electromagnetics 6

8/13/2019 Electromagnetics 6

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C H A T E R

l r lchdging B field

rangmg

Maxwell'sEquationsfor Time-Varying ielDynamicsFields6-l Faraday's aw6-2 Stationary rop in a Time VaryingMagnericField6 3 The Ideal Tiansformer6-4 Moving Conductor n a StaticMagneticField6 5 The ElectromagneticGenerator6 6 Moving Conductor n a Time-VaryingMagneticField6-7 DisplacementCurrent6-8 BoundaryConditions or Electtomagnetics6-9 Charge-CuirentContinuityRelation6 l0 Electromagnetic otentials


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OYNAMICIELDSE,ectric charges nduceelectdc fields and electnc cur-E:ts induce magnetic ields. Thoseare the lessonswe:lined in the prece.dingwo chapters.As long as the::ge andcurrent istributionsemain onstantn time,r aill tbe ields hat they nduce.Howeve! if the charge-l crmeotsourceswerc o vary with time,, notonlywilltieldsalsovary wift time.but muchmorehappens.1e electdcand magnetic ields becomenterconnecled,: $e couplingbetweenhemproduces lectromagnetic. s capable f raveling through ree space nd n ma-

media.Elecromagneticwaves,which nclude ight.5. x-mys, nfraredwaves, amma ays,and adio.:5 lseeFi8. l-9], arean mportant anof ourphys-*orld, and heir usesaremanifestedn many ieldsr.rence and technology:. study time-varying electrornagnetic henomeDa,ed to us Maxwell's equationsas an integratedThese 4uations.which were irst innpducedn theseclion f Chapter , aregiven n bot-tr ifler-: and integml form in Table 6-1. whereas n the: czie (010t : 0) we wereable o use he firstr{ Maxwell's equationso studyelectricalphenom-: Chapter4 and the secondpair to studymagneticn]rrrcna in Chapter5, in the dynamic casewe havetd wilh the coupling that exists between heelec-Ed magneticields, asexpressed y the secondandequationsn Table Gl. The first equation epre-Causs'saw, and t is equallyvalid or staticandfields. The same is Eue for the third equa-

(Faiaday'saw) and, conversely, time-varyingelectricfield gives ise to a rnagnetic ield (Ampare'saw).Someof the results we will obtain in ihis and suc-ceedingchaptersmight contradict statementsmadeandconclusions cached n Chapier4 and 5. This 1sbecausethe earlier materialpertainso the specialcaseof steadycunenti and static charges.When A/at is set equal tozero, he resultsand expressionsor the fields underdy-namicconditions ill reduce o thoseapplicable nderstaticconditions.Webegin hischapter ithexaminationsf Faraday'sand Ampdre'saws and someof their practical ppli-calions.We will thencombineMa,\.well's quationsoobtain relationsamong he chargeand cufent sources,pv and J, the scalar and vector potentials,y and A.and the electromagneticields, E, D, H, and B, for thetime-varying ase ngeneraland or sinusoidal-time ari-ations n particular.

6-1 Faraday'sawThe closeconnection etweenelectricityandmagnetismwasestablished y Oersted,whodemonstratedhata wirecarying an electdc cudent exertsa force on a compassnedleand hat the needlealways ums so as o point inlhed-directionwhen he current s along he -direction.The orceactingon thecompass eedle sdue o themag-netic ield produced y the current n the wire. Followingthis discovery,Michael Faraday evelopedhe ollowinghypothesis:f a currentcanproducea magnetic ield, thentheconveneshouldalsobe true:a magnetic ield shouldproducea current in a wire. To provehis hypothesis,he conducted umerous xperimentsn his laboratoryn

. ; . B = 0, which basicallystates hat therearcirn lhingsas magnetic harges. he second ndequations,however,exhibit different meaningssoc and dynamic fields. ln the dynamic case, ae-1 ng magnetic6eldgives ise to an electric ield

169


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C}IAPTER 6 MAXWELL'S EQUA'IIONS FOR TIME-VARYING

Tbble 6-t Marwell's couations.

London over a period of about 10 years,all aimed atmaking nagnetic ields nducecurrentsn wires. Similarwork was beingconductedby JosephHenry in Albany,New York. Wires wercplacednext to permanentmag-netsof all difrerent sizes,butno curents werc detectedin the wires. Current waspassedhrough a wire whileanotherwire wasplacdparallel o it, with lhe expecta-tion hat he magneticieldof thecurrenl-can)ing irewould induce a cu.rent n the other wire, but again heresultwas negative.Evenoally, these ypesof experi-ments ed to the trueanswer which both FaradayandHenrydiscoveredndepetdendyat about he same ime(1831), They discovered l\at indeginagne icJieldscanproducean ele(tri( current in a closetl loop, but onlyif the nagnetic w linking thz suface areaof the loopchangestirh rirk. The ke) ro rhe nduction rocesssc/rrrg.To explain how the nductionprocessworks, etus consider he arrangement hownn Fig. 6- . A squareconductingoop connectedo agalvanometer, hich s asensitivenstnment used n the 1800s o detect heffowof cu[ent in a circuit, is placednext o a conducting oil

connectedo a batlery. hecurrent n thecoila magnetic reld B whose ines pass hrough heas shown n Fis. 6-1. In Section -5. we defincd

Gaussts awFaraday's lawNo magntic charges(Gfi$s's law fo. magntism)

O t " : - f ,V - B = 0

o " = r * S

{ o . a " =f"".d,:_"#.,"$u . a " :

/ I ' "= / ( r .#) ^


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STAIIONARY OOP N A TIME-VARYINGMAGNETICFIELD t7t

' : / n ' ' . (wb). (6.s)

lignetic ffux O passinghrougha oop as he ntegralofia nolmal component f the magnetic lux densityoverar surface reaof lhe loop, S, or

- Ser stationary onditions,hedr currentn thecoil pro-rEcs a constantmagneticield B, which n tumproducesr .rostanrlux hroughheoop.Whenhenux sconstant,rt iunentisdetectedbythegalvanometerHowever,whenc botlerysdisconnected,herebynteruptingthe low of,.rrEnt n thecoil, themagneticielddrops ozero,and he:6quent changen magneticlux causes momentaryof theSalvanomekreedle.When heba er)I =.onnected,hegalvanometergainexhibitsa mo&en..,deflection, ut n theopposite irection.Thus,curenriducedjnthe oopwhenhemagneticlux changes,nddirectionof the curcDt depends n whelher he flux:;reasing (aswhen hebattery s being connected) ring (aswhenthebattery s being disconnected).ras furtherdiscoveredhat currentcan also low in theg.while hebattery sconnectedo hecoil, fthe looD saround uddenlyorwhilemovingtclosertoorawaythecoil.Thephysicalovemc of heloopchangesmount of ffux linkhg its sudaceS, even hough heB due o lhe coil hasnot changed.{ Ealvanometersdrepredecessorf thevoltmeterand. Whenagalvanomete.etectshe lowofcurrcnrghthecoil. it meanshal a vo[agehasbeen nducedthegalvanometererminals.This voltase s called.lectromotiveforce \enf). ydr, and the process\elerEonagnetic iiduction. Theemf nducedn aconductingoopof N tums sgivenby

of d|enegativeign n Eq.(6.6)will beexplainedn thenextsection.Weoote that the derivativn Eq. (6.6) s atotal riDededvative hatoper'atesn the magnetic ield B, aswellas the difrerentialsurfacea.ea s. Accordingly,an emfcan be generatedn a closedconductingoop undetanyofthe following thiee conditior$:l. A time-varyin\magneticeld linkingd stationary/trop; he nducedemf s tjrcn called.theransformere,nJ, !^t2. A moing loop witha time var.ying reu reunve nthe normal component of B) in a static field B; rJltr-induced emf is then called the motional emf , VJd.3. A moing loop in a time-vaning lield B.

