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Ch29HW
Due:11:59pmonFriday,November25,2016
Youwillreceivenocreditforitemsyoucompleteaftertheassignmentisdue.GradingPolicy

TheAmpreMaxwellLaw
LearningGoal:
ToshowthatdisplacementcurrentisnecessarytomakeAmpre'slawconsistentforachargingcapacitor
Ampre'slawrelatesthelineintegralofthemagneticfieldaroundaclosedlooptothetotalcurrentpassingthroughthatloop.Thislawwasextendedby
Maxwelltoincludeanewtypeof"current"thatisduetochangingelectricfields:

B dl = (Icharge + Idisplacement )
0

Thefirsttermontherighthandside,Icharge ,describestheeffectsoftheusualelectriccurrentduetomovingcharge.Inthisproblem,thatcurrentis
designatedI (t)asusual.Thesecondterm,Idisplacement

= 0

d E
dt

,iscalledthedisplacementcurrentitwasrecognizedasnecessarybyMaxwell.His

motivationwaslargelytomakeAmpre'slawsymmetricwithFaraday'slawofinductionwhentheelectricfieldsandmagneticfieldsarereversed.Bycalling
fortheproductionofamagneticfieldduetoachangeinelectricfield,thislawlaysthegroundworkforelectromagneticwavesinwhichachangingmagnetic
fieldgeneratesanelectricfieldwhosechange,inturn,sustainsthemagneticfield.Wewilldiscusstheseissueslater.(Incidentally,athirdtypeof"current,"
calledmagnetizingcurrent,shouldalsobeaddedtoaccountforthepresenceofchangingmagneticmaterials,butitwillbeneglected,asithasbeeninthe
equationabove.)
ThepurposeofthisproblemistoconsideraclassicillustrationoftheneedfortheadditionaldisplacementcurrentterminAmpre'slaw.Considertheproblem
offindingthemagneticfieldthatloopsaroundjustoutsidethecircularplateofachargingcapacitor.Theconeshapedsurfaceshowninthefigurehasa
currentI (t)passingthroughit,soAmpre'slawindicatesafinitevalueforthefieldintegralaroundthisloop.However,aslightlydifferentsurfaceborderedby
thesamelooppassesthroughthecenterofthecapacitor,wherethereisnocurrentduetomovingcharge.Togetthesameloopintegralindependentofthe
surfaceitmustbetruethateitheracurrentoranincreasingelectricfieldthatpassesthroughtheAmpreansurfacewillgeneratealoopingmagneticfield
arounditsedge.Theobjectiveofthisexampleistointroducethedisplacementcurrent,showhowtocalculateit,andthentoshowthatthedisplacement
currentIdisplacement (t)isidenticaltotheconductioncurrentIcharge (t).AssumethatthecapacitorhasplateareaAandanelectricfieldE(t)betweenthe
plates.Take0 tobethepermeabilityoffreespaceand0 tobethepermittivityoffreespace.

PartA

Firstfind R B dl ,thelineintegralof
Baroundaloopofradius Rlocatedjustoutsidetheleftcapacitorplate.Thiscanbefoundfromtheusualcurrent
duetomovingchargeinAmpre'slaw,thatis,withoutthedisplacementcurrent.

FindanexpressionforthisintegralinvolvingthecurrentI (t)andanyneededconstantsgivenintheintroduction.
ANSWER:

B dl

0 I (t)

Correct

PartB

Nowfindanexpressionfor R B dl ,thesamelineintegralof
Baroundthesameloopofradius Rlocatedjustoutsidetheleftcapacitorplateasbefore.
Usethesurfacethatpassesbetweentheplatesofthecapacitor,wherethereisnoconductioncurrent.Thisshouldbefoundbyevaluatingtheamountof
displacementcurrentintheAmpreMaxwelllawabove.

ExpressyouranswerintermsoftheelectricfieldbetweentheplatesE(t),dE(t)/dt,the
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ExpressyouranswerintermsoftheelectricfieldbetweentheplatesE(t),dE(t)/dt,the
plateareaA,andanyneededconstantsgivenintheintroduction.

