. veg=—-e209 =-¥.(2) o f ) (27-36) Allmaterials become nonlinear atveryhighfieldstrengths. a ot (2)Wededuced Gauss's lawforelectric fields Hereweusedfirsttheconservation offreeelectric charge (Sec.4.2)and veE=", or[e-aa=2. (27-32) thenEq.27-20. & Le € Ittherefore occurred toMaxwellthatthecorrectrelationmustbe fromCoulomb’s lawforstationary electric charges situated inavacuum sD inSec. 3.7.InSec. 9.5wefound thatthesame relation applies tocharges VxH=+=, (27-37) lying’inside matter, ifpandQinclude bothfreeandbound charges. o Nonetheless, Gauss’s lawismore fundamental than Coulomb's law asabove because itapplies eventomoving charges (Sec.17.1), while Coulomb's (6)Observe thatthelineintegral ofB-dl isrelated tothecurrent lawisstrictly validonlyforstationary charges (Secs.3.1and16.5). density Jplus€09E/at inEq.27-31. Thisisremarkable because the (3)OneofMaxwell's equations states that integral forB(Sec.20.3)doesnotinvolve€,3E/3t. (7)Thedisplacement currentdensity31D/3tofSec.9.10isthesumofV-B=0, oriB-dA=0, (27-33) ‘(woterms:“ 3D_3 3E,oP where.fistheareaofanyclosedsurface.Wededucedthisrelationin BtCOE+P=05+Zp (27-38) Sec. 18.3 from the law of Biot-Savart for the Bfield of atime- independent current distribution. This relation is,infact, general (Sec. Inamaterial ofconductivity oandrelative permittivity ¢,subjected toan 17.2), but onone condition: thedensity ofmagnetic monopoles (Sec. alternating electric field, 18.1) must bezero. Since magnetic monopoles have never been observed E,Bfieldandforanycurrent distribution, evenifthese aretime- hasbeenobserved. SeeD.F.Bartlett andP.R.Cotle,Physical Review Leters, vol.5S,dependent. 1-59(1985), 00 MAXWELL'S EQUATIONS 276DUALITY S01 - 3D_. SoV+Bisaconstant ateverypointinspace,ThenwecansetV+B=0 Jy=Onexpjot,Fp=JoeB expjot, (27-39) everywhere andforalltimeifweassumethat,foreachpointinspace, V+Biszeroatsometime,inthepast,atpresent,orinthefuture,With 24=WEE (27-40) thisassumption, Eq.27-3followsfromEq.27-2. J° Similarly, taking thedivergence ofEq.27-4andapplying thelawof aiofchia Inagoodconductor, andatfrequencies lowerthanroughly 1gigahertz, conservation ofcharge, wefind 0~10"siemens/meter, €,~1(Sec.9.9),theaboveratioisoftheorder eeEeype BP (27-44)off/10" atafrequency f,andthedisplacement current isnegligible oror ot" compared totheconduction current. Athigherfrequencies both€,ando 5 Sip >varyerratically withfrequency becauseofatomicandmolecular SAVB)=5(2). ve=Psc (27-45)resonances. ‘Theconstant ofintegration Ccanbeafunction ofthecoordinates. 27.4THELAWOFCONSERVATION OFCHARGE Ifwenowassumethat,ateverypointinspace,atsometime,V-Eand . paresimultaneously equaltozero,thenCiszeroandwehaveEq.27-1. InSec.4.2wesawthatfreecharges areconserved. Atthattimewewere Sotherearereallyonlytwoindependent equations,using thesymbol Jforthecurrent density offreecharges instead ofJ. Letuscalculate thedivergence ofJasdefined inSec.27.1. Wewill 27.6 DUALITY need thevalue ofthisdivergence inthenext section. First, op 3 Imagine afieldE,BthatsatisfiesMaxwell's equations withp,=0,J;=0 veu=F- (+P,vxM)=+5(FP), 2741) inagivenregion.Themedium ishomogeneous, isotropic, linear,and at stationary(HILS).Nowimagineadifferentfield thedivergence ofacurl being equaltozero.Thus E'=-KB=—Kut, (27-46) veg=21300 __Hoy+O)__90 (27-42) H’=+KD=+KeE, (27-47) a ot 3 3t wheretheconstant Khasthedimensions ofavelocity andisindependent Thisisamoregeneralformofthelawofconservation ofchargeofSec. ofx,y,z,f.ThisotherfieldalsosatisfiesMaxwell's equations, asyoucan 42.check bysubstitution into Eqs. 27-20 to27-23. . Figure 27-5illustrates thisduality property ofelectromagnetic fields. 27.5MAXWELL’S EQUATIONS AREREDUNDANT ‘Onefieldissaidtobethedual,otthedualfield,oftheother.Therefore. iffieldcanexist,thenitsdual ilsoexist. Maxwell's fourequations areredundant. WesawinSecs.17.3and17.4 oneFel ° renitsdualcanalsoexi that theequation forVxEfollowsfromtheoneforV-B,andthe Example THEFIELDSOFELECTRIC AND equation for7xBfromtheoneforVE. These are,respectively, the MAGNETIC DIPOLES first and second pairs. Thetwoequations ofthefirstpairarealsorelatedasfollows. Ifwe tmSec,$1wefoundthatinthefelofaneectdipoleoftakethedivergence ofEq.27-2andremember thatthedivergence ofa moment p curliszero,wefindthat £=PGeos0F+sin06). (27-48) yeBig orfeeR=0. (27-43) Later,inthefourthexampleinSec.18.4,weshowedthatinthe3 o fieldofamagnetic dipole ofmoment m 502 277 LORENT2’S RECIPROCITY THEOREM 503 iy ae oP,eo f V=(BXBy~ByXBy)=~Hoa(Jn+2+1XM,) La. AL oP,GL “ga +Hobin=(So+2+XM),(27-53) — he? ae Since,byhypothesis themediumislinearandisotropic, Pis€gx.Eandtheabove time derivatives cancel. Finally, ifthepoint considered isnot Fig.27-5. Pairofdualfields. Lines ofEaresolid, andlinesofHfdashed. insideasource,Ohm’slawapplies,J,=oF,andtheJ;termscancel.We are then left with the VXMterms.Ifthemediumisnonmagnetic, then Mis zero and bom . 5 B=HE(200s07+sin86), (27-49) V-(E,XB,~E,XB,)=0. (27-54) or Ifthemedium ismagnetic, wecanperform asimilarcalculation by =Z25(2c0s 07+sin6) (27-50) usingH’sinstead ofB’sand V+(E,XH, —E, XH,)=0. (2755) Wehave set€,=1, 4,=1.Ifthefield oftheelectric dipole isthe unprimed field,andthatofthemagnetic dipoletheprimedfield, Insummary, therefore, iftwofieldsaandaresinusoidal andofthenK=m/p, thesamefrequency, ifalsothemedium islinear andisotropic, andifthe : pointconsidered isnotinsideasource, thenEq.27-54applies. Ifthe 27.1LORENTZ’S RECIPROCITY THEOREM medium ismagnetic, thenwehaveEq.27-55. rs ‘ingthedir the yields Consider twofieldsE,,B,,andE,,Byinalinearandisotropic medium. Applying thedivergence theoremyields Because oftheprinciple ofsuperposition, these twofields cancither existseparately orbesuperimposed withoutdisturbing eachother,givinga [xt ~Byxt)dst=0, (27-56) third field E,+Ey, B,+By. Wenow usethevector identity where ofisthearea ofany closed surface, with theabove restrictions. _Thisresultisknown asLorentz’s reciprocity theorem. Itisparadoxical V«(EqXBy—EyXB,)=By«(VXE.)~Ey«(7XBy) hecauseitestablishes arelationbetween twounrelated electromagnetic —B,+(VXE,) +E,(FXBy).(27-51) fields. Then, from theMaxwell equations forthecurls ofEandB, Examples|(1)Iftheafieldispurelyelectricandthebfieldpurelymagnetic, 2B,3E,) hen V-(E,XB,—E,XB.)=—B,*BeB«us(dy+€0E*) ‘he a ot V-(E,XH,)=0, (27-57) +B,Been, (I+co") (27-52) subjectonlytotheabovelimitations. Thissurprising statement ot ar becomes obvious after expanding thedivergence: Ifthetwo fields areharmonic functions ofthetime and ifthey areofthe V-(E, XH.) =H, (VXE,)—E,(VXH,) (27-58) same frequency, the3/3t operators canbereplaced byjw,andthetime B, 2D,derivatives cancel.Substituting thevalueofthetotalcurrentdensityJ, =OHSP Ea(%+2) (27-59) sou MAXWELL'S EQUATIONS 279SUMMARY 0s ‘This lastquantity isidentically equal tozero because, by Similarly, taking thecurlofEq.27-4 andsubstituting Eqs27-2 and hypothesis, B,and Eyareboth zero, Jp=0B,=0, and D,= 27-3, wefind that ot PB~Cltyao=UyPXI 27-68) (2)Thefieldofasmallpermanent magnetisthatofamagnetic ~CelloSe=—MoXJ, (27-68) dipole, asinthelastexample inSec. 18.4. Ifthemagnet carries an electric charge Q,italsohasanelectric field which isthenonhomogeneous wave equation forB.Thesource term is oF again ontheright.Ere” (27-60) ‘Outsidethesources, wn 52 PBety=O. (27-69)766 3r Vi(EXH)= V+)Ey00}=0+(Extop) (27-61) ‘According totherulegiven inSec.27-1,thewaveequations fora turHe0 HILS medium areasfollows (2 Hom 2 =v(Pern og)=0. 27462) SEVp.ol, Anerdar PEeyme=Pyyh 7-1 VE eu sree, (27-70) 7.8 THE WAVE EQUATIONS FOR EAND FOR B |278 Q vBeuB==nXd, (27-71) Taking thecurlofEq.27-2 andremembering that .Wetherefore haveawaveequation forthefieldE,andaseparate VXVXE=-VE+V(V-E),(27-63) waveequation forB.Withinthewave,however, EandBareinextricably fromFa,27-4 related through Maxwell’s equations, Inotherwords, purely electric, or then,from Eq.27-4, purely magnetic, waves areimpossible. Thefactremains that, insome a 3 3E. waves,theenergydensitycanbeeithermostlymagnetic ormostlyVE-V(V-E)=5VXB== (ol+eet). (27-64) clectrie Ifaisconstant, Substituting now thevalue ofthedivergence ofEfrom Eq.27-1 and . in 7 SE 3E_Y¥py rearranging, . VE~ cuss —ous =k, (27-72)PE—coltea+py2. (27-65) .oF eo 7 eB. OB PBewe on=0. 27-73) ‘Thisisthenonhomogeneous wave equation forE.Thesource terms are " o ontheright 5Outside thesources, 27.9 SUMMARY PE—cote2E=0. (27-66) Weexpressed Maxwell's equations inseveraldifferentforms.Seethear hack cover. Allthequantities thatappear therearedefined inSec.27.1. ‘Thisistheusualwaveequation. Thespeed ofpropagation, which isthe ‘Thedisplacement current density is speed oflight, is 1 aD_. 3B aP — 27-67 FOS +5 27.Cr ore) a a (138) 506 MAXWELL EQUATIONS PROBLEMS 507 ‘Thesecond termontherightisthepolarization current density. Thefirst PROBLEMS term can exist even inavacuum. The lawofconservation ofcharge, initsgeneral form, takes into 27.1. (27.1) Superconductivty account bothfreeandbound charges: ‘Asuperconductor offerszeroresistance tothemotion ofsuperconductingcharge carriers. These arepais ofelectrons that move as@unit. 3 (@)Ifthere are Nsuch carriers percubic meter ofmass m'and chargevs=-2, (27-2) showthat . ee where pandJaredefined inSec.27.1. Ne"dt ‘The equations forthecurl ofEandforthedivergence ofBare “Thisisthe frstLondon equation. Note thatEiszeroonlyifJisconstant. redundant since onefollows from theother. They areoccasionally called (b)SetK=m'/(Ne”). Show thatinanalternating field thefirstpair.Theother twoaresimilarly related andformthesecond 1 ipair. °jak~~oKForeveryfieldE,HtherecanexistadualfieldE’,H’suchthat )Shhat (©) Show that ”= =- ts =+, 17-46), (27-47) E'=-KB=~-KuH, H'=+KD=+KeE. (27-46), (27-47)vx(x2!)«28 This istheduality property ofelectromagnetic fields. a) ar ‘TheLorentz reciprocity theorem states that, foranytwofields E,,H, “Theequation andEy,H,thataresinusoidal andofthesamefrequency, inlinear and reisotropic mediaawayfromsources, =-B isthe second London equation. Itdoes not followF-E.XH,-E,xH,)=0, (755) Hs qiqa‘mathematically fromthe (d) Show that under steady-state conditions ‘Thenonhomogeneous waveequations forEandforBareasfollows: ve="p, FE_Yo, kK VE~ety=Ospy, 27-65 whe3e,MOSe @745) Inonedimension, thismeansthat OB ae=oyB=-7 7: Boe VB~CottoSa=—HoXJ. (27-71) eK? Outside thesources, thetermsontheright-hand sidearezero.Thespeed orchat lightisthus =Rep— oflig BaBeeRe 1 a en (27-67) where (K'/1ts)"* isthedepth ofpenetration ofthefieldCT) (¢)Calculate thevalue ofthedepth ofpenetration, seting m’equal to ; twicethemassofanelectron,e'=2e,andN=10".Thedepthof InaHILSmedium,ifoisconstant, penetration is,infact,afewtimeslarger.()Much beyond thedepth ofpenetration, E=0, J=0, B=0. «BE E_ Sp Showthatjustoutside asuperconductor Bistangential tothesurface andVEeuSaou =oan (27-72) ‘equalinmagnitude tothesurfacecurrentdensity. oe3E (27-2.(27.1)Experiments onmodels: vBcpeBoyE=o, 7-73) ‘Aproblemthatcannotbesolvedonpapercanoftenbesolvedintheor or laboratory. However, there are instances where afull-scale experiment 508 MAXWELL'S OVATIONS eRopLEMS 509 would betoocostly. Onesuchproblem isthatofthedesign ofanMHD there areonly three independent scale factors. Then ofo'= r/f*and generator (Sec. 22.1) owL?=o'w'L”, where Lisalength Insuch cases itissometimes useful toperform experiments onamodel of Ifthefelds areinavacuum, é,/e! =1andth=el.Then e=hand r=1. convenient size. One then hastherealsystem, forwhich thevariables are Note thatif€/e!=1iuslandofo,=1then 4,9.2,t,EyBy€ntin0,R,C,L,etc.,andthemodel,whosevariables fest nds amen arex’,y',2,etc.Theratios x/x’, y/y’, z/2', ete.arethescale factors. thoesh Notallthese factors canbechosen arbitrarily because boththeunprimed eh et andtheprimed variables must satisfy Maxwell's equations. Thenumber of and e=h, 1=1, [=1. Then themodel isthesame sizeastheoriginal! If arbitrary scale factors isequal to4,thenumber offundamental units themodel istobeadifferent size, then either ¢,,or0,orboth, must be (meter, kilogram, second, ampere). With mechanical systems there are different. This condition isoften impossible tosatisty only three arbitrary scale factors. ‘Atlow frequencies one candisregard thedisplacement current, and Letusset hence attribute anyvalue totheratio ¢,/€!, ortoth/el eye, tl, EL, Hy 27-3,(27.2). Paralle-plate capacitor fedatoneendbyatime-dependent sourceyoyrr hb pee Be pe Figure 27-6 shows aparallel-plate capacitor connected atoneendtoa source whose voltage increases slowly and linearly with time: dVa/dt=k “The other scale factors follow from Maxwell's equations and from other Edge effects arenegligible: a>>s, b>>s:relations. (@)Findthecurrent/asafunctionofx.(a)Use Maxwel’s equations toshow that (b)Find Binside intwodifferent ways. Find Ainside wer «th oh (6)FindBandAoutside.Thecapacitorplatesarethin. Os,EB GyF04. (@)Drawalargecrosssectionofthecapacitorinthemidplaneparallelto bad «el ° thexz-plane, showing /andthevectors A,B,¥XB,E,3E/3t nearboth ()Show that ends. Usearrows ofdifferent sizes toindicate qualitatively howthese vectors vary with xandwith 2 ()Zac, wy2-4, GyLen, Wy2-2, wyLeet 27-4,(27.1) Atransformation thatleavesMaxwell's equations invariantgi OpAp Wp Ro OW (a)ShowthatMaxwell'sequationsforfreespaceareinvariantunderthetransformation (6)Showthat »Been WwBakGaySect,wyLoft B=ak+bcR, B=—()E +B, )Saeu WZ=F, Gi) L-c4, WEAF whereaandbareconstants andcisthespeedoflight. (4)Show that (b)Under what condition aretheenergy density €jE*/2+B?/(ja)and ePoyntingvectorEXHinvariant? ofa4, Sans,Gi)2-8,Go)Paner hePoomingNestorEXerent FoeDOgeake CO=pOp (6)Show that, ifLisalength, then fuol _| , Fuol® — caCZ SF2 6 Frew = ws Isheright? lL -.—|_|Inpractice, these relations aresimplified bytheexclusion offerromag- netic materials because oftheir nonlinearity. Then j,/u;~1, et=lh,and Fig.27-6, si0 MAXWELL'S EQUATIONS PROBLEMS sit 2745(27.3) ‘Theskineffec 2, 4-2‘AsweshallseeinSec.29.1,ahigh-frequency fielddoesnotpenetrate Braet Ay significantly into thebody ofagood conductor. Also, both EandBinside areapproximately tangent tothesurface. Letthez-axis benormal tothe (©)Calculate theenergy acquired byamagnetic monopole thatacceler- surface, pointing outward, with Einthedirection ofthex-axis andBinthe atesover adistance of1000 kilometers intheearth's magnetic field (=10-* direction ofthey-axis. Set3/3x =0, 3/3y =0 tesla) (a)Show that, inside, 3/92 =~38/3. (d)Magnetic monopoles gothrough aloop ofcopper wire (b)Show that, justoutside theconductor, Bistangent, ornearly so Show that theinduced electromotance isequal tominus themagnetic(c)Letthecurrentdensitynearthesurfacebeaamperes/meter current,withtheright-hand screwconvention. ThisisonemethodofShow that, justoutside theconductor, B=yaX2 detectingmagneticmonopoles snl),Dosshsastresultdependonhowthecurrentvaiswithdepth 278.(273)Theconinvous-reation theoryandMaxwell'sequationsImagine anexpanding spherically symmetric universe inwhich there is 27-6, (273). Themagnetic field ofa leaky capacitor continuous creation ofcharge attherate ofqcoulombs/meter-second. ‘Acharged capacitor whose electrodes areparallel and circular lesina Creation ofelectric charge occurs through thecreation ofhydrogen atoms large volume ofdielectric that isslightly conducting, ‘The capacitor carryingaslightexcesschargeyeasinProb,3-15. discharges,‘The rate ofmass creation Qisproportional to9:Q=[m/(ye)la, where (@)Calculate thevalue oftheratio J,/(3D/3t) atany point inthe mis themass oftheproton, theuniverse being mostly hydrogen. dielectric interms oftheresistance andthecapacitance (2)Bysymmetry, thevector potential canonly beradial(b)Showthatthisratioisequalto~1atthesurfaceofanelectrode Showthatundersteady-state conditions thecurrentdensityJis(6)Show thatBiszeroeverywhereinthedielectric.Thismeansthatthe everywherezero,accordingtoMaxwell'sequations. magnetic field oftheconduction and polarization (not displacement) Lyttleton and’Bondi (see Prob. 3-15) suggested that, ifcontinuouscurrentsinthefringingfieldexactlycancelsthemagneticfieldofthe creationdoesexist,thenMaxwell'sequationsmustbemodifiedasfollowsconduction andpolarization currents intheregion between theplates.(4)ShowthatBisalsozeroforelectrodesofanyshape. 13E_f1 _eftItiatetic scapesonlaofteldocpactor, aythe reammssa5 [pal 7e-Z [phregionbetweentheplatesofaparallel-plate capacitor,thenthevalueoftheratio calculated under (a)applies. However, atthe surface ofanelectrode, where thenewterms areenclosed inbrackets. Thequantities VandAare this ratio isnotequal to—1because charge migrates from theouter theusual scalar andvector potentials: surface ofan electrode totheinner surface, where itleaks out. Then, inthe dielectric, Jj]>|8D/3t\, J,~3D/3t points inthedirection ofJ,andthus g--w—- peexa. ofE,andthere isanazimuthal magnetic field 3 27-7. (27.3) Magnetic monopoles ‘Theother twoequations ofMaxwell for0XEand¥-Bremainun- Ifmonopoles exist,thenMaxwell's equations requiretwomoreterms,to changed. Lyttleton andBondisuggested thattheconstant ,whichhasthe takeintoaccount magnetic charges andmagnetic currents. Itisthecustom dimensions ofalength, would beofthe order oftheradius ofthe universe. towrite theequations inthefollowing form ‘Thenewterms would therefore benegligible inallbutcosmological p op or problems.vee2,vxee229,pepaps,vxpen(es% +3) {D)ifthesemodifiedMaxwellequationsarecorrect,areVandA & ot Ld measurable, inprinciple? Remember that,withtheabove equations forE wirepsthenape chargedeny, expres inwebersnete and 494intermsofVanAolytheatesofchangeofVandAdetermine q‘isnanes surestdensity,inwebers/second-meter (c)Writeouttheequation fortheconservation ofthetotalcharge(Sec. (®) Show that . 274),vesta- * (4)WouldtheLorentzcondition (Secs.17.9and37.1)stillbevalid? (c)Now set-A=A'r, where A’is@constant, and assume Vtobe This isthe equation ofconservation formagnetic monopoles. constant. Show thatB=0,E=0, J=(q/3)r, p=VIE: (b)Show that, byanalogy with electric fields, near point magnetic ‘Assuming that thevelocity oftheoutward flow ofmatter isthesame as charge Q* that ofthe charge, namely J/p, itfollows that the radial velocity is si2 MAXWELL'S EQUATIONS PROBLEMS 313 proportional tor,which isconsistent with the linear velocity-distance (©)Sketch thevalue of relation observed byastronomers: v=r/T, where T~3x10”secondsis a theHubbleconstant. Gfpease,(H)Show thatp=g7/3. “ {g)Nowthespace-charge densitypisnye/m,where9,themassdensity whereofistheareaofthespherical segment ofradiusY,asafunction of oftheuniverse,isabout10*kilogram/meter’, ‘ ‘Yt,WhenthechargeisjusttotheleftofY=X,thefluxofDthroughShow that,ifthistheory iscorrect, thenQ~1/(2% 10")hydrogen points totherightandis+Q/2. Immediately afterward, thefluxpointsto ston/meter”-second. theleftandis~Q/2, sothere isadiscontinuity inthiscurve. 27-9. (27.3) Another transformation thatleaves Maxwell's equations invariant (d)Theintegral ofH-dlaroundthecircleshowninthefigureisequal (@)Show thatMaxwell's equations forfreespace areinvariant under the to22YB/jg. Calculate theintegral ofthisquantity overtimefrom minus transformation infinity toplusinfinity. SetX'=0tosimplify thecalculation. Explain your result. E'=Ecos0+cBsin 9, B’="sin 8+Bos8 I-11. (27.9) TheWatson theory ofcontinuous charge creation € Problems 3-15and27-8sketch theLyttleton theory, according towhichionE'=KB,H'= hydrogen atomsarecontinuously createdintheuniverse, eachatomey cntoedalaswoncenainn tePotwal bearingaslightpositivecharge.W.H.Watsonhadproposedasimilar Gnrapecvalys 7Ne‘Perielcuescorresponding 609=2/21 theoryseveralyearsbefore.’Watsonpostulated ascalarpotentialNsuch (b)Showthattheenergy density €o£?/2+ B°/(2ua) andthePoynting that vector ExHarealso invariant under thistransformation. on _ VeD=p, +e, VXH=J,+2—BN. 27-10.(27.3) Themagnetic fieldofapointcharge thatmovesataconstant D=py+Eales 3 velocity Figure 27-7shows apoint charge Qthattravels along thex-axis ata (2)Findthenonhomogeneous wave equation forNfrom theequationvelocityVf.Itspositionattime1is(77,0,0). forthenonconservation ofcharge.Settherateofcharge creation equalto (a)FindBatapoint P(X, ¥,0), notattheorigin, attheinstant that 4coulombs/meter’-second thecharge passes through theorigin. Theparticle travels inavacuum, (b)Findthenonhomogeneous wave equations forEandforH. andv'<e?, where cisthe speed oflight _ sahhf2mhavestaiedChap6,compare yourrestwiththatofSe “SecL.G.Chamber,lournalofMathematicalPhysic,vo,1373(1963) oNfi Al \ cl \ fd H\ ei * . 1| \ef2 \,/NA Fig.27-7. 282 UNIFORM PLANE ELECTROMAGNETIC WAVES sis Ifyou arenotfamiliar with wave propagation you would bewell 28 advisedtoreadApp.Cnow. CHAPTER28.1 THE ELECTROMAGNETIC SPECTRUM PLANE ELECTROMAGNETIC Maxwell’s equations impose nolimitonthefrequency ofelectromagneticwaves. The known spectrum extends continuously from thelong radio WAVES I waves totheveryhigh-energy gamma raysofcosmic radiation, asinFig. 28-1.Intheformer,thefrequencies areoftheorderof100hertzandthe UniformPlane Waves inFreeSpace, in wavelengths about3megameters; inthelatter,thefrequencies areoftheNonconductors, and inConductors order of10*hertz andthewavelengths lessthan 1femtometer. ‘The known spectrum thus covers arange of22orders ofmagnitude. Radio, heat waves, light, x-rays, andgamma raysareallelectromagnetic, although the sources and the detectors, aswell asthe modes of interaction withmatter, varywidely asthefrequency changes byorders 28.1THEELECTROMAGNETICSPECTRUM SiS ofmagnitude. 28.2UNIFORM PLANE ELECTROMAGNETIC WAVES INAGENERAL Many experiments demonstrate thefundamental identity ofalltheseMEDIUM sis . waves. Inparticular, theyarealltransverse, andtheyalltravel atthe 282.1THERELATIVEORIENTATIONS OFE,HANDS17 speedcinfreespace,exceptinspecialcicumstances. Forexample, 28.22THECHARACTERISTIC IMPEDANCE ZOFAMEDIUM S18 J 2823THEWAVENUMBER 518 simultaneous radioandopticalobservations onstarsshowthatthe 2824 THEWAVEEQUATIONS Sif velocity ofpropagation isthesame, within experimental error, for 28.3 UNIFORM PLANE WAVES INFREE SPACE 520 wavelengths differing bymore than 6orders ofmagnitude. Example: THEEAND BFIELDS INALASER BEAM 522 WeuseHrather than Bindiscussing electromagnetic waves, inspite of 28.4UNIFORM PLANEWAVESINNONCONDUCTORS —522 thefactthatuntilnowwehaveusedHonlyformagnetic materials. Example; THEBANDBFIELDSOFALASERBEAMINGLASS —524 ‘TherearetworeasonsforusingH,insteadofB,indealingwith 28.5UNIFORM PLANE WAVES INCONDUCTORS 524 electromagnetic waves: one isthat EXHisapowerdensity,andthe 28.5.1 THE COMPLEX WAVE NUMBERk=f-ja524 otheristhatE/Hisanimpedance. Thesetwoconceptshavegreat 28.5.2THECHARACTERISTIC IMPEDANCE ZOFACONDUCTOR 526 practical value 285.3 THE ENERGY DENSITIES 527 286THEPOYNTING THEOREM 528 28.2UNIFORM PLANE ELECTROMAGNETIC2uMMARY 5 WAVES INAGENERAL MEDIUMPROBLEMS 530 Awave front isasurface ofuniform phase. The wave fronts ofaplane wave areplanar. Awave isuniform ifawave front isasurface ofuniform phase anduniform amplitude. Weshall notbeconcerned withnonuni- Wenowgoontothestudyofelectromagnetic waves. Thischapter formwavesuntilChap.31,concerns thepropagation ofuniform planewavesinunbounded media, Uniform planeelectromagnetic wavesinunbounded mediapossessfirstageneral medium, thenfreespace,thennonconductors, andthen several general properties thatapplywhether thewavetravels infreeconductors. Goodconductors andionized gasesfollow inthenext spaceorinmatter. Toavoidneedless repetition, westartwithageneralchapter. Laterweshallstudyreflection andrefraction inChaps. 30to32, medium €,,1,0thatishomogeneous, isotropic, linear,andstationaryguidedwavesinChaps.33to36,andfinallytheprocessofradiation in (HILS), Chaps. 37to39. 242 UNIFORM PLANE ELECTROMAQNETIC WAVES si7 Weassume asinusoidal wave traveling inthepositive direction ofthe2 2. . z-axis, Wealso assume that theEvectors areallparallel toagiven. 2 Lal 3 direction, Inotherwords,weassume thatthewaveislinearlypolarized.' a epee Iftheplane wave isnotlinearly polarized, thenitisthesumoflinearly« =. 5 2g2s polarized waves.’Theplaneofpolarization isparalleltoE. = a : Srez Inalinearly polarized planewave, EandHareoftheform .; é eePo ael 33% E=E,,expj(wt-kz), H=H,,expj(wt—kz)' (281) ° al 7 ¢ eae? a =* 2 $832 whereE,,andH,,arevectors thatareindependent ofthetimeandofthe a—s =é3£ coordinates. Ifthereisnoattenuation, thewavenumberisreal: *;5 gse2 ‘Fs z 2352 2a_1 2z cgse k=2a Set .F : g22 viak oo a«= 5 2228 ‘< 235 where visthephase velocity, 2isthewavelength, andX(pronounced - 5 | ali’ “lambda bar")istheradianlength. Youcaneasilyshowbysubstitution2 ere thatEqs.28-1aresolutions ofEqs.27-66and27-69. -* 338s “ -* 5 E53 ee : ”~ #. 3Bees 28.2.1. TheRelative Orientations ofE,H,andk 2. 2. 22 2225332 Forthisparticular field, | z Po b258 a 3.,9,,8, 9z. - 2 g2ee Saja, ve2e+ 5422-2 ¢2-ine, 5.;fs gize grim VHSetSotsensesike(283) ' ‘ 23 . ree Wesetpy,=0.Wealsoset -> Jj 4288 ; ote ts 2i253s J,=08, (28-4) ’epooryan eee * 35-2: se ontheassumption thatwxBisnegligiblecomparedtoF,wherewisthe in-. :——F- 52Pe velocityofaconduction electron. * “1 i______ £3ye —. z 5Pre "Forexample,onecanaddtwolinearlypolarizedwavesE,andE,,withE, 2 “ rd BeBe perpendicular toE's,thatdifferinphase.Then,atagivenpoint,themaximaofE,andof : wd g = $o52 ,donotoccuratthesametime,andtheirsumEdescribesaneliseaboutthe:-anis.We 2 - ——++t eGkze thenhaveanellipically polarised wave.IE,andEhaveequalamplitudes butarex/27 id£3282 radiansoutofphase,theellipsebecomesacircleandthewaveiscircularlypolarized. The =“ ——+,— $£538 Polarization isrightoFleft-handed according 10whetherthevectorsBandHtrotate- 5 Beee clockwise orcounterclockwise foranobserver looking towardthesource s °+ 22252 *Weshallusethisparticular notation forawavetraveling inthepositive direction ofthe .= Feeig z-axisbecause itisthemostcommon. Someauthors usetheopposite signintheexponent 2- t 3isd2 fandwriteexpi(kz~a),usuallywith/insteadofjforthesquarerootof—1.This '- RE228 expression hastheadvantage ofreducing 10expikzaftertheexp(—iat) factoris 2» pEEEE suppressed. Withthatconvention, onemustsubstitute ~7forjinallphasorcakeulations° £2shs Forexample,theimpedance ofaninductorbecomesR—iwL, insteadofR+jul,Electricalengineersuseexp(jaw~yz),wherethepropagation constant,isequal10Jk S18 PLANEELECTROMAGNETICWAVESI| 519 ‘Then Maxwell's equations 27-24 to27-27 reduce to { , —jk? -E=0, —jk? xE=—jouH, (28-5) ~jké-H=0, -jkEXH= 0B+jock (28-6) 'andthento No k 2-E=0,E=-———ixH, (28-7) éweayet™™ 8-7) k 2-H=0,H=—2xE. (28-8) (~ouItfollowsthatEandHaretransverseandorthogonal.Figure28-2shows <THSGS therelative orientations ofE,H,andk=kz.Observe thatEXHpoints >. inthedirection ofpropagation 28.2.2 The Characteristic Impedance ZofaMedium | Lq™N The ratio E/H isthe characteristic impedance Zofthe medium of propagation: bx “Hoes gees EAT EY 28.2.3 The Wave Numberk 5:|ABRET Je=: ‘Thevalueofk?followsfromtheaboveequation: Be, ee wee!=wren-joou=o*en(1-7), (28-10) Bulo|.«|ecetele|=ae:: °=wPeauoeate(1- 7-2). 28-11 (OFCalto€rt(Joe (28:11) Fig.28-2,TheEandHvectorsforaplaneelectromagnetic wavetravelinginfree ‘space inthepositive direction along thez-axis. (a)Thefields EandHas ‘The oterms account forJoule losses andattenuation functions ofzataparticular moment. Thetwovectors areorthogonal andin. phase.(b)LinesofF(arrowheads), asseenwhenlookingdownonthexz-plane. 28.2.4 The Wave Equations ‘Thedotsrepresent linesofHcoming outofthepaper, andthecrosses linesgoingintoth‘ThevectorEX: eve inet Wefoundthenonhomogenous waveequations forEandBinSec Propagatine TheVectorEX41pomntseverywhereinthedirectionof 27.9: PEcaoe+yB, (28-12) Weeu2ayygSE os~MSHS =Mos (28-14) VB—€oltosz=MoVXJ. (28-13) OBor VBeuPaWoWXE=noe (28-15) Wenow apply therule ofSec. 27.1 and Eq.28-4 toobtain theequivalent equations foramedium ¢,4,0.Weagainsetp,=0.Then Itisthecustomtowritethesewaveequations intheform 2 OE =3.767303 x10°~377ohms. PE~uFpo=0, (28-16) | ms. (28-25)" " Thus, since B=oH infree space, 2; eB B i}ied no=0, (28-17) BE1 on t BrGay? FOE Be. (28-26) oH on ‘TheEandHvectors infreespaceareinphasebecause thech 2 SHHcharacteristic WHeuarHOG, =O (28:18) impedance offreespaceisreal. “Then, from Sec. 28.2.1,‘The electric andmagnetic energy densities’ areequal: 2ew E7/2_&—k?+wen—jwou)E=0, (28-19) oF2ooHe)= r 1jou ) cep ale)! (28:27) andsimilarly forH.Theexpression enclosed inparentheses isequal to ero. from Eq. 28-10Atanyinstant thetotal energy density fluctuates with zasinFig, 28-3, and itstime-averaged value atany point is 28.3UNIFORM PLANE WAVES INFREESPACE CoE2e-Holt?Ee = CoEime=HoHime (28-28) Infreespace, €,=1, u,=1,0=0,thereisnoattenuation, andfromEq.28-19, : Abandoning thephasornotationforamoment,ket (28:20)Io E=E,,cos(wt—kz), H=H,cos(wt—kz). (29-29) - 2 =@(€otto)' (28-21) ‘Themagnitude ofthePoynting vectoris FromEq.28-2thespeedoflightis 191=1BHt=EH,cos?(wot—hz). (23:0) e=9eTar=2.99792458 x10°meters/second. (28-22) cs ‘Thisequation isremarkable. Itlinks three basic constants ofelectro- magnetism: thespeed oflightc,thepermittivity offreespace €that appears intheexpression fortheCoulomb force, andthepermeability of freespace Mofrom themagnetic force law. Since Uois,bydefinition, exactly equal to41x10-7, thevalue of& follows from the value ofc:- 1 Fa.283,Theenergydensity€oE?,oFoH,asafunctionofz,at1=0,foracogi8BSAIBTBIT X10°!farad/meter. (28-23) planewavetravellingalongthez-axisinfreespace. ‘The characteristic impedance ofthevacuum is *Wehave shown thatthe electric andmagnetic energy denis are 9/2 andpol?/2 komt sa roger son) mripemnea Wen matehaeeEK MD gem(Me applytoany are,infact, inagreement withexperiments ontheenergy fxineRaanEee" (a) es24) Compete " 522 PLANE ELECTROMAGNETIC WAVES I 284 UNIFORM PLANE WAVES INNONCONDUCTORS 323 ‘Weshall seeinSec. 28.6 that thePoynting vector, when integrated over a 1 coesurface, yieldsthepowerflowthrough thatsurface. Powerflowsinthe ®Cen (eu)am (28-38) direction ofS. Returning tophasors, thetime-averaged Poynting vector is(Sec. 2.4) where mistheindex ofrefraction: P=} Re(EXH") (28-31) n=(ep,)'?. (28-39) and,forauniform planewaveinfreespace, ‘Thephase velocity vislessthaninfreespace, sinceboth¢,and1,arelarger than unity. Innonmagnetic media, $uy= 4Re(EH*)E (28:32) hs (EH")s :n=e)?, (28-40) 25Emm, =$c€[Eml?2=C€oEime=ZR? (28-33) AswesawinSec.10.1.2,¢,isafunction ofthefrequency, sonisalso p frequency-dependent. Asarule,tablesofnapplytooptical frequencies~ant watts/meter’. (28-34) (=10"°hertz),whereas tablesof€,applyatmuchlowerfrequencies, at3 bestuptoabout 10"°hertz. Pairsofvalues drawn fromsuchtables donot i 2 7 therefore satisfy theabove equationThis isthetime-averaged total energy density €E%,., multiplied bythe hea ;speedoflightc. Thecharacteristic impedance ofthemedium is E (py? 2Example |THEEANDBFIELDS INALASER BEAM z=7-(4) =3r7(“") ohms, (28-41) Most lasers operate atpowers oftheorder ofmilliwatts. However,. there exist afewlasers thatarevastly more powerful. One of Theelectric andmagnetic energy densities areagain equal: these supplies apulsed beam of27terawatts (27,000 gigawatts!) of peak power overacirculararea0.1millimeterinradius.Then ER“eel (28-42) nx108 uP GP=ZW 29x10watts/meter’, (2835) x(10'Y andthetime-averaged energy density is Ema=(377 *9X10)? =6x10"volts/meter. (28-36)»_Eine,WH 22 “Thisisanenormousfield:60voltsoverthediameterofanatom boy=Ei4i2=UH (28-43) (=10°" meter)! Air breaks down atfields ofabout 3x10° volts/meter. Also, ThePoynting vector EXHpoints again inthedirection ofpropaga- Byys=Ee~2x10?teslas, (28:37) ‘ton,and eviFu=SRe(EH*)= (5)Einat (28-44)orabout 2000times thefieldbetween thepolepieces ofa us werfulelectromagnet, |)""Bag pe In~(miEies2watts/meter? (28-45) 28.4 UNIFORM PLANE WAVES IN : NONCONDUCTORS =euEom=VEmt. (28-46) ‘Thesituationhereisthesameasinfreespace,with€andyreplacing€ ‘Thetime-averaged Poynting vectorisagainequaltothephasevelocity ‘andjlo.The phase velocity isnow multiplied bythetime-averaged energy density. sz PLANE FLECTROMAGNETIC WAVES1 245UNIFORMPLANEWAVESINCONDUCTORS 32 Example |THEEANDBFIELDSOFA andthephasevelocityisLASER BEAM INGLASS o v=o (28-56) There isnopoint inreferring here tothelaser beam ofSec B 28.3 because such abeam would instantly vaporize glas. Say we have a1.0-milliwatt beam with adiameter of1.0millimeter in Let usfind aand iinterms of€,,1,0,and Ay.First weset lass whose index ofrefraction is1.5.Then p=—5=1,310"watts/meter® 28. - - andlor (28-5 Gmgn 13X10" watts/meter, (28.47) we \eak/ar |@D/a wy ye axxo? a This isthemagnitude oftheconduction current density, divided bytheFm=[(2) 9]=e 3)x13%10°] magnitudeofthedisplacement currentdensity.Asarule,@(for(2-48) dissipation”), iswrittentan/,asinSec.10.1.1: =5.7%10?volts/meter, (28-49) o=tanl, (8-58) BimHola=Ha)Erma=(Et)Em (28-50) where|isherethelossangleofthemedium,butweusefor =(165° 8,85107? 30107)! 5.710 concisenessSOLSBESxOEAeLOY ‘Thepermittivity €thatappearsaboveistherealparte/ey(Sec.=2.9x 10°tesla (28-51) 10.1.1). Onecanaccount forconductivity either bymeans ofacomplex permittivity (€;~je!)éy orbymeans ofareal permittivity and a 28.5 UNIFORM PLANE WAVES INCONDUCTORS conductivity ,where o=wee, again asinSec. 10.1.1. Thus WestartatthepointwhereweleftoffinSec.28.2.4. 9=f (28-59) 28.5.1 The Complex Wave Number k=B—jar ; : This quantity, like €/and €!,isalways positive. InSec.28.2.3 wefound thatinaconducting medium If®<1, themedium isagood dielectric; if>>1,themedium isa . + goodconductor. Forcommon typesofgoodconductor, ~107(0= w=(1-j-2), (28-52) 5.8107forcopper)and€,~1(Sec.4.3.6).Youwillremember from % we Sec. 4.3.6 that €/o istherelaxation time ofamedium. sokiscomplex. Itisthecustom toset Thus 2=(B-jay=(SH)-j9 7 k=B-ja andthen E=E,,exp(~az) exp(wt—Bz),(28-53) k=(B—jay=(7B)a-i9), (28-60) where both aandfarepositive and Thequantity 1/aristheattenuation distance ortheskindepth 8over 1(ene ogwhichtheamplitude decreases byafactorofe.Therealpartfofkisthe a=(SY) lato, (28-61) inverse ofA: : 1 ia=t, (8-4) pa}(SH)"ara, (302) 1_20 en)?5-3, (28-55) n=lwe)(149%) exp(Jarctan5)(28-63) 526 PLANE ELECTROMAGNETIC WAVES I 527 ‘Theargument oftheexponential function iscorrect because fispositive as (Sec. 2.1). Inalow-loss dielectric Dissmall, and w a <)ae (28-64) Z (ety) o cer 65 ona Coe ese) on 0 00Insuchmediatheconductivity hardlyaffectsthephasevelocity, butit 0° x gives risetoanattenuation thatisindependent ofthefrequency. Fig.28-4,ThTnagoodconductor9>>1and Rie,284.ThephaseOfwithrespecttoHasafunctionofinawavetraveling Ko j9 Hj ensv%eaty= jouw, (28-66)i o=arctan. (28-73)k=(4°)a-), (28-67) ; 2 Figure28-4shows0asafunctionof2. Therefore, ou)" 7a=p=(4") . 28-68) - Bay (28-68) E=E,, exp(—az)expj(wt—Bz), (28-74) ‘Theindexofrefraction ofagoodconductor H=H,exp(—az)expj(wt—Bz—@), (28-75) with ¢ _B_ (on)'?