The total emf is given by%Dr=y*r+y#r , (6.7)

\ i th y.k - 0 i f rhe oop s stal ionaDcaset.))andy.".r = 0 it B is static case 2),. Forcase j). neirherterm is zero_Each of the threecaseswill be examinedsepamtelyn the ollowingsections.6-2 Slationary00pna Time-VaryingMagneticieldThe single{um, conducting,circular loop with con-tour C and surfaceareaS shown n Fig. 6-2(a) is in atrme-varyingmagnetic ield B(r). As wasstatede3dier,theemfinducedwhenS s stationaryand he ield is timevarying is called the transformer e4f and s denotedy$r- Since he oop s stationary /dl in Eq.(6.6)nowopemtesnB(l) only.Hence,

where the fulf deiyative d/dt has beenmoved nsidethe ntegralandchangednro apanial derivativeA/Ar to

- . d o' ^ t = - N E : (v). (6.6)r4 [s .a 'thoughheresulrsading oEq.(6.6)werealsodis-independently y Henry,Eq.(6.6) s anributedoy and s ktown asFaradal t tan. Thesignificance

-"1#' (6.8)


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CHAPTER 6 MAXWELL'S EQUATIONSFOR TIME.VARYING

Increasins (, )

1 1 1(a) t oop in a changingB field

v.S,(r)

(b) EquivalentcircuitFigurr 6-2: (a) Stationarycircular loop in a changingmagnetic6eld B(t), and(b) its quivalentcircuit.

signify that t oprates n B only.The hansformeremf istlrvoltagedifferencehat would appear crosshe sfiullopenilrgbetwenerminals aIId2, evenn the absence fthercsistorR. That s, vjdr = y12,whereyr2is theoPen-circuitvoltageacrosshe operendsof the oop.Underd-cconditions,y*r : 0. For lhe loop shown n Fi8. 6-2(a)and he associatedefinition or VSr givenbl E4.(6.8),thedirectionof ds, the oop'sdiffercntialsurface ormal,can be chosen o be eithet upwardor downward.Thetwochoicesareassociated ith oppositedesignations fthe polarities of terminals I and 2 in Fig. 6-2(a).Theconnectionbetween hedirction of ds and thepolantyof y"k is govemedby thefollo,ting righlhand rule: ifds pointr along the thumbof the right hand,then thedireclion of the contourC indicatedby the four fingelsis slch that t alwayspasses crossheopening rom thepositive erminatof V*r to the negativeerminal-

Ifthe loop hasan ntemalrcsistanceRi, theFig. G2(a) canbe rcpresented y the equivalentshown n Fig, 62(b), in which care hc cuntrttkougi the circuit is givenby

R + R j

in the loop is always in such a direction as tothe changeof mdgnetictLt O(t) thatP,Irdu.:edt

Forgoodconductors,l?iusually s very small,and tbe gnored n comparisonwith practicalvaluesof R,ceptwhen R = 0 (loop s shortedacrossts ends).Thepolarity of V*r ald hence he directionofEo,refiedby lznz's tar, which staies hat tie

cunent 1 induces a magnetic field of its own, Bind,co[esponding flux Oind.The direction of Bi.d lsby ie right-handute: f ,l s n aclockwicedirectionBiodpointsdownwmd hrcugh,Sand,conve$ely, fin a counterclockwise irection, henBindpointstbroughS. If the originalfreld B(t) is increasing,means hat do/dt > 0, then accordingo lsz'shas o be n the directioD hown n Fig. 6-2(a)mBid to be noppositionoB( l. Consequenllr.would b at a higherpotential han tenninal l, andwould havea negative alue.However. f Bfr)remain n thesame ircctionbut o decreasen magnithen do/dt would bcomenegative, he currenthaveo reverse irection,and ts induced ieldBio6be n the same irectionasB(r) soas o opposehe(decreasetf B{r). ln thatcase.V$t wouldbeIt is mDortant o rememberhatBi'd servesochengen B(tr, nd notnecessa.ily(t) itseh.Despite he presence f the small openingterminals and2 of the oop in Fig. 6-2(a),we shallthe loop as a closedpathwith contow C. We doin orderto establish he link betweenB and hefield E associated ith the nduced nf, y":r. Also,pointalong he oop, he ield E is related o thecuflowing through he oop. For contourC, y* is


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STAIIONARY LOOP IN A TIMFVARYING MAGNETIC FIELD t't3r E b Y

5- N = I (a loop with one tum), equatingEqs.(6.8)(6.10) ivesf n.ar=-|,ff.*, (6.r

i-rch is the ntegral form of Faraday'saw given n Ta-:e 6'1. We shouldkeep n mind hat he direction f!c contourC and the direction ofls are rclatedby thed9t-hatd rule.By applying Stokes's heorem o the eft-handsideof-t 16.11), e have

Figure 6-3: Circul& loop wilh ,V tums n the r-], plane.The magnetic ield is B = BoO2 + 23) sin.d, (Example6-1).

vk= {,n.ar (6.10)

l ,o"u,t"=- | , f f .0, (6.r2t

.|!j in order or the two integrals o be equal, heir inte-l|]rds have o be equal,whichgives

aj differcntial form of Faraday's aw states hat aE-\"rying magnetic ield inducsatrelectric field E4(r5e cud is equal o the negative f the timederivativer B Even though the derivation eadingto Faraday'st srarted ut by considering he ield associated ith ar.rical circuit, Eq.(6.13)appliesar anypoinr n space,.r.$er or not aphysicalcircuii existsat thatpoint.EEmple -1 Inductorn a Chan0inglaot!tic ield

\n inductor s formed by winding N tums of a thin:r,lucting wAe into a circular loop of radiusc. Ther:a-tor loop s in the, ) planewith its center t thernn, and t is connectedo a resistor R, as shown n:_,: G3. ln the presenceof a magnetic ield given by| = a002+i3) sin@t,where@ s theangular requency,ir:

(a) the magnetic lux linking a single urn of the nduc-tor,(b) the transformeremf, given that N = 10, Bo :0.2 T, d = l0 cm,and@ 103 acVs,(c) thepolariiy of Y*r at I : 0, ard(d) the inducedcurent in the circuit for R : 1 kQ(assumehe wire resistanco benegligibly small).

