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A·IA

mxonat



Wadhwa

TRIBUTORSew Delhi - 110 002 (India)



sian edition

hlCllI3HlllKO.ll3>> ,1983rPublishers,1986

hed inIndiabyarrangement with

ation,Mir Publishers, 1986of this book may bereproduced or trans-ymeans,electronicor mechanical, includ-,or any infomation storageand retrievalnwriting, from thepublisher.

ublishers& Distributors.w Delhi- l 10 002 (india)

06



-ptsof thetheory and thepracticalmsinonebook.Therefore, each

riptionof the theory of thelustratedby concreteexamples)oblemswithsolutions. Theprob-etext and often complement it.sed together with thetext. Inedproblemsshould enabletheerstandi ng of many importantnwithout solving the problemsem) thewiderangeof applica-n thisbook.st important lawsof electro-oclarifythemost difficult top-redtoexcludethelessimpor-escribethemain ideasconcise-mecorrectly,thetext hasbeenmathematical formulas, and thethephysical aspectsof thephe-in view,variousmodel repre-rs.special cases, symmetry con-

mployedwherever possible.eusedthroughout thebook.eGaussian system of unitsiscludedin Appendices3 and 4e most important quantitiesussian units.ntsand termsaregiven in

aterial and problemsinvolvingalculationsareset in brevier

mittedon hrst reading withoutrevier typeisalso used for



xtbook for undergraduatestu-(intheframework of thecourseso be used by university

. A.A. Detlaf and Readerdthemanuscript. and madeantsand suggestions.



cuum 11

16ssTheorem 19GaussTheorem 23. Potential . . 25tial and Vector E 30

staticField 45. . 45

deaConductor . . . 47rfaceof a Conductor 49onducting Shell .. . 51trostatics.ImageMethod 54s57

cs67cs67

f P 72

. 80sDielectric84

4tem of Charges. . 94onductor and aCharged

d. . . 100edBodies105ctric106



nuityEquation. . 116ogeneousConductor 119w... ...122hhoffsLaws1269

Capacitor Circuit 133

um 14141. 145cField ......... 148orem on Circulation of Vec-

asicLawsof MagneticField 15455rrent Loop. . .. . 159acement of Current Loop 161

ance172tance.Magnetization Vec-

76

r B and H182usMagnetic185

cand MagneticFields198ChargeInvariance198nfor FieldsE and B.... 200of Field Transformation 206nvariants203

n217romagneticInduction. Lenz’s

ticInduction 220

...... 234o Current Loops...... 238



agneticField 240

nergy of ElectromagneticField 253253

. 257’sEquations. . 261x. Poynting`sVector 264Momentum 268

tory Circuit 277ions. 280ations2850

asurement . 299tsof Measurement . .299Electricand MagneticQuan-stems.. .. . 299-cityand Magnetism in Sl

304305



old-faceupright lettersnthestandard-typeprint (r, E)e,vector.edby anglebrackets( ),

quantitiesdenote:hequantity, viz.thedifferencevalues,e.g.AE = E2-—-El,

simal increment), e.g. dB, dep;f aquantity,e.g. 6Aistheele-

nit vectorsof Cartesian coordi-

of cylindrical coordinates

mal to a surfaceelement;gent to thecontour or to an

aryfunction f isdenoted byheletter denoting thefunction, f.aredenoted by asinglesym-otation of theelement of inte-ement,dS isthesurfaceele-f length.Symbol denotesthentour or aclosed surface.heoperationsinvolving this

ows: Vqv isthegradient of rgenceof E (div E), and V>< E

.b], wherea and b are



t isknown that diversephe-on four fundamental interactions,viz.strong,electromagnetic,ractions.Each type of inter-rtaincharacteristicof a particle.al interactiondependson the

electromagneticinteraction ises.icleisoneof itsbasiccharac-g fundamental properties:wo forms,i.e.it can beposi-

gesin any electrically insu-e(thisstatement expressestheiccharge);visticinvariant: itsmagni-referencesystem, in other words,er thechargemovesor isfixed.esefundamental properties

quences.ewith modern theory, inter-omplished through a field. Anyertainway thepropertiesof createsan electricfield'. This

hargeplaced at somepoint. of ionof aforce.rceF actingon afixed testerepresented in theform

ntensityof theelectricfield at1) showsthat vector E can ben a positivefixed unit charge.chargeq' is sufficiently small so



noticeablydistort thefieldult of possibleredistribution of

It followsdirectly from exper-theintensity of thefield of atancer from it can berepresent-

ant and e,. isthe unit vector of m thecentreof the iield, where

epoint weare interested in.SI. Herethe coefficient

oulombs(C) and thefield inten-m).Vector E isdirected along rgon thesign of thechargeq.ul0mb’slawin the“iield"form.sityE of thefield created by a

oportional to thesquareof theesultsindicatethat thislaw*3 cm to several kilometres,xpect that thislawwill bes.heforceactingon atest

yafixedpoint chargedoesnothargeis at rest or moves.Thisargesaswell.

Besidesthelawexpressed byperimentsthat theintensity of dpoint chargesisequal to theof thefields that would beately:



eenthechargeq,and thepoint

rincipleof superposizion. of neof themost remarkableprop-to calculatefield intensity of resenting thissystem asan ag-oseindividual contributionsare

to simplify mathematicalenient to ignorethefact thature(electrons,nuclei) and"in acertain wayin space. Ino replacetheactual distribu-sby a fictitiouscontinuousossibletosimplify calculationsucingany significant error.nuousdistribution,theconcepted,viz. thevolumedensity p,

density X. By definition,

inedin thevolumedV, on theh dl respectively.toconsideration wecan repre-

ent form. For example, if theevolume,wemust replaceq,nby integration.Thisgives

Ts’ormed over theentirespacewith

ionof charges, wecan complete-ing theelectricfield intensitybution isdiscreteor byformu-ulaif it iscontinuous. In theinvolvescertain difficultiesnciplenature). Indeed, in ordert calculateitsprojectionsEl,t wemust takethreeintegrals



heproblem becomesmuchem of chargeshasn certain 8`ym-xamples.axisof a thinuniformly charged ring.distributed over athin ring of radiusa.sityE on theaxisof the ring asafunctiontre.

ector E in thiscasemust bedirectedg.1.1). Let usisolateelement dl on the

A. Wewritetheexpression for thecomp-dby thiselement at apoint C:

esof r and czwill bethesamefor all thecetheintegration of thisequation issim-ent of hdl by q. Asaresult, weobtain

hefieldE z q/4ne0z’, i.e. at largedis-sapoint charge.niformly ehlnlged straight Element. Ath2l isunifo y charged bya chaq.

a point separated byadistancexrgxtéomand locatedsymmetrically with respect

nsiderationsthat vector E must be. This showsthewayof solving this



omponent dEx of thefield created by theaving thechargedq,and then integratentsof thefilament. In thiscase*1,

hargedensity.Let us reducethisequationntegration.Figure1.2 showsthatot; consequently, ..

sily integrated:

valueof tl1eanglecz, sin cz0= l/ }//l‘·i —}— ag?.

5/F-i—a:*'·'z°’ for z> l asthefield of a point charge.lectricField. lf weknow

ectrictield can bevisually re-ldlines, or linesof E. Such ant to it at each point coincides

E.Thedensity of thelines,nit areanormal to thelinesisdeof vector E. Besides,the lineshispattern givestheideaaboutelectricfield,i.e. about. thedi-tor E at each point of theiield.f FieldE.Theiield E definednt properties.Theknowledgeof eeper understand thevery con-teitslaws, and also madeit pos-

oblemsin a simpleand elegantheGausstheorem and thetheo-E, are associated with twomostacteristicsof all vector fields:It will beshown belowthat in



not only all thelawsof electri-magnetism can bedescribed. Letescriptionof theseproperties.

arity,weshall usethegeomet-ield (with thehelp of linesthehat thequalor E.Thenthe

mal- ntoector E,S cosat (seest theaelement dS. In a morecompact

fvector E onto thenormal n toisthevector whosemagnitudeioncoincideswith thedirectionoted that thechoiceof thedi-) is arbitrary. Thisvector could

ceS, theflux of E through

y. sinceit dependsnot only onE but alsoon thechoiceof thedit iscustomary to direct theenveloped by thissurface, i.e.al.Henceforth weshall always

etheflux of E, theconcept of tor fieldaswell.



x of E through an a1·bitraryrkableproperty: it dependsonlyhargesembraced by thissur-

ral symbol indicatesthat theer aclosedsurface.lytheGausstheorem; thefluxof

equal to thealgebraicsum of theface, divided by eo.heiieldof a singlepoint chargeanarbitrary closed surfaceSE through theareaelement dS:

S coso.- MSOdS2,(1.8)esting on theareaelement dSpoint wherethechargeq islo-sexpression over theentireeintegration over theentireement of dQby 4at. Thusweedby formula(1.7).morecomplicated shapeof amay begreater than at/2, andgenerally assumeeither positive! isan algebraicquantity: if f thesurfaceS, (YQ> O,whileif 2 < O.



__following conclusion: if deaclosed surfaceS,thefluxqual tozero.In order to provehrough thechargeq aconicald surfaceS. Then the integrationS isequivalent to theintegra-ter sideof the surfaceS will be

ngleQ> O.whiletheinner side,esbeingequal in magnitude).dCD =-= 0, which_ also agreesinesor lines·of E,thismeansringthevolumeenclosed byumber of linesemerging from

ewhen theelectricfield ischargesql, gz, . In thisprincipleof superposition E =, isthefieldcreated by theof E can bewritten i`n theform

m,ssaidabove, each integral on the/so if thechargeq, isinsideual to zero -if it isoutsidethendsidewill contain thealge-rgesthat lieinsidethesurfaceS.etheorem,it remainsfor us

chargesaredistributed contin-ity dependingon coordinates.that each volumeelement dVVThen on the right-hand side

ormed only over thevolumesurfaceS.efollowing important circum-



lf dependson themutual con-,theflux of E through an arbi-rminedby thealgebraicsumfaceS. Thismeansthat if wedE will bechanged everywhere,aceS.Generally, theflux of Esochange.However, if thed not involvetheir crossing of rough thissurfacewould re-

westress again,thefield E itself What aremarkableproperty of

ssTheoremtheconfiguration of all charges,ydoesnot allowusto deter-certain casestheGausstheoremeanalytical instrument sinceitnciplequestionswithout solving

determinethefield E in averysomeexamplesand then formu-onsabout thecaseswhen appli-isthemost expedient.ilityof stableequilibrium of achargethat wehavein vacuum asystem of fixed.Let us consider oneof thesecharges,le?ion, let usenvelop thechargeq by a

1.6). For thesake of detiniteness, weas-uilibrium of thischargeto bestable, it isatedby all theremaining chargesof theesurfaceS hedirected towardsthechargel displacement of thechargeq from theeriseto arestoring force, and theequilib-able.But such a configuration of the

in contradiction to theGauss theorem:rfaceS isnegative, whilein accordanceust beequal to zero sinceit iscreated byfaceS. Un theother hand, thefact thathat at somepointsof thesurfaceS vectoromeother pointsit is directed outside.



_electrostatic,field_achargecannot bein

niformly charged Elaine. Supposethato. It isc ear from t esymmetry of thely benormal to thecharged plane. More-threspect to thisplane, vectorsE obvi-

udebut oppositedirections. Such a configu-hat aright cylinder should bechosen forin Fig. 1.7, whereweassumethat o > 0.

surfaceof thiscylinder is equal tothrough theentirecylindrical surface

ea of each endface. Achargeo AS is.Accordingto theGauss. theorem, 2lJAS =o. In a moreexact form, thisexpression

fvector E onto thenormal n to the chargedrectedaway from thisplane. If o > 0or E is directed away from thechargedOntheother hand, if o < 0 then En < U,ardsthechargedplane. Thefact that E is

m theplaneindicatesthatj thecorre-form (both on theright and on theleft of

only for an infiniteplanesurface,nuse thesymmetry considerationsdis-sresult isapproximately valid for theregioniformlycharged planesurfacefar from

parallel planescharged uniformly withkecharges.das superposition of theEeldscreatedg.1.8). Heretheupper arrows correspond



_elycharged plane, whilethelower arrows,harged plane. ln thespacebetween theieldsbeingadded havethesamedirec-

will bedoubled, and theresultant held

tudeof thesurfacechargedensity. It canhisspacethefield isequal to zero. Thus,

ocatedbetweentheplanesand isuni-

yvalid for theplatesof finitedimen-aration between theplatesis considerablyensions(parallel-platecapacitor). ln thisf thefie d from uniformity areobservedates(thesedistortionsareoften ignored in

nfinitecircular cylinder uniformlyhat thechargeKcorrespondsto itsunit

symmetry considerations, thefieldor E at eacigoint iifperpendicular to

gnitudedepen sonly on thedistancerpoint.Thisindicatesthat aclosedsur-heform of acoaxial right cylinder

E through theendfacesof thecylinder isxthrough thelateral surfaceisE, 2 mhvector E onto theradiusvector r coincid-ateral surfaceof the cylinder of radiustheGausstheorem, E,2mh M/z~·,, for



e.vector E isdirected away fromceversa.oesnot contain any chasinceiniveof r. Thus, insideacircmr infinite

over the surfacetheheld isabsent.pherical surfaceuniformly charged by

allysymmetric: vector E from anytreof the sphere, whileitsmagnitudetancer from thepoint to thecentreof theh a conhguration of theheld weshouldclosedsurface. Let theradiusof thisordancewith theGausstheo1~em,'E,4nr' =

13)of vector E onto theradiusvector r coincidingaceat each of its points. Thesign of then of theprojection E,. in thiscaseaswell.ctionof vector E itself: either away fromardsit (for q< 0).esnot contain any chargeand henceywhere.Inother words,insideauni-rfacetheelectricheld isabsent. Outsideeswith thedistancer in accordancewithcharge.niformly charged sphere. Supposethatributed over asphereof radiusa. Obvious-is centrallysymmetric, and hencefor de-takeaconcentricsphereasaclosed sur-at for thefield outsidethesphereweobtainviousexample[see(1.13)]. However,sionfor theheld will bedifferent. Theesthechargeq' = q (r/a)’ sincein ourual to theratio of volumesandispro-

hird power.Hence, in accordancewith

edspheretheheld intensity growslinear·scentre.The curverepresenting thede-in Fig. 1.10.sultsobtainedin theabove



direct integration (1.5) as well.seproblemscan besolved in aheGausstheorem.

problemsconsidered abovemayn about thestrength of the method

dh,

af

g_ M0

g thefieldof asymmetricchargechargeddisc. In thiscase, thecomplicated,and aclosedionof thefluxof E cannot be

fectivelyapplied to calcula-eld hasaspecial symmetry (iul,or circular).Thesymmetry,ation,must besuch that, firstly,surfaceS can bef ound and,he' flux of E can bereducedf E (or E,.) by thearea of theconditionsarenot satisfied,eldshould besolved either di-using other methodswhich

GaussTheoremectricfield expressed by theGaussheorem berepresented in adifferent formsibilitiesasan instrument for analysis

calledtheintegral form weshall seek Gausstheorem,which establishestherela-

argedensity p and thechangesin thefieldagiven point in space.



resent thechargeq in thevolumeVceS in theform qm; = (p) V,where{p),averaged over the-volumeV. Then weo Eq.(1.7) and divideboth itssidesby

tend to zero by contracting it to thethiscase, (p) will obviously tend to the

of theheld,and hence theratio on theill tend to p ls,.

mit of theratio of § E dS to VasV -» 0heheldE andisdenoted by div E.Thus,

vector held isdetermined in asimilaron(1.16) that divergenceisascalar

sion for thedivergenceof the held E,(1.16), take an inhnitely small volumeV,

ughthe closed surfaceenveloping thist isfluxto thevolume. Theexpression

will depend on thechoiceof thecoordi-temsof coordinatesit turnsout to beartesiancoordinatesit isgiven by

V —» 0 in (1.15), itsright·hand sidehandside tendsto div E. Consequently,s related to thechargedensity at theion

Gausstheorem in thedifferential form.nsand their applicationscan beconsid-ducethevector differential operator v.eoperator v hastheform

)vectorsof theX-, Y-,and Z-axes. Thee any meaning. it becomesmeaningful



__calar or vector function by which it isr example,if weform thescalar producteobtain

$Ex+·bE/··Ey+··b;· Ez-hisisjust thedivergenceof E.eldE can bewritten asdiv E or v·Ee divergenceof E").Weshall heusingotation.Then, for example, theGaussform

ferential form isa local theorem:at a given point dependsonly on thehispoint.This isoneof theremarkableor example,theheld E of apoint charges. Generally,thisrefers to thespatialand6E,/6: aswell. However,theGaussf thesederivatives,which determinesthebeequal to zero at all pointsof theheld

re thedivergenceof E ispositive, we(positivecharges), whileat the(pointsvesinks(negativecharges). Thehel linessand terminateat thesinks.Potentialector E.It isknown from

ry heldof central forcesiscon-by the forcesof thisheld isdependsonly on thepositionThisproperty is inherent ineheld createdby asystem olit positivechargefor thetestnt 1of agiven held E to point 2.forcesof theheld done over theandthetotal work of the heldweenpoints1 and 2isd ehnedas



acertain line(path) and isgral.

m theindependenceoi lineetween two pointsit followsitrary closed path, thisinte-l (1.21) over aclosed contour istor E and isdenoted bynof vector E in any electro-e.

heorem on circulation of

m,webreak an arbitrary closedb1

gral (1.21)ot`pointsI

n the

ent b but taken in theopposite

2) is calledthepotential field.disapotential field.of vector E makesit possiblent conclusionswithout re-sconsider two examples}an electrostaticfield E cannot be



trueand somelinesof held Ewerelationof vector E along thislinewewoulddictionwith theorem (1.22). Thismeanssedlinesof E in an electrostaticheld;vechargesnes(or go

tionof an I Iig.1.12 l

elybecomesm on circula—,_.{.....,....l_.

contour ,_____________|shedline.ndicatethe——-——-—-—¤·—·—··—····With such aur,thecontri- Fi 112m its ver- g' ‘

nceinO. It remainsfor us to consider thequal lengths.Thehgureshowsthat theonfrom theseregionsareoppositein

ude(thecontribution from theupper seg-linesaredenser, and hencethevalueof rculationof Bdiffersfrom zero, which

deredthedescription of elec-tor E. However, thereexistscribingit by using potential qayoutset that thereisa one-to-nthetwo methods). It will beodhasa number of signihcant

21) representing thework·of splacement of a unit positive2 doesnot depend on thepathectricheld thereexistsacertaindinatessuch that itsdecreaseis

uesof the function qi at theqa(r) dehned in thiswayis



comparison of (1.23) with theeby theforcesof thepotentialo thedecreasein the potentialeld) leadsto theconclusiontitynumerically equal to thepo-ivechargeat a given point of the

eto an arbitrary point 0 of thetential.Then thepotentialsof

will beunambiguously deter-wechangecp,by a certain valueer pointsof thefield will change

nedto within an arbitraryof thisconstant doesnot playhenomenadepend only on theetermined,as.will be"shownagiven point but by thefpoten-hbouringpointsof theheld.olt (V).oint Charge.Formula(1.23)efinition of potential qa,theion.For thispurpose, it issuf-

ral S E dl over any path be-present theobtained result astionwhich isjust cp (r). Wecan- usethefact that formulainitedisplacementsbutffor ele-well.Then,in accordancewithdecreasein thepotential over

hefield E (r), then to find rphthehelp of appropriatetrans-a certainfunction. Thisfunction

finding thepotential of the



. . . .‘ -._.__ _ ’

that e,dl 1-(dl),. s: dr, sinceandhenceon r, is equal to theoi vector r,i.e. dr. Thequantitysunder thedifferential isexact-onstant contained in thefor-sical role,it isusually omitted inionfor qaThus, the potential

eisgivenby

constant in this expressionnallyassumethat thepotentialor r —» oo).stem of Charges. Let asystem

esq1, gz, .... In accordanceosition,thefield intensity at anyy E = E,+ E2 -i— ., wherem thechargeq1, etc.Byusingwrite= E,·dl + E2·dl +.ncipleof superposition turnsoutell.Thus, thepotential of asisgiven by

thepoint chargeqi to the pointalso omitted an arbitrary con—

eement with thefact that anyunded in space, and henceitsto zero at infinity.



ystem aredistributed contin-ssumethat each volumeele-hargep dV, wherep isthechargeaking thisinto consideration,in adifferent form:

ormed either over theentireningthecharges.If thechargesceS, wecan write

edensity and dS istheelementxpression correspondsto thealinear distribution.distribution(discreteor con-e, find thepotential of any

tial and Vector Eis completelydescribed byng thisfunction, wecan findnder investigation at any pointrk of field forcesfor any displace-n. And what doweget byof all,it turnsout that if weagiven electricheld, wecan re-eeasily. Let usconsider this

d E can be established with thesplacement dl be parallel towherei is theunit vector alongrement of thecoordinatex.In

fvector E onto theunit vector it dll).Acomparison of thisex-



__ _ · _ _ ___•_j) gives

_derivativeemphasizesthat thedil'lerent.iated only with 1·espectare constant in thiscase.ainthecorresponding expres-

andE,. Having determinedyfind vector E itself;

esesisthegradient of thepo-Weshall beusing thelatter, more

formally consider vqn as ther Vand thescalar cp. Thendin theform

ual to thepotential gradientexactly theformulathat canhefield E if weknowthe

nsityE if thefield potential hasthey, whereaisa constant; (2) tp (r) ——a·r,and1- istheradiusvector of a point under

; weobtain E = a(yi -{- mj).unctiontp ascp -—a,,.·r --- uyy --areconstants.Then with thehelp of for-i A- ayj -+- azk = a. lt can beseen that inm.ul formula.Wewritetheright-rm E·dl ll, dl,wheredl =cement and 15; isthepr»;jec—placement dl. Hence

E onto the direction of thedis-



directional derivativeof thedby thesymbol of partial de-

usintroducetheconcept of esurfaceat all pointsof whichlue.Weshall showthat vector Esdirectedalong thenormal totowardsthedecreasein the

32) that theonto

atagivenThisorm-.Fur--o theingnd accord-

decreasingcp, or in thediroc·ector vcp._potential surfacesin such ance between two neighbouringhedensity of equipotentialtethemagnitudesof field inten-ldintensitywill behigher inntial surfacesaredenser ("the

anequipotential surfaceevery-hogonal to thesesurfaces.imen.siona1pattern of an electricfield.

d to equipotential surfaces, whilethe solidarepresentation aan beeasily visualized.rectionof vector E, theregionswherefieldeit is lower, aswell astheregionswithntial relief.Such apattern aan he used to

oanumber of questions, suht as“In whatedat acertainpoint move?Whereistheradient higher?At which point willgebegreater."etc.



_tial.lt wasnoted earlier thatelycharacterized by vectortheuseof introducing potential?onsfor doing that. The conceptseful,andit isnot bychancesed not only in physicsbut in

l tp(r),wecan easily calculateeinthedisplacement of apointnt 2:

tentialsat points1and 2. Thiskisequal tothedecreasein the

geq' upon its displacement fromhework of thefield forceswith

not just very simple, but isbleresort.ibuted over athin ring of radiusa.rcesdonein the displacement of apointhe ring to infinity.chargeq over thering isunknown,weeabout theintensity E of thefield createdhat wecannot calculatethework by eval-nthiscase. Thisproblem can beeasilyntial.Indeed, sinceall elementsof theafrom thecentre of thering, thepoten-s,,e. And weknowthat rp = U at infinity.q’cp,,= q’q/Zirrsoa.hat in order to find electricfirst to calculatethepotentialthanto calculatethevalueof E

bleadvantageof potential. ln-must evaluateonly oneintegral,

must taket/zreeintegrals(sinceeintegralsfor calculating qu arer Ex, 13,,, and E,.snot apply to a comparative-swith high symmetry, infield E directly or with thehelput to bemuch simpler.



agesin using potential which

lectricdipoleisasystem of twoarges+ q and—— q,separated

nthe dipolefield isconsidered,

itself ispointlike,i.e. thedis-epointsunder consideration ishan l.

metric.Therefore, in any planeaxisthepatternof theiield ishisplane.ial of the dipolefield andthen25),thepotential of thedipole4a) isdefinedas

y;. + °rom Fig.1.14a that r- ... r+ =er is thedistancefrom theintlikel).Taking thisinto

moment of thedipole. Thisquan-irectedalong thedipoleaxis



tivecharge:

tor directed asp.1.34) that thedipoleheldent p. It will be shownbelowolein anexternal held alsoy,p isan important characteristic

hepotential of the dipoleheldr faster than thepotential of theoportion to 1/rzinstead of 1/r).eld, weshall useformula(1.32)of vector E onto two mutuallyngtheunit vectorse,. and eg

or E will be1-+3c¤s2o.(1.37)il = rt/2weobtain theex-ityon thedipoleaxis(Eu) and

, (1-38)ityE u istwiceashigh asE _,_.ipole.Let usplacea dipoleintoSupposethat El. and E- arel held at thepointswherethechargesarelocated. Then the

hedipoleis(Fig. 1.15a):— E-)-s theincrement AE of vector Eipolelength l in thedirection



of thissegment issmall, wecan

into theformulafor F, wehed ipoleisequal to

ectricmoment.Thederivativeiscalledthe directional deriva-

mbol of atesthat itertainioncoincid-

tyof ve: takingsarather com-cal operation.squestioniontodresult.t in auniformThis

ceactsorm field.Next, in thegener-ncidesneither with vector Eoincidesin direction onlyent of vector E, taken along theb).onsof theforce F acting on adipoleq for three different dipoleorientations.ove independently that it isreally so.jection of forceF onto aficient to writeequation (1.39)ntothisdirection, and weget



eof thecorresponding projec-hedirectionof vector l or p.beorientedalong thesymme-

orm fieldE. Wetakethepositivexample,etheincre- q•_______;=___?___hedirec- Ftive,F,,< 0,edtotheeldin- 4* _———— lpshown in ·

at thedipoleFmmetry°°""""'”E${silyseenig- U6ng on aaviour of adipolein an exter--of-masssystem and find outteor not.For thispurpose, weernal forceswith respect to the

of forcesF+ = qE+ and F- =ecentreof massC (Fig.1.18) is

r+ XqE+] — [r- XqE-1,svectorsof thecharges+ qt C. For a sufficiently small di-= [(r+ --r_) XqEl. It re-ount that ro, -r_ = I and

o rotatethedipole sothatntedalong theexternal field E.eisstable.ricfield adipolebehavesasrespect to thecentreof mass in orderertial forces.



themoment of force(1.41),tedalongthefield (p E),resultant force(1.39) it is

- C Po

wherethefield E haslargernsare simultaneous.

nExternal Field. Weknowargeq in an external field iseldpotential at thepoint of lo-oleisthe system of two charges,

xternal heldis+ ——w-) ,tentialsof theexternal field atharges+q and -q. To withiner of smallness,wecan write

veof thepotential in thedirec-ngto (1.32), Ocp/0l = —E;,

= —E·l, from which weget

that thedipolehasthemini-pE) in theposition pM E (them).If it isdisplaced from this



ernal forceswill return thedi-on.

iformly chargedwith surfacechargericfield intensity E on theaxisof thishthediscisseen at an angleQ.

metry considerationsthat onthediscwith thedirection of thisaxis(Fig. 1.19).hecomponent dE, from thecargeof oint Aand then integratetheobtainedurfaceof {thedisc. It can he easily seen

0is thesolid angleat which theareapoint A, and expression (1) can bewrit-

yis

gedistancesfrom thedisc, Q= S/r*•scand E = q/4nc,r' just asthefield of the

mmediatevicinity of thepoint 0, the/2:;,,.



gring of radiusRischarged with awhere X., isa positiveconstant and qaeelectricheld intensity E at the centre

distributionisshownin Fig. 1.20.utionimpliesthat vector E at thepoint 0s magnitudeisequal to thesum of thenof E of vectorsJE from elementary

of vector dE onto vector E iscosqa, (ilosqa dsp. Integrating (1) over rp between 0deof thevector E:

tegral isevaluated in themost simplethat (coszcp) = 1/2. Then

—=n.

ht uniformlycharged iilament hasand t emagnitudeand t edirection of theparated from thefilament by adistanceyefilament,passing through itsend.

duced to finding Exand Ev, viz.the1.21, whereit isassumed that Z.> 0).ntributionto Ex from tbe"chargeelement

n tothe form convenient for integration.,r = y/cosot. Then

ver ot between 0 and ::}.2, weEndnEv,it issufiicient to recall thatsina Qin (1) issimply replaced by cosq.



and EU= Z./4:re,,y.stingresult: Ex = Evindependently of y,

eangleof 45° to thefilament. Themodu-

oy.heintensityof anelectricfield de-es.1: and y asfo lows:

andj aretheunit vectorsof theX- andhin asphereof radius` Rwith thecentre

htheGausstheorem, therequiredE through thissphere, divided bye,.

netheflux asfollows. Sincethefield Eisfauniformlycharged filament), wearrivex throughthesphereof radius_f_Risequall surfaceof acylinder having thesamedarranged asshown in Fig. 1.22.Then

•2R= 4nR’.Finally, weget

uniformly charged sphereof radiusm filledby achargewith thevolumeden-ositiveconstant and r isthedistancefrom



thechargeo f thespherefor which thesidethesphereis independent of r.Find

rgeof the spherebeq. Then, using the

ethe following expression for asphericalhespherewit thechargeq):

E·4IIT2dT.

rm thisequation to+ émxrz/2z-:0.end on r when theexpression in theparen-e

ntensityE in theregion of inter-rmly chargedby unlikechargeswith theif t edistancebetween thecentresof thetor l (Fig. 1.23).eorem,wecan easily showthat theauniformly charged spliereis

Ct0l‘ 1'€l3l.lV€ to l.l”l€ COIll.l`€ of {lit! S]'!l'iE'l'€. W9regionof intersection of thespheresasthetwouniforlmy charged spheres. Thenat4) of thisregion wehave

—- r,.)/3s,, = pl/3s,,.ection of thesespheresthefield isuni-idregardlessof theratio between theradii



tancebetweentheir centres. In particular,scompletelywithin theother or, in otherical cavityin asphere(Fig. 1.25).hepreviousproblem, find thefieldver whichachargeisdistributed with

s6, whereon is aconstant and 0 is

o spheresof thesameradius, havingechargeswith thedensitiespand —p.hespheresaresep-1.26); Then,inn of t epreviousTZon of intersectionorm: _ _

gedihersfrom zero / \ r avery small l,t of thesurfaceThethicknessof \ / sdeterminedby ~` I';o l cosil.Hence,*·§=§unit areaiso0=pl,and_entedin theform Fig. 1.26

theZ-axisfrom which theanglefi is

ial of acertain electricfield hasbthedtheprojection of vector E onto thedi-(— 3k at thepoint M (2, 1, -3).ctor E:2*)-



