Basic lawsof electromagnetismIf. E. I I P O ) ~ O BOCHOBIII>I E3AHOllbI9JIEKTPOi\lAI'HETH3:\IAM3AaTeJlbCTEO (BLICmaH umoaasMOCKBaI.E.IrodoYBasic lawsofelectromagnetismTranslated fromRussianbyNatasha Deineko and Ram\Vadhwalir Pnblisllers IBow',-CBSPUBLISHERS&. DISTRIBUTORS4596/1.. A, 11DaryaGanj, NewDelhi 110002(India)First puhlished Hevised fromthe Russian edition@J 1I3,naren he'1 BCl (i Rhl Cnraa III KOJla)), 1983EngIi 5ht fa n51 ali on. Mi r Pub lis hers,First Indian Edition :1994Reprint: 1996Reprint: 1998Reprint: 1999Reprint: 200 1Ired8YRasie L8""Thiseditionhas beenpublishedinIndiabyarrangement withMirPublishers, MoscowCopyright (g EnglishTranslation, Mir Publishers, 1986All rightsreserved. Nopart of this book maybe reproducedor trans-mitted inanyformorbyanymeans, electronic ormechanical, includ-ingphotocopying, recording, oranyinformationstorage andretrievalsystem without permission, inwriting, fromthepublisher.Publishedby:SatishKumar Jainfor CBSPublishers&Distributors,4596/1-A, 11 Darya Ganj, New Delhi.. 110002 (India)For saleinIndia onlyPrinted at :Nazia Printers, Delhi .. 110006ISBN: 81-239-0306-5PrefaceThe mainideabehind this bookistoamalgamate thede-scriptionof the basic concepts ofthe theoryand the practicalmethods of solvingproblems in one book. Therefore, eachchapter contains first a description of the theory of thesubject beingconsidered (illustrated byconcreteexamples)andthenaset of selectedproblems with solutions. The prob-lemsarecloselyrelatedtothe textandoften complement it.Hence theyshould be analysed together withthe text. Inauthor's opinion, the selected problems should enable thereader to attainadeeper understandingofmanyimportanttopics and "to"'visualize (evenwithout solving theproblemsbut just bygoingthroughthem) the wide rangeof applica-tions of the ideas presented in this book.Inorder toemphasizethemostimportantlawsofelectro-magnetism, and especially toclarify themost difficult top-ics, the author bas endeavouredtoexcludethelessimpor-tant topics. In anattempt todescribe the mainideasconcise-ly, clearlyand at thesame time correctly ~ the text has heenkept free fromsuperfluous mathematical formulas, andthemain stress hasheenlaidonthephysical aspectsof thephe-nomena. Withthesameend in view, variousmodel repre-sentations, simplifying factors. special cases, symmetry con-siderations, etc. have been employed wherever possible.SI units of measurements are used throughout the book.However, consideringthat the Gaussian systemof units isstill widelyused, wehaveincluded in Appendices 3and4the tables of conversion of the most important quantitiesand formulas from SI to Gaussian units.The most important statements and terms are given initalics. More compl icatedmaterial and problems involvingcumbersome mathematical calculations are set in breviertype. Thismaterial canbe omittedonfirst readingwithoutany loss of cont inuity. The brevier type is also used forproblems and examples.P r t ! 1 4 C ~'Thebookisintendedas atextbookfor undergraduate stu-dents specializinginphysics(in theframeworkofthecourseon general physics). It can also be used by universityteachers.Theauthor is grateful to Prof. A.A. Detlaf and ReaderL.N. Kaptsov who reviewed the manuscript and made anumber of valuablecomments ann suggestions.I. IrodovContents.Preface ....List of Notations . .t. Elfttrostatic Field in a Vaeuum .1.t. ElectricField .......1.2. TheGauss Theorem . . . . . . . . .1.3: Applicationsof theGauss Theorem . .1.4. Differential Formof the GaussTheorem1.5. Circulation of Vector E. Potential . . . .t.6. RelationBetweenPotential andVector E .....1.7. Electric Dipole . . . . . . .Problems .........2. ACondudor inan EleeteestatieField2. t. Field in a Substance . . . . . . 2.2. Fields Inside and Outside a Conductor . . .2.3. Forces Actingon theSurface of a Conductor .2.4. Properties of a Closed Conducting Shell .2.5. General Problem of Electrostatics. Image Method2.6. Capacitance. CapacitorsProblems .....3. Eleetrir Field in Dielectrics .3. t .Polarizationof Dielectrics .3.2. Polarization .3.3. Properties of the Field ofP .3.4. Vector D . . . . . . ... ..... ..3.5. BoundaryConditions . . . .3.6. Field ina HomogeneousDielectricProblems ...4. EDftIYof Electric Field . .4.t. ElectricEnergyof a Systemof Charges .....4.2. Energies of a Charged Conductor and a Charged'Capacitor .4.3 ...Energyof Electric Field . . . . .4.4. ASystemof 1'\\0 Charged Bodies4.5. ForcesActingina DielectricProblems .............510it1116192325303439454547495154576067677072768084869494991001051061tO85. Direct Current . . 5. t. Current Density. Continuity Equation . . .5.2. Ohm's Lawfor a Homogeneous Conductor 5.3. Generalized Ohm's Law . . . . . . . . .5:4. Branched Circuits. Kirchhoff's Laws .....5.5. Joule's ~ . "'w . . 5.6. Transient 1'1.icesses in a Capacitor Circuit .Problems ...........6. MepeticFieldinaVaeaum . .6.1. Lorentz Force. Field B .6.2. The Biot-Savart Law . . . . . . .6.3. Basic Laws of Magnetic Field . . .6.4. Applications of theTheoremon Circulationof Vec-tor B ................6.5. Differential Forms of Basic Laws of Magnetic Field6.6. Ampere'sForce . . . . . . . . . .6.7. Torque Acting on a Current Loop 6.8. \Vork Done upon Displacement of Current LoopProblems .....7. Magnetic Fteldin a Substance ....7.1. Magnetization of a Substance. Magnetization Vec-tor J ~ . . . . . . . . . . . . . .7.2u Circulation of Vector J . . . . . .7.3. Vector H ............7.4. Boundary Conditionsfor BandH . . .7.5. Field in a Homogeneous Magnetic7.6. Ferromagnetism ........Problems .8. RelativeNature ofElectricandMagnetic Fields .8.1. Electromagnetic Field. Charge Invariance . .8.2. Laws of Transformationfor FieldsEandB . 8.3. Corollaries of the Laws of Field Transformation8.4. Electromagnetic Field Invariants .Problems .9. Electromagnetic Induction .9.t. Faraday'sLaw of ElectromagneticInduction.. Lenz'sLaw .............9.2. Origin of Electromagnetic Induction .9.3. Self-induction ..9.4. Mutual Induction . . . . . . . . .9.5. MagneticField Energy . . . . . ..9.6. Magnetic Energyof Two Current Loops .lt6nefi9122126129t33136141t4tf4S. f48150154155i59161163172272f761781821851881921981982002062082092172172202262312342382402440thenEn >0, andhencevector Eis directedawayfromt.hechargedplane, asshown inFig. 