SolutioD:(a)The magnetic lux linking each um of theinductor so = [ t . a "

I:I | B o t y 2 + i 3 t s i n r r . t | 1d s : 3 r a z B o s i n @ t .(b)To ind V"fu,wecanapplyEq.(6.8)or wecanapplythegeneral xpressionivenby Eq.(6.6)dircctly.Thelatterapproachives

d "- . . t t3nNut Bosinut t - 3- tN' . ,azB0co\tDt.ForN = 10, a=0.1f i ,@= 103 ad/s, nd Bo = 0.2T,

Y*r = -188.5cos03. (v).

(6.13)


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CI]APTER6 MAXWELL'SEQUATIONSORTIME-VARYINC(c)Att=0, d@/dt>0 rd,V:nr -188.5V Sincethe flux is inqeasing, the curcnt I mustbe in the di-Fction shorn id Fig. 63 in o.derto satisry rDz's lawConsqueDdy,oint 2 is at a higherpoEntial thanpoint Iand

vS= y1 y, = -188.5(d)Thecurrent isgivenby

(v).v - v ,. R ffcoslo3t = o.lgcoso3r (A)I

and he correspondingransformer rnf sIXERCISE.1 For the oop shownn Fi8. 6-3,what sY& if B = iBo cos rt? Explain.Ans. yh : 0 becauseB is orthogonal o the loop'ssurfacenormalds. (SeeA)IXERCISE.2 Supposehat he oopof Example -1 srcplacedwith a lo-tum squareoopcntercd t theoriginandhaving 20-cmsidesorientedparallelto the i- andy-axes.fB = iBsx2cos103tandBe= 100T, find hecurrentn the circuit.Am. 1 = -133 sin103r mA). (SeeO)

Lcnz's af,Detemine the voltagesyt andy2across he2-Q and4-Q resistorsshown n Fig. 64. The loop is located nth.r-) plane,ts area s4 m2, hemagneticlux densityis B = -i0.3t (T), and he ntemalresistance f thewiremaybe gnorcd.

Solulion:The lux nowtnghrougi he oop so = l n . a s : l t - n 3 t ) i d sJs ../s

= - 0 . 3 t x 4 = - 1 . 2 t ( w b ) ,

: t.2 (v).

t = - - : : L : l j a = o ) A .R r * R ) 2 + 4and

V : I & = 0 . 2 x 2 : O . 4 v ,Y 2 = I R 2 : 0 - 2 x 4 = 0 . 8 V .

daaa

Since the magnetic lux through tbe loop is along-z-direcl ionintohepager nd t i . increasingonitudewith time t. ldz's law states hat the icurrent1 shouldbe n a direction such hat theflux densiry .d inducd y / counleractsheofchanee of O. Hence, t has to be in the directionin hecrrcuil ecausere oresponding 'nds+z-dirertion in the region nside the loop area.Thi t:tum, meanshat yl ard y2arepositivevoltages.The total voltage of 1.2V is distributedacrossresistors n se.ies.Consequently, (L

BEVIEWUESTIONSQ6.1 ExplainFaraday'saw and he functionoflaw.

Vz vr:@ +@ @ @

Figrrc 6-4: Circuit for Example6-2.


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:-] T]IE IDEAL TRANSFORMER 175lal Under what circumstances s the Det voltagerurnd a closed oopequal o zrro?ta-l Suppose dle magnetic flux dedsity linking!. loop of Fig. 6-4 (Example 62) is givcn by| = -i0.3e-t (T). Wlat would the dirction of thetrent be,relativcto that shown n Fig. 6-4, for t > 0?itplain.

i3 Thedeal ranslormer_!ctransfonner hown n Fig. 6-5(a)consists ftwo coilsr-{]ndarounda commonmagneticcore.The coil of theElary cacuit has Nl tums and that of the secondaryr:uil hasN2 ums. Theprimarycoil is connecled o ants: voltagesourceq(t) and he secondary oil is con,F-1edo a load resistorRL. In an ideal transformerhe:re has nfinitepermeabilityp = oo1,2n4 hg m.r..ti.lll is confinedwithin rhecorc.The dircctionsofthe cur-tlts flowing in the wo coils, 11and 12,aredefnedsuchlza when t and 12arebothpositive, he ffux generatedF ! is opposite hat generated y 11.The ransformert'; its namercm the act that it is used o transform1 .ents, oltages,nd mpedancesetueen tsprimary;: secondaft ircuits-Cn tbe primary side of the transformer, he voltageruce h generatescurent 1l in theprimarycoil, whichfiblishes a flux O in the magnetic ore.The lur O andE \oltage yt arcrelatedby Faraday's aw:

, , , , d 4vl : - /vr - - - ,ar similarly, on the secondary ide,

(6.14)

Figure 6-5: ln a transformeahe directionsof /r and 12are such h3t the flux O gnerated y one of them s op-positethat genemtedby the other. The directiofl of thesecondarywinding in (b) is opposite hat in (a), and soar he directionof 12and hepobdty of y2.

Thecombinationf Eqs. 6.14) nd 6.15) rves

da

j = ; . ( 6 . 1 6 )

In an ideal lossless ransformer,all the insranBneouspowersuppliedby thesourceconnected o theprimarycoil is delivered o lhe loadon the secondary ide.Thus,noDowers lost in the core.and(6.1s) (6.17)


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CHAPTER6 MAXWELL S EQUATIONSFOR TIME.VARYING

Figur 6-6: Equivalent ircuir for theprirnrry sideof the

Since Pr = IrVr nd Pz = /:Vz, and in view ofEq- 6.16),we have! : I ] . , o . te ,h N t

Thus,whereashe atioofthe voltages ivenby E4.(6. 6)is proportional o the correspondingumsratio, lhe ratioof thecurrents s equal o the inverseof the tums ratio.lf Nt/N2 = 0.1, V2of the secondaryircuitwouldbel0 times yr of theprimarycircuit but 12would beonlyL/to.The ransformer bown nFig. G5(b) is dentical o thatin Fig. 6-5(a) except or the directionof lhe windingsofthesecondary oil. Because f thischange, he directionof 12and he polarityof V2 n Fig-6-5(b) are he revefteoflhose n Fig.6-5(a).The voltage and cunent in the secondary ircuit inFig. 6-5(a)arc relatedby y2 = I2RL.To the nputcircuit,the ransfomermaybe epresentedy anequivalntnputresistanceRin, asshown n Fig. 6-6, delinedas

Whenrhe foad is an impdanceZL af.d vt is asoidal source, he nput resista.trceepreseDtationa!axtendedo aDcquirdent input idFdance Zi! giva!

6-4 Moving onduclorna StaticMagnel icieldconsider a wire of lenglhI moving acrossa staticnetic field B = iBo at a constantvelocity u, asin fig. 6-7.ThecondLrcungirecontaiDsreeFrom Eq. (5.3), the magnetic orce Fn acting oncharged afticle4 movingwith a velocity u in afield B is givenby

F m = 4 ( u x B ) .This magnetic orce is equivalento the electricalthat would be exerted on the Darticle bv an

v,nr : f , .Useof Eqs. 6.16)and 6.18)gives

f (f)'=(t)'"(6.19)

(6.20)

(#)"^- = (6.2r)


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MOVINGCONDUCTORN A STAI'ICMAGNETIC IBLD

E givenby

:.ld En generatedby tbe motion of the chargedi\ called a motional electri. frew, and t is nion perpendicularo the planecontainingu andiir d wire shown n Fig. G7, E- is along -i. Theforceactingon thc electrons o the wirecauses',. move n the directionof -En; that is, towardlabeled in Fig. G7. This io rum nducesa volt-fu-erencebetweenendsI and 2, with end 2 beinqrgher potential. he nduced ohages calledaernl, yctf, and s dennedas he ine inregralof

Ingeneral,fanysegmentofaclosedcircuitith contou.rCmoveswith a velocityu acrcssastatic magnetic ield B,then he nducedmotionalemf s eivenbv

Onlt- those segmentsof the circuit that cross nagneticrteA nesco tribute to V*r.