J·—2¤k)(i+3k)__ —¤¤(v-—6¤>

at theedgeof athin discof radius.}!utedover oneof itssideswith thesnr-

otential in thecaseof asurfacechargeegral (1.28). In order to simplify in-earea element dS in theform of a partdthfdr (Fig.1.27). ThendS == 20:- dr,n 0 dd. After substituting theseex-), weobtain theexpression for qi at the

6.

noting 1*} = •

os*}—|—sin 0,tingthelimitsof integration. Asare-

ield insidea charged spheredependsts, centreto thepoint under considerationr”+ b, whereaand _b areconstants.olumechargep (r) within the{sphere.fieldintensity. According to (1.32),

em: 4m·’E,.= q/so. Thediherential of

nr* dr,



inedbetween thespheresof radii r and

przdr,(%€{·-l-·—§—·E, == TBP-;.equation,weobtain

duniformly within thesphere.eof interaction between two pointpl, if thevectorspl and pzare directedcting thedipolesand thedistancebe-

9), wehave

yfrom thedipolepl, determined by the

sexpression with respect to l and sub-for F,weobtain

poleswill beattracted when pl 11 pl and

ostaticField

elds.Thereal electricfield inedthemicroscopicjield) variesn time.It isdifferent at differ-einterstices.In order to iindat a certainpoint at agivenheintensitiesof the fieldsof allof thesubstance, viz. electronsis problem isobviously notult would beso complicated



to useit.Moreover, theknowledger the solutionof macroscop-t is sufficient to haveasimplerch weshall beusing henceforth.asubstance(which iscalledall understand themicroscopicnthiscasetimeaveraging isisperformed over what iscalledlume,viz. the volumecontain-and having thedimensionshanthedistancesover whichablychanges.Theaveraging overirregular and rapidly varyingpicheld over thedistancesof retainssmooth variationsof

macroscopicdistances.ceis

eon aField.lf anysubstanceiseld,thepositiveand negatives) aredisplaced, which in turn

of thesecharges. In certainompensatedchargesof differentnoniscalledtheelectrostaticappearing asaresult of sepa-arges.ditional electricheld whichial (external) field formstheexternal field and thedistribu-canforget about thepresenceof culatingtheresultant held,cehasalready been taken intoucedcharges

hepresenceof asubstanceisperposition of theexternal fieldges. However, inmany casesbythefact that wedo not knowargesaredistributed in space,benot assimpleas it could

shown later that thedistri-



mainly determined by theprop-s physical natureand thel haveto consider theseques-

ea Conductort usplacea metallicconduc-aticfieldor impart acertaineelectricheld will act on allr,andasaresult all thenegativesplacedin the direction againstcurrent) will continueuntill fractionof asecond) acer-in, at which the electricfieldnductor vanishes. Thus, in thensideaconductor isabsent

herein theconductor, thensated) chargesinsidetheconduc-l points(p - ·= 0).Thiscanbepof theGausstheorem. Indeed,

E = O, theflux of E through anynductor isalso equal to zero.eno excesschargesinsidethe

on theconductor surfacewithenerally different for different

uld benoted that theexcessvery thin surfacelayer (whosewo interatomicdistances).a conductor indicates, inpotential cp in theconductorpoints, i.e.anyconductor inuipotential region, itssurfacece.conductor isequipotentialevicinity of thissurfacethetedalong thenormal to thesur-ue,thetangential component of

moveover thesurfaceof the



:brium would beimpossible.of an uncharged conducting sphereq iséocatgdiiit adistancer from its centre

for all pointswe can calculate

the sphere,int it canmost simpleway:s thepotential of ndisthe

ucedont sinceall induced chargesareat thesamendthetotal induced chargeisequal tothiscase thepotential of thespherewillrst term in tl).dthechargedistributionsfor

onducting spheresoneof whichof electricinduction, thechargeson thesurfaceof theright_of these__chargeswill in turn

argeson thesurfaceof thelefttribution will becomenonuni-figurearethelinesof E,whileersectionof equipotential sur-gure.Aswemoveaway from

al surfacesbecomecloser_andld linesbecomecloser to ra-aseresemblesmoreand morethethetotal chargeof the given

Surface.Weshall showthattl1eimmediatevicinity of thenectedwith thelocal chargeacethrough asimplerelation.hedwith thehelp of theGauss

theconductor surfacewearecuum.’l`hefield linesarenor-e.Hencefor aclosed surfacewe



ndarrangeit asisshowninhrough thissurfacewill beghthe"outer"endfaceof thethelateral surfaceand the

ero). Thusweobtain E,, AS =

projectionof {vector E ontoespect to theconductor), ASthecylinder and o isthe local

conductor.Cancelling bothAS,weget

tor E isdirected from theeindirection with thenormal n).ctor E is directed towardsthe

heerroneousconclusion thataconductor dependsonly onhisisnot so. Theintensity Egesof thesystem under con-

ueof o itself.rfaceot a Conductorn acharged region of thesur-

on avacuum. Theforceactingnductor surfaceis



thiselement and E0 isthefieldgesof thesystem 'Qintheregion

cated.It should benoted at theal to the field intensity E inaceelement of theconductor,ain relation between them. LetressE0 throughE.efieldcreated by thechargehepointsthat arevery closetoit behavesasan infiniteuni-in accordancewith (1.10), E0=

ideand outsidetheconductorsthesuperposition of thefieldshe areaelement AS thefield E0etheheld E0hasoppo sitedi-for thesake of definitenessit isheconditionE = Oinsidethe0== E0, and then outsidetheE = E0 + E0 = 2E0.Thus,

hisnthengctor:

expression

in it areinterconnected. Indeed,= 0/e0, or E = (0/e0) n, whereesurface element at agivene



that o = e,E,,§and E3,= E2.surfacedensity of force. Equa-lessof thesign of o,and hencee Fu isalwaysdirected outsideetchit.on for theelectricforceacting in awhole,assuming that theheld intensitieshevicinity of theconductor surface.obtain theexpression for theforcedF

nt dS:

ant forceacting on theentireconductorhisequation over theentireconductor

Conducting Shellrium there_areno excesscharges

material insidetheconduc-nsequently,if thesubstanceismeinsideaconductor (aclosedot changetheheld anywhere,ibrium distribution of charges.hargeisdistributed on aesameway as onauniformurface.ricchargeswithin thecavityero in it. External charges,outer surfaceof theconductor, do

nthe cavityinsidetheconductor.trostaticshielding, i.e. thescreeningtruments,from theinfluence

ds.In practice,asolid conduct-a sufficiently densemetallic

insideanempty cavity can. Let ustakeaclosed surfaceSingcompletely in thematerial



eld E isequal to zero insidetheough theturfaceS isalso equalewith theGausstheorem, thel to zero aswell. ThisdoesnotedinFig 2.5, when thesurface

qual quantitiesof positivever, thisassumption isprohibit-hetheorem on circulation of ntour l` crossthecavity alonglosedin theconductor material.al of vector E along this contourn contradiction with thetheorem

ewhen thecavity isnot emptycchargeg (or several charges).external spaceisfilled by alibrium,thefield in thismediumsthat the medium iselectrical-cesscharges.uctor,thefield flux through ahecavity isalso equal to zero.orem,thismeansthat thealge-hinthisclosed surfaceisequal toraicsum of thechargesinducedl in magnitudeand oppositeinhechargesinsidethecavity.

nduced on thesurfaceof thecompensatecompletely, in the



field createdby thecharges

m iselectrically neutral every-theelectricfield in any way.medium,leaving only aconduc-thefield will not bechangedqual to zero beyond thisshell.

essurrounded by aconductingucedon thesurfaceof thecavityhell) is equal to zero in theen-

mportant conclusion: aclosedentirespaceinto theinner .andetelyindependent of one anotherhismust be interpreted asfollows:

of chargesinsidetheshell doesthefield of theouter space,ution on theouter surfaceof theesamerefers to thefield insideges) and to the distribution of ywalls.They will also remainement of chargesoutsidethergumentsareapplicableonlystatics.within an electrically neutral shellerical shape(Fig. 2.6). Find the potentialdetheshell at adistancerfrom the.terminedonly by chargesinduced onesince, aswasshown above, thefield of theargesinduced ontheinner surfaceof thewhereoutsidethecavity. Next, in viewof eouter surfaceof theshell isdistributed

eisa special caseof aclosedon one sideof thisplaneishespaceon itsother side.property of a closed con-



clroslatics.

roblemsin which thechargethepotentialsof conductors,ngement aregiven. Wemusty point of thefield between thealledthat if weknowthepoten-f canbe easily reconstructed

mediatevicinity of theconductorrminingthesurfacechargeors.uations. Let usderivethe differentialpotential).For thispurpose,we substitute20) theexpression for E in termsof cp,wo obtain thegeneral differential equationthePoisson equation:

rator (Laplacian). In Cartesian coordinates

see(1.19)].eentheconductors(p = 0), Eq.(2.8)requation,viz. theLaplaceequation:

must find a function cp which satisfiesre` spacebetween theconductorsand ac-cp,,on thesurfacesof the conductors.ythat thisproblem hasauni-t iscalled the uniquenesstheo-t of view, this conclusion isorethan onesolution, there1iefs",and hencetheheld E atasinglevalue.Thuswearriveusion.m,wecan state that in a



ibutedover tbhesurfaceof awell.Indeed, thiereisaone-to-

tweenthecharges: on theconduc-hevicinity of ittssurface: 0=

lyfollowsthat theuniquenesseuniquenessof the chargetor surface.d (2.9) in thegenerail caseisa compli-

m.Theanalyticsolutionssof theseequationsparticular cases. As ffor theuniquenessution of a number of eelectrostaticprob-

blem satisfiestheLaiplace(or Poisson)onditions, wecan stateethat it is correctmethodsby which itt wasobtained (if

mptycavityof a cconductor thefield

y theLaplaceequaticon (2.9) insidethevalueq>0 at the’cavity’swwal s. The solutionfying thiscondition caan immediately beordancewith theuniquemesstheorem, thereence, E == —V<p= t0.ificial methoad that makesit

mpleway theelectricfield inses. Let us consiider thismethodf apoint chargeqq near an infinite.n that wermust find another

ysolved and whtosesolution or aproblem.In ourr casesuch asim-bout two chanrges: q and ——-q.



ell known (itsequipotentialhown in Fig.2.7b).planecoincidewith themid-potential cp == 0) and removeingto theuniquenesstheorem, theewill remain unchanged. Indeed,

planeand everywhereat infinity.nsideredto bethelimiting caseor whoseradiustendsto zeros, theboundary conditionsforlf-spaceremain thesame,andn isalso thesame(Fig. 2.70).n arriveat thisconclusioniesof a closedconducting shellf—spacesseparated by thecon-lyindependent of oneanother,ge—-q will not affect t.hefield

deration thefield differsfrom—space.lnorder to calculatethis

uceafictitiousimagechargeto thechargeq, by placing it ontingplaneat thesame distanceplane.Thefictitiouschargeq'pacethesamefield asthat. of thee. Thisisprecisely what isictitiouschargeproducesthesamearges.Wemust only bear inictitiouschargeextendsonly t.ol chargeq islocated.In anothernt.t theimagemethod isessen-

otential to theboundary con-another problem (configura-field configuration in theregionisthesame. Theimagemethod

fthis"can bedonewith thehelpgurations.Let usconsider one

placed between two mutually perpen-a). Findthelocation of fictitiouspoint



echargeq isequivalent to theaction of alllf planes. *(point chargesfor which theequipoten—ulg(a) (lz)tious_his’ lf Q·q qsuch a|of four ’ I- _|_reA" "' """'- lal to lhar- q -qd|e lQì r .,8on i"“'

tionof point charges(another problem),er of other questions, for example, {ind

sity on any point withinthe"right angle"on thechargeq.

osConductor. Let usconsider aonductor removed from otherrges.Experimentsshowthat thedirectlyproportional to itsat at iniinitypotential isequal to,theratioq/cp doesnot dependrtainvaluefor each solitary

pacitanceof an isolated conductornumericallyequal tothechargeonductor in order to increasepacitancedependson thesize

nceof an isolated conductor which hasusR.2.10) that for this lpurposewemustor by a chargeq and caculateits poten-



1.23), thepotential of asphereis

2.10), wefind

e capacitanceof a conductor1 Vwhen achargeof 1 Cispacitanceiscalled thefa-

ntity.It correspondsto thephere9 >< 10° km in radius,usof the Earth (thecapacitancetual practice, weencounterand 1 pF.not isolated,itscapacitancesother bodiesapproach it. Thisldof thegiven conductor causesnthesurrounding bodies, i.e.t thechargeof the conductorucedchargeswill benearer toivecharges. For thisreason, thehichisthe algebraicsum of theandof thechargesinduced onhen other uncharged bodiesitscapacitanceincreases.ossibleto createthesystemonsiderablyhigher capacitanceuctor.Moreover, thecapacitancend on surrounding bodies.pacitor.Thesimplest capacitor(plates) separated by asmall

of external bodieson thetsplatesarearranged with re-away that thefield created bythem isconcentrated almosttor.Thismeansthat thelines

must terminateon theother,



must beequal in magnitude——q).capacitor isits capacitance.isolatedconductor, thecapaci-edastheratio of itschargetoeentheplates(thisdifferenceis

isthechargeof itspositively

fa capacitor isalso measured

or dependson itsgeometrythegap between theplates,

ecapacitor.Let usderiveitancesof some capacitors

uum betweentheir plates.ateCapacitor. Thiscapacitoresseparated by agap of width h-risq, then, according to (1.11),weenitsplatesisE == 0/e,,eaof each` plate. Consequent-latesis

into (2.12),we obtain

without taking into accountgesof theplates(edgeeffects).necapacitor isdetermined bytelythesmaller thegap h inimensionsof theplates.

Capacitor.Let the radii of theatesbeaand b respectively. I1q, field intensity between theausstheorem:



is

capacitanceof aspherical

egap between theplatesis(or b), thisexpression isre-

ression for thecapacitanceof a

Capacitor.By using thesamese of aspherical capacitor, we

gth, aand b aretheradii of thelates. Likein thepreviousn isreduced to (2.13) when theall.on thecapacitanceof aca-ec. 3.7.

noi potential.Apoint chargeq isatOof an uncharged spherical conductingradii areequal to aand b respectively.nt Oif r ·; a.rostaticinduction, say,negativechargessurfaceof thelayer and positivechar-2.9). According to theprincipleof super-l at thepoint Ocanbe represented in

_b )’kenover all thechargesinduced on theile thesecond integral, over all the



It followsfrom thisexpression that

tential canbefound in such asimpleeall thelikeinduced chargesareat the

olilnt and theiro us) ·

wo"@.sphereof · ,/‘What cha- / %heouter •,,/·thepoten- / )Ml to zero?f · / eof poten- / /

m thecen- %matically< ,eassuming

essionsstem (q>H) and in theregion between the

+q)°’ant.Itsvaluecan be easily found from theR2, cpu = q>I. Hence

= 0 wefind that gz = —q1R2/R1.2.10) will havetheform:

RT) ’surfacecharge. Anuncharged me-lacedinto an external uniform field, asahargeappearson thespherewith surface0,, isapositiveconstant and 0 isapolarf theresultant electricforceacting on

theforce acting on theareaelement

nsiderationsthat theresultant forceFs(Fig.2.11), andhenceit can berepresented



rojectionsof elementary forces(1) onto

areaelement dS a spherical zonedS =

ingthat E = o/eo, wetransform (2) as

} di} = — (:wgR’/F0) coss1*} d(cos0),ver thehalf-sphere(i.e. with respect totain

nt chargeq isat adistancel from annd thedensity of surface chargesinducedthedistancer from thebaseof theper-chargeq onto theplane., theQsurfacechargefdensity on aconduc-ctricfieldnear itssurface (in vacuumg,. Consequently, theproblem isreducethevicinity of theconducting plane.

efind that thefield at thepoint Pancer from thepoint Ois



testhat theinduced chargeis opposite

a distancel from an inhniteconduc-theelectricforceacting on thechargeqtoavery largedistancefrom the plane.

work of thisforcedoneupon an ele-ig. 2.13) isgiven by

eforceisobtained with thehelp of theationover :: between I and oo, we

ethisproblem in a different way (throughousresult which differsfrom whator of two.Thisisbecausetherelationnlyfor potential helds. However, in theconducting plane, theelectricheld of ntial held: a displacem_ent of the chargeheinduced charges, andtheir held turns

g of radiusR,having achargeq, isto an inhniteconducting planeat a dis-surfacechargedensity at apoint of theithrespect to thering and (2) theelectrichering.nthat in accordancewith theimageqmust belocated on asimilar ring but



ucting plane(Fig. 2.14). Indeed, only inmidp anebetween theseringsis equal toepotential of the conducting plane. Letalready know.oint O, wemust, according to (2.2),Fig. 2.14). Theexpression for E on the

n Example1 (seep. 14). In our case, thisAsaresult, weobtain

reof thering is equal to thealgebraicpoint createdy thechargesq and --q:

argesarearranged asshown inright angleformed by twoconductingachof thechargesis I q land thedis-wn in thefigure. Find (1) thetotal charge

alf—planesand (2) the forceacting on t e

forming theangleAOB go to infinity,= 0. It can beeasily seen that asystem

eswith tp = 0 coinciding with theconduc-shown in Fig. 2.152:. Hencetheactionconducting hali—planesisequivalent to

arge-·-q placed in thelower left corner of

red thefirst question; -—q.ur point charges, wecan easily find



.15b)

tion.el wires. Two long straight wiresarearrangedin air parallel to oneanother.

iresis1] timeslarger than theradiusof theecapacitanceof thewiresper unit length

hargethetwo wiresby chargesof theitesignsso that thechargeper unit lengthnition,therequired capacitanceis

hepotential differencebetween thewires.owing thedependencesof thepotentialstween theplatesthat thesought potential

+l·(2)held created by oneof thewiresat adis-easily found with thehelp of theGaussn'

wires’ crosssection and b istheseparations.



3) .nat

that b >> a.icplatesarearranged in air at thether. Theouter platesareconnected byaplateisequal to S. Find thecapacitancents1 and 2, Fig.2.17).lates1 and2 by chargesqc and—q,,.pation field appearing between thesewill movein theconnecting wire, after

argednegatively whiletheplateBwilln electricfield appearsin thegapsbe-edby thecorresponding distribution of

ould benoted that as followsfrom thepotentialsat the[middleof thesystem asequal to zero.

nceof thesystem in thiscaseis

ntial differencebetween thepoints1t thepotential differenceUbetween theas thepotential differencebetween theon theright and on theleft). Thisalso

atethat according to (2) thechargeqo onarts: qa/3 on theleft sideof theplate1encestem (between points1 and 2) is

duced charge. Apoint chargeq israllel conducting plates1 and 2 sepa-

etotal chargesq1and q2 induced on eachctedby awireand thechargeq islocatedlate1 (Fig. 2.19a).rposition principle. WementallyeP thesamechargeq. Clearly, thiswilln eachplate.If we nowdistributeuniformlyargewithsurfacedensity 0, theelectricd(Fig.2.19b).thewire, and hencethepotential



qual to zero.Consequently,

ojectionsof vector E onto theX-axistoeplaneP (Fig.2.19b).that

.2), ol = e0E,,,1 = coli]; and 0, =nussign indicatesthat thenormal n2unit vector of theX-axis).

m these equations, we obtain/l.and q2 in termsof q haveasimilar form.er,to solvethisproblem with thehelp

would requirean infiniteseriesof ficti-oth sidesof t.hechargeq, and to find themplicatedproblem.icsricssulators) are substancesthatectriccurrent.Thismeansthatconductorsdielectricsdo notveover considerabledistances



_ricisintroduced into an exter-echangesareobserved in the

elf. Thisis becausethedielec-thecapacitanceof acapacitoradielectric,and so on.tureof these phenomena, we

n that dielectricsconsist eitherargedionslocated at the sitesstals, for example, of theNaCleither polar or nonpolar. In af "mass”of thenegativechargeentreof "mass"of thepositiveeculeacquiresan intrinsicdipoleulesdo not haveintrinsicdipoleof mass"of thepositiveand ne-ide.nof an external electricThisphenomenon consistsinismadeup by nonpolar molecu-chmoleculeisshifted along theeoppositedirection. lf a dielec-les,then in theabsenceof thereorientedat random (duetoactionof an external held, thedominant orientation in thed. Finally,in dielectriccrystalsheld displacesall thepositiveegativeions, against the held.*rizationdependson thestruc-

her discussion it isonly impor-larizationmechanism, all theprocessaredisplaced along therges,against theheld. It shouldonditionsthedisplacementsof n comparison with thedimen-isdueto the fact that theinten-

ngon thedielectricisconsider-olarized evenin theabsenceof anisinherent in dielectricswhich arelepermanent magnets).



esof internal electricfieldsin

arges.Asaresult of polariza-esappear on the dielectricsurfaceerstand better themechanism

es(and especially bulk charges),gmodelf Supposethat wehaveamogeneousdielectric(Fig. 3.10.)hthecoordinatex accordingby pi and p; themagnitudesof ositiveand negativechargesinre associatedwithtnuclei and

l field,pQ_ = p; at each pointelectriciselectrically neutral.

ncreasewith x dueto inhomoge-.1b). Thisfigureshowsthat inthesetwo distributionsex-ionof p;_(a:) isshown by the:), bythedashed line).fieldleadsto adisplacementgthefield and of the negativedthetwo distributionswill beher (Fig. 3.1c). Asaresult, un-pear on thedielectricsurfaceas1 an uncompensated negativeIt should benoted that there-



nchangesthesign of all thesenthat in thecaseof aplatemadec,thedistributionspl (x) andndonly uncompensated surfaceheir relativedisplacement in

pearing asaresult of polariza-dpolarization,or bound,charges.hat the displacementsof thesen moveonly within electricallydenotebound chargesby aprime

polarizationof a dielectricrfaceand bulk bound charges

t do not constitutedielectricarges.* Thesechargesmay beidethedielectric.

hefield E in adielectricistheition of thefield E0 of extra-

E' of bound charges:

opicfields, i.e.the microscopicnd charges, averaged over a

ume.Clearly,thefield E in theisalso a macroscopicfield.

scribepolarization of a dielec-lemoment of aunit volume.ctric(or both) are nonuniform,

different at different pointsof racterizethepolarization at alyisolatean infinitesimal volumedthen find the vector sum of thequentlycalled [reecharges, but this

somecasessinceextraneouschargesmay be



culesin thisvolumeand write

is called thepolarization of americallyequal to thedipolemo-esubstance.presentationsof vector P.Ndipoles.Wemultiply and di-3.2) by AN. Then wecan write

ncentrationof molecules(theirnd (pl = (Ep,)/ANisthemeane.rrespondsto themodel of ositiveandnegative"fluids”.olumeAVinsidethedielectric.ivechargepy AVcontainedcedrelativeto the negativeesechargeswill acquirethe. Dividing both sidesof thiseexpression for thedipolemo-

ector P:

he coulomb per squaremeter

xperimentsshowthat forand a broad classof phenomena,

ndson thefield E in adielectric.ndfor not very largeE, there

quantity called thedielectric.Thisquantity is independent

opertiesof the dielectricitself..snot stipulated, weshall



ctricsfor which relation (3.5)

tricsfor which (3.5) isnotnic crystals(seefootnoteon

Ther elationbetween P and Er and dependson thehistoryreviousvaluesof (thislphe-).l PField of P.Weshall showlowingremarkableand impor-

at theflux of P through an

equal to theexcessbound chargeedielectricin thevolumeen-

Gausstheorem for vector P.arbitraryclosed surfaceS

ic(Fig.3.2a, thedielectriciselectricfieldisswitched on,tspositivechargesare displacedes. Let us{ind thechargewhichS of theclosed surfaceS in the).racterizingthedisplacement



bound chargesasaresult of hat thepositivecharge

e"inner”part of theobliqueheareaelement dS from theb). Besides,thenegativechargehe"outer”part of theobliquecyl-

through theareaelement dS.ort of anegativechargein ant tothetransport of the posi-rection.Taking thisinto account,for thetotal bound chargepass-nt dS of thesurfaceS in theout-

_ dS coscx.

pll dS cosoz,

ivedisplacement of positivein the dielectricduringpolari-

P anddq' = P dS cos cz,

ver theentireclosed surfacehat left thevolumeenclosedzation.Thischargeisequal toexcessbound chargeq' willlearly,the chargeleaving theexcessbound chargeremainingith theoppositesign. Thus,we

). Equation (3.6), viz. theGaussor P, can bewritten in thedifferential form



:dof vector P isequal to thevolumedensitythesamepoint, but taken with theoppo-

beobtainedfrom @6) in thesamemannervector E waso tained (seep. 24). Foro replaceE by P and p by p'..

Dielectric?Weshall showxcessbound chargesin adielec-nditionsaresimultaneouslyhomogeneousand (2) thereithin it (p = O).

mainproperty (3.6) of theaseof ahomogeneousdielectric,inaccordancewith (3.5), takerite

st thealgebraicsum of all thebound—insidetheclosed surfaces equal to q + q' Hence,which weobtain

cessbound chargeq' andalidfor any volumeinsidethephysically infinitesimal vol-and q -—> dq = pdV.Then, after) becomes

mogeneousdielectricp' = 0

neousisotropicdielectricof lectricheld,wecan besureeriseonly to thesurfaceboundssbound chargewill hezero atc.ctor P.Let usconsider thenterfacebetween two homo-s,Wehavejust shownthat in



xcessbound bulk chargeandeappearsasaresult of polari-

eenpolarization P and thend chargeat theinterfacebe-ispurpose,let ususeproperty. WeutheS

eend- Df the2l as- I lyl- *d the"'o small Fig- 3·3llalso refersto thesurface densityn bethecommon normal to the

Weshall alwaysdrawvector nc2.rough thelateral surface,in accordancewith (3.6):

rojectionsof vector P in dielec-ndielectric1 onto thenormal

at theprojection of vector P ontoprojectionof thisvector ontomal n, takenwith theoppositen writethepreviousequationcancellingAS):

acebetween dielectricstherP hasadiscontinuity, whosearticular,if medium 2is aondition(3.12) acquiresa

vector P onto theoutward nor-dielectric.Thesign of thepro-



gn of thesurfacebound charge3.13) can bewritten in ad iffer-h(3.5), wecan write

vector E (insidethedielectricce) onto theoutward normal.terminesthesign of o'.Vector P.Relations(3.6)roneous conclusi_on that thely on thebound charge. Actually,ector P,as well asthefield of E,oth bound and extraneous.y thefact that vectorsP andrelationP = xe,,E.Theboundf vector P through aclosed SUI'-f P.Moreover, thisflux isde-u ndchargebut byits part

ldD. Sincethesourcesof anctriccharges-—extraneousandss theorem for thefield E in

eous and boundcharges en-appearanceof thebound chargeand formula(3.15) turnsouthe field E in adielectricevengood" symmetry. Indeed, thisrtiesof unknown field E in termsin turn isdetermined by un-

nbeovercomeb y expressinglux of P by formula(3.6). Then



nsformed asfollows:6)sesin theintegrand isdenoted

danauxiliary vector D

traryclosed surfaceisequal to thechargesenclosed by thissur-

Gausstheorem for field D.or Disthesum of two complete-

and P.For thisr eason, it iswhich doesnot haveany deep

theproperty of thefield of ation(3.18),justihestheintro-ny cases it considerably simpli-n dielectrics.*arevalid for any dielectric,c.t thedimensionsof vector

ector P. Thequantity Disquaremetre(C/mz).8) is

dDis equal to thevolumedensity of anmepoint.

edfrom (3.18) in thesameway asitp. 24). It sufficesto replaceE by Dandneouscharges.lleddielectricdisplacement, or electro-t beusing thisterm, however,in order toreof vector D.



rgenceof vector Dispositivewehave> 0), whileat thepointswherethediver-of thefield D(p < U).and E. In thecaseof isotrop-= x,,E.Substituting thistain D= so (1 + x) E, or

nstant of asubstance:

well asoi) isthebasicelectric.For materialse> 1, whileof edependson thenatureof

we.enthevaluesslightly differ-d several thousands(for somewater israther high (e= 81).

nisotropicdielectricsvectoror anisotropicdielectrics, thesellinear.llyrepresented by thelinesnand density a1·edetermined in. Thelinesof E may emergeandell asbound charges. Wesaysourcesand sinksof vector E.dD, however, areonly extraneousechargesthelinesof Demergeandsswithout discontinuitieseldcontaining bound charges.Vector D. Thefield of vectoraneousaswell asboundchargesThis followsif only from the

r, incertaincasesthefield of by extraneouscharges. It isor D isespecially useful. At`to theerroneousconclusionsonly on extraneouschargesationof thelaws(3.18) andnly acertainproperty of iieldeldproper.



d aboveby several examples.oint. chargeq islocated at thecentreeof anisotropicdielectricwith adielectric

ectionli,.of field intensity li asafunction of eof thissphere.

m allowsusto usethe Gausstheoremproblein(wec’annot useherethesimilarhebound chargeisunknown to us). l·òr antreat thepoint of location of thechargegrelation: 4:tr”U, = q. liencewecan find.20),the required quantity I;`,.:

— •?,80 ri ·D(r) and li (r).ystem consistsof apoint chargeq > 0homogeneousisotropicdielectricainclosed surface. Findout what will hap-and D (and to their fluxesthrough theremoved.paceisdetermined bythechargeqpolarizeddielectric. Sincein our caseieldDaswell: it isalso determined byby thebound chargesof the dielectric.ricwill changethefield E,and hencethe

E through thesurfaceS will also change,swill vanish from insidethissurface.DthroughthesurfaceS will remain theinthefieldD.asystem containing no extraneousund charges.Such a system can, forf anelectret (seep. 68).1·ìgure3.6a shows

What can wesay about thefield D?



traneouschargesmeansthat thefieldllIl€S of D(ll') HOL 0IIl0I‘g0 0I` i.0I`flìlil3lL(2 HI1yWll(‘l°(:.

ndisshown inFig. 3.Ub.The linesel E

phere,but insidethespherethey haveretherelationD= e0eE isnolonger va-

aviour of vectorsE andDo homogeneousisotropicdielec-er generality,an extraneousnterfacebetween thesedielec-nscan beeasily obtained withetheorem on circulation of rem for vector D:

m,.tor E.Let theHeld near the andE2 in dielectric2.We

ctangular contour and orient it asof thecontour parallel to thength that thefield E over thisbeassumed constant. Thet benegligiblysmall. Then, in

on circulation of vector E,



tor E aretaken on thedirectionontour,shown by arrowsinion of thecontour wetakethe

nto theunit vector 1:' but ontohen EW= —-— EH, and it fol-nthat

nt of vector E turnsout to beeinterface(it doesnot havea

tor D. Let us takeacylinderrangeit at theinterfacebe-8). Thecrosssection of theector Disthesamewithin each

ordancewith theGausstheorem

ty of theextraneouschargeatrojectionsof vector Donto thedirectedfrom dielectric1 to== ——D1,,,and theprevious

heform

that thenormal componentiscontinuitywhen passing



wever,thereareno extraneous0), weobtain

onentsof vector Ddo not haveobe thesameon different

neouschargesat theinter-eousisotropicdielectrics, the

y continuously during atransi-hilethecomponentsEnand

Theboundary conditionsomponentsof vectorsE and Ddielectricsindicate(aswillctorshaveabreak at thisin-

g. 3.9). Let us find therelationtz.chargesat theinterface,

h (3.22) and (3.24), E2.E1.,9 showsthat

veconditions,weobtain theandhenceof linesD: `

ndE will form alarger anglefacein thedielectricwith aez> el).raphicallythefieldsE and D at theeneousdielectrics1 and 2, assuming thatouschargeonthissurface.ewith (3.25) org > oil (Fig. 3.10).tial component of vector E remainsun-wecan easily showthat [C2 < lc', in