== 8. (% 28 rr "=OBo(35) (28-69) f=9a(H+)1(tsamoa Hyk\e)(+9)"~a)é)Gray 8-76) isalarge quantity. Itis1.1x10°forcopper at1megahertz, uy? 4~377() Log ohms :28.5.2TheCharacteristic Impedance ZofaConductor «)(+9) (28-77) ‘Thecharacteristic impedance ofaconducting medium iscomplex: FromEq.28-8,£andHareorthogonal inalinearly polarized wave. Ifthewave isnotlinearly polarized, then thevectors Eand Harenot geen Kk_ee (28-70) necessarily orthogonal. Hwe-jo k og5.3 iti=(0)eeiantanal.a(t)"801880019pang) ‘TheEnergyDensities <)asa" @ asa)" Thetime-averaged electric andmagnetic energy densities areintheratio aswesawinSec.28.2.2.ThismeansthatEandHarenotinphase: 6._eBim/2 1 5 @,wHe,/2 +? (28-78) E__on (28-72) . Baie (28-72) Thereislesselectric energythanmagnetic energybecause theconduc-tivitybothdecreases Eandaddsaconduction current tothedisplacement where@andfarebothpositive. SoEleadsHbytheangle current, whichincreases H. 528 PLANEELECTROMAGNETIC WAVES1 47SUMMARY 29 ‘Thetime-averaged totalenergy density is ThePoynting theorem therefore simply states thatthere isconserva- . tion ofenergy inelectromagnetic fields. Itisaproof ofthevalidity ofEq. M€E2a+WHgs)EXP(—2az)=M(€Eim,)[1 +(1+9?)'"]exp(—2az). 27-23,andhenceofEq.27-4 (28-79) For auniform, plane, and linearly polarized wave inconducting material, the time-averaged magnitude ofthe Poynting vector is28.6THE POYNTING THEOREM ged magr yynting Sou.=4Re{[Emxp(—az)expj(wt—Bz) Wereferred tothePoynting vector *Hyexp(—az) exp j(—at +Bz+8))) (28-86) S=EXH (28-80) =4EmH»cos8exp(—2az), (28-87) inprevious sections, butwesaidvery little about it.Weonly stated thatit where 0isdefined asinSec.28.5.2 and isequal tothepower density inanelectromagnetic wave, and that it points inthedirection ofpropagation. _ 6BThePoynting vector isofgreattheoretical andpractical interest. Its 00sO=ry BE (28-88) significance follows from thePoynting theorem that wenow prove First, wehavethevector identity Wefound theratioE,,/H,, intheprevious section. Ifweeliminate H,,, then V-(EXH)=H-(¥XE)-E-(¥ XH). (2881) ar)=(5)(1+9°)"Eing 6058exp(—2az) (28-89) InaHILS medium, Eqs. 27-20 to27-23 apply, and then Ue ; 1ey"? 2)14 2vexH)=-Hu tg.(Ess) (28-82) ~m(e) (1+9°)E2,,.cos8exp(-2az). (28-90) --2(€4 ME)pas, 883) Youcaneasilyshowthata,=(time-averaged energy density) X(phase velocity) (28-91) Wenow change thesigns, integrate over avolume voffinite extent and ofsurface area of,andfinally apply thedivergence theorem ontheleft. 28.7 SUMMARY This yields thePoynting theorem: a(ee?pi Inauniformplanewavethewavefronts,whicharesurfacesofuniform ~[exa-at =F|(Fe) avsfe-sdv. esas phase,arealsosurfacesofuniformamplitude, . Linearly polarized uniform plane electromagnetic waves traveling either ‘The first integral ontheright gives theincrease intheelectric and infree space orinaHILS nonconductor orconductor possess the magnetic energy densities inside thevolume v,perunittime, Thesecond following properties. gives that part oftheficld energy that dissipates asheat, again perunit time. Then theterm ontheleft,withitsnegative sign, must represent the (1).Thevectors EandHaretransverse andorthogonal. eaieinee etenergyflowsintothevolumev. (2)ThePoyntingvectorEXHpointsinthedirectionofpropagation. (3)‘ThemagnitudeofthePoyntingvector,averagedovertime,gives iS-dt= [(EXH)-ast (28-85) thepowerflowpersquaremeterinthewave: isthetotalpower flowing outofaclosed surface ofareasf. Su=hRe(E XH"), (28-31) 530 PLANEELECTROMAGNETIC WAVES1 PROBLEMS so (4)Thepowerdensityisequaltotheenergydensitymultiplied bythe P(EXE+ eRxt)=~ B-i,phase velocity. 7 (5)TheratioE/Hisequaltothecharacteristic impedance Zof werethedotsabove andBindatepartialcliferentiaton withrespect i to time, media. ‘Thisequation hastheformofaconservation law.Thespatial density of Infreespace, theconserved quantity appears between parentheses ontheright. Thisis 1 expressed involts squared percubic meter. Thefluxdensity ofthis c= 75=2.99792458 x10%meters/second, (28-22) conserved quantityisthequantitybetweenparentheses ontheleft.(€otto)’ (b)Showthatthefluxvanishes inthefieldofalinearlypolarized wave.Z~=377ohms, (28-25) (©)Showthat,inthefieldofacircularlypolarizedwave,thefluxdoes> notvanish, thatitisproportional tothefrequeney, andthatitcontrary toy,,~=2 2watts/meter. (28-34) thedirectionofpropagation iftheEandBvectorsrotateclockwiseforan“377 observer who looks atthesource, andinthedirection ofpropagation ifE Innonconductors, andBrotateintheopposite direction. ce (2838) 28-2.(284) Thephaseandgroupindices ofrefractionScreed Thephase indexofrefraction, oftencalled thephaseindex, is/v¥,,a where v,isthephase velocity. Thegroup index issimilarly c/v, where where1istheindexofrefraction. Also, Maaltaplaen))%thedeasMama(ee.29.2.6). 2z~=3(#) ohms. (28-41) 28-3.(28.5)‘Theskindepthasafunctionoffrequencyinlow-conductivity © materials . . (a)Plot onasingle graph the log-log curves oftheskin depth asaInconductors thereisattenuation andk=6—ja,where function ofthefrequency fromf=1tof=10",foroequalto10-7,10-*, 1» 10andfor€,=1and¢,=10.Setu,=1.Theskindepthwillvaryfrom ant(#)(a+93"2-17, (28-61) about10to10°meters. V2 (b)Show that, innonmagnetic good conductors forwhich 0=S0ee, _ =503/(fo)"* p=2(“) (a+r)? 41), (28-62) (c)Showthat,innonmagnetic poorconductorsforwhich0<0.lwe, X\2 6~5.3%10 ee. e 2X4,(285.1)Theopticalproperties ofmetalsarr sn Atopticalfrequencies (f~10"hertz)andabove,thevaluesof0,¢,and 14bearnorelationtothevaluesmeasuredatlowerfrequencies. Formetals, ou om) bothfandaareoftheorderof3/4,withinapproximately afactorof10 z=either way, and f#a. For aluminum at4,=650 nanometers, Bhy= 1.3 . and ah =7.11 i i (a)Calculate 4and6. ‘Theelectric andmagnetic energy densities areequalinfreespaceandin : ‘ ,nonconductors. Inconductors themagneticenergydensityislargerthan (b)Calculate -$fromSec,28.5.1andTable29-1. °N-S.(285.2)2,6,andZinpoorconductors theelectric energy density. ‘ThePoynting theorem isastatement oftheconservation ofenergyin Showthat,nahomtatclysightconducting(91),anelectromagnetic field. ()d=2°K/2 (©)Z=(1 =9°/4)expfarctan(2/2)]Z,-0 PROBLEMS2-6. (28.5.2) Alternate expressions for the characteristic impedance ofa conducting medium 28-1.(282.1) Ageneral theorem forelectromagnetic fieldsinfreespace Showthatthecharacteristic impedance E/Hofaconducting medium is""(@) Showthat,foranyelectromagnetic fieldinavacuum alsogivenbythesetwootherexpressions: 532 PLANE ELECTROMAGNETIC WAVES 1 33 Btia jou_\"? 287.(28.5.2)Thecharacteristic impedance ofultra-low-loss polyethylene §9essranssniscises ce}ae,‘Theultra-low-loss polyethylene thatservesasinsulator insubmarine mscoaxial cableshasalossangleof50microradians andarelative permittivity r 2, of2.26. Calculate itscharacteristic impedance at45megahertz 28-8, (28.6) ThePoynting vector inthefieldofaresistive wire carrying a 9,0 current = ‘Along, straight wireofradius @andresistance R’ohms/meter carries & vig.285currentJ hs dex “os (a)CalculatethePoyntingvectoratthesurface,andexplain 28-11.(286)Energyandpowerinaproton(b)CalculatethePoyntingvectorbothoutsideandinsidethewire. pally28-8shows2highly<eyatedtingofaprotonaccelerator. Explain Agasdischarge within thesource $ionizes hydrogen gastoproduce 28-9.(28.6) ThePoynting vector inacapacitor Protons. Someoftheprotons emerge through aholeandarefocused intoAthin,air-insulated parallel-plate capacitorhascircularplatesofradius abeamBofradiusR;insideaconducting tubeofradiusR:.ThesourceR,separatedbyadistances.AconstantcurrentJchargestheplates isatapotentialV,andthetargetisgrounded.through thinwiresalongtheaxisofsymmetry. Toavoidneedless complications, weassume thatthecharge density in(a)Findthevalue of£between theplates asafunction ofthetime. thebeam isuniform. Wealsoassume thatthevelocity oftheprotons isAssume auniform E.Show thedirection ofEonafigure. ‘much lessthan¢:v?«e*. (b)Themagnetic fieldisthesumoftwoterms, H.,related tothe Calculate, intermsofthecurrent Jandthevelocity v:current inthewire,andH,,related tothecurrent intheplates. Thelatter (a)theelectric energy permeter €;; current deposits charges ontheinside surfaces oftheplates (b)themagnetic energy permeter @2;FindH.,H,,andH.Usecylindrical coordinates withthez-axis along (€)theenergy fluxassociated withthePoynting vector Pe;thewireandinthedirectionofthecurrent.TocalculateH,,apply (@)thekineticpowerPorthefluxofkineticenergy,disregarding P,.‘Ampere’s circuital lawtoeachplate. Youshould findthatthemagnetic ‘Theexistence ofthisPoynting vector isinteresting. Because ofthefields tendtoinfinity asp-+0. Thisissimply because wehaveassumed radial E,thevoltage inside thebeam isslightly positive. Sotheprotons infinitely thinwiresandplates. Show thedirections ofH.,H,,andHon arenotaccelerated tothefullvoltage V,andthekinetic energy inthe yourfigure. beamisslightly lowerthanV/.Mostofthepower flowsdownthetubeas(©)DoEandHsatisfyMaxwell's equations? Youshouldfindthatoneof kineticenergy,andtherestflowsaselectromagnetic energy.Thetotal‘ourassumptions isincorrect Power atanypointalong thetubeandonthetarget isVi.(@)FindEX. (©)Findthenumerical valuesofthesequantities fora1.00-{@)Findtheelectric andmagnetic energy densities inside aradius p. milliampere, 1.00-MeV (megaelectronvolt) proton beam, withR,=1.00 You should findthatthemagnetic energy density isnegligible if millimeter andR.=50.00millimeters. plc’. This condition applies because wehave assumed thatthe 28:12. (28.6) Thesolarwind ‘capacitor charges upslowly. Ifitcharged veryquickly, thentherewould be "Thesolar wind iformed ofhighly ionized, andhewave for i s ancehighlyconduct-awaveofEsandHfinthecapacitor,Ewouldnotbeuniform,andtheabove ing,hydrogenthatevaporates fromthesurfaceofthesun.Intheplaneof(0)NowrelatethePoyntingvectorattotheelectricenergyinsidep. piningcuseardimoatnpioussntrvardincdkomSineshesec __(g)Drawasketch showing E,H,andEXivectors atvarious points fotates (period of27days), whiletheplasma hasaradialvelocity, the inside andaround thecapacitor. lines ofBareArchimedes spirals. This isthegarden-hose effect. Atthe 28-10. (28.6) ThePoynting vector inasolenoid ‘earth, thelines ofBform anangle ofabout 45°with thesun-carth ‘Alongsolenoid ofradius RandN’turns permeter carries acurrent J direction. (a)Thecurrent increases. Calculate F=EXH.(Seeexample inSec. Attheorbit oftheearth thesolar wind hasadensity ofabout 107 19.1.) proton masses per cubic meter and avelocity ofabout 4%10° ‘Sketch across section ofthesolenoid, showing thedirection ofthe ‘meters/second. Themagnetic field ofthesunisabout 5x10-* tesla current and of$.Explain (a)Showthat,inaneutral(p=0)plasmaofconductivity oand (b)Repeat withadecreasing current, velocity v,Maxwel’s equations become 334 PLANEELECTROMAGNETIC WAVES1 535 vee=0, Pxe=-22, ¥-B=0, aor / \ @or / \wwvxBanfo+exB)+o%| / a \ a KE\ Inamedium ofinfiniteconductivity 0,E=~vXB.Thisisasatisfactory (SiGe) Ow|approximation forthesolarwind. X4 }(b)Showthatthecomponent oftheplasmavelocityvthatisnormalto .—Bisgivenbyv,=[BX(vXB))/B™ \ J(©)ShowthatthePoynting vector isgivenbyS=Bv,/j4o,oFabout6 NN 7icrowatts/meter’. Thisisabout4x10°"timesthePoynting vectorof —— Fig.28-7. solar radiation, which isabout 1.4kilowatts/meter’, ThePoynting vector‘ofthesolarwindisnormal tothelocalB (€)Showthedirection ofthemagnetic forcesonthebars. (d)Showthatthekinetic, magnetic, andelectric energy densities ofthe 28-14, (28.6) ThePoynting vector inatransformersolar wind arerelated asfollows: #2 8» bi Figure 287 shows, insimplified form, across section ofatransformer 28-13.(286) ThePoynting vectorinaninduction motor secondary. ThefieldBinsidethecoreC,andtheleakagefieldH,outside,Tnaninduction motor, thestatorgenerates amagnetic ficldthatis bothresultfromthecurrents inthewindings andfromtheequivalentperpendicular to,andthatrotatesabout,theaxisofsymmetry (Prob. currents inthecore.Thesecondary winding isW.Assume thatBandH,18-6).Therotorisacylinder oflaminated iron(Prob,25-7),withcopper incresss.barsparallel totheaxisandsetingrooves inthecylindrical surface (a)Drawalargerfigureandshow,atonepointbetween CandW,Copper ringsateachendoftherotorconnect allthecopperbars,The vectorsA,3A/3t,andE=—3A/3t, disregarding thecurrentinW.Showrotorisnotconnected tothesourceofelectric current thatfeedsthe @vectorEatonepointoutsideW. motor(b)Assume that theimpedance ofthesecondary isapure resistance R. ‘ASweshallsee,therotortendstofollowtherotating magnetic field. ‘Showthedirection ofthe current JinW. Figure28-6showstheprinciple ofoperation. Tosimplify theanalysis, we (c)Showthedirection ofitsfieldHatpointsbetween CandWandsuppose thattherotorisstationary andthatarotating clectromagnet, outsideW.represented herebyitspolesNandS,provides therotating magnetic ()ShowvectorsEXH. fad.(c)How would thedirections oftheEX Hvectors beaffected ifBand (a)Drawalargerfigurewithwideairgaps,showing thedirection of H,decreased?theinducedcurrentsinthebarsandthedirectionofEintheairgaps. (Q)Whatisthetime-averaged valueofavectorEXH;?(b)Thecurrent intherotorgenerates amagnetic field.Addarrows (2)Nowletuscalculate thepowerflowintothesecondary. Assumeshowing thedirection ofsharHf,insidetherotorandintheairgaps. thatthesecondary isalongsolenoid ofNturnsandoflengthL.(c)NowshowPoynting vectors ExHintheairgaps.Thefieldfeeds Disregard H,andset©=®,,expjutinthecore.Integrate thePoynting power into therotorvector over aeylindrical surface situated between the core and the (a)Nowdrawanother figureshowing thecurrents inthebarsanda winding, andshowthatthepower flowing intothewinding islineofBforthesumofthetwomagnetic fields, (reo—t /R=Ving/R,whereVisthevoltageinducedinthesecondary / o\s4/ . seip 7aya\ )*s\\e= < y,——__ Fig.28-6. \UNIFORM PLANE ELECTROMAGNETIC WAVES INGOOD CONDUCTORS 537 29.1 UNIFORM PLANE ELECTROMAGNETIC 29 WAVESINGOODCONDUCTORS. cuarrerTHE SKIN EFFECT Recall from Sec. 28.5.1 that, inalinearly polarized, uniform plane wave *PLANE ELECTROMAGNETIC ‘ropagating inaconductor inthepositive direction ofthez-axis, WAVES II E=E,,exp(~az)exp(ot~fz). (291) - , conductor asamaterialst. ession WavesinGoodConductors andinPlasmas teef9edcond xaerialrchUninhexpress (+99)? +1?=9"? (29-2) 29.1 UNIFORM PLANE ELECTROMAGNETIC WAVES INGOOD, vonie watisfi o%CONDUCTORS. THESKINEFFECT 537 hiscondition issatisfied within 1%if Example: PROPAGATION INCOPPER AT1MEGAHERTZ 540 gu2m[8 ) Example: JOULE LOSSESINGOODCONDUCTORS 540 Gm2m|-8F|250, (29) 29 PLANE ELECTROMAGNETIC WAVES INPLASMAS S12 29.2.1 THE CONDUCTIVITYOFAPLASMA542 +iftheconductioncurrentdensityisatleast50timeslargerthanthe Example:THE IMPEDANCEOFAUNITCUBEOFPLASMA 543 lisplacement currentdensity.Butnoteherethatoand€arefunctionsof 29.2.2 THE CONDUCTION AND DISPLACEMENT CURRENT DENSITIESAND ».especiallyatopticalandx-rayfrequencies, So@doesnotdecrease THEPLASMA ANGULAR FREQUENCY ayS44 adefinitely as1/f,astheabove equation appears toindicate Example: PLASMA-FILLED PARALLEL-PLATE CAPACITOR S85 Tngoodconductors thewaveequation 28-16reduces to! 29.2.3 THE WAVE NUMBERk54S 29.2.4PROPAGATION AT/>/, 546 . OE 29.25THEFIELDAT/ =f, 547 VE- yo =0, (29-4)29.26THEFIELDAT/<j, SAT Example: THE TELECOMMUNICATION BLACKOUT UPON THE andEq.28-10 forthewave number to REENTRY OF ASPACE SHUTTLE S48 . Example: THEIONOSPHERE S88 = —-jwou (29-5) 293. SUMMARY S49 thus PROBLEMS 550 /won\"? 1- ce hk=p-ja=(S) a-aes nao==hk=3(1-)) InChap. 28wededuced themain characteristics ofauniform plane (29-6) electromagnetic wavetraveling through amedium €,,44,0.Wenow 1 1/wou\"applythisknowledge totwosimplemedia,namelygoodconductors and* Bayra=5- (S*). (29-7)low-density plasmas. ‘The propagation ofelectromagnetic waves ingood conductors is wheremistheindexofrefraction, A=A/2:r,asusual,andwhere8is the peculiar inthat theamplitude ofthewave decreases byafactor ofein «attenuation distance, defined inSec. 28.5.1 asthedistance over which the oneradian length A=/2. Theattenuation issolarge thatthewave is mplitude decreases byafactor ofe. hardly discernible. i Westudy plasmas oflowdensity soastobeable todisregard energy :lossesarisingfromcollisions ‘Anequationofthesimeformappliestoheatconduction. SeeProb.29-4 538 “PLANE ELECTROMAGNETIC WAVES It 539 From Eq.28-70, thecharacteristic impedance ofagoodconductor is " HELom (OH) pitZaFen(EYexp, (298) os and Eleads Hby7/4 radian. Compare with nonconductors inwhich E andHareinphase (Sec. 28.4). The difference comes from thefactthat y thecurrent thatisassociated with Hingood conductors istheconduction ope. current, which isinphase withE,andnotthedisplacement current of eet ms nonconductors, which leadsEby90°. 4 RA Therefore ra SY £=Fyew|i(or—2) -{] 99) , E=Emexp|i(wr—-5)-5 |. . o\? zm) 2H=(2)"E,exp [i(we-3-2)-3| (29-10) Vig.291,TheratiosE/E,andH/H,att=0asfunetionsof2/Kforan oulcctromagnetic wave propagating inagood conductor. ‘The vectors Eand Haretransverse and orthogonal, sayEisparallel to thex-axis andHtothey-axis. Interms ofcosine functions, 5, © ww"vp=F=wh=(= (29-14)E=Enexp(-5)cos(w-3), (29-11) 6 on: 'sproportional tothesquare rootofthefrequency._(ey 2 2ox Ingoodconductors thegroupvelocity(App.C)istwiceaslargeastheH=(sa)Enexp(-7)cos(wr-2-7) (29-12) phasevelocity: z zm 1 = 2)cos(ws2-7). 29-13) "Tada 2 9-15 Hmexp(5)0s(wr5*) (29-13) “aaa (29-15) Figure 29-1shows E/E,, andH/H,, asfunctions of2/Aat¢=0. ioandqarenotfrequency-dependent, ‘Theamplitude ofthewave decreases byafactor of(I/e)** =2x10° Theratioofthetime-averaged electric tothetime-averaged magnetic inonewavelength, andthePoynting vector by(1/e)** ~3x10°*. Thisis cnergy densities is theskin effect. 2Theattenuation distance6inconductorsistermedtheskindepth,or caelanted (29-16)thedepthofpenetration. Theskindepthdecreases iftheconductivity, the HHiml2 9B50 relativepermeability, orthefrequencyincreasesGoodconductors “ Theenergyisthusessentially allmagnetic,Thisresultsfromthelargetherefore opaque tolight,except intheformofextremelythinfilms.Isconductivity0,whichcausesE/J,tobesmall,Theelectricfieldstrength foesnotoieorascatatsubstances thatarenonconducting atlow 'sweak,butthecurrentdensityandhenceHarerelativelylarge.Soeeaaees ane Sane Stepten Sequences. FromEqs.29-11and29-13,thetime-averaged valueofthePoTable29-1showstheskindepth4forvarious conductors atfour sectors eragedvalueofthePoynting typical frequencies. Theattenuation inironismuch larger than insilver, despite thefactthatironisarelatively poorconductor. v.<1R 1/0 \? 22)09 , wePRe(ExH*)=5(2 -S)re 29 sw si Table 29-1 Skin depths 6forconductors 5Table29-1Skindepths8forconductorsCONDUCTOR ° o or? oneRTZ TKILOHERT2 —1MEGAHERTZ —3GIGAHERTZ Aluminum 354%10" 1.00 0.0846 10.9 2.67 36 14Brass(65.8Cu,34.2Zn) 1.59x10°1.00 0.126 16.3 3.99 16 230 Chromium 3.81071.000.0816 103 258 a16 149 Copper 5.80%1071000.0661 853 2.09 66.1 121 Gold4.5010"1.000.0750, 9.69 237 75.0 137 Graphite1.0x10° 1.00, 1.59 21x10? so 1.6.x 10° »Softiron 10%10 2x10"=0.011 1 04 10 02Mumetal 1.6%10° 2x10" 0.0028 oF 0.09 3 0.05Nickel 13x10 1x10" O04 2 ot 0 03‘Seawater s 1.00 2310 3x10" 7x10 2x10° 4x10"Silver 6.15x10’ 1.00 0.0642 8.29 2.03 64.2 LyTin 870%10° 100 0.71 2.0 540 m1 3.62Zinc 1.86%10 100 0.7 15.1 308 u7 23 *AtB=0.002 tesla. *Atthisfrequency, ¢,~35,9~1,andseawater isnotagood conductor, Example PROPAGATION INCOPPER AT1MEGAHERTZ Copper hasaconductivity of5.80x10”siemens/meter. Then, at1 conductor, perpendicular tothedirection ofpropagation, asinmegahertz, Fig.29-2.Iftheamplitude ofEontheleft-hand faceisE,,,then7 5.80 107 a fontheright itisE,exp(~Az/d) and,within theslice, theoe ee (29.18) time-averaged Poynting vector decreases by and p=!(2)ef 2a2)a 3 , ate=3(525)#2[1exn(-722)]. 922) oh(Sie RSaxxa I =66micrometers (29-19) 7 ‘Thewavelength isabout 0.4millimeter incopper, while itis300 | meters inair.The phase velocity iscorrespondingly low: : u,=@k=415meters/second, (29-20) which isabout 10times less than thevelocity ofsound incopper (3.6kilometers/second). ‘n ‘Thecharacteristic impedanceis /fi IE]_(22x10x4107)" 4 s ai-|F-(A ea )"=3.7%10-ohm, 29-21) hj Example |JOULE LOSSES INGOOD CONDUCTORS Fo leas Letuscompare thepower lostbythewave andthatgained bythe .29-2. Element ofvolume normal tothedirection ofmedium throughJouleloses.Weconsider athinsliceinside@ Prbparation i'sgoodconductor, restonl 32 “PLANE ELECTROMAGNETIC WAVES IE 29.2PLANE ELECTROMAGNETIC WAVES INPLASMAS. 543 Thepower P,lostbythewaveisabtimesthis.If22/3 <1, Wenowrequire thevelocity vofafreeelectron ofmassmsubjected to ab/ 0\"., 2Az 1nalternating electric field. Sincepat(52)ee obinaba2 029-28)2(op 3 dv mee=mjov=CE»expjot, (29-26) Thisisjostthepower dissipated perunitvolume byJoule losses dt (Sec. 4.3.7). then e v=j——E,, expjot. (29-27) 29.2 PLANE ELECTROMAGNETIC WAVES fom INPLASMAS Thevelocity leadsthefieldby90°,Substituting thisvalueofvinthe ibov forthe conductivity gives’ Aplasma isanionized gas.Asarule,plasmas contain freeelectrons and OveEXPRESSION FOANECONENAINTEY BINS positive ions. Since theions aremore massive than theelectrons, the jNe? 4.48 10-%N current iscarried almost exclusively bytheelectrons. Wetherefore oeomF~CSS@mens/meter. (29-28) disregard ionic currents. Weassume that thegaspressure islow. This will permit ustoignore The average power dissipation percubic meter is, collisions between thefreeelectrons andthegasmolecules, andhence to a a . »ignore energy losses. Then thegasoffers noresistance tothemotion of P'=2Re (E-J7)=}Re(aEE*) =0, (29-29) freeelectronsandactssomewhat likeasuperconductor. Also,wemayset 1accordance withourassumption ofzerolosses.6=1 Finally,wedisregard thermalagitation. Ineffect,wesetthetempera- Example |THEIMPEDANCE OFAUNITCUBEOFPLASMAture equal tozero. First, imagine acube ofsome resistive material such ascarbon, 29.2.1 TheConductivity ofaPlasma withcopper electrodes deposited onopposite faces. Ifthecubehas avolume of 1cubic meter, then the resistance between For afixed observer, afree electron situated inanelectromagnetic wave ‘opposite faces is issubjected toaLorentz force ~e(E +vXB), and theelectron current 14 density inaplasma is Rass o (29-30) Jp=AE+0XB). (29-24) ‘Thustheconductivity ofamedium istheconductance G'=1/R', -or,moregenerally, theadmittance ¥",of1/2",between opposite WecandisregardthevXBterminthisequationforthefollowing facesofaunitcube reason. Asweshall seebelow, theratio E/H forawave inaplasma is With aplasma, always larger than infreespace, where E/H =(t4o/€y)"? (Sec. 28.3). So, 1 minaplasma,E/H>(do/€q)'? andE/B>c,Also,wemaysafelyassume 2!=1=jo(gs)=jol', (29:31) that thevelocity vofafree electron willbelessthan cbymany orders of magnitude. Sowemay set where L’isthe“inductance” of1cubic meter ofplasma. This quantity isexpressed inhenry-meters and isindependent ofthe Nev frequen Jj=-Nev=0E, o=-, (29-25) ‘quency whereNisthenumberoffreeelectrons percubicmeter,eisthe “Thisexpressionismeaningless atzerofrequencywheretheconductionelectronswouldmagnitude oftheelectronic charge,andoistheconductivity realm eereachaninevelo.Equations225,however,applytosteady sas “PLANEELECTROMAGNETIC WAVESIt ss 29.2.2 The Conduction and Displacement Current Densities and thePlasma Angular Frequency w, Inaplasma, ae aviano 2D Ne =esOB=jweoE +OE=joeE —EE (29-32)ot om q a + 1 Net~ = jweoE(1- 29-33)re arr aay s The displacement current density 3D/3t leads Eby90°, while the it conduction current density oElags by90°. The term between parentheses isthe equivalent relative permittivity ofthe plasma because, inadielectric, Hig.293.Thecurrentdensities3/3,oF,andthersum,afunctionsofw/a,wwplasma 8D SraHOKE: (29-34) Theequivalent relativepermittivity canbeeitherpositive ornegative. txample |PLASMA-FILLED PARALLEL-PLATE CAPACITOR Rewriting Eq.29-33, Ifthecapacitor plates have anareasiandareseparated by@ distance s,then,uponapplication ofavoltage V, apwi SptoF=jocvE(1 ~22), (29-35) or o). si alls),12(jwc+02)r=(joc+24)y (29:39)where thesecond termbetween theparentheses istheratioofconduction a currentdensitytodisplacement currentdensity,andwhere ‘Thepresenceoftheplasmathereforehasthesameeffectasifone . hhadaninductance sL’/sfconnected inparallelwiththecapacitor. Ne?\!2 ‘Thenetcurrentiszerofor o,=(%2) (2936) currentiszer0fem SL’ wont (29-40)istheplasma angular frequency. More simply, a ®, teoqne Inactualfact,theplasmawouldnotbeuniformandNwouldbe f=S2=8.98N'*=9N'? hertz (29-37) smaller neartheelectrodes thaninthebodyoftheplasma, Moreover, wehave ignored edge effects. Sothisexample isnot Also, from thevalue ofothatwefound above, realistic 2 9.2.3.7 jo0% (29.38) 29.2.3 TheWave Number k oe @ Recall fromSec.28.2.3 that,inaconductor, thewavenumber kisgiven Figure 29-3shows aE,3D/3t, andtheirsumasfunctions of@/W,. by Intheplasma ofagasdischarge, Nvaries widely butistypically ofthe aEth oorderof10"electrons percubicmeter,andf,isabout10gigahertz. In ane (1-i). (29-41) theionosphere, Nisoftheorder of10"electrons percubic meter, andf, about 3megahertz. Seebelow Then, from Eq.29-38 andsetting ¢,=1,yt,=1, 546 <PLANEELECTROMAGNETIC WAVESIt so 21 ow} ' 224 ()-22 29-42)Rag(1-25) (29-42) i “Thewavenumber iseitherrealorimaginary inalosslessplasma. HH H 29.2.4Propagation atf>f, Hse [Athighfrequencies suchthat@>@,,orf>f,.kisreal.Thenthereis toh----4---~-- SSSzeroattenuation, asweassumed atthebeginning !“TheEandHvectorsforauniform, plane,linearlypolarized waveare Haad transverse andorthogonal, asinanyhomogeneous andisotropic medium. ‘Also, from thegeneral expression forthecharacteristic impedance aE/HthatwefoundinSec.28.2.2foranylinearlypolarizeduniformplane °‘25ry 4 wave,“ E_om bole _\" 377 Fig,29-4,Thephaseandgroupvel1 pL wm(tales) 3 77ohms, (2943 groupvelocitiesu,andv,ofanelectromagneticwave 2a (;ora) Goa)? TTohms.(29-43) *Plasma, = ‘Thewavelength 2=27/k islonger thaninfreespace, andthephase Wemaysetthesignalvelocity equaltothegroupvelocity (App.C).velocity islargerthanc: Then c 1rod = — as29. =--1_.¢ acGao)? (29-44) “Edam y, Ueteme (29-46) c c“G=R06NIPY? 1-8INIPY® meters/second, Figure29-4showsthegroupvelocityasafunctionoftheratiow/w,.. (29-45) 29.2.5TheFieldatf=f, asinFig. 29-4 Wethef‘Thephasevelocityincreases withincreasing electrondensity.Waves tenau oftheappliedfieldisequaltotheplasmafrequency f,, therefore tend tobend away from more highly ionized regions, justasien k=0,5+ inEqs. 29-11 and 29-13 forEand H,and lighttends tobend away from low-density airinamirage .Thisrequiressomediscussionbecause,accordingtorelativity,asignal E=Eqcost,—H=Hyc0s(or-2), (29-47) cannot travel atavelocity greater than «.“Thephasevelocity isthevelocity ofpropagation ofagivenphase,and Thereisnowave,and9D/3t+oFiszero. notthevelocity ofpropagation ofasignal. Thereason isthatawavecantransmit asignalonlyifitismodulated, inamplitude, infrequency, or 29.2.6 TheFieldatf<f,otherwise. Nowanymodulation, andhenceanysignal,involvesfre- |Atlowfrequencies, kisima a‘quenciesotherthanthecarrierfrequency. Thephasevelocityinan quencies,kisimaginary. Settingk=—jk’gives ionized gasbeing frequency-dependent, thevarious frequency com- E=E, exp(—k'z) expjoti E=En 2 , HE -k’ ponents ofasignaltravelatdifferent velocities. Thenetresultisthatthe P(-K'2)expjar Hyexp(~k'z)expja.(29-48) envelopeofthewave, which carries thesignal, travels atavelocity thatis, There isnowave, andchefielddecreases exponentially withz. different fromthoseofthevarious components ofthewave, andthis ‘Theaverage Poynting vector isclearly zerobecause thereiszeropower signalvelocity islessthan¢ flow,Wecanalsoprovethisformally asfollows. FromEq.29-43. sas. *PLANEELECTROMAGNETIC WAVESIt 93SUMMARY 549 E_jaug Otiohy Ho"? 1 Ourassumption thattherearenocollisions between theAve eile (2)Gro (29-49) electronsandthegasatomsormoleculesisnotsatisfactoryinthe I 5 lower regions oftheionosphere, where thepressure ishighest, at Thenoustt® frequencies oftheorderof1megahertz orlower. =1Re(oHolme -2k'2)= : ‘Theearth'smagnetic fieldrenderstheplasmadoublyrefracting, m=Re(NGr=)erp(-2#')=0. 9.8) andthreavtedstpcevlc deping 0mwate theEvector ofthewave isparallel orperpendicular totheBof Anelectromagnetic wave therefore travels through alow-density theearth. plasma without attenuation atf>f,. Atlower frequencies there isno ropagation.propa 29.3 SUMMARY Example THETELECOMMUNICATION BLACKOUT UPONTHE REENTRY OFASPACE SHUTTLE Wedefine agood conductor asamediumforwhich Inthecourseofonereentrytherewascompleteradioblackoutat 0|o£acertainmomentatallfrequenciesupto10gigahertz.Thenf,was gute|soral=50. (29:3)10gigahertz andtheelectron densitywastherefore about10" melaD/ar electrons percubie meter then Example |THEIONOSPHERE 1-j hyk=p-ja=—*, n=S(1-j), (29-6) Intheupper atmosphere, ataltitudes ranging fromabout50t0 3 ° several thousand kilometers, thefreeelectron density issuficient wou"?tSimterfee withthe propagation ofradiowaves,Themainsource v=(22), oonoftheionization istheultraviolet radiation ofthesun.Onthe 2‘whole,theelectrondensityincreaseswithaltitudeuptoabout300 owHtkilometers, butitshows fourledges where theelectron density yO 2(20) y=2u .increasesmoreslowly,Theseledgesarecommonlycalledlayers, 8B(aa)+Ue2p, (29-14),(29-15) Beyond about 300 kilometers theelectron density decreases slowly. Theexistence ofthese layers isascribable tothefactthat Mostofthefieldenergy isinmagnetic form. both the spectrum ofthe solar radiation and thechemical Theconductivity ofalow-density,andhencelossless,plasmais composition oftheatmosphere change with altitude Both theheights andtheionization densities ofthelayers Ne 48x 10°5N change with latitude, longitude, hour oftheday, season, and oa SSXTYsicmens/meter, (29-28)sunspot cycle.Attheloweraltitudes, partoftheconductivity jom f comes from thepresence ofpostive ions. “Thefreeelectron density Varies from about 10*to10"percubie whereNisthenumberofelectronspercubicmeterandmisthemassofmeter. Thedegree ofionization increases rapidly withaltitude, theelectron. Also, butitremains low,about 0.1%forthehighest layer. ForN=10", 3p wo?theplasma frequency isabout1megahertz. 5)+0=joeok(1~<t), (29-35) Uptoabout300kilometers theelectron density andthephase or w velocity increase withheight. Amplitude-modulated radio waves here (535 kilohertz to1.605 megahertz) andthelower-frequency short ‘waves (6to26megahertz) bend backtoward theearth inthesame Ne\iawaythatlightwavesbendinamirage. However, frequency- o,=( u) (29-36)modulated (88to108 megahertz) and television (54to890 em megahertz) waves bend slightly ingoing through theionosphere, andtheyescape. 1stheplasma angular frequency 350 *PLANE ELECTROMAGNETIC WAVES tt sst Inalossless plasma, Tete292 1h ad PROPERTY COPPER IRONw=,(1-28) (29-42) — BV oF ° 38x10 10x10 ¢ 385 460 If@>»,kisreal,thereisnoattenuation, andthephasevelocity is a Som Text larger than c.Thephase velocity increases with increasing electron * 1 100 density. If@<o,, there isnowave and the field attenuates i exponentially. zy) 2 ny? : roteewpli(or-z)-z]- d=(ZR) PROBLEMSin) Ou px ‘Compare thevelocities ofpropagation ofTandofBincopper andiniron 20-1,(29.1) Goodconductors SeeTable29-2, Show thatforagood conductor 5.(29.1) Thesurface impedance ofaconductor . . Bydefinition, thesurface impedance ofaconductor istheratio E,/H, at(a)S/y€1, (0)BF=2eM(uo"), —()6°=(Es). thesurface,thesubscriptrindicating atangential component. Itisshownin29-2.(29.1) Thedamped transmitted wave Prob. 19-4thatH,isnumerically equaltothecurrent perunitwidthintheconductor Draw curves for£,similar tothose ofFig. 29-1, for wr=0 to2xat intervals of27/4. ® (a)Show thatthesurface impedance ofagoodconductor is 29-3,(29.1)Designing busbars asa(ety" 14y Youareasked todesign copper busbarsthatcancarry 5000 amperes at DOSE) or ee 60hertz over adistance of5meters. The total length ofbus is10meters Thepowerdissipation inthelineshouldnotexceed1kilowatt.Suggesta whereaistheconductivity and6istheskindepth,Thequantity1/08isplausible cross section, thesurface resistance, Thesurface impedance andthesurface resistance are 29-4,(29.1) Heatpropagation expressed inohms/square. SeeProb. 4-9.Forexample, thesurface Itisinteresting todrawaparallel between theflowofheatinathermally resistance ofcopper at3gigahertz is14.4miliohms/square. ‘conducting medium andthepropagation ofanelectric ormagnetic fieldin (b)Show that,ifthetangential magnetic fieldisH,,thenthepoweranelectrically conducting medium. dissipated persquare meter intheconductor isgivenbyH?y/(08). This Let&betheheatfluxdensity inwattspersquare meter and =~AVT, means thatthepower dissipated isthesameasifthesurface current (ofwhereAisthethermalconductivity inwattspermeter-kelvin andTisthe densitynumerically equaltoH,)weredistributed uniformlyoverathicknesstemperature inkelvins.Then,forconservation ofenergy, 4oftheconductor. or %-6,(29.1) Induction heating. V-@=-pc+9, Induction heating consists inexciting eddycurrents inaconductor by" exposing itt0analternating magnetic field, Themethod serves formelting, where pisthemassdensity inkilograms/meter’, cisthespecific heatin inaninduction furnace, forheating before aforging operation, orforjoules/kilogram-Kelvin, andQistheheatproducedwithinthemediumin hardening. Aninductionfurnacecomprisesacruciblesurrounded byacol.owtmeter” ‘Thelargest furnaces havecapacities oftensoftonsandpowers uptoafew Toro, megawatts, Once theloadhasmelted, magnetic forces within theliquidwpcoT provide stiringvr-Eo=0 Thecoilisusually asingle layer ofwater-cooled copper pipethatsurrounds theobject tobeheated. Iftheobject isferromagnetic, asmall ‘Thisequation isidentical informtothatforanelectromagnetic waveina Partoftheheating comes fromhysteresis losses. goodconductor, withpc/2corresponding to10.Itssolution forheatflow Induction heating hastheadvantage ofconvenience andofnotcon-inonedimension issimilar: taminating themetalwithcombustion gases.Also,bychoosingthe s2 MANE BLECTROMAGNENC WAVES sont 53 frequency correctly, itispossible toapplyaheattreatment downtoa Draw thew-fdiagram foralow-density ionized gaswithf,=3known depth. Forexample, plowshares require ahard,heat-treated skin megahertz forfrequencies ranging from3to30megahertz. thatresists abrasion andasoftcorethatresists breakage. 29-11, (29.2.4) Pulsars andsatellites‘ctlcanalanbeheattreatedwithalaserbeam,butwsualytoadepth 22-4)_Polarsandeliegravitational colpee,ornauron ofonlyafraction ofamillimeter stars,andthatrotaterapidlywhileemitting anarrowbeamofradiation.Here isasimple example ofinduction heating. Asteelrodofcircular ‘Thepulse lengths, attheearth, areoftheorder of1millisecond, andthe cross section liesinside asolenoid thatapplies anaxial andtangential periodsoftheorderof1second.Neutronstarsconsistmostlyofneutrons magneticfeld1WeshowedinProb,194thatthenetsurfacecurrent iatsomeelectronsandsomeionsTheitmasesareoftheOrdeofthat density isequal toH,.Awave penetrates normally totheconductor, and ofthesun,buttheirradiiareonlyoftheorder of10kilometers. thepower dissipation intheconductor ithesame asifthe current were Within afew months after thediscovery ofpubars, stance estimates distributed uniformly over athickness equal totheskindepth. Seethe were obtained inthefollowing manner. Itwasobserved thatthearrival preceding problem. time ofapulse depends onthefrequency ofobservation, thearrival timeThesolenoidhas100turnspermeterandcarriesacurrentof600 beinglateratlowerfrequencies. Thisdelayisattributed todispersion inamperes rmsat100kiihertz theinterstellar medium, which isionized hydrogen with anlectionCalculate theskindepthandthepowerP’dissipated intheironper densityNofabout10°/meter’.square meter. Set0=10°andj,,=100.Neglect endeffects, andneglect the (a)Show that,i">08,aplotofthetimedelay Arasafunction of‘at'thattherelativepermeability decreasestounitywhenthesteel fhStamtatata,ploofthetinedelay Aas9fonction of becomes red-hot. distance tothepulsar. 