1.7. Onthe other hand, if a (it shouldbenotedattheveryoutset thatthereisaone-to-one correspondence between the two methods). It will beshownthat thesecond methodhas anumber ofsignificantadvantages.The fact thatlineintegral (1.21)representing thework-ofthefieldforces done in thedisplacement of aunit positivecharge from point 1topoint 2 does not dependon thepathallowsusto!state that forelectric fieldthere exists a certainscalarfunction 1 and qJ2 are the values of the function cp at thepoints 1 and 2. The quantity =--: EdI. (1.24)In other words, if we knowthe field E(r}, then to find q>wemustrepresent Edl (withthe helpofappropriatetrans-formations)as adecrease ina certain function. This functionwill bethepotential (x, y) = -axy, where ais a constant; (2) z)" (1.33)where 2 arethepotentials at points1and2. Thismeans that therequiredworkisequal to thedecreaseinthepotentialenergyof thechargeq' upon its displacementfrompoint 1to2. Calculationoftheworkoftheheldforceswiththe help of formula (1.33) is not just very simple, but isin some cases the only possible resort.Example, ,\qis distributed over a thinringof radius a.Findtheworkof theheldforces dunein the displacement of apointcharge s' fromthe centreof tJ1eringtoi.alinity, .Sincethe distributionof the chargeq overthe ringisunknown, wecannot sayanythingdetini teabout the intensi ty Eof thefieldcreatedbythischarge. Thisrneausthat wecannot calculate theworkbyeval-uating the integralq'Edl in thiscascoThis problemcanheeasilysolved with the help of potential. indeed, since all elements of thering areat the samedistanceaIromthecentreof the ring,' the poten-tial ofthis point isq>o ;:.:= lJ/4it:':'ou.Andwe know that is thefield potential at the point of lo-cation of the charge q. Adipole is the system of two charges,andhenceitsenergyin an external field isW= q+dq>=4neoR J 4!oRoItshouldbenotedthat thisIntegral isevaluatedinthe! most simpleway if wetake into account that (cos!.where thefirst term is the potential ofthecharge C], while the secondis thepotential of the charges induced onthe surfaceof the sphere. But since aUinducedcharges are atthesamedistancea froruthepoint 0 andthe tolal induced chargeisequal tozero, cp'=Oas well. Thus, inthiscasethe potentialof the spherewillbe determinedonlybythe first terruill (1).Figure 2.2showsthefieldandthechargedistributionsfora systemconsisting of twoconductingspheres one of which(left) is charged. As aresult of electric induction,thechargesof the opposite sign appear on the surface of the rightuncharged sphere. The field_of these.charges will in turncause a redistribution of charges on thesurface of the leftsphere, and their surface distribution will become nonuni-form. Thesolidlinesinthefigure arethelinesof E, whilethedashedlinesshowtheintersectionof equipotential sur-faces withtheplane of thefigure. As \ve move awayfromthis~ system, the equipotentiaJ surfaces become closer_andcloser tospherical, and thefield lines becomecloser tora-dial.'The fielditself inthis case resembles more and more the'fieldof a point chargeq, viz. thetotal chargeof thegivensystem.The FieldNear aConductor Surface, \Ve shall show thattheelectricfield intensityintheimmediatevicinityof thesurface of a conductor is connected with the local chargedensityat theconductorsurfacethroughasimplerelation.This relationcanbeestablishedwiththehelpof theGausstheorem., Suppose that the region of the conductor surface we areinterestedinborders on a vacuum. Thefield lines are nor-mal totheconductorsurface, Hencefor. aclosedsurfaceweForcesActing onthe of II Conductor 49shall take a small cylinder and arrangeit as is showninFig. 2.3. Then the flux of E through this surface will beequal onlyto the flux throughthe "outer" endfaceof thecylinder- (the fluxes through the lateral surface and theinner endfaceareequal to zero). Thuswe obtain Eft =n.1:('Fig. 2.2 Fig. 2.3= CJAS/8otwhere En' is .the projection of [vector E ontotheoutwardnormal n(withrespect to theconductor), asislthecross-sectional areaof the cylinder andais the localsurface charge density of the conductor. Cancelling bothsides of thisexpression bywegetI En = a/eo- I (2.2)Ifa >0, thenEn >0, Le, vector Eis directed fromtheconductor surface (coincides indirection with the normal n).IfCJ whichsatisfiesEqs. (2.8) or (2;9) intheentire" space between the conductors andac-quires the givenvaluescl).Example, Let us represent graphically the fields Eand Dat theill i!.rface betweentwohomogeneousdielectrics1and2, assumingthat::t> F1an.lis JlO extraneouschargeon thissurface.Since F2 > ;"h in accordance with (3.25) ((2>((l (Fig. 3.10).Cousirlering that the tangential component of vector Eremains un-changed and using Fig. ;-J.9, we can easily showthat h'2 < inmagnitude, i.e. the lines of Eill dielectric 1 must be denser than indielectric2, as is shownin Fig. 3.10. The fact that the normal COm-3.5. Bound4ryConditions83pouents .o{vcc!.ors J) al:(\ equal leads to the conclusion that /)2>VI10 magnitude, r.e. the Iines of D must be denserindielectric2.FieldEField DFig 3.9 Fig. 3.10\Ve see that in the case under consideration, the lines of E arerefracted and undergo discontinuities (due to the presence of boundcharges), while-shelinesof D areonlyrefracted, without discontinu-ities (since there are no extraneous charges a t the interface).Boundary Condition "on the Conductor-Dielectric Inter-face. If medium1isaconductorand medium2 is a dielectric(see Fig. 3.8), it follows fromformula (3.23) that(3.26)wheren is theconductor'soutwardnormal (weomitted thesubscript 2since it isinessential in the givencase). Let usverify formula F ~ . 2 6 ) . In equilibrium, the electric fieldinside a conductor is E : . ; : ~ 0, and hence the polarizationp= O. Thismeans, according to (;3.17), that vector D=--: 0inside theconductor, i.e. in the notationsof Iormula(::).23)D1= 0 and IJl n=O. lIenee .D2n~ c.Bound Charge at theConduclor Surface. Ifahomogeneousdielectric adjoins a charged. region of the surface ofa COIl-ductor, boundcharges of acertaindensitya' appear at theconductor-dielectric interface (recall that the vol ume den-sityof hound charges p' === 0for ahomogeneous dielectric).Let usnowapply theGausstheoremtovectorEin thesameway as it was done whilederivingformula (2.2). Consider-ingthat there are both bound and extraneous charges (084 3. ElectricFieldinDielectric,and 0') at the couductor-dlelectrlc interface we arrive atthe following expression: En = (0+O')/EoOn the otherhand , according to (3.26) E,l =Dnleeo = 0/e80 Combin-ing these two equations, we obtain ale == (J --{-0', whenceI, B-1 IC1 == --e-o. (3.27)-I t can be seen that the surface density a' of the boundcharge in the dielectric is unambiguously connected withthe surface densitya of theextraneouschargeon thecon-ductor, thesignsof thesechargesbeingopposite.3.6. Fieldina HoMogeneous DielectricIt was noted in Sec. 2.1 that the determination of theresultant field Einasubstanceisassociatedwithconsider-abledifficulties, sincethedistributionofinducedchargesinthe substance is not known beforehand. It is only clearthat thedistributionof -thesechargesdependson thenatureandshape of thesubstanceas well as on theconfigurationof theexternal field Eo- .Consequently, inthe general case, whilesolving theprob-lemabout theresultant field Ein adielectric, we encounterserious difficulties: determination of the macroscopic fieldE' of boundcharges ineachspecificcaseis generallya com-plicated independent problem, since unfortunately thereisnouniversalformulafor findingE'_Anexceptionisthe case whenthe entire space where thereis a field Eois filledbyahomogeneousisotropicdielectric.Let us considerthiscaseingreaterdetail. Supposethat wehavea chargedconductor (or several conductors) ina va-cuum. Normally, extraneouschargesarelocatedon conduc-tors. As wealreadyknow, inequilibriumthe field Einsidethe conductor is zero, which corresponds to a certainunique distribution of the surface charge a. Let thefieldcreated in the space surrounding the conductor be Eo-Let us nowfill theentirespaceof thefield withahomo-geneous dielectric. Asaresult of polarization, onlysurfacebound charges (J will appear in this dielectricat theinter-face with the conductor. According to (3.27) the charges3.6. Field inGHomogeneouDielectric 850' are unambiguously connected withthe extraneous chargesoonthe surfaceofthe conductor.Asbefore, there willbenofieldinside the conductor (E == 0). This means that the distributionofsurfacecharges(extraneous charges 0 andbound charges 0') at the conductor-dielectric interface will be similar to the previous distri-bution of extraneous charges (0), and the configuration oftheresultant field Ein thedielectricwill remainthesameas in theabsence of the dielectric. Onlythemagnitudeofthefield at eachpoint will be different.Inaccordance with the Gauss theorem, 0'+ 0'=BoEn ,whereEn =Dn/ef!,o =(J188o, .andhencea +a' =a/e. (3.28)But if the charges creating the fieldhave decreased bya factorofe everywhere at theinterface,thefieldEitself has, becomelessthanthefield Eo bythesamefactor: ., ~ . ' E= EJe. (3.29)Multiplyingbothsidesofthisequationby EEo, weobtain~ D =Do, (3.30)i.e. thefield of vectorDdoes not changeinthiscase.Itturns outthatformulas (3.29) and(3.30) arealsovalidin a moregeneral casewhena homogeneous dielectricfillsthe volume enclosed between the equipotential surfacesof thefieldEo of extraneous charges(orofanexternalfield).Inthis ea. also E= Eo/e andD= Doinside thedielectric.Intheeasesindicatedabove, theintensityEof thefieldof bound charges is connected by a simple relation withthepolarizationPof thedielectric, namely,E' =-vt; (3.3t)Thisrelationcanbe easilyobtainedfromtheformula E ==Eo +E' if! we take into account that Eo =BE andP=xoE.As was mentioned above, in other cases the situationis muchmorecomplicated, and formulas'[ (3.29)-(3.31) areinapplicable.Corollaries. Thus, if a homogeneous dielectric fills theentire spaceoccupiedbyafield, theintensityEofthefield86 9. ElectricFIeldin Dieleetrteswill belowerthantheintensity Eo ofthefieldofthe sameextraneous charges, but in the absence of dielectric, byafactor of e. Henceit follows that potential q> at all pointswill alsodecreasebya factor of e:q> = CPol e, (3.32)whereCPo is the field potential inthe Jib s ce oftJae dielectric.The same appliesto the potential difference:U= Uo/e, (3.. 33)where U0 is the potential difference in a vacuum, intheabsence of dielectric.In the simplest ease, when a homogeneous dielectricfills theentirespacebetweenthe plates of a capacitor, thepotential differenceUbetweenitsplates will be bya factorof eless thanthat in the absence of dielectric (naturally,at thesamemagnitudeof thechargeqontheplates). Andsince it is so, the capacitance C =qlUof the capacitorfilled by dielectric .will increaseetimesC' =c. (3.34)whereCisthecapacitance of thecapacitor in. theabsenceof dielectric. I t shouldbe notedthat thisformula is val idwhen theentirespace betweentheplates is filled andedgeeffects are ignored.Problems 3.1. Polarization of a dielectric andtheboundcharge. An extra-neous point chargeq is at thecentre of a spherical layer of a heteroge-neousisotropicdielectricwhosedielectricconstant variesonlyintheradial direction as e =alr, where a. is a constant andr is the distancefromthecentre ofthesystem. Find thevolumedensityp' of a houndcharge asafunctionofrwithinthelayer.Solution, Weshall use Eq. (3.6), takingasphere of radius r astheclosedsurface, thecentreofthespherecoinciding withthecentreof thesystem. Then4nr2 Pr = -q'(r),where q' (r) is the bound charge inside the sphere. Let us take thedifferential of this expression:4", d(rtPr) = - dq", (1)Problems 87Here dq' is the boundcharge inathin layer betweenthespheres withradii randr + dr. Considering thatdq'=p'4nr2dr, wetransform(1)asfollows:r2dPr + 2rPrdr = -p'rsdr,whencep' = - (dPr-t-2-Pr) dr r(2)Inthecase under considerationwe havee-1 e-1 qPr=xeoEr == -e-Dr=-e- 4nr2and after certain transformation expression (2) will have the form-1 qp'=4na rtwhichisjust the requiredresultxE ~\ .,fJ\\JO ~ aEx fJ/-0 3.2. The Gauss theoremfor vector D. AninfInitely largeplatemade of a hsmogeneous dielectric with the dielectric constant e isuniformlychargedby an extraneouscharge withvolume densityp >o.The thicknessof the plateis 2a. (1)Findthemagnitude of vectorEandthepotentialcpasfunctionsofthedi-stance1fromthemiddle of theplate(assume thatthe potential is zero inthe middle oftheplate), choosing theX-axis perpendicular to the plate.Plot schematic curves- for the pro-jectionEx(x) of vector Eandthe po-tential (r) isalsoshowninthis figure. Thecurve q>(r) must havesuchashapethatthederivativeocpl8r taken with theopposite signcorresponds tothecurveofthefunctionE(r). Besides, we must take intoaccount thenormali-zation condition: cp ~ as r -+- 00.Itshouldbe noted that the curvecorresponding to the functioncp(r) is continuous. At the pointswhere the function E(r) has finitediscontinuities, the function of the conductor, supplying mentally a charge q toTt:00 b 00\( e 1 \ q I 1 \ qcp== E; dr.= -- -2- dr-t---- -2" dr. 4no ... t r 4nEo... ron' hIntegrating this expression, we obtain at the point oflocation of one of thecreated by the Held of all the 0 l hercharges, is q === 3q/4ncoa.lieneeTotal Energyof Interaction. If the charges are arrangedcontinuously, then, representing the systemof charges asa combination of elementary charges dq== r dV and goingover in (4.:1) fromsummation to integrat ion , we obtainw'.\ plpdV, I7-0181984. Energy of ElectricFieldwhere q> is the potential created by all the charges of thesystemin thevolumeelement dV. Asimilarexpressioncanbe written, forexample,for a surface distributionof charges.Forthispurpose, wemust replacein (4.4) pby {J and dV byas.Expression (4.4) can be erroneously interpreted (andthis often leads to misunderstanding) as a modification ofexpression (4.3), corresponding to the replacement of theconcept of point charges bythat ofacontinuouslydistrib-utedcharge. Actually, this is not so sincethetwoexpres-sions differ essentially. The origin of this difference is indifferent meaningsofthepotential q> appearingintheseex-pressions. Let us explain this difference with the help ofthe following example.]Suppose a system consists of two small balls havingcharges qlandq2. The. distancebetweentheballsisconsi-derablylarger than theirdimensions, henceql and ill canbeassumed to be point charges. Let us find theenergyWofthe given systembyusingbothformulas.Accordingto (4.3), we have1W =2(qt2_ -1:_ I_n 2 .- 2 -. 2C.Thesethreeexpressionsarewrittenassumingthat C== s!w.Energy of a Capacitor. Let qand ({)+ be the charge andpotential of the positively charged plate of a capacitor.Accordingto (4.4), the integral canbe split intotwoparts(for the two plates). Then .1W=== T (q+fP+ +q-q>-).Since q: ==- q+, we have1 tW=T q;(rp; - R1Problems fitSolution. In accordance with (4.6), the intrinsic energy of eachshell isequal toqff>/2, where q> is thepotential of ashell createdonlybythechargeq onit, i.e. (p =q/4:teoR, where His theshell rad ius.Thus, the intrinsicenergyof eachshell is12WI 2=- - -.- - .'4rco 2R 1, 2Theenergyof interaction between thecharged shells isequal to thecharge q of one shell multipliedby the potentialR), we have1 Q2 1 q1Q'2W12r.:: q l - - - - =:..= - - - - .4neO R 24neO u,The total electric energy of the systemis. 1 (qi . . qt q2 )W=Wl --1 -W2--1-lVI 2= -4- - nco .... 11 I.J:! 12 4.3. Two smallmetallicballs ofradii IIIandR2arein VaCUUll1at a distance considerably exceeding their dimensions and have acertain total charge. Find the ratio q./q2 between the charges of' theballsat which the energyof thesystemis minimal. \Vhat is the po-tential difference betweenballs in this' case?Solution, The electric energy of this systemis1 (qi ql q2)W= WI -I- 2 -!