E indielectric1 must bedenser than inig. 3.10. 'l`hefact that thenormal com-



al leadsto theconclusion that U2 > U,f Dmust bedenser in dielectric2.

r consideration,thelinesof E arentinuities(dueto thepresenceof bound

areonly retracted, without discontinu-neouschargesat theinterface).Conductor-DielectricInter-ctor and medium 2isadielectric

m formula(3.23) that

utwardnormal (weomitted thential in thegiven case). Let usilibrium, theelectricfieldndhencethepolarizationngto (3.17), that vector DOhenotationsof formula(I. .23)

Dm, = U.ctor Surface. lf ahomogeneousegionof thesurface of acon-ertaindensity o' appear at thece(recall that thevolumedon-for a homogeneousdielectric).heorem to vector E in thesameriving formula(2.2). Consider-

d andextraneouscharges(o



ectricinterfacewearriveat= (0 + 0’)/e0. On theother= D0/ee0 = 0/e0. Combin-obtain 0/e== 0 -{- o', whence

edensity 0' of theboundambiguously connected withxtraneouschargeon thecon-argesbeingopposite.

usDielectricthedetermination of theceisassociated with consider-stributionof induced chargesin

beforehand.It isonly clearchargesdependson thenatureswell ason theconfiguration

l case,whilesolving theprob-E in a dielectric,weencounteration of themacroscopicfieldspecificcaseisgenerally acoin-m,sinceunfortunately therending E'en theentirespacewheretheremogeneousisotropicdielectric.greater detail. Supposethat wer several conductors) in ava-schargesarelocated on conduc-equilibrium thefield E insidecorrespondsto acertainfacecharge0. Let thefielddingtheconductor beE0.ceof thefieldwith ahomo-

ult of polarization, only surfacerin thisdielectricat theinter-cording to (3.27) the charges



ctedwith theextraneouschargesctor.eldinsidetheconductor (E =istribution of surfacechargesound charges0’) at theconductor-milar to thepreviousdistri-s(0),and theconfiguration of

electricwill remain thesamectric.Only themagnitudeof different.sstheorem,0 + 0’ = e0E,,,and hence

hefieldhavedecreased by afactorace,thefieldE itself hasbecomesamefactor:

isequation by seo, we obtain

snot changein thiscase.29) and (3.30) arealso validn ahomogeneousdielectricfillsn theequipotential surfacesscharges(or of an external field).d D = D0 insidethedielectric., theintensity E of the fieldedby asimplerelation withectric,namely,

btained from theformulaE =count that E0 = eE and

other cases thesituationandformulas] (3.29)-(3.31) are

geneousdielectricfillstheeld,theintensity E of thefield



sity E0 of thefield of thesamee absenceof dielectric, by asthat potential cp at all pointsr of ez

al in theabsenceof thedielectric.ential difference:

fferencein avacuum, in the

homogeneousdielectricntheplatesof acapacitor, theenitsplateswill be byafactorenceof dielectric(naturally,echargeq on theplates). AndeC—= q/ Uof thecapacitoraseetimes

f thecapacitor in theabsenceedthat thisformulais validentheplatesisfilled andedge

ctricand thebound charge. An extra-ecentreof a spherical layer of aheterogle-hosedielectricconstant variesonly in t ewherecx isa constant and r isthedistancem. Find thevolumedensity p' of a boundhin thelayer.3.6), taking asphereof radiusr aseof the spherecoinciding with thecentre

rgeinsidethesphere. Let ustaketheon:



ein athin layer between thesphereswithdering that dq' = p'4m·”dr, wetransform (1)

p'r2dr,

tionwehave

E?tionexpression (2) will havetheform

ult.r vector D. An inhnitely largeplateectricwith thedielectricconstant sisraneousP > 0.E v2a.(1) _ I _E and §§§Z§:§:§:§:§:§:§§

ofthe di- ggggigigigggigigg Exheplate§;;;$;g·;§;§;§; ‘zeroin [;§;§;§;2$§;{;§;§;§sing theO —n $:§:§:§]Y{;f:§:§:§: 41plate.gg§:§:§:j;ggj:§§§§§ Xepro- / g:§t§£§:§1;Z§:;Z$;;

d thepo- / 2;;::;;:;:::::;:::::; \ _eand / \€°d charge.;§:§E§:§:§E§:¥:§;§:§ \ rycon- / = 0 in Fi 3 Heat all g' ’rpend-late.In or der to determineE, weshallvector D(sinceweknow thedistribu-ges). Wetakefor theclosed surfaceaneof whoseendfacescoincideswith theross-sectional areaof thiscylinder beS.

(l { a),(l > a).



x (x) and cp (.1:) areshownin Fig. 3.11.graphof Ex (sc) correspondsto thede-

), thesurfacedensity of thebound

‘?.Zi.L pa> 0.idesof theplate. Thus, if the extraneousrgesappearingon both surfacesof the

ensity of thebound charge, weusewill haveasimpler form:

°d chargeisuniformly distributed overositeto that of theextraneouscharge.

ectrichastheshapeof asphericalradii area and b. Plot schematically theential qaof the electricfield asfunctionsntreof thesystem, if thedielectricisneouschargedistributed uniformly (1)layer,(2) over thelayer’sbulk.usstheorem for vector D,takingereof radiusrz

argewithin thissphere. Henceit follows

nr“.

n in Fig.3.12a. Thecurvefor cp (r) isalsorvecp (r) must havesuch a shapethat thetheoppositesign correspondsto thecurve, wemust take into account thenormali--» oo.urvecorresponding to thefunctionointswherethefunction E (r) has finitep(z-) isonly broken.ewith theGausstheorem, wehave



yof theextraneouscharge. Hence

r E (r) and cp (r) areshown in Fig. 3.12b.

uniformly distributed with thevolumef radiuso, madeof ahomogeneousdielec-nd (1) themagnitudeof vector E asam thecentreof thesphereand plot the

and cp (r); (2) thesurfaceand volumeden-

ermineE, weshall usetheGausstheo-nowthedistribution of only extraneous

—-80 3880 r,

··•·*··•co Boo F2 °E (r) and 3 (r) areshown in Fig. 3.13.eboun chargeis

density of bound charge, it issufficient todusto formula(3.11), andweget

n adifferent way,viz. by usingand it docsnot depend on coordinates(in-



ep' == ——>< (p + p'), which givesfor-

uctor. Find thecapacitanceof asa, surrounded by alayer of a homoge-

theouter radiusb andthe dielectriccon-vesfor E (r) and cp (r), wherer isthedis-phere,if thesphereis charged positi-

apacitanceC·—= q!q>. Let usfind theor,supplying mentally achargeq to it:

neOS rz dr`

weobtainaa/b') areshown in Fig. 3.14.citor.A spherical capacitor with, wherea< b, isfilled with an isotropichosepermittivity dependsonthedistancer

m ase== oi/r,whereon isaconstant. Findtor.hthedefinition of thecapacitanceof alem is reduced to determination of theivenchargeq:



argeof theinner plateisq > 0. Let usausstheorem for vector D:

sion into (1) and integrating, wefind

°d theprincipleoi superposition.ectricspherewhich retainspolarization

disswitchedoff. lf thesphereis_polar-nsity insideit isE' = —·P/Seo,wherePethisformulaassuming that thesphereisall displacement of all positivechargesof oall itsnegativecharges. (2) Using this0 of thefieldin thespherical cavity insideectricwith thepermittivity aif thefield

wayfrom thecavity isE.t thissphereas acombination of twoaringuniformly distributed chargeswith

Supposethat asaresult of a small shift,displacedrelativeto oneanother by a

at anarbitrary point Ainsidethesphere

eld intensity insideauniformly chargedchdirectly followsfrom theGausstheorem.account that in accordancewith(3.4),

cal cavity in adielectricisequivalentof a ball madeof apolarized material.



ewith theprincipleof superposition, thecanberepresented asthesum E == E' +

c0E, weobtain

In thevicinity of point A(Fig. 3.16)cuum interface,theelectricfield intensityd thevector E0formsthe anglean withthe given point.Thedielectricpcrmitti v-

whereE istheintensity of thefield in-nityof point A.yinsidethedielectricis

3.24) at theinterfacebetween dielectrics,

== E0,,/s= E0 cosot0/e,mponent of thevector E0 in avacuum.nsinto (1),weobtain

a vacuum at adistancel from theeousdielectricfilling thehalf-spacebelowermittivityisc. Find (1) thesurfaceden-function of thedistancer from thepoint

tainedresult; (2) thetotal bound chargeon

inuity of the normal component of acunm interface(Fig. 3.17):

e0rl cos1 2120 ’component of the electricfield creatednder consideration, wherethesurface

e last equality it followsthat



hat cost} = l/r. Asl-» 0 thequantitys just at theinterface, thereisno surface

rfacea thin ring with thecentreatosethat the inner andouter radii of thishesurfacesringisn1* + r'°,theexpres-th (1) gives

ver r be-

on theplaneseparating avacuumusdielectricwith thedielectricpermittivi·vectorsDand E in theentirespace.wsfrom the continuity of the normal

Em = slim. Only thesurfacechargeo'l component of vector E in thevicinityation.Hencetheaboveequality can be

= O.eisno bound surfacecharge(with theect contact with the extraneouspointeelectricfield in thesurrounding spacegeq+ q',and E dependsonly ontiedis-thechargeq' isunknown,and henceem for vector D. Taking for theclosedwith thecentreat thepoint of location of

nifadesof vector Dat thedistanceruum and in thedielectricrespectively.

yof thetangential component ef vector

tions,wefind

1+s) r2 ’



tyin theentirespaceis

heseformulasarereduced to theal-rl) and E of thepoint chargein avac-

esented graphically in Fig. 3.18. It

dl) in thiscaseis not determined by theerwiseit would havetheform of thelield

tem of Chargesion.Theenergy approachricchargesis,aswill be shown,sapplications.Besides, thiso consider theelectricfield

w.owwecanarriveat thecon-ctionin a system of charges.em of two point charges1braicsum of theelementary

of interaction between thertainsystem of_referenceK

by dll and dlzduring thetime



stem of Charges95work of theseforcesis

ccording to the Newtonhis expression in theform?sesisthedisplacement of 2. In other words, it isthedis-esystem of referenceK', whichdaccomplisheswith it a trans-

e initial referencesystem K.dll of charge1 in system Kcancement dll of system K' plusge1 relativeto system—- dll = dll,andm of theelementary worksbitrary system of referenceKntarywork doneby theforceher referencesystem (K') inrest. In other words,thework echoiceof theinitial system of

e1 from charge2 isconserva-sequently, thework of thegivenl canberepresented asthede-y of interactionbetween theeration:

epending only on thedistance

ystem of threepoint chargesasecanbe easily generalizednumber of charges). Thework nteractionduring elementaryrgescanberepresented asthers of interactions, i.e.6.4l. But asit has just been shown,



6A,_h == — dW,,,,and henceWn) = —dW.

eractionfor thegiven system of

ndson thedistancebetweenencethe energy Wof thegivenion of itsconfiguration.ouslyvalid for asystem of sequently,wecan statethat toitrary system of charges, thereof energy W, and theworkof pon achangein this configu-seintheenergy W

sfind theexpression for thensider asystem of threepointound that W= W1,+ W1, -1-nsformed asfollows. Weasymmetricform: Wu, ==h= Wh;.Th9I1

ai + wasT waz)-imilar first indices:

+ W za) + (Wei ‘l‘ Wall-isthe energy W, of interactionall ther emaining cling.canbewritten in theform

Wi.

ralizedtoasystem of an arbi-cetheabovelineof reasoning



n thenumber of chargescon-heenergy of interaction for

hereq, isthe ith chargeof ential created at thepoint of all the remaining charges, weor theenergyof interaction of

t.chargesq arelocated at the verticesea(Fig. 4.1). l·ìnd theenergy of inter-em.oreach pair of chargesof this system isq2/énroa. It can beseen from thefigureactingpairsissix, and hencetheenergyargesof thesystem is

tion of this(4.3). 'l`he pote-nof oneof t.heof all theotherenceqq

lf thechargesarearrangedingthesystem of chargesasychargesdg = pdVand goingon tointegration, weobtain



atedby all thechargesof thent dVAsimilar expression can

surfacedistribution of charges.eplacein (4.4) pby 0*and dVby

neouslyinterpreted (andtanding) asamodification of ding to the replacement of thehat of a continuously distrib-not so sincethetwo expres-riginof thisdifferenceisintential cp appearing in these ex-sdifferencewith thehelp of

f two small ballshavingncebetween theballsisconsi-

mensions, henceq1 and q2can bes.Let usfind tl1eenergy Wof othformulas.

Pi ·'= q2(p2vtial createdby chargeq2 (q1) at2) is located.gto formula(4.4) wemustntoinfinitelysmall elementsem by thepotential qacreatedentsof another ball but by theaswell. Clearly,theresult will

teractionof thechargeelementsher,W2 thesame but for theergy of interaction between theall and thechargeelementsof

sW1 and W2 arecalled theq1 and q2, whileW12 istheenergygeq1 and chargeq2.W calculated by formula(4.3)



gyWm, whilecalculation byenergy of interaction: in addi-cenergiesW,and W2. Disre-

frequently acauseof gross

ionin Sec.4.4.Now, webtainingseveral important re-

Conductor and

ctor. Let a conductor havep. Sincet.l1evalueof rp isthergeislocated,wecan takecpntegral.Theremaining integral

onductor, and weobtain

l (4.6)written assuming that Cq/qy.and q>+ bethe chargeand

harged plateof acapacitor.gral canbesplit into two parts

yQU»thecapacitor and Uisthesplates. Considering thatwing expression for theenergy



heseformulasdeterminetheviz.not only theenergy of in-esof one plateand thoseof thegyof i11teractio11 of charges

electric?Weshall showthatvalid in thepresence0] a dielectric.er theprocessof charging a

mall portionsof charge(dq')

rmedagainst theforcesof the

erencebetween theplatesatortion of chargedq' is being

ver q' between 0 and q, we

ression for thetotal energy of hework doneagainst theforcesetelyspent for accumulating

dcapacitor.Moreover, theexpres-isalso valid inthecasewhen

theplatesof a capacitor. Thus,of (4.7) in thepresenceof a

(4.6) aswell.

Formula(4.4) defineselectricermsof chargeand potential.ergyWcan also beexpressedracterizingthefield itself,

E. Let usat first demonstratempleof a parallel-platecapaci-nsnear the edgesof theplatesntotheformulaW-= CU2/2



we obtain

V (thevolumebetween the

uniform field whichfillsthe

oved that theenergy WE (in thecaseof an isotropic

ula9)onhasthemeaning of theenergyThisleadsus to avery impor-aabout localization of energynwasconfirmedin experimentsIt isthedomain whereween-be explained with thehelp of tionin thefield. These varyingtlyof electricchargeswhich

aypropagateinspacein theves. Experimentsshowthatyenergy.Thiscircumstancecon-self isacarrier of energy.hat theelectricenergy is

volumedensity

rmulais valid only in thefor whichtherelation P = xe,,E

hesituationismorecomplicat-



ormula(4.9). It isknownthat theedconductor isW= qep/2. Let usshow thatedingfrom theideaof localization of

ositively charged conductor. Wefinitesimal crosssection, whichisbound-d takein it an elementary volumedV==the energy

dl.

ergylocal-e. Forelast ex-theproductsectionsbetaken

Q—€`-S; cp,

ginning of theimaginary tube.last step,i.e. integratetheobtained,and find the energy localized in theentiretial cp isthesameat theendfacesof allteon thesurfaceof theconductor), we

ormed over aclosed surfacecoincidingsurfaces.In accordancewith theGauss

al tothechargeq of theconductor, and

esillustratingtheadvantagesenergylocalization in thefield.is at thecentreof aspherical layerlectricwith thedielectricconstant s. Theyer are equal to aand b r espectively.ainedin thisdielectriclayer.ielectricavery thin concentric_spher-r —}—dr. Theenergy localized in this



atingthisexpression over r between a

hat must be doneagainst theelectricdielectricplatefrom aparallel-platechar-dthat thechargeq of thecapacitor remainsectricfillstheentirespacebetween thetanceof thecapacitor in theabsenceof

cforcesin thissystem_isequal to thergyof thesystem:

hefield betweenthecapacitor platesinicandW2 isthesamequantity in theabsencemind that themagnitudeof vector Df theremoval of theplate, i.e. that D,=

·S being theareaof each plateand h

ng Polarizationol a Dielectric.0) for the volumeenergy densityueof E, thequantity wiseeof adielectricthan when it isayseem strange: field intensityhesame.Asamatter of fact.dielectric,it doesan additionalzation.Therefore, under thelectricwemust understand theroper and an additional work ng polarizationof thedielectric.ubstituteinto(4.10) thewhich gives



andsidecoincideswith thein avacuum.It remainsforonal"energy E·P/2 isassociated

neby theelectricheld foreof thedielectric, i.e. for the

and pf. respectively along andreasein thefield intensity fromsecond-order terms, wewrite

onal displacementsdueto ang. 4.3). Considering that

i dl ·E,eadditional displacement of etothenegativecharges. Accord-weget

)=dfor thepolarization of a

ondterm in formula(4.11).nsity w= E·D/2 includesdproper and theenergy E·P/2tionof the substance.



. . —.gedBodiesem of twocharged bodiesin aeinthesurrounding spacethey, thefieldE2. Theresultantthesquareof thisquant.ity is

), thetotal energy Wof thesum of threeintegrals:

v,(4.14)a(4.5) and revealsthefieldringin thissum. Thefirst tworinsicenergiesof thefirst andandW2), whilethelast integralction(W,2).cumstancesshould bemen-rmula(4.14).chcharged body isan essen-otal energy (4.9) isalso alwaysyseenfrom thefact that thein-positivequantities. However,

nbeeither positiveor negative.odiesremainsconstant uponat do not changetheconfigu-dy. andconsequently thisenergy

ditiveconstant in theexpressionch cases,thechangesin Wareby thechangesintheinterac-lar,thisisjust. themodeof be-ystem consisting of twopointedistancebetween them.gyof theelectricfield ishe energyof afield E which

2is generallynot equal to thefieldsin viewof thepresenceof . Inparticular, if E increasesrz

gyof thefield increasesnztimes.



ctricsshowthat adielectricintheaction of forces (sometimeseromotive).Theseforcesappearl asa whole. Ponderomotive

n dueto theaction of a nonuni-emoleculesof thepolarized dielec-eina nonuniform electricedirected towardstheincreasingsare causedby thenonuniformi-picfield but the microscopicdmainlyby the nearest mole-ricctricforces, thepolarized

phenomenon iscalled electro-trostriction,mechanical stresses

ot only theelectricforceges) actson aconductor in aan additional mechanicalic.In thegeneral case, theeffectnt forceacting on aconductor

by any simple relations, andeforceswith simultaneous

of their appearanceis, as arule,r, in many casestheseforcesientlysimplewaywithout ainby using thelawof conser-

ingForces.Thismethod iSsto takeinto account automat-both electricand mechanical)nceleadsto acorrect result.

of theenergy method for cal-st casecorrespondsto asitua-orsaredisconnected from thehechargeson the conductors

maystate that thework Aof em upon slowdisplacementstricsisdonecompletely at the



electricenergy Wof thesystemethat thesedisplacementsdonof electricenergy into otherprecise,it is assumed that such-lysmall.Thus, for infinitesimal

izesthat thedecreaseinthecalculatedwhen chargeson

.equationfor determiningorsand dielectricsin theelectriclows. Supposethat wearein-n agivenbody (aconductor orthisbody byan infinitesimalXweareinterested in. TheneF over the distanceda: isprojection of theforceF ontoX—axis.Substituting this ex-nddividing both partsof (4.15)

efollowing circumstance.e dependsonly on thepositiontion of chargesat a given in-

owtheenergy processwill deve-oveunder theaction of forces.tocalculateFxby formulaect conditionsunder which allarenecessarily constantfind theincrement dWunderwhich isa purely mathemati-

isplacement isperformed atnductors,thecorrespondinglexpression for theforce: Fx =-·d it isimportant!) theresult



h thehelpof thisformulaord beexpected.Therefore, hence-vesto the application of onlyt for any conditions, includingsmall displacements. WemustveOW/0.2: will be calculatedwell.ingon oneof theplatesof aparallel-electric,if the distancebetween thef thecapacitor under given conditionsistainedacrossitsplates.

movetheplatesapart, thevoltageUchargeq of thecapacitor changes(this= q/ U). In spiteof this, weshall calculateon that q =

p of formulaonvenient expres-n for theenergyof thecapacitor is

vityof thedielec-hplate,and :: iser = h).ivedirection of theX-axisasisshown16),theforceacting on theupper plate

laindicatesthat vector F isdirectedon theX-axis,i.e. theforceisattractiveq= US = DS = ee0ES and E = Ufh,

c.Formula(1) of thelasteof interaction between theacitor in a liquid dielectricnding forcein avacuum bya).Experiment showsthat thisheentirespaceoccupied by aaseousdielectric, theforcesof dconductors(at constant charges



r of cz

int chargesql and gzseparat-dintoan infiniteliquid or gas-theforce

rceinavacuum by afactor of eCoulomb lawfor point charges

ntionto thefact that pointneouschargesconcentrated onimensionsaresmall incomparisonhem.Thus, thelaw(4.18) has,acuum,avery narrowfieldcmust behomogeneous, infinite,nteracting bodiesmust becsense.heelectricfield intensityngon apoint chargeq in aousdielectricfilling theentirector of s smaller than thevaluesof dielectric.Thismeansthatnt chargeqin thiscaseisde-laasin vacuum:

yin thedielectricat thepointeqisplaced.Only in thiscaseibleto determinethefield Eown forceF. It should benotedrom thefield in the dielectricouschargeitself (which is. Nevertheless, formula(4.19)

ngeasit may seem.Weshall bespeaking of theeareaof a charged conductorric.For thispurpose, let usacitor in aliquid dielectric.



charged and then disconnectedintainthechargeand thefield

when theplatesaremoved apart.Fig. 4.4.Theenergy of thefield withinit. In accordance= (1/2) EDSx,whereS isthed scis thedistancebetween themythefield). By formula(4.16),

rplateis

of theforceis

tingand important r esultuidor gaseousdielectric). It

nsityof theforceacting on aumedensity of theelectricsforceisdirected alwaysout-surfaceof theconductor (tend-

of thesign of thesurfacecharge.

n.Apoint chargeq isat adistanceZplane.Find theenergy W of interactionchargesinduced on theplane.eeze"thechargedistributed over theseconditionsthepoint chargeq to in-q will movein thepotential eld which

fixed fictitiouspoint charge--q, locatedher sideof theplane. Wecan write

total energies. Asystem consistshellsof radii R1and H,with chargesItrinsicenergiesW, and W., of each shell,onbetween theshells, and thetotal elec-if Hg > R1,



h(4.6), theintrinsicenergy of eachecp isthepotential of ashell created only= q/4:te,,R,whereIl isthe shell radius.eachshell is

tweenthecharged shellsisequal to theiedby thepotential cp created by the charget of location of thechargeqi W12 = qrp.

ve

lf.2’hesystem is

TtiiH‘t ii.allsof radii H1 and R2 arein vacuumxceedingtheir dimensionsand haveaeratio q1/q2between thechargesof the1thesystem isminimal. What isthepo-heballsin thiscase?y of thissystem is

L - --1 )21:211 ’ntrinsicelectricenergiesof theballs (qtp/2),teraction(q1cp2 or q2q>1), and l isthedis-ceq2 q —- q1, whereq is thetotal charge

hen OW/0q1 U.Hencei?—-

that R1 and R2 are considerably smaller

ll(3}`(ìll'1heconsidered isolated) is cp gxq/R,overelation that q.-, q=2, i.e.thepoten-ofor sneh atlistrilnilioii.



n thefield. Achargeq isuniformlyf radiusR. Assuming that thedielectricrywhere,find the intrinsicelectricenergythe energyW1 localized insidethesphereounding space.efieldsinsideand outsidethesphereeorem:

trinsicelectricenergy of thesphere:

—|»—§—-T2- nr dr=§;t—:;-3-1T(§-+1),

heratio W1/W,doesnot depend on

iformlycharged by achargeq.tre.Find thework of electricforcesuponrom radiusR1 to R2.cforcesis equal to thedecreaseintem:

ceW,— W2,we notethat upon theexpan-heelectricfield, and hencetheenergyin thehatched spherical layer. Conse-

41ITzdr,

dintensities(in thehatched region at athesystem) beforeand after theexpan-heGausstheorem, wefind



eobtain

ethework in terms of thepotential.wherecp isthepotential created by thechargeq,,

chargeq, theanswer would bedifferent

thefact that thisapproach doesnot takework performed by electricforcesuponionof thechargeq located on theexpand-

he centreof asphcrical unchargedI’1Il€I` Ellld Ol1l.(2I' I°&(lll {IFB0 Eilltl b 1‘€5[}()Ci.lVCly.ricforcesin thissystem upontheremov-ginal position through asmall holetancefrom thespherical layer.from thefact that the work of electricCI‘€3S€ lfl the0l0Cl.I‘lC0I10I'gy of Ill0 SySf.0II1. Asocalizedinthefield itself. Thus, theprob-tionof thechangein thefield asaresult

fieldaround thechargeq will changeyer with theinner radiuso and theouteral position of thecharge,therewasnothehnal position thereisacertain fieldSpl'l€I`lC3l l3y0l` isf ill` lÌ‘OI'Il thi} Cl'l£1I`g0 q). COIlS€·s

r2 and dV= 4m?dr, and integrating,



moving capacitor platesapart.r hastheplatesoi areaS each. Find theforces,done to increasethedistancebe-2, ii (1) the chargeq of thecapacitor and

nedconstant. Find theincrementsof theor in thetwo cases.

ork is

1}•hefield createsl by oneplate(E = c:}2e0).rgelocatedon theother platemoves.nt for increasing theelectricenergy:

ingon eachcapacitor platewill de-nthep ates.Let uswritetheelementaryaplateduring itsdisplacement. over adis-plate:

that q = CU,E, = U/2;::, and Ceo.?/x.

pacitor is

= ——A'part,we plerform apositivework Theenergy of t e capacitor decreasesinand this, wemust consider asovrce main-nceof thecapacitor at aconstant value.esthework AS.According to thelawol

= AW, whenceAS = AW-——- A' =

nconductorsin a dielectric. Aparal-ersed, in thehorizontal positien, into aliq—tricconstant c, filling thegap of width hecapacitor isconnected to asourceof



heforcej' acting on aunit surfaceof the

[ acting per unit area of each plate

eacting per unit areaof aplatefrom theeper unit areawhenthed ielectricise¤» (2)in theregion occupied by oneplate, creat-r plate.Considering that 0 -= D=) into (1),weobtaine0U2/2h2.h1.0 mm and e= 81 (water), weget

dielectric.Acylindrical layer of athedielectricconstant eisintroduced

r so that thelayer fillsthegap of width dradius of theplatesisll such that R> tl.

p asourceof a permanent voltageU. Findicinsidethecapacitor.W= q2/2Cfo r theenergy of acapac-ncewith (4.16), therequired forceis

C(1)

of thegiven capacitor can becalculated—p1atecapacitor. Therefore, if thediele-x and thecapacitor length isl, wehave

—$)·2JTRT: 80·2JtR(8x+l__x)·

obtain

of two fixed platesin theform of amovableplateof thicknesshmade of aconstant e,placed between them. 'I‘heabout theaxisO (Fig. 4.7) and practical-nthefixed plates. Aconstant voltageUises.Find themoment M about the axisOblep latewhen it isplaced asshown in the

edby themoment of forcesM upon



gh an raglejfelement dot isequal to thegyoi the system at q = const [see(4.16)]:

ion,C= C1 + CE,whereC, andC8rts oi thecapacitor with and without ther with an angleot is determined asS =

) R2/2h,tngwith respect to ot, we {ind70C/oct = (e0R”/2h) (1-e). Substituting

pression into formulaQ) and con-gthat C= q/ U,we0 tain

icatesthat themoment of theforce isyto the_positivedirection of theangleot;tendsto pull thedielectricinsidetheca-

sindependent of theanglecz. However,themoment Mz== 0. T isdiscrepancyall valuesof ct wecannot ignore edgeeffectsof thisproblem.

uityEquationpter,weshall confineour-uction current in aconductings.It iswell known that electric

rgethrough a certain surfaceconductor).rrent canbecarried by elec-ctrolytes),or someother parti-icfield,current carriersperformgethe samenumber of carrierseach sideof any imaginaryIll. passing through S in thiscase



lectricfieldisapplied, an orde-eragevelocit.y uisimposed onrriers,and acurrent flowsan electriccurrent is essentiallyccharges.electriccurrent isintensitysferredacrossthesurfaceS in

pere(A).rrent may be distributed non-

hrough which it passes. Hence,urrent in greater detail, currentd.Themagnitudeof thisvectorurrent dl through asurface

directionof motion of chargethiselement: j dl/dSl. Forkethe direction of velocityion of positivecarriers(or thethevelocity vector of theor-rriers). If thecurrent. consistsof s, itsdensity is defined by the

umedensitiesof positiveandpectively,and uq- and u- aretheotion.In conductors, wherethectrons(p_ < Oand u+ O),y

raphicallyrepresented withinesof vector j), which arer vector E.y vector at each point of theon,wecan also find thecurrent.eastheflux of vector j:



cquantity.It can beseendesother factors,thesign of thechoiceof thedirection of thesurfaceS, i.e. by the choiceS. If the directionsof all thecurrent I changessign.imagineaclosed surfaceS

ough which acurrent is passing.outward direction for vectorsndhence,for vectorsdS.Con-S givesthechargeleavingy thesurfaceS) per unit time.servation of charge, thisintegralechargeinsidevolumeVper

ntinuity equation. It is an ex-ervation of charge.(direct) current. thecharge

mainunchanged.In otherll(l sideof Eq. (5.4) dq/dt =< O. Con—wehave.

thelinesof vector j do not.Thefield of vector j issaideof direct current.tinuity equation. Let us writeEqs.ntial form. For thispurpose, weexpresst neht--hand sideof Eq.(5.4) as—--1% Spdi'partial timederivativeof p,l fllllO{IS \V(‘ll RS OUCOOI'(lll'lIltl‘.. TlI\IS,



_ Way 38 f()l` théflux of VOCl0I` E ill SOC. Wt}ector j at a certain point. isequal to thelP"Sll§y PGP l1l'lllS time at theS3lll€ pOlIll.Z

econdition(when dp/Ot = 0):

rrent theiield of vector j doesnot have

geneousConductorovered experimentally, states:

homogeneousconductor ispro-erenceacrossitsterminals(or

nceof theconductor.ohm (S2).eshapeand sizeof the conduct-rature,andaboveall on the

of thecurrent through thecon-tanceisquiteclear when wemoregeneral caseof thevolume

meaninglessto speak of re-er theposition of theleadsat-e current configuration.ogeneouscylindrical con-y

onductor,S is itscross-sectional,which dependson thematerialerature.Resistivity ismeasured

ctorslikecopper and alumi-8 Q·m at room temperature.