.{pTgl!.6'q stidinoBarThe rectangularoop shown n Fig. 6-8 hasa constantwidthl, but ts ength,ronseaseswithtime asaconductingbar slidesata uniform velocityu in a static nagDeticieldB = iBor. Note hatB increasesinearlywith.r.Thebarstarts romi = 0 at = 0. Find hemotionalemf betweenterminals I and 2 and he cunent 1 flowing thmugh heresistorR. Assumehat he oopresistance


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CHAPIER6 MAXWELL'SEQUATIONSORTIME-VARYINGThe slidingbar, being he only part of the circuit thatcrosseshe inesof the ield B, ir thc onlypart ofcontoui2341 hatco{tributcs o y$i. Hcnce,atr = .ro,

(u x B) .dlvJ*=vQ=v4x=Ir'

The engthofthe loop s related o Ir by.t0 : &t.HeDce,(6.21)

sinceBis static,v*r = 0 aIIdy.di = y#only. To verifythatthesameesultcanbe obtained y thegeneralorm ofFaraday'saw,westartby findingttle flux O through hesudaceofthe loop. Thus,

=lro6u x enoto)9al= -uBoxot.

v#tr= -Bou2lt (v).

: not!o'r ar BYio : / n . a "

= [,Qno' \ 'za'a,

trimplg6'a liloving-oopThe ectangularoopshownnFig.6-9 sx-y planeand movesaway rom theorigin at au = 95(rn/s) n a magneticieldgivenby

(6.28)

Substituting 0 = l.t in E4. (6.28)and thentaking thenegative erivativewith respct otimegives, , dA d /Bo tu ' t ' \v . n f - - : - l : l - la I a t \ z , /

:,Bou2lt (V), (6.29)which s identical ith Eq. 6.2?). inceq2 is negatrve,the current1 : Bou2lr/Rflows n thedirectionshown nFig.6-8. I

BO) = a.o.2e-oi' (T).If R = 5 O, find the cunent 1sides re t 1 :2fia d12:maybe gnored.

at the nstant hat lE2.5m.The oopSolution: Sinceu x B is along , voltagesareacrossonly the sidesorientedalong , namelybetweenpointsand2 and besidebetween ointsHad B beenuniform, the inducedvoltageswouldbeeD he sa(neand he net voltage across he rcsistdhavebeenzero. n thepresent ase,however,Bexponentiallywith t, therebyassuming differettoverside -2 thanoverside3-4.Side -2 is at,r =and he corresFxdingmagnetic ield is

Btr,): iO.Z"ot) - i}.2e-42 {T).


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MOVING CONDUCTOR N A ST]A(IICMAGNETICFIELD

- u B ( y ) l = - 5 x 0 . 2 e o t x 2: _1.558 (V).

:Lrequentlt the current s in the directionshownn the3 and tsmagnirudeis, Vq - Vo O.O79R 5

F5 lvlovin0od{s{ toa Wire--c wireshown n Fig.6- 0 is in freespace ndcame\rnl I = l0 A. A 30{mlong meral odmoves rat velocityu = i5 tn/s.Fird yrr.

iDduced oltage%2 sthengivetrbyvtz tr'r. x,,or)r.dl=

l-"'{gs,eo.x-o\.;a,

: -e-o21= -2"-oz - -t.Ur, (V).

Solution:Thecurrent inducesamagneticield

r = C 3 ,z1twherer s heradialdistancfromhewire andhedirectionofC is into lhe pageat rhemetal cd sideofthe wire.Themovement f the odin thepresencef the ield B inducesa motionalenrfgivenby

Vn= I (u x B) .dl. l 0 r m ,= (zs ' i ' ! ! l+a,

J40cm \ t t t I J__5po l f to " d rzit J4an r

5 x 4 n x l 0 7 x l 0

B(ERCISE.3 For the moving oopof Fie.6-9, find 1at the instaDt hat the loop sidesare at y1 = 4 m and)2 = 4.5 m. Also, reversehdirectionof motionsuchthatu:-i5(m/s).Ans. 1: -13(rnA). (Seed)EXERCISE.4 Supposehatwe um the oopof Fig.6,9so hat ts surfaces parallel o the -z plane.What would/ be n thatcase?Ans. 1= 0. (SeeO)

15.8mA). r

" r"(*!)=r:.et,vr..


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CHAPTER MAX!!'ELL'SEQUATIONSORTIME-VARYING

BEVTEWUESTloltSQ6.4 Supposehatno ftiction is involved n slidhg theconducting ar of Fig. 6 8 atd that he horizontalarmsofIhc circuit arcvery ong.Hercq ifthe baI sgivenaDnitialpush, t shouldcontinuc moving at a constantvelocity,and ts movement eDerateslectricalenergy n the ormof an nducedemf, ndefinitely. s this a valid ar8umcnt?If not, why not? How canwe generate lectricalenergywithouthaving to supplyan equalamountof eneryybyothermeans?Q6.5 Is hecunentflowingn he odofFig.6-10asteadycurrent?Examine he orceon a chargeq at ends and2


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TIIE ELBCTROMAGNETICGENERA]OR 181Segment + moveswith a vetocity -u. ApplicationofEq.(6.26),consistentwith our choiceof i, gives

t t t tv h = v a = , ( u B ) . d+ / r u x B r . / lJ2 Jat t l z r , r t ,= I ll iu- ) x LBel.itdxr ' t12

I t / 2. ,I I l ( - n o ; ) x t 8 o l . i d r . { 6 .3 3rt/r 'Usitrg q. 6.32)n Eq. 6.33),weobrainhe esult

V"T,t 11rlar8osinc:AarBosina, (6.34)whereA : ul is the surface reaof the oop. The angledisrelated o o by

r:'!re 6-12:A loop otatingn a magneticield nduces

- a{isof rotationofthe conducting oop salong he-js. Seghents -2 and 3-4 ofthe loopareof length,andboth cross he magnetic lux lines as the oop. The other two segmentsare each of width l,, andcrosseshe B Lineswhen he oop rotales.Hence.,gments -2and3-4 contribute o thegeneration fe.uonalemf,Y*.rj irc loop rotateswith an angularvelocity@about tsi].i s-segment -2 moveswith a velocity u givenby

(6.3s)whereC0 s a constant eterminedy initialconditions.For example,fa = 0 at : 0, thenCo= 0. In gene.al,

Y* = A.,Bosin(@t Co) (V). (6.36)This resultcanalsobobtaiDd y applying hegeneralform ofFaraday'sawgivenby Eq.(6.6).The lux tinkingthesurface fthe loop s

f fo = r B . d s : l i B s . n d sJ s J s

--n- he surface ormal o the oop, makes n angled:r:-aiis. Hence,(6.32\

-- EoAcos(.rt+ Co), (6.37)and

- . d o d - -V.nr ---- = - -, lAo,4 os(dr Co)l= A.d8o in(@t Co), (6.38)

which s denticalwith the resultgivenby Eq.(6.36).