Hint1.Findtheelectricflux
WhatistheelectricfluxE (t)throughthissurface?
ExpressyouranswerintermsofE(t)andothervariablesgivenintheintroduction.

Hint1.Thedefinitionofelectricflux
TheelectricfluxE throughasurfaceisdefinedby

E = E dA,

wheretheintegralistakenovertheentiresurface.
ANSWER:
E (t)

AE(t)

Hint2.ExpressR B dl intermsof
E (t)

Find R B dl intermsoftheelectricflux
E (t)betweentheplates.

ExpressyouranswerintermsofE (t),dE (t)/dt ,andanyneededconstantsgivenintheintroduction.

Hint1.Howtoapproachtheproblem
UsetheextensiontoAmpre'slawandthedefinitionofdisplacementcurrentgivenintheproblemintroduction.
ANSWER:

B dl

0 0 d E (t)
dt

ANSWER:

B dl

0 A 0

dE(t)
dt

Correct

Anecessaryconsistencycheck
PartC

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Wenowhavetwoquitedifferentexpressionsforthelineintegralofthemagneticfieldaroundthesameloop.Thepointhereistoseethattheybothare
intimatelyrelatedtothechargeq(t)ontheleftcapacitorplate.FirstfindthedisplacementcurrentIdisplacement (t)intermsofq(t).
Expressyouranswerintermsofq(t),dq(t)/dt ,andanyneededconstantsgivenintheintroduction.

Hint1.FindthefluxusingGauss'slaw
AssignthecapacitoranarbitraryplateareaA.IfweapplyGauss'slawforelectricfluxtothetheleftplateofthecapacitor,wefindthat

E dA =

qencl
0

,wheretheelectricfieldisentirelyinsidethecapacitor.WhatistheelectricfluxE (t)?

Expressyouranswerintermsofq(t)andanyneededconstantsgivenintheintroduction.

Hint1.Gauss'slaw
Gauss'slawstatesthattheelectricfluxthroughaclosedsurfaceequalstheenclosedchargedividedby0 .
ANSWER:

E (t)

q(t)
0

Hint2.Findthedisplacementcurrent
WhatisthedisplacementcurrentIdisplacement (t)?
ExpressyouranswerintermsofE (t),dE (t)/dt ,andanyneededconstantsgivenintheintroduction.
ANSWER:

Idisplacement (t)

0 d E (t)
dt

ANSWER:

Idisplacement (t)

dq(t)
dt

Correct

PartD
NowexpressthenormalcurrentIcharge (t)intermsofthechargeonthecapacitorplateq(t).
Expressyouranswerintermsofq(t),dq(t)/dt ,andanyneededconstantsgivenintheintroduction.
ANSWER:

Icharge (t)

dq(t)
dt

Correct
UsingGauss'slaw,youhaveshownthatthedisplacementcurrentfromthechangingelectricfieldbetweentheplatesequalsthecurrentfromtheflow
ofchargethroughthewireontothatplate.ThismeansthattheAmpreMaxwelllawcanconsistentlytreatcasesinwhichthenormalcurrentdueto
theflowofchargeisnotcontinuous.ThisrealizationwasagreatboosttoMaxwell'sconfidenceinthephysicalvalidityofhisnewdisplacement
currentterm.

TheMagneticFieldinaChargingCapacitor
Whenacapacitorischarged,theelectricfieldE ,andhencetheelectricflux,betweentheplateschanges.Thischangeinfluxinducesamagneticfield,
accordingtoAmpre'slawasextendedbyMaxwell:

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B dl = (I + 0
0

d
dt

).

Youwillcalculatethismagneticfieldinthespacebetweencapacitorplates,wheretheelectricfluxchangesbuttheconductioncurrentI iszero.