Law. Let usestablish are-ensity j and thefield E at am. Weshall confineourselves

whichthedirectionsof vectors

ndacertainpoint in acon-l volumeelement with. genera-dhenceto vector E. If thecylind-adS and length dl, wecan write) thefollowingexpression for

ntermswe obtain thefollowing

tivityof themedium. Theunitda mlw(Siemens). Consequently,t of 0 ismho per metre.ential form·of Ohm’slaw.rentials(derivatives), but isnceit establishesarelationnding to thesamepoint in aq. (5.10) isan expression for

sistance(R). Thereareseveraltanceand all of them are ulti-tionof relations(5.8)-(5.10).rticular method is determinedblem and by its symmetry.these methodsaredescribed in

gConductor. If thecurrent ishargeinsidea homogeneousverywhere. Indeed, Eq.(5.5)akinginto account Ohm’slaw



write(5.5) asfollows:

ormed over any closed surfaceahomogeneousconductor, thef the integral:

rem,theremaining integral issum of thechargesinsidetheional toface. How- 1;,, ‘"' Bfrom theIal isaIThismeansIequal to FT

hosen ar- Jnder theseI ·‘—"* ‘

zeroin-

ar only 5.1eouscon-

mesincontact with otherionswherethe conductor

carrying Conductor. Thus,sschargeappearson thesurfaceityregion).In accordancewithmal component of vector Ef theconductor. Further, fromtial component of vector E,al component of thisvector alsoconductor.Thus,near thesur-r E forms(in thepresenceof the

with thevertical (Fig. 5.1). When

dystate, thedistributionnductingmedium (generallyoesnot changeintime, although



eachpoint, newchargescon-ingones.Thesemoving chargesh isidentical to theonecreatedameconfiguration. Thismeansonary currentsisa potential

ld in thecaseof stationaryfrom electrostatic, or Coulomb,rgesarein equilibrium, the

onductor is alwaysequal tozero.rycurrentsisalso an electro-ughthechargesinducing this

hefield E of steady-statecurrents—carrying conductor.

forcesactingonchargetrostaticforces, thepositiveheaction of theseforcesfromial to thosewith a lower poten-erswould movein theoppositeotheequalization of potentials,uallyconnected conductorssterminatingthecurrent. Innof Coulombforcesalone, aticfield.rect.current circuit must

uitsin which positivecarriersreasingpotential tp, subcircuitsoveinthedirection of increas-cfield forces.In thesesubcircuits.ssibleonly through forcesthatll call such forcesextraneous.esustained only with thehichareapplied either to cer-theentirecircuit. The

usforcesmay bequitedive1·se.y physical or chemical inho-for example,forcesresultingrsof different types(galvanic



_ _ _ ’ductorsat different temperatures

order to describeextraneousnceptsof extraneousforcefielduced.Thelatter is numerically

eactingon a unit positivecharge.ensity.If an electricfield Eurrent of density j = oE, itmbinedactionof thefield EdE*,thecurrent density will

elaw(5.10) to thecase of conductingmedium, and is

fslawinthedifferential form.orm Subcircuit. Nonuniiornibythepresenceof extraneous

t practicallyquiteimportantongathin.wire. In thiscase, thecoincidewith theaxisof thej canbeassumed to bethesameionof thewire. Supposethat.hewire isequal to S (S need notngth of thewire).

5.11) by o, forming ascalaression with theelement dl

m cross section1 to crosssectione positivedirection) and in-hewirebetween thetwo cross

12)

ndin thefirst integral: wedl,where j, istheprojectionof vector dl. Further, wenoteyand dependson theorienta-



if j d l then j I >_0; if j fl dl,ceil by I/S,whereI, theuantity (likeil). SinceI isecircuit when wearedealingntitycanbetaken out of thet

estheresistanceof thesub-eintegral of this expression iscircuit between crosssections

-hand sideof (5.12). Here,en-differencecp,-— q>,, whileelec-ct-t:

eelectro-

sthemotion of positivecarriers12> 0; if,thee.m.f. hinders

mationsdescribed above,form:

form.of Ohm'slawfor anon-5.11) which describesthesame

rcuit shown in Fig. 5.2. Theresistance€gll'lOlllZ R.l0WGI’ p3I‘l» of l.hCllg'\ll'CSll0WSover thisregion.Let usanalysethecourse



sover thesegment Rfrom left to right,rrent flowsin thepositivedirectioncase, Q1 < Q2, but thecurrent flowsfrom

dsincreasing potential. Thisispossiblecting in thiscircuit from 1 to 2, i.e. in

followsfrom thisequationcal for aclosedcircuit, i.e.uation acquiresasimpler form:

ceof theclosed circuit, and 8ee.m.f.’sin thecircuit.rcuit containingthesourceals1 and 2. Then, Rin (5.15)eof thee.m.f. sourcein theon,whileQ, — Q2 the potentials.If thesourceisdisconnected,her words, thee.m.f. of he potential differenceacrossuit.osstheterminalsof an e.m.f.nal resistanceisalwayslessn theload.tanceof acircuit isq timeshigher thanesource. Findtheratio of thepotentiallsof thesourceto itse.m.f.anceof thesourceand Ratheexternalording to (5.15),’Q2 —-— cg, = (5** — R2],

R2-1- Ra) I = (9.From t esetwo equa-

a: *1

higher thevalueof n, thecloser thepo-current terminalsto itse.m.f., and vice

usconsider apictureillus-ent in a closed circuit. Fig-n of potential along aclosed



ourceon segment AB. Forentation,potential tp isplottedcylindri-

urrent con-correspond toerminalsanbe seen from the

wof current can besitiveong thein-point q>Ato. pointcircunt,iers"rise"from point tp Bto cp),orceshown by an arrow.

hholfsLawsrcuits, for example, determina-branches,can be considerablyowingtwo Kirciihojfslaws.nstothejunctions, i.e.d statesthat thealgebraicat ajunction isequal to zero:

t ajunctionand diverging fromsitesigns; for example, wecantiveand thelatter to be negativeerial).When applied to Fig. 5.4,I1 —- I2 —}— IB0.enceof thesteady—statecon—hargeat ajunction would changeeinsteady state.hislawisapplicableto anydc.ira:ui.t. andstatesthat thetsof cumerzt and rcsista.ncein eachthealgebraicsum of e.m..j`.’sin



Kirclzho/fh Laws127s sulticient to consider thetsol threesuhcircuits(Fig. 5,5).not' circinnyention t.o he clockwise,

ying ()hn1’slaw(5.15) to eachbtain

cancellingall potentials, weKirchholf`ssecond law.

uenceof (_)llltl’S lawfor non-

uations.In each specificcase,ompletesystem of at;_{eln·a.ir

say, for tinding all theun-.heform (5.17) and (5.18)r of unknown quantities. tiareat noneol theequationsisaonin the system:junctions,independentheset,up only lor N1 junc-elast junction will hea corol-ons;ntainsseveral cle.loops.f type(3.18) can heup onlyhtaineil hy snperiiuposin;5l·òr eyample,such equa-rloopsZ2.! ; sl 234 ol` the



uation for theloop 1234cedingones. lt is possibletonsfor two other loops, say,ation for contour 234 will bea

quations. The number of auto-18) will beequal to thesmallesthich must becreated in a cir-oops. Thisnumber isrcuitsbounded byconductorsnaplanewithout intersections.to set up threeequationsof (5.18) for a circuit (Fig. 5.7)

hisis so becausethenumber of rossesin thecircuit) breakingmber of subcircuitsìsalsof weassumethecurrentstoiscontinuitieswill besix inrof subcircuitsbetween junctions,mber of independent equations.ould be adopted whileset-17) and (5.18).are marked hypotheticallyr thedirectionof thesearrows.to bepositiveasaresult of

thedirection chosen for it isent isfound to benegative, itsositeto theonepointed by the

aryclosed loop, wecir-irection,say, clockwise. lf y current coincideswith the



hezcorresponding {term IR inhe positivesign; in theoppositebeused. Thesameprocedureisncreasespotential in thedi-hould beassigned thepositiveld beused in theoppositecase.deand direction of thecurrent passinguit shown in Fig. 5.8. All resistancesandnown.Rndhence,

and I, inarilly) thesup-rentsby ar- X-B1 R, I1Inctions.·

yoneinde- I7): ’

sof type8 mber of sub- ’ R2o of them.Let F. 5 8ningRand lg' ‘,. Takingrcumventingeach of theseloops, we

ponding equation for theloop withom thesetwoequations. Solving theeobtain

substitution of numerical valuesintont actuallyflowsasshownin Fig. 5.8.theoppositedirection.

ghaconductor having ampaniedby liberation of heattask isto find the quantity

meinacertainpart of thecir-



caseswhich arepossible, viz.circuits.Theproblem is consid-of conservation of energy and

sethat weareinterested inctions1 and 2 of aconductorrk time12.con-

geheachuctorn particular,s

rgewill leavecrosssection 2.heconductor remainsunchanged

rect),thewholeprocesschargedq from section 1to1 and cp, respectively.field in such achargetrans-

— <1>i>d¢·servation of energy, an equiv-t beliberatedin another form.yand no chemical transforma-

ergymust beliberated in theergy.,Consequently, thecon-hanism of thistransformationthework doneby thefield,,electronsin metals) acquire

ywhich isthen spent on excitingisionsof thecarrierswith the

elawof conservation of ener-= Qdt, whereQis theheatmal power). Acomparison of ingonegives



hm’slawcpl — rpg = RI,

well-known Joule’slaw.ession for thedifferentialzingtheliberation of heat atngmedium. For thispurpose,t of thismedium in theform of cesparallel to vector j, viz.en point.Let dS and dljbethegth of thesmall cylinder re-eat liberatedin thisvolumeduringccordancewith Joule’slaw, bydt=pi2dVdt,

umeof thesmall cylinder. Di-Vdt, we obtain aformulaforper unit timein aunit volume

or thethermal power density of

e’slawin the di]j“erential form:f current at anypoint is propor-rent density and to theresistivity

eneral form,of Joule’slaw,irrespectiveof their shapeornatureof forceswhich induceargecarriersaresubjected towrite,on thebasisof Ohm’slaw

neral naturethan (5.20).subcircuit containsasource



will besubjected not only toneousforcesaswell.In accordanceof energy,the amount of heatequal to thealgebraicsum of ricand extraneousforces. Thending powers: thethermal powericsum of thepowersduetos. Thiscan beeasily verified

-22)xpressionisthe thermal powert.Inthepresenceof extraneouserminedby thesameformulauniform subcircuit. Thelaststhe power generated by ex-cuit under consideration. Itlast term (S5!) isan algebraicreversesitssign when thedi-versed.'eheat liberated in there-sequal to thealgebraicsum of ewers.Thesum of thesepowers,hisequation,iscalled theelectricircuit.It can then be statedit,thethermal power evolveddevelopedin thesubcircuit

unbranched circuit (cpl =

e heat liberated per unit timetot.hepower- developed bymeans that heat isgeneratedTheroleof theelectricfieldf thisheat over different parts

2) in thedifferential form.yboth sidesof Eq.(5.11) by



d pjzOd (see(5.20)).f current in an inhomogeneous

nintheform

Capacitor Circuitrethe processesinvolvingaryregimeof thecircuit toaprocess-

gingof a Iideredin

onlydi- ""Rwever, that Kthiscase

atingcur-esin which Via- 5-lllt too rapid.current will then practicallyonsof thecircuit. Suchcurrentsem arecalledquasistutionaryuasistationary currentsand

be describedby thelawstsby applying theselawstontities.harging and charging of acurrentsin theseprocesses

f the platesof achargedare connected through aresis-

ough theresistor. Let I, qvaluesof thecurrent, chargepotential differencebetween

ispositivewhen it flowsnegativeplate(Fig.5.10),



t.According to Ohm’slaw,wesionfor theexternal subcircuit

—·dq/di and U== q/C, wecannasfollows:

sinthisdifferential equation

eon thecapacitor and 1: isaof time:

ationtime.It can beseen fromringwhich thecapacitor chargel value.spect to time,weobtain

nt:

t at theinstant i = 0.ndenceof thecapacitor chargendencealso hasthesame- form.nsider acircuit in whichd asourceof e.m.f. are2). To begin with, thecapacitorn). At theinstant t == Othetartsto flowthrough thecircuit,Theincreasingchargeonthe

ct thepassageof thecurrent

l newbe assumed positiveeplateof tho. capacitor: I

wfor anonuniform subcircuit



ceof thecircuit, includinge.m.f. source.Considering

, Uq/C, wecan rewrite

get I

dert := 0, we obtain

ngvalueof thecapacitor charge

accordingto thefollowing

on t are shown in Fig. 5.13.



tingmedium. Ametallicsphereof concentricthin metallicshell of radiusb.o electrodesisfilled with a homogeneousf resistivity p.Findtheresistanceof the

des.olateathin spherical layer betweenof current at all pointsof thislayer areper-esuch a layer can betreated asacylin-and across·sectional area4::r*. Using

ithrespect to r between aand b, we get

cballsof radiusa areplaced in auctingmedium with resistivity p.Findtheweentheballsunder thecondition thatsmuch larger than their size.

mpart charges+q and—q to the balls.distancefrom oneanother, electricfieldispracticallydetermined only by the,and itschargecan beconsidered to beesurface. Surrounding thepositivelysphereadjoining directly theball’ssur-n for the current through thissphere;

y.Using Ohm’slaw(j = E/p) and thebtain

al differencebetween theballs:ZoG.enby

sof the magnitudeof the dielectriccon-

bitrary shapearepllaced into an in-conductingmedium wit theresistivity pFind t evalueof theproduct RCforesistanceof themedium between the



ual capacitanceof theconductorsin the

gply thecharges-{-q and ·-q to theum etween them ispoorly conducting,orsareequipotential, and thefield configu-bsenceof themedium.e,thepositivelycharged conductor bydjoining theconductor and calculateR

S

nover thegiven surfaceS.Whilecalcu-win theform j = oE, andC wascalcula-usstheorem.ns, weobtain

ndary of a conductor. Aconductora dielectricwhosedielectricconstant isacertainpoint Aat thesurfaceof thecon-

beingdirected away from theconductorhenormal to thesurface. Find thesurfacector and thecurrent density near the

edensity on theconductor isgiven by

ound with thehelp of 0lun’slaw: i =ntinuityequation (5.5) that the normalqual,andsincein thedielectricin =·— 0onductor wealso havein = 0. Hence,ngent to itssurface. Thesame appliesctor.from the theorem oncirculationl componentson different sidesof theceE = E,== Dsin cz/ce•, whereE.,isvector E in thedielectric. Taking thesebtain

platesof a parallel-platecapacitor isielectriclayers1 and 2 with thicknessestsel and ez, and resistivitiespl and p,.

manent voltageU,theelectricfield being



indthesurfacedensity of extraneousweenthedielectriclayers.

cechargedensity isgiven byE1· (1)d E2, weshall makeuse of two conditions:1/pl == E,/p2, and besidesEli, + E,l2=

ons,wefind E1 and E,.Substituting these

for, slp,= czp,.nductor.Along conductor of acir-s madeof amaterial whoseresistivity

cer from theconductor axisasp = on/rz,onductor carriescurrent I. Find (1) f ieldand(2) theresistanceof the unit length

with Ohm’slaw, field intensity E ishilej isrelated to current I. Hencewe

ameat all poi tsof thecross section of ndependent of r. Wecan easily verifyar contour in theconductor so that oneswith, for example, theconductor’saxis,tour thetheorem on circulation of

heintegral,and asa result of inte-

ngth of theconductor can befoundH== U/1.Dividing both sidesof thisction of theconductor, having resistance

iform subcircuit. ln thecircuit.’s6 and 5% of sources, the resistancesceCof acapacitor areknown.Theinter-enegligibly small. Find thechargeon

hOhm’slawforaclosed circuit con-



wecanwrite

nischosenclockwise. On the other hand, forRbof thecircuit wehave

wehave

quationsR0

4rminedby theformulaq, = C (cpl — q2,)-

6 > go and viceversa.source.Aglass platecompletelyesof aparallel-platecapacitor whosehentheplateisabsent. The capacitor ismanent voltageU.Find the mechanicalainst electricforcesfor extracting the

w of conservation of energy,wecan

al work accomplished by extraneousforcesthework of thevoltagesourcein this

esponding increment in theenergy of theontributionsof other forms of energy tohe system isnegligibly small).followsfrom theformulaW= CU2/2 =·-pacitor that for U= const)capacitor decreasesupontheremoval of eof the capacitor aso decreases(Aq <rgehas passed through thesourceagainstf extraneousforces, and thesourcehas



(2), weobtain

oninto (1) givesc—- 1) C5U*.t of thecapacitor, we(extraneousgainst electricforces). Thee.m.f. sourceinative work,and theenergy of thecapaci-

Acircuit consistsof apermanentstor Rand capacitor Cconnected in se-of thesourceisnegligibly small. At the

ceof the capacitor wasabruptly (jump-of 1]. Find thecurrent in thecircuit asa

lawfor theof thecircuit

ereC’

)nwith respect to time, considering thatdt = -1:

ives

ondition (1).Indeed, we can write

e of thecapacitor beforeitscapacitance

lied to a capacitor with capacitance= 0wasshunted by resistor R. Findin theresistor asafunction of timet.



nt of heat isgiven by

rst of all wemust find the timedependenceall useOhm`slawfor the part IR2 oi

respect to Q1C

weobtain

ndition(2) for q == qu, i.e. Ig <= qu/RC.integratingover time, weget

uum

sshowthat theforceF actingydependsnot only ontheposi-n it.svelocity v.Accordingly,nto two components, viz. thech doesnot depend onthemotionticcomponent Fm (which

city).Thedirection and magni-any point of spacedependge,thisforcebeing alwayssides, at any point, themagne-hedirectionspecified at this

deof thisforceisproportionalhich isperpe ndicular to this



eticforcecan bedescribedf magneticfield. Characterizingdeterminesthespecificdirectionnwritetheexpression for

eticforceacting on chargeq

Expression((5.2) isuniversal:l asfor varying electricandcityv of thecharge.rceon achargecanin pri_n—

ngthemagnitudesand directionsheexpression for theLorentzedefinitionof electricand

einthecase of electricfield).*t magneticfielddoesnot aclIn thisrespect; magneticfieldtricfield.Magneticfield acts

forceactingdueto magneticnd hencein thisrespect it isterizingtheforceacting due

neticforceisthat it isalwaysyvector of thecharge. There-hecharge.Thismeansthat themoving in apermanent magne-hangedirrespectiveof the

imation,theLorentzforceesnot depend onthechoiceringfield Bhavebeen worked out. Inebasedon thephenomenathat can be



tem. On theother hand, theorentzforcevaries upon atran-renceto another (becauseof v).

ponent qE must also change.ompositionof the total forceFectricand magneticcompo-of ther eferencesystem. Withouttem, such a decomposition is

mly Moving (Charge. Experi-eld.isgenerated by movinglt of thegeneralization of expe-tarylawdehning thefield Bat aconstant nonrelativisticslawis written in theform*

nstant,thecoefficient

m the point chargeq to theil of theradiusvector r ise-th B

nt

ig_ 61°ning vectorsv and r, therota-rectionof vector Bforming

din thecasewhenthechargemovesly at sufficiently small distancesr fromvelocity v of thechargedoesnot noti-

mer/e).



-__,hvector v (l·ìg. 6.1). lt shouldal (pscudovector).ncficinduction.redin teslas(T).hargeq moving at anon--

bedby the samelaw(1.2).bewritten in theform,(6.4)

micconstant (c1/l/conn), equal

cuum (thiscoincidenceisnot

rcesof magneticand electricinteractiont two point chargesq of asufficiently largeeanother with thesamenonrelativistic

6.2.Find theratio between themagneticing,for example, from chargeI on

Band Fc= qi} wherev isthevelocitytheinduction of themagneticand thescreatedby charge1 at the point of loca-

ccording to (6.4), in our caseB==

locities,e.g. v = 300 km/s, this rationeticcomponent of theforceisa milli-component and constitutesanegligiblece.o thefollowing question:tigating?It turnsout that. t.heyd reasonsbehind this.



eamsof particlesmovingocityof light,for which thiscforcebecomecomparable(itn(6.5) isalso valid for relati-

ay,of electrons, along wires,ountsto several tensof a mil-densities,whiletheratio

negligiblecorrection to theer of fact,in thiscasethe magne-ly forcesinceelectricforcesalmost ideal balance(muchnegativeand positivechargesonof avast number of chargesatesfor thesmallnessof this

hargeson thewiresarenegli-nwith thetotal chargeof carriers.rcesin thiscase considerablytingou excesschargesof the

on. Experimentsshowthatectricfields,obey theprincipleticfield created by severalisequal to thevector sum of theachchargeor current separately:

consider theproblemcfieldcreatedby adirect electricproblem on thebasisof law(6.3)of the field of auniformlybstituteinto (6.,3) thechargeolumeelement and pistheake into account that in accord-



___ormula(6.3) becomes

6.7)thinwirewith thecross-

length of thewire. Intro-tionof thecurrent I, wecan

ows:

edvolumeand linear currentcingin formula(6.7) thevol-inear one,weobtain

——r, . (6.9)resstheBiol-Savart lawipleof superposition theult of integration of Eq. (6.7)

ments:g>_§Ildl><rl)

emagneticinduction ot"figurationby theseformulascalculationscan beconsider-tributionhasacertain sym-

ral simpleexamplesof deter-onof current.oi thelinecurrent. i.e. thecurrentt wireof iniinitelength (Fig. 6.3).ctorsdBfrom all current elementshaveion,viz.aredirected behind theplaneof mmationof vectorsdBcan bereplaced byitudesdB, where



at dl cosez== r da and r = b/cosct.l·Ience

ver all current elements, which isequiva-n between-::1/2 and sr/2, wefind

t theaxisof circular current. Figu-thecurrent element I dl located to the

ill form theconeof vectorsdB, and it canant ve¢·lor _Bat point A will bedirectedhismeansthat in order to find themagni-ient tosum up the componentsol` vectorsuchprojectionhastheform

that theangltebetween theelement dlual to n/2,and hencethesineisequal tosion over dl (which gives2nR) and takingndr == (zi Ii’2)‘/2, weget

agnitudeol` vector Bat thecentreof thedistancez>> Ris given by



cFieldheld hastwo very important

s, whicharealso related to theor held,expressthebasiclaws

s, weshould consider thed B. Just asany vector held,entedwith thehelp of therawn in aconventional way,

eselinesat any point coincidesB, andthedensity of the

magnitudeof vector Bat agiven

ainedin thisway makesitut theconhguration of agivenablysimplihestheanalysisof

clawsof magneticheld,e theorem on circulation.dB. Theflux of B throughozero:

generalizationof experience.postulatetheexperimentalB haveneither beginningmber of linesof vector B, emerg-dedby aclosed surfaceS, isf linesentering thisvolume.corollary which will berepeat-f Bthrough aclosed surfaceSur doesnot depend on theshape

easilygrasped with thehelpvector B. Sincetheselinesere, their number through then contour (i.e. the[lux of B)of the shapeof thesurfaceS.efact that thereareno magne-ch thelinesof vector Bbegin



.magneticfield hasno sources.

Vector B(for themagneticvacuum).Circulation of vector Bl` isequal to theproduct of ui,currentsenveloped by thecontour l`:

braicquantities. Thecur-direction isconnected withen-hecur-ec-ve. { e7 ». Fresare,3 > 0of ,11),,ghI 2 0hileFig.6.5

((5.15) can beproved onthe. Inthegeneral caseof arbitra-er cumbersomeand will notl treat statement (fi.15) as

mentallyk.If current I in ((3.15) iswherecontour l` located,orm

ral istaken over an arbitraryur I` Current density j inthepoint wherethearear dS forming theright—handedthecontour circumvention.. (6.15) can bewritten as



6.17)or B generally differsontrast tn electrostaticfield,ld. Suchfieldsare called

Bisproportional to thecur-our, in general wecannotalar potential whichwould ben expression similar to E =dnot besingle-valued: uponurrent and return to theinitialcquirean increment equal toential rpm can beintroduced andnof spacewherethecurrents

Circulationof Vector B.hesameroleastheGauss.It is well known that lield B,whilecirculation of vector B,

pedby thegiven contour. Ins(inthepresenceof aspecialrculationprovesto bequiteodetermineBin avery simple

when thecalculation of ereduced,by an appropriateproduct of B(or B,) by theart. Otherwise, field Bmustmethods, for example, witht. lawor by solving thecor-tions,andthecalculation be-

orem

icallyimportant examplesof the application of thetheo-



Band then seewhether this

oi§astraight current. Direct current!ngstraight wirewith acircular cross sectioneticheld induction Boutsideand insidethe

yof theproblem that thelinesof eform of circleswiththecentreat the

vector Bmust bethesame at all pointsof thewire. Therefore, in accordancewithf vector Bfor a circular contour 1Ì

= pal, whenceit followsthat outsidethe

)ct solutionof thisproblem (with thehelpout to be morecomplicated.nsiderationsit fol owsthat insidethe

also circles.According to thetheorem oncircular contour P2 (seeFig. 6.6),r/cz)?isthecurrent enveloped bythegi-d that insidethewire

own in Fig.6.7. _tube, theinduction Boutsideit is8),whileinsidethetubemagneticfield islyshown with thehelp of thetheorem on

of asolenoid.Let current I flowalong an thesurfaceof acylinder. Such cylindercalledasolenoid. Supposethat aunitnsnturnsof theconductor. If thepitch

mall,each turn of thesolenoid canbeapprox-d loop.Weshall also assumethat thecon-



sso small that thecurrent in the solenoidver itssurface.nsshowthat thelonger thesolenoid,ionoutsideit. For an infinitely long sole-,it isabsent at all.yconsiderationsthat thelinesof vector Bedalong itsaxis, vector Bforming thehthedirectionof thecurrent in thesolenoid.

I (gc";.