(6.31)


15/28

REVIEWUESTIONSQ6.6 Contrasthe operation f ana-cmotorwith that ofana-cgnelator.Q6,7 The rctatiq loop of Fig. Gl2 had a single um.Whatwouldbe heemfgenemted y a oop with 10 ums?Q6,8 The magnetic lux linking the loop shown inFig.G 12sa maiimumwhen = 0 (loop n r-y plane),andyetaccording o Eq. (6-34), he nducedemf is zerowheo@= 0. Conversely, hena = 90', the lux linkingdeloop s zcfo,buty* is atamaximum. s thisconsistentwithyourexpctations? hy?

6-6 Movingonductorna Time-VaryingMagnel icieldForthe eneralase f a siDgle-tumonductingoopmov-ing na ime-varyingmagnetic ield, he nduced mfis thesumof a uansformer omponent nd a motionalcompo-nent.Thus,hesumofEqs.(6.8)and 6.26)gives

CHAPTER6 MAXWELLS EQUATIONSFOR TIME.VARYINGof which s theeasiero apply. f lhe oopconsiststums, he ems on the right-handsi&s of Eqs. 6.(6.,10)horld bemuttipliadby iv.

=- [P o, -d (uxs r .a t . (6 . re )Js dt Jc%o{ s alsogivetrby thegeneralexpression f Faraday'slaw:

da = -L I s. as. 6.qot

i${W.!N1i Et.ctmmagmttccnrntorFind the nduced oltagewhen he rotating oopelectromagneticenerator f Section6-5 s in afreldB = i8n cos rl.Assumehatd = 0at, = 0.

Solution: n his asehe lux O sgiven yEq. 6.Bo eplaced ith [email protected],

and

ao%.f =_ ,d .= -- (roAcos'(,t)

= 2BoAacos t sinat = BoAasin2aDt

6-7 DisplacemenlCurenlFromTable6- . AmDdre'saw n differential ormby

v x H = r + #

1",,l[ \ e rale he5urfacenregral f bothsides fE+overanarbitary opensurface twithcontourC,

vj,f + v#$u '0 ,

(Amperc'saw).

' o .a " / r .a .+ S.a"nfacl. t canbeshownmathematicallyhat hedght-handsideof Bq.(6.39) s equivalent o the right-handside ofEq. 6.40).Foraparticularpmblem, he choicebetweenusing q. 6.39) rEq. 6.40)s usuallymade n hebasis

ThesurfacentegralofJ is equal o theconductro/c lowing hroughS,and hesurfacentegralof v


16/28

DISPI-ACEMENTURRENT 183

'ure 6-'.: Thedisplacementurrent 2d n the nsuratingmateriarof the capacitors equal o theconducring unent1lc n

r ,'onverted to a line integralof Il overthecontourC-. lvoking Stokes'sheorem.Hence.

. scondermon the righFhandsideof Eq.(6.43)has:lve the sameunit (ampcres)asthecurrent1c,andit is proportional o the time derivative f theic f,ux densityD (which is alsocalled heelectncement), t is calledthedt:splacementanent IA.

- la = 3D/0t represen,s displacementcurrent4.In viewof Eq. 6.44),

Maxwell in I873 in his successful lremF ro eslablishaunrfiedconnectionbel\reenelectricand magnetic ield\under ime-varyingconditions.The larallel-plate capacitor is commonly used as acoDvenrentxample o illullrare lhe phy!ical meaningofInecu_splacementurTentd. The simplecircuit shown nrrg, b-tJ consist\of a capacitorand an a< sourcewtthvoltage%(r)givenby

%(r)= Yocos.dr (V). (6.46)

-.- is he otalcurreot.nelecEostatics,D/at = 0ierefore 1d = 0 and I = 1". Theconceptof dis

AccordingoEq. 6.45),he otal urrentlowing lroughanysudace onsists,n genera], f a conduction urent Icandadisplacementunent1d.Let us ind Ic and1d hrouglteach fthe following two maginarysurfaces:I ) thecrosssection fthe onducting ire,Sl and 2) he ross ectronof thecapacirorsurface 2 nFig.6_13).Weshajl enotetheconductionanddisplacementufients n thewire 1,^and /'dand hosehroughhe apacilor r.and 2d.. In aperlecr onductor. = E = 0i hence. q. 6.44rgtves1td= 0 in tlrcwire.As to 1lc,we knowfromcircuit

theoryhat t is relatedo hevoltageacrosshecapacitor "

f u .a r= + o- , (6 .4s )=mentcunent wasfirst introducedby JamesClerk

r^1I, l^.d"="T.^, {6.44)


17/28

CHAPIER6 MAXWELL S EQUATIONSFOR TIME-VARYINGwith hatgivenby Eq.(6.47) or thecooductiony

t," = c * : c { V"cos@t)= -cvo@ srn@.,' dt dt " $.4i)wherc we used the fact that % = %(t). Wilh1ta= 0, the total current in the wire is simplyIt = Ib = -Cvo@[email protected] kough surface 2in Fig.6'13,which s an maginary,opensurface arallelto thecapacitor latesand siluatedsomewhere etweenthem.Thespace etween he woplates,eachof which sof areaA, is filled wid| aperfectdielectricmatrialwithpemittivity. Sinceelectricalcharges annotphysicallymovehrcugh dielectncmedium, onduction annot akeplacebelweenheconductingplatesof the capacitor, ndtherefore 2"= 0. To determine12d,we ned o applyEq. 6.44).FromExample4- I I , the electric ield E in thedielectric$pacings related o the voltage% across hecapacitor y

(6.48)*here d is the spacingbetweenheplatesand is the di-rectionrom he higher-potential late oward he ower-potential late.The displacement urent 12dn thedirec-tionshownn Fig.6-13 s obtained y applying q. 6.44)withds idr:

whereweusedhe clation C : A d for thecapacitanceoftheparallel-plateapacitor. heexpressionor /2d n thedielectric egion etweenheconducting latess dentical

,t. in thewile. Thcfact thatrhese wo currcntsar!cnsu&6 he coothdty of currEnt lor through hcciEven though the disphtement current does ot cantcharge, it nonethelessbehaves ike a real cunent.In thecapacitor xarnple,we ieated hewire asaconductor, nd we assumedhat fte spacingcapacitorplates s aperfectdielectric. f the wtrefiniteconductivityc'", thenD n thewirc wouldnotand hereforehecurrent1t would consistof acurrent11"aswell as a displacement urrent 1td;1r = 1r"* /ra. By lhesame oken, f thedielectdcmaterial asanonzero onductivitydd, henchargesbe able to flow betweD he two plates aDd 12cbe zero. In that case, he total cunent ffowingcapacitorwould be 12 12" /zo,and t would bcto he otal urrentn thewire.Thatis,lt = 1r.

