09 向量势

109TheVectorPotentialUptonow,wehavemetthedescriptionofthemagneticfieldbytwodifferentialequations.Thesetwoequationsarediv0B(9.1)4curlcBj.(9.2)Ifthecurrentdensityjvanishes,equation(9.2)demonstratesthatthemagneticfieldcanberepresentedasthegradientofascalarpotential:gradMB(9.3)sincecurlgrad()isalwaysequaltozero.Fromequation(9.1)weconcludethatthemagneticfieldBmayberepresentedaswellintheform(0)curlBA.(9.4)ThequantityAiscalledavectorpotential.Ascalarpotentialisdeterminedalwaysonlyuptoanadditiveconstant.Thevectorpotentialisdeterminedonlyuptothegradientofascalarfunction,sincewecanalwaysperformthegaugetransformation:gradAAwithoutchangingthemagneticfield.Thisissimilartothecaseoftheelectricpoten-tial,whichisalsodeterminedonlyuptoaconstant.ButtheelectricfieldintensityisalwaysthesamewhetherthepotentialisorC.Toachieveuniquenessofthevectorpotential,weneedanadditionalconstraint.Wechoosediv0A.ThisiscalledtheCoulombgauge.Thegeneralproblemsconnectedwiththegaugewillbediscussedinmoredetailinanothercontext(Chapter20).Withthevectorequation2()()BAAAafurtherrelationshipbetweenthemagneticfieldandthevectorpotentialcanbederived.DuetotheCoulombgaugethefirsttermontheright-handsidevanishes,andweobtaincurlBA.(9.5)Withequation(9.2)wehaveforthevectorpotential4cAj.(9.6)Inequation(8.15)themagneticfieldwasrepresentedintermsofthecurlofavectorfield.Hence,1()()||dVcjrArrr.(9.7)Also,0A.Thisfollowsdirectlyfromequation(9.7)byapplyingGausstheorem.ThisexpressionforthevectorpotentialcorrespondsalsotoasolutionofPoisson'sequation(9.6)encounteredanalogouslyforthescalarpotentialofelectrostatics(seeequations(1.31),(1.60),(1.61)).Therefore,theuniquenesstheoremsdiscussedinChapter2arealsovalid.1Example9.1:Thefieldofacurrent-carryingcircularconductorWeplaceaconductingloopinthe,xy-planeasinFigure9.1.Letthecentercoincidewiththeorigin.Forthecalculationwewillusesphericalcoordinates(,,)r.Then,thecurrenthasacomponentin-directiononly,asinFigure9.2.Forthecurrentdensity,wewrite(cos)()()rajIajree(9.8)whereaistheradiusofthecircularconductor.Thefactor1acomesfromthevolumeelementinsphericalcoordinates.Notethataccordingtoourconsiderationsonthepropertiesofthe-functioninChapter1,1(cos)sin22.So,wecanseethat2000()22jdVIddrardraI.Duetoequation(9.7),wecanwriteforthevectorpotential()()(sin,cos,0)11()||||jjdVdVccrerArrrrr.(9.9)Sincetheunitvectorsinsphericalcoordinatesarenotconstant,fortheintegrationitissuitabletogotoCartesiancoordinates.Wehaveplacedtheconductorinthe,xy-planesothatthecurrentdensityhasonly,xy-components.Becauseofequation(9.9),thesameholdsalsoforthevectorpotential,()1()||xxjAdVcrrrr()1()||yyjAdVcrrrr()0zAr.(9.10)Cartesiancoordinatesandsphericalcoordinatesarerelatedaccordingto()()sinxjjrr,()()cosyjjrr.Wearefreetochoosetheoriginoftheangle:setting0,then()sin1()0||xjAdVcrrrrandweneedtoconsideronlythey-component.Sincetheproblemiscylindricallysymmetric,fortheobservationpointwecansetalways0.Thus,()cos1()()||yjAAdVcrrrrr.(9.11)2Theunitvectorsrrandrrcanbeexpressedbytheanglesofthepolarcoordi-nates,(sincos,sinsin,cos)rrand(sincos,sinsin,cos)rr.Substitutingequation(9.8)into(9.11),then(with0)22(cos)()cos()2IraAdVacrrrrr222(cos)()cossin2(sinsincoscoscos)Irardrddacrrrrandaftertheintegrationoverthetwo-functions2220cos()2sincosIadAcrarar12222021sincoscosIaaadcrrr.Restrictingourdiscussiontolargedistancesfromtheconductingloop,ra,wemayexpandthesquareroot:22222012sin()1sincoscos2IaaaIaAdcrrrcrr.(9.12)ThetwoothercomponentsAandrAvanish(becauseofthespecialchoice0andequations(9.10)).Now,fromBAwecalculatethecomponentsoftheB-field.With0rAAweobtainforthecomponentsofthecurlinsphericalcoordinates,accordingtotheformulaknownfromthelecturesonmechanics(Chapter11),22sinsin11sinsinsin00sinrrrrrrrrrrrArArArAeeeeeeA.Hence,1(curl)(sin)sinrArA,1(curl)()rArrA,(curl)0A.Substitutingequation(9.12),then23cos2rIaBcr,23sinIaBcr,0B.