4 _t:e(`:,., ..,,,;;u_ "l

OOOOOOOOOI

onsabout themagneticfield configura-that wemust choosearectangular contourtionof vector Baround this contour is

envelopsthe current nl] According toBl (uml],whenceit follows that inside

enoid isuniform (with theexception of enoid’sendfaces, which isoften ignoredct nf iscalledthenumber of anipere-turns.2 A, themagnetich eld insidetheso-

f atoroid.A toroid isawire wound

mmetry considerationsthat the lineswhosecentreslieon theaxis OO' oi thefor thecontour wemust takeoneof such

oid, itenvelopsthecurrent N1,wherehetoroidal coil and I isthecurrentadiusbe r; then, according to thetheoremI,whenceit followsthat insidethetoroid

6.18) shows that themagneticfield insidemagneticheld N] ol. thestraight cur-xis. TendingNand radius Hol thetoroid



idcrosssection), weobtain in the limitmagneticfield of an infinitely long so-

ur passesoutsidethetoroid, it doesd henceB·2nr = 0 for such a contour.oroid magneticfield isabsent.t was assumed that thelinesof cur-.e. theplanespassing `through the'faxisoids,thelinesof cur-yin theseplanes, andponent around §`nt createsan addi- idof a circular cur- BE

ofacurre¤t··car·2 y 3an infiniteconduct- )l current uniformlyigure6.10 shows1 ; 4lanewhen thecur- Qf thefigure(this ! Bintroducethecon-y i directedalong theRudeof thisvector Fi 610which playstheg' 'l area"_nt—carryingplaneinto thin currenthat theresultant field B will bedirected

wardsto theright of theplaneand upwardshesedirectionscanbeeasily establishedndscrewrule.ionBof thefield, weshall usethector B. Knowing thearrangement of thehoosethecontour in theform of rectangleccordancewith thetheorem on circulation,hof the contou1·sideparallel to the

magneticfield isuniform on both sidessovalid for a bounded current-carryinglyingnear theplaneand far from its

resultsobtained in theaboveirectlywith thehelp of the

hetheorem on circulationheseresultsmuch moresimply

th which thefield wascal-pl8S IIIUSI 110l. pI`0dUCB311 BITOHBOIIS



ialitiesof the method basedorem on circulation. Just asrem for electricfield, the num-easily solved by using thetheo-r Bisquitelimited. It issuffi-chasymmetriccurrent c0nhgu·theorem on circulation be-pparentlyhigh symmetry, thendoesnot allowus, however,iredfor calculations, and theother, much morecumber-

asicLaws

ifferential form of theGausstheoremtenas

sEqual to ZETOeverywhere. AsWHS IDBI]-t magneticfield hasno sources(magneticenerated by electriccurrentsand not byot exist in nature.al nature: it is valid not only for con-aswell.t propertyof magneticfield expressed

onof vector Bmotivatestherepresentationntial form which broadensitspotential-onsand calculations.er theratio of circulation of vector Bontour.It turns out that thisratio tends. Thislimit dependson thecontour orienta-e.Thecontour orientation isspecified bylaneof thecontour around which the

directionof nbeing connected withtheentionthrough theright-hand screwrule.t of thisoperation isascalar quantitytionof acertain vector onto thedirectionof thecontour around which thecircula-alledthecurl nf yield Band is denoted as



we havetheprojection oi curl Bonto

fieldBcan becharacterized by curl Budearedetermined by thepropertiesof t. Thedirection of curl Bisdeterminedtothe surfaceelement S, for which theesthemagnitudeof curl B,attainsits

ained in coordinaterepresenta-osesanother fact ismoreimportant: itBcan beconsidered asthe crossproductB, i.e.VX B. Weshall beusing thetion sinceit allowsusto represent theehelp of thedeterminant:25)

vectorsof Cartesian coordinates. Thisrl of not only field B, but of any otherar,of field E.orem on circulation of vector B.Accord-berepresented in theform

he theorem oncirculationy: thedirection of. curl_ Bcoincideswith that of sityat thispoint, whilethemagnitude

_ _ulationof vector E isequal to zero,

equal to zeroeverywhereisa potentialhesotenoidal field. Consequently, electro-d,whilemagneticfield isa solenoidal neld.

carrier experiencestheTheactionof thisforce istrans-h which chargesaremoving.



tswith acertain forceon acur-us find thisforce.nsityof chargeswhich areectronsin ametal) is equalolumeelement. dVin the

harge(current carrier) equalgon thevolumeelement dVten by formula(6.1) asfollows:

hinconductor,then, ac-nd

idingin direction withthen element of length of thethin

xpressAmpére’slaw. Inte-er thecurrent elements(volumeagneticforceacting on acer-r on its linear part.sinamagneticheld are

actionbetween parallel curients.th whichtwo infinitely longwireswitha vacuum,if thedistancebetween themnit length of thesystem.sin themagneticfield of the current I 1,ieldB, == (al/im) 2I1/b. Theanglebe-t I 2_and vector B, isst/2. Hence, it ollowsforceactingper unit length of thecon-== I2B,, or

ion can beobtained lor theforce actingctor with current. I I.



nthat currentsof thesameetheoppositecurrentsrepellly about amagneticforce.er, that besidesmagneticforcestheforcesduetoexcesscharges. Consequently, if wespeak aboutbetweenthewires, it can eitherpending on theratio betweenmponentsof thetotal force

nt Loop.The resultant Ampe-loop in a magneticfield is(6.29),by

ormed along thegiven loop

rm, vector Bcanbe takenoblem is reduced to thecalcu-

Thisintegral isaclosed chainhenceisequal to zero. Conse-ltant Arnpere’sforcein auni-to zero.ldisnonuniform, theresul-iffersfrom zero and in eachwith thehelp of (6.31}. Theher analysisisthat of aplanemall size.Such acurrent loop

tarycurrent loop can beh thehelp of magneticmoment pu,.ment pm isgiven by

heareabounded by theloopoopwhosedirection isconnectedent in _tho leop through theill). Intermsof magneticheld,

mpletelycliaracterized by its



hthehelp of formula(6.31),ll sizeof theloop, yieldstheforceacting on an elementary

m magneticfield:

of themagneticmoment of theativeof vector Balong the

gthedirection of vector pm.pression(1.39) for theforcen anelectricfield.3) that,likein thecaseof

eld F = Osince0B/6:2 = O;generally doesnot coincidepm; vector F·coincidesin

mentaryincrement of vector Btor pm at thelocusof theloop.2, wherethreearrangementseld of straight current I0resultant forceF acting on thewn in thefigure(it isusefult is reallyso).

ojection of forceF ontofficient to writeexpression



tionsonto thisdirection. This

veof thecorresponding pro-h re-

espect .B

rrent loopm isar- .—>...Xesymmetry - pmticfield,vec-

g vector B. Letnof the

13. SinceBxwill beF. 613vector pm, lg' ‘ectedtoeBis greater. If werotatethecur-ough 90° so that theloop centrecoin-isof held B, in thisposition Fx = O,andicularlyto theX-axisand to thesame

rent Loopwith acurrent I in auniformwn above(seep. 157) thatingon thecurrent loop inqual to zero. It is known fromnt of theforcesacting on anyesultant moment of theseforcesion of point. Orelativetowhich

Therefore,wecan speak aboutmpére`sforcesin our case.moment.of Ampere’sforces

ula(6.29). If weperform cal-t iscumbersomeand hardlyhere),it turnsout that the



traryshapeof the loop can

oment of thecurrent loop (pm =

hemom-cting onmagne-ular both to vec-

'lhemag-,,.Bsin cn,een vec-pm B, M :. 0

hat.thepisstable.Whent suchable: theslightest deviationceof themoment of forcethat

m the initial position still

alidity of formula(6.36) by usingngular current loop (Fig. 6.14).ethat theforcesacting on sidesaarevector B. 1-timetheseforcesare directed

renot shown in theligere) and only striveoop.Sidesii areperpendicular to B,tedupon by theforce

eloop so that vector pm becomesdirectedop isactedupon by acoupleof forcespr duct of thearm asin ir. of the couple

;ne;i=; s{im == I*.-er thesurfaceei stretcheei over thecur-snot depend on thechoiceof thesurfaceSp over which it is stretched.



a bounded by theloop and Iba= pm,

written as(6.36).notethat expression (6.36)neticheldsaswell. Theonly

of thecurrent loop must beehect of nonuniformity on theM can be ignored. Thispre-tarycurrent loop.behavesi n anonuniformay asan electric{dipoleintricheld: it will be rotatedeequilibrium (for whichr theactionof theresultanttheregion whereinduction B

acement ot Current Loopn external magneticheld (we

constant),certain elementsAmpere’sforceswhich hence

lacement of theloop. Weshalleby Ampere’sforcesupon anhe

byI

of theBmops.Fig. 6.15

cular case: acircuit (Fig. 6.15)ngthZ isin a uniform magneticcuit planeand directed behindcordancewith (6.29), thejumperorceF IZB. Displacing theisforceaccomplishespositive



, (6.38)thearea bounded by thecon-

n of magneticflux (D,weal n to thesurfacebounded byaway that it formswiththe

circuit aright—handed systemurrent I will alwaysbeaposi-and, theflux (Dmay beeitherase,however,both (Dandtities(if thefield Bweremper weredisplaced tothee casesexpression (6.38) can.37).d for an arbitrary directionvethis,let us decomposevec-: BBn + B, + Bx.Theg,thejumper is parallel to the

oduceany forceacting onthealong thedisplacement) is

pendicular to thedisplacement,nywork.The onlyremainingormal to theplanein whichnformula(6.38) instead of B,,_ dS = dd), and we again

urrent loop whichisarbi-nt nonnniform magneticfield

ybe arbitrarily deformed inividethegiven loop into infinite-

dconsider their infinitesimalonditions,themagneticfieldrrent isdisplaced can beas-splacement,wecan apply toexpression dA= I d'iDfor theDexpressesthecontribution of o theincrement of thefluxngup elementary worksfor

we again obtain expression



ment of themagneticflux

neby Ampére’sforcesuponcurrent loopfrom theinitial

on2, it is sufficient to integrate

dconstant during thisdis-

neticfluxesthrough thecontourons. Thus, thework of Ampere’sotheproduct of thecurrentneticilux through thecircuit.onlythemagnitudebut alsowork.current I is rotated in magnetic

which n it Bto theposition inwhichtheloop (it should berecalled that the

nnected withthedirection of thecurrentwrule).Thearea bounded by theloop isS.

orcesupon such a displacement, assumingedconstant.ehave

A> 0, whileuponthereverserotation

(6.40) is accomplished not at theexternal magneticheld (which doesnotf thee.m.f. sourcemaintaining thecurrentill beconsidered in greater detail in

nductionB. Current I flowsalongown in Fig. 6.16.Find themagneticequired dataarepresented in the Figure.tity B= B- —|— BV, whereB_isthe



erectilinear part of theloop and BV,ding to the Biot-Savart law(seeExamplei

G"

pol __

2na.for cz,·—» 0,wearriveat thefamiliar

ormsa planespiral consisting of packedturnsthrough which adirect cur-f theinternal an external loopsareae magneticinduction Bat thecentre2) themagneticmoment of thespiral for

6.13), thecontribution of oneturn of uction isequal to

urnsin theinterval (r, r+ dr),



(2) and integrating theresult over r

f aturn of radiusr ispm, = Irrrz,pm, dN, heredNis defined by formula

¤b+b2>.gstraight conductor having thes section in theform of athin half-ring

urrent is directed from thereader behindthemagneticinduction BontheaxisO.

eterminethedirection of vector Bwe mentally`dividetheentireconductorth currentsdI. Then it isclear that thelamentsgivesvector dBdirected to thently,vector B will alsobedirected to the

hefield Bat point Oit issufticient toonsof elementary vectorsdBfrom eachectionof vector B:

ehave

g. 6.19). Substituting (2) into (1),weget



onof Band princifileol sup¢·rposi·longstraight wireof circu ar crosssection,l cavity whoseaxisisparallel to the con-lativeto it by a distancel. Adirect cur-thewire. Find magneticinduction B

htheprincipleof superposition, theresentedasfollows:

ductionof the conductor without cavity,

uction of thefield at thesamepoint duegh thepart of theconductor which hasatethecavity.rst of all thecalculation of magneticconductor at a distancer from itsaxis.ation,wecanwrite2nrB== gr5nr*j, whenceoncan berepresented with thehelp of

Bo and B', wefind their difference(1):><w1=-%*ln><<»-——·-·>1.i— r', whencer -— r' == l, and

ticfield Bin thecavity is uniform, andardsus(Fig. 6.21), thefield Bliesin the

rectedupwards.



ition. Current I [lowsthroufgh alongonal areaisS and thenumber o turnsn.ghtheendfaceof this solenoid.

hrough thesolenoid endfacebeQ. If enoid in contact with theendfaceof ourEli .,.

hecontactingendfaceswill beQ+ Q=hrough thesolenoid’scrosssection farve

entionto thefollowing propertiesof along solenoid.das sown in Fig.6.22. Thiscan bef superposition principle: if we placeonfield Boutsidethethusformed solenoidleonly with the field configuration shown

rposition, it followsthat thenormalmeover the areaof theendface, sincewhenoid,Bn -|— Bn = Bo,whereBsis thefieldm itsendfac-es. At thecentreof theendfaceBo/2.d. Thewinding of along solenoid of band of width h, woundin onelayer prac-urrent I flowsalong theband. Findoutsidethesolenoid asafunction of dis-

ar current density l can berepresentednts:

nd luisclear from Fig. 6.23b.In ourevectorscanbefound with thehelp of



}/i—-(h/2na)’.

nsidethesolenoid isdetermined, ine quantityi L , whileoutsidethesolenoid,

h/2M)“( r < ¤>.

n circulationwhilecalculating Baout-¤2:riaiH.ecurrent in thesolenoid asthesuper-"longitudinal"components, wearriveatongitudinal component of the field Bandonly thetransversecomponent out-aight current).andwidthmaintaining thecurrenth -» O, but I/h = const. In thiscase,enoid remains, i.e. the solenoid becomes

el currents.Two long wireswithnted- at their endsby resistor Rand atdtoatage.Theeach

distancector of of there-ultanteen the

sssurirrespec-rrent isflowing through them) (Fig. 6.24)-agneticforceFm, wemust takeinto accountethat an excesscharge1.correspondsto

entheelectricforceexerted per unitr wire canbefound with thehelp of the

: —--—-

eentheaxesof thewires. Themagneticof thewirecan hefound with thehelp



nof vector B:

wire.oforces, electricand magnetic, are

ctricforceisresponsiblefor the,attractionmagneticforcecausestheir repulsion.forces:

tween thequantities1 and 74.(see

owsfrom relation (2) that

obtain

tionvanisheswhenthisratio isequalenR== R0, where

thewires repel each other. lf, on8, and thewiresattract each other.mentally.ent-carryingwiresattract each othern theelectriccomponent of theinterac-r asufficiently small resistanceRinthe

rceof interaction between thewireswegenerallycannot determinecurrent Id to avoid confusion.re'sforces. Aloopwith current Ight wirewith current I0 (Fig. 6.25). 'I`hecular to thestraight wire.Find the moment

n thisloop.The required dimensionsof thee.tingon curvilinear parts of thelooper hand, theforcesacting on ther ectilinears. Wemust calculatethetorqueof this

mentsof theloop (Fig. 6.26). It can betorqueof thecoupleof forces correspond



ere’sforceisgiven by

neticinduction B on thedistancer

found with thehelp of thetheorem on

o (2),then (2) into (1) and, conside-a, integratetheobtained expressiongivesp,heleft (Fig.6.26).ent,having themagneticmoment

circular loop of radiusR,along whicheforceF acting on thecoil if its` distancesl and vector pm isoriented asisshown

), therequired forceis defined as

uctionof thefield created by theloops select Z-axisin thedirection of thevec-of (1) onto thisaxiswill bez,that BZ = Bfor the given direction of neticinduction Bisdefined by formula



tionof theforceFl < 0,i.e. vector Fwith current I . Theobtained result canform as follows:

m(and hence

edoppositely, then Iand henceFZ >Oedto theright, Fig.6.27irectionof pm.onfor Fisvalid for both orientationsofpmong thin-walled circular cylindereexertedon thecylinder walls.urfacecurrent element ldS, wherey and dS isthesurfaceelement. Weshall

esurfaceand volumeelementsof thecur-

esappearing in thisrelation is clarifiedrm,wecan write

n thesurfacecurrent element. in thismulaobtained from (6.28) with thehelp

duction of thefield at thepoint of location,but createdbyall other current elementsrder tofind B', weproceed asfor cal-2.3). Let B, bethemagneticinduction



rfaceelement itself at apoint very closewhereit isassumed that thecurrent isgto (6.22),Bi = (1/2) poi.n circulationof vector Band symmetry

yseethat themagneticinduction of thear itssurfaceis equal to

efieldisabsent.thefieldB"from the remaining cur-d2verycloseto the cylinder surface(seeandsatisfythefollowing conditionsinsideace:'

2), we obtain thefollowing expression

-—•-·

inally get

hat the cylinder experienceslateral com-

anceance.

ainsubstanceisintroducedby currentsin conductors,explained by thefact that eacht is magnetized (acquires

heaction of magneticfield.atesitsown magneticfield B',

withtheprimary field B0 createdesultant field

ieldsaveragedover aume.



.0 of conduction currents,arges).Hence, for theresultant

magnetic,theGausstheorem

esof Bremain continuousof a substance.n.At present, it isestab-y substanceshaveintrinsicemotion of intrinsiccharges

ment correspondsto a circularfieldin thesurrounding space.agneticfield,themagneticriented at random, and henceddueto thesemomentsisequal

magneticmoment of thesub-hesubstancesthat havenosenceof an external field.an external magneticfield,themagneticmomentsof the

minant.orientation, and thesub-resultant magneticmomentthiscase,themagneticmo-esno longer compensateeachd B' appears.ewhosemoleculeshavenosenceof external field proceeds

erialsareintroduced into ancular currentsareinducediresubstanceacquiresamag-sultsin thegeneration of the

ymagnetizedwhenintroducedferromagneticmaterialssucheir manyalloyshaveclearlyerties.f magnetization of amagnet-gneticmoment of aunit



______ledmagnetization and denoted

nitelysmall volumesurround--hemagneticmoment of anmmationisperformed over

onefor polarization Pan berepresented in theform

ntrationand (pm) istheaveragecule.Thisformulashowsthatyto theaveragevector (pm).issufficient to knowthebehav-me that all themoleculesthesamemagneticmomentsimplify theunderstanding of phenomenon of magnetiza-

sein magnetization Jof a mate-ng increasein vector (pm): if

h point of asubstance, it isformlymagnetized.

s wasmentioned above, mag-ausedby the preferential orienta-findividual moleculesin oneaid about theelementary circularhmoleculeand called molecularat such a behaviour of molecularnceof macroscopiccurrentsl's. Ordinary currentsflowing inedwith themotion of chargerecalled conduction currents(I).

magnetizationcurrentsappear,r madeof a homogeneousmag-is uniform and directed alongin the magnetized magnetic7.1. Molecular currents



. _intsof their contact haveop-nsateeach other. Theonly uncom-are thoseemerging on thelateralecurrentsform macroscopicnt I' circulating over the lateral

current I ' inducesthesamemac-hat of all molecular currents

case: amagnetized magneticr instance,themolecular currents7.2, wheretheline thicknessof molecular currents. lt followss directedbehind theplanemagnitudewith thecoordi-emolecular currentsarermtmogeneousmagnetic, and aslumemagnetization current l'positivedirection of theY-axis.bout thelinear i' and surfacej'easuredi n A/m and j' in A/mzagnetic. It can bestated that

om a magnetized magnetichat would becreated by thetI' in avacuum.ln otheredistribution of magnetizationhehelp of the Biot-

sponding to them, and thenBby formula(7.1).



onof currentsI' dependsnand thepropertiesof amag-ll. For thisreason, in thegeneralBina magneticcannot besolvedtry and find another approachm. Thefirst step in thisdirectionportant relation between mag-rtainproperty of vector J,

arycasethecirculation of bitrarycontour I` isequal

netizationcurrentsI' enveloped

ionisperformed over ann thecontour Fm,weshall calculatetheurrentsenveloped by contour I`.rfaceS on the contour I`

m thefigurethat somemoleculareS twice(oncein one directionheoppositedirection). Hencebution to theresultant mag-hesurfaceS.rentsthat arewound around



urfaceS only once. Such molec-scopicmagnetization current I'

eequal to 1,,, and thearea,asisshown in Fig. 7.4, theiswound by thosemolecularnto theobliquecylinder with

whereot istheanglebetweenhedirection of vector Jat theAll thesemolecular currentsand their contribution to the= Imn dV, whererzisthe molec-tinginto thisformulathenJcosot dl = Jdl,that I ,,,8 _,, pm is themagnetic

molecular current, and ImSmnaunit volumeof thematerial.ver the entirecontour I`,m isproved.

if amagneticisnonhomo-1·ent l' generally piercesthend not only theregion near itstour F Thisis why wecan useereintegrationisperformedunded by thecontour I`. Inedasif to "drive"theentirethesurfaceS. Theonly goalthecalculation of thecurrent I':

nJis equal tothemagnetization currentpace.f J. Thepropertiesof they Eqs.(7.5) and (7.6),. do notself is determined onlyby thehich isbounded only bythe



etjc) dependson all currents,oncurrentsI' and the conduc-somecasesythesituationor JwereI'

magnetizationength of a cylinder madeagnetic,if its magnet-r J being directed every-s.(7.5) to the contourg.7.5. It canbeonof vector Jsequal to theproductnetizationcurrent

denoteitslinearnder consideration embracesthemagne-wsfrom theequality Jl = i’l that

ay, that vectorsi‘”and Jare mutually

Vector H(for magneticfieldsticsplacedintoan externalncurrentsareinduced. Henceowbe determined not onlyy magnetization currentsas

ction.and magnetization cur-ontour l`rrentsI' in thegeneral casermula(7.8) becomesinap-.It turnsout,however, thataryvector whosecirculationheconductioncurrentsembracedweknow, thecurrent I ’ is



hemagnetization:

nof vectorsBand J istakenexpressI' in Eq. (7.8) through

of thisexpression byH.liaryvector H,

arbitrarycontour l` isequalonductioncurrentsI embraced

heorem on circulation of vec-or Haround an arbitrary closedraicsum of theconduction currents

thesameasin thecaseof p. 149).r His acombination of two

noand J. Hence, vector Hr which doesnot haveanydeeper,theimportant property of orem on itscirculation justiiiesr: in many cases, it consider-

ationof thelield in magnetics.arevalid for any magnetics,

edmagneticheld intensity, but weo emphasizeagain itsauxiliary nature.



_ _2) that theagnitudeof of thecurrent per unit length.peresper metre(A/m).rem on circulation of vector H:

qual to thedensity of theconductionthesubstance.and H. lt.wasshownabovedson themagneticinduction Bnce.However, customarilybut with H. Weshall confinenly thosemagneticsfor whichdH islinear, viz.

ceptibility.It isa dimensionlessnetic(that Xisdimensionlesscordingto (7.11), thedimen-e).ywewhich isalwayspositive,aybeeither positiveor negative.atisfying formula(7.14) areX> 0) anddiamagrzetius(yg < U).ileintliamaoguetnzJit H.ddition to theenagne tics,r which thedependenceJ(H)

m: it is not linear, and besides,ndson theprehistory of amag-bedescribed in greater detail

For magneticsolwyjng (7.11),eform (1 §— X) II B.#'u,,.

of themedium.



hilefor diamagneticsu < 1.thcasesit differsfrom unityicpropertiesof thesemagnetics.H. Let us consider aquestion

derstanding: which currentsr H?Generally,theheld of Hdson all currents, both con-Thiscan beseen even fromsesthefield His determinedVector His very helpful forthesametime,thismakes

onclusion that thefield of Hioncurrentsand for incorrect

m on circulation of vector Hm expressesonlyacertain prop-ut definingthe'field itself.of along straight wirewith current Igamagnetic(Fig. 7.6). Let usfind thetorsBand llor It around I.

on the

spaceisyn current! 9ioncurrents1 / nour case, = B/nun, _

of vectorconduction Fi 7 (tioncur- g' °)

paramagneticwill lead to a changehefield ll. Thecirculation of vector Bochange, sincethesurfacestretched onr bepiercedby themagnetization currents.will remain.However,thecirculation.ur I` will remain unchanged in spiteof elf.

etic?Weshall shownowthatdeamagneticareabsent if (1)usand (2) therearenoconductione.for any shapeof a mag-of themagneticfield, wecan be



tioncurrentsareequal to zero,tioncurrentsremain.all usethetheorem oncir-dan arbitrary contour l` com-netic.In thecaseof ahomo-ubstituting XHfor J, takeXal andwrite

heremainingintegral isequalonduction currentsI embracedahomogeneousmagnetic, we

ecurrentsI' and I is valid fornetic,in particular, for avery> d1' = j§,dS and I ·—>- dI = j,,dS.dafter cancelling dS, weobtainfor any orientation of therection of thenormal n to it.

emselvesarerelated through

neousmagneticj' 0

r B and I-Iionsfor vectorsBand H atomogeneousmagnetics. Theseust as in thecaseoi dielectrics,eorem and thetheorem oncir-

vectorsBandHthesetheorems

(7.19)usimagineavery lowcylindereentwo···rutagneticsasisshown



B in theoutward direction {weateral surface) can bewritten

f vector I? onto thenormalBU; = ——-5,,,,and after can-ation becomes

of vector Bturnsout to bethe

terface.Thisquantity doesnot

shall assume, for higherductioncurrent with linearacial.surfaceof the magnetics.

circulationof vector H toour whoseheight isnegligiblylengthlj (thearrangement of 7.8). Neglecting thecontribu-e contour to thecirculation,ntourf vector i onto thenormal N

msa right-handeil system withumvention).Taking thetwoo thecommon unit vector of theeobtain NW-—H,,. Can-equation,weget



nt of vector H generally hastionthrough theinterface,eof conduction currents.nductioncurrentsat theinter-O), thetangential componenthesameon both sidesof the

ioncurrent at theinterfacemagnetics,thecomponentsBn

without ajump) uponan transitioneother hand, thecomponentsdiscontinuities.r Bbehavesat theinterface

HbehaveslikeE.elinesof B undergo refractionmagnetics(Fig. 7.9). Asinall find theratio of thetangents

tothecasewhen thereisno con—ce.According to (7.22) and

Bin·



account,weobtain the lawof tor B(and henceof vector H

sof vectorsBand Hnear thenetics(in theabsenceof con-it,.Acomparison of theden-B2>Bl, whileH2 < H,.

discontinuity upon atransitionheHlinesdo (due to thesur-).eldlinesisused in magneticprotection.nshell (layer) isintroduced into an exter-lineswill beconcentrated (condensed)idetheshell (in thecavity) themagnetic

derablyweakened in comparison with theds, theiron shell actslikeascreen. Thisnsitivedevicesfrom external magnetic

Magneticthat determination of the

hepresenceof arbitrary mag-catedproblem.Indeed, for thish (7.1),thefield B0 of conductiontedby themacroscopicfield B'rrents.Theproblem isthat weeconfiguration of magnetizationhat thedistribution of theseureand configuration of thenfiguration of theexternalnductioncurrents. And sinceweof magnetization currents, we

sewhen theentirespaceedby ahomogeneousisotropiciscasein greater detail. lintthephenomenaobserved when

ong a homogeneousconductor



ductor isa magnetic, magneti-oughit, viz.volumecurrentscurrents.Let ustakeacontouringconductor. In accordanceionof vector J(7.5), the alge-volumeand surface) currentsisnceJ¤= 0 at all pointsof theQ== 0. HenceIQ.= -1;,i.e. theizationcurrentsareequal inirection.an ordinarycase, when cur-

ythin wires,themagneticfieldavacuum) isdetermined onlyemagnetization currentscom-r, perhaps,thepointslying

rounding theconductor byingmagnetic(for thesakeof it is aparaniagnetic, X>0).smagneticand thewire, surfaceappear.It can beeasily seendirectionasthe conduction

econductioncurrent 1. thezationcurrentsin the conductorecurrentscompensateoneanotherd), and the surfacemagnetiza-nconducting magnetic. For sufficient-eld Bin themagneticwill beecurrent I +I'dto findingthecurrent I'd theconductor byacontourof thenonconducting magnetic.e perpendicular to thewire

oncurrents.Then-, taking (7.7)n,wecanwrite

owsthat I' XImagnetizationcurrent I' and



racticallycoincide(thewiresintstheinduction B' of ther ntsdiffersfrom the induc-nduction currentsonly in mag-relatedin thesameway as the

sultant iieldB= B0B'

tirespaceisfilled with ahomo-esit times.In other words,thesein themagneticiield Buponpiedby thetield with a magnetic.25) by uno, weobtain

tion,lieldHturns out to be

also validwhen a homogeneousumebounded by thesurfacesformedf theconduction current). InnductionBinsidet.hemagnetic

e, themagneticinduction B'currentsisconnected witheticthrough thesimplerelation

yobtainedfrom theformulaunt that li' XB,) and

nedabove, in other :.asestheated,and formulas(7.24)-

consider two simpleexam-

enoid. Asolenoid having nl ampere—witha homogeneousmagneticof the



magneticinduction P of field in

bsenceof magneticthemagneticise ual to pon!.Sincethemagneticfills

elddiffersfrom zero (weignoretheedgetionBmust be p timeslarger:

or Hremainsthesameasin the absence.sedby theappearanceof magnetizationrfaceof theonasthe{

olenoid 'Iever, p< 1, Hntswill be'

o valid yorm of a/ ’ Bdethesole- / jhesole- / it- A‘

ght.magnetic.salonghoseaxisFig 7 Hpermea·` Find the_function of thedistancer from thecylin-

heorem on circulation_of vector Bsareunknown. I nthissituation, vector llnisdeterminedonly by conduction cur-wehave2nrH= I, whence

emagnetic-vacuum interface, theeH, undergoesadiscontinuity

magneticiscaused by surfacemagnetiza-coincidei..direction wifh_thecur1·ent Isystem and hence"aniplify’ thiscurrent.

hecylinder thesurfacemagnetizationly, but it doesnot produceany effectOutsidethemagnetic, themagneticfieldsoneanother.

crespect,all substancescannetic(paramagneticsand dna-



gnetic(ierromagnetics). lt isnceof magneticlield, para-gnetizedand arecharacterizedencebetween magnetization J

ances(solids) that. may possess.e. which aremagnetized even

agneticfield.Typical ferro-nd manyof their alloys.Atypical featureof ferro-

onlinear dependenceJ(H) ormagnetization curvefor a

tizationfor H0is also zero.magnetization curve. Even atof H, magnetization J attainsduction Bpo (II J) alsoningsaturation,li continuesthelinear lawBpoll —§·

-S. Figure7.13 representsthebasicB-Hdiagram.endenceB(H), thepermeabil—not bedefined asa constantpropertiesof a` specilicferro-re, it'isassumed that it l>’,¤;,:,,H,Fig. 7.14) Thevalueu,,,,,,

gneticsmay he very large. For5,000, whilefor supermalloy

oncept of permeability is



____magnetizationcurvesince, aspendenceisnouuuique.sthenonlinear dependenceeticsalso exhibitmagnetichystere-andHor Jand Hturnsout to

inedby thehistory of theferro-f wemagnetizean initiallyticby increasing Hfrom zeroionsetsin (point 1 in Fig.7.15),

H, to -—-H1,themagnetization0, but will takepath 1 2 3 4.