Errkiplc -? oisplacementurrent ensityTheconduction urrent lowing hroughawireducdviryo 2 \ l0?S/mandrelat ivepe rminivi tyis givenby1c= 2 sin.,)r mA). f ., = 10s ad/s,displacement urrent,

Solution: Tbe conductioncurrent Ic : ./A =whereA is thecrosssectionof the wire. Hence.2 x l0-3 sin @t

2 x 1 0 ] AI r l 0 - ro= - stn ut (v/m).Application fBq. 6.44),with D = et,leadsto

:AAE/A ta / t , 1 0 - r o \= eA- | - sm rl IA r \ A )

:ero x l0 10cos ..,t 0.885x l0-l2cosor

Lr

a=il =t! ,"" . ' .

L

u=|,fr"=/f*srf*'-)] r^'= -' jvo, "inrt = -'Cvol;.sior-,l t, 6.49)


18/28

BOIJNDARYCONDMONSELECTROMAGNETICS

weused@= 10erad/sand= sn = 8.85 x 10-12.- \ote that c and d are n phas uadrature90"phase:betweenhem).Also, Id s aboutnine rders fmagni-imaj er ian /c.whichswhy hedisplace e tcurre t-l! is gnoredngoodconductors.

6.5 A poorconductorisharaclerizedy acon-lityo = 100S/m) ndpemfttivitye4ro.Atwhat@is heaDplitudeofheconduc!ioncur-J.nsityJ equal o lhe amplitudeof thdisplacementdensityd?@= 2.82 x lor' (.adls). (Seea)

BorndaryorditionstorElectromagnelics::rpters 4 and 5 we applied the integral form ofell's equationsunder static conditionsto obtain:arl conditionshat le tangentialandnormalcom-as f E, D,B, andH mustsatisfyatthe ntedacebe-twocontiguousmedia.Theseconditionsategiven4-9 for E andD andn Section5-7 or B andH.

In thedynamic ase,Maxwell's quationsTable -l] in,clude wo new erms, B/at in Faraday'sawandaD/atin Ampere'saw.Nevenheless,heboundaryconditiotlsderiyedpreyiouslJfor electrostatics nd magnetostaticremain alidfor time-tarylng.fellsas well This s be-cause,f we were o apply he sameprccedures utlinedin the above-referencedections,wewould 6ndthat thecombinationof the aforementionederms vanishas theareasof the rectangularoops n Figs. 4-18alrd5-24aremade o approach ero.For easyaccess,he combinedsetof elecmxnagncocboundary onditionssgiven n Table6-2.

REVTEWUtSTt0ilSQ6,9 Wlenconductionculrent lows hrough material,a certainnumberof charges ntet hematerialon o.e endand an equalnumber eaveoll theother end. W]at's thesituation ikeforthedisplacementcunenttbroughperfectdielectric?Q6.10 Verify that the integral form of Amtsre's lawgivenby Eq. (6.43) eads o the boundarycondition hatthe tangentialcomponent f H is continuousacrcss heboundary etweenwo dielectricmedia.

Tsble 62: Boundaryconditions or thelectric andmasneric6elds.Iidd CodporeDts Genertl Form Mdium r Mdium2Dielectic Dielectric Medium I Medium 2Dielectric ConductorTrDgentlrlENonnal DTatrgntidHNomal B

A 2 x ( B r - 8 2 ) = 0i z ' @ r - D ) = p .i 2 x ( H r - H 2 ) = J sf i 2 . ( B r - E z ) : 0Er = EzLBn - Bz"

Er= EtDzn oH x : 0

rotcs: (l) ps rs th surtaccchargedensityat rhe roundary; 2) Js is the surface unent aensityat rfe boundary;(3) mrmal componcntsof all 6lds arc alonga2, the outwardunit vcrorof medium 2; (4) Erl : t2i impliesthat he tangential omponents re quarn rnagnitude ndparauel n direction; (5) direction of J. isortboqonallo (IIr Htr.


19/28

EMF ensorsAnelectromotiveorce(emt)sensor s a device hatcln gen|aic afl irducad rctbgp ir rEFCtr|g oan sxternalslinulus. Tfuee typ of srd sonsorsare profiled n lhis lechnicalbriel: lhe piezoelectrictransducer,he Faradaymagneticlux sensor, ndthe hermocguDle.PiezoeleGtricransducersPiezoeleqlricity efers to the property of certaincrystals, uchas quartz,o become lectricallyo-larizedwhen he crystals subjeciedo mechanicalpressure,hrebyexhibiting \roltage cross t.The crystal onsists f polardomains epresentedby equivalent ipoles A). Under he absenceotan externalqrce,he polardomains re randomlyoriented hroughoul he material A1),but whencompressivertensile stretching)tresss appliedto the crystal,he polardomains lign hemselvesalong one of the principalaxes ol lhe crystal,leadingo a netpolarizationelectricharge) t thecrystalsurfaces A2 and A3). Compression ndstretchinggeneratevoltages of oppositepolarity.The piezoelectricetfect (piezeinmeanslo pressor squeezen Greek)wasdiscovered ytheCuriebrothers,Piere and Paul"Jacques,n 1880,anda year later, Lippmannpredicted he converseproperty, amely hat, if subjectedo an electricfield, he crystalwouldchangen shape. hus, hepiezoelectricetfect is a reversible(bidirectional)electromechanicalrocess.Piezoelectricrystalsare used n microphoneso convertmechanical i-brationsof he crystal urtace) aused y acousticwaves ntoa correspondinglectrical ignal,andthe converse rocess s used n loudspeakersoconvert lectrical ignalsntosound B). nadditionto havingstitfness aluescomparableo that ofsteel, some piezoelectricmaterialsexhibitveryhighsensitivityo the orceapplied pon hem,withexcellenlinearity vera widedynamicange. heycan be used o measure urface elormationss

TECHNOIOCY BREFi EMF

smallas anometers10-'m),makinghemularly ttractivespositioningensorsntunnelingmicroscopes. s accelerometers,can measure ccelerationevelsas low as Ilo as high as 100g (whereg is the accedue ogravity). iezoelectricrystals nd


20/28

0

BRIEP:EMFSENSORS

used in cigarette ightersand gas grills asgenerators,ncloc*s ndelectronicircuitryprecisionoscillators, n medical ultrasoundnosticequipment s UansducersB), and inrous ther pplications.

Famdayagneliclux ensorAccordingo Faraday'saw(Eq.6.6), he eml vrclt-age inducdacross hdiienirlnalSol a conduc{-ing loop s dkectly roportionalo the time rate olchangeof the magneticlux passinghrough heloop. or heconligurationn C),

V^t = -u Bolwhereu = dxldt is thevelocity f the oop inandout of lhe magnet'scavity), with the direction of ndefined spositive hen he oop s movingnwardinto hecavity,0 islhemagneticieldofthemagnet,and I is the loopwidth.With Bo and I beingcon-stant,he variation f %nr(r)with ime becomesdirect ndicator f he imevariation t ,l(r). he imederivative f r/0) provideshe acceleration(r).ThermocoupleIn 1821,ThomasSeebeck iscoveredhatwhenajunctionmadeottwo ditferentconduc.tingmaterials,such as bismuthand coppe( is heated, t generatesa thermallynduced mf,whichwe nowcalltheSee-beckpotential s (D).Whenconnectedo a resis-tor,a currentwill low hroughhe resistor, ilen byI = V s / R .This eaturewasadvanced y A. C. Becquereln1826 as a means o measure he unknown emora-ture f2 of a unction elative o a temperature4 ,ol a(cold)eferencejunclion.oday,uchagenerator lthermoeleclricitys calleda thermocouple.niiially,an ice bath wasused o maintain r at 0" C, bdin today'semperatureensordesigns, n artificialcold unctions used nstead. he artificialunctionis an electric ircuithatgenerates potenlial qualto thal expected rom a reference unctionat tem-oerature'.


