工程电磁场第八版课后答案第09章.pdf

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1、CHAPTER99.1.InFig.9.4,letB=0.2cos120⇡tT,andassumethattheconductorjoiningthetwoendsoftheresistorisperfect.ItmaybeassumedthatthemagneticfieldproducedbyI(t)isnegligible.Find:a)V(t):SinceBisconstantoverthelooparea,thefluxis=⇡(0.15)2B=1.41⇥ab102cos120⇡tWb.Now,emf=V(t)=d/dt=(120⇡)(1

2、.41⇥102)sin120⇡t.baThenVab(t)=Vba(t)=5.33sin120⇡tV.b)I(t)=Vba(t)/R=5.33sin(120⇡t)/250=21.3sin(120⇡t)mA9.2.IntheexampledescribedbyFig.9.1,replacetheconstantmagneticfluxdensitybythetime-varyingquantityB=B0sin!taz.Assumethatvisconstantandthatthedisplacementyofthebariszeroatt=0.Fi

3、ndtheemfatanytime,t.ThemagneticfluxthroughtheloopareaisZZvtZdm=B·dS=B0sin!t(az·az)dxdy=B0vtdsin!ts00ThentheemfisIdmemf=E·dL==B0dv[sin!t+!tcos!t]Vdt9.3.GivenH=300acos(3⇥108ty)A/minfreespace,findtheemfdevelopedinthegeneralzadirectionabouttheclosedpathhavingcornersata)(0,0,0),(

4、1,0,0),(1,1,0),and(0,1,0):Themagneticfluxwillbe:Z1Z1=300µcos(3⇥108ty)dxdy=300µsin(3⇥108ty)

5、100000⇥⇤=300µsin(3⇥108t1)sin(3⇥108t)Wb0Thend⇥⇤emf==300(3⇥108)(4⇡⇥107)cos(3⇥108t1)cos(3⇥108t)dt⇥⇤=1.13⇥105cos(3⇥108t1)cos(3⇥108t)Vb)cornersat(0,0,0),(2⇡,0,0),(2⇡,2⇡,0),(0,2⇡,0)

6、:Inthiscase,thefluxis=2⇡⇥300µsin(3⇥108ty)

7、2⇡=000Theemfistherefore0.1649.4.Arectangularloopofwirecontainingahigh-resistancevoltmeterhascornersinitiallyat(a/2,b/2,0),(a/2,b/2,0),(a/2,b/2,0),and(a/2,b/2,0).Theloopbeginstorotateaboutthexaxisatconstantangularvelocity!,withthefirs

8、t-namedcornermovingintheazdirectionatt=0.AssumeauniformmagneticfluxdensityB=B0az.Determinetheinducedemfintherotatingloopandspecifythedirectionofthecurrent.Themagneticfluxthoughtheloopisfound(asusual)throughZm=B·dS,whereS=ndasBecausetheloopisrotating,thedirectionofthenormal,n,chan

9、ging,andisinthiscasegivenbyn=cos!tazsin!tayTherefore,Zb/2Za/2m=B0az·(cos!tazsin!tay)dxdy=abB0cos!tb/2a/2Theintegralistakenovertheentirelooparea(regardlessofitsimmediateorientation).TheimportantresultisthatthecomponentofBthatisnormaltotheloopareaisvaryingsinusoidally,andsoit

10、isfinetothinkoftheBfielditselfrotatingaboutthexax