PartA
AparallelplatecapacitorofcapacitanceC withcircularplatesischargedbyaconstantcurrentI .Theradiusaoftheplatesismuchlargerthanthe
distancedbetweenthem,sofringingeffectsarenegligible.CalculateB(r),themagnitudeofthemagneticfieldinsidethecapacitorasafunctionof
distancefromtheaxisjoiningthecenterpointsofthecircularplates.
Expressyouranswerintermsof0 andgivenquantities.

Hint1.Howtoapproachtheproblem
TheAmpreMaxwelllawallowsyoutofindthemagneticfieldincasesofhighsymmetry.Suchacaseisgiveninthisproblem.Youshould
convinceyourselfthatthemagneticfieldvectoriseverywheretangentialtocirclesaroundanimaginedaxisjoiningthecentersofthetwocircular
plates.Hencewecanconvenientlyintegratearoundsuchacircle.Wecalltheradiusofthiscircler,andrealizethatonlythechangeofflux
throughtheareawithinthecirclewillcontributetothemagnitudeofthemagneticfieldvectorBonitsperimeter.Butthechangeofelectricfluxwill
bethesameeverywherebetweentheplates.Therefore,allweneedtodoistocalculatethechangeoffluxacrossallthespacebetweenthe
capacitorplatesandthenmultiplybyascalingfactor.

Hint2.Whatistheelectricfluxbetweentheplates?
Expresstheelectricfluxtot
chargeqoneachplate.

= a E

betweentheplatesofthecapacitorintermsoftheparametersoftheproblemandthemagnitudeofthe

Expressyouranswerintermsof0 andgivenquantities.
ANSWER:
tot

Hint3.Whatistherateofchangeoftheelectricflux?
Whatistherateofchangeofelectricflux

d tot
dt

betweenthecapacitorplates?Recallthat

dq
dt

=I .

ANSWER:
d tot
dt

I
0

Hint4.ScalingfortheareawithintheAmprianloop
ThechangeoffluxthroughtheAmprianloopofradiusr,L ,isafractionofthechangeoffluxacrossthewholeareaofthecapacitor,tot .Find
thescalingfactorf

L
tot

,whichreducestotheratiooftheareas.

ANSWER:

a2

Hint5.Evaluatetheintegral

SincewetakethescalarproductofBwith dl ,theonlycomponentofthemagneticfieldthatmattersisB ,whichisconstantaroundtheloop.

B dl

evaluatedabouttheAmperianloopdefinedinthefirsthint?Rememberthatthemagneticfieldiseverywheretangent

Expressyouranswerintermsof,theradiusr,andB(r).
ANSWER:

B dl

B(r)2r

ANSWER:

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B(r)

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=

0 I

r
2a2

Correct

MagneticFluxandInducedEMFinaCoil
Inaphysicslaboratoryexperiment,acoilwith250turnsenclosinganareaof12.7cm2 isrotatedduringthetimeinterval4.00102sfromapositioninwhich
itsplaneisperpendiculartoEarth'smagneticfieldtooneinwhichitsplaneisparalleltothefield.ThemagnitudeofEarth'smagneticfieldatthelablocationis
5.10105T.

PartA
Whatisthetotalmagnitudeofthemagneticflux(initial )throughthecoilbeforeitisrotated?
Expressyouranswernumerically,inwebers,toatleastthreesignificantfigures.
Youdidnotopenhintsforthispart.
ANSWER:
|initial |

1.2510

Wb

IncorrectTryAgain10attemptsremaining

PartB
Whatisthemagnitudeofthetotalmagneticfluxf inal throughthecoilafteritisrotated?
Expressyouranswernumerically,inwebers,toatleastthreesignificantfigures.

Hint1.Theanglebetweenthemagneticfieldandtheareavector
Theanglebetweenthemagneticfieldandtheareavectorafterrotationis90degrees,sincetheplaneofthecoilisparalleltothemagneticfield
andtheareavectorvectorisnormaltotheplaneofthecoil.
ANSWER:
|f inal |

= 0 Wb

Correct

PartC
Whatisthemagnitudeoftheaverageemfinducedinthecoil?
Expressyouranswernumerically(involts)toatleastthreesignificantfigures.