29-7. (29.2) Wave propagation inaplasma (b)Inthecaseofpulsar CP0328, arrival times measured at151,408,pomthatinapleaUB<&E,ofE/B>>,wherevistheveloityof and610megahertzgavethefollowingresults:between610and408anelectron megahertz, thedelay was0.367 second; between 408and151megahertz, thedelay was4.18 seconds, 29-8, (29.2.5) Plasmas compared tometallic conductors Find thedistance toCP0328 inparsecs where 1parsec is3.086x10" (a)Find thevalue of@,asdefined inSec. 29.1, foralow-density meters. Itisthedistance from which theradius oftheearth's orbit,plasma, . 1.495x10"meters,subtendsanangleof1’,(b)Show thatthevalue ofk?thatwefound foraconductor inSec. ‘Thefactthatsuchplots givestraight lines passing through theorigin 282.3 agrees wththatofSee,20.25 indicates thattheasumption that05> ofiscorect. Thedelay therefore ‘Thevalues of andoffthatwefound inSee.285.1 arenotvali occurs overlage distances ininterstcllar space, andnotinside thepulsar however forimaginary values of9. Pet 294,(28.2.4) Thephaseandgroupvelocities inaplasma ‘Similar methods areusedtoreduce satellite ranging errorsduetothe“Twouniform planeelectromagnetic wavesofequalamplitudes propagate ionosphere intheionosphere where thefreeelectron density isNpercubic meter. One 29-12, (29.2.4) The energy densities, thePoynting vector, andthegroup wave hasacircular frequency «,andacorresponding wavelength 2,;the velocity inaplasma other hasaslightly different circular frequency w,andawavelength 2: ‘Anelectromagnetic wave travels inalow-density plasma. Theelectric (a)Atagiventime#thereexistvaluesofzforwhichthetwowavesare fieldstrengthisEcoswt. inphase andother valuesofzforwhichtheyareoppositeinphase.Whatis (a)Calculatethesumoftheelectric,magnetic,andkineticenergy the distance between the maxima? densities. {(b)What istheir velocity? This isthegroup velocity v,. (b)Calculate thePoynting vector. (©)Show that, inthelimit, v,=1/(dk/da). (c)Calculate thegroup velocity.(d)Calculate ‘thephasevelocities andthegroupvelocityforf=5.3 (d)Findtherelationsbetweenthesequantities.megahertz, f,=5.4megahertz, andN=5x10”electrons/meter’. (e)Calculate the distance and the number ofwaves between two anim 29-10. (29.2.4) The w-fdiagram foralow-density ionized gas ‘Thew-Bdiagram isacurve ofwasafunction ofB.Theratio w/Bis ‘equal tothephase velocity, while theslope dao/dB isequal tothegroup velocity. 555 30CHAPTER .EHSiSS oN 2 ,PLANE ELECTROMAGNETIC we OR7asWAVES III co ospiReflectionandRefractionA:TheBasicLawsand 2<qFresnel’s Equations , Fig,30-1.Anelectromagnetic waveincidentonthe interfac $01REFLECTION ANDREFRACTION 555 thetrespective wavefontsandpointinthedirection ofpropagation Theanges 302 SNELL'SLAW $57 In8,Brate,respectively, theanglesofincidence, reflection,andrefraction. 303. FRESNEL'S EQUATIONS 558 30.3.1 ENORMAL TOTHEPLANE OFINCIDENCE 589) (2) The .eee neONNCIDENCE st 2neemediarehomogeneous, isotropic, linear,stationary (HILS) 304. REFLECTION AND REFRACTION ATTHE INTERFACE BETWEEN TWO NONMAGNETIC NONCONDUCTORS S61 (3)Theinterface isinfinitely thin.Inother words, thereflection is30STHEBREWSTER ANGLE=565, specular.EranMEASURING ueRELATIVEPERMETTIVITYOFTHEMOON'S (4)Theincidentwaveisplaneanduniform. 306 THE COEFFICIENTS OFREFLECTION RAND OFTRANSMISSION T 37 Anelectromagnetic wave incident onaninterface usually gives riseto +307REFLECTION BYANIONIZED GAS 569 bothareflected andatransmitted wave. Thistypeofphenomenon is 308SUMMARY 572 vommon. Forexample, asound waveincident uponawallgivesbotha PROBLEMS 573 reflected wavethatcomes backintotheroomandanother thatproceeds intothewall. Awave propagating along atransmission lineispartly reflected andpartly transmitted atadiscontinuity. Waves onstrings show Chapters 28and29concerned thepropagation ofelectromagnetic waves thesametypeofbehavior, asshowninApp.C. inunbounded media. We now investigate thebehavior ofawave . encountering adiscontinuity, asinFig.30-1.Themediawillbethesame 30.1REFLECTION AND REFRACTION asthose ofChaps. 28and29,namely dielectrics, good conductors, andionized gases. Aspreviously, wedefine dielectrics asnonconductors, Medium 1carries theincident andreflected waves. Medium 2carries thewhether magnetic ornot. |___efracted wave.Forsimplicity, weassume inSecs.30-1and30-2thattheInthisfirstchapter outofthreeonreflection andrefraction, wemake ‘incident waveislinearly polarized. Then,intheincident wave, several simplifying assumptions: E,=Einexp(wt—ky), (30-1) (1)Themedia extend toinfinity oneither sideoftheinterface. Thisavoidsmultiple reflections. wherethevector wavenumber &,isrealandpointsinthedirection of 556 PLANEELECTROMAGNETIC WAVESmIff|302SNELL'SLAW 557 propagation oftheincident wave.Themagnitude ofkyisnko,OF":/a, Allthreewavesareofthesamefrequency. Thisisobvious because thehrbeingtheindexofrefractionofmedium1and2ptheradianlengthofa} wavesareallsuperpositions ofthewaveemittedbythesourceandofwvaveofthesamefrequency inavacuum. Forconvenience, wesetthe ||thosewavesemitted bytheelectrons executing forcedvibrations inmedia origin ofrintheinterface, asinFig.30-1,andwetakeE,,,tobereal. |1and2.Recall frommechanics thatforced vibrations areofthesame Thisequation defines aplanewaveforallvalues ofrandr,andthusa frequency astheapplied force. wave thatextends throughout alltime andspace. However, itapplies Also, from theabove equations fortheE’s, only inmedium 1incetheincident waveisplane,alltheincident raysareparallel. By KeyBn=heeFw=KerPa (30-8)hypothesis, theinterface isplane.Nowthelawsofreflection andof Th . enthe k’ a >refraction mustbethesameatallpointsontheinterface. Itfollowsthat ee ee eeay(natheircomponents parallelsthereflectedraysareparalleltoeachother.Similarly, therefractedrays aval.Inpe+ifkry=0asinFig,30-1,then areparallel toeachother. Further, sinceawave frontisbydefinition ky=0, ky=0, (30.9) perpendicular toaray,wecanexpect thereflected andtransmitted wavestobeoftheform andk,,ke,kyarecoplarar. Theplanecontaining thesethreevectors isvalledtheple ce ci sa a EqEnaexpileogt ket), (02) saletheplaneofincidence. Thexcomponents ofthek’sarethusall Ex=Ermxpj(Wrt~kr°1). (30-3) Kee=kre=kre=kySinBp, (30-10) What doweknow about kyandky?From thewave equation 27-72 where 6,istheangle ofincidence shown inFig.30-1 applied tomedium 1,witho=0,py=0, Itisnoweasytofindky VEx+C1w2ER=VEQ+ER=0, (30-4) Keg+eg=ke+kis=kD (30-11) where ndliom Khe=Kienkee=—kre 30-12) kyapaB=mky=oe) (30-5)wo eon) hy hy Wechoose thenegative sign because the reflected wave travels away _ 7iromtheinterface.Itfollowsthat,ifk,isreal,asweassumedatthe Asimilar string ofequations applies tok,.Also, beginning (there iszero attenuation inmedium 1),then kyisalsoreal, K+, +R =Raatdy+hc=, HR+= 30-6) thereflected waveisuniform, and ‘Thewave numbers k,andk,arereal,butkeandkyarevectors thatcan 8,=On. (30-13) becomplex.‘Thetangential component ofEiscontinuous attheinterface, This theangleofreflection isequaltotheangleofincidence. means thatthetangential component ofE,+E,inmedium 1,atthe ‘Therefore theincident, reflected, andtransmitted raysarecoplanar, interface, isequaltothetangential component ofE,inmedium 2,atthe andtheangleofreflection isequaltotheangleofincidence. These are interface. Thesameapplies toH.These continuity conditions willpermit thelawsofreflection ustofind alltheunknowns inEqs. 30-2 and 30-3. Some relation mustexistbetween E;,Ex,Erattheinterface forall¢ 30,2SNELL’S LAW andforallpoints rayontheinterface. Such arelation ispossible onlyif thethreevectors areidentical functions of¢andrq.Then Nowreturn toEq.30-10. Itsaysthat 0=Op=Wr (30-7) ky,=kysin@, (30-14) 558 PLANE ELECTROMAGNETIC WAVES It 59 Then aN Kp=B=ke,=KE—Kisin?6,=kin}—njsin?@,).30-15) €\7~wem Iftheterm inparentheses isnegative, thenthere istotalreflection. We “ey NNdisregard thispossibility untilChap.31,Otherwise, ky.isreal,krisreal, ee afandthetransmittedwaveisplaneanduniform.If@,istheangleof gor Voogacti wae LOrefraction asinthefigure, ee Pg kre=—kyc0s Or,ers=keinBy (30-16) Ne A FromEqs.30-14and30-16, NJ asin, =k,sin@,, or —nzsin@p =n,sin). (30-17). Fig.30-2,Theincident,reflected,andtransmitted wavesforanincidentwave‘Whenanelectromagnetic wavecrossesaninterface,thereisconserva~ iglarizedwithiesSeldnormaltotheplaneofincidence.Thearrowsshowthe ‘Therefore, choosing axes asinFig.30-1, wefindthat E,=Eyexpjot—ky(xsin8,~z088,)), (30-18) Hy+Hpe=Hr Hiy+Hey=Hy (30-22) Ex=Exmexpjlot—k(xsin6,+zcos@,)], (30-19) Since therelation Ey=Ermexpjot~k(xsin87~200887)] (30-20) KXEAoon (30-23) ‘The laws ofreflection andSnell’s lawaregeneral. They apply toany twohomogeneous, isotropic, linear, andstationary (HILS) media, volSec.28.2.1 applies toallthree waves, wefirstfindEyandEyandthen whether conducting ornot,witheither realorcomplex &’s,provided that deduce HpandH, oneallows complex angles asinthenextchapter. Itwillbeconvenient todivide thediscussion intotwoparts. We onsider successively incident waves polarized with their Evectors 30.3 FRESNEL’S EQUATIONS wormal andthen parallel totheplane ofincidence. Any uniform plane sncident wave isthesum oftwo such components. Wenowrequire relations between Em,Erm, andErmthatwillensure Wenowdefine oursignconventions. SeeFigs.30-2and30-3.Observe continuity ofthetangential components ofFandHattheinterface.” iatthetwofigures agree atnormal incidence. Weutilize thecontinuity Inthischapter k;,ke,Krareallreal.Thus allthree waves areplane EyandH,inFig.30-2, andthecontinuity ofE,andH,inFig,30-3. anduniform, andhence transverse (Sec. 28.2.1). TheEvector ofthe Ihiswillyield relations thatapply again toanypairofHILS media and incident wave canpoint inanydirection perpendicular toky twanyangle ofincidence Then theconditions ofcontinuity attheinterface require that bth <F tha <b 50210.3.1 ENormal tothe Plane ofIncidence etEmEm By+BayBry (3021) Fine £andHvectors oftheincident wavepointasinFig.30-2.The ——sediabeingisotropic,theEvectorsoftheothertwowavesarealso 1Wecouldalsoapplythecontinuityofthenormalcomponents ofDandofB.Butthen swormaltotheplaneofincidence. Thisisbecausetheelectrons inbothoutte policetoseflectionfromthesaeOakwasteonlycnne‘mediaoscillateinthedirectionnormaltotheplaneofincidenceandoneten aangin * * teradiate wavespolarized withFnormal totheplaneofincidence. 560 +4THEINTERFACE BETWEENTWONONMAGNETIC NONCONDUCTORS 561 K Thesubscript 1indicates thattheEvectors areperpendicular totheiS~ planeofincidence.’ThesearetwoofFresnel’sequations,8 Pi 10.3.2EParallel tothePlaneofIncidence oe x 2 heE'sarenowallintheplaneofincidence, asinFig.30-3,andSRE. 8 oweya.Zs|eo Hin~Ham=Hrs (30-30)SitiyWg w bokae Fim=Enm_Erm . SL A ZF (30231) 7 ~ Also, y (Eim +Em) 60S8;=Erm608Or. (30-32) Fig.30-3.ThisfigureissimilartoFig.30-2,exceptthatnowtheEfieldsareall then paraile!totheplaneofincidence. (Em)_Zc08Oy~Z,cos8; (3035) IftheEvectors pointinthedirections shown, attheinterface, thenthe Eim!y Z_008 Or+Z,cos8; Hivectors point asshown, toorient thePoynting vectors ExH(Sec. Erm 22,008 6,28.6)intheproperdirections. (F).=Zocos0;+Z,0080," (30-34) The continuity ofthe tangential component ofEatthe interface requires that [hisisthesecond pair ofFresnel's equations. _Atnormal incidence 6;=4,=6,;=0, theplane ofincidence is Eim+Enm=Erm (30-24) undefined, andthetwopairsofFresnel’s equations areidentical: atanygiven point ontheinterface. Similarly, forcontinuity ofthe Enn_ ZZ, tangential component ofH, Em=Zitz,, (30-35) Him,COS8;—HemCOS8;=Hm,COSOy (30-25) Em=Ls. (30-36) or,from Sec.28.4, tim Lat dy (Eim =Erm) 6088; _Erm608Or (30.26) 0.4REFLECTION ANDREFRACTION ATTHE Z Zo’ - INTERFACE BETWEEN TWO NONMAGNETIC NONCONDUCTORS where Zisthecharacteristic impedance ofamedium Eon on on ce Wecontinue todisregard totalreflection. Fornonmagnetic non-aEoHou_on_ conductor 2a Take Mol a (30-27) nductors, Z,=Ho _Sho sho :nbeingtheindexofrefraction. aakom) a? G37) Solving, and Enm\_Z:0080;~Z,cos6 (==(ulna)€080,—cosOr 30-38) (EE),~Zeee+Zon6," (30-28) Ea).(mlnaoosFear" (30-38) Erm) ____2Z,008 0, 562 563 N ; . = =\f* 3 \7 if wh \¢, x (abe Z yihe & 7 orSET 7A a 4 aOS >L~« Ne,Lo | |Se 7| oe| Pa | | / \ rye ey °/ ky tig,30-4, (a)Therelative phases, attheinterface, oftheB'sinthe reflected and x wismitted waves forn,<n andform,>ns,with E,,,normal totheplane of / sulenee. On the left, the reflected wave is2radiansoutofphasewithrespectto At licineident wave. Thetransmitted wave isinphase,inbothinstances.(b) z . Gist Cicsts” ofthe fieldEatsomeparticulartime.Thecrestsareonewavelength PoeF .jurtandtravelinthedirectionsofthearrows.Notethephaseshiftofupon ,“a stlectionfromaglasssurface.Notealsotheinterference patternthatresults“4kygel / ‘ ‘woonthesuperposition oftheincidentandreflectedwaves.(c)TheEvectorsata cPa we ‘veninstantintheineident,reflected,andtransmitted wavesatnormalincidence“ tvglassairinterface,Ontheright,Er,islargerthanE},,However,sservation ofenergy still applies. (Erm 2ny/n3)cosO, Ge) =e 30-39) a A ,a y ‘serve thatthesecond ratioisalways realandpositive. Thismeans Ps, je * viuat,attheinterface, thetransmitted wave isalways inphase withthe . a , rncidentwave. sid Thefirst ratio can, however, beeither positive ornegative, depending ro. Bm F 1thevalue ofm/ny. Ifmy/nz<1, then 8<8, and cos8;<cos8; 2 is rdGia. whereas ifn,/n:>1, then 07>, and cos0,>cos 6.The reflected - —¥— — + save isthus either 2radians outofphase with theincident wave atthe / vuterface ifmy<z,oFinphaseifm,> ‘ , Figure 30-4 illustrates theEvectors forboth types ofreflection. Figure |i 45shows theabove ratios forn,/n2=1/1.5.Thiscorresponds toalight / wf save incident inaironaglass whose index ofrefraction is1.5 10, ‘Thesecondratioisalwayspositive.ThenEy,andEy,areinphaseat“l theinterface, asinFig.30-3,whichweusedtoarriveatthisresult. °However, theratio forEp, can beeither positive ornegative, which oe indicates thatEg,canpoint either asinFig.30-3orintheopposite ol direction. The tangential components ofEyqandofEg, canthus be 0 cither inphase orxradians outofphase. They areinphase if oo 1 - .02 eo, 0887>c088, (30-42) os Exe a06 (=)‘ orif oa sin6;cos6,~sin6,cos@,>0, (30-43) 10! sin26,~sin26,>0, (30-44) .Reflection andrefractionwhen,/n2=1/1.5,forexample,whenlight - FaeaaceofnS.TheEeldisnormaltotheplaneofincidence sin(8~6)608(87+8,)>0. (30-45) Thisinequality requires thateither Foranincidentwavepolarized withitsEvectorparalleltotheplaneof 6;>6,and6+8,<x (30-46)incidence, 2 (2)7008+(m/s)608Br oa) ” : Eim/y 0088;+(my/n2)cosOy Or<6;and—Or+8>5. (30-47) Emm 2(ns/n3)cos8, | (®)=S58,4(n,/n,)c0s 8; (30-41) ThephaseofthereflectedwavewhenEisparalleltotheplaneof ave tee incidence does nottherefore depend onlyontheratio m,/n,; itdepends sceFig.304 ‘onboth6,[email protected]~m/Ermcanbeeitherpositive ornegative,both forn,>n, and forn,<n;. Figure 30-6 shows the above ratios for n/n=UL, . Ea W.5 THE BREWSTER ANGLEos (E): Wehaveseenthat,whenEisparalleltotheplaneofincidence,Epmis oe) wither inphase or2radians outofphase with theincident wave,o slepending onwhethersin(8,~6,)cos(8,+6,)isgreaterorlessthansero. Itfollows that there isnoreflected wave when this expression is 02 equal tozero, that is,when 6,=6,=0orwhen 6,+6,=2/2. The firstol 4 condition isincorrect. Itarisesfromthefactthatwehavemultiplied the -02' my7,6088r~cos8,>0 (30-48) Fg,366,Reletionndeftionwhenm,/n:=1/1.5, asinFig.30-5,butwith hysin8,whichisequaltozeroat6,=0. parallelto i 366 116COEFFICIENTS OFREFLECTION RANDOFTRANSMISSION 1 $67 , The Brewster angle iscommonly used tomeasure the index of i g refraction ofasubstance byreflecting arayoflightfromitssurface. The » #LZ .ou measurement canbeaccurate tofivesignificant figures. = i Sha | Aplane wave incident onaplateofglassattheBrewster angle meetsge~ fh: eea thesecondinterfacealsoatitsBrewsterangle.Sothereisnoreflection, eS~ Ke gS citheratthefirstoratthesecondinterface. &ONO SPWE op ss 51Example MEASURING THERELATIVE PERMITTIVITY Sag<AE| OFTHEMOON'SSURFACE .AT RADIO FREQUENCIES, ‘The nature ofthemoon's surface can beinferred, tosome extent, , > from thevalue ofitsrelative permittivity ¢,=n° (Sec. 28.4) at Fig.30-7.IftheangleofincidenceisequaltotheBrewsterangleandifEliesin aee ortneecanservetomensarethis theplaneofincidence, thereisnoreflected wave.Thepositionofthemissing Wtradio.weve.oninat feficedsayisat0°tothetransmitted ray @radiowaveoriginating fromasatelliteinlunarorbit y illuminates themoon, thereflection observed attheearth is similar tothe reflection ofsunlight from the surface ofalake: ‘Thus for most ofthelight comes from theregions that happen tobe correctlyorientedforspecularreflection.Thesurfaceofthemoon 0,+0,=2 (30-49) thusglistens overanareaoftheorderof100kilometers in 2 diameter, thearea depending ontheheight ofthesatelite above thesurface ofthemoon and ontheroughness ofthesurface. there isnoreflected wave when theincident wave ispolarized with itsE Ifthedetector ontheearth receives boththereflected wave and vector parallel totheplane ofincidence. Thisisremarkable because the ‘adirect wavefromthesatellite, itispossible todiscriminate: ~ : between thetwobyusingthefactthattheDoppler effect makes wavethengoesthrough aninterface without reflection. SeeFig.30-7. thetworadiofrequencies slightly different. Aplotoftheintensity This angle ofincidence istheBrewster angle. Itisalsocalled the ofthereflected wave asafunction oftheangle ofincidence when polarizing angle, since anunpolarized wave incident onaninterface at theEvector liesintheplane ofincidence shows zero reflection at thisangle isreflected asapolarized wave with itsEvector normal tothe ‘theBrewster angle.planeofincidence.* joanesuchmeararoment, performedat‘trequeneyof megahertz, the Brewster angle was 601°inthemare AttheBrewsterangle, northwestofHansen.Thisivesaneof30202, 5 Itisibletoperformsimilarmeasurements atother points my_SinOr_sin(2/2=810)_ootyy (30-50) conthesurfaceofthemoonbecauseoftherelativemotionsofthe nzsinOp sinO15 three bodies involved, namely thesatellite, themoon, andthe earth For light incident inaironaglass whose index ofrefraction is1.5,On=56.3". 0.6THECOEFFICIENTS OFREFLECTION RANDOF TRANSMISSION T "The Brewster angleisoftenexplained incorrectly asfollows.Forthisparticular angleof thecoefficients ofreflection andoftransmission concerntheflowof incidence, themissing reflected rayisatXPt0thetransmitted ray.Itargucd thatthe cemergy across theinterface. The average energy flux perunit area ina thereisaBrewster angleevenwhenmedium 2isavacuum. Alo,withmagnetic media, 1X31.Weexclude totalreflection aswellasreflection fromconducting there can existaBrewsteranglewhenEy,isperpendicular totheplaneofincidence media.Setting4,=1,wefindthat 568 PLANEELECTROMAGNETIC WAVESIt 569 1e\ 1.04 =1(2)"" Be 7 Pin=3(G)Ein (3051) Vie"os 6)". Fan=3(2)Eknfizs (30-52) . l(a e2g Srv=2(2)Ekimbir, (30-53) os wherefi,isnormaltoawavefrontoftheincident wave: aa| Ly a=K, (30-54) ——— Aky ; ci 7 0 andsimilarly forfigandfir. - theaverage energy fluxes perunittimeandperunitareaattheinterface: fi)_ER Then,fromFresnel'sequationsfornonconductors, R=[gax-4|=oie, (30-55) fayBTSim (n,/n2)6086,—00s6? R=|cevco8|" (30-57) wherefistheunitvectornormaltotheinterface; (n,/n2)cos@,+cosOr. 4(n,/n3)cos8,cosOr Practfi|_(€\'7EhmC08Or_M12Em608Or T= Sa (30-58) =|Zan|_(Sa)ErwcosOr_mErmc08Or(39.56) 1/n)60s8;+€0s8 rlFeal(S) Eeaee Enea, 059 Kerncos6x08Br =C0s6;+(m/z)Cos2] R,=|Bt s/s)c08Or . ‘[‘cos6,+(m,/n2) cos6J” G059) 10 4(n/n2)cos8,cosOr Tp=Melita)£0881608Or Yow {eos6,+(n/m)c0sOP (60) Inboth instances, R+T=1, asexpected from the conservation of os nergy. AttheBrewster angle defined above, Ry=0and7,=1,again as expected. SeeFigs. 30-8 to30-10. os ‘30.7 REFLECTION BY AN IONIZED GAS. 02 We saw inSec. 29.2.4 that, inalow-density ionized gas, thephase ook =r a ar velocity islarger thaninfreespace andtheindex ofrefraction 7isless 4 ‘han unity: , 22/1 g0.6N\" Fig.30-8.Thecoeticients ofreflectionR,andoftransmission T,asfunctionsof =£2[1-@,|—(1580.61 theangleofincidence6,formy/n2=1/1.5. my (%)(fe),(0-61) , 370 sm ; | lenired« xi — Fig.30-10, Thecoetficients ofreflection Randoftransmission Tatnormal. incidence, asfunctions oftheratio m,/n>. Fig. 30-11. Reflection from anionized gasinwhich theelectron density increases with increasing 2,orupward. where cisthespeed oflight, v,isthephase velocity, «,istheplasma angular frequency, istheangular frequency, Nisthenumber offree ltm=1, thenelectrons percubicmeter, andfisthefrequency. v=,Iftheionizedgascouldhaveadefiniteboundary andauniformfree nsin0=sin6, (30-63) electron density, then reflection andrefraction atitssurface would besimple: thegaswould actasadielectric withnmlessthanunity.In Differentiating withrespect tothedistance /measured alongaray,we practice, neither assumption isvalidandreflection occursgradually, asin findthat amirage. do dn‘Wecouldcalculate thepathofarayasinProb.30-1.However, Snell's annan (30-64) law(Sec. 30.2) provides themain features ofthereflection.Weselect coordinates asinFig.30-11andassume thattheindexof Iftheraypenetrates intoanionized region where theiondensity refraction nvariesslowlywithz,butnotwiththeothertwocoordinates. increases withz,theindexofrefraction ndecreases with|andtheTobemorespecific, weassume thatnvariesbyanegligible amount over derivative dn/dlisnegative, sotheangle@increases withdistance, asin ‘onewavelength. Thenagivenraygraduallybendsdowntoanangle8,atFig.30-11.Aftersomedistance,ifNbecomessufficientlylarge,@ apoint wheretheindexofrefraction isn becomes equalto90°,thetangent of@becomes infinite, anddn/diWecancalculate @inthefollowing way.When refraction occurs atthe becomes zero.Afterthis,tan@becomes negative, whereas thederivativeinterface between anytwomedian,andnz,thequantity nsin@is 4dn/dlbecomes positive, and@keepsincreasing untiltherayescapesback conserved ingoingfromonesideoftheinterface totheother. Thisis fromtheionized region atanangleequaltotheangleofincidence 6. Snell's law.Ifnvaries gradually withzbutnotwithxandy,thenwecan >Atthetopofthetrajectory imagine themedium tobestratified inthinlayers, andthevalue of nin@remainsthesameallalongtheray.‘Thus sin=1,Manor=sinOy (30-65) nsin0=n)sin6, coe Thisischeindexofrefractionrequiredforreflectionwhentheangleof sm PLANE ELECTROMAGNETIC WAVES ft PROBLEMS 573 30.8 SUMMARY __4(ny/nz) cos8,cosOy qe[cos6,+(m,/n2)cosOr]°* (30-60) Attheplane interface between two homogeneous, isotropic, linear, | stationary (HILS) andlossless media, (1)theincident, reflected, and Reflection fromanionized gasoccurs gradually asinamirage. transmitted waves allhavethesame frequency, (2)thethree vector wave ,numbers ky,ke,krarecoplanar, and(3)theangleofreflection isequal PROBLEMS totheangle ofincidence. These arethelaws ofreflection. .‘Snell's lawstates that W-1.(30.2) Therayequation‘Awave travels inastratified medium whose index ofrefraction isa =msi function onlyofthecoordinate y. nzsinOr=n,sin6,, (30-17) (a)Showthattheangle@betweenarayandthey-axisobeysthe following law’ where m;istheindex ofrefraction ofthefirstmedium andnisthat of 46 ldn thesecond. annay® ‘Fremef's equations arewsfollows: where thedistance /ismeasured alongtheray. E,(n/n;)088,~00s8 (b)Youcannowverifytherayequation (®)={tr/nz)0sBy—C08Or (30-38) d Em) x(m,/n;)cos6;+00sOy qnd=Fn,Em 2(y/n3)cos8(EZ),~Germjeae 8,+058," 0-39) where7isaunitvectortangenttotherayatapointwheretheindexof tee ume t r refraction is1. ()=0088+(i/nz)cosOr (30-40) 2,(30.3)Reflection andrefractionatthesurfaceofadensemedium Em)» 6088;+(;/n3) cosOy” Write down Fresnel’s equations forthecase where jy=1, a= 1, n.2>n;.Youwillfindasurprisingresult:iftheEvectoroftheincident (2)=—_2(m/nz)cos8 (30-41) waveisparalleltotheplaneofincidence,theamplitudeofthereflected Em)» 6088;+(ri/n2)008OF wave isindependent oftheangle ofincidence! For what range of8,areyour formulas valid? ‘AttheBrewster angle ofincidence 6,5, 3. (30.3) Fresnel’s equations expressed intermsof8,and@,alone (a) First show that nmFmcotBi, (30-50) sin(8,~87)c0s(0,+87)=sin8,60s6,~sin8,c0s8, sin(0,+87)cos(8,~8)=sin8,cos0,+sinO,cosOy. and there isnoreflected wave ifEliesintheplane ofincidence. ‘Thecoefficients ofreflection Randoftransmission Taretheratiosof (b)Showthat,fornonmagnetic nonconductors, theaverageenergyfluxesperunittimeandperunitareaattheinterface: o(=)=200) Gy(Erm)=2emeersin gr oy Em). sin(@+0,) VE), ~sinG+Or)*a,=[ulnsdens OrconBr].(30-57) @(=)2120(1=01)Gy(E).2c0s6,sinOy (rains)cos6,+008Or. Emm)tan(0,+87) Bn),~in(0,+@,)608(8,~Ar) 4(n,/n;)cos6,cosOr 1Gea eeoF (30-58) W-4,(30.3)Measuringanindexofrefraction 1/3)6088,a Set —cos8;+(m;/n2)cosory. Ene Exe) =fxe08.r+(rls)cosOr]? p=(2),s=(Ee a-|cos8;+(m%/n3)608O; 0-59) (2), (), S78 PLANEELECTROMAGNETICWAVESIt|PROBLEMS 315Show that, withalaserbeamincidentat45°inaironamediumofindexof| supportforvarioustypesofcoating.Thefilmissothinthatmultiple refractionn, o-pxt- ; reflections insideitdonotgiverisetoghostimages.ye==Py=s) Nowithaslongbeenknownthataseriesofparallelglassplatessetat(+p\ +s) Brewster's angle filters outwaves polarized withEnormal totheplane of incidence. Seethetwopreceding problems. Thesamecanbedonewith Here, sisnegative, from Fig, 30. The ratiopisalsonegative,fromFig, pelliclesinlessspaceandwithouttheinconvenience ofghostimages 30-6s : Inpractice, instruments measure abeampower.Sopisequaltominus izPellicies arevirmally losslessandcanthuspolarize high-power laser theStsddailonereflected toincident powerswithparallelpolariza- (a)Find(R/T), forapelliclesetatBrewster's angle,Takebothtio,andsimilarlyfor interfacesintoaccount,butdistegardmultiplereflections. 30-5.(30.5)TheBrewsterangle (b)Calculate thisratioforapelliclewhosemis1.5inair.Calculate theBrewster angles forthefollowing cases: (©)Find ageneral expression fortheratio (R/T), forNinterfaces. (a)light incident onaglass whose index ofrefraction is1.6, This result isgrossly wrong because wehave neglected multiple (b)light emerging from thesame type ofglass, reflections inside thepellicles. Inactual fact, theratio isapproximately (©)@radio frequency wave incident onwater (n=9 atradio fre- equal toNR. With40interfaces,theaboveresultistoolargeby2orders quencies),‘ofmagnitude! 30-6, (30.5) TheBrewster angle andtheratio n/m, W-11. (30.5) TheBrewster angle formagnetic media (a)Show that, ifns>m, then Bry>45°. ‘Awave isincident inaironanonconducting magnetic medium such as (b)Show that, ifm,<7, then 8j»-<45°. ferrite30-7.(30.5).TheBrewsterangle (a)Showthattheratio(Enn/Eim); iszerofor (a)ShowthatBrewster's angleisalsogivenby sano, £46) sin?8p= end Tenn ° There isaBrewster angle only if,>4, Itfollows thatthere exists aBrewster angle onlyifthe ratiomy/n: isreal (b)Show that(E_q/ Er). 82erowhen(b)Showthatsin8p=€0s8. Hel=6)30-8.(30.5) Brewster windows forlasers tantOT“Themirrors ofsome gaslasers areoutside theglass tubethatcontains the te discharge. Then thetubeissealed atboth ends withwindows setatthe Nowthere isaBrewster angle, butonlyifp,>€,. wwster angle.Browsethatthere isnoreflection fromsuchawindow aslongastheE° ‘W-12,(30.6) ‘Thecondition thatmakesR=T’atnormalincidence vector oftheincident waveliesintheplaneofincidence Findtheration/n;thatmakes R=T=0.5atnormal incidence,WB,(30.6)E,H,R,andTatnormalincidenceonawatersurface valueofR,atthefacesofadielectricplatesetattheBrewster 309.oe‘ThevalueofR,atthefacesof diclectricplate ‘A60-wattlightbulbissituatedinair1meteraboveawatersurface."Abeamoflightinamedium ofindexofrefraction n;fallsonaplateof (a)Calculate therootmeansquare (rms)values ofEandHforthe Sicam,atheBrowserange " incident,reflected,andrefractedraysatthesurfaceofthewaterdirectlya : underthebulb.Assume thatallthepower isdissipated aselectromag (2)Showthat,atthe firstinterface, neticradiation. Theindexofrefraction ofwateris1.33. B=os28=(Late) (b)Calculate thecoefficients ofreflection andtransmission,A+mijn} W-14, (30.6) Antireflection coatings forphotographic lenses andsolarcells ‘Thereareinstances wherethereflection coefficient ofadielectric must (b)Show thatR,hasthesame numerical value atthesecond interface. ee erates ereae reflection coefficient© (©)FindthevalueofR,forglasswhosenis1.5,inair. beclose{0Zer0.Thebeatknown examplesarephotographic lensesand 30-10, (30.5) A“pile ofplates” polarizer with pellicles Clearly, theway toeliminate thereflected wave isbyinterferenceApellicleisaverythinfilmofcellulosenitratethatisstretchedtaut Coatingthedielectric withathinfilmofanothertypeofdielectric‘over flatring. The cellulose nitrate istransparent and canserve asa provides two reflected waves that cancancel, The situation is,however, “* H Thisratioismuchlargerthanunity.Forexample,with@,=30"oe i (n=1.414),itisequalto13.93‘ (b)Show that &—<~:— ypalleapr ms f 7 where Ta.—<— <sin29, =Ru/Ri=2Jasin20r=RR+1 :‘The signs before thesquare roots arepositive net eW-16. (30.7) Ducting intheionosphere Fig.30-12. Under certain circumstances, theindex ofrefraction oftheionospherevaries with altitude insuch away that araythat starts out horizontally follows apath ataconstant altitude above the earth's surface. The complicated bythepresence ofmultiple reflections inthefilm.Also,the ionosphere thenactsasaduct,andthephenomenon iscalledducting.OfSaat recsioe saveswiththe angleofincidence andwiththe course,therequired condition appliesonlyoveracertaindistance. When a atthe ray emerges from this region, itisdeflected cither upward or er arthereisnoreflected waveatnormalincidence inai downward. Radarsignalsoccasionally traveloverlargedistances inthis(n,=1) whenthedielectric ofindexofrefraction n,iscoatedwitha wy.qrarterwavelength filmofadielectricn:~n\".Takemultiplereflections (a)Howmusttheindexofrefractionvarywithaltitude?intoaccount, andusethenotation ofFig.30-12. (b)Howmusttheplasmafrequency varywithaltitude? (b)Calculate andsumtheamplitudes ofthefistfourreflected waves when n,=4, tofour significant figures (©)Asilicon solar cellhasanindex ofrefraction of3.9at 600nanometers. Calculate thereflection coefficient fornormal incidence atthat wavelength. (a)Calculate thethickness andtheindex ofrefraction ofacoating that would eliminate reflection atnormal incidence atthatwavelength. "Attheinterface between airandglass, R= 0.04.Incomplex optical systems with many interfaces, thelossisimportant. Moreover, stray reflections reduce contrast intheimage. Good-quality lenses arecoated ‘vithmagnesium fluoride (n=1.38at550nanometers). Thisreduces Rto 0.015, onaverage, over thevisible spectrum. 30-15. (30.6) Asimple andaccurate method formeasuring anindex of refraction Possibly themost practical andmost accurate wayofmeasuring an index ofrefraction istomeasure theratio R,/R, forabeam incident on thematerial inairat45° (a)Show that, if=45", 1=sin26, 1sin20,)*Raatzanzey ®-(Tsinzn) Itfollowsthat R._1+sin20, Ry 1sin20, q! : 31.1 NONUNIFORM PLANE WAVES 579 no, it!31 sin0,=Fisin8,>1. GL) CHAPTER ‘Then thecoefficient ofreflection isequal tounity Total refiection isfascinating because oftheunusual features ofthe PLANE ELECTROMAGNETIC 1 transmitted wave. Also, optical waveguides (Chaps. 35and36)usethe factthattotalreflection islossless.Theanalysisofthosewaveguides rests WAVES IV | onsomeofthekeyresults foundhere. Reflection andRefraction B:Nonuniform Plane iaotfirstwemuststudybrieflynonuniform planewavesandcomplex Waves and Total Reflection* | |31dNONUNIFORM PLANE WAVES | Ina plane wave theequiphase surfaces areplane. Inauniform plane 31.1 NONUNIFORM PLANE WAVES 579. wave, theamplitude isuniform throughout anygiven plane equiphase31.11COMPLEX ANGLES Si |surface. Thewavesthatwehavereferred tountilnowareofthistype 312 TOTALREFLECTION ssiH Inanonuniform planewavetheequiphase surfaces areagainplane. 31.2.1THEREFLECTED WAVE —5&3 However, theamplitude overagivenequiphase surface isnotuniform. 31.22THETRANSMITTED WAVE S&3 |Indeed, propagation occursinonedirection, andtheamplitude oftheExample: LIGHT EMISSION FROM ACATHODE RAYTUBE 588 wave decreases exponentially inanother direction. InFig.31-5, forExauple:THECRITICAL ANGLEANDTHEBREWSTER ANGLE=589 example, thetransmitted wavetravelsfromlefttoright,butitsamplitude 313°SUMMARY 5% decreases exponentially downward. PROBLEMS 59 Withnonuniform planewaveswecanstillwritethat E=E,,expj(wt—k-r), (31-2) InChap. 30weestablished thelawsofreflection andofrefraction aswell H=H,,exp (ot—k-r), G13) asFresnel’s equations. The laws concern therelative orientations ofthe vector wave numbers k,,ky,andky,while Fresnel’s equations concern where theamplitudes E,,andH,,may becomplex. However, thewave the relative amplitudes and phases ofthe incident, reflected, and vector then hastheform transmitted waves. Weexcluded totalreflection aswellasreflection bya k=B-ja, or)conducting medium, This chapter concerns total reflection, andthenextonereflection bya where thetworealvectors @andBpoint indifferent directions. Then conductor. The laws and equations that wefound inChap. 30also apply tothose cases, even though thewave number k;ofthetransmitted wave E=E,, exp(—a-r)expj(ot ~Br), G15) isthencomplex - ~a- -B-Totalreflection occursatlargeanglesofincidence, whentheincident H=Hoexp(—a-r)expj(ot— Br). G16) waveliesinamedium whoseindexofrefraction islargerthanthatofthe 7 ‘Theseequations defineawavethatpropagates inthepositive direction second medium, More precisely, totalreflection occurs when j ofthevector Batthephase velocity _ ®*Allthematerialinthischapterisessentialonlyifyouwishtostudyopticalwaveguides| va G17) inChaps. 35and 36 $80 PLANEELECTROMAGNETIC WAVESIV s12TOTALREFLECTION $81 andwhose wavelength is Byextension, Acanbeanycomplex number. Ifweapply Eq.2-7, a-%, hencep=4. G18) expjA=cosA+jsinA, 1-13) then ‘The amplitude ofthewave decreases exponentially inthepositive ; direction ofa,andtheattenuation distance 5,overwhichtheamplitude sina=SP/A—exp(TIA), APIAFEMPL-IA) (414) decreases bythefactorofe,isgivenby 2 2 6-1 (31-9) Clearly, sinAcannowbeanycomplexnumber.Thesameappliesto a cosA.Aswithrealangles, Equiphase surfaces areperpendicular to,and equal-amplitude sin?A+cos*A=1. (31-15) surfaces areperpendicular [email protected] wave ofFig.31-5, B points totheright, andapoints downward. Wehave generalized theconcept ofangle soastorender itmore Ifwesubstitute theabove expression forEinthegeneral wave useful, Butwenowhaveanabstract quantity thatdoesnotlenditselftoa equation 28-14, wefindthat simple geometric interpretation. : : Remember thatexp2)isequaltounity.Youcaneasilycheckthat ke=(B—jay’ =B= 0?