-lV12 4neo2R1-t- 2112' -t- -l- ,whereWI and W2aretheintrinsic electric energiesof theballs(qrp/2),W12 is theenergyoftheir interaction (Ql.0; if j Uell,thent, 0; if, thee.m.f. hindersthis motion (:12 2 thepotentialdifferenceacrossitsterminals. If thesourceis disconnected,I ~ - - = 0 and t"==q>2 - 0 >80and viceversa . 5.8.. The workof ane.m.I, source. Aglass platecompletelyfills thegap between the plates of a parallel-plate capacitor whosecapacitanceisequal toC. whenthe plateisabsent. The capacitor isconnectedto asource of permanent voltage U. Find the mechanicalwork whichmust be done against electric forces for extracting theplateoutof thecapacitor.Solution. Accordingto thelawofconservationof 'energy, we canwriteAm + As= AW, (1)whereAmis. the mechanical workaccomplishedbyextraneousforcesagainst electric forces, A 8 is the work of the voltage source in thisprocess, andAWis the correspondingincrement inthe energyof thecapacitor (weassume that contributionsof other forms of energy tothechangein the energyof the systemis negligiblysmall).Let usfind!1WandA s- I t follows fromthe formulaW= CU2/2 ==qU/2 for the energyof acapacitor that for U= constAlV== scU2/2=&qU/2. (2)Since the capacitanceof the capacitordecreases uponthe removal ofthe platea). (6.t8)It shouldbenotedthat adirect solution of this problem(withthehelpof the Biot-Savartlaw) turnsout tohemore complicated.Fromthesamesymmetryconsiderations it follows that insidethewirethelines ofvector B arealsocircles. Accordingtothetheoremoncirculation of vector nfor a circular contour r 2 (see Fig. 6.6),B 21lr== r' whereI r = 1 (rla)2 isthecurrentenvelopedby the gi-ven cont.our. Whence we lind that insirle the wireB /rla2(r R, Fm0) and diamagnetic (z 1, whilefor diamagnetics f.1 nt, while 0).At theinterfacebetweenthis magneticandthewire, surfacemagnet.izat ion current If wil l appear. It can be easilyseenthat this current has thesame direction as the conductioncurrent I (when X>'0).I\Sa result, weshall havetheconductioncurrent 1II thesurfaceandvolumemagnetization currentsintheconductor(themagnetic. fields ofthese currents compensate one anotherandhencecan be disregarded), andthesurfacemagnetiza-t io n current 1/ on the nonconducting magnetic. For sufficient-ly thinwires, themagneticfieldBin themagneticwill hedeterruiued as the field of the current I + I'.Thus, the problemis reduced to finding the current I',For this purpose, we surround theconductor bya contourarrangedinthe surface layer ofthenonconducting magnetic.Let the plane of the contour be perpendicular to thewireax i.u. to the magnetizationcurrents, 'I'hen, taking(7.7)and (7.1.-'1) into considerut.ion, we can writeI' = i' dl =.r dl= YoII .u.;\cc,ording1.0 (7.12), it Iollowsthat Ifzl.T! ie eo I"atio IlS of t lif' iza t io Il currenl I' and7.5. Field in aHomogeneous Magnetic 187of the cond uctioncurrent I practicallycoincide (the wiresarc thin), and hence at all points the induction B' of thefield of the magnetizationcu -its differs fromthe induc-tion' Bo of thefieldofthecunuuctioncurrentsonlyinmag-nitude , these vectors beingrelated in thesamewayas thecorresponding currents, viz.B'XBo. (7.24)Thenthe induction of the resultant field B= Bo-;-. B' ='=(1+X) BoorB1l,Bo"(i .25)Thismeans thatwhenthe entire spaceisfilledwithahomo-geneousmagnetic, Bincreases fJ, times. In otherwords, thequantity showstheincreaseinthemagneticfield BuponIilling the entire space occupiedbythe fieldwith a magnetic.If \\'Cdivide bothpartsof (7.25) by f.tfAo, we obt.aiuH ::=: H, (7.2{))(in the case under consideration, field II turns out to bethe SHIue as in a vacuum).Formulas (7.24)-(7.26) are also validLoken a. homogeneousmagneticfills theentirevolumeboundedbythesurfaces [armedby the lines of Bo(the field of the. conductioncurrent). Inthiscasetoothemagnetic. inductionBinsidethemagnet.icis t-t times larger thanBo.Inthecasesconsideredabove, themagneticinduction13'of the Held of magnetization currents is connected withmuguet.izatiouJof themagneticthroughthe simple relationB' == !loJ. (7.27)Th is expression can be eusil y obtained fromthe formulaB._-=:: Do-+- B' if we take into account that B' :--= %B.oandB :..---- f-llloII, where H J/X.As hasalreadybeenmeutioneu above. ill other easesthesituation is much THOrf-) compl icatcd , and Iorrnulas (7.24)-(7.2i) become iuu ppi icable.Concl nd ingi.ito let us considertwosimpleexam-ples,Example L Field Hin a solenoid. /\ solenoid havingnl ampere-turns per uni t length is Iillcd wi th a homogeneous magnetic of thef88 7. Magnetic Fieldin (l. Substanccnermeabtlity ,....> 1. Find the magnetic induction 01tnt>Held inthe magnetic,According to (6.20), in the absence' of magnetic the magneticinduction Dointhesolenoidis equal to!.lonI. Sincethe magneticfillstheentirespacewhere thefield differsfromzero (weignore the edgeeffects), tho magnetic inductionB must be ,.... times larger:B=(7.28)Inthis case, thefieldof vector1{ remains the as intheabsenceof themagnetic, i.e. H=Ho.The change infield B is causedbythe appearance of rnagnetizationcurrents flowing over the surface of the .magneticin the samedirectionas the ,conduction currents in the solenoid .,winding when p c-1. If. however.jr-c f ,the directionsof these currentswill beopposite.The obtained results arealso validwhenthe magnetic has the formof averylong rodarrangedinside thesole-noidso thatitis paraIlel to thesole-noid's axis. rExample 2. The field of straight.correat ia thepre!eIIee 01 a Inagnetic.Suppose thata magnetic fills alongof radius a alongwhose axis Fig. 7.11agrvencurrent I flows. The permea-bility of the magnetic f.4 >1. Find themagneticinduction Bas a functionof thedistancer fromthe cylin-deraxis.We cannot use directly the theoremon circulation of vector Bsince maznetizatlon currentsare unknown. Inthis situation, vectorIIis veryhelpful: itscirculationis determinedonlybyconductioncur-rents. Foracircleof radiusr, wehave2nrH =I, whenceB ===- f.1floII21tr.Upon a ,. transition through the lnagnetic-vacuuminterface, themagnetic induction B, unlike Ht undergoes a discontinuity(Fig. 7.11).Anincrease inBinsidethe magnetic is caused by surface magnetiza-tioncurrents. Thesecurrentscoincidei .. direction withthecurrent Iin thewireontheaxis of the systemandhence "amplify"this current.On the other hand, outside the cylinder the surface magnetizationcurrent is directed oppositely, but it does not produce any effecton field Bin themag-netic. Outsidethemagnetic, the magneticfieldsof bothcurrentscompensateone another.7.6. FerromagnetismFerromagnetics. III magnetic respect, all suhstances canbe divided into weakly magnetic (paramagnctics and diu-iBn7.6. Fcrnmagnettsm--_. ----_..._----=--------,.------magnet.ics) and stl'ollg'ly m.urnot ir (Ierromagnet ics), I t iswel l known that in the absence of magnetic field, para-anddiamagueticsare 110t mngnet.izcd and arecharacterizedby a one-to-nne correspoudence between magnet.iznt.ion Jand vector H.Ferronuignetics arethesubstances(solids) that. maypossessspontaneous magnetizaiion , i.e. which are magnetized eveno 7.12HBoFig. 7,13Hin the absence of external magnetic field. Typical fel-ro-magnet ics are iron, cobalt, and many of their alloys. .Basic Magnetization Curve. 