—H1 to +H,_ themagnetizationepath 45 61.calledthehystere.¤i.—r loop.

oop is obtained when;at.nrationIf, however, thereis no satu-hysteresisloopshavesimilarscribedinto themaximum

HO magnetization doesnotcterizedby thequantity 1},on.It correspondsto theresidualnceof permanent magnetsismagnetization.Thequantity Ber theaction of thefield H6ositeto that of thefield causingntityNciscalled thecoercive



different ferroniagneticsvaryformer steel, thehysteresis,whilefor ferromagneticsusedt magnetsit is wide(H,. is

coalloy,HC50,000 A/m

etizationcurvesareused in a convenientnetizingferromagnetics. Amagnetizedhroughwhich an alternating current isgraduallyreduced to zero. In thiscase,edto multiplecyclicreversemagnetiza-psgradually decrease, contracting to theszero.omagneticisheated uponbe shown that in thiscasetheper unit volumeof the ferromag-

o the"area"Sn of thehysteresis

ureincreases, theability of etizedbecomesweaker,and inizationbecomeslower. At aeCuriepoint, ferromagneticappear.uriepoint, a ferromagnetic

netism.Thephysical natureexplainedonly with thehelp ol-called exchangeforcesmayainconditions.Theseforcesthemagneticmomentsof elec-10 pm in size) of spontaneousleddomains, appear in crystals.main,aferromagneticismag-a certain magneticmoment

mentsaredifferent for differentnceof external field, theresultant

mpleisequal to zero, and thescopicallynonmagnetized.fieldisswitched on, the



eldgrowat theexpenm of field. ln weak fields, thisonger [ fields, thesimultaneous{Ìl0llt‘ lIlUIllt‘|IlS Wllllill l.llOClllll'0 (l0lIl&\l|Irreversible,which explainsnetization.

face.In thevicinity of point Aacuum interface, themagneticinduction

0 forming an anglean with thenormalbilityof themagneticisii. Find themagneticin thevicinity of point A.ntity

0) and (7.22) at theinterface, wefind

omponent of vector H0 in theViltlllllll.nsinto (t),weobtainn`; 0.0.ncurrent.Along thin current—carry—isarranged perpemlicularly to aplane(Fig. ill?} Thep:~rnieabi1ity of themag-rrent.density i' of the surfacemagnetiza-ceasafunction of thedistancer from the

r theconfiguration of thesurfacemag—rom Fig. 7.·!7 that thiscurrent hasthe



ethetheorem on circulation of magnetiz-ntour asmall rectanglewhosep aneisizationcurrent in thisregion. 'l`hearrange-n in Fig.7.18, wherethecrossesindicate

magnetization current. From theequality

whereH can bedetermined from thedthecircleof r adiusr with thecentre

H= I (it isclear from symmetry consider-t becircleslying in theplanesper-

withthecurrent I). Thisgives

H. A long thinconductor with cur-atingthespacefilled withanonconduct-tyit from avacuum. Findthemagneticceas afunction of thedistancer fromdthat thelinesof Bare circleswith centres

learly also circles. Unlikevector B,ntinuityat thevacuumanagneticinterface.neticfieldsin themagnetie· and vacuumdancewith thetheorem on circultaionour in theform of a circleof radiusr withaxis,wehave

e,orther,weobtain

r '



ldsB and ll in thiscaseareshown inder to convincehimself that for it == 1,formulasfor Band II i n avacuum.

sHand J. ¢\‘direct current I flowscylindrical wirewith acircular cross sectiondeof aparamagneticwith thesusceptibil-unctionof thedistancer from theaxiszationcurrent density j' insidethewire.onof vector Haround thecircleof ewireaxis, it followsthatr),

of thedependencesH(r) and B(r).ncircu ation of magnetization J(seeFig. 7.20): 2m-J= I', whereI' is

mbraced by this contour. Wefind theon(going over from r to r + dr):

quationcan hetransformed as

XH= (XI/2nR“) r.Thisgives

current hasthesame direction asthehesurfacemagnetization current which

led withanonhomogeneousisotropictibility dependsonly on thedistancer== arz,whereais aconstant. Themagneticisis B0. Find (1) magnetization Jand (2)ty j' as functionsof r.case,His independent of r (thisdi-onof vector Haround the contour shown

HenceH = Hu on thesolenoid’saxis,

culationof magnetization Jaroundshown in Fig. 7.21 on theright,it f ollows



ht and dr itswidth. Hence

at vector j' isdirected against thenormalht—handsystem with thedirection of cir-

n other words, j' isdirected towardsusntour in thefigure, i.e. volumemagnetiza-or B_,theleft-hand system.hastheshapeof aring with anarrowmiddlediameter of thering isrl. Thegapuctionof thefield in thegap isB. Findand Jinsidethe magnet, ignoring theedgesof thegap.

m on circulationol vector Il around theFig.7.22) and considering that thereisnwriteO,

of vector ll onto thedirection of contourenso that it coincideswith thedirectione

—— ———-—— _ {

at thedirectionof vector ll insidethetetothedirection ol vector Bat thesamet asb » O, II -—» Uaswell.ationJcan be found by formula(7.11),



· ·

ctorsB/po, ll, and J at any pointn Fig.7.23.

gNturnsiswound on an iron corehemiddlediameter d. Anarrowtransversecut in the core. When current. I ispassedgneticinduction in theslot isB. Finder theseconditions, neglecting thedis-gesof theslot..hthetheorem on circulation ol vector Iler d (seeFig.7.22), wehave

gnitudesof ll in iron and in theslot re-nceof lielddissipation at theedgesmeans

hat B ==

amagnetic.'l`hedeviceshown inng(withthehelp ol` abalance) alorre withcsphereol` volumeVisattracted to theneticinductionon theaxisof thepoleasBB0e‘*¤x2,whereBn and an arecon-

m at which thespheremust beplacedemaximum and (2) themagneticsus-et.icil` thennaximum forceoi attraction



keof deiiniteness, vector Bon theaxisncidewiththeX-axis. 'l`hen, in accordanceherewe took into account that thectedasvector B(for paramagnetics) and

Vandehave

ativero,weob-or deter-hence

we M

that it y. 1 for paramagnetics.ical rod madeof a paramagneticwith

Xand cross-sectional areaS isarranged

rrying coil.Oneend of therod isat themagneticfield isB,whiletheother end

magneticheld ispractically absent. Findactson therod._

olatean element of therod ol length d.·ron thiselement isgiven by

eaxisoi thecoil is directed to therightsitivevaluesof :1:. Then Bx= B,dn = 01,



HS dx, weget

dB.

ion,weobtain

at vector F isdirected to theleft, i.e. thet-carryingcoil.lumeV, madeof a paramagneticwithasdisplaced along theaxisof acurrent-wherethemagneticinduction isBto theieldispractically absent. Find thework accomplishedin thiscase.—axisalong theaxisof thecoil. Then

gainst themagneticforcesupon thedis-da: isequal to

z·:~;·dI,f themagneticforce(6.34) onto theX-axisat thework isdoneagainst thisforce.axisis directed towardspositive

ndOn == 6.r (otherwise, Bx== ——Band0Bx/dn doesnot depend on thedirectionat pu] = JV== XII V, werewriteEq. (1)

u.

ween Band 0, weobtain

c

ChargeInvarianceelectricand magneticfieldsingany clear relation between



ly becausethetwo fieldswereer,it is impossible.and magneticfieldsmust

her asa singletotal electromagneticout that electricand magnetic

hecomponentsof asinglephysi-electromagneticyield.agneticfield into electricativenaturesinceit dependstoerencesystem in which theThefield which is constant inneral caseis found to vary in

les.l system Kat a constant velocity v.rveboth the electricand magneticfieldsvarying with time. If, however, wegomoving together with thecharge,thes’ system,and weshall observeonly the

rgesmovein thesystem KtowardseachI nthissystem,weshall observeboth,both thesefieldsvarying. In thiscaseitystem K' whereonly oneof thefields

sapermanent nonuniform magneticdpermanent magnet). Then in thesystemtem Kwe shall observeavarying mag-

own below,an electricfield aswell.ent referenceframesdifferentenelectricand magneticfields.

n pointsof thischapter, viz.f fieldsupon a transition fromher,wemust answer thefor further discussion: howdodtheGausstheorem for vector Es?nt,thereisexhaustiveevi-f anisolated system doesnothemotion of chargecarriers.eutralityof agas consistingtronsin thesemoleculesmove

anprotons. Therefore, if the



ocity,thechargesof theelec-compensateeach other and theever, no charge(to 20decimaln hydrogen gas.edwhileobserving the heat—.Sincetheelectron massis

emass of nuclei, thevelocityust increasemorethan that of

pended on thevelocity, thesub-edupon heating. Nothing fof .gedepended on its velocity,stancewould changeasaresultheaveragevelocitiesof electronschemical composition. Cal-

weakdependenceof the chargeinextremelystrong electricemical reactions. But nothingn. thiscaseeither.ationof all modern particleassumption that a particle’ssvelocityischanged.usion that thechargeof anyriant quantityand doesnotocityor on thechoiceof the

orem for Field E.lt turnseralizationof experimentalem ®EdS = q/eo is validut for movingchargesalso. Integral must betaken for adef-nreferencesystem.tial referencesystemsarecordancewith therelativityhe(lausstheorem validerence.nfor FieldsE and Beferencesystem to another,edin acertain.way. Thelaws



ablished in thespecial theorylicatedway. For thisreason,corresponding derivationsandecontent of theselaws, thehem, and theapplication of pecificproblems,.Supposewehavetwo inertialstem Kand thesystem K’ movingavelocity vo. WeknowthendBat a certain point in spacehat arethemagnitudesof theoint in spaceand timein theamepoint in space and timesandtimeinthetwo referencetheLorentztransformationsf)eanswer to thisquestion isityfrom which it followsthatf fieldsareexpressed by the

_:];c= (8.1)‘_"*‘.—;‘*"‘_-".

mark thelongitudinal and trans-vo) componentsof theelectric, and cis thevelocity of light

ons, theseformulashavetheform+-¤>,,H,/cz)

Ey(Cz’

X- and X'- axesof coordinatesaredirected.Li#£lL_v



xisisparallel totheY·axisand theZ'-axis

d (8.2) that each of thevectorsntermsof E and B.Thisistm·eof electricand magneticchof thesefieldshasno absoluteicand magneticfields, wemustncein which thesefieldsare

thepropertiesof electro-thelawsof itstransformationdB' at acertain point in spaceeuniquely determined only bysamepoint in spaceand time

elawsof field transformation

ponentsof E and B,whichm onereferencesystem to an-

mponentsdo not changeand remain

ent systemsareconnectedhly symmetricrelat ions. Thisheform of thelawsof trans-

nsof thefieldslsee(8.2)].mulasof theinversetrans-

s sufficient toreplaceiu for-medquantitiesby unprimedl asthesign of vo.formation(oo Qc). If thethesystem Kat a velocitynominatorsof formulas(8.1)dweobtain

y —-[vo >< E]/cz.(8.3)

-iv.,>< Eye.! isa)



rst formulain (8.4) can bempleway. Let a chargeqcityvo at acertain instant t.thischargeisF zzqE »(—r to theinertial system K' moving

hesamevelocity asthechargeqcityvo. At thisinstant, theem K', and theforceactingurely electric: F' qE' If onsideration),thisforceiseobtain thefirst formulain

nformulasfor magneticholdehelp of thetheory of relativity

omecalculations.mpleof theapplication of

platemovesat a constant nonrelativisticgneticfield B(Fig. 8.1). Find thesurfacen thesurfacesof theplateasa result of

cesys- Bcordance4), in thisnt uni-

edtowardsxternal Fin 81splaced so °` 'pear onandnegativecharges, on thehind sur-

sechargeswill besuch that thefieldidetheplatecompletely compensatestheuilibrium theresultant electricfield insideo. Taking into account relation (1.11),

esolving thisproblem, wecould followz. from thepoint of viewof thereferencevesat thevelocity v.ln thissystem theredetheplate.lt appearsasaresult of theonent of theLorentzforce causing the



ronsin theplatebehind theplaneof nt surfaceof theplatewill bechargedce,negatively, and the electricfieldesuch that theelectricforceqE compen-ent of the Lorentzforce, q [v >< Bl, whenceldis connected with thesurfacechargedensity

= e0vB.ion of thegiven problem areequally

etism.Formulas(8.1) andof fieldslead to aremarkableof the magneticfield isapurelyfrom theexistencein natureual tothevelocity of light in

evelocity of propagation of o magnetism would exist at all.

electriccharge. Only electricwherethischargeisat rest. Con-

no magneticf ield wouldK' if ctended to infinity. This

fact that cis finite, i.e. in theiceffect.gnetism isauniversal physicalatedwith theabsenceof mag-

ffects,magnetism can hees, for example, themagneticconductor.Thereason behindcesisthat themagneticfield

genumber of moving electricsof thealmost completevanish-

practically ideal balanceof dprotonsin conductors. lnctionispredominant.nsationof electricchargeststo study relativisticeffectser correct laws.For thisreason,m,unlikeNewton’slaws, haveationof thetheory of relativity.ves. Sincetherelationsbe--fieldsvaryupon atransition



,thefieldsE and Bmust bece,even theconcept of thehargeby a moving magneticisemeaning.Theforceisdeter-uantitiesE and Bat thepointlf asa result of motion of thetheir valuesat thispoint change,.Otherwisethemotion of theagnitudeof theforce.

blem about theforceactingE and B at thepoint of locationelocityv. All thesequantitiesinertial referenceframeweare

'|

vo

____

ctithat upon atransitionanother thelield must be

eisat rest between thepolesof amag-t usgo over to the system K' that moves

nrelativisticvelocity vo relativeto thehat thecharged particlemovesin themag-2) Findthe forceacting on thisparticle

UCS lll them.dg}lc?UCIt SllOUlll be DOl.C(l,emagneticheld and not relativeto thegful to say that a particlemovesrelativegnet,or other objects. However, wecannotlativeto themagneticileld. Thelattereaning.All thisrefers not only to magnetic

wemust takeinto account that therectedtowardsus(Fig. 8.2), willissystem, thechargewill moveto theand thismotion will occur in crosseilelectricsakeof deliniteness, wesupposethat the



0. Then theLorentzforcein thesystem K'

Bl“qlVoXBl=0llowsfrom the invarianceof theforceupononsfrom onereferenceframeto theother.of Field Transformationeveral simpleand useful rela-om transformation formulas

ispresent in thesystem Kthefollowing relation existsn thesystem K'

/1— [52 and B], : 0,— BZ —lv0 ><i *E'l/cz, wherewe

crossproduct wecan writeiesto theprimed quantities).B1, wearriveat for-

neispresent inthesystem KO), then for thesystem K'

y/l/1-—[52 and E], == 0,SubstitutingBl for B and thenduct,wearriveat formula

dtothefollowing importante yields(E or B) is present in theagneticfieldsin thesystem K'E' _L B'). It should be notedstrueonly under certain ad-on the magnitudesof thevec-

(8.6) contain only thequan-stem of reference, theseequa-



.o thefieldsvarying in spaceandleisthefield of a uniformly

ativisticCharge. Theformulasnterestingaboveall bythatpertiesof electromagneticfield.from awsinceproblems

deter-mlymov- led asaq Vpurelyervedin theecharge.lem 8.10)reatedbyvetheform

theve- Fig- 8-3terncor-ure"of the electricfield config-nt P of thereferencesystem, vec-diusvector r drawn from thegeq to point P.determinedby t.heformula

/=·(SJ)between theradiusvector ring to thevelocity of thecharge.d"in thedirection of motion

he degreeof flattening beingcityvof thechargeto the veloc-o bornein mind that thefieldtogether with thecharge. That

renceframerelativeto whichh time.find thefield Bin thesame



f relation (8.5) in which wehaveiesby unprimed onesand v

ons(8.7) and (8.8) aretrans-pectively.

nvarianiscterizingelectromagneticfielderence(at thesamepoint inuestion arisesconcerning thecfield,i.e.thequantitative

of thereferencesystem.xist two such invariantswhichtorsE and B, viz.v.| (se)ntities(relativeto theLorentzuenceof field transformationquestion isconsidered in greater

ariantsallowsusin someckly and simply and makethend predictions.Weshall listons.mediatelyimpliesthat if E _L Be. E ·B = 0, in all other inertial systems

anceof E2 — c2B2that if E = cBany other inertial system E' = cB’ aswell.ystem theanglebetween vectorsEmeansthat E •Bisgreater (less) than zero),

m theanglebetween vectorsE' and B'l.stem F ;> cB(or E < cB). (whicl:eater (or less) than (zero), in any otherr E' { cB"·as well.l tozero, thenE _1_ Band le?== cBill beshown later that thisispreciselywaves.

eisequal to zero, wecan hudarefer-E' = 0 or B' = O.Thesign of theotherof thesevectorsis equato zero. The



or B= 0 in somereferencesystem,stem. (Thisconclusion hasalready been

k: it should bebornein minddepend both on coordinates

eachinvariant of (8.9) refersdtimeof thefield, thecoordi-n different systemsof referenceLorentztransformations.

ranslòrmation.Anonrelativisticonstant velocity v. Usingthetransforma-eticf`1eldBof thischargeat a point whoseeis defined by theradiusvector r.hesystem of referenceK’ fixed to thehe Coulomb field with theintensity

oaccount that theradiusvector r' = rsticcase). Let usnowreturn from theat movesrelativeto thesystem K' aturpose, weshall usetheformulafor theheroleof primed quantitieswill beplayed

dviceversa),and replacethevelocity Voseunder consideration, Vo= v, and henceonsidering that in thesystem K' B' = 0

6.3) which wasearlier postulated asof experimental data.



homogeneousdielectricwith thenstant nonrelativisticvelocity v in auni-wn in Fig.8.5. Find thepolarization P

acedensity o' of boundcharges.stem fixed to theplate, in addition tol bean electricfield which weshall denotermulas(8.4) of field transformation, we

ctricisgiven by

that accordingto_(3.29), E' == Eg/e.Thergesis

ate(Fig.8.5), o' > U, whileon theopposite

nchargedlong straight wirewithper unit length of thiswirein the referencelywithanonrelativisticvelocity vorectionof the current I .hthetransformation formulas(8.4),< B] will appear in themoving reference

thewireaxis.ormulawas obtained with thehelp

on.gto theGausstheorem (in the moving

it length of thewire.gives

rigin of this chargeisassociated with thencedby the"chains"of positiveand nega-erent.velocities!).m of protonsmoving with arelativis-f referenceK. At a certain distancefromeelectricfield isequal to E. Find the



field at thesame distancefrom thebeamrelativeto thesystem Kat the velocity voeam.e solved in themost simplewaywithut first we must find theinduction Binstancefrom thebeam wheretheintensity E

ationof vector Band theGausstheo-

thebeam,I = kv isthecurrent, and Aof thebeam. It followsfrom theseformulas

tingtheexpression for Bfrom thisfor-mationformulas(8.1), weobtain

esof B' form theright-handed systemthey form theleft—handed system (sinceK' in thiscasewill flowin the opposite

particlemovesin thespaceoccupieddicuar electricandmagneticlieldsE

ectilinearlyin thedirection perpendicularE' and B' in thereferencesystem movingcle.description of motion of theparticlehe condition

formationformulas(8.1), wehave

derationt.heLorentzforce(and henceequal to zero.formation formulas, themagnetic

isshown in Fig. 8.6, from which it followstly,considering that according to (1)



fy that theobtained expressionssatisfy

e in crossed fieldsE and B.Anon-ecificchargeq/m movesin aregion where

cular fieldsE and Bhavebeen created0 theparticlewas at point Oand its

ind thelawof motion of theparticle,

sunder theaction of theLorentzforce.eparticlealwaysremainsin theplaneXY.most easilyin acertain system K' wheresent.Let usfind thisreferencesystem.ons(8.4) that E' = 0 in areferencelocity Vo satisfying therelation E ==nient to choosethesystem K'whoserdsthepositivevalues ontheX-axistem theparticlewill moveperpendicularlyill bethesimplest.

movesto theright at thevelocityd only thefield B' isobserved. In accord-wehave1 -—vg/cz).rticlevo < c,wecan assumethat B' = B.will moveonly in themagnetic field,

ular to thisfléld. Theequation of motionm K' will havetheform



heinstant ¢= 0, when theparticleshown in Fig. 8.8. Since theLorentzular to thevelocity of theparticle, vo =1) that in the system K' theparticles

formly with thevelocity vo in acircleurn, movesuniformly to theright withThismotion can becompared with theim of a wheel (Fig. 8.9) rolling with theqB/m. _t thecoordinatesof theparticleq

sinmi),ot),qB/m.rm electricfield E in an inertial sysand directionsof vectorsE' and B' in

veto thesystem K at aconstant relativisticovector E.formationformulas(8.1) and taking

esystem K, weobtain\/1-8*,B=v,/c.tor E' isgiven by*cos’ on)/(1-8*)ctorsE' and Vucan bedetermined from

- 8%nof vector B' can befound in a similar

2 I/i—B°)» B' =Bl·nd

nd magneticfieldsE and Bof thesamef referenceK.Find. themagnitudesof glebetweenthesevectorsin thesystem K’sticvelocityvain the direction perpen-

hformulas(8.1), in thesystem K'



l bealso perpendicular to thevector voof thevectorsE.' and B' can be found by

(voE`/¢’)’

orsE"and B' can befound fromitstan-

ak —|· tan ab)/(1 —— tan ak tan cab).n ab= vol?/c’B(Fig.8.10),we obtain

o -» c(B -+ 1), theanglea' —» n/2. The

nowE and Bin onereferencesystem, andtorsislessthan 90°, thereexist r eferenceorsE' and B' are parallel to oneanother.ransformation formulas(8.1)san invariant.isproduct will be

Ei-Bi»+¤1·Bl ·inth thehelpof formulas(8.1):

ox ll/°)_ (2)EL and BL are perpendicular to thenumerator of (2) to theforml·Bi (l -—tl“).(3)



vo XB_L]•[vo XEL] = v§BJ_EJ_X8.11). Theremaining two scalar productsethevectorsare mutually perpendicular.(1) becomes

~|—E_L·Bl =E·B

mlymoving charge. Apoint chargeqnearlywith arelativisticvelocity v.

eldcreatedby thischargeat a point whosehargeis equal to r andformsan angle0

hargemovesin the positivedirectionof referenceK. Let usgo over to thesys-chargeisat rest (theaxesX' and Xof letheY'- and Y-axesareparallel). In ‘thechargehasasimpler form:

ve

·¤ ”(1)se transition to theinitial system K,system K' with thevelocity ·—-v. At thessesthrough theorigin of thesystem K,ctor r areconnected with thepro]ectionsthefollowing relations:0=y' (2)

kinto account that thelongitudinal dimen-ontraction,whilethetransversedimensionscordancewith thetransformationsinverse



ssions(1) wherez' and y' arereplacedssionsfrom formulas(2), weget

—---—·——·-—·-· 7;,- -—-_— -...-.— ..........;.............r’° l/1..52°v= zr/y,i.e.vector E is directed

Thesituationissuch asif theeffect of laghisis trueonly when v = const. If,witihian acceleration, thefield E isno longer

nd themagni-

___/7·"+!l’

rzand

lectricfield intensity will begiven by

"/2°two moving charges. The relativisticargeq moveparallel to each other at thein Fig.8.12. Theseparation between theon(8.7), find theforce of interaction be-

glebetween thevector vof oneof theowardstheother particleis0 = 90°.ormula(8.7),theelectriccomponent of

nent of theLorentzforceis

that in thecaseunder consideration Bula(8.5) from which it followsthat B=



ldbenoted that theratio

case(6.5). It can beseen that asv -» e,heforceFm tendsto Fe.ction(repulsiveforce) isgiven by

i/1-pa.nromagneticInduction.

asestablished that therelationof an electromagneticfield,viz.,is

m. In

nentsIanni -in- I§}=!=l.—_.L-ere` .:|:I=.`n

andIdinticZ Q

1 Fi8· 9-istlectroilynninics-thepheno-

nduction.It consistsin thateinduced current) appearsinonachangein themagneticflux

bythisloop.edcurrent indicatesthat theY', appearsin the loop asaresultflux.It isimportant herethatn howthechangein themagneticeterminedonly bytherateof Besides, achangein thesign



sto achangein thesign or

nduced current may begener-hisis illustrated in Fig.9.1,

hcurrent I (thecoil createstheL connected to thegalvano-einduced current.displacement of theloop Lieldof thefixed coil C.s fixed but themagneticmotion of thecoil Cor asaresultn it, or dueto both reasons.ometer Gindicatesthepres-eloop L.f theinduced current (andde.m.f.) is determined by Lenz’sirectedso that it counteractsthewords, theinduced current

ch preventsthevariation of theeinduced .e.m.f.

Fig.9.1) is drawn to theoughtheloop increases. Then.thiscaseisin the clockwiseoop from theright). Thiscurrentirected"to theleft, which pre-neticflux, which causesthe

aceif weincreasethecurrenthecoil and theloop L.On the

hecurrent in thecoil C, thecur-ll reverseits direction (it willckwiseif we look from theright).ortant physical fact, viz. the

nteract thechangein itsstate

agneticInduction. Accordingd ina loop isdefined by the



' :evariation of themagnetic{luxuctingloop.Theminussign

dwith acertain sign rule. Theisdeterminedby thechoiceof

bounded by theloop undern of 8, dependson thechoicercumvention of theloop.1edirectionof thenormal llitivedirectionof theloop cir-hrough the right—harzd screwrule*hen wechoosehenormal,

xcb aswell asction")of the

vedirectionst-handscrewFig- 9-2d)/dt have

uxistheweber (Wb). If thenet-icflux is1 Wb/s, thee.m.f.to 1 V[see( 9.1)].etic-fluxLinkage). If aclosed

nduced containsnot onebut Nqual to thesum ofthee.m.f.shemagneticflux embraced by

ual to tbl, thetotal flux cb throughch acomplex loopcan be

tal magnetic{lux, or _thescase,the e.m.f. induced inancewith (9.1), by theformula

reconnected throughtheleft·handot containtheminussign.



ticInductionical reasonsbehind theandtry to derivethelawof ealready know. Let usconsider

nt MagneticField.Let usblejumper of length l (Fig. 9.3)m magdicular to the planeindtarttheright withcarriersins,willameeach electronneticdalong thejumper. The

downwardsalong thejumper,ccurrent directed upwards.ent.Thechargesredistributed

nductorswill createan electricrent in theremaining partsof

heroleof anextraneousforce.is E* = F/(-——e) == lv >< Bl.

xpression can also heobtainedrmation formulas(8.4).

undaclosed contour gives,of the induced e.m.f. Intheehave

n inaccordancewith theadoptedesurfacestretched over thedirectedbehind theplaneof , and hence, according toepositivedirection of circum-ection,asis shown inthefigure.eldE* is directed against the



rcumvention,and hence8,

increment of thearea bounded/dt). HencevBl -.-= BdS/dt =crement of themagneticfluxin our case,d€D>0). Thus,

al form that law(9.1) isbitrarilyin apermanent non-Problem 9.2).ede.m.f.during the motiongneticheld isexplained bytheproportional to [v >< Bl,otion of theconductor.y that theideaof thecircuitbasisof all induction generatorsdingrotatesin an external mag-

ing MagneticField. In thisnt isan evidenceof thefact thatthtimecreatesextraneousaretheseforces?What istheirt e magneticforcesproportionals cannot set in motion thest (v = O). But therearenodqlv >< Bl! Thisleavestheonlycurrent isdueto the appearanceire. It isthisheld which ise.m.f.in ahxed loop placedgwith time.gnetic held varying with timeaceof an electricheld regardlesstingloop. Thelatter just allowshiselectricheld dueto the

l,amagneticheld varyingricheld.Circulation of vector Eloop is dehnedas



erivativewith respect totimeat theloop and thesurface

ncetheflux (D= 5 BdS (theundan arbitrary sur ace stretch-ested in),wehave

edtheorder of differentiationgrationover thesurface, whichdthesurfaceare fixed. Thenin theform

7)tructureasEq. (6.17) theroleof thevector -08/Ot. Conseque ntly, it canform thesameasEq.(6.26), i.e.

cal relationbetween electricand magneticmagneticfield Bat a given point determinesmepoint.Thefact that Vx E differsfromesenceof theelectricheld itself.of theelectricheld induceddiffersfrom zero indicatesthatential held.Likemagneticfield,tricheld can beeither apotentialortex held.heldE may bethesum of theldinduced by amagnetichecirculation of the electrostatic.6)-(9.8) turn out to bevalidwhen theheld E is thevector sum

cheld has found abrilliant applicationelerator,viz. bemzron. This acceleratortedchamber arranged between thepoles4). Thevariation of the current in theesa varying magneticheld which inducestter acceleratestheelectronsand simul-equilibrium circular orbit of a certain



_ncetheelectricfield isavortexltield,ingon theelectronsalwayscoincideswithheelectronsenergy.During the_ ,dincreases(~1makeabout aNXQQJW

energy up to4001ectrons’ veloc- Jocityof light ...zI'iE$\\.

etatron) re- lichtherole of __ { / sting oi asingleMg- 3-*beam.electromagneticinductioneticfluxthrough aloop varies

heloop or thetimevariationtoho lh reasons). However,

edijfereut phenomenafor ex-wo cases: for themoving loop weticforcepropo rtional to lv )>< BIarying withtime, OB/Ot,cfieldE wasemployed.profound principlewhich

mena,we must interpret thelawnasthecombined effect of twoa. Thesephenomenaaregen-rtheless(which isastonishing)s alwaysequal to therateof va-through thisloop.dB varieswith timeaswellgement of theloop inthefield,calculatedby formula(9.1)sidethetotal timederivativekesinto consideration thetwoaw(9.1) can herepresented

—hand sideof this equation isHerethefirst term. isduetoneticiield,while thesecond



op. Theoriginof theseconddetail in Problem 9.2.meswecomeacross _asitu-omagneticinduction in theformcable(mainlybecauseof thehechoiceof thecontour itself).

o resort to thebasiclaws, viz.ressthephysical meaning of thection in all cases.iveexamples.ipisdisplaced at avelocity v throughcfield Bisapplied (Fig. 9.5).ln theby thedotted circlewheretheheld Bisnometer Gisconnected to fixed contacts

ng strip.Will thegalvanometer indicate

eemsnot so simple, sincein thiscaseittour itself: it isnot clear whereit shouldowthispart of thecontour will behavep. If, however, weresort to theLorentzelyclear that theelectronsin themovingrds,which will createan electriccurrentdirectedclockwise.eaof thisexperiment formed thebasistheinternal (thermal) energy isdirectlygy. Instead of aconducting strip, aplasmaositiveions) is blownat ahigh speedl other respects, thesituation isthesametingstrip.einFig. 9.6. showstheregion in which



Bislocalized (it isdirected perpendicularlyhisregion is encircled by afixed metallicg contactsto theother sideof theringhemagneticflux tb into theclosed con-eter G(1-—-initial position, 2—f1nal posi·r showacurrent pulse?9.1), weshould concludethat thereBut it isnot so. Thereis no current sinceheLorentzforceareequal to zero: theand theclosed loop movesin the regionield.Thus, wehavenoneof the physicalf electromagneticinduction.eknowthat theforceacting

agneticheld isperpendicularomplishesno work. Meanwhile,nt-carryingconductor (movinges undoubtedly accomplish somet isthematter?

disappearsif wetakeintohecurrent-carrying conductorvitablyaccompanied by electro-cethee.m.f. induced in thethecharges,thetotal work of

eld(thework doneby Ampere’sm.f.) isequal to zero. Indeed,ement of a current loop in thercesperform thework (see

complishesduringthesametime

.11)that 5,= —-cZ<D/dt. It followst thetotal work is

of themagneticfield includesrk (dueto Ampére’sforces)ee.1n.f. induced during theorksareequal in magnitudeandheir sum isequal to zero.



sisdonenot at theexpensel magneticfield but at theexpense

maintainsthecurrent in theloop.rmsadditional work against

GAM >= —-2?,I dt = I dd),measthework GAAof Ampére’s

duringthedisplacement of gAmpére’sforces(whichuced in accordancewith Lenz’swork of theinduced e.m.f.:)

nergy,thisistheessenceof thenerators.

ppearsin all caseswhenthepchanges.It isnot importantonof thisflux. If in a certaingcurrent themagneticfield of .And thisleadsto thevariation

hthecontour,and henceto thef.nt inacircuit leadsto them.f. in thisgircyit. Thisphe-

ducti0n._"”'rromagneticsin thespaceslocated,field B and hencegh the contour areproportional

write

alledtheinductanceof thecircuit.tedsign rulefor thequantities

DandI alwayshavethesameductanceL isessentially a

n theshapeandsizeof`-theeticpropertiesof thesurrounding



ndthereare no ferromagneticsuctanceisaconstant quantity

nry (H). In accordancewithcorrespondsto theloopthe

ch isequal to 1 Wb at thecurrentH= 1 Wb/A.ceof a solenoid, neglecting edgeeffects.

oidbeV, thenumber of turnsper unitermeabilityof asubstanceinsidethe

I. Consequently, theproblem isre-of thetotal magneticflux CD, assumingr thecurrent I , themagneticfield in themagneticflux through aturn of thesole-whilethetotal magneticflux piercing N

.p°n2VI,ductanceof thesolenoid is

houldbenoted that thedeter-thehelp of formulaL = CD/Iulties.No matter howthin theional areaisfinite, andwesimplythebody of theconductor ageo-r calculatingLD. Theresultufficiently thin wirethisambigu-for thickwiresthesituation ise, theresult of calculation of Letothe indeterminacy in thentour. Thisshould bealwaysshow(seeSec. 9.5) that thereerminingL, which is not subject

gto (9.1), avariationeadsto theappearanceof the

onstant upon avariation of onof thecircuit doesnot change



tics),then7)

esthat $8 isalwaysdirectednthecurrent (in accordancem.f. tendsto keep thecurrenturrent when it increasesand.In self·induction phenomena,efore, theinduction effectsfluxconstant just in thesamerivestopreserve thevelocity