~2ja- B=wen jou, (31-10) B-a?=wreu, 2aB=won, Gl-11) sin(A+2m)=sin A,—cos(A+22)=cosA. (31-16) where €,4,0,andwareallrealandpositive. These lasttwoequations Weselect thatvalue ofAwhose realpartliesbetween 0and2. areinteresting. 31.2. TOTAL REFLECTION (1) The first one shows that 6>a.Then A<é (1-12) ‘AswesawinEq.31-1,totalreflection leadstoavalueofsin6,thatis real and larger than unity. Then @;iscomplex. Snell’s lawand Fresnel’s (2) According tothesecond one, B-a=0. Thus theangle @between equations stillapply. thedirection ofpropagation andthedirection ofattenuation isatmost ‘Thecritical angle ofincidence, forwhich @,=9°,isgivenby #. nm (3) Inconductors, ¢#0. Then @#0. Ingood conductors 6?~a*<« sinOe= 1-17) 2a Foraglasswhoseindexofrefraction n,isequalto1.5,andwithn,=1, (4)Innonconductors ¢=0and@isequal to90°.Theattenuation ofa thecritical angleis41.8°, nonuniform planewaveinadielectric canoccuronlyinadirection Atangles@,>@).,sin@rislargerthanunityand6;iscomplex. ThenPerpendicular tothepropagation. thewave istotally reflected asinFig.31-1. Thisphenomenon is independent oftheorientation oftheEvector intheincident wave. Nonuniformplanewavesarenottransverse. IfEistransverse, thenH Therenonetheless existsatransmitted wave.Medium2actslikeapure isnot,andinversely. SeeProb. 31-1 inductance fedbyasource ofalternating voltage: theaverage power flow iszero,with erflowing alternately onewayandthentheother. 31.1.1ComplexAngles Totalreflectionhasmanyuses,mostlybasedonthefactthatthe ‘The magnitude ofanangle A,expressed inradians, isequal tothepure coefficient ofreflection (Sec. 30.6) isthen equal tounity iftheinterface isnumbera/R,whereaisthelengthofthearcofacircleofradiusRwhose clean.(SeeChaps.35and36onopticalwaveguides.) Onerelatively center isattheapex. little-known application isinternal-reflection spectroscopy, inwhich one se 2TOTAL REFLECTION 583 Also, a £058,=(1~sin?6)! =~j(sin?0,~1)", (31-23) xBe ~ Wehavechosen anegative signbefore jtoagree withEq.31-22. e4 Forexample, ifm,=1.5 andn,=1.0, thenthecritical angle of tt incidence is41.8°. Ifnow6,=60°, thensin6,=1.299, b=0.755, we weay « Oy=(21/2)+0.755),andcosOy=-0.829}.S NA Theincident, reflected, andtransmitted waves areofthesameformas se inSec. 30.2: Ey=Eimexpjet—k,(xsin6,—2cos6,)), (31-24) - y Ex=Enm expjot—ky(xsin8,+zc0s0,)), (31-25) Fig.31-1. Forangles ofincidence @,larger thanthecritical angle @,.,the Ex= Ermexpilot~k(xsin8,~z0s87)] (31-26) interface actsasaperfect mirror. This istotal reflection. observes thespectrum ofthereflected wavewhenthesecond medium is 31.2.1 TheReflected Wave iklyabsorbing. Themethod isconvenient forinvestigating thesecond . ;Tnedium, formaterials thatdonotlendthemselves toconventional Applying Fresnel’s equations fordielectrics giveninSec.30.4,wefind transmission orreflection spectroscopy. that Se Exm\_(1/3)cos6,+j(sin’67~1)'? et (Gar)=emcee rN =expj®,, (31-27) Op=a+jb. (31-18) Emm!(mi/nz)cos8,—j(sin®Br~1) Thewhere en 2 v2(sin? 6, 1)—exp(- ®,=2arctan 227)” -sin6;aseiletp)epCites) 22aTTcosOy (31-28) Thisisthephase ofthereflected wave with respect totheincident wave =b) ~exp (~ja) exp ,~Seep(Ch)exp pt (1-19) ‘atanypointontheinterface.ThereflectedwaveleadstheincidentwaveJ torthispolarization. SeeFig.31-2. Since thisquantity isreal, @must beequal to27/2and Observe that theincident and reflected waves areofthesame amplitude: total reflection islossless. ;expb+exp(—b) 531-20) Thephaseshiftisdifferent when£isparalleltotheplaneofincidence.sin87=~ =coshb. 1-20) (SeeProb.31-6.) Totalreflection ofawavethatispolarized inan So ubitrary direction yields areflected wave thatiselliptically polarized. Fortheaboveexample, ®,=95.7°. or=2+jp G1-21) i 31.2.2 The Transmitted Wave Then cos8,=oxPila+ib)+exp(~i)(a+jb) Thevectorwavenumberforthetransmitted waveis 0s0,=<Po r 2 T=Br—jar=kfI7&—COsO72) (31-29) _ifexp (=) -exp 5)_stp (1-22) kr=Br—jar=ka{sin6s—cosOrt) G12) FZ ysinh0. =k,(sin@,£+jsinhba). (31-30) ‘584 585 oo 4 or]+, 4 4 | = ca 5 aow(we rs2 ww Fig.31-2. Thephases ®,and®,ofthereflected wavewithrespect totheineident waveatapointOntheinterface, fortotalreflection whentheEofthe dineidentwaveisnormalandparalleltotheplaneofincidence.Theratiom,/nzis orweor7we~ equal to1.304 Fig.31-3.Theratio5/,wheredisthedepthof flb ofpenetration forthe transmit‘Thefirstterminsidetheparenthesis showsthatthewavetravelsinthe waveandJisthewavelength inmedium2dividedby2,asafinctonofthe positive direction ofthex-axis. Thesecond termprovides attenuation in. thenegative direction ofthez-axis ifbispositive. Ifbwerenegative, the‘waveamplitude wouldgrowexponentially withdepthinsidethesecond Figure31-5shows“crests” ofEforthethreewavesinonespecificmedium, which isabsurd. case.‘Theattenuation distance inthedirection perpendicular totheinterface Letuscalculate Hy.Sincethetransmitted waveisnotuniform, H,isis 1kah nottransverse(Prob.31-1).ThenFig.302isofnowsetocalculateHrj-_t_-—&_ 2»_ 131 ievectorEisnormaltotheplaneofincidence.FromSecs.28a Kesinhb~ mysinhb- 2xn;sinhb 2xsinhb 28.2.2,andwithaxeschosenasinFig.31-1, 2and Fortheabove example, 5,=A;/5.21. Then thewave amplitude decreases byafactor of¢overadistance ofabout A,/5.Thetransmitted wave barely penetrates intothesecond medium. Theamplitude decreases bya bo ofactorof183overadistanceequaltoAs!SeeFig.31-3. we 4-0 ‘Applying againFresnel’s equations fromSec.30.4, as Em 20s8 sErm)____2c0s) ___ 31. " (=), 088;~jKra/m,)(sin® 8;—1) G132) , 2cos 6, ®.a exp 22 2c0s6 (2. I = 2081expBt 31-34) —Gin 2 G13) 00 os 10 is zo t Figure31-4showshowthisratiovariesinamplitude andinphasewith Fig.31-4,Theratio(Eym/Emm), =&+jn,plottedi‘ma Erm) plotted inthe lane fortheangleofincidence 8.Itsmagnitude isequalto1.34fortheabove satiousanglesofincidence6ylargerthantheericaangle,tadfrar1.50 nrTheamplitude ofthetransmitted waveislargest atthecritical angle."The | transmitted wave leads theincident wave bytheangle 9. 387 586 ] A sehlUcFTtlCOSe aan Po , 4 | 2 we Enf-ies FA FS 2 ae see 7 Fig.31-5. “Crests” ofEfortheincident, reflected, andtransmitted waves fory=3.0,m=10,6,=75. . Fig.31-6.TheHvectorofthetransmitted waverotatesinthedirectionshown, wheXEr_ hrXEr_ krEro 313s Thetransmitted wavetravelstotheleft,andtherotation itheeverseofthatof fyEAEeB ors) eters hieeeee _(Gin7#~cosOr2)xErf_(008 Or+sin8rBEr Gyap= Ze=Ze+6136) ‘Thereexistsapowerflowparalleltotheinterfaceinthepositive ‘Thendlireetion ofthex-axis. The flux isafunction of6),asinFig. 31-7, for E Hac== 0080, Hogs==Psin Op, ens) normal totheplaneofincidence: forn=1, a a €0)!myc0s6,sin26, 2 SinceFrom =(A) eee exp|2(nisin?6)—12 Em rms(ie)ing? EmenPgesins—0) Hm=F (31-38) ian) then Heme =Hiry 60887, Hrs =HmSin8, (1-39) Allthisapplies toanincident wave ofinfinite extent, butwhat happens ifthe incident wave has afinite cross section? Our analysis cannot precisely aswhen 07isreal andH,transverse. Also, provide ananswer tothisquestion. What happens isthis: anincident ray Fo ; . penetrates intomedium 2andreturns tomedium 1abitfarther along theHime+Himz=Hitm(COS* Or+sin?Br)=Him (31-40) v-axis.ThisistheGoos-Haenchen shift.’ IfE;isparallel tothe plane ofincidence, the formalism isagain the same aswhen 87isreal _ Since sin8,isreal and positive while cos@,isimaginary and negative, *SeeHelmutK.V,Lotsch,Optik,vol.32,pp.116189,299,$83(1970and1971).There thexcomponent ofHylagsthezcomponent by/2.TheHyvector cystaalps phenomenon oases HLBernt andTTami,Applied rotates attheangular velocity «,asinFig.31-6. Physics, vol.2.p.197(197). ' 588 312TOTALREFLECTION 589 04 4.allthelight crosses theglass-air interface. This isnotabad approximation because thecoefficient oftransmission isclose to unity, except near thecritical angle. Then Fisthesolid angle 96 corresponding to6, divided by2x(not 4x, because ofthe < mirror). Since theconedefines asolid angle equal totheareaof & thespherical segment, shown asadashed lineinthefigure, §o ; divided byR°, é i 1("2a sin@Rd0akfe2aRsinORcosa, Be os il Foz - rere (31-42) For aglass whose index ofrefraction is1.5, @,=41.8° andow ur or 7 wrra F=0.255. ThefractionFis,infact,evensmallerbecauseofour “ | approximation,Fig.31-7.Thetime-averaged Poyntingvectorparalleltotheinterfaceinmedium —|)Fyample|THECRITICAL ANGLEANDTEproportional tocos6sin28,fortotalreflection.Thevectorisnormalto THEBREWSTER ANGLE theplane ofincidence. Wehaveset =1.50andn;=1.00 ‘The critical angle (Sec. 31.2) issomewhat larger than theBrewster angle (Sec. 30.5). Forexample, again forlight propagating inside a SION FROM A aglasswithanindexofrefraction m,of1,5,thewaveistotally Prample|UNTHODE RAYTUBE tamuitoteaiat.glasaievace whentheangeof incidence isthe Brewster angle, 33.7", and itistotally reflected Inacathode raytube theelectron beam generates light ina backintotheglass beyond thecritical angle of41.8°, fluorescent coating deposited ontheback ofthetube face. A Figure 31-9shows these twoangles asfunctions oftheratio givenpontinthefurescent material radiates inalldirections mlnForlargevaluesofryfma,thatis,forlightincident in&butathinlayer ofaluminum, asinFig.31-8, doubles thelight output. Eventhen,mostofthe lightstaystrapped inside theglass o bytotalreflection andtravels backtothegunendofthetube ‘What fraction Fofthelightcomes outthrough thetubeface? Thisiseasytocalculate ifweassume thatinsidetheconeofangle K oTal reflection Fvorescent Gistobe— tase itangReecting icc ~ ' 5 a ee Fig. 319. The critical angle and theBrewster angle asfunctionsof theratio m,/ns. The incident wave islinearly polarized with theE vector parallel tothe plane ofincidence fortheBrewster angle Fig.31-8. Section through thefaceofacathode raytube carve 60 PLANEELECTROMAGNETIC WAVES1V|roms so Then latively“dense”medium,8,isnearlyequalto8p.Formedia in Withmoresimilarindicesofrefraction. theBrewsteranglei cos6,=—jsinhb=~j(sin?6,~1) (31-22),(31-23) approaches 45°whereas thecritical angle approaches 9X incidence,theamplitudeofthereflectedwavechangesrapidly (sin?6,=)!"whentheangleofincidence liesbetween theBrewster angleand ®,=2arctan“ —t (31-28)thecriticalangle.Thispeculiarbehavior ofthereflected and (n,/nz)cos6;mitted waves could beuseful formeasuring small angular; —-, The transmitted wave isnonuniform. Itpropagates parallel tothe interface anditsamplitude decreases exponentially perpendicular tothe_|interfacewithanattenuation distance 31.3SUMMARY ; ; i 6,22 (31-31) Inanonuniform plane wave, propagation occurs inthedirection} 2asinhb perpendicular totheequiphase planes,andtheamplitude desenses Also,exponentially inadifferent direction. Then thevector wavenumber isof (Eis)=298ay®sine theform { Em) (=@ainyyyeOP2 G134) k=B-ja, G14) where@andBpointindifferent directions, with PROBLEMS 1 1 3-1. (31.1) Anonuniform plane electromagnetic wave isnottransverse. a ere G18),G1-9) Wedefineatransverse waveasoneinwhichEandHarebothperpendicular tothetwovectors @and. . “ (a)Write outMaxwell's equations foraplanesinusoidal waveinfree 6beingthedistance overwhichtheamplitude decreases byafactorote space,replacing Pbyfeand3/31byJonNotethatEosalwaneevalto Normally, themagnitude Aofanangleisarealnumber.Byextension, zero.Sincethevector&iscomplex,ithasnospecieorientationinspace, Acanbeanycomplex number. Then except thatitliesintheplane defined bythevectors @andB. 'yPl cia) (b)Suppose thatHistransverse: H'=Ha,where a=0and@- =0. xpiAexp(~iA) _expjA+exp(i : ShowthatEisthennottransverse. Similarly,ifEistransverse, thenHis sina=Pe PITA) cosa=PEEP, GLI) nottransverse Atos A= 31-15 81-2,(31.2) Total reflection asinFig,31-5 sin’A+cos’A=1, G5) ‘Anelectromagnetic wavepolarized withitsEvectornormaltotheplane ;i ofincidence istotally reflected asinFig.31-5attheinterface between a Totalreflectionoccursatanglesofincidencelargerthanthecritical dielectricwhoseindexofrefractionis3.0andair.Theangleofincidenceis angle given by 15° ny (a)Calculate 6,/2, and8./%, sin<2 @u7) (8)Cale thepes ofthe rested andtransmited waveswith 1 respect totheincident wave atanypoint ontheinterface. Then 6;iscomplex: (c)Check thecontinuity ofEacross theinterface. x 5.2 81-3,(31.2)ThePoyntingvectorfortnetransmitted waveOr=5 +b G12) CheckthevalueofFn...giveninSec,31.2.2,form= 81-4,(31.2) Totalreflection onaplasma whenw/a,<1 wherebisdefined by Show thatawaveincident onanionized region istotally reflected if sin0,=cosh b (31-20) ©<w,, where ,istheplasma angular frequency (Sec. 29.2.3). ‘a so PLANEELECTROMAGNETIC WAVESIV PROBLEMS 39 315,(31.2) Thephaseshifts©,and@,intotalreflection (©)Plotcurves ofT;and7,form/n;=3.5. Notethatthe t(2)Showthat cosficients areequalandapproximately independent oftheangleofy=zaraye inceencwhen6,smal.denttheBrewserangle =imInn8,Tm) €thatthefaceofthesemiconductor isflatandparal wine Tn) lSAssumethathefaceofhes orislatandparalleltothe (b)Showthat ii CalculatetheSectioné‘ofthe light emitted atthesource that reaches the . surfaceatananglesmallerthantheeiticalangle.Sh tan tn(2-2)= } (a)ShowthatFT~1/[n(n+1)". we.ShowthatF=1/(4n’).22)cos@,(sin®8,~n3/ni) i (e)CalculateF,T,andFTforn=3.5. i ( Calculate FTfor an LED's(6)Plot®,~©)asafunctionof8,between40°and90°form=1.5and woveindeofreactionithesameasthalofthesemiconducor fs n=] imprctialbecause shaping theemicondactor expensiveas 31-6,(31.2)ThetransmittedwavewhenEisparalleltotheplaneofincidence (2)LEDsareusuallycoveredwith&hemisphe f Oe)int ban , whosemsaout1.6 1hemisphericaltransparentsesin culate the two transmission coeficients and the effi (=)-amperS-7) efficiencyisimproved,butitisstilverylow=eficiensy.The 2, (312) Seinvt,‘ke. 31-9.G12)JonBivocterclhntnaemtoedwore 31-7.(31.2) Scintillation particle detector fefoundthatthexand2cony 5,tuallymadeoutofasinglecrystalofsodiumiodideorofasuitable Evectornormaltotheplaneofincidence mwinatransparent plasticembeddedinareflectorR,emitslightwhenitis ShowthatthevectorrotatesinthedirectionshowninFig.31-6traversedbyanionizingparticlesuchasanelectron.Aphotomultiplier PM ‘s. detects theemitted light “Thescintillator hasanindex ofrefraction m,andisfixedtothefaceofthe photomultiplier withacement Cofindexmm. Lightisemitted inalldirections inthescintillator, butonly afraction Freaches the photomultiplier.(a)Calculate Fasafunction ofn/n,assuming thatT=Iforanglesofinduence smaller than thecritical angle andthatthescintillator is Surrounded by@nonreflecting substance. (5)Drawagraph ofFforvalues ofa/nyranging from0.1t01.0. 31-8.(31.2) Totalreflection inlight-emitting diodesInlight-emitting diodes (LEDs), radiation occurs inajunction planewithin -asemiconductor whose index ofrefraction isquite large. For Cxample, withGaAsP, n=3.5. Totalreflection atthesemiconductor-it mertace imits theeficency ofLEDs toafewpercent (a)Calculate thecriticalangle. Vj f Vv s AY N CMM Fig.31-10. 395 §) cuarreR32 j od ™~5 ”.) | e oii el : PLANE ELECTROMAGNETIC rs aSoNee& Ay AG WAVES V wteg Reflection andRefraction C:Reflection and aeRefraction attheSurface ofaGoodConductor* _” Fig,32-1.Theinci ec ra CONDUCTOR 595theplane ofincidence ectoohthetncideatwaveisnormal86 32.1.1ENORMALTO THEPLANEOFINCIDENCE $97 32.12 EPARALLEL TOTHEPLANE OFINCIDENCE $97 Example; COMMUNICATING WITHSUBMARINES ATSEA598 32.1REFLECTION ANDREFRACTION ATTHEExample:STANDING WAVESATNORMALINCIDENCE ONAGOOD SURFACE OFAGOOD CONDUCTOR CONDUCTOR 598 +322. RADIATION PRESSURE ONANONMAGNETIC GOOD Aspreviously, theincidentandreflected wavesli iieinmedium 1.Here, CONDUCTOR 399 medium1isadielectric, whilemedium2isa i7 +32.2.1 ENORMALTOTHEPLANEOFINCIDENCE 59 32-1and32-2isagood conductor, asinFigs. #3.2.2EPARALLEL TOTHEPLANEOFINCIDENCE 602 Equations 30-17to30. i #3223THEMOMENTUM FLUXDENSITYANDTHEMOMENTUM DENSITYIN media.Wemaythjaeiytoanypairoflinearandisotropic [ANELECTROMAGNETIC WAVE 62rite that +3224THEcos"&,TERMINTHEEXPRESSIONFORTHERADIATION E,=Einexpj(cot—kxsin6,+k,z6086) PRESSURE 603" “he G21) Examples 6034323.THEELECTROMAGNETIC MOMENTUM OFSTATICFIELDS 64324SUMMARY 604 PlPROBLEMS —6s ;x 3 Wn ee ex, Reflection andrefraction atthesurfaceofagoodconductor are oeMowByasomewhat similartototalreflection inthattheangleofrefraction isagain wee ng Sscomplex.However,withgoodconductors,theimaginarypartofOis Caieonegligible and8,~0.Thenthetransmitted waveisapproximately "gE‘uniform, buthighlyattenuated. <Radiation pressureonaconductor resultsfromaHalleffectonthe fNN conduction electrons moving inthemagnetic fieldofthewave.5 Fig.32-2.Reflection andrefractionatthesurf oe Ma, 322. Rect atthesurfaceofagoodconductor. TheEof mischapterisaprerequisiteonlyforChap.34. identwaveisparalleltotheplaneofincidence. 596 PLANEELECTROMAGNETIC WAVESVv 121THESURFACEOFAGOODCONDUCTOR sor Exq=Enn€xpj(wt~kxsin8,~kzc0561), (322) andthetransmitted wave propagates intotheconductor along thenormal . totheinterface, whatever theangle ofincidence. Also, theamplitude of - - 7 32-3) . Ey=Ermexpj(t~kyxsin8,+kzz€0s67) .(23) thetransmitted wavedecreases byafactorofeoveroneskindepth6. ~Emexpj{oryxsin,hs2|1~(%) sin?o,)"}2-4) 32.1.1ENormaltothePlaneofIncidence Now, from Sec. 29.1 Refer toFig.32-1. From Fresnel’s equations with |n,/n,| <1, my_ky_(ems) Enm)_(tu/n2)6088,~cos6 mahyolemy"S 32-5) San)(rlns)cos6cosOr - mk, Inj oe (), (m/nz)cos0,+cos 8; G24) -_> (32-6) foranyangleofincidence. Reflection fromasuperconductor (n,—>=)is a-p, lossless. Thenegative signmeans thattheEvector ofthereflected wave isinthedirection shown inFig.32-1,opposite tothatshown inFig.30-2, ‘Weshellessumethat because webasedourcalculation onthelatterfigure 6«1orthat [<a (2-7) Also,fromSec.30.3andfromthefactthatcos6;~1,24, ing| ~ Em’ 2 eo, A: (Ezz)=2inslns)e089 966,00, (3218) SeeSec. 29.1. Then theexpression inbrackets inEq.32-4 isapproxi- Em (n,/nz) cos8,+cos8,“in; matelyequaltounity.Then againforanyangleofincidence 8,. 5-«E,expi(wt—kyxsin9,+2=p) (32-8) Atthesurfaceofadielectricsuchthatn,>>m,onealsohasthat 1=Erm 1 3 (tpz (2)=-1, (2)=O. (32-16) =Emexp[itor~kyxsinoy20*P2) (2.9) Em Emm! Weselecttheplussignbefore theztermsothatE,willtendtozeroasz 32.1.2 EParallel tothePlane ofIncidence tends tominus infinity. Soweneed aplus signinEq.32-4, andfor Refer nowtoFig.32-2. From Fresnel’s equationsNy. q reflection from agood conductor, .te (2)_(tulns)608Br~cos8,cos6=+1=(™Ysinea<1,6r=0. 210) Eim)y (M/nz)€0s87+605O; ny 2 _(auln2)~cos6, Therefore (yin) +6050,” (32-17) z),2 Ex~Ermexp[i(we~kyxsin8,+5)+3.G2-11) Thelastapproximation isnorvalidatgrazingincidence,where6,isclose 1090°,Also, But Erm 2(n,/ns) cos8, , te Erm) ___2(ni/n2)cos6,_ - sshd ki(so)=20Ble (32-12) (z2),086;+(n:/n3)€0sOy (218)2 2m,_2nd ‘Then wt an 81+f). (32-19) 242] (3213) ny ~+2) 42], Br™Brmexpli(+5) +5 Theapproximation isagaininvalidatgrazingincidencePP 8 grazing 1 598 PLANE ELECTROMAGNETIC WAVES V so Example COMMUNICATING WITH SUBMARINES ATSEA For shore-to-ship communication with thesubmarine antenna submerged, theefficiency isvery low, first, because ofthelarge coefficient ofreflection atthesurface oftheseaand, second, because ofthehigh attenuation inseawater. The attenuation in seawater isabout 172decibels/meter at20megahertz, 5.5at 20kilohertz, and0.33 at75hertz. One solution istooperate atlow frequencies (about 75hertzand17to25kilohertz) andveryhigh | power,withhugetransmitngantenss, manyKometers omthe side. ‘Another solution forshore-to-ship communication istomodu- lateslaserbeamemitted byasatelite, seawater beingquite | wt transparent toblue-green light. Remember thatourdiscussion on << atomicandmolecular phenomena andisvalidonlyuptoroughly >gigahertz, Opticalfrequencies areoftheorderof10"hertz. | 7 )Ship-to-shore communication atlowfrequencies isimpossible | y swwithlongradio waves because asubmarine canneither supply the q: required power nordeploy along enough antenna. Two-way. ‘communication takes place atafewmegahertz withthesubmarine antenna projecting above thewater. | Fig.32-4. Thestanding wave pattern forreflection atnormal incidence onagood conductor, ataparticular time. Nodes ofE Example STANDING WAVES ATNORMAL INCIDENCE ON | andofHarespaced 4/4apart. AGOOD CONDUCTOR Figure 32-3shows theincident, reflected, andtransmitted waves. Since thedirection ofpropagation ofthereflected wave isopposite tothat oftheincident wave, and since EXH points inthe inFig. 32-4. The nodes ofEand Hare thus one-quarter direction ofpropagation, theHofthereflected wave isinphase ‘wave-length apart. Theenergy density isuniform. with that oftheincident wave attheinterface, asinthe figure. At ‘Asimilar situation exists forreflection from anysurface. Either thereflecting surface, theelectric fields nearly cancel andthere is theEortheHvector must change direction onreflection, inorder anode ofE;themagnetic fields add, andthere isaloopofH,a tochange thedirection ofthePoynting vector EXH. 7 .*32.2 RADIATION PRESSURE ON A .iLKe NONMAGNETIC GOODCONDUCTOR>< 7 32.2.1 ENormal tothePlane ofIncidence Ni Le WithEnormal totheplaneofincidence, Eistangent totheinterface,tat > i“ ']therearenosurfacecharges,andthereisnosurfaceforce.aati ws Inthebodyoftheconductor thecurrentdensityJisoF.ThusJisa < parallel toE,andperpendicular toHy.Itturnsout,asweshallsee,that 2 ‘ itheQuXuoHforcepushes theconduction electrons awayfromthesurface. This isjust another manifestation oftheHall effect ofthefirst Fig.32-3.Reflection atnormalincidencefromthesurfaceofa exampleinSec.22.1.1.Theresultingelectricforceperunitareaistheg00d conductor: Em~~Eim andErm&Em radiation pressure. ‘Thesituation wouldbedifferent iftheconduction clectrons were n~2)" x1-j entirelyfreetomovethroughthemetal:theconductivity wouldthenbe T(omexp(-13)Er=ome“Er (32-23) imaginary, asinthelow-pressure ionized gases ofChap. 29,the conduction current would lagthefieldby/2radians, andthere would Thevalue ofErmdepends onthepolarization oftheincident wave. bezeroradiation pressure. Inasuperconductor oisrealandtendsto ‘Thetime-averaged valueoftheelement ofpressure onasheetofinfinity, 5tends tozero, andthefollowing discussion applies. } thickness dzinside theconductor isthus Weset $ n!4p...=tomeRe(EH)de=jouRe(EAtiE;)ds(224 Fal«1 (3220); ~ Tomod &*) . i] =—Z IE,exp==‘asinSec.32.1except that,now,u2=1. Weconsider anelement of} JosEri"exp5dz (32-25) volume, asinFig.32-5, parallel totheinterface, ofareaabandthickness J! ang dz.Itcarries acurrent oE;bdz andissubjected toamagnetic force} > a ‘oE;abdzoH;inthenegativedirectionofthez-axis.Thenthe{ PA=5Glentexpdr=EpoeEnel.(62-26)instantaneous pressure exerted ontheelement ofthickness dzis in a jn62d 24o dp=oEzig,dz, (221) However, weneedp.intermsoftheinputpowerflux(Sec.28.4) withthepositive directions forEyandHychosen asinFig.325.A| Say=WEEin=$CEEHElm (32-27) positive result willshow thattheincident wave pushes ontheconductor. FromEq,32-13,thephasorforEyis Someexpress p.fistintermsofEqandthenintermsof9.From €. 32.1.1, 2) ,2Er=Emexp|i(or+2) +3], (222) Emm)_5M 7 (3)6 (=),2,0On (32-28) while thecorresponding phasor forH,follows from Sec.29.1: evenatgrazing incidence. Then a \sm e ahs Pam4[2508,En| (32-29) a with nie oy mi (ery!iey.HoHe-) z(o). (32-30)XG Soh = TARTS bothmedia 1and2beingnonmagnetic, Then :|| =2(£1)cos?6,£3,=2¢,cos?o,(H)"”, eines Paw=2(2)cos6,Bin=2¢,c0864(2!)So~APe =2(€44,)'?COS?Fr. (231) * y Substituting theindex ofrefraction n,ofmedium 1for€}andsetting Fig.32-5.Element ofvolume ofthickness dzinside aconductor. Hy=Hoyields iu ' onPLANE ELECTROMAGNETIC WAVES V 7322 RADIATION PRESSURE ON ANONMAGNETIC GOOD CONDUCTOR 603 2m to would notbeconservation ofmomentum. This isaproof that thePsav=1008" O1Fiau=2E"608"81, (32-32) magneticforceisindependent ofuu,forslowelectrons. One interesting application ofthemomentum fluxinanelectromag: where @’istheenergy density inthewave, netic wave isthelevitation oftransparent particles inavertical light |beam, See Prob. 32-14." *32.2.2 EParallel tothePlane ofIncidence 7 . Wedonotgothroughthecalculation here,Asonemightguess,the|32.2.4wees oncemintheExpression forpressure isthesameasabove. However, thecalculation issomewhat { adiation Pressure tricky: there arenow twoforces, amagnetic force (asabove) plusan Itiseasy toexplain thepresence ofthecos”@,term intheradiation electric force exerted bytheelectric field inthedielectric onthesurface pressure ifoneimagines photons raining onthesurface oftheconductor. charge. Letmedium 1beavacuum. Saytheir energy density is&’joules/meter’. ;Then¥isequalto€'c,theirmomentum densityisé'/c,andtheir *32.2.3 TheMomentum FluxDensity andtheMomentum momentum fluxdensity is',Thechange inthecomponent oftheDensity inanElectromagnetic Wave |momentum fluxdensity normal totheinterface is26”cos6, Suppose medium 1isavacuum. Thenatnormal incidence theradiation | ‘Suppose theincident beamhasacross-sectional areaof1meter’. Itpressure exerted onmedium 2,theconductor, is2Y;./c. Sincethe illuminates anareaof1/cos@,meters*. Then conducting surface actsasanear-perfect reflector, thepressure cor- y responds toachange inthemomentum ofthewave of2%;,./¢ perunit | Pav=(2€'c0s8,)cos8,=28"cos?0,=27"cos?8,(32-37)timeandperunitarea.Then,intheincidentwave, Hl c ¥, |Examples|Underordinarycircumstances, radiationpressureisweakand Momentum fiuxdensity =". (32-33), difficulttoobserve. Insunlight, atthetopoftheatmosphere, Syn.© isabout1.4kilowatts/meter’, andtheradiationpressureona Now metallic reflector isabout 10°*pascal, orabout 10°" atmosphere Momentum fluxdensity =momentum volume density xc.(32-34) echhesucofthesun,teradiation pressure slang bythe Itfollows that,inauniform planeelectromagnetic wavepropagating ina (Distance fromsuntoearth)” _(1.5108)vacuum, Radiusofson) *(peigr) =46%10" iow—ErmsHime _energydensity (32-38) ‘Momentum volumedensity=2=Ermffems _enerByCENSIY (39.35) |¢ ¢ ¢ This gives aradiation pressure ofonly 5x10° atmosphere Radiation pressure isunimportant even intheinterior ofthesun, These results agreewiththoseofatomic physics, where weassociate butitmayplayanimportant roleinthemoreluminous stars electromagnetic waves with photons ofenergy hw(R=1.05x10" is| Comet tails point predominantly away from thesun. ‘ThisPlanck’sconstantdividedby2x)andmomentumh/Ztravelingataspeed phenomenon resultpay{romfadiationpressureandpartlyromthesolarwind(Prob.28- c.Thus,foronephoton, Aphoton-drag detector consists ofacrystalofgermanium with Momentum _A/K_ 1_1 electrodes plated oneach end. When thebeam ofapowerfulCO, as hh At (32-36) laserpassesthrough thecrystalalongitsaxis,theconduction nergyelectrons drift forward. ‘The voltage difference between the Atthebeginning ofSec.32.2werestricted ourselves tononmagnetic electrodes isameasure ofthebeampower.SeeProb.32-12. media andassumed amagnetic force oftheform QuXoH. What ifthe conductor ismagnetic? The pressure isthesame, forotherwise there A.Askin, Science, vol.210,p.1081(1980) os 32.3THEELECTROMAGNETIC MOMENTUM OF Alternatively, radiationpressureresultsfromthefactthatthenormal ‘ATIC FIEI component ofthemomentum oftheincident wave reverses upon reflection, Wehave seen above thatanelectromagnetic wave possesses amomen- Inanyelectromagnetic field, static ornot,themomentun density ina tum density that isproportional toEXH, atleast inavacuum. Two vacuum isEXH/c?, questions come tomind. (1)What ifthefrequency iszero? Thevector product makes noreference tofrequency. Does thisexpression apply to static fields? (2)Inanelectromagnetic wave, there exists amathematicalrelationbetweenEandH:givenE,Hfollows,andinversely. However, PROBLEMS instaticfields, Ebearsnorelation toH.Forexample, theelectric field 32-1,(32.1) Reflection fi 7]mightbethatofsomechargedbodyandthemagneticfieldthatofa onorawtwofiguresninethanefb larto if.30-4, showing Eand Hforan permanent magnet. Does theabove expression apply toanypairof electromagnetic wave incident onagoodconductor. Youwill,ofcourse unrelated Eand Hfields? Sayone brings apermanent magnet near a have toexaggerate thevalues ofE,,andof4intheconductor. Besuretocharged capacitor. Does thefieldpossess momentum? What ifthe showthephases correctly. Showx-,y-,z-axes onbothfigures torelateone voltage across thecapacitor istime-dependent? Isthefieldmomentum withtheother. alsotime-dependent? 32-2.(32.1) Reflection fromagoodconductor‘Theanswer tothesetwoquestions isbynowwellestablished: the ‘Showthat,foragoodnonmagnetic conductor inair, momentum density ofanyelectromagnetic fieldinavacuum isEXH/c’. ‘Thisfacthasbeendemonstrated bythoughtexperiments inwhichone ()|~1-2e050, —(b)EI~1-—4establishes afieldwhere EXHisnotzero, taking intoaccount the Emi, hy” Emly %yc088,” magnetic forces andtorques, aswellasthemechanicalforcesandtorques islatterrel validaiswf requiredtokeepthesystemimmobile,Tosatisfythelawofconservation sro. Mastionnetwadotgacngincidence,wenocon6;tansto ‘ofangular momentum, themechanical torque integrated overtimeis Agoodconductor isabetter reflector whenEisnormal totheplaneof ‘equaltotheangular momentum ofthefield.Inthiswayonealways finds incidence. High-quality metallic reflectors havecoefficients ofreflection ofthattheelectromagnetic momentum density isEXH/c?inavacuum. about90%nearnormalincidence inthevisible,withunpolarized light. SeeProb, 32-15 32:3.(32.1) |Enn/Em| aS.afunction oftheangleofincidence forreflection onaconductor For agood conductor, ¢/we >50.Then 32.4SUMMARY Boa(208)(SBE) wesc=Sean)" Reflection atthesurface ofagoodconductor isslightly lossy.The So4/210if=4andp,=1 transmitted wave isweak, highly damped, andtravels inadirection Plot|Epm/Enm|, and|Ewn/Eiml; a8functions of8,foranonmagnetic nearly perpendicular totheinterface 00dconductor inairandforA,/3=10.Youwillfindthat,when Eisinthe When electromagnetic radiation illuminates aconductor, thetrans- Plane ofincidence, thereexists apseudo-Brewster angle forwhich the mitted wave’s electric fieldgivesrisetoaconduction current thatflowsin amplitude ofthereflected waveisminimum.thewave'smagnetic field,withtheresultthatthecloudofconduction 12-4.(32.1)Liquid-crystal displays (LCDs) electrons ispushed back.Thisisessentially aHalleffect andshows upas Inliquid-crystal displays theliquidissandwiched between atransparentradiation pressure: ‘multiple electrode infrontandasingleblackelectrode intheback.Uponspricationof2voltageseenofthefrontwindow,therodlike aeculesofthenematicfuidinthat i" PawPaw 05"Fis (32-32) window, andonecanseetheblackelectrodeintheFach;Elsewhere,the moleculesreflect light because their orientations arehaphazard. ' 06 PLANEELECTROMAGNETIC WAVESV «wn ‘Thetransparent multiple electrode isathincoating either ofasemiconductingmetaloxide,suchastinoxide,orofgold.Thesurfaceresistance i(Prob. 49)isoftheorderof10to100ohmspersquare. At tt600nanometers, andforgold,BX,~1.29andah,=2.59.Theconductivity [|taserbeam‘ofgoldintheformofathinfilmis4.2610”siemens/meter. Lt(a)Calculatetheskindepth6. tt(b)Bywhatfactordoestheamplitudedecreaseinthegoldfilmifits => im thickness sis0.055? |1 (©).What isthesurface resistance?(a)Calculate thethickness ofthefilminwavelengths 2o i ‘Aproper calculation ofthetransmission would takeintoaccountmultiple reflections. Theeffectofmultiple reflections is,however, much | Silpa Fig.32-6. {essthaninProb. 30-10 because oftheattenuation inthefilm. 32-5. (32.1) Thesurface impedance ofaconductorBydefinition, thesurfaceimpedance ofaconductor isequaltotheratio 32-9.(32.2)Theradiation forceonasphereofthetangential components ofEandHatthesurface, ortoE,/H, Calculate theradiation forceonareflecting sphereofradiusRinterms(a)Showthatthesurfaceimpedance ofagoodconductor isgivenby ofthePoynting vectoroftheincidentradiation vw32-10. (32.2) The radiation force on Lionacylinder 2=($) ara [Catsiate theradiation foreperunitlengthonaoyinder ofradiusR whose axis isperpendicular tothe Poynting vector incomi“Thequantity1/06correspondstothesurfaceresistanceofProb.4-9.For radiation. ‘oyntingvectoroftheincoming come,1/06isequalto0.261miliohmpersquareatImegahertz, from ‘2:11,(222)Radiationpeesureanccomettails Table 29-1.morn OsShowthatthepowerdissipatedpersquaremeterintheconductori| (a)Comparethegravitationalandradiationforcesexertedbythesun FEog]08. mesphericalparticleofradiusawhosedensityis5000kilograms/meter’.‘NowwesawinProb.19-4thatH,isequaltothecurrentperunitwidthin ‘Thesunradiates3.8x10%wats.SeetheTableofPhyialConstantsattheconductor.Itfollowsthatthepowerdissipatedintheconductoristhe =endofthe book.Assume thatthepares ack.aoecorethecurrentwereuniformlydistributedthroughoutthethickness Cbaleltetevaofaforwhichthetwofoesreequal :shoul icles smaller than about 0.1 micrometer in ;radius arerepelled atany distance from thesun.This explains wl 3246,(321)CuttingtealatewihIsertem Hernebay SnneeetinetAlertgfstaee ‘gure 32-6showsalaserbeamcutting asteel plate, aresaidtobeType2.We hav resun (a)Whydoesthebeamcutatafaster rate when theEvectorliesinthe importantwhenai,[chavedisregardeddifraction,whichis planeofthepaperthanwhenitisperpendicular? "ThetailsofType1comets aregaseous. Th{b)Roughlywhatpercentageofthebeampowerservestoheatthestee! sunbutforadhforemreson,Theforcethesarebomttcresten intheformer case? between thisgasandthesolar windwhich consbts of mC (Prob.28-12), whichconsist{)CanyouexplainwhytheKerfisnarrowerandmoreevenwhenEis fonizedhydrogenthtevaporatesfomthesua, nenmsof intheplane ofthepaper?Iftherequiredkerfisnotstraight,thenthelasershouldrotatetokeep 32-12.(32.2)Photon-drag radiationmonitortheEvectorofthebeamparalleltothepath,Asimplersolutionistousea Figure32-7showsaschematic diagramofaphoton-drag radiationey polarized beam moni,Theedeveaewedmonotnintenofovert er ebeam enters ontheleft, through anantiref 327.(42.1) Thestanding waveatnormal incidence onagoodconductor arenereflection coating 3 atransparent electrode (Prob. 32-4). Thebo ‘Anelectromagnetic wavefallsatanangleof6,onaslabofdielectricthat Radiationpressureinthesemiconductor propelsthechargecarrierstoisbacked byagoodconductor. theright.Ifthecarriers areelectrons, theelectrodes become charged‘Under whatcondition isthereasingle reflected wave? inthefigure, andthevoltage Visameasureofthebeampower. ws PROBLEMS 09 oe particle frombelow, onetotheleftoftheparticles’ center andonetotheCo right.Refraction deflectstheraysandhencechangestheitmomenta. (b)Show that foragiven beam intensity thevertical position ofthe beamisalsostable.Areflectingparticleisejectedlaterally. vw™) 7 vw 32-15.(32.3)‘Theangularmomentum ofanelectrically chargedpermanent magnet Thefieldofanelectrically chargedpermanent magnetpossessesan Le, | angular momentum because Eisradial,whileHpointsapproximately in b t Fig. 32-7. the@direction, sothat EXH isazimuthal. | Wefirstcalculate thevalueofthemomentum fromtheknown values of i Eand H, and we then show that itsexistence follows from the law ofThesemonitors haveashortresponse time,oftheorderof|| conservation ofmomentum,nanosecond. Theyaremadeinvarious sizes,withcrystals oftheorder | Imagine aconducting sphere ofradius Rwhose magnetization MisfofIcentimeter indiameter andafewcentimeters long.Thecrystal || uniform. Youmaytakeforgranted thatoutsidethespherethemagneticabsorbs aboutone-quarter ofthepulseenergy. Thepeakpower density fieldisthesameasthatofasmallmagnetic dipole ofmoment $Mcanbeashighas20megawatts/centimeter’ | situated atthecenter.ThespherecarriesachargeQ. FindtheratioV/Jy.SetY=Jexp(ar)insidethecrystal. | (a)Findtheangularmomentum ofthefield.; (b)Calculate thevalueoftheangular momentum LforR= 32-13.