1\ typical fea t ure of ferro-magnetics is the complex nonl inear dependence J (H) orB(H). Figure 7.12 shows the magnetization curve for aIerrornaguet ic whose rnagnetization for H =-:= 0 is also zero.This curve is called the basic magnetizationcurve. Even atcomparativelysmall values of II, mugnet izut.ion J attainssaturation (J Magnetic inti uction B ==(II J) alsoincreases with H. After attainingsaturation, Hcontinuesto growwith II according to the linear lawH:== tlo11 _.\-+const , where const Figure 7.13 represents the basicmugnetiznt iou curve on the IJ-Hdiagram.Inviewof thenonlinear dependence B(If). thepermeabil-ity Il for Ierromagnetics cannot be defined as a constantcharacterizing' the magnetic properties of : l ve x HI directed towards us {Fig. 8.2), wil!appear in thesvst-m1('. III thissystem, thechargewill move totheleft attheveloci t,: --Vo, alldthismotiouwilloccur ill crossed electricandmagneticfields. FOl' the sakeof definiteness, wesupposethat the206 8. RelativeNature0/ElectricandFieldscharge ofthe particle isq >o. Then theLorentz forcein the system K'isF' =-= qE' +q l-vaXB] =q lvoX B]-q[voX 8]=0which.. however, directlyfollowsfromthe invarianceof theforce uponnonrelati vistictransformationsfromone referenceframe to the other.8.3. Corollaries of the Laws of Field TransformationSome Simple Corollaries. Several simple anduseful rela-tions follow in some cases fromtransformation formulas(8.1).'. I f theelectricfield Ealone is present in thesystemK(the magnetic field B==0), the following relation existsbetween the fields E' and H' in thesystemK':H' == -[Yo X E/]/C2 (8.5)Indeed, if B==0, then .::= E.lJV1- and B'I == 0, -[Yo X E]/c2Vi- -rvoXcB' (or E' .... ~ cH') as well.5. If bothinvariants are equal to zero, thenE1. Rand 1:: ~ cBin all referencesystems. I t will beshownlater that thisis preciselyobserved in electromagnetic waves.6. If the invariant I: B aloueisequal tozero, W ~ canfinda r e f ~ r encesysteminwhicheither E' =0or H' =o. The sign01 theotherinvariant determines which of these vectors is equal to zero. TheProblems 209opposite is also true: if E =0or B= 0 in some reference system,then E' 1.. H' inanyothersystem. (Thisconclusionhasalreadybeendrawn in Sec. 8.3.)Andfinallyone moreremark: itshouldbe borne in mindthat generally fields E and B depend both on coordinatesK X-Ir/

qFig. 8.1IB,c

-I 8.5and on time. Consequently, each invariunt of (S.U) referstothesamepoint inspaceandtime" of the field, thecoordi-natesand time.of thepoint indifferent systemsof referencebeingconnected through the Lorentz transformations.Problems 8.t. Special case of Held t ranstormatlon.A nonrelativisticpoint chargeqmoves at aconstant velocity v. Usingthetransforma-. tion formulas, findthemagneticfield B ofthis charge atapointwhosepositionrelative to thechargeis defined bythe radius vector r.Solution. Let usgoover to the systemof reference K' fixed to thecharge. In this system, onlytheCoulombfieldwiththeintensityE'- _1_!Lr- 4nBorS'is present, where we tookintoaccount that theradius vector r' = rinthe systemK' (nonrelativisticcase). Let us nowreturnfrom thesystemK' to the systemKthat moves relative to thesystemK' atthe velocity -v. For this purpose, weshall use theformula for thefield B from(8.4), inwhichthe role of primed quantities will be playedby unprimedquanti lies (and vice versa), andreplace the velocityVoby -Vo(Fig. 8.4). Inthe caseunder consideration, VI) = v, andhenceB = B'+[v X E']/c2 Considering that in the system K' B' =0and that c2=we findB==hq[vXrJ4n ,3 \Ve have obtained formula (6.3) which was earlier postulated asa result of the generalization of experimental nata.1'-01812iO 8. RelativeNature0/ ElectricandMagneticFields 8.2. Alarge plate made of homogeneous dielectric with thepermittivity 1l10VeS ataconstant nonrelativisticvelocityvin auni-formmagnetic Held Bas shownin Fig. 8.5. Find the polarizationPof the xllelectricand thesurface densitya' of hound charges.S olution, In thereferencesystemfixedtothe plate, inadditiontothemagneticfield, therewill beanelectricfieldwhich we shall denotebyIn accordance withformulas (8.4) of field transformntion, wehave [v XB).Thepolarizationof thediplectri(' isgiv(,11 byP, e-1l B)'=== xeoEo-e- v X twhere wetook intoaccount that accordingto'E' == Eo!t.:. Thesurface densi ty of bound charges is 8-110'1 eAt thefront surface of theplate (Fig. 8.5), 0'>0, while on the oppositeface a'v, theyformthe left-handedsystem(sincethecurrent 1/ inthesystemK' in thiscase will flowin the oppositedirection) . 8.5. Arelativistic chargedparticlemoves inthespace occupiedby. uniformmutually perpendicular electric and magnetic fields EandB. Theparticlemoves rectilinearlyinthe directionperpendiculartothevectors E andB. Find E' and B' inthereference system movingtranslationall y wi th the particle.Solution. Itfollowsfromthedescriptionof motionof the particlethat itsvelocitymust satisfytheconditionvB= E. (1)Inaccordance with the' transformation formulas (8. t), we haveE' == E+[v X B] =0 tsince in the case under consideration the Lorentz force (and hencethequantityE +1v X B]) is equal to zero.According to the same transformation formulas, the magneticfieldis given byH'==: B-l.vXEJ/c' .}'/ The arrangement ofvectors is showninFig. 8.6, fromwhich it followsthat l v X El tt B. Consequently, considering that according to (1).,.212 8. RelativeNature0/ Electric andMagneticField,v = EtB, we can writeB' =B-E2/Bc" _y't pior in the vector formB' = B Y1-(ElcB)I.We advise the reader toverifythat the obtainedexpressionssatisfybothfield invariants. M.G. Themot iunor achargein crossedfields J l ~ andR. A non-relativisticparticlewi th aspecificchargeql mmovesinaregionwhereEYEvXB qB'[V"E]ZFig. 8.6 Fig. 8.7uniform, mutually perpendicular fields Eand Bhave beencreated(Fig. 8.7). At themoment t =0the particlewasat point 0 and itsveloci tywas equal to zero. Find the lawof motionof the particle,z(t) and y(t).Solution, Theparticle moves undertheaction of theLorentz force.I tcanbe easily seenthat the particle always remains intheplane XY.Itsmotion canbedescribedmosteasily inacertain systemK'. whereonlythe magneticfieldis present. Let us findthisreferencesystem.It follows fromtransformations (8.4) that E' = 0ina referencesystem-that moves with a velocity Vo satisfying the relation E== -[voX B). Itismore convenient to choose the systemK' whosevelocity vo is directed towards the positive values on the X-axis(Fig. 8.7), since in such a system the particle will moveperpendicularlyto vector H' and its motionwill be the simplest.Thus, .in the systemK' that moves to the right at the velocityVo=EIB,the field E' = 0 andonly the fieldB'is observed. Inaccord-ance with (8.4) and Fig. 8.7, we have _.B'= B- [vo XE]/c 2= B(1- v3/c2).Sincefor a nonrelativistic particleVo C,we canassume thatB'=B.InthesystemK', theparticle will moveonly inthemagnetic field,itsvelocitybeingperpendiculartothisfield. Theequationofmotionfor this particlein the systemK' will have the formmv31R =qvoB. (1)Problems 213This equation is written for the instant t =0, when the particlemovedin thesystemK' as is shown in Fig. 8.8. Since the Lorentzforce Fis always perpendicular to the velocityof the particle, vo ==const, and it follows from(1) that in the systemK' the particlewill move in a circleof radiusR=mv.lqB.. Thus, theparticle movesuniformly withthevelocityVo inacircleinthesystemK' which, in turn, moves uniformlyto the right withthesamevelocityVo = EIB. Thismotioncanbecomparedwith themotion ofthepointq at therimofawheel (Fig. 8.9) rollingwiththeangular velocity co= voiR =qB/m. .Figure 8.9 readily shows that the coordinates of the particle qat the instant tare gi yen byx=vot-R sinoot=a (rot-sinCUi),1J=R-RcosCI>t=a (i-coswt),where a =mE/qBI and w =qBlm 8.7. ThereisonlyauniformelectricfieldEinaninertial systern K. Findthe magnitudes and directionsof vectors E' and B' inthe system K' IIWvingrelative to the system K at a constant relativisticvelocity Vo at an angle a to vector E.Solution. According to tzansformation formulas (8.t) and takingintoaccount that B== 0inthe systemK, weobtainE'II= E cos a, = E sin a/ Hence the magnitudeof vector E' is givenbyE' "--= V f- HI:- E 1/(1- cosS a)/(t - and the anglea' betweenvectors E' and Vo canbedetermiued Iromthe formula .tana' == Ell = tan a! Vt - pl.Themagnitude anddirection of vector D' canbe Iouud ina similarway: .BII=0, BJ. =-(vOXEl/ (c2 D'This means that H' 1- ve, andB'== voE sina/(c2. 