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CHAPTER88.1.Apointcharge,Q=−0.3µCandm=3×10−16kg,ismovingthroughthefieldE=30azV/m.UseEq.(1)andNewton’slawstodeveloptheappropriatedifferentialequationsandsolvethem,subjecttotheinitialconditionsatt=0:v=3×105axm/sattheorigin.Att=3µs,find:a)thepositionP(x,y,z)ofthecharge:TheforceonthechargeisgivenbyF=qE,andNewton’ssecondlawbecomes:F=ma=md2zdt2=qE=(−0.3×10−6)(30az)describingmotionofthechargeinthezdirection.Theinitialvelocityinxisconstant,andsonoforceisappliedinthatdirection.Weintegrateonce:dzdt=vz=qEmt+C1Theinitialvelocityalongz,vz(0)iszero,andsoC1=0.Integratingasecondtimeyieldsthezcoordinate:z=qE2mt2+C2Thechargeliesattheoriginatt=0,andsoC2=0.Introducingthegivenvalues,wefindz=(−0.3×10−6)(30)2×3×10−16t2=−1.5×1010t2mAtt=3µs,z=−(1.5×1010)(3×10−6)2=−.135cm.Now,consideringtheinitialconstantvelocityinx,thechargein3µsattainsanxcoordinateofx=vt=(3×105)(3×10−6)=.90m.Insummary,att=3µswehaveP(x,y,z)=(.90,0,−.135).b)thevelocity,v:Afterthefirstintegrationinparta,wefindvz=qEmt=−(3×1010)(3×10−6)=−9×104m/sIncludingtheintialx-directedvelocity,wefinallyobtainv=3×105ax−9×104azm/s.c)thekineticenergyofthecharge:HaveK.E.=12m|v|2=12(3×10−16)(1.13×105)2=1.5×10−5J1388.2.Comparethemagnitudesoftheelectricandmagneticforcesonanelectronthathasattainedavelocityof107m/s.Assumeanelectricfieldintensityof105V/m,andamagneticfluxdensityassociatedwiththatoftheEarth’smagneticfieldintemperatelatitudes,0.5gauss.WeusetheLorentzLaw,F=Fe+Fm=q(E+v×B),where|B|=0.5G=5.0×10−5T.Wefind|Fe|=(1.6×10−19C)(105V/m)=1.6×10−14N|Fm|=(1.6×10−19C)(107m/s)(5.0×10−5T)=8.0×10−17N=0.005|Fe|8.3.ApointchargeforwhichQ=2×10−16Candm=5×10−26kgismovinginthecombinedfieldsE=100ax−200ay+300azV/mandB=−3ax+2ay−azmT.Ifthechargevelocityatt=0isv(0)=(2ax−3ay−4az)×105m/s:a)givetheunitvectorshowingthedirectioninwhichthechargeisacceleratingatt=0:UseF(t=0)=q[E+(v(0)×B)],wherev(0)×B=(2ax−3ay−4az)105×(−3ax+2ay−az)10−3=1100ax+1400ay−500azSotheforceinnewtonsbecomesF(0)=(2×10−16)[(100+1100)ax+(1400−200)ay+(300−500)az]=4×10−14[6ax+6ay−az]TheunitvectorthatgivestheaccelerationdirectionisfoundfromtheforcetobeaF=6ax+6ay−az√73=.70ax+.70ay−.12azb)findthekineticenergyofthechargeatt=0:K.E.