21/28

CI1APTER MAXWEII'S EQUATIONSORTIME.VARYINC

Fignr 6-14: Total current lowing out of a volume v isequal o the fux ofcurrentdensityJ through hesurfaceS,which n tum s equal o the rate of deqeaseof the chargenclosednr).

6-g Charge-Currenl0nlinuityelali0nUnderstaticconditions,hecha-rgeensityp" and hecur-rentdensityJ at agivenpoint n a materialare otally inde-pendent fone anotherThis s not rue n the ime-varyingcase.To show lre connectionbetweenp" andJ, we startby considering narbifary volumev bounded y a closedsurfacS,assho\a'nn Fig. 6-14.The nelpositivechargecontaioedtr v is 0. Since,according o the aw ofconser-valionofelectricchargeSection -3.2],charge anneitherbecreated ordestroyed.heonlyway can ncRasesasaresultof a net nward flux of positivecharge nto thevolume ard, by hesameoken, or Ctodecrcasethereasto bea netoutward lux of charge rom v. The nwardandoutwatd luxesofcharge onstitutecurents l owingacrossthe surfaceSinto andout of v, respectively.We define1asthenet urrentfowing cro$ ,S rtoly. Accordingly,1 s equal o thenegdrtveateof change f 0:

Hence,

By applyingdrc divergenceheoremgivcnby Eq.(3we canconveft he surfacentegralof J into a volunelegralofits divergencE .J, which thengives

f; 0"

o r: -? . te.s+r

v . J = 0 ,

/ r r "=

=-*I"^o'

l ,".to,=-fi1"0"0,

l"r.to":l,#*

FI

Fora staiionaryolume . the imederivativpv only. Hence,we can move t inside the integralexpresst asapaftialderivative f p":

In order for the volume intesals on the two sid6Eq. 6.53) c beequal or anyvolume . rheirhave o be equalat everypointwithin v. Hence,

(6.50)whereA is the volume chargedensity n v. Accordingto Eq.(4.12), he current1 is alsodefinedas he outwardflux of the curent densityvectorJ through he surfaceS.

, : #: *1"^0",which s knownas hecr,arg-c nent continairyor simply as hecrdrg? continuit! equation.If the volumechargedensitywithin an elementalumeAr(suchasasmall ylinder)snota unction fG.e., p, At : 0), t meanshat henetcunentout of Ay is zroor, equivalently,hat hecurrentinto Av is equal o thecurrent lowingoutofit. In this84.(6.54) ecornes


22/28

.-IO EI-ECTROMACNETICPOTENTIALS t89

y'r.a"=o

FigurGl5: Kirchloff's current aw slares h3r healgbraicsum of aI the cunents lowing out of a unction is

-:J ts ntegml-form quivalentfromEq. 6.51)]s

(Kirchhoff scurrent aw). (6.56)

,:t us examine he meaningof Eq.(6.56)by consideringr.$ction (or node)connecting wo or morebra{ches n. .lectriccircuit. No matterhow small, hejunctionhasaaiume v enclosed y asurface t. Thejunction shown n:: 6-l5hasbeendnwnasacube,anditsdimensionshaveEn artificially enlarged o facilitate thepresent iscus-in. The unctionhassix faces surfaces),whichcollec-r.ely constitutehe surfaceS associated ith the closed,:L1-acentegmtiongivenby Eq.(6.56).For each ace, ted--fmtionreprcsentshecurrent lowingout hrough hatr-.j. Thus,Eq. 6.56) anbe ewritten s

Irr=O (Kirchloff's urrentaw), (6.57).-rre 1i is the curent flowing outward hrough he ithrs. For hejunctionof Fig. 6-15,Eq.(6.57)aanslates

tnto(/r a 4 * 1: = 0. n itr generalorm,F4.(6.57)sanexpression flfr:rcrrtoJft curerl lar, which stateshatitrar electriccircuit t re algebraicsum of all the currentsfowing out of a unction is zerc.

REVIEWUESTIONSQ6.11 Explain ow hechargeontinuity quationeadsto Kirchhoff'scuftent aw.

6-10 ElectromagnelicotenlialsThrough urdiscussionf Faraday'sndAmpere'saws,weexaminedwo aspects fthe nterconnectionhatexistsbetweenheelectdcandmagneticieldswhen he ieldsaretimevalying.Wewill nowexaminehe mplicationsofrhisinterconnection itb regardo theelectdcscalar otentialy and he vectormagneticpotentialA.ForA/at : 0, Fanday's awrcduces o

V x E :0 (elecrrostatics), (6.58)whichstateshat heelectrostaiicieldE is conservarive.According o the ulesofvectorcalculus, fa vector ield Eis conseflative, t can be expressed sthegmdientof ascalar.Hence, n Chapter4 wedefinedE as

E: VY (electrostatics). (6.59)In thedynamiccase,Faraday'saw is givenb/

V x E : - # , (6.60)


23/28

CHAPIER6 MAXWELL'SEQUATIONSFORTIME-VARYINGandn view ofthe rclationa = V x A, Eq.(6.60)becomes When hescalarpotentialV and hevectorareklown, E canb obtained rom Eq.(6.66),andBb obtained tom(6.61)x r = - f t v x . l l ,whichcanbe rewrittenas

o" (e - a^1 ) :o {dynamiccase r .6 .6 ) )\ d r llrt us or themomentdefine

B = v x A . ( 6 . 6 7 )

at the nterfacebetweenwo different mediasameorboth staticaoddynamicconditions.

Next we examine he relations etwern hepolenrialaandA, and heirsources,he charge ndcunenttions vand , n he ime-varyingase.

CHAPTERIGIITIGHTS. Faraday'saw states hat a voltage s inducedthe erminalsof a oop f themagnetic lux lisurface hanges ith time.. Inan dealransformer.theraliosofheprimaryondary oltages.urrents. nd mpedance(reernedby the ums ratio.. DisDlacement urrent accounts or the "flow of charges hrough a dielectric. Inchargesof oppositepolarity accumulate longtwo endsofa dielectric,giving heappearancefrent low through t.

Boundarv onditionsor theelectromaenetic

",=* ,in whichcase q. 6.62) ecomes (6.63)V x E':0. (6.64)

Following tbe same ogic that led to F4. (6.59) fromEq. 6.58),wedefine

(6.65)Upon substitutingF4. (6.63) or E' in Eq.(6.65)and hensolvingorE, wehave

E' = -Vy.

r = -ou $ (dynamicase). (6.66)

The chargecontinuity equation s astatemenl of the law of conservation ofcharge.. In the dynamiccase, heelectric ield E isboth he scalar lectricpotentialy and hevector otential .