Hint1.Formulafortheaverageemfinducedinacoil(Faraday'slaw)
TheformulafortheaverageemfE inducedinacoilis
E = N

d 1
dt

= N

1
t

where1 isthetotalchangeinmagneticfluxthrougheachloop,N isthetotalnumberofloops,andtisthetimeintervaloverwhichthis

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averageinducedemf= 4.05104 V

Correct

UnderstandingChangingFlux
Inthisproblem,youwilluseLenz'slawtoexplorewhathappenswhenanelectromagnetisactivatedashortdistancefromawireloop.
Youwillneedtousetherighthandruletofindthedirectionoftheinducedcurrent.

Considerthearrangementshownin.

PartA
Whentheswitchisopen,whichofthefollowingstatementsaboutthemagneticfluxthroughthewireloopistrue?Assumethatthedirectionofthevector
areaofthewireloopistotheright.
ANSWER:
Thereisnomagneticfluxthroughthewireloop.
Thereisapositivefluxthroughthewireloop.
Thereisanegativefluxthroughthewireloop.

Correct
Whentheswitchisopen,thereisnocurrentflowinginthecircuit.Thus,theelectromagnetdoesnotproduceamagneticfield,andthefluxiszero.

PartB
Whatisthedirectionoftheinducedcurrentinthewireloop(asseenfromtheleft)whentheswitchisopen?
ANSWER:
Thereisnoinducedcurrent.
Theinducedcurrentisclockwise.
Theinducedcurrentiscounterclockwise.

Correct

PartC
Nowtheswitchontheelectromagnetisclosed.Whatisthedirectionoftheinducedcurrentinthewireloopimmediatelyaftertheswitchisclosed(as
seenfromtheleft)?

Hint1.Howtoapproachtheproblem
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Whentheswitchisopen,themagneticfieldproducedbytheelectromagnetiszero.Whentheswitchisclosed,acurrentflowsthroughthecoilof
theelectromagnet,andamagneticfieldisproduced.Determinehowthisfieldchangesthefluxthroughthewireloop.ThenuseLenz'slawto
determinethedirectionoftheinducedcurrent.

Hint2.Thefieldproducedbytheelectromagnet
Thecurrentflowsfromthepositivebatteryterminaltothenegativeone.Therefore,whentheswitchisclosed,thecurrentintheelectromagnetis
clockwiseasseenfromtheleft.Thefigureshowssomeofthemagneticfieldlinesproducedbytheelectromagnet.

Hint3.ApplyLenz'slaw
Lenz'slawstatesthatthecurrentinducedinthewireloophasthedirectionsuchthatthemagneticfluxcreatedbythecurrentopposesthechange
intheexternalmagneticfluxthatcausedthecurrent.
Usetherighthandruletofindthedirectionoftheinducedcurrentthatwouldgenerateafieldopposingthechangeinflux.
ANSWER:
Thereisnoinducedcurrent.
Theinducedcurrentisclockwise.
Theinducedcurrentiscounterclockwise.

Correct

PartD
Finally,theswitchontheelectromagnetisreopened.Themagnitudeoftheexternalmagneticfluxthroughthewireloop______(A.increases,B.
decreases,C.remainsconstant),andthereis_______(A.zero,B.aclockwise,C.acounterclockwise)currentinducedintheloop(asseenfromthe
left).
Entertheletterscorrespondingtotheresponsesthatcorrectlycompletethestatementabove.Forexample,ifthecorrectanswersareAandC,
typeA,C

Hint1.Howtoapproachtheproblem
Whentheswitchisclosed,thereisanonzeromagneticfieldproducedbytheelectromagnet.Whentheswitchisopened,currentstopsflowing
throughtheelectromagnet,andthemagneticfielddisappears.DeterminehowthischangesthefluxthroughthewireloopanduseLenz'slawto
determinethedirectionoftheinducedcurrent.