(32.2.2) Radiation pressure withEintheplaneofincidence 20millimeters andM=10°amperes/meter whenthesphereischarged to‘Showthattheradiationpressureonanonmagnetic goodconductor, || ‘potentialof1000volts.Couldthespherebeusefulasagyroscope?whenEliesintheplaneofincidence,isthesameasinSec,32.2.1.Inthis (ec)Nowletusstartwithanunchargedsphereandgradually,deposit instancethereisbothamagnetic forcewithintheconductor andan chargeonitbymeansofanaxialjonbeam.Chargefowsinatthenorth electricforceonthesurface charges. UseGauss's lawtofind0 poleanddistributes itselfuniformly overthesurface ofthesphere. The ‘magnetic field exerts atorque Taye On the charging current. To 32-14. (32.2.3) The levitationoftransparent particlesinalaserbeam thenherefromturning.ve[bool opposing fit.weir Figure 32-8showsasimplified diagram ofadevice forlevitating agenTeochthettransparent particles inalaserbeam. Theparticles canrange from 1to a 100micrometers indiameter. Thelightintensity ismaximum ontheaxis Tae -T,,ateofthebeam andtapers offoneither side.Theparticle staysontheaxisof ee at smatafixedheight. ‘ ‘MCshowqualitatively that,iftheparticlestraysawayfromtheaxis,it Choosepolarcoordinates withthenorthpoleat@=0. suffers arestoring force. Theaxisistherefore aposition ofequilibrium ‘Show thatthedownward surface current density at0is Youcanshowthisbysketching thepathsoftworaysthatenterthe Lasoo OOGRR sin dt” (@)NowshowthatTau,=—dL/dt,asabove. \Ce, tec ‘There isnotorque exerted bythecurrent onthemagnet forthefollowing ° reason.Theequivalent currents onthespherical surfaceofthemagnet *‘areazimuthal, andanyforceexertedonthemhasazeroazimuthal | ‘component. ||Fig.32-8, 1331 GENERAL PROPERTIES OF AN ELECTROMAGNETIC WAVE on transmitters and receivers, inradar sets, forexample. Metallic guides are lossy. Dielectric waveguides are nearly lossless and serve totransmit CHAPTER ‘Asweshallsee,thecoaxialandmicrostrip linescantransmit wavesofanyfrequency, fromzerotoabout10"hertz,whilehollowmetallic GUIDED WAVES I guidesarenarow-band devicesthatoperateatfrequencies oftheorderof 10”hertz andhigher. General Principles. TheCoaxial and Microstrip Lines , 33.1GENERAL PROPERTIES OFAN ' ELECTROMAGNETIC WAVE PROPAGATING IN ASTRAIGHT LINE 331 GENERAL PROPERTIES OFANELECTROMAGNETIC WAVE PROPAGATING INASTRAIGHTLINE61 Tosimplify,weassumethesixfollowingconditions. 3B.L1 THE TRANSVERSE COMPONENTS ARE FUNCTIONS OFTHE LONGITUDINAL COMPONENTS 612 (1)Themediumofpropagation ishomogencous, isotropic, linear,and 3.12TEANDTMWAVES 6 stationary (FILS). 331.3 TEM WAVE 33.1.3.1 VINTEM WAVES —616 (2)Itisnonconducting. Thisdoesnotexclude metallic guides, because33.1.3.2 AINTEM WAVES 617 thewave propagates along ametallic guide 3.14 BOUNDARY CONDITIONS ATTHESURFACE OFASTRAIGHT iy msMETALLIC WAVEGUIDE 618 vl (3)Thefreecharge density iszero.Thismakes ¥-E=0. 332 THECOAXIALLINE 619 (4) Propagation occurs inastraight line, inthepositive direction ofthe 333 THEMICROSTRIP LINE 621 z-axis. There isnoreflected wave traveling inthe~zdirection. 334 SUMMARY @22 PROBLEMS 24 (5)Thewave issinusoidal. (6) There iszero attenuation. Ifthe guide ismetallic, then its InChaps.28and29westudiedthepropagation ofelectromagnetic waves goncuctvity mustbeinfinitetoavoidJoulelosses.WeshallseeinSec.inanunbounded region.TheninChaps.30to32weinvestigated the 34.8howtocalculate attenuation withrealconductors. a vesattheinterfacebetweentwo retinandtherefraction ofplane waves attheinte Wemaytherefore writethat mediWenowstudyhowelectromagnetic wavescanbeguidedinprescribed E-=Eqexp(ot~kt)=(Emi+Ey5+Emad)exp(otk.2),(33-1)directions bywaveguides, firstmetallic guidesinChaps. 33and34,and: 7 y » 5thendielectric guidesinChaps. 35and36. H=H,expj(wt—k-2)=(Hyuk+Hny9+Hms®)expj(@t—kz),(33-2) Inthischapter wefirstinvestigate some general properties ofwaves .propagating inastraightline,butwithoutmakinganyassumption asto wherethecoeficients FsEnEmerHas«»af0unspecified functionsthewayinwhichtheficlddepends onthetransverse coordinates. Then ofxandy.Thedependence onzand¢appears onlyintheexponentialalyth |andmictostrip lines. function. Thewavenumber k,fortheguidedwaveisreal,sincethereiswhoreexistmontyiypesofwaveguiidesThemostcommonisthecoaxialjzeroattenuation. Itisequalto27/A,,whereA,isthewavelength ofthe- " .+ guided wave.line. Itserves tointerconnect electronic instruments and, inolder . ; .systems,isusedforlong-distance telephony. Hollowmetallicguidescan 7betussabato aboveexpressions forEandHintoMaxwell'soperateathighpower.Theyservemostlyforconnecting antennas to ‘a , = 612 GUIDED WAVES 1 23.1GENERAL PROPERTIES OFANELECTROMAGNETIC WAVE 613 Em,2Eny, Wehaveassumedthatk,#kforthemoment.Both&andk,arereal“xtytikzEmz=. (33:3) andpositive. Similarly, . iEm: ,OH: -B=0, and syes (hyee—coptae) . Similarly, V-B=0,an EwwEB(«Son ) (313) BH,3H,Ins,Hm_ig1,=0 (334) i [BEmeHa) ax*ay Hs=aE(wethea). (33-14) From the fact that VxE=98/31, 3E, oH,Hyy=pot(weSte+k,Ee) (33-15) aE.i RAK ox ay "+ ikEmy=—jOUHmes (33-5) ey |Weusethesubscript1toidentifycomponents thatareperpendicular ' tothe direction of tion. Th“ihebsEas=jot, 36) 0thedirectionofpropagation. ThusEm. =Emad +Emy$s Hs =Hy +Hyy9(33-16) BEmy_9Ems iST iene: on | Moresuccinctly, From Vx H=aD/at,. Ee=pega(keVEms+OUTXHoes (33-17) oH, Ale+IKHy=jOCEms (338) ay7!q Hn.Sree =WEVXEct). (33-18) Hine :: “IkHyg—He=jE (33-9) Soweneedtosolvethewaveequation andapplytheboundary conditions onlyforthetwolongitudinal components. Oncethatisdone, BHiny PHms—iyeE,, (33-10) theotherfourcomponents willfollowimmediately. oxey Thelongitudinal component ofEsatisfies thewave equation 27-70 with , =0,= 33.1.1 TheTransverse Components areFunctions ofthe P¢=0, Jy=0.So Longitudinal Components PEms PEm 12St A-KEne=—€UOEm:=—KEmes(33-19) Wecannowshowthatthefourtransverse components EneEmmy»Hm» a*ay Hyyarefunctions ofthelongitudinal components Ez,Hm.-From Eqs. or 33-6and33-8, 2Ene,&E,ont428sPVE: =0, (33-20) Ealia(kB+onHe), (33-11) ax?7ay? mR Oe ay 24tp (Vi+R=KEme=0. (33-21) Here 1 Similarly, k=olen)? => (33-12) > .x FHins, PHBetay+PKHm:=0, (33-22) isthewave number ofauniform plane wave ofwavelength 2traveling in themedium. (V5+k KDAms=0. (33-23) ois uIwED WAVES1 {0.1GENERALPROPERTIES OFANELECTROMAGNETIC WAVE ois Applying theproperboundary conditions (Sec.33.1.4)yieldsthevalueof 33.1.3 TEMWaves ke Ifk,=kinEqs.33-11 to33-15, theitems inparentheses must bezero. The simplest way ofsatisfying thiscondition istosetboth Ey and Hyg, 33.1.2 TEand TM Waves equal tozero. Wethen have aTEM wave. Itisconvenient toconsider separately threetypesofwave:(1)transverse WithTEMwavesthewavelength ,oftheguided waveisthesameaselectric (TE)waves, inwhichE,.=0; (2)transverse magnetic (TM) thatofauniform planewaveinthesamemedium ofpropagation becausewaves,withH.,,.=0;(3)transverse electric andmagnetic (TEM) waves, |r isequaltok,so WithEn:=0,Hyg=O. \ aah (3331) WitheitherTEorTMwaves,itfollowsfromEqs.33-11to33-15that Ifthemedium isair,thenthephasevelocity isc,whatever thegeometry Ems__Eny (3.24) oftheguideandwhatever thefrequency. SuchaguideisdistortionlessFy Hm because thevarious frequency components ofacomplex waveform all ' travel atthesame velocity.” Ifk,isrealandpositive, aswehaveassumed, theseratiosarealsoreal Setting En,=0,Hy:=0inEqs.33-6and33-9givesandpositive. Thenthecomponents ‘| be \e=(# =-(% E,=Emeexpj(ot—k.z) and H,=Hyyexpj(wt~k,2) (33-25) ty Em=(2)Hayemy=—(E)Hoe 3332) areinphase, andsoareE,and—H,.Thisfact,together withEq.33-24, ‘Thewaveimpedance isnow implies that E, (E+E) ”Ree fo=Getal, =(2) 333) ReE,__ReF, (33-26) In (Hine+Hay)?\€ ReH, Re(~H,) =377Tohms =(€,=1,u,=1). (33-34) andthattherealpartsofE,andH,aremutually orthogonal inbothTE Theratio(u/e)!” isthecharacteristic impedance ofthemedium (Sec. and TM waves. .28.5.2). Theratio Em,/Hm. isthewave impedance. Thisisarealpositive ‘Theelectric andmagnetic energy densities areequal: quantityifthereisnodissipation: fee (3335) Zp=Entttn(He (33-27) 22 : Hy. ke \€) hy .Im. Allso,theaveragePoyntingvectoris4, A, =3.76731x10°4£=377% ohms. (€,=1=1),33-28) bye dyhy a=1RE(EXH") =5(5)Ez (3336) aEmske(HY!ho (33-29) ” n tH, we(2)i, =(5)Elyd=2.65441x107Ge)Elggwatts/meter? (33-37) ~se ohms (€,=1,4,=1). (33-30) =VEERg=UHHead, (33-38) Here AandAyarethewavelengths u/fofaplane wave ofthesame "Thisionlyapproximately true.Theconductivity ofmetallic waveguides beingfinite, frequency f,andA,isthewavelength oftheguided wave. thereattenuation anddispersion 616 GUIDEDWAVES1 7 where NXve—tae tae (33-39) “at(ny? (en) x isthephase velocity.Ve: ) ‘Themagnitude ofthetime-averaged Poynting vector isequal totheQ ap energydensitymultiplied bythephasevelocity. s a} 3.1.3.1 VinTEM waves ‘Theuniform plane waves thatwestudied inChap. 28areTEM waves. |‘WenowstudyTEMwavesthatfollowaconducting guide. ! 5 InaTEMguided wavetheelectric fieldinaplaneperpendicular tothe}directionofpropagation isderivable fromapotential, inthesamewayas Fig.33-1.Hollowconducting waveguide.anelectrostatic field.Wecanshowthisasfollows. |Insideaconductor ofinfinite conductivity, E=0, forotherwise J|wouldbeinfinite.But7XE=—jwB,andthusBandHarealsozero |Thisisnotrigorously truebecauseTEMwavesareallowedifthe insidetheconductor. NowletCbeanarbitrary closedcurvesituatedina|wavelength ismuchlessthanthecross-sectional dimensions. For¢x-planeperpendicular tothez-axis.SinceHistransverseinthemediumof ample,lightgoesthroughastraightlengthofmetalpipe.Weshallseeinpropagation andzeroinsidetheconductor, thereiszerolongitudinal H Sec.34.4thattheTEMwaveisthenalimiting caseofaTEwave.andthemagneticfluxlinkingCiszero.Thus Intheshielded-pair andparallel-wire linesofFig.33-2and“inthecoaxial line ofFig. 33-4 the conductors need not allbeatthesame fe«dl=expj(wt—k,2){E,,-dl=0, (33-40) potential,V,,isafunctionofxandyaswhenthefieldisstatic,andE kcf need notbezero, soTEM waves arepossible. 33.132 final=0, (3341) 33.1.3.2AinTEMwavesApplying thegeneral expression forEthatwefoundinSees.17.7and andEisderivable from apotential:23-5 gives oA ev ov ov rc)-=—MingVn Ba-w BaNeNy(he Eq=—Wy=SESH: (33-42) Ox” By”esPaar)+(33-45) ov ovBo-w= -a-Z5, 33-4 artay” a) with V=V,,expj(wt~k,2). (33-44) la: Ifthewaveguide is@hollowconducting tube,asinFig.33-1,the \\iltangential component ofEatitssurfaceiszero,Vj»is@constant all \eaearound thetube,andtheonlypossible solution insideisV,,=constant, YS ar Now, ifVjisconstant throughout theinside oftheguide, E,,iszer0,E-=0, andsince0XE=—9B/61, thereisnoHwaveeither, Therefore Fig.33-2,(a)Shielded-pair line.Theoutercylinderisgrounded, andthereisa‘TEMwavescannottravelinsideahollowconducting tube. potentialdifferencehetweenthetwowires,(bh)Purallel-wire line. res 61s GUIDEDWAVES1 «9 where theexpression between parentheses isequal tozero, from Eq 33-43. Since Aisofthesame form asV(Eq. 33-44), k—jk.V2+joA=0 orA=oOVi. (33-46) ‘Thevector potential Aistherefore proportional tothescalar potential VinaTEMwave. Also,Aislongitudinal, which means thatthecurrents oN inaconducting guide forTEM waves arelongitudinal ‘ 33.1.4 Boundary Conditions attheSurface a ofaStraight Metallic Waveguide fc Byhypothesis, theguiding structure comprises straight conductors of“a infinite conductivity, parallel tothez-axis. Thecross section oftheguide Fig.333. Portion of j i aig. 33-3.Portion ofarectangular istherefore uniform. waveguide. With TEwaves VH,, is (1)Withanytypeofelectromagnetic wave, Evanishes inside aperfect tangent tothewall conductor. Then, because ofthecontinuity ofthetangential component ofEataninterface (Sec. 10.2.3), that component ofEiszerocloseto 33.2THECOAXIAL LINE theguide.Forguidesoffiniteconductivity, seeSec.34.8. |(2)Againforanytypeofwave,PX=0 insidetheconductor, Inthecoaxial lineillustrated inFig.33-4,thewavepropagates inthejoB=0andB=0,H'=0.Then,because ofthecontinuity ofthenormal annular spacebetween theconductors, andthereiszerofieldoutsidecomponent ofBataninterface (Sec.20.8),thenormal component ofB, Themedium ofpropagation isusually alow-loss dielectric.closetoaperfectly conducting guide,iszero. ‘Thistypeofguidenormally carriesTEMwaves. Various TEandTM‘Oncethetangential Hisknown,thesurfacecurrentdensityafollows modesarealsoallowed,butonlyatwavelengths thatareofthesame from therelation H=@X~n (Prob. 19-4), where aiistheunit normal vector pointing away fromtheguide. Thesurface current density canbe *}different fromzerowithE=0ando—=. |Inother words, closetotheguide, Histangent tothesurface, |gtorthogonaltoa,[email protected],withTMand a lsTEMwaves,Hiseverywhere transverse andthecurrents intheguideare Ll G longitudinal = (3)Finally,forTEwaves,evenwithimperfectly conducting guides, i‘Bsa,’NH -:«-ito(BH2Hm: en) Ems=Emct+Emyd=3(St 33-47) Sie \i a End+Emi=p(Gy far3)GH \\aNGeie aa | pat) a adagiee =OPH) x2 (33-48) K NoeaR Mma)XE :|bs With aperfectly conducting guide, E,,. isnormal tothesurface and Hp: istangent, asinFig.33-3. Thus, therateofchange ofH.inthe > direction normal tothesurface, atthesurface, iszero, Fig.33.4. TheE,H,andEXHvectors inside acoaxial line oy OUIDED Waves1 393THEMICROSTRIP LINE oat order ofmagnitude as,orsmaller than, thediameter oftheline enc(.=10"2meter, f=10"hertz).Thenthevariousmodestravelat 1=2a0(5) ppt kez) (33.55) different velocities, andthelinedistorts complex waveforms. Inpractice, a ‘oneoperates atfrequencies wellbelow thethreshold fortheTEandTM we expi(wt—k _modes,andcoaxiallinesarethennearlydistortionless. goPiokez)(y=1), (63-56) ThusEn=0, My=0, (63-49) ‘Anequalcurrent flowsintheopposite direction alongtheinnersurfaceofthe outer conductor. Ifthe medium ofpropagation isair, then and,fromEq.33-42, C=60,V5..Ve ‘ThePoyntingvectorEXHpointsinthedirectionofpropagation, asinp=-(S8+ x§)exp(or~k.2). (33-50) Fig,33-4,andtheaveragetransmitted poweris Foragiven¢andagivenz,Evarieswithxandyexactlyaswhenthe n-[r=} Syw2n » : field isstatic. Itisradial andvaries as1/p(Prob. 3-8): |,Pu2ae do, (33-57) cwhere £=Cexpj(ot~k:2)d,(3351) arep a=RE(EXH")=(T) apt (33-58)Cisaconstantand . whereCisaconstanta :Thus 1 ot 2 k=p=(eu)?=F, (33.52)| weKini i i PrmerS5inb2 watts =1). (33-59) asforauniform planewaveinthemedium ofpropagation, Wehaveset Also, 41,=1sincemagneticmaterials arelossy PredRe(VI*) =AVsh=Vl‘Thevelocity ofpropagation isequal to«/k. r=Re(V1)=Vuln=Vonaleme (33-60) The potential oftheinner conductor with respect totheouter Thecharacteristic impedance ofacoaxial lineistheratio V/Iwhen conductor is there isnoreflected wave traveling inthe~2direction: vfEdp=Cinexpj(wt-k.z), sy fl aU Dy2 , ps pranine ohms(y=1). (3361) where p,andpzaretheinner andouter radii oftheannular region between theconductors. This isthelinevoltage 33.3 THE MICROSTRIP LINE From theprevious section, Hisorthogonal toEand. neo. . Figure33-5showsacrosssectionofamicrostrip line.Itcomprises @ H=()Gewpilor— kez)(i=1)(33-54) groundedconducting plane,aninsulatingsheet,andaconducting strip.The strip isgold orcopper, plated onto theinsulator, which iseither The vectorsEandHareasinFig.33-4. aluminaorfusedquartz Thelinecurrent flowing alongthesurface oftheinnerconductor is Microstrip linescarryTEMwaves.’ Theyareparticularly usefulinrelated toHthrough thecircuital lawofSec.20.6: printed andintegrated circuits thatoperate atfrequencies ofaboutoneto 1=Igexpf(t~k.2)=25P\H yap, "Thissanapproximation; EandHalo havelongitudinal components m134 SUMMARY oo s Wig.355,Crosesecton HereA,isthewavelengthoftheguidedwave,and2isthewavelengthof> oesapline aplanewaveofthesamefrequency inthemediumofpropagation|G,ground-plane layers ForTEMwavesthewavelength 4,oftheguidedwaveisthesameas 3 Biles sara thatofauniformplanewaveinthesamemediumofpropagation,Aah, (33-31) tensofgigahertz. Theyhavetheadvantage ofbeingmuchlesscostlythan andthewaveimpedance is either thecoaxial line ortherectangular waveguide. Their maindisadvantage liesinthefactthattheirfieldisnotstrictlylimitedtothe Em(#)'737 ohms (ie= 3regionimmediately belowtheupperelectrode, Microstrip linesare H,\e) s(Hr=1)(33-33) therefore lossyandcaninteract withother elements inacircuit, unlesstheyareeitherspacedorshieldedproperly. Lossesarehigh,oftheorder Theelectricandmagnetic energiesareequal,andthemagnitude ofthe of05decibel millimeter.Poynting vector isequal tothetotal energy density multiplied bythei iplryphase velocity. 33.4 SUMMARYInaTEMwave,Eisderivable fromapotential: E=—VV.Also,the currents flowing inaconducting guide arelongitudinal. Inanyelectromagnetic wavepropagating inastraightline,thetransverse Theboundary conditions atthesurfaceofaperfectly conducting guidecomponents ofEandHaresimplefunctions ofthelongitudinal areasfollows: Evsnscnist =)Brormat=0sHrangensiat =@Xf,VHmsis‘components. Usingthesubscript 1forcomponents thatareperpendicu- tangentiallartothedirection ofpropagation, Figure33-3showsaportionofacoaxialline.Inthefield, c Eg=pela(heWEms+OUPXHyeé), (3317) : B=expi(ot—k.2) be 3-51) Hg=ygReHe~EVXEm8)(33-18) ! u=(*)postkez)o(33-54) ‘Thelongitudinal components themselves satisfythewaveequations: whereCisaconstant thatfixestheamplitude ofthefield.The voltage, current, and power flow are (5+= K3)Ems =05 (33-21) . >v=cine - 7 (+B K)Hye=0 (33-23) mp,PHC—Kez), (33-53) ‘Thereexistthreetypesofwave:TE(Ey:=0),TM(Hm:=0),and 1=2ac(“) *expi(wt—k,z), (33-55) TEM (Em: =0,Hm: =0)waves.Ho! InTEandTMwaves, therealparts ofE,andofH,aremutually Pp=Voousbons: 33-60) orthogonal. : ‘The wave impedances areThe characteristic impedance ofacoaxial line is E, Mo"?A,_377A, 7) pr a Ems_(Mo)AsS77As = . Zap? hr 7 ce eldaceCO Sant ohms 6361 _(0)ALTA Ger (3329) Thevelocityofpropagation ofaTEMwaveinacoaxiallineisthesame e)2,eh, a asthatofauniform planewaveinthemedium ofpropagation. 2a VIDED WAVES 1 os PROBLEMS 33-1, (332) Thefieldinside acoaxial linea)”Sketch arather largecross-sectional viewofacoaxial lineinplane (acontaining theaxis.ShowlinesofEandofHatagiveninstantoveratleast \f \Sewavelength,Thelinesshouldbemostcloselyspacedwherethefieldis wa) 10) \strongest.Indicatethedirectionsofthefieldsbymeansofarrowheads.The \Se jA q direction ofpropagation should point totheright ; {b)Addarrows, atvarious points forepresent Poynting vectors, usingD longerarrowswherethepowerflowislarger.Assume thtthelengthofthe :arrowrepresents themagnitude ofthePoynting vectoratitsmidpoint ® o i(@)Sketch across-sectional viewofthecoaxial lineinaplane perpendicular totheaxis,andshow linesofEandofHataparticular instantig. 33-6. Relate thisplane tothefigure youdrew in(a)! <a)Addplusandminussignstobothfigurestoshowthesurfacecharges. ‘Showthat,ifthematerial hasaconductivity 0andathickness s,andif‘Thespacingbetween thesignsshouldindicate qualitatively therelative thepermittivity ofthedielectric is¢,thenR,C’=€/s0.“™ magnitude ofthesurfacechargedensity (©)Onecanmeasure LasinFig.33-6(c)bymeasuring theresistance R%€)Nowaddatrowsofvariouslengthstoyourfirstfiguretorepresent between electrodes CandD.2 surface current densities.‘Show that RiL’ =su/s0. Thus Z,=(w/e)!(R/R.)" (f),Howdothecurrentpatternschangewithtime’ 536.(633)Themicrostrip line4332.(33.2)Thecurrentandthechargedensityin coaxilline | Figure33.5showsacrosssectionofamicrostriplineShowthat,inacoaxiallinofinfiniteconductivity, thecurrentisequalto (a)SketchlinesofFandofH.Usearrowstoshowthedirections ofFthelinearchargedensitymultipliedbythespeedofpropagation sndifasBentine,Showthedectonofpropagationin practice. the wi Ttisknownfromtransmission-line theorythatthecharacteristic im- instantancous value’/Yoftheaneeited powerfequaltthePoyntingpedanceofalineisgivenbyZ=(L'/C)", whereL’andC’are, wectorintegratedoverthecrosssectionbh,AssumethattheremeFespectively, theinductance andcapacitance permeter.Showthatthis reflected wave,ssume that there isno appliestothecoaxialline (e)Showthatthe characteristic impedance 11'sequa0(ule)"Ab ‘Anair-insulated coaxiallineisterminated byasheetwhosesurface (c)Showthattheadditionof»s3 eae pet uae (10 9) caf,SHO thattheationofasecond grounded plane paced symmetr ‘Showthattheresistanceofthetermination isequaltothecharacteristic iy istreducesthecharacteristic impedancebyafactorof2.impedanceoftheline.Thereisthennoreflectionattheendoftheline. 33-7.(333)ieIicrostripline.aleulate thetransmitted power w voltage across , 38.5,(33.1.3) Predicting thecharacteristic impedance ofaTEMguide lieofFigs35.616¥"Diaregardagecerts,andecuWtthereee ‘Oneimportant parameter ofaTEMwaveguide isitscharacteristic reflected wave.+andassume thatthereisn0 impedance Z=(L'/C)", where L!andC’are,respectively, theinduc:tance andthecapacitance permeter ofguide (Prob. 33-3) Tedesigning suchlinesitisimportant topredict thevalues ofL'andofCifthegeometry issuchthatthesequantities aredifficult tocalculate, as ihFig,33-6(a), onecanperform thefollowing measurements ona resistance-sheet analog. a)Onecanfindthevalue ofC”bycutting outasheetofresistivematerial intheshapeofthecrosssection ofthedielectric asinFig,33-6(b) lindbymeasuring theresistance R,between electrodes AandB.SeeProb. 9:10 a 7 Table34-1Characteristics ofafewstandard rectangular waveguides (TE,mode) eee eee CHAPTER 34 INSIDE OPERATINGDIMENSIONS |CUTOFF WAVELENGTH rowera@xb WAVELENGTH RANGE ATTENUATION RATINGEL NEENGTH, RANGEATTENUATION RATING GUIDED WAVES II nidineters “B/meter ‘mezawatts . 721x340 aa TheHollow Rectangular Waveguide a5x20 a Bos 0207-0. 2MRx 15.8 07 366-512 0.0755-0.0982 0.6422.910.2 457 242-366 ©0.147-0.212 0.25 j1587.9 316 16.7-24.2 0273-0312 3M.THEFIELDCOMPONENTS OFATEWAVEINARECTANGULAR Bare 62 omosMETALLIC WAVEGUIDE 028 ; M2 THE CUTOFF WAVELENGTH. NONPROPAGATING FIELDS 630 !343THETE,MODE 631 |_howsuchaguidecanbeconnected toacoaxialline.Table34-1listssomeSEEMULTIPLEREFLECTIONS 638 |common standard sizes, butthere aremany more, down toawidth ofa M.STHEPHASE, SIGNAL, ANDGROUP VELOCITIES 635 |fewmillimetersM6THETRANSMITTED POWER 636 Weassumethatthemediumofpropagation isair,asisusuallythe947THESTORED ENERGYDENSITY 637 |case.Wealsoassume,forthemoment,thattheguideisperfectM8ATTENUATION 637 ,conducting. y M9SUMMARY 681 |Weconsider onlytheTransverse Electric (TE)mode(Ey,=0)thatPROBLEMS 642 |_Fesultsfromthemultiplereflectionofaplanewaveonthefacesparallel totheyz-plane inFig. 34-2, Ahollow metallic waveguide issimply ametallic pipe inside which an electromagnetic wave can propagate byreflection ontheinner surfaces, inmuch thesame way asasound wave. There cxist_many types ofhollow metallic waveguides, butwe 4 raconcentrate onguidesofrectangular crosssection, asinFig.34-1.Thisis | L mae themostcommon type,anditisalsothesimplest one.Thefigureshows a Ja ow 7RS 7 A |IPS. SSeS fawn 7 ]| ae4 Sa |Fil SS.a F|a e @ ® i} NI -34-2.Typicalwavefrontinsidea Fig.34:1.Rectangular metallicwaveguides fedbyacoaxialline.(a)A anaes Teelgvfrootinside quarter-waveantenna injectsanelectricfieldatadistance of2./4fromthecloed . Oocurt’on thenattowfaresneueis endoftheguide.(b)Asmallloopantennainjectsamagneticfield. : cvrywhere tnaswene. ne os OUIDEDwaves1 M4THEFIELDCOMPONENTS OFATEWAVE 629 The Transverse Magnetic (TM) waves inhollow rectangular wave- Thus guidesarenotusefulbecausetheyrequirelargerguides.Aswesawin ~Moos(kx+ 4-8)"Sec.33.1.3.1, TEMmodesareforbidden. Hs=Mcos(kx +),ky=+(ko~ki)", (34-5) whereMisanarbitrary constant thatdefinestheamplitude ofthewave. 34.1 THE FIELD COMPONENTS OF ATE WAVE IN Wenowapply theboundary conditions ofSec,33.1.4: ARECTANGULAR METALLIC WAVEGUIDE ou,Sea0 atx=0,a, 34-6) Weusethecoordinate system ofFig.34-2. Thewave propagates inthe ox positive direction ofthez-axis bymultiple reflection ontheupper and Hye lower walls. Thefigure alsoshows awavefrontofaplane waveincident By70 aty=0d. G47) onthetopface attheangle . ° With thismode ofpropagation, ‘Thesecond condition isalready satisfied because 3/3y =0.From thefirst 3 condition, Eme=0,Eme=0, Hy=O,5=0. G41) k,sina=0 andik,sin(k,a+a)=0. (34-8) Werequire thethree other components Ey, Hes Hon: Now k,isapositive number. Therefore Weproceed asindicated attheendofSec.33.1.1. Firstwesolvethe =0 and =na, 34.9)waveequation forH,.forthegivenboundary conditions. Thiswillgive; “ and kewns, Ge) usboth H,.andk,.Then thevalues ofE,,,andH,,,willfollow, from where 1isaninteger. Eqs.33-13and33-14. i Observe thatk,acantakeononlydiscrete oreigenvalues andthatFrom Eq.33-22, n= isforbidden: FH,. | n=1,2,3, (34-10)Sis—(42—12)Hpe, 34-2 ar sie ated So where i na 126 2a@ Hye=Mcos (411)kyepape ee (343)odoce and,fromEq.34-5, isknown, foragivenfrequency. However, nx\2)?_ (1~[ndg/2a)R}"? bap Gey fyg Then, from Eqs. 33-13 and33-14, remembering that both E,,,and 3/3y isunknown, 2,being thewavelength oftheguided wave. i areZer0, Weexpect aninterference pattern ofsome sortinthexdirection. So « Timo BHme_foto Hyaisasinusoidal function ofx,andthisrequires thattheexpression in| Ey=e ope OAM sinket) O41)parentheses inEq.34-2benegative. Soweknowthatk.<koandhence | thatA,>A,orthatthewavelength measured alongtheguideislonger =10Mgsinkx=—22H4sin (4-14)thanthewavelength ofaplane wave inair,Thismakes thephase velocity ke an a larger thanc,which iscorrect; thegroup velocity willturnouttobe ja nme smaller thanc. Hage=e Msin (415) oo MaTHE TE,MODE os ty ee Thewavelength 2a/n isthecutoff wavelength fortheTE,mode. This corresponds tothecondition «=«w,forpropagation inanionized gas. Atthat wavelength k,=and 2,2. Atwavelengths larger than2a/n, k,isimaginary, there isnowave, and ° 7 ol thefielddecreases exponentially withz.There iszeropower flowonce thefield isestablished. Atthese longer wavelengths thefieldamplitude decreases rapidly with z.Forexample, attwice thecutoff wavelength, where thefrequency istoolow byafactor of2, w_nn oF207}?_ey, Fig.343.TheamplitudeF,=FayoftheelectricfieldstrengthforaTEwave 2-3, a=[(2)-(F2)]"=-723% oaay (a)TE,mode.(b)TE:mode " ©2a (2)(<) ¢ Wechoose thenegative signbefore thesquare rootsothattheamplitude Figure 34-3(a) shows E,,,asafunction ofxforn=1:Eiszeroalong the willdecrease exponentially withz,andthen walls and maximum intheplane x=a/2. With n=2,Ey,iszeroat ans" : x=a/2exp(=jkez)=exp(~=)=exp(10.88=). (3422) The various values ofnthus correspond todifferent modes of do de propagation, denoted asTE),TE,etc.Asweshallseebelow,TE,isthe ‘Theamplitude decreases byafactorof5%10* inonefree-spaceonlyusefulmode, wavelength 2!Summarizing, ‘Thewaveguide thusactsasahigh-pass filter,withthelowerfrequency Emc=0, Ey,=2H sin™, 20, 34-16) limitfixedbythewidtha,andnotbyb. ;nt a Thefree-space wavelength 4pmust beshorter than twice thedistance jka nm k hax ;between thereflecting walls. Forexample, if@=100millimeters, theng Hg,eM sin = Egy, Hyy=O,Hye=Mcos", mustbelessthan200millimeters andthefrequency mustbehigher thanun aOho “ 1.5gigahertz (1.5x10°hertz). (34-17) k=LPaleot ye (34-18) 34.3THETE,MODE .‘ Inpractice, oneselects firsttheoperating frequency, andthenaguide 34.2THECUTOFF WAVELENGTH. whosedimensions aresuchthatitcancarryonlythen=1mode.ThisNONPROPAGATING FIELDS condition requires that2abelargerthanAp,asabove.Butamustbeless thanAytomake TE:,TEs,... forbidden modes. Thusthedimension a FromEq.3+18, mustbesuchthat fee(2)?_(ey?_(yey? a<ho<2a (34-23) a=[@-(Z)]"-[()-()] ous) For With single-mode propagation thefield configuration iswell defined.nae a Rectangular metallicwaveguides arenarrowbanddevices:foragivena, -@> or <tr, (3420) |Aycanvarybyatmostafactor of2. ‘Theantennas ofFig.34-1launch anassortment ofmodes, butonlythe &,isrealandawave canpropagate unattenuated down theguide. ‘TE,survives. 62 UIDED waves 1 M44MULTIPLE REFLECTIONS 633 Wenowwrite outthefieldcomponents forthen=1 mode. We 34.4 MULTIPLE REFLECTIONS simplify thenotation bysetting u ‘Themultiplereflections thattakeplacebetweenthetopandbottomwalls EL,=—nea | (34-24) areinteresting. Weneed onlyconsider theEfield. Wesubstitute k,for 7 t/a,inagreementwithEq.34-9. ThenFirst recall that E,=0, E,=Elysin™expj(or—k.z), E,=0, (34-25) expikex—exp(—jk,x)120,Ey=Eqsinexpj(at—k,z), E,=0,(342 sinkx=SPR (3430) t=—Ek:in™expj(wt-k,z), Hy=0, (34-26) ‘Thenwecanrewrite theexpression forE,asfollows Hy a gEF! ay p,=Em +ka—k,2)— kx 34 H=thsgos™exp|or—k,(2-*)]. (3427) E,2[exp(ot+kx—k.2)—exp(or—k,x—k,z)]-(3431) &omgaa 4 where Thefirsttermbetweenthebracketsrepresentsawavewhosevectorwave > o/(2ayey! ay?" numberis~k,&+k.(App.C).ItisorientedasinFig.34-5(a).The k,2=olay) [1-(a)]ky,(3428) secondtermcorresponds tothewaveinFig.34-5(b).Sothesinetermina do i Eq.34-25forE,doescomefromtheinterference between anup-goingdo 34.29 andadown-goingwave. =alQahy? (3429) Similarly, theH_vectoristhesumoftwotermsoftheform AXE/(op), corresponding totheup-going anddown-going waves. Figure 34-4shows lines ofEandofHforthisTE,mode Mathematically, there isasingle up-going wave andasingle down- goingwave UB ‘Theangleofincidence@isgivenby Ay} k dy)! oa| sino=F=[1-(32)|:(3432) fya i}“4 do Vj.ERA| cos@=<2 (34-33) Ca)IV\ z 433)(CA |. r :ZF, Ee, [ee E a i4» »MA.LinesofE(dotsan EctandofHf(ovals)fortheTE, Fig.348.WavefrontsandtheK'sforthe(a)down-going and(b)up-going. mode waves.Observethathy=ki+A3,asinEq,MS.Theangleofincidenceis0 636 ourpenWaves1 aesAPTENUATION o7 This isinagreement with thefactthatasignal cannot propagate ata —pezab/€0) g\?)" speed larger than¢(Sec.13.5.3). Pray=Ey(2)[}-(8)|. oH Thegroup velocity v,isequal tothesignal velocity v,.SeeProb. 34-8. 34.7 THE STORED ENERGY DENSITY 34.6 THE TRANSMITTED POWER ‘The instantaneous electric energy density is}€,|Re E|’, and itstime- Tocalculate theaverage power transmitted through theguide we averaged value is integrate thetime-averaged Poynting vector overthecrosssection ofthe « aguide: w=PeR=SESsin, (34-48) Pra[fe(adedy=ifIRe(EXH")\drdy. (4-41) Thetime-averaged electricenergystoredperunitlengthofguideisthus Here fope[ga(™ = aneoe e=$e3,[sin(2)oae=Sane, G449) ey2 wasno£0|-isereE,Hz2),(34-42) ‘Similarly,theinstantaneousmagneticenergydensityis}uo|ReHl, H: 0H? with atime-averaged value of where thecomponents ofEandofHareasinEqs.34-25to34-27.After Eng=Mo2,=88(He?+He?) (34-50) substituting 2fork.A., theexpression forH,becomes moan gm g Wim ime 7 yt Egks)? 5|(Ey)? 5x AIAmy0g _ =te(Fat 2(Emi22). 34-51 1,=Acosexpj(ot—h:2) oes) fy f(a) sin+(Zee)co]. etsy ‘Thusthexcomponent ofthetime-averaged Poyntingvectoriszero.The Theelectricenergydensityishighestnearthecenter,whilethenetpower flows inthedirection ofthez-axis and ; magnetic energy density ishighest nearthesides. Upon integrating over thecrosssectionoftheguideasabove,onefindsthat,perunitlengthof Eke5% guide. Gu=Foesin8 Cn © : <one Einag=C=aDEny (34-52) Weassume that E',,isreal. ‘Theaverage power density S,isindependent ofy,asexpected, since } Asonemight expect, thetransmitted powerisequaltotheenergy perboth#andHfareindependent ofy.Itiszeroatx=0,x=awhere Bis} saittengthmultiplied bythesignalvelocity 1‘ inezeroandmaximum atx=a/2. | ‘Thetime-averaged transmitted power isthus i Pray=(61+ Grange (34-53) Ek, ax | me 2 (34-45) PraFae[sinEde GH) 1)34.8ATTENUATION _Emk.ab_Enykoab[-(e)} (34-46) Wehaveassumeduntilnowthatthewaveguidewallsareperfectly sony doo 2a conducting; itistimetoconsider realwaveguides offiniteconductivity. 638 GUIDED WAVES IT 34.8ATTENUATION 639 Intheprocess ofguiding electromagnetic waves, conductors dissipate inadistance Az.Theapproximation isexcellent forstandard guides. partofthewave energy intheformofJoule losses, because thewave Then induces electric currents intheguide. Arigorous calculation ofthefield 4_ PLforaguideoffiniteconductivity isdifficult,butfortunately unnecessary. PLAz=(QaAz)Pr, a=>5 (457) ‘The procedure forcalculating theJoule losses isthefollowing. Asa firstapproximation, weconsider theguide conductivity otobeinfinite. TherealpartBofk,isapproximately equal tothek,obtained onthe ‘Then thefieldistheonethatwefound above. Asasecond approxima- assumption ofperfectly conducting walls. You willrecall that we tion,wemake @large butnotinfinite. Arefracted wave penetrates the calculated Py,,,inSec,34.6, conductor, andatinypartoftheelectromagnetic energy islostasheat. Wecancalculate Pi,asfollows. Along thefacex=0, andatz=0, ‘Asaresult, theamplitude oftheguided wave decreases withincreas- fromEq.34-27, ingz. Eby‘Theattenuation isexponential forthefollowing reason. Theamplitude BeHe sg|oPJon. (34-58) ofthewave refracted into theguide isproportional totheamplitude of theincident wave, ortotheamplitude oftheguided wave. Then the Thisisindependent ofy.Now,fromProb.29-5,thepowerdissipated per power lostintheguide isproportional tothepower inthewave, which is square meter isthesame asifonehadacurrent /ofthesame magnitudethecondition requiredforanexponential attenuation. Seealsobelow. asH,spreaduniformly overacrosssection1x4,where6istheskinInthissecond approximation, witholarge butnotinfinite, EandH depth. Sothepower dissipation persquare meter isI?m,X1/(0X1X6),arethesameaspreviously, except thatk,isnowcomplex: or(H;,/2)/(06). Then, forthefacex=0andatz=0, -12 +\3/upow\!2 (aE Ja)? k,=B-ja. (34-54),Fux}(222)ae(z=)(“g2)-(Ela (34-59) Letuscalculate a,Thevector Histangential atthesurface ofthe | Hema! 0820\owa OMCua) guide.Sincetheguidematerialhasafinite0,theskindepthisfinite,|ThepowerlostoveralengthofImeterinonefaceistimeslarger,there arenotrue surface currents, and thetangential Hfjust inside the and forthetwofaces atx=Oandx=a, itis guide isthesame asthetangential Hjustoutside. Awave penetrates into theconductor and,fromSec.29.1, Phac=En/ay2b(facesx=Oandx=a). (3460) E [uot on2Qugi) anho)PI 455) |Forthefacesaty=0andy=b,Hhastwocomponents. Thepowerlost ‘This small Eexists onboth sides ofthe interface. We assume that the.Over Tmeter ofBuide inthese twofaces proves tobe guide isnonmagnetic. This Eisaperturbation oftheideal field thatwe , (aE), y/2a\?derivedforperfectconductors. Themethodissatisfactory becausethisj Poav=(seal) (facesy=Oandb). (34-61) ‘small Ehardly disturbs thewave. Wethushaveatangential F,atangential H,andaPoynting vector |~~Summing thesetwopowers, thatisnormal totheconducting surface anddirected intothemetal. aWecanthuscalculatetheaveragepowerPi,thatleaksintothe Pin[pe |2+(2) (6402) conductor permeteroflength.Wethenrequiretheattenuation constant | o'Quow) alla©\Ao/J° a.This constant must besuch that multiplying both theEandtheHof 7theguidedwavethatwefoundabovebyexp(—az) decreases the Finally,usingEqs.34-47and34.57, averagetransmitted powerPy...byafactorof wel[xtclua® r1+Q2b/a)[ao/(2a)P aexp(~2aAz)~1-2aAz (3456) boy) =[aseaey ore) 640 49SUMMARY oat “ connecting atransmitter orreceiver toanantenna, either forradar offor b microwave links. Table34-1givesthemaincharacteristics ofafewstandardtypesof ‘ol rectangular waveguide. 