Equanon6.66reduceso Eq. 6.59 n thesrrric a5e.


24/28

{CBLEMS 191

EOBTEMS

?:1ion$ -1o6-fi Faraday'sawrr itsApplicationsr: Theswitchn hebottomoopofFig.G 16 s closedr = 0 and hen openedat a later ime tr. What s the=--:tion of the curnt I in the top loop (clockwiseor: 'rterclockwise)at eachof thesewo times?lj The loop in Fig. 6-17 is in the rrl plane andI = i80 sin(,t with Bopositive.Wlat is thedirectionof1

L A coil consists f 100 urnsof wire wrapped round- i rre frameof sides0.25 m. The coil is centered t he. ::i witheach f tssides aralleltoher-orJ]-tu\is.indt iducedemf acrosshe open-circuited ndsof the coil: :e magneticield sgivenby'$swe(s) availabler Appendix .

a iolurion available n CD,ROM.

Figure6-17: -oopof Problem .2.

( a )B = i 1 0 e " ( T )(b) B = i locosr cosl0r,(T)(c) B = i locosr sin2) cos 0rr (T)

6.4 A stationary onducting oop with an ntemat esls-tanceof 0.5 Q is placed n a time-varyingmagnetic ield-When he oop is closed,a currentof 2.5 A flows hroughit. Whatwill the currentbe f the oop is openedo createa smallgapanda 2-Q resistor sconnected crossts openends?6.5* A circularloopTV antenna ith 0.01-m2 rea sin thepresence i a uniform-amplitude30GMHZsignal.Whenoriented or maximum esponse,he oop developsan emf wiih a peakvalueof 20 (mV). What s the pcakmagnitude f B of the ncidentwave?6.6 Thesquareoopshownn Fig.6-18 s coplanar itha ong, straightwirecarrying a curent

i(t) = 2.5cos2r x loal (A)

figure 6-16: l-oops of Prob]em


25/28

CHAPTER6 MAXWELL'S EQUATIONSFORTIME.VARYING

(4,(b)

a

DetermirFheemfinduced cmss smallgapqeatedin the oop.Detrmincbe dirction and magdtudeof the cur-rEnt hatwould low hrougha +Q tcsistorconnectedacrosshegap.The oop hasan ntemal esistancefl a .6,? The rectangularonductingooP shown nFig.6-19 rotatesat 6,000 cvolutlonsprminute na uniformmagnetic lLtxdensitygivenby

B : i50 (nr)Determinehecurrent nducedn the oop l its inter-nal esistances 0.5O.

6.8 A rcctangular onductingoop 5cm x l0 cm with asmallairgap n one of its sides s sPinningat 7200 evo_lutions erminute. fthe fieldB isnormal o the oopaxisandtsmagnitudes 5 x t0 6T, what s rllepeakvoltageinduced crosshe airgap?

F- 10cn -l

TlOcmI

Figure -18: rop coplanarwithongwire Problem.6).

6.9* A 50-cm-longmetal od rotatesabout hezI 80 revolutionsDerminute.with end I fixedat theas shownn Fig. 6-20. Determinehe nducedemf VlrB : i 3 x 10 -aT .

at ltEalF.


26/28

4.OBLEMS 193

r l0 The oop hownnFig.6-21moveswayfromawire::'.! ingacurrent t = 10Aataconstantvelocityu i5:ts).IfR = l0 Q and hedirection f 12 s asdefinedn= figure, ind 12asa unctionof ]0, thedistance etween:r wireand he oop. gnore he n0emalesistancef theroPr ll' TheconductingcylindershowninFig.6-22roratesFoutilsaxisat,200revolutionsperminuteinaradialfild: 'en by

B = i 6 ( T ):\. cylinder,whosemdius s 5 cm ard height s l0 cm,::s slidingcontactsat its top and bottomconnectdo a: Eneter.Deteminethe nducedvoltage.' ll Theelectromagneticenemtor hown n Fig.6-12.onnectedo anelectricbulb with a resistancef 100Q.' :heoop area s0. 1 m2and t rotates t 3,600 evolutions:d minuten a uniformmagnetic luxdensityBo= 0.2 T,-rermine theamplitudeof ihe currentgenemtedn the-:ht bulb.

TOcmIFigure 6-22: Rotatingcylinder n amagnetic ield (Prob-lem6.1l).

6.13* The ircular iskshownn Fig.6-23 ies n he ),planeandrotateswith uniformangularvelocity,) about

Figur6-23: Rotaiing irculardisk n a magneric eld(Problern.13 .

a

Figure6-21:Moving oopof ftoblem6.10.


27/28

194. CHAPTER6 MAXWELL'SEQUATIONSORTIME-VARYING IELUI Pthez-ards. hedisk s ofradiusa andisprcsentn a uniformmagneticlux densityB = i80. Obtainanexpressionortllc cmfinduccd t the im relativc odlcetrGroftlrcdisk

Section-7:0isplacennturrenl6.14 Theplatesof aparallel-plale apacitorhaveareasof l0 cm2eacb ndareseparatedy 1 cm.Thecapacitoris filled wifi a dielectric materialwith s = 4eo,and hevoltageacrossitisgivenby(t) = 20cos2r x l06t(v).Find hedisplacement urrent.6.15t A coaxial apacitor flength : 6 cm uses ninsulatingdielectricmaterialwith , = 9.The radiiofthecylindricalconducto$are0.5 cm and cm. fthe voltageappliedacrosshe capacitor s

Irct l lFa

Y(r): l00sin(l207rr)

what s dledisplaaementunent?

(v)

6.16 Theparallel-plate apacitorshown n Fig. 6-24 s6lled with a ossydielectricmaterialofrelativepermittiv-ity randconductivityd. Theseparationbetweenheplatesis d andeachplate s of area 4.Thecapacitors connected!oa ime-varyingvoltagesource (r).(a) Obtain an expressionor 1", he conductioncufientflowing between he plates nside the capacitor, nte.ms f thegiven uantities.(b) Obtainanexpressionor Id, hedisplacementurrentflowing nside he capacitor.(c) Basedonyourexpressionsor parts a)and b),give

anequivalent-circuiteprcsentationor thecapacitor.

(d) Evaluate he valuesof the circuit elements or A2 cm2, = 0.5 cm, e. = 4, o : 2.5(Sim),Y(.) : l0cos(32 103t)(V).6.17 An electromagneticwavepropagatingn seawr-ter hasan electric fieid with a time variationgivenb1rE = iEo cos rf. If thepermittivity of water s 81roadits conductivity s 4(S/m), ind he ratio of themagnitud.tof the conduction urrentdensity o displacement ufiEdtdensityat eachof the ollowing fiequencies:(a) l kHz(b) I MHz(c) I GHz(d) 100GHz


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PROBLEMS 19s

IIll

I

Seclion-9:Conliruily qualion6.1E* If the curent density n a conductingmedium sEiven y

J(r, ), a; t) : (i. - i3y2 + i2x1as-tllermine the corresponding charge distribution.1(ir, , z; ).

6.1q.21 Additional SolvedProbiemscomDlete so-lutions n l.

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electromagnetics 6

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