=12m|v(0)|2=12(5×10−26kg)(5.39×105m/s)2=7.25×10−15J=7.25fJ8.4.Showthatachargedparticleinauniformmagneticfielddescribesacircularorbitwithanorbitalperiodthatisindependentoftheradius.Findtherelationshipbetweentheangularvelocityandmagneticfluxdensityforanelectron(thecyclotronfrequency).Acircularorbitcanbeestablishedifthemagneticforceontheparticleisbalancedbythecentripitalforceassociatedwiththecircularpath.WeassumeacircularpathofradiusR,inwhichB=B0azisnormaltotheplaneofthepath.Then,withparticleangularvelocityΩ,thevelocityisv=RΩaφ.ThemagneticforceisthenFm=qv×B=qRΩaφ×B0az=qRΩB0aρ.Thisforcewillbenegative(pullingtheparticletowardthecenterofthepath)ifthechargeispositiveandmotionisinthe−aφdirection,orifthechargeisnegative,andmotionisinpositiveaφ.Ineithercase,thecentripitalforcemustcounteractthemagneticforce.Assumingparticlemassm,theforcebalanceequationisqRΩB0=mΩ2R,fromwhichΩ=qB0/m.TherevolutionperiodisT=2π/Ω=2πm/(qB0),whichisindependentofR.Foranelectron,wehaveq=1.6×10−9C,andm=9.1×1031kg.ThecyclotronfrequencyisthereforeΩc=qmB0=1.76×1011B0s−11398.5.ArectangularloopofwireinfreespacejoinspointsA(1,0,1)toB(3,0,1)toC(3,0,4)toD(1,0,4)toA.Thewirecarriesacurrentof6mA,flowingintheazdirectionfromBtoC.Afilamentarycurrentof15Aflowsalongtheentirezaxisintheazdirection.a)FindFonsideBC:FBC=ZCBIloopdL×BfromwireatBCThusFBC=Z41(6×10−3)dzaz×15µ02π(3)ay=−1.8×10−8axN=−18axnNb)FindFonsideAB:Thefieldfromthelongwirenowvarieswithpositionalongtheloopsegment.WeincludethatdependenceandwriteFAB=Z31(6×10−3)dxax×15µ02πxay=45×10−3πµ0ln3az=19.8aznNc)FindFtotalontheloop:Thiswillbethevectorsumoftheforcesonthefoursides.Notethatbysymmetry,theforcesonsidesABandCDwillbeequalandopposite,andsowillcancel.ThisleavesthesumofforcesonsidesBC(parta)andDA,whereFDA=Z41−(6×10−3)dzaz×15µ02π(1)ay=54axnNThetotalforceisthenFtotal=FDA+FBC=(54−18)ax=36axnN8.6.ShowthatthedifferentialworkinmovingacurrentelementIdLthroughadistancedlinamagneticfieldBisthenegativeofthatdoneinmovingtheelementIdlthroughadistancedLinthesamefield:Thetwodifferentialworkquantitiesarewrittenas:dW=(IdL×B)·dlanddW0=(Idl×B)·dLWenowapplythevectoridentity,Eq.(A.6),AppendixA:(A×B)·C=(B×C)·A,andwrite:(IdL×B)·dl=(B×dl)·IdL=−(Idl×B)·dLQED8.7.Uniformcurrentsheetsarelocatedinfreespaceasfollows:8azA/maty=0,−4azA/maty=1,and−4azA/maty=−1.Findthevectorforcepermeterlengthexertedonacurrentfilamentcarrying7mAintheaLdirectionifthefilamentislocatedat:a)x=0,y=0.5,andaL=az:Wefirstnotethatwithintheregion−1y1,themagneticfieldsfromthetwooutersheets(carrying−4azA/m)cancel,leavingonlythefieldfromthecentersheet.Therefore,H=−4axA/m(0y1)andH=4axA/m(−1y0).Outside(y1andy−1)thefieldsfromallthreesheetscancel,leavingH=0(y1,y−1).Soatx=0,y=.5,theforcepermeterlengthwillbeF/m=Iaz×B=(7×10−3)az×−4µ0ax=−35.2aynN/mb.)y=0.5,z=0,andaL=ax:F/m=Iax×−4µ0ax=0.c)x=0,y=1.5,aL=az:Sincey=1.5,weareintheregioninwhichB=0,andsotheforceiszero.1408.8.Twoconductingstrips,havinginfinitelengthinthezdir