24 x 34.9 SUMMARY 56 a Atransverse electric (TE)wavecanpropagate inside arectangular waveguide byreflection from twoopposite sides, asinFig.34-2. Then y ° Eme=0, —Emy=~/2HOugsin™™, =0,(34-16) 0 or 08 a6 oF v0 nt a eo% ik.a nxXe HyeEM sin, Hy=0, Hye=Mcos™, (34-17)nx @ @ Fig.349.Theproduct aa”asafunction oftheratio4s/2afor6=a/2andfor 1=[hol2a)"capper. k=Hebelenry" (4-18) 2 Here nisthemode order, and, asusual, thesubscript mstands foraw 11+ Cb/a)lAolCad (34-64) “maximumvalueof.” b(1200%0)'"{1—[Ao/(2a)}} Inpractice, oneusestheTE,mode, andthen Inpractice, 2b/a~ 1.This makes themagnitudes ofthepower densities om aboutequalonallfoursides.Thisalsoexcludes aTE,modewith \ E,=0, Ey=Enysin expj(wt—k.2), E,=0, (34-25)reflection ontheotherpairofsides. H rokFigure 34-9shows aaasafunction of4o/(2a) forcopper andfor 1,=—E2k:gn™expotk),Hy=0,3426) b=a/2.IntherangeAg=0.2x2ato0.7x2a, { Oo a Ey mx A 4x10-5 H,=7EmZexpl ~k(z—2 ' one 4.65) cos expjlor~k.(z—%)], 427) 2a_{1[o/(2a)P)? Theoptimumvalueof4pisabout0.42a,butactualvaluesare ayaleeSee (34-28)somewhat larger soastoachieve strong attenuation forthe1=2mode: * Theattenuation ofthen=1mode incommercial guides isoftheorder} where A.isthewavelength oftheguided wave. of0.1decibel/meter' atfrequencies ofafewgigahertz, Withthismode theangle ofincidence (Fig.34-2) isgiven by ‘Thisisahighrateofattenuation. Forexample, over100meters the conattenuationisabout10decibels,whichmeansthattheamplitudeofthe sin=[i_(Call: (34-32) wave decreases byafactor of3.Such guides transmit high-power signals 2a, ve a fe overdistances oftheorderoftensofmeters orless,usually for “Thephaseandgroupvelocities are a ft c"2010(amplitudeattheourpa)/(amplitueattheinput]=0.1|Mape Me=U=CsINO<c,(34-39),(34-40) on GUIDEDWAVES11 PROBLEMS 683 . vector is 72.1millimeters extends from 2.61to3.95gigahertz. ‘Thetime-averaged Poynting Calculate thevaluesof@atbothendsofthisrange. Enks 25 x 9=Ettkssin? (34-44) 34-4,(34.3)Measuring thestandingwaveratioinarectangular guideemo a Ifaloadisnotproperly matched toawaveguide, partoftheincident : wave turns back andthere isastanding wave along theguide. Then only a andthetime-averaged transmitted power is fraction ofthepower available atthesource reaches theload. Itis vy ayia therefore useful tobeabletomove asmall probe along alongitudinal slotPra=E22(2)[1.(2)i}.(34-47) tosamplethefieldinsidetheguide.Thevoltagestandingwaveratio raeomeNy 2a (VSWR)istheratioofthemaximum totheminimum time-averaged rmsvoltage measured attheprobe. Under ideal conditions there isnoreflected where (jto/€,)"2= 377. wave, andtheVSWR isequal tounity. Inrealwaveguides offinite conductivity 0,thefieldishardly different, ‘Theprobe canbeashortlength ofwirethatresponds totheelectric field ‘orasmallloopcoupled tothemagnetic field.Theprobeprojects intothe exceptthat fieldbyabout1millimeter.k,=B-ja (34-54) (a)WiththeTE,mode,whereshouldtheslotbecuttodisturbthewave ;aslitteaspossible? withBequaltotheabove k,and (b)Iftheprobeisaloop,howshould itbeoriented? 114 2b/a)[o/2a)P (34-64) (6)Howwouldyouproceedtomeasurethewavelength oftheguided ©52008) (T=ola)PY? wave?(@)IftheVSWR isequal to2,what isthe value oftheratio ExereaealEweston? PROBLEMS,34-5. (34.3) An artificial dielectric Figure 34-11 shows asetofparallel conducting plates uniformly spaced34-1.(343)Wavelength andfrequency inarectangular waveguide byadistances.Ifs/Ahasthecorrectvalueandiftheincidentwaveis‘Anelectromagnetic wave propagating intheTE,mode ina34.0x correctly polarized, thismedium actsasanartificial dielectric whose index 72.1millimeter rectangular waveguide has awavelength 2,of | ofrefraction isfessthan unity 138millimeters. Calculateitsfrequency, 8netee‘heorrentation otrfuyth Mh ind the indexofrefraction asafunctionoftheratios/Ko 342,(343)ThesurtacecentsandthePoynting vectorinarectangular {©)Showraysdefected by()'speismcod(a)«sonectene cylindricalctallic wave guide eamepigure 34-10showsthreesidesofarectangular guidethatissplitopen tensmadeinthisway. andflattened, Draw afigure likethisandshow, onface B,lines ofH, eleccharges, andvectorsBAYatagiveninstant,Thenadinesof7)currentonallthreefaces | Ay 34-3. (34.3) Theangle ofincidence @inarectangular guide ‘According toTable 4-1, therecommended range ofoperating fre- quenciesforarectangularwaveguidewithacrosssectionof34.0|A ’ a a ES‘ aw ‘wy Fig.4-10. Irenene Fig.Me. ons UIDeDWavesiL is 346. (34.3) Dielectric-filled rectangular waveguides ‘Arectangular metallic waveguide Aisair-filled, anditscross section is any a,b.Another rectangular metallic waveguide Disfilled withadielectric ¢,, anditscross section isa/e!”, b/€!". (a)Show thatwaveguides AandDhave thesame cutoff frequeney and that, atagiven frequency, A.p=A.a/€!. Thus, foragiven operating ——— O_Ofrequency, adielectricilled guideissmaller thananair-filled one.With causes (ySeeders Teflon (€,=2.1), bothdimensions aandbatesmaller byafactor of1.45, ‘Also, thephase andsignal velocities aresmaller byafactor of1.45 (b)Compare thepower ratings ofAandDatagiven frequency. ‘The dielectric strength ofamateriaisthemaximumpermissiblevalueof iin nat Ebefore breakdown. Thedielectric strength ofagooddielectric suchas See Fig.412, ‘Teflon isoftheorder of10times thatofair.Setthisratio equal toR.The valueofRincreasesasthethicknessdecreases. approximately 1~(2¢,w/o)"*c0s @,ifEisnormaltotheplaneof34-7,(34.5) Thephaseandgroupvelocities inarectangular waveguide incidence.(a)FindAy/A,asafunction ofAo/A,forarectangular waveguide, where (b)Showthatthisisinagreement withtheattenuation calculated in2,isthecutoffwavelength Sec.34.8, (b)Findo%./easafunction ofk,X,.Foragivenfrequency, M4,(48)“Therelativepowerratingsofacoaxiallineandofrectangularyo _oh guideera (a)Calculate themaximum powerthatcanbecarriedbyacoaxialline* andbyaestangular metalic waveguide at2.00gigahertz, Thecoaiagroup,orvelocityisgiven linehasadiameterp;of25.0millimeters, andtheguidehasaninside ‘Thegroup osignal, velocity sven by cxoss section of375%75,0millimeters. The.coaxialinesatstheue_doh.fe) condition formaximum powertransfer, namely, p;/p;= 1.65,anditseaeoy outside radiusissmallenough toensuretheattenuation ofhigher-ordermodes. Both lines areairfilled, andthecurrent-carrying surfaces are 34-8. (44.5) Thegroup orsignal velocity inarectangular waveguide silver-plated. Themaximum allowed Eis1.5megavolts/meter. (Under Showthat,inahollow rectangular waveguide, i ideal conditions thebreakdown field at3gigahertz isabout, 10*volts/meter.) Thereisnoreflectedwave.y=. (b)Calculate thepower dissipation permeterinbothcases.Clearly, i[do theslinescanoperateathesepowerlevelsonyduringshortpulses. One canmeasure thepower transmitted down arectangular metallicwaveguide byreading thevoltage induced inatinyloopprojecting intothe 3412,(34.8) Decibels andnepersguide,asinFig.34-12.Theloopissituated atx=a/2, y=0,anditliesin Attenuation onatransmission lineisexpressed indecibels permeter.aeys-plane ‘Thenumber ofdecibels permeteris20timesthelogarithm tothebase10Showthat oftheratiooftheE's(ortheH's,orthevoltages, orthecurrents) atthe abo ') twoendsofalinemeterlong.Thedegreeofattenuation isalso Prov =8.40 10-6 —2, | expressed inneperspermeter,andthisissimplythevalueoff. Gay] Show that 1neper/meter isequivalent to8.686 decibels/meter. where Visthermsvoltage induced intheloop and.fistheareaofthe Joop. Theeffective value ofsfisunknown, butthemethod issatisfactory formeasuring relative values ofPr. 34-10, (34.8) Attenuation inrectangular guides (a)Show that, upon reflection from agood nonmagnetic conductor, in air,theamplitude ofanelectromagnetic wave decreases byafactor of OuIDED Waves tt 67 Fiber bundles aresimilar tolight guides except that their ends are coordinated, whichmakesthemsuitablefortransmitting images. 35Fiberscopes areuseful,amongotherthings,formedicalexaminations. CHAPTER ‘They comprise alight guide, forillumination, and afiber bundle equippedwithlensesateachend.Someoscilloscopes havefiber-optic, or GUIDED WAVES III ‘microchannel, faceplates. These consist ofastackofparallel fibers, with ; bothfacespolished, andphosphor ontheinnerside.Thefibersconvey ThePlanarOpticalWaveguide A. thelightemittedbythephosphormoreefficientlythanaglassplate.See TheEigenvalue Equation thefirstexample inSec.31.2.2.Thispermits theobservation ofveryfasttransients. One commercial oscilloscope hasawriting speed of2 10"meters/second. Optical fibers aremade ofeither silica orplastic, andtheir diameters range from that ofahuman hair toabout 0.5millimeter. The index of 35.1.THEPLANAR DIELECTRIC GUIDE 648 refraction decreases withtheradius, either gradually orabruptly. The 382RELATIONS BETWEEN THEFIELD COMPONENTS 650 smaller diameter fibers cantransmit digital information atratesupto383.THEFIELD COMPONENTS INTHE THREE MEDIA 651 sigabits persecond. 354 THE CONDITIONSOFCONTINUITY ATTHEINTERFACES —654 Totalreflection hastheadvantage ofbeinglossless.Forexample,the 35.4.1CONTINUITY “meedciutke ncea attenuation ofalightwaveinanopticalfibercanbeaslowas0.1decibel 35.4.2CONTINUITY ATTHEx=@ —— (98.9%transmission) perkilometer, whiletheattenuation inahollow 38.5THEFIELD COMPONENTSINTHEeeeAws conductingguideisofthesameorderover1meter.Theattenuationin i eee couwiew ne opticalfiberscomeslargelyfromabsorption, sincenomaterialisperfectly 358 ATHIRD EXPRESSION FORa69 transparent _ , 359SUMMARY 659 Metallicguidesserveatfrequencies ofseveral gigahertz, usually athigh PROBLEMS 660 ' power, whiledielectric guides serveatoptical frequencies andlowpower. ‘Thetwotypes aretherefore complementary, Rather than gothrough thefairly abstruse mathematics offibers, we study thesimplest form ofoptical guide, namely, theplanar optical Chapters 35and36arethelasttwochapters onguided waves. They i waveguide, composed ofthree layers ofdielectric: asubstrate, asheet, aguidingstructure thatcomprises onlydielectrics. Thewave 4 andacover.Theindices ofrefraction ofthesubstrate andofthecover againfollows12th,asinthehollowconducting waveguides of areonlyslightlylowerthanthatofthesheet.Similarplanarguidesarethence ‘patguidanceresultsfromtotalreflection {usedinmillimeter-wave integratedcircuits.ThephysicsofplanarguidesInoptical waveguides totalreflection occursinsideadielectric flanked isbasically thesameasthatoffibers. byanother oneofalower index ofrefraction. Although mostofthe i Planar optical waveguides haveoverall thicknesses oftheorder of power flowsthrough theinnermedium, some power alsoflowsthrough 10micrometers andwidths about 10times larger. These guides arethe theouterone,andthewaveisnottightly confined aswithahollow basiecomponents ofvarious integrated opticsdevices thatare,inaway, licguide. theopticalanalogsofintegratedcircuits. eeiclecnic‘waveguides servemostlyintheformofopticalfibersforthe Figure35-1showstwomethodsforlaunchingalightwaveintoaplanar transmission ofinformation. However, therearemanyotherforms. Clad dielectric guide. rods,eitherstraight orcurved anduptoafewmillimeters indiameter, Ourdiscussion parallels thatoftherectangular guidesofChap.4.Weservetoconvey lightfromonepointtoanother. Lightguides consist of studyonlythetransverse electric (TE)waves inwhichtheelectric vectorbundles offibers inside aflexible sheath. They serve thesame purpose. isnormal tothedirection ofpropagation oss 619 4 a ee; <| Fig.35-2.Coordinatesystemforthe® planaropticalwaveguide. Thethreemedia areinfinite intheyandz Giretons, Media1and3extendto ee interfaces makes thewaypropagate inZigwee <= 2agfashionalongthepostivedirectionofee 4 thez-axis. tb) ofrefraction, and total reflection occurs atthex=—-a and x=a Fig.35-1. Two methods forlaunching anelectromagneticwaveinaplanaroptical interfaces.Thenthewaveabovex=aandbelowx=—aspillsoverinto waveguide.Inbothcasesthecouplingefficiencyisabout80%.(a)Theprism theshadedregion.Thefieldinthecorewillalsopenetrate they=—b ‘The angleOis,infact,closeto97.(b)Gratingcoupler.Thegroovesonthe normalandtangentialcomponents ofEandofH.Sothefieldofa surfaceoftheguide diffract abeam attheappropriate angle 0. rectangular dielectric waveguide offinite cross section isquite complex. It iswiser forustodisregard these edge effects andmake theguide infinite 35.1 THE PLANAR DIELECTRIC GUIDE inthe+ydirection. Ouranalysis willbesatisfactory aslong asthe breadth ofthecore ismuch larger than itsheight. That isusually theWeselectaxesasinFig.35-2.Medium 1isthesubstrate. Thesheet,or|case. slab, ismedium 2.Medium 3,thecover, canbeair.Total reflection; occurs attheinterfaces 2,1and 2,3. i , — ‘Weassume thatallthreemedia arehomogeneous, isotropic, linear, a4 x Bae and stationary (HILS) and that they arenonmagnetic (x,=1),lossless of hoe (o=0),andnondispersive (theindexofrefractionnisindependent of| an|pee thefrequency). i ia al eeTosimplifythecalculations, wealsoassumethatmedium1extends *| eSdownward toinfinity, while medium 3extends upward toinfinity. In an ae practice, media 1and3needonlybeseveral wavelengths thick ed Wedisregard reflected waves, which means thattheguide extends to L oe infinity inthe+2direction. i a edFinally, weassume thatallthree media extend toinfinity intheplus ; aes andminus ydirections. Thereason forthisassumption isthefollowing in eas Figure35-3showsacrosssection ofaguidewhosecoreoccupies only 1 ! flee Fig.35-3.Crosssectionofanopticaltheregion between x=+aandy=+b.Thecladding hasalower index ' ‘ waveguide ofrectangular crosssection. 0 cunpeD Waves nt 363THEFIELDCOMPONENTS INTHE THREE MEDIA 6st ‘Theelectromagnetic fieldpropagates totherightatsomephase 2H,PH, oH,velocityv,,andthefieldcomponents inallthreemediatherefore have atHet=OfolloGa=~w*eeotoll,=KH, (35-9) the form where E=E, exp(wt~k.2)9, (35-1) on,Sp Kile ; where E»isafunction ofxonlyand o (35-10) yah, 52)Wemust therefore solve thedifferential equation i, hy oH, 4,beingthewavelength oftheguidedwave,Thewavenumberk,isreal ae(OH, (5-1) because wehave assumed zero attenuation. We setforeach medium insuccession with, again, kequal toky,ks,oFky 1 2 byabatealeutd, (35:3) 35.3THEFIELDCOMPONENTS IN . THE THREE MEDIA l_o_o > byapata8,=mk=O(€€otto)", 35-4) , Ge (€n€oto) (354) Inmedium 1werequireanexponential decrease inthe—xdirectionTherefore, inmedium 1,theexpression between parentheses inEq. andsimilarly fork;andky.Thewavelengths 2o,Ay,42,4)applyto 35-11ispositive, k,>ky,and uniform planewaves inavacuum andinmedia 1,2,and3,respectively. . Hs=Kexp(2—Ki)! =Kexpkat, (35-12) 35.2.RELATIONS BETWEEN whereKisaconstant and& THE FIELD COMPONENTSynstant and ky,isreal and positive: =4(2-8)". [Asstated previously, weconsider onlyTEwaves. YouwillrecallfromKi=+(k5—ki) (35-13) ‘Sec.33.1.1 thatthetransverse components ofthefieldinaguided wave ‘Thevalueofk,isasyetunknown. aresimple functions ofthelongitudinal components. Herek,isequalto “Then, fromEqs.35-6and35-7,inmedium 1neither k,norkznorks. 2!° 7 Byhypothesis, 3/3y=0.Also,inTEwaves,E,.,=0. Then,fromEqs. c,,=i! y, 33-11 to33:15, setting #,=1, 1me (5-14) En=0, (355) k| Haz=——Enmy- (35-15) Goto FHmeOy Enga Gs | 2ox Inmedium 3werequire anexponential decrease inthe+xdirection nw theSls esr, |Tonstheparsotbesin inEg,35-1lsagainponive,b>,ndwesetme IRE Bx Oly. Hye=Lexp[=(k2=k3)!7x]=Lek * Hy=0. 65)x]=Lexp (kay), (35-16) HereLisanother constant, andk,.is ‘itive: “These equations applytoallthreemedia, withkequaltoky,ka,oFky constant,andky,isrealandpositive: TofindH,,wemustsolveitswaveequation key=(KE—3), (35-17) 62 GUIDED waves tt 353 THE FIELD COMPONENTS INTHE THREE MEDIA 653 Also,inmedium3,fromEqs.35-6and35-7. Wehaveaddedaphaseangleatosatisfythecontinuity conditions atthe interface. From Sec. 35.2 Egy=+OMHy (35-18)L Egy=—PEMsin(kx+0), (35-23) Hy=—=Emy (35-19) * My k Hy = =~ Ey (35-24) Inmedium 2, omy on, Grouping Eqs.35-13, 35-21, and35-17gives Sot=(k?—K)Hys=KH (35-20) “ Kizki-ki, 24K, B= 35-25) Wehave putanegative signbefore thelastterm forthefollowing reason. Bothk,andk,arereal.Then43,isreal,Ifk,,isrealwithk?<3,then Allthek’sarerealandpositive. Therefore k,<k: andk,<k,, in Hy.isasinusoidal function ofxandtherearetwouniform plane waves agreement withtherequirements n,<1,andn,<ny fortotalreflection inmedium 2,zigzagging upanddown theguide asinSec.34.4,Thisis attheinterfaces. thetypeofpropagation thatweareinterested inhere.So Thesituation isasfollows WSK, ka=(KE RY! (35-21) Hy: =Lexp (~kseX) (35-26) Wearbitrarily choose aplussignbefore theparenthesis. F,,=J!H,., szObserve thatthe&’sarerelated asinFig.35-4, Then Medium 3: kw =Mo 22 k Hyys=M008(kaeX+@). (35-22) Hy™~—>Ey=Hh, (35-28)Ou ky Hys=M608(kX+@), (35-29) =1004i Medium2: Ey=—TpMsin(kaak+2), (35-30) k Ha =~ Ew (35-31) Hyg =Kexp kk (35-32) __jewtoMedium 1 Emp==Ames (35-33) k ik Ha=——=Egy=HHing (35-34) Fig.35-4.Rayoflightpropagating downanopticalwaveguide bytotalreflection Thewavenumbers k,,,k2.,andky,areallsimplefunctions ofk,,aswe‘Actualvaluesof@arecloseto9° haveseen. 654 GUIDED WAVESIT 35.5THEFIPLDCOMPONENTS INTHETHREEMEDIA 655 35.4 THE CONDITIONS OFCONTINUITY 35.4.2 Continuity atthex=aInterface ATTHE INTERFACES Because ofthecontinuity ofHm, There aretwointerfaces, andateachonethetangential components Hy, Moos(b +a)=Lexp(-kya), L=Mcos(b+a)expkya (35-42) and Ey, arecontinuous. This provides uswith four independent | : equations. Thenormal component H,,,isalsocontinuous, butits Proceeding similarly forE,,.,, continuity isalready ensured bythatofE,,., which isamultiple ofHm. jou_joy ‘Thesefourconditions ofcontinuitywillfirstgiveustheratiosK/Mand ~FaeMsin(b+a)="PELexp(kya), (35-43) L/M,interms ofthek’sandofa.Theywillalsoprovide two kindependent expressions fora.Laterthesetwoexpressions forawill L=-MBsn (b+a)expkya (35-44lead ustothe values ofthe four k’s. We use Masameasure ofthe ae amplitude ofthewave, which is,ofcourse, arbitrary. Equating nowthetwoexpressions forL, 35.4.1 Continuity atthex=—aInterface tan(b+a)= —ka (545) Tosimplify thecalculations, set ks. kaa. (35.35) Remember thatallthe&’sarerealandpositive.Thentheangleb+@ BecauseofthecontinuityofHm,atx=a, lesineitherthesecondorthefourthquadrant,likeb—a. . ka es Mcos(—b +a@)=Kexp(-k,,a), K=Mcos(b ~a)expkya. a=~arctan@—b+mx, (35-46)(35:36) "bei Me.We tm” "i ‘Also,fromthecontinuityofEn, asi‘cingagainaninteger./ecannotsetm”arbitrarilyequaltozero, to joo —EEREMsin(6+a)=="22Kexp(hua), (537) 35.5THEFIELD COMPONENTS* ‘" INTHE THREE MEDIA K=—M~ sin(b-a)expky.a. (35-38)k,n ep Nowthefieldcomponents areasfollows. Equating thetwovaluesofK,wefindthat 2 Hy.=Mcos(b +a)exp[kula—)) 57) - ky 7-_jws tan(ba)=~. (35-39) Medium 3 EyEeHas (35-48) Sincebothk'sarepositive, theangleb~aisineitherthesecondorthe t=-*e,,- 1, ass)fourth quadrant. Also, ony ky x=arctan +b+m'x, (35-40) ' Hy,=M008(ka,+@), (35-50) Ey=—22HMsin(ka,x+a”) 35-51 wherem'isanyinteger.Settingm'=0gives Medium2: mykoe a; (35-51) r=arctan24.6, (35-41) Hy=—**By, (35-52) ky omy 656 GUIDEDWAVEStt 357THEEIGENVALUE EQUATION 657 Hg.Moos(b~a)exp{kula+)h (35-53) 35.7THEEIGENVALUE EQUATION F,,=—/2!, (35-54) Equating thetwoexpressions forafromEqs.35-41and35-46gives Medium1: a kyo k kk ik (03555) arctan +b=~arctan =~b+m", (35-61)--‘-¢,,-u,, s _ Hin Oo kur k k . 2b=~arctan*—arctan+m’'n 35-62) WesetMequaltoarealnumber. Thewavenumber kandm”arenow arctanpoarctan ma. 5-62) theonlyunknowns. But,from thedefinition ofthetangent function, arctan2=—arctank#* 35-63) 35.6THE PHASE SHIFTS ONTOTAL REFLECTION arctan[=~ arctan (35-63) ky ky Weshallneedthephaseshiftsatthetwointerfaces inamoment. arctan ==~arctan == (35-64)Figure 35-4shows araythatzigzags down theguide. Theangleof ky2 keincidence is0(the 8,ofChap. 30).Also, Setting Kscsing, +=cos6 (35-56) meam'=l1, 05-65)ka ka wheremisagainaninteger, Observe thatk.<k2,Since k=1/2,2,>Az,where2,isthewavelength b bsoftheguided waveandA;isthewavelength ofauniform planewavein 2b=arctan +arctan +mat (35-66)medium2.Thenthephasevelocityoftheguidedwaveislargerthan Now ae a In ;.WefoundthephaseshiftuponreflectioninSec.31.2.1.Remember b=kya=kya6080=ngkyacos0=20cos6,(567)thatthefirstmedium isnowmedium 2,whilethesecondismedium 3. 2 . 2,3 interface isSubstituting (t2/ns)sin@forsin7,thephaseshiftatthe2,3int whereA;=to/nsisthewavelength ofauniformplanewaveinmedium2. > 212 Substituting the©’sfrom Eqs. 35-58 and35-60 into Eq.35-66, and 2aye )?—(klk) a y©,=2arctan(ent@—tealnay" =2arctan(atta)=est »multiplyingbothsidesby2,weendupwiththerelation 7 cos ‘a : 557) ¢ 8xpos0=0,,40,542mz, (35-68) 2 ke (35-58) =2arctan , Thisistheeigenvalue equation for8,the®’sbeingfunctions of6.The angle ofincidence @ofFig.35-4cantakeonlythose values thatsatisfytan223k (35-59) theaboveequation, cachintegral valueofmcorresponding toan2ky eigenvalue of6andtoaparticular propagation mode. Thusmisthe 7 identrayliesinmedium 2,here.Similarly, |modeorder.AsweshallseeinSec.36.1,m=0.Thevarious modesare Rememberthattheincident ray denoted asTE»,TE,,TE,etc. k 24 kin Wecanalsodeduce theeigenvalue equation directly, asinSec. 34.4.o=2aretan -,tanSt (35-60) Figure35-5showstwowavefronts,AandC,oftheup-going wave.Ata 359 SUMMARY 659 658 j Before going ontothat, letusfindathird expression forthephase » angle a. — a ,Net xiNA 35.8ATHIRD EXPRESSION FORa QY \ ‘, Combining Eqs.35-41,35-63,35-66,35-59,and35-60,wefindthat© Aas nN ' a=arctan&+b (35-73) 3 No HO) Os)symxva ararcararcae (35-74) N ©,o, x Fig.38-5.Geometric construction fordeducing theeigenvalue equation iereeie wadGna (35-75) Inasymmetric guide,medium 1andmedium 3areidentical, the'sare given time1,setE,=expjor forthatwave along A.Atthesame equal, and time t, oa a=(m+v5. (35-76) Ec=exp,i(o+=). (35-69) ‘AtpointD,reflection ontheupper interface givesaphaseshiftof 35.9SUMMARY ,s.Thus,alongB,Ep=expj(wt+®>,s)- ;‘Similarly, asaresult ofthereflection atE, Aplanar optical waveguide comprises asubstrate, asheet, andacover, with m<z, ms<ms, mbeing anindex ofrefraction, Total reflection Ee=expj(wt+©,5+21) (35-70) ‘occursattheinterfaces 2,1and2,3. and Weproceed asfollows tofindthefield equations foratraveling wave. 2al : (1)Weapply thetheory ofSec. 33.1. (2)Thefield inmedium 1mustFret Os+2m. @5-71) decreaseexponentially with—x,andthefieldinmedium3exponentially with x.(3)Inmedium 2wechoose uniform plane waves thatzigzag down Ifm=0, then /isminimum andtheangle ofincidence 6ismaximum. theguide. (4)Weapply theconditions ofcontinuity forthetangential ‘Asmincreases to1,2,3,...,/ becomes progressively larger andthe components ofEandHattheinterfaces. angle ofincidence decreases. ThisleadstoEqs.35-47 to35-55 forthefield. Now, from Fig.35-5, ‘Thefactor Misameasure oftheamplitude ofthewave, 1=4acos0. (35-72) b=kza, (35-67) Substituting intoEq.35-71givestheeigenvalue equation, a=arctan +=—arctan®*— 5+x, (35-41),(35-46)Weshallsolve theeigenvalue equation graphically inthenextchapter. ky [email protected] ie.fevalues of@andofthevarious k’swillfollow where m”isaninteger, oo GUIDED WAVES IH 1 ky=HEKID! =(KE3? Kg=(KE3)", ==(35-13), (35-21), (35-17) » ky=f=olencatn)™, (35-4)‘ ee andsimilarly forkyandky. “Theconstants ky_sKoesKassandaareallfunctions ofk,,which isitself related totheangle ofincidence @: 3 k,=kysin0. (35-56) —_____—_— Fig.56. Equating thetwovaluesofaleadstothenvaluc tion 35-2.(35.3)Thek's Equating thetwovaluesof@leadstotheeigenvalue equatio SD Yetketl a (b)Show alsothatif,inasymmetric guide, m,=m)=n, 1:= BF6080=Way4BsIma (35-68) Ann,thenK+ki,=(@0nAnyi paneam35-3. (35.5) Multiple reflections inthesheet thatfixesthevalues of@forthevarious modes m.Here ‘ShowthattheEfieldinmedium 2isthesuperposition ofanup-going andadown-going wave, atthecorrect angle ofincidence k ky 35-4. (35.7) tan2. .= 4 =2 2}7-! S 2k,,aimtermsofthek's y=2aretang andByy=arctan, (35-60),(35-58) eladha=(1Bakke) arethephase shifts upon total reflection.taha oka, ‘The phase angle is. oo +Recall that k,,=k,cos8,wherecos@<1andkz=n:ko=Ms/Ao.a=80m413 (35-75) 35-5.(35.8)‘Thefieldcomponents inasymmetric opticalguide2 (a)Show that, foreven modes inasymmetric guide, En,=M'cosbexp[k,,(a~x)]inmedit PROBLEMS: P[ksA(6i] redium3, =Mcoskay inmedium 2, 35-1, (35.1) Thenumerical aperture ofanoptical fiber° =M’cosb exp[k,,(a+x)] inmedium 1, Figure 35-6shows alongitudinal section through anoptical fiber. Arayemanating fromasource§entersthefiberatananglesuchthat@isthe whereM’=(=1)"?*YouoM /(ks,)criticalangle. (b)Showthat,foroddmodesinasymmetric guide,(a)Showthatsing=(nj—n3)'*. Thequantitysin@isthenumerical ..‘aperture(NA)ofthefiber.Thisexpression isvalidonlyifm3—mi=1.The Em=~M"sinbexp[ky.(a~x)] inmedium3,maximum possible value of9is90°.Iftheangle @increases beyond the =OMsin haat inmedium 2 value defined bytheabove equation, then@becomes smaller thanthe“ 7 criticalangleandtheraydoesnotpropagate downtheguide.Thisequation =+M"sinb exp[k,.(a+x)] inmedium 1, therefore defines anupper limit for9. nasi(b)Showthat,ifthesourceradiatesisotropically. thenthefractionof whereMo=(—1)"""!jeopoM [(k>.). thetotalavailable lightthatiscollected bythefiberisabout (NA)'/4, or i about n;An/2, where An=n;—1).Ifn:=2andm,=1.98,then@=II.Ss andF=0.01.Thelightcollectionefficiencyisthusverylow i 361 SOLVING THE EIGENVALUE EQUATION FOR@ 663 Here aishalfthethickness ofmedium 2(Fig. 35-4), Ayisthewavelength ofauniformplanewaveinmedium2,the’sarethephaseshiftsupon 36total reflection (Sec. 35.6), CHAPTER sin?0—n2/n3)!2Oy)=2aretan NO=min)" (36-2) cos @ in?@—n/n)! GUIDED WAVES IV ,5=2arctan Si Oil (363) ical Waveguide Bot tical Waveguide B. ThePlanarOpi 8! andthemodeordermisaninteger.Forcachallowedvalueofmthere The Guided Wave corresponds onemode ofpropagation. Equation 36-1 does notseem topossess ananalytical solution. Weshall solve itgraphically, butIetusfirst examine itcarefully. 36.1 SOLVING THE EIGENVALUE EQUATION FOR @ 662 Example; ASYMMETRIC GUIDE 663 (1) Total reflection occurs ifboth Example: ANASYMMETRIC GUIDE 666 m h 36.2 THE ELECTRIC FIELD STRENGTH EASAFUNCTION OFx666 @>arcsin and 8>aresin? (36-4)36.3LINESOFEANDOFHFORTHEm=0MODEINTHESYMMETRIC m m GUIDE 666 (2) Since @=/2, theleft-hand side ofEq.36-1 ispositive 36.4 THE PHASE VELOCITYv,668 jl 36.5THEGROUP VELOCITY v,670 (3)FromSec.31.2.1, thephaseshifts®,withEnormal totheplaneof36.6THEFIELD ENERGY 670 incidence, liebetween 0andx.Thus 36.7 THE TRANSMITTED POWER 671 0<0,,40,,<2. 368 SUMMARY 671 21FOngSoa G65) PROBLEMS 672 ‘Then theinteger mcannot benegative. | (4) The 'sincrease with 0,while cos@decreases, Then foragiven mWeterminate hereourstudyoftheplanaropticalwaveguide andof thereisonlyonepossible 0 8 guided waves. Thenext, andlast,subject willbetheradiation oftlectromagnetic waves (5)Asmincreases from1to2to3,etc.,thecorresponding valueof8 Youhaveprobably noticed thatthefieldofaplanar optical waveguide decreases. ismore complex thanthatofahollow rectangular metallic guide. Therearetworeasons. First, thefieldextends overthree different regions, WesolveEq.36-1below byplotting bothsidesasfunctions of@onthe instead ofone.Second, thephase shifts attheinterfaces arecomplex samegraph, andnoting where thecurves cross. Thatislessaccurate, but functions oftheindices ofrefraction andoftheangleofincidence. moreinstructive, thananumerical solution. Thevalues shown inTables Wedonotshow thecalculations indetail inthischapter, because they 36-1and36-2werecalculated numerically. reratherinvolved a Example |ASYMMETRIC GUIDE 36.1 SOLVING THE EIGENVALUE EQUATION Figure 36-1shows asetofsuchcurves forthesymmetric (1,=13) Re guideofTable36-1.Thecurves marked m=0,1,2,3arecurvesoftheright-hand side ofEq.36-1. They start at InSee.35.7 wefound thattheangle ofincidence 0ofFig.35.4 satisfies 1m ms' @aresin =arcsin™, (366) theequation ms nysacon0=5)$0,442 Got)il beingequaltoninthisxe, 664 665 ‘Table 36-2 Parameters foranasymmetric guide woo one CHARACTERISTICS QUANTITY 5 1 ’ 3 a=1hm o 1.5172 1.4649eobee —88 ny=1.9800 sin@ 0.99856 0.99440XXxxxgikxx n=2.0000 k, 2.5097x10"2.4992x10” “lekSzase uFosn y=1.0000 2 2.5036x10-7 2.5141x10-7 HSritiamamcom |enn ke=1.2566x10" key 1.3458x10° 2.6573x10° ky=2.4881x10” ka 2.1724x10" 2.1603x10” ky=2.5133 x10" o 1.3458, 2.6573, ky=1.2566x10° o,,' 2.3629 1.4470sin6.,=0.9900 o, 3.0179 2.8968, Fd. 8.,=81.890" a 1.7351 3.5046 3 Scotts esb 2=0.000" b-a’ 0.3893 0.84733\.| Senex $8 ex et aterS|" |aeeebazennteens = ettB| [RRSSeReRe ers3ee|= "radians AScidciciciciaaRaa |& ba 2 Curve Aisacosine function ofamplitude 8x(a/A,), near 2/2. 2 Decreasing theratioa/A,sweeps Adown inthedirection ofthe 5 arrow,itsright-hand extremity remaining fixed[cos(x/2)=0]3 ‘Thiseliminates thehigher modes (m=3,2,1) onebyone. 3 However, modem=0remains,evenatlongwavelengths. et2 i “Thusaverythinguides monemode: itsupports onlythem=0 °este Be2e|2 mode.Thentheangleofincidence isonlyslightlylargerthantheBxxxcuasax|£ criticalangle. ||\gheeeeasssnezbe |b T]|JBSRReasseecaes | *ee £ » ioe, ee a 2 S|Z|.g. seddeltae| 2 : BGecdddadsesies| 2 & Fs El 2 At|& Bee i bs<= xxxx z m=Oz Egeag 28g|&g228222022 rf ;Z| ziprireaiiist 22 oa « 6 8 7- Clertivgeesie « Fig.36-1. Graphical solution oftheeigenvalue equation forthesymmetric guide ofTable 36-1. Curve Aisaplotoftheleft-hand sideoftheequation. Theotherturvesareplosoftherightbandsideforvariousvuesoftheaeodes al 666 667 —F “ « 0 cs ; “| Fig.36-2. Graphical solution oftheeigenvalue equation forthe° a asymmetric guideofTable36-2, ne Psa ace an Table36-1shows theguideparameters fortheallowed modes aa noo m=0,1,2 “ o Example |ANASYMMETRIC GUIDE ;Nowletmedium3beair.Thenn=1asinTable36:2,withns=2 a ae and 1,=1.98 asabove. The critical angle atthe2,3interface is aresin0.5, of0.52radian, but total refletion atthe2,1interface = fanonly occur for0=arcsin0.99, or1.43radians. The allowed range for8istherefore thesame asforthesymmetric guide a Figure 36-2 shows thegraphical solution forthisguide. Only the‘m-=0 andm=1modes areallowed. Notice that, astheratio 4/2 a eereases, curve Aswings along thearrow, aspreviously, and Pm modo0eveneeallyGaappeer. Soanarymmetric guideersoot Fig.36:3.(a)(b),(c)CurvesofEn,asafunctionofxforthesymmetricguideof igwavelengt Table36-1.(d),(C)CurvesofE,,fortheasymmetric guideofTable36-2. 36.2THEELECTRIC FIELDSTRENGTH EASA Table36-1.WesetM=.Thenwemultiplyeachequationbyexpj(ot— FUNCTION OF x k.z) toobtain thephasors, take therealparts, anddisregard constant factors, while preserving thesigns. The field components arenow as Figure 36-3(a) to(¢)shows E,,asafunction ofthevertical coordinate x follows.forthefivemodesofTables36-1and36-2andFigs.36-1and36-2 H,exp[kss(a~x)]sin (wt~kz),Recall that, fortheTEmodes that weareconsidering, Eiseverywhere 7* paralleltothey-axisofFig.35-2. Medium 3:4E,xexp[ks.(a—x)]cos(wt~k.z), (36-7) H,x~exp[ks,(a-x)]60s(wt~k,2). 36.3 LINES OF EAND OF HFOR THE m=0 MODE INTHE SYMMETRIC GUIDE Hsin kasx sin(cot k2), LetusseewhatthelinesofEandHlooklikewiththem=0modeinthe Medium2[resto to, (368) symmetric guide.StartingwithEqs.35-47to35-55,weseta=2/2,from gl H,*~c0sks,60s(wt~k,2). 08 64 THEPHASE VELOCITY« 669 ; 802,=Ao/(nzsin8)>Ag.Thephasevelocity isgivenby | c cVy=fd,=—— >—.. 36-11Y=She=in” ny Gott) .Forthem=0mode ofthesymmetric guide ofTable 36-1, sin@=0.9988. Thedispersionrelationforawaveis«@expressedasafunctionofk. IF-‘The dispersion relation foraplanar optical waveguide ishidden inthe SZ eigenvalue equation, 36-1. Itsleft-hand side is4an,(w/c)cos8,whileits LiZ right-hand sideisafunction of,m,,na,ma,andm.Thus, foragivenmeet(iiy modeinagivenguide,Eq.36-1expresses«wasafunctionof8inimplicitZaWZ * form.But6isitselfafunctionofk.andw:fromEq.35-56, AYauUi sing==ke (36-12) ARitp kyngwle) iLZ Sotheeigenvalueequationgiveswasafunctionofk,,andinversely.Ty, Toplotwasafunctionofk,weproceedasfollows.Wefirstselecta Avalueform.Then,foreachvalueofw,wesolveEq.36-1for@,andthen L ‘ wededuce k,from theabove equation.” For thesymmetric guide ofTable 36-1, 2 OOw=raf, (36-13) y o 2Bak.=ny2sinO =nyFsin6 (36-14) 7 mi ical 0 Fig.36-4.LinesofE(dotsandcrosses)andofHfforthesymmetrical opt1.Thewavefrom ht.Thefiguredoesnot , ; waveguide ofTable361Taccusetheleceldtheresmuchweaker Now@canvaryonlybetween6,and90F,withsin8,=1.98/2=0.99.So, thaninmedium 2.Compare withFig.34-4 throughout thepermissible range of8,sin8varies atmost between 0.99 and 1.0, and H,«~exp[k,.(@+x)]sin(wt—k,z), ae (36-15) Medium 1:4E,xexp[kie(a+.2)]cos(ot~k,2), (36-9) °Hx~exp[ky,(a+x)]c0s(wt~k.2). ‘Thephasevelocityoftheguidedwaveisapproximately thesameasthat : thFig.34-4 ofauniformplanewaveinmedium2. Figure 36-4 shows lines ofEand lines ofH.Compare withFig.34-4. Forthissymmetric guide,thecurveof«asafunction ofk,isvery nearly astraight line through theorigin with aslope ofc/n=c/2. 36.4 THE PHASE VELOCITY vu, ‘Thewavelength oftheguided wave isgiven by o ‘thewavenumberk,isthefofChap.28andthecurveofwasafunctionoffsthen =ke=asin0=aosin6=Ein, (36-10)ileavapnams1OCCh8P.28andtecareoF383fanevionofBsthe 60 uIpED waves WV sossummary on 36.5 THE GROUP VELOCITY v, Wehaveset (L2)=ki thin(2,3)=k+ (36-24) veloci 29.2.4 andApp.C)isdefinedas Thegroupvelocity(See asPP.C)isdefine andwehaveassumedthatMisarealnumber.y=@ot (36-16) ‘Thetotaltime-averaged energyperunitlengthandperunitwidthis6dk, dk./do thus: This istheslope ofthecurve of asafunction ofk,.Itisshown inProb. SHE 4Bt (36-25) 36.5that,’forasymmetric guidewithny=m;, apaMell|B(Legh+20)+20]. (36-26)»,—£sin 9—Lsi?@=ni/n3)"?+Gala) (6.17) 4Mikay ma(Sin?O—ifn)+(Bala)sin’@ Thisenergyishalfelectricandhalfmagnetic, withaslightexcessof Forthesymmetric guide ofTable 36-1, magnetic energy inmedia 1and3andaslight deficiency inmedium 2. For the symmetric guide ofTable 36-1 and for the mode m=0, vemupmS. (36-18) i=65=0.013683. 36.6 THE FIELD ENERGY 36.7 THE TRANSMITTED POWER Wenowcalculate thetime-averaged electromagnetic field energy inthe Toobtain thepower transmitted perunit width, weintegrate the guide perunitlength andperunitwidth: time-averaged Poynting vector over awidth of1meter: e=f(6.€0E2mm+Holly)X1Xde (36-19) PratRe|-E,H?dx. (36-27) with , , Integrating andsimplifying asinProb.36-8,thevaluesofP’forthethree Boge=41EmlHoge=MUHol?+[me (36-20) mediaare,respectively, ‘Theaverage energies perunitlengthandperunitwidthinthethree py=etekMP 6628)media are,respectively, 1bki(2,1)" 1»__HoMPKE fy. M? Kicksa=tee 36-21 y=OokeM™ (9gyKixKae) " 4k,.2.1) 6621) Pag, (22tanta 5) (36-29) ”welt 2Kukikeke] ttok,M?5=HOME|nana4Risks4Keke) 36-22) 5=StoME 36-30) aa, PFO) 3) Po3) 66-20) ‘HoM?k? Then =pe (36-23) comokM?/1014k(2,3) P=Pi4pigpy=VMoKM (11)36-31 : PitP+PySETS(Etat (3631) "When wedscused the group, orsignal, velocity in rectangular metalic waveguide.wwesawthatwassimpytheaxialcomponent ofanindividual planewavethat2gzaB5 36.8SUMMARYdowntheguide’Tnthatease,v,~(c/n)sin®.Herethegroupvelocityisslightylargerthanthisaxialcomponent, because oftheGoos-Haenchen shift(See. 31.2.2). Agiven ray Wecansolve theeigenvalue equation for8,Eq.36-1, byplotting thetwo tneprotien mediand3Tseenaewhendefenhe‘|tidesseparuclyasfunctionsof8andlnotingwerethesorveacen om GUIDED waves 1v prowess on With asymmetric guide whose parameters arethose ofTable 36-1, we narrow lightpulse broadens asittravels down theguide. Thisismodal have thecurves ofFigs. 