Eiiects.Typical manifesta-observed at themomentsof con-nelectriccircuit. Theincrease

hesthestead y·statevalueaftercreasein thecurrent whencur gradually and not instant-g arethemorepronounced theircuit.salargeinductance. If itsm thesource, thecurrent rapidlyesduring thisprocessa hugeeappearanceof aVoltaicarcwitchwhich isnot only verygnet winding but may even prove

with theresistanceof thethat of thewireisnormallylectromagnet winding. In thising decreasesslowly andis not

er detail themodeof vanishingrent in acircuit.euponthe disconnection of acircuit.tsof acoil of constant inductanceL,ane.m.f.source2 5 and aspecial key KKisin thelower position (Fig. 9.7b)in thecircuit (lweassumethat there-f. it isnegliiib y small). _lyturn ey K clockwisefrom theFig.9.7a). Thisleadsto thefollowing:



urcefor avery short timeand then dis-ithout breaking thelatter.

uctancecoil L startsto decrease, whichlf-inductione.m.f. 8, = —L dI/dt

Lenz’slaw,counteractsthedecreaseinof time,thecurrent in thecircuit will be= 5,/R,or

obtain

over I (between Ig and I) and over t/Is) = ——-RL/t, or

gthedimension of time:

t (or therelaxaton time). Thisquantityreaseinthecurrent: it followsfrom (9.19)ichthecurrent decreasesto (1/e) timesitsalueof ·r, theslower thedecreasein theecurveof the dependenceI (t) describingtime(curve1).current upon closureof acircuit.lyturn theswitch S counterclock-wer position (Fig. 9.7b). Wethuscon-

nductancecoil L. Thecurrent in thecir--inductione.m.f., counteracting thisIn accordancewith Ohm’slaw, RI ==



_

andsideof thisequation and introducedu == RdI.After that, wetransform

o

onstant.— if and RI - 25) andover t (be-8)/(—‘é)] = —-t/1, or

e.of thesteady—statecurrent (for t —» cc).herateof stabilization of thecurrentisstant 1. ThecurveI (t) characterizing theimeisshown in Fig. 9.8 (curve2).neticFlux.Let acurrentn an arbitrary external mag-rying).Thecurrent induced inby

t R= 0, dd)/dt must also bent I cannot beinfinitely large.onst.ngloopmoves in amagnetic

ugh itscontour remainsconstant.isensured by induced currentsaw, prevent any changein theontour.emagneticflux through

exhibited in t.heclearest formctors.ring of radiusrzand inductanceLd B.In theinitial position, theplaneof B, and thecurrent in the ring isequal

theposition perpendicular to vector B.n thefinal position and themagnetic

thering doesnot changeupon itsrota-ro.This meansthat themagneticfluxes



urrent andof theexternal current throughudeand oppositein sign.HenceLI =

with(6.13), createsafield BI = np0aB/2Lresultant magneticinduction at this

2L).

onsider two fixed loops1fficientlycloseto each other.t createsthrough loop 2oportional (in the absenceof nt 1,:

sin _

,2 Fig. 9.9duc-mutual inductanceisnu-eticflux through oneof thent in theother loop. Thecoef-

ontheshape, size, and mutualwell as on themagneticperme-nding theloops.Thesecoef-sameunitsastheinductanceL.ulationsshow(and experimentsof ferroinagnetics, thecoef-al:

fmutual inductanceisusuallym.Owing to thistheorem, we



tween L12 and L2,an d cannductanceof two circuits.25) isthat in any casethe

oop 1, createdby current Igneticflux (D, through loop 2,in loop 1. Thiscircumstance

ablysimplify the calculation,xes.Herearetwo examples.

ops1deliein a

oftheloopsinflux (D,em-f al < a,.eflux (D,atedproblemefieldver,thetheoremolution of thesthesameenthey thiscur-asily< agi it issuf-emagneticinduc-loop (B=thecircle

D, = (D, in accordancewith thereci-

u·rent I hastheshapeof arectangle.rough thehatched half-plane(Fig. 9.11)ndistancefrom thecontour.Assumethatarein thesameplane.cfieldof current I hasa complex con-rydifficult to directly calculatet ef lux (D

However, the solution canbeconsiderablyprocitytheorem.not around therectangular contourehalf—plane,enveloping it at infinity.y thiscurrent in theregion of therectan-guration-thisisthefield of a straightind themagneticflux (D' through theehelpof simpleintegration). In accordance

m,therequired flux (D(D' andthe

rromagneticschangesthe



theorem becomesinapplicable.lpof thefollowing concrete

eticcylinder of volumeVhastwo wind-e winding containsn, turns(per unitinsn, turns. Find their mutual in uctance

,/I,. Thismeansthat wemust createlcula-tethetotal magneticflux throughwinding 2 containsN, turns, weobtain

al areaof thecylinder. Considering thatnder length,and B, = p.,usn,I,, wherelityfor current I,, wewrite(D,Hence

2 in thelast two expressionsaregenerallyrentsI, and I2 in ferromagnetics), thecoincide.

enceof amagneticcouplingedin that any variation of leadsto theappearance of r circuit.Thisphenomenon

thee.m.f.’sappearing iny

}-L , (9.2b)ethat thecontoursare fixedticsinthe surroundings.nduct.ion,thecurrent. appearing,ge in currentsof both circuits)hm’slawthrough theformula



m.f. in contour 1 (besideseinductanceof contour 1.ritten for thecurrent I 2 in

formers,devicesintendeddvoltages, arebased on the

nductanceL which,aswasallyapositivequantity, mutualicquantity (which may,inThisisnce,indI1 referediatelyhesignflux (D2 for agivenepend on

mal to thesur-(or on thevedirection of cir-.urrents(and e.m.f.s) in bothenarbitrarily (and the positiveventionisuniquely (throughlatedto thedirection of theded by thecontour, i.e. in themagneticflux). Assoon asd,thequantity L,2 should befluxesof mutual inductionitivefor positivecurrents,n with self-induction fluxes.r positivecurrentsthecon-er. Otherwise,L1, < O.In specialsof circumvention of thecon-

orehandso that therequirede obtained (Fig. 9.12.).

t. Let usconnect afixed circuit-ndresistor Rto asourceof



w, thecurrent in thecircuitadsto theappearance of aself—·ancewith Ohm’slaw, RI ==

work doneby extraneousforcestimedt. For thispu rpose, webyI dt:

ning of each term and the rela-writeesso f current stabilization,d€D> O(if I > O), thework s out to begreater than the

rcuit.A part of thiswork edagainst theself—inducedafter thecurrent hasbeenentirework of thesourceof {E0onof theJouleheat.complishedby extraneousucede.m.f.in theprocessof current

ature.It is valid in theaswell,sincewhilederiving it

madeconcerningthe 1nagnet.icgs.assumethat thereareno ferroand

ion,weobtain Aaddwof conservation of energy,ement of any kind of energy.work doneby extraneousforces



___theinternal energy of conductorsrationof theJouleheat), whiletabilization) isspent onsome-"isjust themagneticfield, sincewiththeappearanceof the

usion that in the absenceof with theinduction coil L

2-E.l (9.29)gnetic,or intrinsic, energy of yconvertedintotheinternalisconnect thesourceof E0Fig. 9.7, i.e. if werapidly turnposition a.mula(9.29) expressesthentermsof inductanceand cur-

magnetics).In this case, however,nergy of charged bodies, can beof magneticinduction B.Letlecaseof along solenoid, ignoringdgeeffects).Substituting theo(9.29) weobtain

“I”V,.wehave

uniform fieldfilling thevolumeVdunder consideration).hat theenergy Wcan beex-inany case(but in theabsence

heformula

onhasthemeaning of theumeelement dV.



_ectricfieldenergy, wearrivegneticenergyisalso localizedagneticf:`1eld.30) and (9.31) that themag-spacewith thevolumedensity

xpression can beappliedtheBvs. II dependenceis

Bupoli isindependent of H.9.31) and (9.32) areapplicableiainngnetics.ln the caseof sionsareinapplicable*

sen- dyhis""ast

- I. Letormula

y round", F. 9 1.,,a(9.32) lg' ’ yof the

er themagneticfield of an arbitrary3). Let usimaginethat theentirefield isswhosegeneratricesaretheheld linestubea volumeelement dV= dl dS.the energy -E-BHdl dS is localized iu

Win the volumeof theentireelementa-eintegratethelatter expression alongdS through thetubecross-section isin thelong run expressions(9.3i)ncesof theformula6Aadd =¤ I d<i> and of of hysteresis,thework 6Aadd is spent onlynergy dW. For aferromagneticmedium,work 6A“ddisalso spent onincreasingdium,i.e. on itswheating.



hencedfb can betaken out of theintegral:

n circulationof vector H(in our case, pro-

yof all elementary tubes:

magneticflux enveloped by thecurrent

efrom theExpression fordtheinductanceL as thepro-hetotal magneticflux (Dandanother possibility of calculat-

or energy. Indeed, comparingweseethat iu· theabsenceof

way is freeof indeterminacyionof the magneticflux (D

7). Thediscrepancy appearingthrough formula(9.33) and

or theflux isillustrated inl cable.o Current Loopses.Let usconsider two fixedoneanother at asufficientlys amagneticcoupling betweenloop includesitsownsourceseeach loop at theinstanturrent will start being estab-cede.m.f. 2*,,and thee.n1.f. 5,ear.Theadditional work doneconstant e.m.f. against 55,

n above,for creating themag-



-etimedt:Fez`l" gis) I2 dt 3 dW·./dt, K,. = —-L., dl./dt,laasfollows:dl. LMI.dl..that L.2 = L2., werepresent

2.)

_.&4) F·1g.9.14xpression arecalled theintrinsicwhilethelast term, the mutual

ketheintrinsicenergiesof cur-algebraicquantity. Achange

ecurrentsleadsto thereversaln(9.34), viz.themutual energy.avetwo concentricloopswith currentsshown in Fig. 9.14. Themutual energyI2}1dependson threealgebraicquantitiesby t echoiceof positivedirectionsof cir-s. It isuseful to verify, _however, that the> O) dependsonly onthemutual orien-lvesand isindependent of thechoiceof cumventionof the loops. Werecall thatasconsidered in Sec.9.4.9.34). Therearesomemorebesolved by calculating thepsin adifferent way,viz, fromtionof energy in thefield.of current I.,and B2theaccordancewith theprincipleeachpoint isB B.ii- B2,fieldenergy of thissystem of ) dVSubstituting into this

B.B., weobtain



_—*`—KTìT*`_"—?———dV5-—!—-?>-dV.| 9.35( )nindividual termsin formulasdoubt.ead to thefollowing important

system of two (or more)itivequantity (W>0). ThisOC\ B2 dl',whoretheintegrandtanonadditivequantity (dueergy).isproportional to thepro-, OCI,and /}*2 OC/2. Thepro—emainingintegral) turnsoutt to indices1and 2 and hence, inaccordancewith formula

another definition of mutualomparison of (9.35) and (9.34)

agneticFieldr determiningtheforcesactingrgymethod,in which theexpres-nergy isused.tothecasewhen thesystemurrentsI1 and I2. Themagneticberepresented in theform.37)

otal magneticfluxespiercinghisexpression can beeasily4) by representing thelast termz A- (1x2) LMIQII andthen taking



ervation of energy, thework e.m.f.included into circuits1eat 6Qand is spent to_ increaseystem by dW(dueto the motionof currentsin them aswell asal work 6Am (asaresult of aof the loops):. (9.39)tanceof theloopsisnegligiblyenergy can beignored.restednot in theentirework ut only in its part which isdone-inducede.m.f.s(in each loop).dtheadditional n=ul.·) isequal) It di -—(@32 + ¢‘ I

dd)/dt fo: loop, wetheadditional wor theformk,. (9.40)

k of theeaonrces(thed self-inllnen nits) thatonof thelluxes ¢l>, and (D2

magneticenergy of thesystemthemechanical work:

dW—l- 6.1,,,. (9.41)for calculatingthemechanical

esacting in amagneticfield.for obtaining simpler expres-that either all themagneticfluxessflowing in them remain un-ent.Let usconsider thisin

03;, == const), it immediately



_

atesthat theincrement of them must be calculated at constanteobtainedformulaissimilarla(4.15) for thework in an elec-

nt (Ik = const), wehave

owsfrom formula(9.37) that

cewith (9.40), theincrementesystem isequal to half of thesourcesof e.m.f. Another half ng mechanical work. In othereconstant, dW|, = 6A,,,,

t thetwo expressions(9.42)finethemechanical work of rite9.44)ewith thelielp of thesef course,to choosearegimeinsor currentswouldnecessarilymply find theincrement dWstem provided that either (Dk =-apurely mathematical

pressions(9.42) and (9.43)areapplicableto systemscon-tours-one,two,or more.

mplesillustrating theapplica-

of one current loop.SupposethatreABisa movablejumper (Fig; 9.15).dependsin acertain way on t ecoor-



n. Find theAmpere`sforceacting on theconst.and lor ¢D= const.

ion,themagneticenergy of thesystemdancewith (9.29), asfol ows:

cetl1ejumper,for example, to therightwehave

out with thehelp of thetwo formulas44),thesameresult.eentwo current-carrying coils. Two

1 and I2aremounted on amagneticcoremutual inductanceof the coilsdependsingto a known lawL12 (x). Findtheforceils.system of two coilsisgiven by formulaorceof interaction, weshall useexpres-oil 2 through adistancedn: at constantspondingincrement of themagnetic

anical work 6Am = l·`2x dr, according to

etizeeach other. ThenL12 > O,and lorU, i.e. [*2,, -< U.Consequently, thelorcethe attractiveforce: vector l·`2is directed



reacting on solenoid winding.radiusof the solenoid crosssectionthrough thewinding constant. Then

mplishthework 6Am = d W|;. In the casee

sureand S isthelateral surfaceof the

hefact that when I = const,B =·= constxpressions, weobtain

ression for pressureobtainedizedfor thecasewhen themag-

ndB2) on different sidesof asur-n current or magnetization our-ticpressureisgiven by

45)eregion with ahigher magneticonof higher pressure.basicrelationsin magneto-sthebehaviour of electroconduct-ineeringand astrophysics).

n theshapeof aparabolay =-·k:*ld Bperpendicular to the planeXY.initial velocity andat aconstant acceler-arabola(Fig. 9.17). Find thee.m.f.ur asafunctionof the coordinatey.

= d<I>/dt.Having chosen thenormal nn thedirection of vector B, wewritedib =onsidering nowthat sr = Vy/k, we

constant acceleration, thevelocity dy/dt =·—·



Ky. The minussign indicatesthat 6;ckwiseily.A closed conducting loop ismovedrmation) inaconstant nonuniform mag--ay`slaw(9.1) will befulfilled in thiscase.

element dl of loop length, which atlocity v in themagneticfield B.In4) of field transformation, in the reference

ement theelectricfield E = [v X B] willedthat thisexpression can bealso ob-orentzforce,asit wasdoneat thebeginn-

ndtheentire contour is, bydefinition,

ndingincrement of themagneticfluxrpose, weturn to Fig. 9.18.Supposethat

m position P1 to I`, uring timedt. lf inicflux throughthesurfaceS1 stretchedresponding magneticflux in thesecondas(D1 —|—dd), i.e. asthe flux through the

herequired increment of themagneticp dS betweencontoursI`1 and P2.e—<S)[dr><B]dl. (2)mal n ismatched with thedirection of ur,viz.with vector dl (right—handed sys-ctor ds, viz.thearea element of thestrip,f normalsn, and (3) thefollowing cycliccalar tripleproduct:a>< bl = -{b >< a]·c,



t, wefind

r us to compare(3) with (1), which

largenumber Nof tightly wound turnseldperpendicular to theplaneof thecoil

diusof thecoil isequal to o. The magneticgtothelawB = B0 sin, cot. Find thef. inducedin thecoil.thecoilom acircle.theycosmt,urn under con-urnscorrespondingesof theradiusreconnectedinnduced in the

ssion, we obtain thefollowing for theede.m.f.:

the cross~sectional areaS isplacedcoil isrotated at a constant angular velocitygl with itsdiameter and perpendiculare magneticHeld in thesolenoid variesn cot.Find thee.m.f. induced intheecoil axiscoincided with theaxisof

etotal magneticflux through the

sin mt·coswt = (1/2)NB0S sin 2mt.y’slaw,wehave

NB5S·2o> cos2oit === ——NB0Scocos2mt.on. Showthat electronsin a betatronnstant radiusro provided that themagneticto half of thevalue(B) of themagneticnsidetheorbit, i.e. B0== ig)



ativisticequation for motion of an

l, (1)clield, in termsof theprojectionsonton to thetrajectory. For thispurpose,

ntum asp pt and find itstimederivative

(2)that p = mv,m being therelativisticscanbeeasily seen from Fig. 9.20).,)n, and therest is obvious.ay°slaw, 2m·0E |d<l>/dtl, where

kinginto account (2) and (3), in termsngent and thenormal to thetrajectory:

tten, after cancelling v, in theform

onwith respect to time, considering



_d (4) gives

H.

onwill besatisfied when

manufacturingpolepiecesin a specialnosed cones).

quarewireframewith sideaandh adirect current In liein thesameplanef t.heframeis L and itsresistanceisH.hrough18U° around theaxisOO' andnt of electricity that haspassed throughweentheaxisOO' and the straight wire

'slaw, thecurrent I appearing in theeterminedby t.heformula

of electricity(charge) is

—: ——-T?-(Aff) LAI).pped after rotation, thecurrent in itremainsfor usto find the increment.rame(MD(D2 —- (D,).o t.heplaneof theframe, for instance,n isdirectedbehind theplaneof thefigureilyseen that in thefinal position (D2 > 0,D, < 0 (thenormal isopelositeto B),plyequal to theflux throug thesurfacetial positionsof theframe:

whoseform can beeasily found withthelation.sign, weobtain



nedquantityisindependent of theinduct-ionwould bedifferent if the circuit were

m slideswithout friction along twolongy adistanceI (Fig. 9.22). 'l he systemusperpendicular to theplaneof thecircuit.

shunted through resistor H. At theinstanteinitial velocity vodirected to theright.per asa function of time, ignoring theof therailsas well asself—inductanceof

positivedirection of thenormal nay from us. Thismeansthat the positivef thecircuit (for induced e.1n.f. and cur-accordancewith theright—hand screwawthat

——-t"'.‘;——]}’ '

that when thejumper movesto theright,law,the induced current! causesthethemotion, directed to theleft.to theright, wewritetheequation of

theprojectionof theAmpéreforceontoegative,but weomit theminussign1), current I < 0).and(2), weobtainmi-?.ssion,takinginto account theinitial

*



processes.lin thecircuit shown inesource,its internal resistanceRand the

erconductingcoilsareknown.Find theilsafter key Khasbeen closed.ofi’slawsfor electriccircuitsESL,

essionsshowsthat L, dll = L2 dI2, whileve

d

ance.Acoaxial cableconsistsof adius aand external thin—wall conductinguctanceof aunit length of thecable,distributionover thecross section of them. Thepermeability isequal to unity

onsideration,theinner conductor isctanceshould bedetermined in termsof gh themagneticflux. In accordancewith

he cableaxis. In order to evaluatethispendenceB(r). Using thetheorem on

T •Br>h:0· (2)ncesisshown in Fig. 9.24.On account of wo parts.Asa result of integration, we

empt to determinethisquantity ineformulaLn = (Du/I leadsto adifferent



teadof 1/4,1/2 isobtained in thearen-

l wire, 1.C. the larger theratio bla,rencein theresultso f calculation by theseheenergy and through theflux.Along straight wireisarranged alongal coil of rectangular crosssection, whose

9..25. '1`henumber of turnson_ thecoilf thesurrounding medium isunity. lrindnduced in thiscoil if thecurrent I =traight wire.f'.8, —-dd)/dt,. where(D7 Nd),hrough thecross section of thecoil:

the help of thetheorem on circulationderivativeof (D,and multiplying the

d thefollowing expression for theamplitude

l inductance.Twosolenoidsof thellythesamecrosssection arecompletelyTheinductancesof solenoidsareL1and L2.dtheir mutual inductance(modulo).mutual inductance

icflux throughall theturnsof solenoid1sin solenoid 2. The flux (Di Z 1\i!?;·Sirnsin solenoid 1, S isthc<*F0$S‘S°"t'°"°



= up.°n2I2. Henceformula (1) can beasfo lows:2V. (2)that N1 = nil, wherel isthesolenoide.Expression (2) can berepresentedows:== I/Lile-pression defines thelimiting (maximum)| I.-,2] <A small cylindrical magnet M iscoil of radiusa, containing Nturnsctedto a ballisticgalvanometer. TheAfter themagnet had beenrapidly re-

geqpassed through thegalvanometer.f themagnet.moval of themagnet,the total mag-aschanging, which resulted in theemerg-nedby thefollowing equation:

of thisinto

ssion, weobtain Hq = —A€D—L AI.(thecurrent wasequal to zero at the

ndof theprocess), weobtain

rough the coil at thebeginning of theussign sinceit isimmaterial).dto thecalculation of theflux CDthrough

bed etermined directly. Thisdifficulty,y using thereciprocity theorem. Wemen-small current loop creating in the sur-gneticfield asthemagnet. I f theareaof isI , their product will givethemagneticm IS. According to thereciprocityeproblem isreduced to determining theeaS of theloop, which createsthesame

ecoil.Assuming thefield to beuniform



e(2) into (1) and recall that IS = pm.

ield

heory of electromagneticfieldthematicallycompleted by

S {'\\ S+,, ` n

&•_,·/ P

1

ield, it should beexpected thatd (GE/Gt) createsamagneticfield.istenceof anew in principle

n be approached, for example,oning. Weknowthat according

nof vector H,

thecasewhen ap reliminarilyitor isbeing discharged throughe(Fig.10.1a). For thecontour lè wire. Wecan stretch variousexample,S andS'. ~ThesetwoHowever,current I flowsthrough

rfaceS' thereisno current!onof vector Hdependsoniven contour (?l), which is



.er happened in thecase of

howtheright-hand sideof nience?Fortunately, thiscany."cuts"only the electriceGausstheorem, theflux of urfaceisDdS = q, whence

gto thecontinuity equation

ht-handsidesof Eqs. (10.2)ain

hecontinuityequation forthat in addition to thedensity jereis onemoreterm 0D/dtmeasfor current density. Max-

density of displacement current:

and displacement currentsisensity isgiven by

esof thetotal current are con-esof conduction current. If losed, displacement currents

ntroduction of theconcept of ifficulty associated wit.h thevector Hon thechoiceof theour I`.It turnsout that for this



n theright—l1and sideof Eq.dof conduction current, viz.

of (10.7) isthesum of conduc-ent current Idzlt = I + Id.rrent It will bethesame forver thesamecontour I`. Forula(10.4) to theclosed surfaceS' (Fig. 10.1b). Taking into ac-edsurfacethenormal n isdi-

l n' for surfaceS' to thesamet term in thisexpression willin

ationof vector H,which wass,can begeneralized for an ar-form:

)circulationof vector Hisal-medby theagreement of thisxperimentsin all cases without

8):

determinedby thedensity] of conductionflISpI8C0ll'l0IlI, Cll·l'I`0TlI. at tht} S3m0[IOIIIIQ.lacement Current. It should



.ement current isequivalent tom thepoint of viewof itsability

only when an electricfieldics,displacement current con-reutector D=hat theden-

ment cur-iesurrent,rizationantityof hing un-ct that polari-gnetico not differ in naturefrom con-rinciplestatement consistsinhe displacement current,ctedwith any motion of chargesonof theelectricfield, alsoen in avacuum, any temporalcitesin thesurrounding space

menon wasthemost essentialMaxwell whileconstructing the

This discovery wasasvaluablemagneticinduction, according toeldexcitesavortex electricfield.hediscoveryof displacementurelytheoretical discovery of

in which displacement eur·

velycharged sphereis in animiaitcmedium (Fig.fU.2).ElectriccurrentsBening

xciteamagneticfield. Let usfind therbitrary point P.tor B cannot havearadial compo-Bthrough thesurfaceS of thesphere



m zero, whichisin contradiction withctor Bmust be perpendicular to theradialhisis also impossible, sinceall directionsrectionareabsolutely equivalent. Ithat magnetictield isequal to zero every-

ldin thepresenceof electriccurrentadditionto conduction current j, displace-nt intl1esystem. Thiscurrent issuch thateroeverywhere, i.e. jd == -— j at each

.that I) = q/4:1:-2 in accordancewith the

Integral Form. Theintroductionliantlycompleted themacro-neticheld. The discovery of dis-allowed Maxwell to create amagneticphenomena. Maxwell’s

l individual phenomenaofelec-asinglepoint of view,but alsophenomenawhoseexistencewas

aratepartsof thistheory.ntiretheory in theform of a sys-nsinelectrodynamics, called

ationarymedia. In all, therearewearealreadyacquainted withratelyfrom previoussections;ogether).In theintegral form,ationsiswritten asfollows:

(10.10)

dS=O,(10.11).._.....yof extraneouschargesand j

current.



aconciseform all our know-cfield.Thecontent of theseequa-ng.

ound any closed contour isthetimederivativeof themagnet-urfacebounded by thegiven

vortex electricfield butalso elec-ationisknown to beequal to

closed surfaceisequal toouschargesembraced by this

oundan arbitrary closed con-ent (conduction current plusgh anarbitrary surface bounded

bitrary closedsurfaceis

quationsfor circulationsof vec-dmagneticfieldscannot bengein timein oneof these fields

heother. Henceit isonly acombi-bingauniqueelectromagnetic

=—.const and Bconst),otwo groupsof independent

magneticfieldsareindepen-llowedusto study first a con-independently, constant mag-

t thelineof reasoning whichationscanby no meansbecon-uationscannot be"derived"sinceoms, or postulatesof electro-helpof generalization of exper-



_latesplay in electrodynamicsawsin classical mechanicsor the

ell’sEquations.Equationspresented indifferential form,f differential equations, viz.3)

10.14)theremay betwocausesforurcesareelectriccharges, bothh followsfrom the equation

ount that D== e0E + P and). Second,thefield Ealwayseldva1·iesin time(which ex-ctromagneticinduction).

14) indicatethat themagnet-movingelectriccharges(elec-ectricfields, or byboth factorssfrom theequation V>< Hccount that HB/n0—Jand+j' —~l— OP/Ot r-l- e00E/Ot,ncurrent density and 6P/dt isity.Thefirst threecurrentsareharges,whilethelast currenteldE).Asfollowsfrom theequa-urcesof magneticfield in na-gneticcharges) that would be

sequationsin differentialt they expressthebasic lawsalso in that thefieldsE and Bsolving (integrating) theseequu-

of motion of charged particlesntzforce,(10.15)



.fundamental system of equa-m issufficient for describing allain which quantum effectsare

well’sequationsin integralnaturethan their differentialin validin thepresenceof athesurfaceon which theproper-changeabruptly.On theothern thedifferential form presumeinuously in spaceand time.rm of Maxwell’sequationsneralityby supplementing themswhich must beobserved for annterfacebetween two media.ed in theintegral form of

alreadyfamiliar to us:2ntH1t :—-HZT

tionsrefer to thecaseswhenthereesnor conduction currentson thenoted that theseboundary con-nstant and for varying fields.l’sfundamental equationssystem of equationsfor theelec-ationsareinsufficient for de-en distributionsof chargesand

be supplemented with rela-e quantitiescharacterizing in-dium.These relationsarecalledly, theseequationsarerather com-al and fundamental asMax-

implest form for sufficientlyry in timeandspaceat acom-

ase,for isotropicmediaandferromagnetics, thematerialg (alreadyfamiliar to us) form:(E —\— E*), (i0.i7'



ll-knownconstantscharacteriz;cpropertiesof themediumand electroconductivity), andaneousforce field dueto chemi-

sEquationsnear. They contain only thefirstwith respect to timeand spa-t powersof densitiesp and j

nts. Thelinearity of Maxwell’stedwith theprincipleof super-tisfy Maxwell’sequations,lsosatisfy theseequations.ethe Continuity Equation. Thisationof electriccharge. Inet ustakean infinitely smallarbitraryfinitesurfaceSt thiscontour toa point, soIn thelimit,<$H dlvanishes,d thefirst of Eqs. (10.11) be-

quation(5.4). 'l`hiS Gq\¤8ti0¤ ingfrom avolumeVthrou8h 8



.hedecreasein chargein this

tinuityequation) can also befferential equations. For this

kethedivergenceof both sidesndmakeuseof thesecond of Eqs.dp/Ot.

isfiedin All Inertia! Systemsof sticinvariants.Thisisaconse-ciple, accordingto which allareequivalent from thephysi-anceof Maxwell’sequations

nsformations) isconfirmed by.Theform of Maxwell’sequa-

on atransitionfrom oneinertialbut the quantitiescontained inordancewith certain rules. Thend Bwasconsidered in Sec. 8.

Newton’sequationsin mechan-correct relativisticequations.ell’sEquations. Maxwell’screlativeto electricandmagnet-thefact that thereare electric

asit isknown at present, mag-owever,for aneutral homogen-m, wherep= 0 and j = 0,easymmetricform, sinceEmeway asB to 6E/Ot:

0.18)elativetoelectricand magnet-lythesign of thederivativesencein signsof thesederivativesvortex electricfield induced

m a left—handscrewsystem withsof the magneticfield inducedght-hand sc1·ewsystem with

Maxwell’sequationslead toclusion about theexistenceof a



'enomenon: electromagneticfieldthout electricchargesand cur-thiscasenecessarily hasa

calledelectromagneticwaves.wayspropagateat a velocityc.cement current (GD/0t)primary role.It isitspresence,

Ot, that makes theappearanceossible.Any timevariation of aectricfield,whileavariation of nducesamagneticfield. Thisor interactionof the fieldspre-

magneticperturbation propa-

redicted theexistenceof elec-madeit possibleto establishy electromagneticwave, regard-anbeaharmonicwaveor annof any form) ischaracterized

gationin anonconductingedium isequal to0.19)evelocity) arematerially perhandscrewsystem (Fig. 10.5)anintrinsicproperty of anpentlent of thechoiceof coordi-



.ectromagneticwavealwaysos- showing a"ph0tograph"of aluesE and Bat any point be-lationE = vB,or

B) simultaneously attain theirc.n the development of theelec-wasdueto hisunderstanding of

f electromagneticwaves, fol-ations(10.18).

ux. Poynting'sVectordingfrom theidea of energyand on thebasis of theprin-

n,weshould concludethat if nregion,thismay onlyoccurugh theboundariesof theregiondium isassumed to be sta-

mal analogy with thelawpressed byEq.(5.4).'l`hemean—at adecreasein chargeper unitual totheflux of vector j throughlume.lawof conservation of energyditionto theenergy density wsacertainvector S characteri-

y.gyof an electromagneticfield,lumewill changedueto itsflow-ll asdueto thefact that thesubstance(charged particles),thesubstance.In macroscopicwrittenasfollows:

nt.