8.8. Uniformelectricand magneticfieldsEand B ofthesamedirectionexist in a systemof reference K. Find the magnitudes ofvectors E' and B' andtheangle between thesevectors inthesystem K'moving at aconstant relativisticvelocityVointhe directionperpen-dicular to vectors Eand B.Solution. In accordance with formulas (8.1), in the system K' 8. RelativeNature 0/ElectricandMagneticFieldsthe two vectorsE' and H' will be alsoperpendicular to the vee tor v 0(Fig. &.10).Themagnitudesof thevectors and B' canbefound bythe formulasE'=-';-EI+(voB)2 B'=";- B'+IIisfraught withcertaindifficulties. Nomatter howthinthewiremaybe, its cross-sectional areaisfinite, andwesimplydonot know howtodrawinthebody of the conductor a geo-metrical contour required for calculating ide.=-lit (1,,1). (9.16)If the inductance L remains.constant upon a variation ofthecurrent (theconfigurationof thecircuit doesnot change15*'228 9. ElectromagneticInductionand there are no Ierromagnetlcs), thenf =- L .E- (L=canst).B dt(9.17)The minus sign here indicates that ~ a is always directedsothat it prevents a change in the current (in accordancewiththeLenz's law). Thise.m.I. tendsto keepthecurrentconstant: it counteracts the current when it increases andsustainsitwhenit decreases. In self-inductionphenomena,the current has "inertia". Therefore, the induction effectsstrivetoretainthe .magneticfluxconstant just in thesameway as mechanical inertia strives to preserve the velocityof a body.Examples of Self-induction Effects. Typical manifesta-tionsof self-inductionareobservedat themomentsof con-nectionor disconnectionof an electriccircuit. 'rhoincreasein the current beforeit reaches thostead y-st ate valueafterclosing the circuit and the decrease in the current whenthecircuit is disconnectedoccur gradual ly andnot instant-aneously. Theseeffectsof lagarethe more pronouncedthelarger-theinductanceof thecircuit.Any large electromagnet has a large inductance. If itswindingisdiseonnectedjfromthesource, thocurrent rapidlydiminishes to zero and creates duringthis process a hugee.m.f. This often leads to the appearance of a Voltaic arcbetween the contacts of theswitchwhich is not onlyverydangerousfor the electromagnet winding but may evenprovefatal. For this reason, a bulb with the resistunce of thesame order of magnitude as that of the wire is normallyconnectedin parallel to theelectromagnet winding. Inthiscase, thecurrent inthewindingdecreasesslowlyandis nota hazard.Let usnowconsider in greater detail the mode of vanishingandestablishment of thecurrent in acircuit.Example f , Current collapseuponthedisconnectionofa circuit.Suppose thatacircuitconsistsofacoil ofconstant inductanceL,aresistorR, anammeter A, ane.m.I. sourceji and aspecial key K(Fig. 9.7a). Initially, the key Kis in the lower position(Fig. 9.7b)andacurrent10= ~ / R flows in the circuit (weassumethat there-sistanceof the sourceof e.m.I, ~ is negligiblysmall).At the instant t =0, werapidlyturnkey Kclockwisefromthelower to the upper position(Fig. 9.7a). Thisleads to the following:9.3. Sell-induction 229thekeyshort-circuits thesourcefor a veryshort timeand thendis-connects it from the circuit without hreaking the latter.The currentthroughthe inductance coil Lstarts todecrease, whichleads to the appearance of self-induction e.m.I,8 == -LdlldtL(a)Fig. 9.7L(b)110oFig. 9.8(9.18)which, in accordance with Lenz's law, counteracts the decrease inthe current. A instantof time, the current inthecircuitwill bedetermined by Ohm's law1 = orRI= -IJ dI_dt Separating the variables, we obtain.!y-- _..!!.- dt;I - LIntegration of this equation over I (between 10and I) and over t(between 0 and t) gives In(111'0) =-RLlt. orI=Ioe-t/T.where't is aconstant havingthedimension uf time:'t = LIR.(9.t9)(9.20)It is called the time constant (or the relaxaton time). This quantitycharacterizestherateofdecreaseinthecurrent: itIollowsfrom(0.19)that Tis thetimeduring whichthecurrent.decreasesto(tIe) timesitsinitial value. Thelargerthevalueof "t, theslowerthedecreaseinthecurrent, Figure9.8showsthecurveof the dependenceI(t) describingthe decrease of current withtime (curve1).Example 2. Stabilizationor current upon closure of acircuit.At theinstant t =0, werapidly turntheswitchS counterclock-wise front the upper to the lower position(Fig. 9.7b). We thuscon-nected t.he sourceto theinductancecoil L. Thecurrent in thecir-cuit starts to grow, and a self-induction e.m.f., counteracting thisincrease, will appear. In accordance with Ohm's law. RI :=230= ~ +I,. or9. ElectromagneticInductiondlRI=qz-L-" dt (9.21)WetransposeI to the left-handsideof thisequationandintroduceanewvariable u =RI - ~ . du= Rdl . After that, wetransformthe equation thus obtained todulu=-sdtl,where 't == LIRis the time constant.Integrationover u(between - ~ andRI- ~ ) andover t (be-tween 0 and t) gives In (RI- 1)/(-"= -stl, or1=10 (t_e-f/'t), (9.22)whereI. =~ / R isthe value. of the steady-statecurrent(for t ~ 00).Equation (9.22) shows that the rate of stabilizationof the currentisdetermined bythe same constant T. The curve I(t) characterizing theincreasein thecurrent withtimeis shownin Fig. 9.8 (curve2).On the Conservation of Magnetic Flux. Let a currentloop 1110Ve and be deformed in an arbitraryexternal mag-neticfield (permanent or v.arying). Thecurrent inducedinthe loop in this caseis given byI _ ~ f + ~ . = _-.!.