36-1 and 36-3(a), (b), (c). dispersion. WesawinProb. 35-1 thatthenumerical aperture isonlyofthe Withtheasymmetric guide ofTable 36-2 wehave Figs, 36-2 and order of1%forasymmetric optical guide withn,andnyslightly smaller 36-3(d) and(e), thann:.Howwould thenumerical aperture andthemodal dispersion for. -1beaffectedifmedia1and3were ? The lines ofEand ofH(Fig.36-4)arereminiscent ofthosefora theguideofTable36-1beaffectedif erebothair hollow rectangular metallic guide (Fig. 34-4). 36-4. (36.4) Thephase velocity v,. The phasevelocity Letmedium 1bedenser thanmedium 3.Lis . Showthat,if@isonlyslightlylargerthanthecriticalangleatthe v.=—— (36-11) interface 2,1,thenthephasevelocity oftheguided waveisapproximately°*nzsin8 equaltothatofauniformplanewavetravelinginmedium1 isonlyslightly larger thanthatofauniform plane wave inmedium 2. 36-5.(36.5) ‘Thegroup velocity v,ywthatthegroupvelocityinasymmetric planaropticalwaveguide is Thedispersion relationisthecircularfrequency wexpressed asa Showthatthegroupvelocityinasymmetricplanaroptical waveg function ofk.. =£sing— Aotold) _ey Thegroup velocity is en ORS (ila) sin@nsind _do_c (sin?@=ni/n3)'? +(As/a) whereA=sin’6~nj/n. Yeak ngGin—n/n)?4Gala sintO17) Showthattheapproximate valueappliestothemodem=1ofTable36-1. You can find thevalue ofd0/dw bydifferentiating theeigenvalue forasymmetric guide ‘equation with respect to«.Thus ‘Thefieldenergy perunitlength andperunitwidth is uyol A+ sla) wee niAl?+(Aa/a)sin’ yMoMPPAE(11 raleG@te+™) +2a], (36-26) 36-6.(366)Thefieldenergy (a)Show that, ina planar optical waveguide, theelectromagnetic energy andthetransmitted power perunitwidth is perunitwidth andperunitlength, inmedium 1,comprises three terms: ipMoM?cos*(b—a) . 2 wok M? (1 1 ej=ki +ki+2). 7=MokeME(A,A : aR, pane (pete t4) (36-31) The first term istheelectric energy; thesecond isthelongitudinal magnetic PROBLEMS energy, orthemagnetic energy associated with thelongitudinal component‘ofHandthethirdtermisthetransverse magneticenergy.FromSec.35.3,Y.(36.1)Negative modeordersareforbidden ekKi‘Thusthereismoremagneticenergythanelectricenergyin InFig.36-1,curvesform=—2,-3,—4,...wouldintersectcurveAat "1:242 wwthat&j=woMPR2/[4K,,(2, 1)},where(2,1)is . anglesofincidence largerthan90°,whichisabsurd.Sothosemodesare 36Tdi ye eeOLne OOMeioferbdden,Butformodem=—1thecurveswouldintersectat@=2/2, ismoremagneticthanelectricenergyWehaveshownthatallnegative valuesofmareforbidden. Showina (©)Showthat ns Keak?kek?diferentwaythatthemoden=fsforbidden aatiagotalHal] 36-2. (36.1) Themaximum value ofthefree-space wavelength A,asafunction . oe ofthemode order m (@)Show that, forthethree media together andforasymmetric guide, Find the maximum permissible value of4,asafunction ofmfora symmetric optical waveguide. iM 2symmetric opticalwaveguide. eat (2)(ra)+e] 36-3. (36.1) Modal dispersion andthenumerical aperture inasymmetric guide 2 Mad! MhInmultimode propagation, eachmodehasitsowngroupvelocity.Thena Asonewouldexpect.4=#, 7 36-7. (36.7) ‘The power density and theelectric field strength inanoptical guide Oneauthorstatesthathehastransmitteda150-milliwatt signalthrough 37optical goide with across section of3by $micrometers(Calculate thespace”andtime-averaged valveofthePoynting vector. CHAPTER ()Calculate thepeak electric field strength. This isthe breakdown field ‘The index ofrefraction is1.5. 3648.(36.7) Powertransmission inthethreemedia RADIATION I _2)Semathepowerstanserterofwthnmeand ThePotentials VandA,andtheFieldsEandB 1 ltak, Ook, Ku, Kee Jayerege SEbarasyeasy where (2,1) and (2,3) aredefined inSec, 36.6. Bysymmetry, thepower transmitted in3is 37.1THELORENTZ CONDITION 676 take 312 THE NONHOMOGENEOUSWAVEEQUATIONFORV67S Pte," 37.3 THE NONHOMOGENEOUS WAVE EQUATION FORA 678 374 THE RETARDED POTENTIALS «80 (c)Calculate theratio P{/P5 forthesymmetric guide ofTable 36-1for Example: THE RETARDED POTENTIALS FORTHEOSCILLATING modes 0,1, and2 ELECTRICDIPOLE 681 369, (36.7) The energy transport velocity Example: THE RETARDED POTENTIALSFORTHEOSCILLATING Bydefinition, theenergy transport velciy isequal tothe ratio P'/@", MAGNETICDIPOLE 684 where P’isthe transmitted power per meter ofwidth, and #°isthe 37.5 ASECOND PAIR OFINTEGRALSFOREANDB686 tlectromagetic energy permeter ofwidth andpermeter oflength, 376ATHIRDPAIR OFINTEGRALS FORE AND B67 Show thatthe energy transport veloiey fora symmetrical guide isequal soGonntany oes tothegroup velocity given inProb. 365 cone We have studied thepropagation ofelectromagnetic waves incon- siderable detail. InChaps. 28through 36westudied successively their propagation infree space and invarious media, across aninterface, and then along various guiding structures. Our final topic, theradiation ofelectromagnetic fields, willoccupy us during thenext three chapters. Here weturn togeneral considerations onthefields oftime-dependent sources. Until now wehave disregarded thetime taken bythefield to propagate from thesource ofradiation tothepoint ofobservation. This islegitimate only ifthetime delay isasmall fraction ofoneperiod. Weshall obtain thecorrect integrals forthe retarded field from the nonhomogeneouswaveequationsforEandB.Theseequationsare |similar totheusual wave equations, except that they include asource term, 6 RADIATION 1 37.THELORENTZ CONDITION on 37.1THELORENTZ CONDITION vitalpyvt 677) InSecs. 9.4 and 20.3 wefound that and Lf Pay: Ho[Zaye vane fLyesaefLav’ 375 vag[Paw. anf[2aw, (37-1) Gahor WwakViedv’. (37-8) whereVandAare,respectively, thescalarandvectorpotentials atthe ‘Thesecond integral isidentical totheonethatwestarted from,exceptfieldpointP(x,y,2,p=py+Psisthetotalchargedensity atthesource thatnowwehavethedivergence ofJ/ratthesourcepointP’instead ofpointP'(x',",2"), dv’istheelement ofvolume dx’dy’ds’atP’,ris atP.Thissecond integral iszeroforthefollowing reason. According to thedistance between Pand P’,andJ=J, +3P/3t+XMisthetotal thedivergence theorem,itisequaltotheintegralof(J/r)-dfoverthecurrent density atP',Thevolume v’encloses allthecharges andallthe surface bounding w'.Now,bydefinition, v’encloses allthecurrents. So,currents. onthesurfaceofv',Jiseitherzeroortangential, and(J/r)+dof=0.NowpandJarenotindependent quantities because theysatisfy the ‘Then‘equationfortheconservation ofcharge(Secs.4.2,16.4,and27.5): veAqtaf!Peerdv" ra) vey--2 (G72) - 3t Here weused aprime ontheVthat appears under theintegral sign a ; because weneeded todistinguish between Vand ¥’.But ¥’+J isreally Letustherefore seckanequation linking VandA.Weshallneedthis thesameastheBJthatappears inEq,37-2.Inthislatterequation we equation inthenext sections. Wehave already found thisequation in were concerned with only thesource point, andaprime onthe¥would Sec.17.9forthefieldofapoint charge moving ataconstant velocity. havebeensuperfluous. Then, applying thelawofconservation ofcharge, Westart with thedivergence ofA.ww[9p/ae Moe[J,,tofot vea=-£/2018py. (37-10) veantey.f Za=HefVed’ (37-3) axle fa Ser aah - Since the distance rbetween Pand P’isnot afunction ofthe time, we Wecaninverttheorderofthe¥andoftheintegral signbecause the ‘canremove thetimederivative fromundertheintegral sign,andformer operates onx,y,zandthelatter onx’,y’,z'.Now Mo2[2iy wv ea=288 [Pay=ecto, - vitetygss ve (37-4) FARratherAY=ee orn) or sohti i bye ae av ‘The first term ontheright iszero becauseJisafunctionofx’,y’,2’and VeA+cateY=o. 7-12) notofthecoordinates x,y,zofthefieldpoint. Also, from identities 15 a and16fromtheinsideofthefrontcover,V(1/r)=~V"(1/r). Thus ThisistheLorentz condition. Itisanidentity withVandAdefined asin vte_yeg or Eq.37-1.TheLorehtz condition isaconsequence oftheconservation of rr charge. Also,This result isgeneral, butourproof isnotbecause, asweshall see, the m1 a above integrals forVandAarevalidonlyforslowly varying fields. TheVira dts Ve. 7-6) Lorentz condition haspractical importance. Ofcourse, thesixcomponents ofEand Bderive from thefour components ofVandA. ‘Addingtheselasttwoequationsandrearranging. wefindthat dlHowever,withtheLorentzcondition,andifVisafunctionof¢,one ons RADIATION 1 19.3"THENONHOMOGENEOUS WAVEEQUATION FORA 69 need onlyknow thethree components ofAtofindVandthesix an components ofEandB. VXB=lod+Cottoa (37-20) For example, one can calculate Eand Hinthefield ofanantenna solely fromA,hence solely fromthecurrents intheantenna. Here oP 37.2THE NONHOMOGENEOUS WAVE EQUATION TRY GtOXM, (7-21) FOR V asinSec.27.1.Thus Wecannowfindthewaveequation forVinjustafewlines,Ifpisthe PX(PXA)= plod+alto(-¥v-A) (37-22)total charge density py+pp,then °Or or]” =f ov FAveeE-£, (37-13) VP: A)—PA=pod— OV|PA ’Ps (F-A)—VA=tod—emoV+53), (37-23) 9A)_Pp ov) ov Ey v-(-py-—)=F, (37-14) Catto) =V°A=tind— ve vA .(4)a r(Elloa)VA=Hod~eltoP5eatSs (3724) yy 2pan -2 5 FAWty rane, @7-15) VPAottoa=Hod (3725) vytoSE=-£ (37-16) WeusedthevectoridentityfortheLaplacianofavectorfromtheinside I" & ofthefront cover andtheLorentz condition ofSec. 37.1. This isthe Thisisthenonhomogeneous wave equation forV. nonhomogeneous wave equation forA. Atpoints where p=0, Outside current distributions, wehavethehomogeneous wave equation ev ay— ve - : oA PPV~catto53=9, (37-17) VPA~cates=O. (37-26) alhome veequation(App.C). 7 witneiphme select waveequation(App.C). Forstaticfields,thehomogeneous waveequationreducesto 4 ons) VA=0, (7-27)(Cotto)! " asinSec,19.2.asforEandB(Sec.27.8). 4ObvervethatwededucedtheexistenceofthepotentialsVandA, IfVisconstant, then lefined by vy=-£, 37-19) --ry-24 =ee ¢ E=-WV-—- and B=XA, (37-28) Sec. 4.1. asin See from Maxwell's equations 37.3 THE NONHOMOGENEOUS WAVE EQUATION op FOR A VeB=0 and PXE=—= (37-29) ' Wecanfindthecorresponding equation forAjustaseasily, starting this Thenwededuced thenonhomogeneous waveequations forVandAfrom timefromtheMaxwell equation forthecurlofB: gf theothertwoequations ofMaxwell 680 RADIATION 1 374THERETARDED POTENTIALS a A SE Similarly,veE=2 and PXB=od+eto. (37-30) Te.y',z'.t~ €3 AG,y,z,.)=42[Jahyr2tr)ay, (37-33) 4x J r 374THE RETARDED POTENTIALS ‘Theretarded potentials satisfy theLorentz condition.. Fortunately, retardation effects areoften negligible. Retardation is “The integrals forVandAofEg,37-1donottakeintoaccountthefinite importantwheneverthedelayisanappreciablefractionofaperiod velocity ofpropagation ofelectric andmagnetic fields.Forexample, if THifthecigs Gstibetiom shiftsnoneregion thentheintegrals implythat Foraninfinite, homogeneous, isotropic, linear, andstationary (HILS) VandAchangesimultaneously throughout allspace. medium €,4, 1 beInfact,thepotentials atagivenpointandatagiveninstantdonot Vay. 08g[ey 2trl), (37.34)correspond tothecharge andcurrent distributions atthatinstant, unless mee r thecharges areallfixed inspace. Theanalogy with astronomy isobvious: wf ey zit-rlv) wecannot seeastarasitisnow,butonlyasitwasmillions orbillions of AGyzOng , dv’, (37-35) years ago. Thestarisnoteven attheplace where weseeit!‘ThecorrectintegralforVisthesolution ofthenonhomogencous wave wherevisthespeedofpropagation 1/(eu)'? (Sec.28.2)inthemedium. equation 37-16 forV.This is ‘This isanother application oftheruleofSec. 27.1. _1fayzst-rlo) Example |THERETARDED POTENTIALS FORTHE Vony20=aeif 7ide G73) OSCILLATING ELECTRIC DIPOLE ThisintegralisidenticaltothatofEq.37-1,exceptthatpistheelectric FreetaweooSectrcipotesieartothatofFieotcharge density attheprevious timet—r/c,theintervalr/cbeingthetime foieote polemomentpI@sinuso taken byawave ofspeed ctotravel thedistance r.This istheretarded O=0.expja. (57-36) scalarpotential. =Qsexpjot=paexpje =O. GrNoticethatthespeedthatwehaveusedisc,thespeedoflightina PoQ.scxpjer=pacpjet, §pe-Qs3757) vacuum, and notthespeed inthemedium ofpropagation. This comes ‘Thissimple-minded model serves tocalculate theradiation fieldof about asfollows. Asweshall seeinChap. 38,electromagnetic radiation anycharge distribution whose dipole moment isasinusoidal occurs whenever electric charges accelerate. Anelectromagnetic wave function ofthetime. travels through spaceatthespeed¢everywhere, Uponpassing through, ‘Theupward current through thewireis say,apiece ofglass, thefluctuating electric field ofthewave polarizes the «a molecules, which radiate inturninalldirections. Thesumofallthese 1=F=joQ,expjor=1,expjot. (67.38) waves isasingle wave that travels ataspeed different from c.Ifwe Thus express VandAinEq.37-1, with 1n=J0Q-, 1nS=joPm, —Is=jop. (37-39) _)e We could first caleulate Aand then deduce Vfrom the Lorentz p=pyt py and Jad +>+0XM (37-32) condition (Sec.37.1).However, itwillbemoreinstructive to calculate Vseparately. Refer toFig. 5-1. under theintegral sign,weconsider, ineffect, thatthemedium of First propagation itselfactsasasource inavacuum. Then theretardation ya Qnewielt=nle) Qnexpjolt—rle) (5749) involves thespeed c. der, ane a i:} yakeexpsultfexp{peos0/20)exp(-is80100)} al t H 4xeor 1=(scos@)/(2r) 1+(scos@)/(2r)© H H (37-44) i braces aspower series, neglecting terms ofthethird order andi H iH higherins/randins/4,andthensubstituting p,,forQ,,5,wefind. { Hl Hi that ,iHi vnPa(AsiJcosderpjalt](er.s'<X). (37-45) |H | =phy(241)00s@expo (scr, s*<cK).(37-45) H H H andX.Ofcourse, -d i Fea(Eet)"exps(arctan)6746) @)i ‘Thus |/i\/ | vapBee(E41)cosexpi{o(s—)+arctns}. (7-47) ‘ ‘ ' Naturally, V=0intheequatorial plane at@=90°,where the «) o potentials ofthecharges +Qand—Qcancel exactly. Figure 37-2 Fig.37-1. Anosilating electric dipole. (a)Thevectors points in Shows aradial plotofVasafunction of@and {nedircetion shown. (b)Thecharges andcurrents sfunctions of ‘Observe thepeculiar wayinwhich theamplitude ofVdecreasesthetime, withthedistance r.Closetothedipolewherer°/#<1,Vfallsoffas1/r, But farther out, where r?/Z*>>1,VfallsoffasI/r. Atzerofrequency,£—>=and[r]=t.ThsreturnstotheVofa where thenumerators arethecharges, asthey appear atPofFig. static field (Sec. 5.1). Sat the timetNoticethatthetwocomponentsofVdifferboth ‘ASt0A,itissimplythevectorpotentialoftheclementof inamplitude and inphase current fs ‘Onecanusually sets*<«<r’ ands’«4°.Thislegitimates a wosimplerexpression forV.Setting a=He_wool) (37-48) femtye8, zoos ¢) Since j : then =, 2.2 2 Ty Fr.scos 0) sMoec!ck @749) (1-2) =0(1-24+2S**) ~ole]+Foose, (37-2) .o%Aallelioexpju(t—5). 7-50) whereAmenchr 4xeqchr © , aierbd teAnt 7-83 Were and moO (743) ‘Thevectorpotential propagates atthespeedceverywhere. Itis parallel tothepolar axis, anditsmagnitude atPdepends solely on Asimilar expression applies to«(t~r,/c), withanegative sign thedistance rtothedipole. SeeFig.37-2. before thecosine term Tapolar coordinates, tre. —HP)(605.6%-sin06 7.Then i AFaeegcHe COSOFSin0) Ges} 684 685 i ce Z / . f aaa is Pig A // \ i\ [ /\ f i / Me le “he JN i\\ 1= _ \ i1—] ‘—— : ' ‘ 1 / ' Ne A ~ Nee 6 . as Fig.37-3.Anoscillating magnetic dipole Fig.37-2. The scalar potential Vandthemagnitude ofthevector potential Aas, functions of@and6,about anoscillating electric dipole oriented 2sshown. The radial distance from thecenter ofthe dipole tothespheres marked Vis proportional tothevalueofVatafixeddistance inthatparticular direction, ‘The u exp (e= PVH) |Scalarpotentialismaximum atthepolesandvanishesattheequator,wherethe A=al,expotfSPUCT cosdg.(37-53) individual potentials ofthecharges —Qand+ofthedipole cancel. Itispositive an » r inthe northern hemisphere, where thefield of+0ispredominant, andnegative inthesouthern hemisphere. Thevector potential Aisindependent, both in Asinthethird example inSec. 18.4, weset ‘magnitude andindirection, ofthecoordinates 8and¢. . Ft(145+ Sos} (7-54) Example |THERETARDED POTENTIALS FORTHE ‘Then,withinthesameapproximation r'>a’,OSCILLATING MAGNETIC DIPOLE ‘Themagnetic dipoleofFig,373identi tothatofFi.189, rar(4 B28), ppBOM rsexcept thatthesource nowsupplies analternating current. By r rr hypothesis, there iszero netcharge and V0. Also, ‘Wenow have that bere 2en ont EareS C75) Ho Pnexpjolt=re) .=a a acesodeo (152) ifa®<<24,Thentheaboveintegralbecomesequalto 686 RADIATION 1 57.AUTHIRDPAIROFINTEGRALS FORBAND# 687 ra2S08)(188 cospt J;beingthefreecurrentdensity,9P/9rthepolarizationdensity,and rJo 2 r dr Ur. VXMthevolume density ofequivalent currents inmagnetized matter at Thisintegral iseasytoevaluate. Firstwedisregard thecosine Pw',y’,2), - andthecosine cubed terms because theirintegrals arezero. Also, Theabove integrals apply only ifretardation effects arenegligible. To theintegral [email protected] findmoregeneral expressions, westartwiththenonhomogeneous wave axl, 1 equations forEandforB.Wefound those equations inSec.27.8. ByAE(b+i) analogywiththesolutionofthecorresponding equationforA,theretarded fields are Substituting nowrsin@forx,andm,,for7a°J,,,m,,beingthe 1vr, PRCUE,‘maximumvalueofthedipolemoment,m=myexpjn, B=- {(ip+eS/20)gy, 37-82) 46 Je r ain (HE . AaHette(24j)sin@expjolt]& (37-57) of(POxa air(7*!) ots[ESgy 7483) juclm|<P()_BSeerae ‘ =1-j- Wer a«we (37-58) tar('I;) , HereEandBarethefieldsat(x,y,2,£),whilethebracketed termsare Compare Eq.37-57withEq,3745frtheVofanlerric dipole. takenat(x’,y’,2’,r/c),rbeingagainthedistancebetweenPandP’. Since V=0, then V+A=0, from theLorentz condition (Sec. You will remember that theintegrals fortheretarded potentials 37.1), involved [p]and [J]; theintegrals fortheretarded fields involve more Forr>>&thevector potential propagates asaspherical wave of complex quantities.wavelengtheffwhoseamplitudeisinverselyproportional 0. WefoundtheaboveintegralforE,butforstaticfields,inSec,12.2.equatorial plane,againaswemightexpect. ‘Theintegral forBisintriguing. Forastaticfieldwenowhavethat pate(Pay 37-64) 37.5 ASECOND PAIR OFINTEGRALS “ah (37-64) FOR EAND B . instead ofEq. 37-60. The source term isnow V’XJ, instead ofJx#, Wefound thefollowing integral forEinSec. 9.4: and theterm inthedenominator isrinstead ofr*. 1 lgpai [Fa", (37-59) 37.6ATHIRD PAIROFINTEGRALSon FOR EAND B whereEistheelectric fieldstrength atP(x,y,z),@=Py+Ppisthetotalcharge density atP'(x’,y',z'), dv'=de' dy'dz’ istheclement of Wecandeduce another pairofintegrals forEandBfromtheretarded volume atP’,andristhedistance between PandP’.Wedisregard potentials ofEqs. 37-31 and37-33. Letusstart withE: surfacechargeshere. 3A 1rhPVyr—2Ho[Ulp=—vy AL pt [lay 2! (ay a7 InSec,20.3wefoundthat Evo Vaed, peoe aG75) Ho[1X? pate|aw, 7-60) 1 0[algaath.7 --[ vielays—efAMSyy, (37-66) 42€o J, r ati or here Bisthemagnetic flux density atthe pointP(x, y,z)andwhereBisthemagi ¥Pot , Wecaninsertthedeloperator underthefirstintegralsignbecauseit Jos,oPvxM, (3761) operatesonx,y,2whiletheintegraloperatesonx’,y',z'.Recallthator thequantities between brackets aremeasured at(x’, y’,2’,(r/c). 77 0 RADIATION 1 PROBLEMS o Show that, atadistance pfrom thewiresuch that4p*< C*,andaway 37-7. (37.5) from theends, Show that tly, © Janttin’. [redar=0 Refer tothefirstexample inSec. 18.4. Wehave disregardedatermthatis 37-8.(37.6)Theelectricfieldoftheelectricdipole,calculatedfromthethird independent ofboth thetime andthecoordinates andthat does not, Sea tors therefore, affect either EorB.This particular Atherefore hasthesame Calculate Einthefieldoftheelectric dipole, starting from theintegral of form asifthecurrent were constant. Sec. 37.6. Tosimplify thecalculation, disregard terms ins°/r* ands"/r' aswellasthehigher-order tX.Thiswillleave radiat 37-2. (37.4) Thepropagation speed ofVandAinthefieldofanelectricdipole termigher-ordertermsins/f.Thiswillleaveyouonlytheradiation Find thepropagation speed ofthe scalar andvector potentials VandAinthefieldofanelectricdipole,ontheassumption thats°<<r°andthat 37-9.(37.5)Thefieldatthecenterofarotatingdiskofcharge.S«K ‘Suppose you have adisk ofradius Rand thickness 2s <R.Itcarries a Observe thattheaddition oftwowaves ofVtraveling atthespeed cgives charge density Q°=K(R~p)(s ~2)WeuseQ'forthecharge density inawavewhosephasespeedislargerthanc.Thisisaninterference effect. ericaiouseforthevadialsoodinste, Observe alsothatthespeed ofpropagation V,close tothedipole, isa You arerequired tofindthevalue ofBatthecenter when thedisk function ofthewavelength andhence ofthefrequency. Thedispersion rotates asasold attheangular velocity «.Ofcourse, Bisnormal tothe originates, notin theproperties ofthemedium, butrather inthegeometry, plane ofthedisk, SeeProbe 18-4, asinrectangular metallic waveguides andindielectric waveguides, Calculate Bfrom Eq,37-60 andthenfrom Eq,37-63. 37.3. (374) CanAbezeroinaradiationfield? Inthefieldofamagnetic dipole, V=0,A#0. Dothere exist radiation fields where the inverse istrue? 37-4. (37.5). Thealternate integral forBt We have shown that Me(U'XI ee eralo ifretardation isnegligible. The volume v'encloses allthecurrents. Show that theterm ontheright isequal toVXAforafinitecurrent distribution. Refer tothe identities onthe inside ofthe front cover 37-5. (37.5) Identity ofthefirst and second integrals forE ‘Show that, foranyfinite charge distribution, theintegrals forEgiven in Sec. 37.5 areequal ge[aw=he[Ea 37-6, (37.5) Identity ofthefistandsecond integrals forB Show that, foranyfinite current distribution, theintegrals forBgiven in Sec. 37.5 areequal Mo[TKF Mo[WxXt a[PXgyatefPOFGy Garni [a Load 38.2ELECTRIC DIPOLERADIATION 07 Ad,_Qsin@vtsin@ Qin?@art_QasinOr989) fcoder rege evr os “0.4 Combining, finally,Eqs.38-4,38-5,and38-9,andsettingr=ct,gives4 win7, 1 A@e_HoQasindEy <Bassin tan crc (8-10) ae ‘Therefore,bothEyandBinthekinkresultfromtheacceleration, they VAsareproportional totheacceleration a,they vary asI/r,andthey are F maximum inthedirection perpendicular totheacceleration. Moreover, E,andBaremutually orthogonal, and orthogonal tothedirection of propagation ofthekink. Finally, E,=Bc.Allthese characteristics apply totheradiation fields thatweshall study inthischapter andthenext." _9 38.2ELECTRIC DIPOLE RADIATION s d & Figure38-5showsanelectric dipole. Aswesawintheexample inSec. 37.4, this simple model serves tocalculate the radiation field ofany chargedistribution whosedipolemoment isasinusoidal function ofthe Fig.38-4. Thecharge isatb,asinFig.38-3. Cisacircle whose plane is perpendicular tothepaper andthatiscentered onthe2-axis. | InFig.38-4thespherical surface bounded ontheleftbyClies,just | before thearrival ofthekink,inthefieldofastationary charge situated | [email protected] JA Q2Q1-cose rhY/| ,- 22 _Ol-cos8 evedeeG2° (8-6) of7 | where Qisthesolidangle subtended byCattheorigin. y | Afterrsecondsthekinkhassweptthroughe.Thepoint¢thenliesin < | thefieldofachargesituated slightlytotherightofb,and®,islarger: x ° 1-cos(8+A8) a a Oy+Ab,=21ser80) cenII ;Ne |. ~ So ~| ; Ade __O -27Dee4-086)=57sin8.8. (38-8) Fig.38-5.Electricdipoleofmomentp=QsandapointPinitsfield. Now thetwocircles inFigs. 38-3and38-4 arevery close together. We a cantherefore setrA@=vtsin8.Then "Theydonotapplytothefelclosetosourceofradiation. 698 699 a ‘ h SN x " iws | ~ , Fig. 38-6. Electric dipole antenna attheendofacoaxial line. The outer conductorisfoldedback. t2 time,disregarding quadrupole, octupole, andhigher-order fields.Inthe ae oscillating dipole, charges oscillate along s,andtheir acceleration isthus > Fig.38-7. TheE,H,andS=Nengthwie \ EXHvectorsforanoscillating. + electric dipole forr>>Aand Wecalculated VandAintheexample, setting s'<r° ands'<¢A,but ‘olt]=0,without limiting therelative magnitudes orrand2,Wenow deduce E, H,,thePoynting vector f,andtheradiated power P. ‘The dipole isinfreespace. Figure 38-6showsanelectric dipole antenna attheendof@coaxial l 2 2=lel{Fi) cosor+(%—4)} line. ene (5+)*)cos87+(5147%)sino}. (38-14) 38.2.1TheElectric FieldStrength E Closetothedipole,wherer<2,oratzerofrequency (A>),onlyFirst, inpolar coordinates, twoterms survive, and av. lav _P 07 _Was rtge (38-11) Bmeg200Or+sindb) (reA,orf=0). (38-15) Vbeing independent [email protected], from Eq.37-45" ‘These arethestatic terms ofSec.5.1.They falloffas1/r? Farawayfromthedipole,forr>>,onlytheradiation termremains, RookRoa 1itfallsoffasI/r vv~PL.{(1-2%24)eosor—(+14)sino}.(38-12) anes wl° E=-Plsino (r>A). (38-16)drehr Also, from Eq.37-51, GA Bl bl SeeFigs.38-7and38-8.Remember thatwehavesets?<r’ and5°<<2. Sioa=mer -Hee(08OFsin06)(38-13) ObservethatEisproportional to1/22,ortow®,oftothesecondtime "eh NTET derivative ofp,ortotheacceleration ofthecharges. and 38.2.2 The Magnetic Field Strength H "Weagainreserve brackets forquantities measured attheprevious time—r/e Intheexample inSec.37.4wefound thatinthefieldofanelectric dipole ! 700 38.2ELECTRIC DIPOLE RADIATION 701 Then fle) (44 3_lt a=(14/7)sin06, (38-21) a \ againwiths*<rand5°2°. Forr<«A, and using Eqs. 37-39, YY. f jolly g_Ulsx*y/ \ H=GpinOo= (ra),(38-22) SaA whichagreeswiththeBiot-SavartlawofSec.18.2. Yi 4 Forr>4, ‘ th j ofp] olplxt__clp]xéN / } =—OUP] inog=—SUPLXT_ _clPIX?4 { / res nad 4qdr ax (A), 3823) \ w \ - asinFigs. 38-7 and 38-8. fy Y For r>>4, thecharacteristic impedance isthesame asthat fora | ee uniform plane wave infree space (Sec. 28.3): Fig.38-8.Polardiagramsofsin@(outersurface)andofsin®@(innersurface) a 'showing,respectively, theangulardistributions ofE,orofH,andofZ,.ata Z=F,=Hoe=377ohms (r>X)- (38-24)distance 7>>Xfrom anoscillating electric dipole situated attheorigin. Theradial distance from thedipole toone ofthesurfaces isproportional tothemagnitude of ithequantity inthecorresponding direction. Thereiszerofieldandzeropower ‘Theelectric andmagnetic energies areequal. flowalong theaxis. Curiously enough, Apropagates everywhere atthespeed c,butH propagates atthat speed only forr>>4. Closer in,itsphase speed is larger than c. Atzero frequency w=0and H’=0, asexpected. ip} ; - A=Fregehr6080F—sin84). 8:17) 38.2.3ThePoynting Vector EXH We lateth = ve ‘ThenAhasonlya¢component: wesnowcaleulate thetime-averaged Poynting vectorandtheradiated 1sa 3A,)5 Foranyr, nadyxans {2¢a,)-Aho (38-18) . HoHor(or Fo=4Re(EXH*)=}Re((E,F+ Eo)xH36}. (38-25) Toworkthisout,werequire thefollowing twoderivatives: Recalling that aa r}jolp)__ile) *xG=-6, OxO=7, (38-26) 2tpy-2 —2))=—/olel, ip) 38-19) alel=5, Pmexp(0‘) ¢ BG19) 3then, from Eqs. 38-14 and 38-21, <Ipl=0. 38-20) 5 5el?l 6820) Sp,=1Re(-E,H30 +Egy) 8-27) bb 702 RADIATION 1 34.2ELECTRICDIPOLERADIATION 703 Hoop}, om? f'pr, aoe Substituting again [n.5/@ fOF Prmss=HoeDesig?gp=MOLDmin?9 (38:28) sabe ”32a%er’ deor horsp? PaEEPEimS?=8.779X10fF? (38-35) =4.137 10°LP sin? —watts/meter’. (38-29) . r 2x10~e (8)? sy; od (3)Figs=19.99($) Figgwatts. (38-36) Or,since J,=OPm=27fPm» A aps Thustheradiatedpowerisproportional to:(1)(s/4)?,whichisasmall 9,=ols”Sn?96 quantityasweassumedinthefirstexampleinSec.37.4;(2)s?,wheres 8eor inthelength ofthedipole; (3)f?foragiven Ljm,; and(4)f*foragiven 1sfms? >p= 2 Pes =1.049610°3"sin?OFwatts/meter? (38-30)ap Example THECOLORS OFTHESKY,OFTHESETTING meNt F 4 .-ue(3)fosin?OF (38-31) SUN,ANDOFTOBACCO SMOKE’a r Dust particles suspended intheatmosphere scatter thelight 5\Bm a, coming from thesun.Thisscattering occurs because theelectric=2.386(5) “sin?0%—watts/meter? (38-32) fieldoftheincidentlightwaveexciteselectronspresentinther particles. These electrons actassmall electric dipoles and av ‘ reradiate. Ifwedisregard resonances, thepyofanoscillating Notethefollowing pointsaboutthistime-averaged Poynting vector. Titan iprovovtonal tthe amplitude oftheincdcae wa (1)Itinvolves only theradiation terms, despite thefactthatour Then thereradiated power isproportional tof*,andthelight calculation isvalidevenifrisnotmuch larger than4.Thetime-averaged Scattered bytheskyisbluerthansunlight.lial,atleastforros,1Po>s,SeeFips Iftheairwerecompletely dust-free,theskywouldstillbeblue, powerfiuxiseverywhere radial,atleastfor1°>>s°, 4°>sBs. butdarker:atomsandmoleculesoftheairalsoabsorband 38-7and38-8 reradiate energy, butmostly intheultraviolet. Thelightfrom the (2) Itvaries as1/r?because, under steady conditions, thepower flow sunthat reaches thecarth, particularly atsunset, isreddish through anygiven solid angle must beindependent ofrtosatisfy the because partofthebluehasbeen diffused out. conservation ofenergy. This1/7?dependence results fromthefactthat Itisforthesamereason thattobacco smoke iseither bluish or theradiation terms forEandforHbothvaryasI/r. reddish, according tothewayyoulookatit,withrespect t0@si sourceoflight.SeeFig.38-9. (3) Since itvaries assin?6,itiszero along theaxis ofthedipole and maximum intheequatorial plane, asinFig.38-8. Anelectric dipole does 38.2.5 Radiation Resistance not radiate along its axis. 5lorradiate alongUsax Wesawabove thattheradiated power isproportional tothesquare of ;thermscurrent. Thefactorofproportionality istheradiation resistance: 38.2.4 The Radiated Power P 5Integrating thetime-averaged Poynting vectoroverasphereofradiusr Res=19.99(5) ohms (38-37) yieldstheradiated power: if(S/AP«1. or®ftp?,, P(* Let(s/4)’=0.01. Then(s/A)?=0.46, andtheradiation resistance is P=tortpiefsin?67°sin6d0dp (38-33) about09ohm. “f*Din=3.466X10"f'p2y, watts, (38-34) jopSttM:Minnen, TheNatureofLightandColorintheOpenAi.Dover,NewYork 7 A 704 342ELECTRICDIPOLERADIATION 0s 1 18 =(sina)ao=-12. 9Sepp tsing) JSdr (38-41) — Multiplying bothsidesbyrsin@gives res SSetsite wv3= 5 5veetyy Jy(Hrsin6)d0+5(Hrsin8)dr=O.(38-42) ThetotaldifferentialofHrsin@isthereforezero,andonagivenline ofE,Hrsin@isaconstant,SubstitutingthevalueofHfromEq.38-21 /|\ afterreplacingjapbyJs,expressingthecomplexfactorinpolarform,and replacing theexponential function byacosine, wehave theequation Bish eh for aline ofE: Fig.38-9. Fine particles, asintobacco smoke, scatter blue light preferentially poy r r‘Thetransmitted lightisthusreddish. sin?(5+1)60s(we—$4arctan1)=Ki. (38-43) *38.2.6 The Lines ofE ‘The parameter Kvaries from one line tothenext. Remember that we , oo have assumed theconditions (s/2)'«1and(s/r)°<1. Ofcourse, there isnosuch thing asadistinct line ofEorofH.Allthat Figure38-10showseightfamiliesoflinsofE." weknow isthatthefields EandHpossess both amagnitude anda Ark direction thatvary from onepoint toanother inspace, andfrom one " instanttoanotherintime,according tocertainlaws.LinesofEandofH, sin?6cos(orary3)=Ki, (38-44)however, providethebestwaytopictureafield. x72)>** Wecanfind anequation forthelines ofEbysetting and thelines ofEtravel outward atspeed wi=c. Closer in,thearctan E,_Eo (38-38) termisafunction ofrandthespeedofthelinesislargerthanc.ar rd6 *38.2.7 The K4 Surface drandrd6areth tsofanelementofalineofE.Hi tiana nn Itsinstructive toplotEq.38-43asathree-dimensional surfaceasinFig. *° Pi ponding phasors. 38-11.TheloopsarebothcontourlinesandlinesofEat=0.Infact, ourcalculation withphasors. nee Fe eneneeeaFirstwenotethat,fromMaxwell's equationforthecurlofH(Eq. theyarethesamelinesofEasthoseofFig.36-10(a).Astimegoeson,nz theargument ofthecosine function increases, andtheripples moveout, asadamped wave, carrying thelines ofEwith them p-—-yxn. 68:39) Letusseehowthelinesbehave. Figure 38-12showstheintersection ofjoe, theKAsurface, again at¢=0, with theplane 8=1/2. These curves lie inside theenvelope Since Hhasonly@@component, from Sec.38.2.2, poykr=4(S41)° Gino=0. (38-45) F-t2sino), Fp---+-1 20m,(38-40) " "Faesrsin 6304S)»—Be=Feay ."Seetheanimated filmlooponDipoleRadiation byR.H.Good,California State ‘Thus thedifferential equation forthelines ofEbecomes, Universiy, Hayward, California 706 707 Ht WAAANN QO } / \ AAN \a\ wo wana Us «2 Astheripples move out, their height decreases rapidly and soon approaches unity.Clearly,linesofEwithKA<1cantraveloutto Mfif \\l] \ infinity.Theyprovidetheradiationfield.Itisalsoclearthat,ifKA>1, i| ))i}ie)) ) theycannotgofar.IfKAisonlyslightlylargerthanunity,aloopshrinks \N my) j\ untilitreachesthetopofarippleandthendisappears. |() SNA Exon ingelectricdipoleforwt=0,2/4,2/2. Fig.38-11.TheparameterKiplottedasafunctionofthecoordinates rand@a ‘wavelength decreases withdistance. ThelinesofHarecircles perpendicular to peaksoscillate inunison from—%to+,andtheripples moveoutradially. Thethepaperandcenteredontheaxisofthedipole loopsarebothlevellinesandlinesofE. rs ue PROBLEMS 709 The radiation power is f P=3.466x10-8f*ping (38-34) i =8.799x10-1f272,5? (38-35) 5) =19,99(5) a| (aTeoywatts. (38-36) kad SSS Theradiation resistance ofanelectric dipoleisabout20(s/2)? ohms.‘=<>PSII ES OnalineofE, J on nzg(# 1) r rif. sin’o(5+1)cos(ar~jtartan?)=KX8-43) “H LinesofHarecirclesperpendicular toandcenteredontheaxisofthe l dipole. “4 PROBLEMS Fig.38-12. This figure illustrates howtheintersection oftheKXsurface with the{6=/2changeswithtime.Thecurvesallliewithintheenvelopeshownas Plane9»Paidchangeswithti‘curvesallliewithin pe 38-1,(38.1)TheradiationfieldofalongwirecarryingastepcurrentFigure 38-13(a) shows alongwirethatcarries acurrent thatvaries asin Fig. 38-13(b). Beyond p:~crthereisnofield.Insidep=c(t~x)thefield +38.2.8 The Lines ofH isthatofasteady current. Intheshaded region, WehaveseeninSec.38.2.2 thatHhasonlya@component. Thusthe Anetals, linesofHarecircles perpendicular to,andcentered on,theaxisofthe 2a"p clectric dipole where Cisthelength ofthewire, asinProb. 37-1 (a)Calculate FandBintheshaded region. 38.3 SUMMARY (b)Show thatMaxwell's equations apply. (6)Sketch curves ofEandBasfunctions ofpatagiven instant,; betweenp=0andp=p;.Therearediscontinuities atp=p,andatp=p. Inthefieldofanoscillating electric dipole, because wehaveassumed thatd°//dris infinite at=Oandatt=. il{Row (Fk x } =P 154+)- S-1+j=) sino 6},(38-14) Crea 25+) 00s0#+(5 14/2)sin(38-14) a — , opel ogo elp]X P eo: lEee i aS(-14+j- 06, 38-21 asp SS fl n=28(-1+;) sino, (38-21) 1 oa fj Gp,=}Re(EXH*)=4.137x10°"PPIsig?(38-29) —*— 1} i r aed Hl =2386(+) “sin?@#—watts/meter™ (38-32) pra °*a re bt Farawayfromthedipole (>>A),Ehasonlya@component andHis Fig.38:13.azimuthal. The time-averaged Poynting vector iseverywhere radial. 7 70 RADIATION 1 PROBLEMS am 38-2. (38.2) Thedipole moment ofanoscillating charge 38-8. (38.2.4) Cosmological evolution Acharge Qoscillates along thez-axis, and2=z,,expjot. Consider aparticular class ofastronomical object lasars. Assume findtheequivalence inthefollowing way.Ifthecurrents arethesame, then Show that,ifisthenumber ofobjects whose radio-frequency fluxistheA’s are thesame. Then the V's are the same, from the Lorentz greater than #attheearth, thenaplotoflog.NagainstlogYshouldbea condition(Sec.37.1). ThentheE’sandH’sarethesame, Straight linewhose slopeis—1.5.Theslopeforquasars is,infact,larger. 38.3. (38.2.1) The three componentsoftheEfieldofanelectricdipole Thisispossiblyameasureofcosmological evolution, — Showthattheelectricfieldofanelectricdipolehasthreecomponents: 38-9,(38.«polarizationofsky one that dependsontepostionsofthecharges,onethatdependsontei raeumes sessateDrawsthshowingth 38-4.(3823)Theradiationpatternoftheelectricdipole polarized.Thelightisonlypartiallypolarizedbecauseitisveattered‘manyWhat fraction ofthe total power inthe field ofanelectric dipote is times. radiated within 45°oftheequatorial plane? 