(9.13)3Thisfieldisanalogoustothefieldoftheelectrostaticdipole(comparethisequationtoExercise1.4).SeeFigure9.3.Thecircularconductorcorrespondstoa(magnetic)dipolewithdipolemoment2Iamcwhere2aistheareaenclosedbytheconductor.Thisfactisofgeneralimportance,aswewillseeinChapter10.Fromthisexercisewelearnthataconductinglooprepresentsamagneticdipole,atleastatlargedistancefromtheloop.SeeFigure9.4.There,thefieldoftheloopisidenticalwiththefieldofabarmagnethavingthemomentm:3333()2cossinrrrmmrrrmeemBee.1Exercise9.2:MagneticfieldofachargedrotatingsphereAsphereofradiusRwithachargequniformlydistributedonitssurfaceisrota-tingaboutadiameterataconstantangularvelocityωasinFigure9.5.CalculatethevectorpotentialAandthemagneticinductioninsideandoutsidethesphere.SolutionThevectorpotentialisgivenby1()()||dVcjrArrr.Thecurrentdensityisgivenby2()()4qrRRjrvωrandthevectorpotentialis2()()4qcRArωFrwith()()||rRdVFrrrr.Sincetheintegraloverthesphereisrotationallysymmetricaboutthedirectionr,rFFe.Wetakether-directionasthez-axis;thenistheanglebetweenrandr.(SeeFigure9.6.)WithdVinsphericalcoordinatesandintegrationover,weget2220()()cos2sin||2cosrRzrRrFdVdrdrrrrrrr133222201cossin222cos2RddRrRrRrRrR424,,34,.3RrrRRrRrThus,thevectorpotentialis2223,,344,.3qrRqqcRFcRcRrqRrRcrωrrAωFωrωWehavecurlBA,andforaconstantangularvelocityω()[()()()()]33qqcRcRBωrωrrωrωωr[()()]3qcRωrωr,rR.Letusconsidertheexpression()ωrinmoredetail.Writtenincomponents,thethjcomponentofthistermreads()iijixx.The33matrix()ijijAxx2appearinginthisexpressioniswrittenfrequentlyindyadicform(withoutdotandwithoutcross!),()ijijjijiAxxr.Sincethedyad(atensoroftherank2)r1and3r,forrRwehave23qcRBω.ForrRweobtain2223333()()333qRqRqRcrcrrcrrrrrBωωωω2233531131()33qRqRcrrcrrωrωrωrrω253()rrmrrm.So,insidethespheretheB-fieldisuniformandproportionalto;thisisplausible.Intheexteriorthespherebehaveslikeadipolewiththemoment23qRcmω.ThisservesasasimplemodelforthemagneticfieldoftheEarth.1Exercise9.3:ThevectorpotentialofparallelconductorsCalculatethevectorpotentialoftwostraight,infinitelylongconductorscarryingsteadyantiparallelcurrentsI.Consideralsothecaseofparallelcurrents.Howcanawell-definedconvergentexpressionbeachieved?SolutionThevectorpotentialis1()()||dVcjrArrr.(9.14)Weutilizethesymmetriesofthearrangementandchoose(,,0)xyr,(,0,)dzr.WithzdVIdzje,equation(9.14)becomes2222221()||()()LLzLLIdzIdzdzccxdyzxdyzArerr.(9.15)Takingintoaccountthattheintegrandisanevenfunctionofz,andintroducingtheabbreviations221()xdy,222()xdy(9.16)wecanwrite222200122()LLzIdzdzczzAre.(9.17)Byacleversubstitutiontheintegralcanbecalculatedsimply.Setting1sinhzu,togetherwith22coshsinh1uu,theunpleasantrootexpressionvanishes,andweobtainfortheintegral112222210000111sinhsinhcoshsinhLLLLdududzduuzu221110arcsinhln1LzLL.(9.18)Substitutingthisinto(9.17),weobtain221222122()lnzLLIcLLAre.(9.19)TakingthelimitLweobtainthefollowingresultforthevectorpotentialofinfinitelylongconductors:212lim()lnzLIcAre.(9.20)Inthe,xy-planetheequipotentiallinesaredefinedby21const.Thevectorpotentialvanishesonthey-axis.Inthecaseofparallelcurrentsthe"minus"in(9.15)becomesa"plus."Withthisreplacement,from(9.18)and(9.19)weobtainforthevectorpotentialoftwoparallel,current-carryingconductors222212122()lnzLLLLIcAre.(9.21)2Obviously,thisexpressionisdivergentforinfinitelylongconductors(L).Werewritethisexpressionsothatthedivergentpartissplitoff:2222122212()()22()lnln(4)4zLLLLIILcLcAre.(9.20)Thelast,divergenttermhasnopositiondependence.Takingthecurlof()Ar,thistermplaysnorole.Hence,wemaysubtractthedivergenttermwithoutchangingsomethinginthephysicallyrelevantfieldintensity.Inphysics,withtheseorsimilarmethods,onecanfrequentlyderivephysicallyrelevantresultsfromformallydiver-gentexpressions.TakingthelimitLinthecorrectedexpression,thevectorpotentialofparallelcurrentsis1221lim()lnzLIcAre.(9.21)Now,theequipotentiallinesinthe,xy-planearedeterminedby12const.