- d/dt must also beequal to zero, sincethecurrentI cannot be infinitely large.Hence it follows that cI> == const.Thus, whenasuperconductingloopmoves in a magneticfield, the magnetic fluxthrough its contour remains constant.This conservationofthefluxisensuredby inducedcurrentswhich, accordingto Lenz'slaw, prevent anychange in themagnetic flux through the contour.Tho tendency to conserve the magnetic flux throughacontoural ways existsbutis exhibitedintheclearest formin the circuits of superconductors.Example. Asuperconductingring of radius a and inductance L'isinauniformmagneticfield B. Inthe initial position, theplaneofthe ringis parallel to vector B, and the current in the ringisequaltozero. Theringis turnedto thepositionperpendicular to vector B.Find the current in the ring in the final position and the magneticinductionat i Ls centre.Themagneticnuxthroughthe ringdoesnot changeuponits rota-tionand remains equal to zero. This means that the magnetic fluxes9.4. Mutual Induction 231ofthefieldof theinducedcurrent andof theexternal currentthroughthe ringare equal in magnitude and opposite in sign. Hence LI== 1f,(J,tB, whenceI = na2BIL.This current,inaccordance with(6.13), createsa field BI= ttf!oaB/2Lat the centreof the ring. The resultant magnetic inductionat thispoint is given byBr es= B-B]= B(1 - 9.4.Mutual Inducti'onMutual Inductance. Let us consider two fixed loops 1and 2 (Fig. 9.9) arrangedsufficiently. close to each other.Ifcurrent /1 flows in loop 1, it creates through loop2thetotal magneticflux2 proportional (in the absence offerromagnetics)to the current It:$2= 421[t (9.23)Similarly, if current 12flows inloop 2, it creates through thecontour 1 the total magneticDux(9.24)The proportionality factors 12and L21 are called mutual induc-tances of the loops. Clearly, mutual' inductance is nu-mericallyequal to themagnetic flux through one of theloopscreatedby aunit current intheotherloop. Thecoef-ficients 21 andL12dependontheshape, size, andmutualarrangementoftheloops, aswell asonthemagneticperme-abilityof the mediumsurrounding the loops. These coef-ficientsaremeasuredinthesameunitsastheinductanceL.ReciprocityTheorem. Calculations show(and experimentsconfirm) that ill the absence of Ierromaguetics, the coef-ficients L12and 21 are equal:(9.25)This remarkable propertyof mutual inductance is usually.called the reciprocity theorem. Owing to this' theorem, we232 9. ElectromagneticInductiondo not have to distinguish between L12and L21and cansimply speak of the mutual Inductanceof two circuits.The meaningof equality (9.25) is that in any case themagnetic flux 1 through loop 1, created by current Iinloop2, isequal to themagneticflux cI>2throughloop2,createdbythesame currentI. in loop1. Thiscircumstanceoften allows us to considerably simplify the calculation,for example, of magnetic fluxes. Here are two examples.&amplet , Twocircularloops 1and2whose centrescoincidelie ina,--Ic::;:::> 1plane (Fig. 9.10). Theradii of theloopsare 41 and a.. Current 1 flows j nloop1. Find themagneticnux2 em-Fig. 9.10 bracedby loop 2, if til ,is clearly a rather complicated problemsince the configuration of the fieldOitself is complicated. However, the1.applicationof the reciprocitytheoremgreatlysimplifies the solution of theproblem. Indeed. let us passthe samecurrent I through loop 2. Thenthemagnetic flux 1 created by this cur-rent through loop 1 can be easilyfound, provided that a1 al:itissuf-F ficient to multiplythe magnetic Indue-ig. 9.1 lion B atthecentre of theloop(B ==.....112a,,)bythe areanaT ofthe circleand takeintoaccountthat throughthehatchedhalf-plane (Fig. 9. t f)whoseboundary. is atagivendistancefromthe contour. Assumethatthis half-plane and the 'loop are in the same plane.10this case too, themagneticfield of current 1hasacomplex con-figuration, andhence it isvery difficult todirectly calculatethe Duxin whichwe areinterested. However, the solutioncanbeconsiderablysimplified by using the reciprocity theorem.Supposethat current I flowsnot around therectangular contourhut along the boundaryof the half-plane, envelopingit at infinity.Themagnetic field created bythis current inthe region ofthe rectan-gular loop has a simple configuration-this is the field of a straightcurrent. Hencewecan easilyfindthemagnetic flux ' through the contour (with the help of simple. integration). InaccordanceWith the reciprocity theorem, the required Oux 0), theworkdone bythessurce of turnsout to be greaterthantheJoule heat liberated in the circuit. Apart of this work(additional work) is performed against the self-inducede.m.f. It should be noted that afterfhe current has beenstabilized, d = 0, andthe entireworkofthe source ofwill be spent for the liberation of the Joule heat.Thus, the additional work accomplished by extraneousforces against the self-induced e.m.f. in theprocess of current, stabilization is given by16Aadd = Ida>.1 (9.27)This relation is of a general nature. I t is valid in thepresence of ferromagnetics as well, since while derivingitno assumptions have been made concerning the magneticproperties of the surroundings.Here(andbelow)weshall assumethattherearenoferromagnetics, Then d == L dl , and6Aadd :-:-= Ll st. (9.28)Having integrated this equation, we obtain A add === L[2/2. According to the lawof conservation of energy,anyworkis equal to the increment of anykindof energy.It is clear that apartoftheworkdonebyextraneous forces236 9. EleetromagnettcInductionis spent onincreasing theinternal energy'of conductors(itisassociated withtheliberationoftheJ oule heat),whileanotherpart (during current stabilization)is spent onsome-thing else. This "something"isjust themagnetic field, sinceits appearance is associated with the appearance of thecurrent.Thus, wearriveat theconclusionthat in the absenceofferromagnetics, the contour with the induction coil Land current I has the energyIW= {LJ2=-}I(J>= *.J (9.29)Thisenergyis calledthemagnetic, or tntrtnsie, energyofcurrent. It can be completely converted into the internalenergy of conductors if we disconnect the source of tofromthe circuitasshowninFig. 9.7, Le. ifwerapidlyturnthe key Kfromposition bto position a.Magnetic Field Energy. Formula (9.29) expresses themagneticenergyof current intermsof inductanceand CUf-rent (inthe absence of ferromagnetics). In this case, however,theenergyliketheelectric energy of chargedbodies, canbedirectly expressed in terms of magnetic induction B. Letus show this first fora simple case of a long solenoid,ignoringfield distortions at itsends (edge effects). Substitutingtheexpression II ==into (9.29) we obtainW =(1/2)12= And. since nI == H=BlJA. .... ot. we have BHV. (9.30)2Jif.1o 2This formulaisvalidIor auniformfield fillingthe volumeV(as in thecaseof thesolenoidunder consideration).The general theory shows that the energyl-Vcan be ex-pressedintermsof BandHinanycase(butin theabsenceof Ierromagnetics) through the formula= JB;JIdv1 (9.31)The integrand in this equation has the meaningof theenergycontained in the volume element dV.237 9.5. MagneticField_E_n_er_'.-;;,y _Thus, as inthecaseof theelectricfieldenergy, we arriveat the conclusionthat themagneticenergyis alsolocalizedin the space occupied by a rnagnetic field.It follows fromformu las (9.30) and (9.31) that the mag-netic. energyisdistributedin spacewiththevolumedensity