38.5. (38.2.3) The electric and magnetic energy densities inthe field ofan electric dipole Calculate the ratio ofthe time-averaged electric energy density tothe time-averaged magnetic energy density inthefield ofanelectric dipole (a) forri,(b)forr=X,and(c)forr>>4. 38-6. (38.2.3) The Poynting vector and theenergy density inthefield ofan electric dipole ‘Show that, for r>>,themagnitude ofthePoynting vector inthefield of anelectric dipole isequal totheenergy density multiplied byc. 38-7. (38.2.4) The light source paradox {Alllight sources should beblack, forthefollowing reason. Take thesun, forexample. Acone intheretina oftheeyecollects radiation emanating from avery large number ofatoms. These sources areincoherent, ‘Atany given instant there isanear-infnite number ofphasors inthe complex plane, allofdifferent magnitudes anddifferent phases, rotating at different velocities. Their vector sum isclearly zero. The same reasoning applies toanyobject, sayawhite wall illuminated with incoherent light. The radiation that reaches agiven eone comes from anarea that ialarge number ofwavelengths indiameter. There again, the netfield atthecone should bezero, and thewall should appear black. ‘Toexplain thisparadox, consider Nwaves ofasingle frequency andofa given linear polarization’ butofrandom amplitudes and phases. The hhumber Nis very large. For the i-th wave, E,= Eqexp(wt ~a)atthe cone,andthenetEisthesumoftheE,’s.Now theeyeissensitive, nottoEbuttoY,andthus toEE*, Show that ¥,,= 5... This means that thenetenergy flux isequal tothesum ofthe ‘energy fluxes oftheindividual waves." *You have probably noticed thatlaser light diffused byawallorasheet ofpaper hasa ranula structure. The pattern moves ifone moves one's head from side t0side. This phenomenon iscalled speckle. Iiswsed forstudying surfaces. Speckle arises inthefollowing way. Each point ontheobject, saythesheet ofpape. produces ontheretina adiffraction pattern whose shape and size depend ontheoptical these patterns andthefield varies from point topoint onthe retina tT 39.1 RADIATION FROM AHALE-WAVE ANTENNA, m3 This chapter ends our study ofelectromagnetic fields and waves. Obviously wehave notexhausted thesubject! Indeed, wehave done no chaprer39 ‘morethanestablishabasefromwhichyoucanexploreonyourown. RADIATION Il39.1 RADIATION FROM AHALF-WAVE ANTENNA .Figure391shows half-wave antenna connected toatransmitter TheHalf-Wave Antenna, Antenna Arrays,and throughaparallel-wire line.Thehalf-waveantennaisessentiallyapairof theMagnetic Dipole Antenna wires, each4/4long,fedwithacurrent /,,coswfatthejunction. HereA isthewavelength ofauniform plane wave inthemedium ofpropagation. Atshort wavelengths one canfold back alength 4/4oftheouter 39.1 RADIATION FROM AHALF-WAVE ANTENNA 713 conductor ofacoaxial line,asinFig.38-6toobtain ahalf-wave antenna, 39.1.1 THEELECTRIC FIELD STRENGTH E714 Roof antennas forautomobiles areonlyone-quarter wavelength long; 39.1.2 THEMAGNETIC FIELD STRENGTH 715 theother halfisareflection inthesheet metal oftheroof. Transmitting 39.1.3 THEPOYNTING VECTOREXH 716 antennas forAMwaves aresimilarly 49/4towers standing onconducting 39.1.4 THERADIATED POWER PAND THE RADIATION RESISTANCE 716 ground 392 ANTENNA ARRAYS 717 , ONE-HALF WAVELENGTH 717is 393 MAGNETIC DIPOLE RADIATION 720 1 39.3.1THEELECTRIC FIELDSTRENGTH E721 1lycosexpjot. (3941) 39.32 THE MAGNETIC FIELD STRENGTH M72239.33THEPOYNTING VECTOR, THERADIATED POWER,ANDTHE RADIATION RESISTANCE 723Each element oflength diradiates asanelectric dipole, 39.3.4 ELECTRIC AND MAGNETIC DIPOLE RADIATION COMPARED 724 394 THEELECTRIC DIPOLE ASARECEIVING ANTENNA 724 39.5THEMAGNETIC DIPOLE ASARECEIVING ANTENNA =724 396 SUMMARY 725ra PROBLEMS 726 Pye. = i Thehalf-wave antennaisalong,straightconductor, one-half wavelength \@\em :long,thatcarries astanding waveofcurrent. Itsradiation pattern is —. aan \ tacossimilartothatofanelectricdipole.However, foragivencurrent,it one -7m~~radiates much moreenergy. Thisisthebuilding block forassembling eee Prd arrays ofantennas. Wededuce itsfieldfrom thatofanelectric dipole. aa a ‘Thedirectivity ofahalf-wave antenna ishardly better thanthatofan oa electric dipole. However, arrays ofsuchantennas, withtheproper Ler™ spacings andtheproper phases, canbehighly directive. Some arrays~ comprise afewantennas, butothers comprise thousands.Wealsocalculate EandBinthefieldofamagnetic dipole,andwe Fig.39-1.Half-wave antenna, Thebrokenlineshowsthestanding waveofdiscuss brieflyelectric andmagnetic dipoles asreceiving antennas. Correntatcoset=1. va 16 RADIATION tt Prone 7 P tatthetime¢—r/eandPistheradiated 39-2.(38.1.4)Theelectricfieldofaradioantenna sareaeeerei aeaadfentistheradia CalculateEatadistanceof1kilometerintheequatorial planeofa power.SeeFig. , : half-waveradioantennaradiating1kilowattofpower.Set4<1kilometer. ‘Theradiation resistance is73.08ohms. 39-3.(39.2) Theimageofahalf-wave antennaArrays ofantennas canbemoredirective thansingle antennas. ‘Anantenna isnormally situated nearaconductor (theearth,anairborne Inthefieldonanoscillating magnetic dipole, vehicle, asatellite, etc.).Energy radiated toward theconductor isreflected, andthetotalfieldisthusthevectorsumofthedirectwaveplusthe gaol) (1-j4)sin06, (39-31) reflectedwave.Itisconvenient toconsiderthatthelatterisgenerated,not aan I, , byreflection, butbyanimage oftheantenna located behind thesurface of fm) (of woo theconductor.ml (8A 1B Dei ' (a)Show thatthecurrent intheimage ofahorizontal half-wave antenna asiar [2(jet/;eosors (1++/7) sino},0934 andthecurrentintheantennaflowinoppositedirections. i (b)Show thatthecurrent intheimage ofavertical half-wave antenna 9,=foMimsdsin2gp (39-37) andthatintheantennaflowinthesamedirection. ar Both rules apply tooblique half-wave antennas : (6)Wehaveshown thattheradiation resistance ofahalf-wave antenna is_ o,f : 73.1ohms. Find theradiation resistance ofaquarter-wave antenna=4.603x107!sin? 8Fwatts/meter’, (39-38) perpendicular toaconducting plane. Snotmeagf* 39-4,(39.2)Theradiationpatternofalineararrayofhalf-wave antennas P=Sheeel (39-39) Alineararrayconsistsofparallelhalf-wave antennaslyinginaplane.Say3e thereareNantennas,uniformlyseparatedbyadistanceDandexcitedin- 2p phase.=3.856% 10-™mzftwatts (39-40) (a)Showthat,intheplaneperpendicular totheantennas, ay‘ sin((ND/24)cos6) =197.3(£) (Nlou)® watts, 39-43} Sn(ND2D 2s9)| 5) (Mn)?wats 6543) Ein(DI2) cos@) ; ; where @istheangle between thedirection ofobservation andtheplane ofifm=Niza®, Nbeing thenumber ofturns andatheradius oftheloop.. hearray.Thebes histosum E 0 aeneoercrkyNtchews tearyTheetapproach osmteindiaphasorsgraphical Anelectric dipole receiving antenna feeding ahigh-resistance receiver (b)Find theangular positions oftheminima andmaxima ofE. generates avoltage equal tothetangential Emultiplied bythelength of Differentiation yields onlythemaxima. thedipole. Amagnetic dipolereceiving antenna undersimilarconditions (©)ShowthatforgivenpacingD,themainlobeat@=x/2becomessak narrowerasNincreases. generatesavotedA/d,whereAitheBakingfur.However,see FoeeeriesoyofFafonctionof@between0and34for -39.5. anarray of30parallel half-wave antennas that areinphase and spaced by WA (©)Now plot thesame function, using Cartesian coordinates, between 0 and180°withalogscalefortheE-anis. PROBLEMS (6)Explain why themain lobe istwice aswide asthetwo neighboring lobes.sonpatternsoftheelectric ; (g)Showthatitshalf-width(theanglebetweenthemaximumandthe 39-1.(321)|Theradiationpatternsoftheelectricdipoleandofthehalf-wave firstminimumononesideortheother)isapproximately equaltoA/I,NarsantinthefrSedofonsacsSiptleX=:71P*da6. whereVisthelengthofthearray. (b)Show that forthehalf-wave antenna 39.5. (39.3) The ditectivity ofanantenna Bydefinition, thedirectivity ofanantenna isequal totheratio ofthe "cos {(x/2)cos6} PoyntingvectoratthemaximumoftheradiationpatterntothePoynting Em7.0222 A086) vectoraveragedoverasphericalsurfacesurrounding theantenna Examples: One meter equals 100centimeters. One volt=10°electromagnetic units ofpotential a CGSSYSTEMSMULTIPLE PREFIX SYMBOL MULTIPLE PREFIX SYMBOL —rrr —_.—.-— ouantiry st ou ema 10"atto a 0 deka’ da a——wo femo of w ecto h Length meter 10"centimeters 10°centimeters7 . Mass kilogram 10°grams. 10°gramswe - , a Kilo k Time ‘second 1second 1second10" ano LJ 10° mega M Force newton 10°dynes 10°dynes.10° ‘micro “ 10" giga G Pressure pascal 10dynes/centimeter? 10dynes/centimeter™10? milli m 10" tera T Energy joule 10ergs 10ergsa " Power watt 10°ergs/second 107ergs/second 0i et nessa “ ol peta P Charge coulomb 3x10 wt 10deci a 10 ox E Electricpotential volt te 0 ,— Electric field strength volt/meter. ——_1/(3 10*) w Thisprefix fswritten décainFrench, Blectric flux coulomb 12xx10” 4xx107" Caution: thesymbol fortheprefixiswritten nexttothatfortheunit BiceGurdeonity soctomb/movar® 12x10 pedalwithoutadot.Forexample, mNstandsformillinewton, whilemNisa Polarization Sxlomb/mene’ 311 wemeter-newton, orajoule. Electriccurrent ampere 3x10 0Conductivity siemens/meter 910" wo" Resistance ohm 19:10") wConductance siemens 9x10" 10Capacitance fared 9x10" 10Magnetic flux weber ve 10°maxwell Magnetic uxdensity tesla 1/310") 10"gausses Magnetic fieldstrength ampere/meter 12% 10" 4x10 oersted Magnetomotance ampere axx10 442/10 gilbert Magnetization ampere/meter —1/(3x10") 0 Inductance henry 1/9x10") 0Reluctance ampere/weber 36x10" axxo? Note: We have set¢=3x10°meters/second. APPENDIX € 733 Thephase ofthewave isthequantity between brackets. Itisconstant forz=uyt.Hence u,isthephase velocity. APPENDIX Itisusually more convenient towrite 5 @=a,,Cos(wt~kz), (C3) where @ _2af_2x_1a2a2} (C4) Ifonedisturbs amedium insome way, thedisturbance travels outward as ‘Ye awave.Forexample, avibrating objectgenerates acoustic wavesinair. inthewavenumber, Notethatthisquantity is2times1/8,or2xtimes C.1PLANE SINUSOIDAL WAVES thewavenumber usedinoptics. Forthisreason, kiscalled thecircular wave number. Suppose thequantity @propagates atthevelocity v,inthe positive Thewavelength 2.isthedistance overwhich kzchanges by2.Theuppes quantity @propagates YUp Pos quantity X=4/(27),read“lambda bar,”isoftenmoreconvenient touse direction ofthez-axis. At z=0and forallxandy,itisgiven by :than 4.This istheradian length. e=e,expju. (cn Inphasor notation (Chap. 2), Then,atanyposition z, 4=Gyexpj(wt—kz) (cs) ; Ifthewave travels inthenegative direction ofthez-axis, za=a,c0s[o(¢-2)] (C2)& =ay,expj(wt+kz). (C6) Thisequation defines anunattenuated planesinusoidal wave.Thewave Ifthereisattenuation, thenthewaveamplitude decreases exponentially fronts aresurfaces ofuniform phase, atagiven time. Here, thewave withsand fronts arenormal tothez-axis. This wave isalso uniform because its amplitude a,isuniform overawavefront. a=aqexp(—az) expj(wt~Bz) (C7) ‘Atagiven z,@isasinusoidal function oft.Atagiven 1,@isa sinusoidal function ofz,asinFig.C-1 =a,expjlat—(B=ja)z]. (C-8) 4| a ‘The wave number isthen complex 2 An /\ k=B~ja. (c-9){\ ftalWoianaanaVY\ Notethenegativesign.Bothaandfarepositive: —tre! ¥ ' 1 1 @ o) Bay ans, (C10) Fig.C-l. Thequantity a=a,,c0s (wt~kz)asa function of=andasafunction wheretheatenuationdistance&isthedistanceoverwhichtheamplitude na avmewonec APPENDIX 735 _ avi At2,1thephases willbe,respectively, wr—kz and(w+ w)t— decreasesbyafactorof¢=2.71828.Thephasevelocity isthen (k+Ak)z.Thetwowavesareinphasewhen =e. (c-11) (Aw)=(Ak)z=0 (C-15) ofatpointssuchthat Weoftenrefertowaves traveling inaspecified direction. Thenthe atpol practiceistouseavectorwavenumber 2Ao cao)keke tktks, (C12) oak — Inother words, thepoints where thetwowaves areinphase travel atthe asinFig.C-2,and groupvelocity 4=a,,exp(wt—k-r) (C13) ay=2 (C-17) =a,,expj(wt—k,x—ky—k,2). (C-14) Ue‘Bk ( ‘Thevector wavenumber canbecomplex. Thesuperposition oftwowaves ofslightly different frequencies gives » ‘anamplitude-modulated wave. Thegroup velocity oftheenvelope. Remember that ‘ 1_2a_@ayoarats (C8) & z Ifu,isindependent ofw,thenkisproportional towandthephaseand A groupvelocitiesareequal: Ya x 4o_oladytaltek (C-19) Ifthephase velocity isfrequency-dependent, then thephase andgroup 4 velocities aredifferent. Inother words, thewavelets inFig,C-3travel either faster orslower than theenvelope, depending onthenature ofthe +2,Thevectorwi 110 paRelaegeelederery dointorome medium.(Forwavesatthesurfaceofwater,v,=2v,.)Inthelimit, Same value anywhere onagiven wave front. dw 1 Yak”did cm C.2 THE PHASE AND GROUP VELOCITIES Letussuperpose twoplanewavesofangularfrequencies oande+Aer Ifonedrawsacurveof«as.afunctionofk,itsslopeisequaltovg,while w/kisequaltovp. Seeas fggAKTheamplitudes areequalandthe Theaboveequation isexacteitherifonehasonlytwowavesofcircular 7 736 APPENDIX € APPENDIX € 737 frequencies wandw+Aworifkisalinear function of«.Inpractice, Foranattenuated wave, neither condition isstrictly true inmatter, and theabove definition ofthe groupvelocity isanapproximation. Afterawhilethegroupspreads out 8a, oaeeiccppears, Pang sathsr (C27) C.3 THE DIFFERENTIAL EQUATION FOR A andEqs. C-23 toC-25 apply. PLANE SINUSOIDAL WAVE C.STHE WAVE PROPAGATION OFA Youcaneasilycheck that,fortheaofEq.C-2, VECTOR QUANTITY @a_id'a e1 ‘a fe. (C-21) Avector quantity, such asanelectric field strength £,canalsopropagatedz’vu,dr asawave.Ifthewaveisuniform andplaneandifitpropagates inthe positive direction ofthez-axis, then ‘This isthedifferential equation foranunattenuated plane wave traveling alongthez-axis. E=E,, expj(wt~kz) (C28) Ifthere isattenuation, thenthedifferential equation is . 7 ;=(Emck +Ey9+Encd)expj(iot—kz). (C29) @a_ da, da aeSanta (C22) Thevector E,,maydepend onxandony,butitdoesnotdepend onz: theonlydependence onzand1appearsintheexponential function.The where wave number kcanbecomplex. ool ee (23)ao" o C.6THE NONHOMOGENEOUS WAVE EQUATION Inversely ; a Intheabsenceofattenuation, thenonhomogeneous waveequationisofol)" Wy ‘ ~ theformp=0(§) [(1+25)+1]. (C24) ~o(8)[uete)=a)” (C25) PaAEByey'.29, (30) a=o3) oF v3dt wherefisthedisturbance atthesource. Thusf=0outside thesource. C.4WAVE PROPAGATION IN Iffisnotafunction of1,thenthereisnowaveand THREE DIMENSIONS Va=f, (C31) More generally, the differential equation forany unattenuated wave propagating inspace atthephase velocity v,is which isPoisson's equation, and Lda Lffey'2')Va=tte (C-26) yzy=ef Day . oae (C26) a(x,y,2)4[ ee (C-32) 72 ANSWERS answers 743 24-1.(a)NaR*yoN’. (b)Noifitsdiameter ismuchsmaller thanthelengthof ORCthesolenoid. 27-6.(a)—~~ 4-3.3.410°’Rhenry ia >* 27-10,(a)OTsin6.()Sety=1,Vle=0.Thetovaluesagree.(@)Q 24-6.(a)R'=24,1=pena’.(b)(Ats)" x=wpioab, .ob” °4ae?) 28-4.(a)5.0x10°”meter,1.5x10°*meter.(b)2.5x10°’meter. 24-8.L=R,RyC, R=RyRi/R.. 28-7.250+6.3x10°9johms, 24-10,(a)6ohms,0.86radian,3.9ohms,4.6ohms.(b)65amperes, 76amperes Pp Ry mt(c)336microfarads. (d)65amperes. rel.@)OPE (144ine)G5#e.Gi)2650.Go)2,onv. 1 RoC 10 o :at) +j(wn-8°), (6)37ohms,57.5° (b)(i)2.010°"joule/meter. (ji)4.2x10°”joule/meter. 4.08(14esi) tHeREee+7) oR teeta (©)2.7x10*siemens, ~58°.(d)0.2watt.(e)No.(f)(i)Atall 29-3.Use6verticalplates,about6millimeters thickand6millimeters apart,frequencies. (ji)Never. (g)0 withatotalwidthofImeter 957,ZeitOM 29-6,5%10"meter;7.2x10°watts/meter,” Z+ +2am” c _ 22 29-9.(8)mes (b) yy 3.2 PaBN,woNxa?|2xa ona IP Gay Gabaer: OM 25.12.()SEP.(bySAEjo, (0)EC)LO, BB=FfWT)=Cole)z z10°meters/second, vj»=3.2x10°meters/second, v,=2.8%10*meters/ (ota?BN% igN?xa?|wsw?N?xa*0,b second.(c)2.8x10°meter,45. 2FS4fgMA(SE4OHAMONTeOot optFT) og Hees 29-11,(b)270parsecs. (©+/90ohums,91+/36obs. 30.2,FE~1and(£2)=2,exceptnear4,=9";LR a ; i im) 25.14, (a)£=—R_.. (b)14+ R2W°C% (0)770picofarads, 150milinenrys.OcriTRee (E22),~2080forany0,0" ampere-turns Fale Me 26°. 29% 10tampere-tumns, 309.(6)0.15 268,(9)2klowat-hoase/ mee(b)70atmoepheres 30413.(8)Eig=42Vols/metr,Hn=011ampere/meter, Egy, ©Howat . 6.0volts/meter, Hess=1.6X10-*ampere/meter, Eyj_,=36volts/meter, 26-12. (a)4atmospheres. (b)Thesame. Hrom=0.17ampere/meter. 5 oB,IxR°dB, 30-16.(a)————-, whereKisaconstant, Aisthealtitude, andRisth 26-16,(6)-$23,mB,(gM. RGR rae .radiusofheeat.(0)[1ga] 2-1.(e)1.210"meter. 5 5 31-2.(a)2=0.37,28= 1.1.(b)150°,74°, ed Cook «4_Eallok . 4i 2733.(a)<a=x),(0)B=BEany,A=SHE(a—xah. uis . Hsia») 31-8.(a)17°,(e)F=0.020,T~0.69,FT~1.4%. (0.Abovetheupperpate,“stot8lowtheupperplate, 32-4.(a)3.710*meter.(b)0.95.(c)13ohms/square.(A)3.1x10°%%y Ahasthesamemagnitude andtheoppositesign. 32-8.Eparalleltotheplaneofincidence and6,equaltotheBrewsterangle. ms ANSWERS BRS woo,SAF. (=exp(al) ze,Peepca), 6)?Vinb 7. (£) nt. x() INDEX 341. 3.0gigahertz. 343,58°,37. Accelerator, 208,280, 435 Capacitance, 108-110, 117,403 By" ‘Admitance, 134,152 Capacitor, paralle-pate. SeeParale-pate 34S.(a)Emustbeparalleltotheplates.(b)(1-3) Alternating currents, 32.127-139, 147, capacitor Ampere, 48,398 Cathode-ray tube, 588 of ‘Ampere’ circuitallaw,382,367,378,496 Causality, 287 347.(@)(LalK,)+elo. (0)=(Chk)?#1) “Ampere-turn, 353 Chambers, L.G.,513, €“Ampetian formulation, 494 Charge. SeealsoElectric eld S411.(a)3.1megawatts. (b)Pico=8.2kilowatts/meter, Pipase= Freaieaemmenbanaielead SemFT ae‘12kilowatts/meter. (c)130amperes,4.0kilovolts. ‘complex,$80 pasnteetinadielectric,186 la ofincidence,$55,81,589 invarianceof,288,290,306 362.am Antenna magnetic.SeeMonopole,magnetica= arrays,717,727 ‘Chargedensity. igawatts/meter’, (b)1.6x10°volts/meter. divectivity, 727 atanimterface,$0 36-7.(2)10gigawatts/meter’. (0)electricdipole.SeeElectricdipole, freeandbound,174,182,183 xoscilting inaconductor,75,402,408 372(144),6 hattwave,712-717,726 total,178,233image,727 ChargedparticleiaE,B,406 37-9.§(upKosR?) magneticdipole.SeeMagneticdipole, Child~Langmuir law,68.oscillating Grcit,letric ae.(a)“Mtn(SeKiet6 receiving,724 active,121aap)*aap Antirefection coatings,$75 bridge,146,167,450 Askin, A,3 <elta-sar transformation, 163 38-4,88%, ‘Attenuation distance, 524,S80,733 diferentiating, 143-144 7 integrating. 164-145, 222 39-2. 0,22 volt/meter ietcrating 44 39-7. (a)Faraway, 6.1x10°/If; closeby,4.3*10°/If. (b)Faraway, Bartlet, DF.499 meshmethod, 122 6.1x10"/If; elose by,8.6% 10'/If Battery, 82,151,167 nodemethod, 126 Bertoni,H.L.,587 passive,121 Yu13 Biot-Savart law,327,334,364 as son,pea Bohm,David.270 Re.24,135,138, 19-171me Bohrmagneton,344,373 Ry467 Bondi, H.,63, Sil Ry,RLC, 482-470 Boundary conditions symbols anddefinitions, 120 fora metallic waveguide, 618 {heorems, 149-167 forBandH,370 Gitcut,magnetic.SeeMagneticcircuit for BandJ,207 Gretalaw.SeeAmpére'scicuitallaw forV,D,andE,197 Circular frequency, 32,130 Brackets, useof,682,658,714 ‘Glock, timereadonarapidly moving, 255Branch,120 Coaxialin,60,190,477.48, 532,Brewsterangle,S6S~S67,574-575,589 619-621,624,645Bridge circuit, 146,167.450 Coeticient ofcoupling, 46,467-368 746 Noe Ispex 77 Coeteive force, 376 polarization, 176 Eddy currents, 425, 432-433, 436, 461, 468 impresence ofdiclectrics, 204, 209-210 Callimator paradox, 264 Current density, 67-72, 329, 676 Edwards, 7.C.,621 fonconductors, 110, 114, 209-210 Comet tails, 603,607 displacement, 186,499 Engenvalucs, 29 Electric polarization, 173 ‘Complementary solution, 125 ‘equivalent, 313,363,373 Einstein, Albert, 254 Electric potential V,46-49, $4,89‘Complexnumbers,30-38 four-current density,285 Electret,17,180,208,379 atthesurfaceofastar,62 Conductance,138 polarization, 175,313 Electriccircuit.SeCircuit,electric average,overasphericalsurface,S4 Conduction, 69,72-77, 80,82 total, 316 Electric dipole, 84-87, 93,98,173,S01 sradient, 47,316,424current,287,306 (Currentsource,120,152 ‘oscillating, 681-684,690,697-708, Poisson'sequationfor.65.189,198,233,Conduction electrons,60,72-82,310 Carveplotiers,141 70 Electricsusceptibility. 181 driftvelocity, 72,80,300 (Cyclotronfrequency,405406 Electricdisplacement, 186.SeeasoElectric Electrolytic tankforplottingmagneticConductivity. 70 fluxdensity D fick, 386 sg£0und, anomalies in,160 9,525,537 Electric energy, 101-107, 118,488-489) Electromagnet, 383,450 ofaplasma,S42 D’Alembertian, 290 associatedwithpolarization, 203 Electromagnetic momentum, 602,604,609Conductor, 64,77.SeealsoElectriceld: Decibel,640,645, inawave,$21,523,$27,$39 Electromagnetic potentials,314-322,324Magnetic field Deloperator, 4,6, 19 interms ofE,105, 107 retarded, 680-686, 689clectticfieldatsurfaceof,110 Delta-startransformation, 163-165,467 intermsofEandD,202 Electromagnetic waves,S14-729electricforeeon,110 Depthofpenetration, $07,$38.Seealso intermsofpandV,101,200 ‘guided.SeeGuidedwaves 00d, 500 ‘Skindepth ofacharged conducting sphere, 105 inageneral medium, 15-520 hollow, 77 Diamagnetism, 361, ofacontinuous charge distribution, 104, inconductors, 524, $26, S31, 537-542 Conjugate, complex, 31 Dielectric constant. SeePermitivity, 16 infreespace, 520-522Conservation lawsforcollidingparticles, relative Blectrcfield inplasmas,542-549276,281 Diclectric strength, 207,684 atthesurface ofaconductor, 110 innonconductors orpoorconductors, Conservation ofcharge,64,69,290,314,Dielectrics, 172-210 averageEoverasphericalvolume,56 $22,526,31 50,676 andmagnetic materials compared, 366 average Vover aspherical surface, $4 nonuniform, 579-S81, 391 Constant ofintegration, 125 anisotropic, 195 ‘ofacharge embedded indielectric, 186 polarization of,$17,527 Constants, physical. Seeinside theback artificial, 643, ‘ofaconducting cylinder inauniform &, propagating in.astraight line,611-618cover lossy,193 235 spectrum,515‘Contact potential, 77 ontiomogencous, 190, ‘ofaconducting sphere inauniform standing, 598Continuity, conditions.SeeBoundary nonlinear,197,207 228 TE,614conditions polarization of,173 ‘ofadielectric sphereinauniformE,231TEM,615-618,624‘Continuous creation theory, S11, $13 Digitalto-analog (D/A) conversion, 142 ofapolarized dielectric, 176, 189 uniform, $15, 520, 522Coordinates Diode,vacuum,66-69 ‘ofasphericalcharge,52,65,24 Electromotance, 413,26 Cartesian, 2,19 Dipole. SeeElectric dipole, Magnetic dipole ofanatomic nucleus, 236 Electron cylindrical, 17-24 Dirac, P.A,M.,236,327 ofancleiric dipole, SeeElectric dipole drift velocity inside aconductor, 72,80 orthogonal curvilinear, 15-17 Direction cosines, 4 ‘ofmacroscopic bodies, 49 emission, 66-69spherical,18-24,225 Displacement currentdensity,186,499 ElectriciidstrengthE,44,49,56,63,89, mass,effective,74,408Corte, P.R.,499 Divergence. 8,20 293,204,430,6865-657, 690-691 pairformation, 275 Coulomb, 43,398 four-dimensional, 289 and thew%Bfield.427 Energy.SeealsoElectricenergy:Magnetic Coulomb'slaw,42,51,20 ofA,351,676 curlof,46,311,314,420,430 eneray Coupling,coefficientof,446,467-468 ofB,310,314,333,364,498 divergenceof.51,319 kinetic,26 Crab nebula, 405 ofD,179 fluxof.0 relativistic, 273-275, 294 Crack detector,469 ofE,151,309,314,498 imtermsofVandA,316,425 rest,74 Curl,1,22,20 ofH,373 induced,13,420 Encraystorage ofA,333 ofJ69,$00,676 insideadielectric,176 electric,27,489 ofB,312-314, 352 Divergence theorem. 9 integrals for,234.236 magnetic, 472,49) ofE,48,311,314,420,499 Domain,electric,196 lineimtegralof,46,198,430,495 Energytheorem,487 ofH,365,98 Domain, magnetic, 374 maximum, inair,118, Equations, differential, 33,38, 124Current Dopplereffect,257,262.281,283 Poisson'sequationfor,233 Equations,sixkey,48)conduction, 69,306 Duality, S01 transformation of,301 Equipotential surface, 90,86,230,232, 318,convection, 306 Ducting,577 ElectricfuxdensityD,179 323displacement, 186,499 Duinker. Simon, 154 divergence of,179 Equivalent currents, 363,373, eddy, 425, 432-433, 436 lines of,232 equivalent, 313,363,373 Earnshaw’s theorem, $4 Electric force. 61,293 Fabry equation, M5,imesh,122 Earth'selectriccharge.60,212 andlinesofB,112.117 Farad,108 750 INDEX INDEX 751 Magnetic field(cont) Magnetic pressure, $83-484, 490‘talongcylindrical conductor, 98,330, Magnetic separation, 490 Mossbauer effec,283 Planaropticalwaveguide, 646-674335,353388 Mapectsshorter,490 Motiontransducer,145 dispersionrelation,669 ‘ofamagneticdipole.SeeMagneticdipoleMagneticsusceptibility. 368 Motor,410,534 ‘cigenvalue equation,657,662-666 (ofamagnetizedrod.364,377-380 Magnetictorque,485,490 omopotar,399-403,411 fieldcomponents,650-686,661,646-668ofaMaxwellpair,346 Magnetizationcarve,309.Seealso Multipoes,electric,88-97 fieldenergy,670,673 ‘ofamovingcharge,291-301,S12 Hysteresis, ferromagnetic modaldispersion, 672‘ofarotatingdiskofcharge,691 Magnetization M.361,373 Neper,645 ‘modeorder,687,672(ofasolenoid,331,346,$49,355,389,367Magnetohydrodynamic (MHD)generator, ‘Neumannequation,437 ‘numericalaperture,640,672‘ofaspinningchargedsphere,47 389-392 Nitrobenzene, 195 ‘haseandgroupvelocities,668-670,673‘ofatoroidal coil,358 Magnetometer, 434.436 Nodemethod, 120,126 Phaseshiftsontotalreflection, 656fofHelmholtzcoils,$46 Magnetomotance, 367.381 Norton'stheorem,152,167,168 transmitted power,671,674‘ofmagnetizedmaterial,342,373 Magnetoressior, 409 Nucleus,atomic,61 Planck'sconstant,ofparallelcurrents,336,383 Marsden,J.E.,18 Number,binary,142 Planeofincidence,$57ofsaddlecoils,343 Mass Plasma,$42-$83,569oftheearth,406,32 elective,74 Observer,239 frequency,$44,547rotating, 345 inertial, 281 ‘Octupote, electric, 89 Pointcharges, 104transformation of,301 relativistic, 20 ‘Oh’ law,70,134, 453, Pointfunction, 3,5uniform, $17 rest,270 ‘Omega-beta diagram, 552,669 Poisson's equation, 737 “Magnetic field strengthM,365-371 Maxwellbridge,450 forB,357 cutof,368 Maxwell, JamesClerk, 492,499 forB,233divergence of,373 ‘Maxwell's equations, 178,308-314, 421, Panofsky, Wolfgang, K.H.,241 forV,65,178,189,233lineintegralof,367 492-500,S11.Seealsoinsidetheback Paradoxoftheperpendicularcapacitors, Polardiclecrics,173 Magneticfix,312,329 cover xa Polarization, electric,173compression, 450 varies,509,512 Parallelplatecapacitor,109,114,118,509,currentdensity,176,313 leakage,381-382 linearity,498 532,545 Polarization, wave.SeeElectromagneticlinkage,349,421 tedundvney, 311-313,$00 diclectrc-nsulated, 184,187,190,193, waves,polarization ofMagnetic uxdensityB,293,294,326, McAlister, S.P.,393 210,203,205 Polarizer, 574686-687,50-691 MeCunig,Malealm,383 magneticfieldincircular,S10 Polanaingangle,Seaverageoverasphere,399 MeKinnon,W.R393 moving,303,321,323 Poles,magnetic,379 cuof,312,400 ‘Meshmethod,120,122 Paramagnetiom, 361 Polyvinylidene fluoride,177divergenceof,10,314,333,364 Metaldetector,$69 Particularsoltion,125 Port,135lineintegral of,352,367,78,430 Metalic pass,377 Peaking strip,434 Positive logic,142linesof,295,329,356,371,378 Mho,70 Penfield,Paul154,307,493, Positron,236,275saturation, 369 Microstripline,621,625 Permalloy,44 Potential ‘surface integralof,333 Millikan,R.A.,63, Permeability,368-370 lectric.SeeElectricpotentialV ‘Magnetic force,293-294,300,326,388,SeeMillan's theorem,153,168 Offreespace,294,328,398,520 four-potential, 319,320,323 also Lorenta force Minkowski diagram,248,250 Permeance,382 lineintegralofvector,388 and linesofB,482 Minkowskiformulation, $95 Permittivity, 181 scalar,316,33between coaxial solenoids, 481 Minnaert, M.,703 ‘complex relative, 19,525 scalarmagnetic, 386‘betweenlongparallelcurrents,397,410Mobility, 73,82,408 frequencyandtemperature dependence, vector.SeeVectorpotential between particles,410 Models,507 195 Potentialdivider,141,145 betweentwoelectriccurrents,396,478,Modesofpropagation,630,663 offreespace,43,398,520 Potentialenergy.SeeElectricencray 481,489 Momentum, 270,282 relative, 75,181,198, 544 Potentials, electromagnetic, 314-322, 324insideferromagnetic materials,392 electromagnetic, 2,604,609 Table,182 retarded,680-686,689 ‘onamagneticdipole,491 fiuxof,02 Perturbation method,638 Potentiometer, 41 fonavolume distibutionofcurrent,398four-momentum, 271-273,276,287, Phaseangle,32 Powderediron,425 onawire,393 0 Phaseshifter. 146 Powerwithin anisolated circuit, 485,489 relativistic, 270 Phasors, 32-40 factor, 138,195,445,450,451 Magnetic induction. SeeMagnet fux Monopole, electric,93 Philips,Melba,241 inaltenating-current circuits,138 densityB Monopole,magnetic,310,327,343.498, Photomultiplier, 406 Power-transfer theorem,161 Magnetic materials, 361-384 S10 Photon, 244,263,277 Poynting theorem, 528Magneticmonopole,310,27,343,498,$10Moon,116,$67 Photon-dragdetector,613,607 Poyatingvector,$19,S21~S23,$32-835, Pincheffect,405.411 39) L INDEX, 753 12oex Sun,344,603 ofBandB,301 Precipitation,electrostatic,62 Relaxationtime,75528 Superconductvity, $07 ofM,307 Pressure eee ‘Superposition, principleof,45,48,149,329ofMaxwellsequations,313, ‘magnetic,483484,490 Remanence, ¥ ' Sutface ofP,307 radiation,99-603,607 Resitanee,7 ! integralofB.SeeGauss’slawforB‘ofquad,289 Propagation constant517 fadlation,703,716,723 integralofB.SeeGauss’slawfor ‘ofthefieldofafongsolenoid,328 Proton beam,61,533, ees |Surface,orientable,14 ‘ofthefieldofaparallel-plate capacitor, Pulsar,533 eassiey, 8iit Susceptance, 135 303,321,323 Pump, electromagnetic,407 Resistivity 8,435,SeealiConductivity HSusceptibility ‘ofthepotentials,319,324 Purcell,E.,174 Resistojet, ' electric,181 ‘ofthespaceandtimepartialderivatives, Resonace,453-456 cet rhsRateativity 0e | Synchrotron radaton, 722,728 ‘ofthespeedoflight,244,260 ofacircuit,454 Rigrhandsew2ie,36,131 ova ‘Quad,289 .OFroot-mean square,Yale, I ofvisiblelightintohigh-energyradiation, ‘Quadrupole, electric,87,89,4-95,99Rowlandring,375.42 \‘Tamir,T.587 soylenin habrencray Rutherfordexperiment, Taylor, EdwinF.,239,242 Transformer, 457,459,469,S35 on “ellegen'stheorem,154-157,168 auto,463 Radar, James,254 curren,49 Radianlength,517,733 Sande,Maanon4 as ‘eal,ota Radiation, 675-729 darpatential.SeeElectricpotential ‘Thermalagitation, 72 ‘magnetic-core, 377,425,460,468 byanacceleratedcharge,692-697 SearpotentSeeElecpote “Thevenin'stheorem,150,167,391 powertransferthrough3,464 byanoscillatingcharge,710 eeeneeledetector,92 Tides,432 Transient,126 electricdipole,697-708 ‘Scatitetion vice,38/ Time ‘Transmission lines,147,489,615,617.See magneticdipole,684-686,720-724 Seicontoore, 70,38,389 constant,125 ‘4lsoCoaxialline;Hollowrectangular pattcra,720-721 Pamearmnpradicvenetrs t dilation,254,257 ‘waveguide; Microstripline;Planar reste,3 Seperationofae inserproper.25 palwevegnce Radiationpressure,599-603,607 7cat,ofrapidlymovingobject, . proper,254,258 ‘Travel,interstellar,282 Rasett,F.,392 ‘Shape,apparent teadonamovingclock,285 Tromba,A.J.,14 Rayequation$73 shield77,450 ‘Transformation,281 Takada,Masanobu,81 Reoperator, 7 ni ailean,240,283 “Twinparador,262 Reactance,134 Si(SystémeInternational) unis,730, { Gallen,0.20 paReciprocity theorems,117,157-161, 7 ! coven aa168-109,.39 rere soa,70 ofacurrentdensity,287 Uniquenesstheorem,211,222,231 Redshift,261,281 Semeas,EW.von, ofaforce,276,28. Units,730.731Resfae29.28 sone Fane atNeer,20.27 ssce Stimdept,32453,538,540 ofafowrmomeatem, 273Brewster'sangle,$65-S67,ST4-S7S,S89Skineffect,359,443,488,510,S37 offourpotential: $2,324 VanBlade,Jean,26 cocticensof,567-569 Sait3.355 |ofaleagih,251 VandeGraifaccelerator,61,388 ‘conductors,394-608 Sociaw.557 ofamagneticlux,429 VanderPauw,169 Fresnel’sequations,$58-S6S,573 SolarwindSS ‘ofamagneticforce,324 Variables,separationof,216,226 m Sourscaretdensityine3,3 smedenny,280 Veco,2-28,267SeaioFourvector lewofeflestion,$57 ee ste " ‘ofamutualinductance,458 <efinitions,identities,andtheorems.Seesonconductors, Speceshart. insidethefrontcover Space-time diagram,244-287,289,266 ‘ofarelativepermittivity, 307 insidethef laneofincidence,$57 we ‘ofatimeinterval,254,258 oneraon,3-24 n7 ofavelocity,259 rotating, Fadiationpressure,$99-603,607 Specie. 710 ofanangle,209 unit,216,17,18,27 rayequation,573 ‘Spence, onicfunctions,215 ofanelectricchargedensity,287 ‘Vectorpotential,314,323,333,347-349,Snell'slaw,$57,570 Spherea 7 ofanelectromotance, 429 47 total reflection, SB1-$93,656 aan ofanclementofarea,253,261 divergence,351,676 Relay,208325 Stokestheo oanelementofwoe,385 Laplacian,31,67 raland , reer lineintegralof, Lenten 7 Substitutiontheorem,150 ofanequation,495 lineintegraof348 Vector definitions, identities, and theorems Definitions Identities Rectangular coordinates rae: 1.(AXB)-C=A-(BXC)oN! 2.AXx(BXC)=B(A-C)~C(A-B) 1v=LerFenLe I )=BAC) CAB)arttay?*3s [os 3.Vue)=/¥e+e¥f 2peaaOdeAy,As 4.(a/b)=(1/b)Fa—(a/b*)¥Ooxbyoz 5.V(A-B)=(B- V)A+(A- V)B+Bx(VXA)+A x(PXB) ‘8A,_9A,’ 9A,_9A,\,|(dAy_9A,), 6.V-(fA)=(Pf)-A+f(P-A) 3.vxA=(SA:2Ar)gy(SASA),(darAs (GS (ee GeSe 1F-(AXB)=B-(PXA)~A-(PXB) 4.wpaSE,oF 8.(FWY=FFBe"By*Be 9.Px(FA)=0 5.WA=V'A8 +04,9+FA,2= O(P-A)— PX(PXA) 10.V-(WxA)=0 Cylindrical coordinates ILWx(A)=(FA) KA+f(P XA) Ff,15,Ff, 12.0x(AXB)=(B- V)A~(A+ VB+(V-B)A~(P- AB. 6=35b+5599*a2? 13.Vx(PXA)=P(P+A)—PPA(Sec.1.11.6) 13 1A,,3A, 3B, 3B, 3B.).-A=-— Sey 14.(A+V)B=| A,"+A,+A,TVART5PA)+og*az (-v)[oxAroyTAP3e| 3B, 3B, 3B,(124:2Ae)y,(Bde2A.)172(yyydele +[a2+4,Besa,By 9.vp=t2(pe)AL62h +[ABa,BsBe p30\Pap)*pag?”dz ©”* : 15.¥'(1/r)=F/r*. Thisisthegradientcalculated at(x’,y’,2"),andristhe 10.V7A= ¥(V-A)— VX(VXA)(Sec.1.11.6), veetor rpointing from(x’,y’,2")to(x,y,2). Spherical coordinates elUn=—F/P.Thisisthegradient calculated at(x,y,2)withthesame unwee 5 17.s@=4$crX dl,wherethesurfaceofareaafisplane.Thevectorrextends or 30" rsind 3g from anarbitrary origin toapoint onthecurve Cthat bounds . 13 13 1_3A 18.f.Wfdu=Sufdst 12,VA=52(PA)+——=(Aysin8)+ Se 19.f\(WXA) dv=—JyAXdsf, whereofistheareaoftheclosedsurface Po rin 036 rain830 thatboundsthevolumew apa _OAs), Lf19A,ary) 20.$efl=~J..¥fXdefwhereCistheclosedcurvethatboundstheopen 8vxralgueto FelsLaaae oe| sacfate . ame ‘Theorems :13/,9 1a of) 1of 4.v= 42 (PD) 454? (sinoZ)+4 St 1,Thedivergence theorem. J,A-dof=,V-Advwhereofistheareaof 15, A= (PA)—PXPXA(See.1.11.6) 2.Stokes'stheorem:Go-dl=f.o(VA)+d. Physical constants’ Maxwell’s equations forstationary media Se A.Differential formwithE,B,P,MElementary charge =1602177x10-"C Electronrestmass m=9.10938x10°"kg pep OP pypyBio, = 5.10999x10°eV © ot Proton rest mass im,=1.67262x10°?’kycpswnsetoe V-B=0, vxBhEau(y+2svxM). Speed oflightinvacuum. ¢=2.99792458 x10°m/s Permittivity ofvacuum €)=8.85418782 x10°"F/m B.Integral form ‘L/4x€o =8.9875518 10°m/F 1Permeability ofvacuum uo4xx10-7H/m [e-at=2 [(o,-7-PyawAvogadroconstant Ny=6.0221310°mot!Boltzmann constant =1.3806«10-™9/K [B-ast =o, Planck constant h=6.62607 x10“J+s“ dh =h/(2x)=1.08457210-5 [7xe)-aet =fB-at=-} Gravitationalconstant G=6,67210-""N-m'/kg? Massofthesun 1.98596x10”kg [crxp)-act=f peatnof(+24vxm+eX)datRadius ofthe sun 6.965 x10"m “ © i) o Mean sun-earth distance 1.495 x10" m C.Differential form with E,D,H,B Earth's mean orbital speed 2.98x10"m/s Massoftheearth 5.974x10"kg VD=py, vxE+Boo, Radiusoftheearth 6.378x10'm a Massofthemoon 7.3305x10kg Re 9D_ Radiusofthemoon 1.74x10'm PBR0PRM Ie Moon-earth distance 5.84395%10"km D.Sinusoidal fieldswithE,D,H,Bandforlinearmedia *CodataBulletin,November1986. V-cE=p, VXE+jouH =0 VewH=0, VXH—jweeoE=J.