. ’ynting’stheorem: adecreaseiven volumeisequal to theenergynding thisvolumeplusthework

which isaccomplished by thefieldtanceinsidethegiven volume.V, wherewis thehelddV, wherej isthecurrentheldstrength.Theaboveex-edasfollows. During thetimeA= qE·u dtoverapoint chargeocity.Hencethepower of forceer tothevolumedistribution of V, wherep isthe volumechargej·E dVIt remainsfor usmeunder consideration.r P in (10.21) can beeitherter takesplacewhen positiveveagainst held E,or negativeedirection.For example, suchpointsof amedium wherein

E,theheld E"‘ of extraneoushpoints, j = o (E + E*),magnitude,theproductrnsout to be negative.ssionsfor theenergy densityxwe1l’sequations(weshall not).If amedium containsno ferro-(i.e. in theabsenceof hysteresis),tromagneticheld isgiven by

idual termsof thisexpres-searlier [see (4.10) and (9.32)].gneticenergy flux,which ishned as



.’sequationscannot giveun-hetwo quantitieswand S.bovearethe simplest from an in-uations. Therefore, weshouldtulateswhosevalidity should

ent with experimentsof their

will demonstratetha t althoughhelpof formulas(10.22)seem strange,wewill not begincredible,any disagreementjust the evidencethat these

nelectromagneticwave(in vacuum).assing during timedt through aunit area

onof propagation of thewave.nd Bin theregion of location of the

y,w == e°E2/2 —i— MHZ/2. In accordancemagneticwavewehave

nergy density in theelectromagnetico the magneticenergy density at theor the energydensity

t.ll obtain withthehelp of PoyntingWcan berepresented in termsof themag-s:cE2 dt.

w and for S) lead to thesame result

t ina conductor. Let acurrent I [lowireof radiusa (Fig. 10.7). Sincethewirectricfield E is acting along it. Thesameesurfacein avacuum. Besides, thepres-

magneticfield.According tothetheoremear the surfaceol thewirewehaversE and Harearranged sothat the



. "nsidethewire normally to itslateral sur-ly, theelectromagneticener y flowsinto

ngspace! But doesit agreewith theamountuctor?Let us.magneticenergy If asection of

RI2,. Berenceacross""`°- §"`Risitsresist- Hconclusion thatlowsintothe·

mpletely con-must agreeedconclu·_

owergainst current If thesource................—.. vt,edoutside:ticenergy flows··········—*·—·—·ipaIn other words, Irom thesourceF. 10 8wiresbut lg' °ngaconductorctromagneticenergy, viz.theflux of vec-

wsasectionof abalanced (twin) line.thewiresare knownaswell asthefactp, < q>2. Can .

or to theleft) theE

anbeob- cpoynting vector. ,_tion,between theHownwardswhile‘helaneof the

E >< I-Il)is direc€.··dourceison theFig.10.9right.pacitor.Let us takeaparallelsplatesof radiusa. Ignoring edgeeffects(fieldomagneticenergy flux through thelateralnceonly inthisregion the Poynting vectortor (Fig.10.9).varying electricfield E and amagnetiction.According to thetheorem on cir-

wsthat 2ncH= rm? OD/Bt,wherethe



.hedisplacement current through thecontourshed line.HenceH== (1/2) a 0D/(it. llatesis h, theflux of vector S through the

lg?V,(1)umeof thecapacitor. Weshall assumethatfor increasing thecapacitor’senergy.

y dt, weobtain theincrement of the capa-medt:Q-V)ion,wellnd theformulafor the energy W, everythingi-sall right in this casetoo.MomentumWave.Maxwell showed theo-icwaves, being reflected or ab-

hey are incident, exert pressurereappearsasaresult of the actionveon theelectriccurrentsof thesamewave.cwavepropagatesin ahomo-absorption. Thepresenceof ab-

heat with thevolumedensitymedium.Henceoak 0,i.e.nductive.eexcitesin thismedium an elec-oE. As aresult, Ampere’swill act onaunit volume

directedtowardsthepropaga-and causesthepressureof the

, theconductivity o = 0theelectromagneticwavedoesmedium.

mentum.Sincean electromag-on asubstance, thelatter ac-

However,if in aclosed systemd an electromagneticwaveonlyomentum, thelawof conserva-eviolated.



tem can beconserved onlygneticfield (wave) also possessesacquiresthemomentum dueeelectromagneticheld.

t of momentum density Gof quantity numerically equal tonaunit volume. Calculationshowthat the momentum density

ing’svector.Just asvectorenerallyafunction of time and

for an electromagneticwave= I/u0H. Hencetheenergy den-f Poynting’svector arere-

S= EH: l/eo/po E2.. And sinceVenn, = 1/c,ght in vacuum, S = wc. Takingain that for an electromagnetic

nergy and momentum isinhe-ry of relativity) in particles

quitenatural since, according to



jns, anelectromagneticwaveisotons, viz.theparticleswith zero

Pressureof Electromagneticng formula(10.25), thepres-agneticwaveon abody when thessurfaceand is partiallyreflect-In accordancewith thelawof po := pg --l— p, wherepo, pg areand reflectedwaves, whilepdto thebody (Fig. 10.11). Hav-ntothedirection of theinci-hequantitiesto unit timeandeobtain

veragevaluesof themomentumreflectedwaves. It remainsforon(10.25) between (G) and

w') : p(w),wherep is theult, thepreviousexpression be-

e pressureof theelectromagnet-ase of total reflection, philefor total absorption,

ressureof electromagneticw(theexception isthepressurenbeams, especially after beamof radiation insidehot stars).solar radiation on the Earthhich issmaller than the atmo-of 10‘°. In spite of insignificantxperimental proof of thewaves-, viz. thepressureof

dev.Theresultsof theseexperi-h theelectromagneticthcory of



nt. Apoint chargeq movesuniformlyonrelativisticvelocity _v. Find thevectordensity at apoint P lying at adistancerht line(1) coinciding with itstrajectory;toryand passing through thecharge.

urrent density jd OD/0t.Hence,sreducedto determining vector Datndingitstimederivative. In both cases,unit vector of r.Let usIind thederivative

whereit isassumed that q > 0), we

that for Eoint P, thederivativeOr/Ot = —v.of thecargc(in the direction of itsmotion)dbedirectedin thesame way and would

v, and viceversa.| dD|/D= v dt/r, and hencewecan

sa.a long straight solenoid with the

varied so that themagneticfield insidet_heaccording to thelawB = Biz, whereBcement current density as:1 function of noidaxis.



:.edisplacement current density, we0.5), first find theelectricheld strength). Using Maxwell’sequation for circu a-

< R);R“Bt/r (r > R).So OE/dt,we can find thedisplacement

/r (r > R).id(r) isshown in Fig. 10.13.acitor isformedby two discsthespaceahomogeneous,poorly conducting medium.nd then disconnected from a power source.hat themagneticfield insidethecapacitor

be absent sincethetotal current (con-cement current) isequal to zero. Let uscurrent density. Supgosethat at a certainuction current isj.Oviously, j Q Dandechargedensity on thepositively charged

g.10.14).ncurrent eadsto a decreasein thesurfaceinDaswell. Thismeansthat conductionbythedisplacement current whoseden-

—·jn == —-—j.d

heplatesof aparallel—platecapacitors filled with ahomogeneouspoor y con-uctivity o andpermittivity e. Ignoringtudeof vector Hbetween theplatesattheelectricfield strength between the

cordancewith thelawE ==

quationfor circulation of vector H,

awin == oE‘,, (t), weobtain—{-gi'}-(o cos 0.¤t·—-8800.) sin wt).



ionintheparenthesesto cosine. For thisvidethisexpression byj == \/0* —{- (ee,,co)”rough theformulas0/f = cos6, cc,,w// =

0c0)2cos(cot—|—6).vesin a vacuum uniformly andvisticvelocity v. Using MaxweIl’s

o*f vector H,obtain theexpression for Hrelativeto the chargeischaracterized by

ymmetryconsiderationsthat for a con-lationof vector Hshould betaken, wentre0(itstraceis shown in Fig.10.16

circle.ctor Dthrough asurfacebounded by thiscity,weshall takea spherical surfacer (Fig. 10.16). Then theflux of D throughherical surfaceis:isin¤'da’,theselected surfaceis

1), wedifferentiate(2) with respect to

·,



hai-gefrom point 1 to point 2 (Fig. 10.17)vev dt·sin cr.= r doc,whence

then (3) into (1), weobtain

that R= r sin cz. Relation (5) in vectorws:

(6.3) which wehavepostulated earliereqmitions.onof an inertial system of referencemagnitudeB= const,rotating at an

XE in thisregion asa function of vectors

equation VXE = —-6B/Ot thatpositely to vector dB. Themagnitudeof

dwith thehelp of Fig. 10.18:w.

rotonshaving thesamevelocity vosssection with current I.Find the direc-ting’svector S outsidethebeam at a



that S it v. Let usfind themag-and Hd epend on r. According to the

t length of thebeam. Besides, it followstionof vector Hthat

from thelast two equationsand takingobtain

rough thewinding of along straight

· . ·

Showthat therateof increasein theenergyolenoid isequal to theflux of Poynting’sace.eases,themagneticfield in thesolenoidortex electricfield appears. Supposethatosssectionis equal to a. Then thestrengthar thelateral surfaceof thesolenoidelp of Maxwell’s equation that expressesnduction:

ateral surfaceof thesolenoid canbere-

hand mzzl isitsvolume.flux through thelateral surfaceof theS) isequal to therateof variation of theolenoid:



.ourceof constant voltageUistrans-ongcoaxial cablewith anegligibly smallecableisI. Find theenergy flux throughming that theouter conducting shell of

er-mula

flux density,of alueof S isconstant,and aand b are the

d of theouter shell of thecable(Fig. 10.20).egral,wemust knowthedependence

eobtain

nit length of thewire. Further, by thetheo-ve

om formulas(2) and (3) into expression

not givenin theproblem. Instead,elationbetween thesequantities:

gives

ueof power liberated in theload.ir capacitor whoseplatesaremadesaare connected to asourceof varyingcyco.Find theratio of themaximum valuesergyinsidethecapacitor.rossthecapacitor vary in accordancet and thedistancebetween thecapacitor



energy of thecapacitor isequal tocos' cot. (1)determinedthrough theformula

valuatingthisintegral can befoundulationof vector H: 2m-H= m-’ OD/6:.B/po andOD/0: = ·—e•(Um/h) oa sin cot,

3)te(3) into (2),wherefor dV wemustn theform of aring for which dV=egration,weobtain

luesof magneticenergy (4) and elec-

ndco = 1000 s·‘, thisratio isequal to

aloryCircuitWhenelectricoscillationsoccur,

with time and,generally speak-at each instant of timein diff-dueto the fact that electromag-ealthoughat avery high but

re,however, many caseswhenent provetobepractically thercuit (such acurrent is callede,all timevariationsshould occuron of electromagneticperturba-



stantaneous. If l is thelength of r an electromagneticpertur-l isof theorder of 1 = l/c.

urrent,quasi—steady condition

ations.length l = 3 m, thetimecanbeassumed to bequasista-

of 10** Hz(which correspondsto

meeverywherethat in theuasi-steadyconditionsareob-istationary. Thiswill allowus toaticfields.In particular, weshallaneousvaluesof quasista-

slaw.uit including acoil of indu-capacitanceC,electricoscilla-ason,such acircuit iscalled and out howelectricoscillationsan oscillatorycircuit.pper plateof thecapacitorelower plate, negatively (Fig.eenergyof theoscillatory cir-pacitor.Let usclosekey K.arge,and acurrent flowsthroughfthe capacitor isconverted in-e_coil.Thisprocessterminatesrgedcompletely,whilecurrent



mum value(Fig. 11.1b).hecurrent beginsto decrease,not,however, ceaseimmediate-yself-induced e.m.f. Thecur-and theappearing electricfieldnt.Finally,thecurrent ceases,acitor attainsitsmaximum value.citor startsto dischargeagain,sitedirection,and theprocess

gthe oscillatory circuit haveicoscillationswill beobservedf theprocess, thechargeon theacrossthecapacitor and thevary periodically. Theoscil-mutual conversion of theenergyds.of conductorsRsk O, then, inbed above, electromagneticto Joule’sheat.ircuit.Let usderivethe equa-n acircuit containing series-ction coil L,resistor R,andg. 11.2).directionof circumvention,

q thechargeon thecapacitorchto theother platecoincidescircumvention,Then current in

dq > Oaswell, and viceversathat of dq).w for section IRL2 of the

M2)ede.m.f.In thecaseunder con-

—— cp,4: q/C



with thesign of thepotentialC>0). HenceEq.(11.2) can be

),

cillatory circuit, which isali-geneousdiff erential equation withthisequation for calculatingevoltageacrossthecapacitor

Cand current I byformula(11.1).rycircuit can begiven adiffer-

nisintroduced:)enatural frequency of thecircuit.Themeaning of thesequantities

sually called freeoscilla-edfor R = 0 and damped fornsecutively all thesecases.on:s.If acircuit containsnoexter-nceR= 0, the oscillationsand undamped.Theequationsisaparticular caseof Eq.(11.5)



nisthefunction)

valueof thechargeon capacitorency of the oscillatory cir-.Thevalueof coo isdeterminedcircuit ·itself, whilethevaluesinitial conditions. For thesecon-mple,thevalueof chargeq

ment t = O.I/LC; hencetheperiod of sgiven by is

ifferentiating (11.8) within mind that the voltageacrosssewith chargeq, wecan easily

scillationscurrent I leadsincapacitor platesby xt/2.ems, energy approach can al-

rcuit,freeundam d oscillationspacitor platesareslowly moved aartllationsdecreasesby afactor of r]. Whichisdonein thiscase?presentedasan increment of theenergy

/E,and hence1] = coé/co•= VC/C',

Everyreal oscillatory circuitrgystored in thecircuit isgrad-oscillationswill bedamped.for a givenoscillatory circuithisgives



ot do it heresinceweareinter-eproblem) that for B2 < mj,neousdiHerential equation has

constantsdetermined from thethefunction (11.11) isshown

at thisis not aperiodicdampedoscillations.= 2:1/coiscalled thepe-

)"( )eundamped oscillations.) is calledtheamplitudeof ndenceon timeis shownin Fig.

nd Current in an Oscillatoryn find thevoltageacrossaca-ircuit.Thevoltageacrossa

c).(11.14)

—l—c1.)—oJsin(wt—!—<1)].n in thebracketsto cosine. Fornd dividethisexpression byntroduceangle6 by theformu-



sin 6. (11.15)rI takesthefollowing form:+ 6).(11.16)ngle6 liesin thesecond quad-ansthat in thecaseof anon-n the circuit lem voltageinphaseby morethan rz/2. It shoulddvanceis6 = nz/2.esUC(t) and I (t) havetheform11.3 for q (t).cuit containsacapacitor of capacitanceRand inductanceL. Find theratio be-gneticand electricfieldsin thecircuit atnt isat amaximum.3) for an oscillatory circuit,

ximum,dI/dt = 0, andRI = -—q/C.

amping.axation time1, viz. thetimeof oscillationsdecreasesby aeen from formula(11.11) that

. of damping.It isdefined as theccessivevaluesof amplitudes

scillations:

hecorresponding quantityrm, wecan write



cillationsduring thetime1:,amplitudeof oscillationsto

value. Thisexpression can beas(11.17) and(11.18). __§), then to z mo = 1/VLC18), wehave20)circuit.By definition,

ecrement.Thesmaller thedamp-For aweak damping (B2 <11.20),wehave

mulafor theQ·factor in thecaseof

din thecircuit and 6Wisthede-heperiod T of oscillations. In-rtional to thesquareof theam-or charge, i.e. WOC_e"’B’.in theenergyover theperiod isnsfor us to takeinto account.21),2. = st/Q.

ould be noted that whenargeof the capacitor will occuresistanceof thecircuit for whichn iscalledthecritical resistance:

es.rcuit hasa capacitanceC, inductanceL,mber of oscillationsafter which theesto 1/eof itsinitial value.d€Cl‘E&S€S to {/B of HS lHll»i8l



. :B. During thistime, Ngoscillationswillampedoscillations, then‘·

and B= R/2L,we obtain

ring which theamplitudeof currentagivenQ-factor wil decreaseto1;.11

quency of damped oscillationsisequ

1,,, 0( e'B’» the timein during whichfactor of 1] isdetermined by theequation

ctor isalso related to B:

wo equations,weobtain

ationst us returnto equations(11.3)circuit andconsider thecasearying external e.m.f. Z3whose

easedby theharmoniclaw:

placeowingto thepropertiesof o retain aharmonicform of os-external harmonice.m.f.

an oscillatory circuit is

cet, (11.26)

mt. (11.27)



_ics,thesolution of thisequa-l solution of thehomogeneoust-hand side) and a particulareouseq uation.i11 steady-stateoscillations,of thisequation (thegeneralsequationexponentially attenu-certaintimeit virtually van-

hat thissolution hastheform1.28)f chargeon thecapacitor and ipcillationsof thechargeand of ._It will beshown that qmropertiesof thecircuit itself andout that ip > 0, and henceqe.

nstantsqm and ip, wemust sub-l equation (11.27) and trans-r thesake of simplicity, weshallrst find current I and then sub-1.26). By theway,weshall sol-ngtheconstantsqm and ip.threspect to time t givesonqm cos(cot -——q> —|— srt/2).asfollows:29)litudeand qaisthephaseshifternal e.m.f.E,. (11.30).For thispurpose, weproceed asinitial equation (11.26) in the

osmt, (11.31)hesum of voltagesacrossinduc-resistor R. Thus, weseethatum of thesevoltagesisequal



g into account relation (11.30),

1.32)p):—é%-cos(mt—(p——%) , (11.33)

t-—q>)4)eeformulasshowthat U RClags behind I by rt/2, and UL

1t (

sually represented with thepictingtheamplitudesof vol-

according to (11.31), isequal

eof thisdiagram, wecanexpressionsfor 1,,, and cp:

ector diagram obtained above



t for solving many specific, easy,and r apid analysisof

e namegiven to theplotsof ng quantitieson thefrequency mI, chargeq on a capacitor, and

definedby formulas(11.32)-

t 1,,, (co) areshown in Fig.wsthat the current amplitude

mL -—— 1/ mC = O.Consequently,current coincideswith thena-atorycircuit:1.37)isthehigher and thesharpertor B= R/2L.argeq,,, (oi) on acapacitor arececurvesfor voltageUC,esameshape). Themaximum of at theresonancefrequency38)ser tomo with decreasing B.emust represent q,,,,in accord-m q,,, == Im/to whereIm is

( )



:onor, whichisthesame, then befound by equating to zerodwith respect to co.Thisgives.38).litudesof voltagesURu UC,pendingonthe frequency totternis depicted in Fig. 11.7.r UR, UCand UL aredeter-ulas:

(11-40)·thecloser aretheresonancefre-o thevaluecoo.factor.Theshapeo f resonanceainway withthe Q-factor of annectionhasthesimplest formg,i.e.for B2 < m§.In this case,

co§, thequantity (oreszcood (11.35),Ucm ,.,,3 = 1,,,/co,_,C=m = VLC/CR= (1/R) I/L/Cformula( 11.22), just theQ-

it (for B2 < wg) showshowalueof thevoltageamplitudeductioncoil) exceedstheampli-

so connectedwith anothere resonancecurve, viz. its2 < cog

equency and owisthe width of ght" equal to 0.7 of thepeak

ecaseunder consideration is



llationsat thefrequency of ex-sfrequencyisequal to thena-orycircuit.Resonanceisusedomponent from acompositevolt-ontechniqueisbased on reso-tha given radio receiver thehereceiver must be tuned. In

nd L of theoscillatory circuit,ncebetween itsnatural frequencyavesemitted by theradio

ceis also associated with acer-.m.f.or voltagemay besmall, butual elementsof the circuit (the

may attain thevaluesdangerousysberemembered.

e).Steady—stateforced elect.ricanalternatingcurrent llow-

acitance,inductance, and resist-xternal voltage(which plays

esaccordingto thelaw4)

terminingthecurrent ampli-ecurrent relativeto U.thecurrent amplitudeIm (oa)asOhm`slawfor theamplitudee. The quantity in thedenomi-ch hasthedimension of resist-

calledthetotal resistance, or



11.46)wo ·= 1/I/I-Z', theimpedanceisequal to theresistanceR.eparenthesesin formulas calledthereactance:

edtheinductivereactance,alledthecapacitivereactance.

XCrespectively.Thus,——X,

ductivereactancegrowswithcapacitivereactancedecreasess said that acircuit hasno ca-erstood so that thereisno capac-al to 1/wCand hencevanishesr isreplacedby ashort—cir-

nceismeasuredlin thesameifierin principle. Thisdifference

stancedeterminesirreversible, for example, theconversionntoJoule’sheat.Circuit.Theinstantaneouse product of instantaneoust:wt —— cp). (11.49)—cp) = cosnot coscp -l—rm (11.49) asfollows:s cp-}- sin mt coswt sin cp).

hevalueof power averagedConsidering that (coszmt) -·=

weobtain.50)



enin a different form if welowsfrom thevector diagram

RI",.Hence,

thedirect current I == Im/L/-2.

an—square) valuesof current andoltmetersaregraduated foroltage.ge power (11.50) in termsof oltagehas theform

ally called thepower factor.n ana.c.circuit dependsnotnt but also on thephaseangle

P} == Oregardlessof the val-he energy transmitted over aenerator to an external circuit

transmittedfrom theexternalng thenext quarter of theperiodillates"uselessly between thercuit.r on cosipshould be taken intosmission linesfor alternatingaveahigh reactanceX,cos cpnone. In such cases, in orderer to aconsumer (for agivensnecessary to increasecurrentof uselessenergy lossesin feed-ctancesand capacitancesshouldmakecoscp ascloseto one

, it issufficient to makereactanceensure theequality of in-ances(XL XC).g an a.c.circuit arein motion,



ancebecom.eswider sincein ad-ectricenergy into Joule’sansformationsarealso possi-ectricenergycan be convertedricmotors).

llations.Freeundamped oscillationst consisting of acapacitor with capacitanceinductanceL. Thevoltageamplitudethee.m.f. of self—induction in thecoil at

eticenergyis equal to theelectricenergy

m`slaw,

sthe capacitor (U== cpl —- tcp,). Inour—U.

ageU at themomentswhentheelectricual to themagneticenergy of thecoil.nwriteCU?

Umf \/2.t consistsof an induction coil withedcapacitor of capacitanceC. The resist-ecoil isin apermanent magnetictield.cingall theturnsof thecoil is(D. At theeld was abruptly switched ofi. Findfunction of timet.tching off of theexternal magneticinduced current appears, but the capac-In accordancewith 0hm’slaw, wehave

hence-§— Ll = 0. '1`hisgivesC D= lgo,nt (immediately after switching off ne

een switchedoff, theprocess isde-ation:



onwith respect to timegives

onicoscillations. Weseek its solution

be found from theinitial conditions

wsfrom Eq. (1), sinceat the initial momentarged).From theseconditionswefindt,we obtainscoat,

An oscillatory circuit with alowCand inductanceL.ln order to sustaincillationswith thevoltageamplitudeUmessarytosupply theaveragepower (P).

uit.ow, wecanmake useof formula(11.23):

= (P) T, T bei_n_g_theperiod of dampedT0 = 2:: }/LC. Having substitutedeobtain

.An oscillatory circuit includesaninduction coil of inductanceL, aresistorhecapacitor wascharged with thekeysed, acurrent beganto flow. Find theratiocitor at timet to thevoltageat theinitiallosing thekey).sthecapacitor dependson timein the. Hencewecan write), theVol U(0) ’=j Vm·¢>¤¤ ¤» ‘Vh¤¥•smomentfqlvemust End U` (0)/ Um, i.e.other initial condition: at the0. Sinceq = CU,it is sufficient to



t totimeand equat.etheobtained expres-ain—-Bcoson —— to sin ot =¤ O,whence

edratio is_..—————=-—;—————— (2)

) areshown in Fig.11.8.B2, wetransform (2) asfollows:

V ‘1-—/f2C/4L.that B= R/2L and cog = 1/LC.cuit with acapacitanceCand induc-

occur,in which thecurrent varies withlawI (t) = Ime·B¢ sin wt. Find thevolt-function of time.lockwisedirection asthepositive

Fig.11.9). According to Ohm’slawforwehaveRI = cp, — q>2+ E5,.In ourl = q/C= UC,whereq isthechargeonnbe written as

formulatheexpression for I (t) and

—o>coson).ionintheparentgzsesto sine. For

nd divideit by )/mz—i— {12 = on, and in-ormulassin 6.(1)



m }/L/Ce"°' sin (mt·—6),ncewith (1),in thesecond quadrant, i.e.rm. Thus, thevoltageacrossthecapacitor

ase.tions. An induction coil of inductanceectedat themoment t = 0to asourceof tot. Findthecurrent in thecircuit as

—— LI,ort.nisthegeneral solution of thehomogeneousolution of thenonhomogeneousequation:

cs cot- .,

stant and anglecp isdefined by condition

m theinitial condition I (0) = 0.¢o°2L2) costp. Thisgives

—·e(R/Ll! cosqa].second term in thebracketsbecomes

ainthesteady-statesolution I (t) Q

sectionof acircuit consisting of and resistor R,is connected to a sourcemplitudeUm.Theamplitudeof theut to be I m. Find thephaseanglebetweenage.onsideration,wt — ap),la(11.36): tantp = -1/¢oCR.citanceCcan befound from theexpres-de: Im = Um! VR2 + (1/wC)’,whence

ession into formulafor tan tp, weobtain*i·



eansthat the;_current leadstheexternal

ningaseries-connected capacitorainresistance, is connected to thesource

.12gewhosefrequency can bealtered withoutequenciesml and cozthecurrent ampli-obethesame. Find thcresonancefrequency

5), the amplitudesareequal under the

l·"’ncecurvefor current corresgndsto theal frequency coo = 1/ \/LC. Further.siteinequalityisalso possible, it willquality (1) can then bewritten in amore

= co, —mg/o>,,or

on both sidesof thisequality, weobtain

cuit consisting of a series—c0nnectedndinduction coil with resistanceRando anexternal voltagewith theamplitudeming that current in thecircuit leadsinonstruct thevector diagram and useit

deof voltageacrossthecoil.m for the caseunder consideration isdilyseen from thisdiagram that the



he coil is

<oL — 1/0JC)’. ln thepresenceof resistanceadsthecurrent by lessthan at/2.it.Acircuit consisting of aseries-

oinductanceand an induction coil withctedto themain system with ther.m.s.power developed in thecoil if ther. m.s.resistor Rand thecoil areequal to U,

or diagram shown inFi . 11.12. Inosinesweobtain from this¢§agramcpl,.(1)coil is

weobtain



sof Measurement* n, nano,10**p, pico-, 10*****

ElectricandMagneticanSystems

**



aa;-·L;”‘_-{i

m CGSE unit il(3 X 10*)GSE unit 1/300SE unit 3x 10**unit 3 x 10*it 121: X10**0*

GSE unit 3x 10*/(9 X10**)nit 1/(9 X10*)t 9X10*1nit 9 X10*0*

10*CGSE unit 10*

unit 10****

cityand IIQIINEIII

m

***

",T‘ q’="}"

V_q), q,1__%= S E dl



m

i;: » M={P><E1.W=—p·E

P = xE

0E-}- P D= E -[-4nP0E D= BEzq § Dds: [mq

C= .}....

/2= C U’/2 == q°/2CD__ E ·D

...2*% V.j = -.%.+$¤¤» _ i-=¤(E+E*>=_. pj:] I F=qE-}—%—[v><B]



ystem[vX;·] B_ [ QUVXT]

>i£;£’l’ dp, = J. Dill}!

<r[

nI B 1 Zn!

—-é-?-—nl>< B] dli = -{-· [dl >< Bl

B]dV

F = 2I 112

m = —E- IS

M=[pm><B]

(])2...([)‘) A:__!_ { my __Q, U

é··· IzB,/po-JH = B-- 4:1.1

r§, [-{ dl .; ........ [

u=1+4nXH



em

XBl *4 :E+·;·lVc><Bl

B—--—;—[v0>(E]VvE2-- B2: inv.- ...1.. .....d(D

”V_ 1 dl

W:T ·§*‘

Hw_ B·H

_ 4 ap

·s@Ed1=-7-5Bds

__ •__ 1- °‘BV XE‘ "`T B

(i+,m)



em

Ht/pp,E Ve==H Vp.

] S= —£t—·[E XH]

3,- [E >< Hl‘ 6:-23}- [E XII

sc= 2.998 X108 m/s*){ 6.67 X104 cms/(gm,)807 m/s’6 .022 X10*3 mole‘*n e= { FCCGSE units.911 X10·°° kg

—;-ng-..{ 527 X101., CGSE units/g72><10·2‘* kg

X10*11 F/rn

57 X10·° H/m

=1/Ve opo



c, 144

192 alternating, 290Iìion,174

c;m;2;’ 253f’ 271magncdzation 174rfaceHg 178 19259 votumé175 ’ulag174 177

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gsstlatiohary278

nt(s)f, linear,M6 ’um°· M6

tion,189288f

0 Damping{actor , 280, 283netism,180ld,45 dielcctrictst

°t"?p’°·wi

86, hq“‘d•*08*



electrostaticshielding, 51

ion,227

dcapacitor,99f, 139uctor,99f

mutual.238field, 94,100, 103,f, 92,105* 107* 2:%% 276<>*¤€¤u9m¤g¤¤*·¤¤¤¤d· 253112netic' 243oops, 238or,Of mt°m°u°n•961*,rges,97. 105. Um

10, 105,110f rzneticfield, 234, 236f,

x’ 266,6*

u°m?"(§)*6, 118, 261form,118aflei 5;;,

zsvr, 2-:2,, 274(?m°r°n;";_l term, iss)

tfggra,_°";,f,,,5wopvmcs0,* 26217fi,Synimgtry of _7, 279f 1, 262f Poisson,54Eotential surfaces, 32,47•



155, 25232 solenoidal, 155urcesof, 25

f,24, 15r.1orce(s),57,159,1cs, 171,

moment Of, 100, 100%,36122

ctric,10611

ous1221 235

ectromotive,217

tionc. ’

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,Gcncmmr MHD,224

2718090



third,95127, 131, 134, 136ff,33. 249.293. 295

rential form, 120'¤li¤6d. 123

124orm subclmult,

transformation for fields E, 206, 213f

manent,68, 195neticw,172, 174, 181475177cmagneticconstant: i43’dintensity,179

_946,

185elat1venatureof,204m¤g¤·=¤¤*·_*·¤¤·*729 *95

90e¤¤¤>hvdr¤dv¤¤¤¤i¤¤. 244

9, m°lh°d•

62f, 67,

ar,68t,magnetic,1571}

n- oscillation(s),294



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230tibility,magnetic, 180

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anics

clawsof mechanics—lawsof mo-nof energy,impulseand momentof their applicationsareindicated.and exercisesare included.accordancewiththesyllabusof for university students. The book entswho intend to specializein



godsky’sHandbook of Higher Math-ngineers,technicians, researchersof physics.Includesdefinitionsrief formulationsof physical laws,

nomena,tablesof physical quanti-nits, universal physical constants,





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