I2 2;.t ....o(9.32)It should be noted that this expression cannppliedonly to the med ia for which the IIII dependence islinear, i.e. intherelationB llJ-loHisindependentof If.Inotherwords, expressions (D.31) and (n.32) areapplicableouly to paramaguet ics and In the case orferroruagnetics, these expressions ,H'P j IlH ppl icable"Jl should also benoted"thatthe{Hug-netic energyis anessen-tially positive Auant.ity. Thiscan heeasilyseenfromthe lasttwo formulas.Another Approach lo the Sub-stantiationof Formula (9.32). Letus provethevalidityof this formulabyproceeding "the other way round!"i.e. let usshow that if formula(9.32) Fig. 9.13is valid, themagnetic energy of thecurrent loop is W =LI2/2.For this purpose, we consider the magnetic field of an arbi traryloopwith current I (Fig. 9.13). Let us imaginethat theentirefield isdivided intoelementary tubes whose generatrices are the field linesof B. \Ve isolate in one such tube a volume element dV = dl d.S:According to formula (9.32), the energyBH dl dS is localized inthis volume.Let usnow findthe energy dJ-V inthevolume ofthe entire elementa-ry tube. For this. purpose, we integrate the latter expression alongthetube axis, The flux dcI> = BdSthroughthetubecross-section is*This is due to thefact that in the longrun expressions (9.31)and (9.32) are theconsequences of the formula 6Aadd =-.= Jd(j) and ofthefactthat, intheabsenceof hysteresis, thework 6A add is spent onlyon increasingthemagneticenergydW. For a ferromagnetic medium,thesituationis different: thework6Aaddis also spent onincreasingthe internal energyof the medium, i.e. on238 9. Electromagnettc Inductionconstant along the tube, andhenced0) depends only onthe mutual orien-tationof thecurrents themselvesand isindependent of thechoice ofthe positivedirectionsofcircumventionof the loops. Werecall thatthesignof thequantitywas consideredin Sec. 9.4.Field Treatment of Energy(9.34). There are some moreimportant problems that can be solved bycalculatingthemagneticenergyof twoloops in a different way, viz. fromthepoint of viewof localizationof energyinthefield. .Let B1be the magnetic field of current II, and B2thefield of current 12 Then, in accordancewiththe principleof superposition, the field at each point is B :"--= B) -i-B2 Jand accordingto (9.31), the field energy of this systemofcurrents is W= dV. Substituting into thisformula ,B'2 ,.=.-+-'-t- 2B1B2, we obtain____.InductionIw=,+\+I'dv.1 (9.35)f__' J J IThe correspondence between individ ual terms in formulas(9.35) and (9.34) is beyond doubt.Formulas(9.:3--1) and(9.35)loadtothe Iol lowing importantsequences.1. The magnetic energy of a systemof two (or more)currents is an essentiallv positive quantity>0). Thisfollows Irorn the Iact that W 0( IF dl',where the iutegraudcontains positive quantities.2. Theenergyof currents is a nonadditivequantity(dueto thp ex istence of mutual energy).3. Tho last integra! inis proportional to the pro-duct. 11/2of currentvsiucc /-/ 1 ex: II anrlex: I:!. 'fhe pro-portionall ty factor (i. e. the rema ining' i utegral) turns outto he symmetr ic withrespect to indices1 and2 andhencecan he denoted by 1J12 or 1..121 ill accordance with formula(9.34). Thus, 12 =-= L21indeed.4. Expression(U.35) leadstoanotherdefiuitionof mutualinductance 1J12. Indeed, a comparison ofand givesl1 \' B1.82d1.T..112=-11 --- t'.1 2 "(9.36)9.7. Energy and Forcesin MagneticFieldThe most general methodfordetermining the forcesactingin amagneticfieldis the energy method, inwhich theexpres-sion for the magnetic field energy is used.We shall confine oursel ves to the ease when the systemconsistsof twoloopswithcurrents II and12 Themagneticenergyof suchasystemcan be representedintheformltV'"(11= R Thus, the problemis solved.It shouldbenotedthatthevector diagramobtained above288 11. ElectricOsctllauon----------proves to be veryconvenient for solvingmanyspecificproblems. It permits a visual, easy, andrapid analysisofvarioussituations. .ResonanceCurves. Thisis thenamegiventothe plotsofdependencesof thefollowingquantitieson thefrequency IDof external e.m.f, current It charge q ona capacitor,andc6Fig. 11.6 Fig. tt.7voltages UR' Ucand UL defined by formulas (1f.32)-(1i.34).Resonance curves for current 1m(ffi) are showninFig.t1.5. Expression (11.35) shows that the current amplitudehas the maximum valuefor wL- 1/wC= O. Consequently,theresonancefrequencyfor current coincides with the na-tural frequencyof theoscillatorycircuit:001 rei =roo = 1/VLe. (t t .37)Themaximumat resonanceisthehigherandthe sharperthe smaller the dampingfactor=R/2L.Resonancecurves for thecharge qm (00) ona capacitor areshowninFig. 11.6(resonance curves for voltage UCM'across the capacitor have thesameshape). Themaximumofcharge amplitude is attained at the resonance frequencyWq rea =.y(11.38)which comes closer and closer to (t)o with decreasing p"Inorder toobtain(11.38), wemust represent qmf in accord-ance with (11.30), in the formqm =I m/(J) where 1fA isdefined by (11.35). l'heflqm= _ (11.39)V(w3 _mt) 2 -J-4P'cul11.3. ForcedElectricOscillation, 289(11.40)Themaximumof thisfunctionor, whichisthesame, theminimumoftheradicandcanbefound byequatingto zerothederivativeoftheradicandwithrespect to (0. Thisgivesthe resonance frequency (11.38).Let us nowsee howtheamplitudesof voltages UUc-and UL are redistributed depending on the frequency (0of theexternal e.m.f. Thispatternis depictedinFig. 11.7.The resonancefrequenciesfor UR' Ucand UL are deter-mined by the following formulas:roc res =(00 1- 2(P/(t}o)2,(l)L res =000 1-2 (P/O)2.Thesmallerthevalueof thecloseraretheresonancefre-quencies for all quantities to thevalue roo.Resenaneec.vesandQ-factor. The shape of resonance.curvesisconnectedinacertain way withtheQ-factorofanoscillatorycircuit. This connectionhas the simplest formfor thecaseof weakdamping, i.e. forInthis case,o.;;== Q (11.41)(Fig. 11.7). Indeed, forthe quantity(Oresrooand, according to (11.33) and(11.35), Uemres =I m/rooC===Cm/woCR.. or Ucmres/cern=LC/CR== (fiR) lfLICwhichis, in accordancewithformula (11.22), just the Q-factor.Thus, the Q-factor of a circuit (for shows howmanytimes the maximumvalue of the voltage amplitudeacrossthecapacitor (andinductioncoil) exceedstheampli-tude of the external e.m.I.TheQ-factor of a circuit is also connectedwithanotherimportant characteristic of the resonance curve, viz. itswidth. It turnsout that for Q= (fJolf)(fJ, (11.42)where 000is theresonancefrequencyand{)roisthewidthoftheresonancecurveat a"height"equal to 0.7of thepeakheight, i.e. at resonance.Resonance. Resonance in the case under considerationis1/2 19-0181290 11. ElectricOscillationstheexcitationof strongoscillationsat thefrequencyof ex-ternale.m.f. or voltage. This frequencyisequalto the na-tural frequency of anoscillatory circuit. Resonance is usedforsingling out arequiredcomponentfroma composite volt-age. Theentireradioreceptiontechniqueis basedonreso-nance. Inorder to receive with a given radioreceiver thestation we are interested in, the receiver mustbetuned. Inother words, byvaryingCandL of the oscil latory circuit,wemustattain the coincidence between its natural frequencyand the frequency of radio waves emitted by the radiostation.The phenomenon of resonanceisalsoassociated withacer-taindanger: the externale.ll1.f. or voltagemaybesmall , butthe voltages across individ ual elements of the circuit (thecapacitor orinductioncoil) may attainthevalues dangerousfor people. This should al ways be remembered.11.4. AlternatingCurrentTotal Heslstance [Impedance}, Steady-stateforced electricoscillations can he treatedas anaIteruat ing currentIlow-ingin acircuit havingacapaci lance, inductance, and resist-ance. L'nder the action of external voltage [which playsthe role of external e.ru. f.)U==Umcos (.Ilt, (11./13)the current in the circuit varies according to t.he IawI = I rn cos ((ut --rp), (11.44)whereIUrn t wL-1/mC (111.5)m =R2+(r)L-1/wC)2, anq:o = R .':1Theproblemis reducedto determining thecurrent ampli-tude and the phase shift of the current relative to U.Theobtainedexpressionfor thecurrent amplitude1m(00)canbeIormallyinterpretedas Ohm'slawfor theamplitudevaluesof current andvoltage. 'The quantityin thedenomi-nator of this expression, wh ich has thedimension of resist-ance, is denoted byZand is calledthetotal resistance, orimpedance:11.4. AlternatingCurrent 291(11.50)z= y R2+(wL -1/roC)2. (11.46)It can be seenthat for ro =