Hint2.Thefieldproducedbytheelectromagnet
Thefiguresshowthemagneticfieldproducedbytheelectromagnetwhentheswitchisclosedandthenthefieldwhentheswitchisopen.

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Hint3.ApplyLenz'slaw
Lenz'slawstatesthatthecurrentinducedinthewireloophasthedirectionsuchthatthemagneticfluxcreatedbythecurrentopposesthechange
intheexternalmagneticfluxthatcausedthecurrent.
Usetherighthandruletofindthedirectionoftheinducedcurrentthatwouldgenerateafieldopposingthechangeinflux.
ANSWER:
B,B

Correct

Nowconsiderthenewarrangementshownin.Notethattheorientationofthebatteryisreversed
withrespecttothefirstarrangementyouconsidered.Answerthefollowingquestionsrelatedtothe
arrangmentwiththenewbatteryorientation.

PartE
Theswitchontheelectromagnet,initiallyopen,isclosed.Whatisthedirectionoftheinducedcurrentinthewireloop(asseenfromtheleft)?

Hint1.Howtoapproachtheproblem
Whentheswitchisopen,themagneticfieldproducedbytheelectromagnetiszero.Whentheswitchisclosed,acurrentflowsthroughthe
electromagnet,andamagneticfieldisproduced.DeterminehowthisfieldchangesthefluxthroughthewireloopanduseLenz'slawtodetermine
thedirectionoftheinducedcurrent.

Hint2.Thefieldproducedbytheelectromagnet
Thecurrentflowsfromthepositivebatteryterminaltothenegativeone.Therefore,whentheswitchcloses,thecurrentintheelectromagnetis
counterclockwiseasseenfromtheleft.Thefigureshowssomeofthemagneticfieldlinesproducedbytheelectromagnet(recallthatthebattery
hastheoppositeorientationtothatofthefirstarrangement).

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Hint3.ApplyLenz'slaw
Lenz'slawstatesthatthecurrentinducedinthewireloophasthedirectionsuchthatthemagneticfluxcreatedbythecurrentopposesthechange
intheexternalmagneticfluxthatcausedthecurrent.
Usetherighthandruletofindthedirectionoftheinducedcurrentthatwouldgenerateafieldopposingthechangeinflux.
ANSWER:
Thereisnoinducedcurrent.
Theinducedcurrentisclockwise.
Theinducedcurrentiscounterclockwise.

Correct

PartF
Nowtheswitchontheelectromagnetisreopened.Themagnitudeoftheexternalmagneticfluxthroughthewireloop______(A.increases,B.decreases,
C.remainsconstant),andthereis_______(A.zero,B.aclockwise,C.acounterclockwise)currentinducedintheloop(asseenfromtheleft.
Entertheletterscorrespondingtotheresponsesthatcorrectlycompletethestatementabove.Forexample:A,C

Hint1.Howtoapproachtheproblem
Whentheswitchisclosed,thereisanonzeromagneticfieldproducedbytheelectromagnet.Whentheswitchisopened,currentstopsflowing
throughtheelectromagnet,andthemagneticfielddisappears.Determinehowthischangesthefluxthroughthewireloop,anduseLenz'slawto
determinethedirectionoftheinducedcurrent.

Hint2.Thefieldproducedbytheelectromagnet
Thefiguresshowthemagneticfieldproducedbytheelectromagnetwhentheswitchisclosedandthenwhentheswitchisopen.

Hint3.ApplyLenz'slaw
Lenz'slawstatesthatthecurrentinducedinthewireloophasthedirectionsuchthatthemagneticfluxcreatedbythecurrentopposesthechange
intheexternalmagneticfluxthatcausedthecurrent.
Usetherighthandruletofindthedirectionoftheinducedcurrentthatwouldgenerateafieldopposingthechangeinflux.

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ANSWER:
B,C

Correct

VideoTutor:EddyCurrentsinDifferentMetals
First,launchthevideobelow.Youwillbeaskedtouseyourknowledgeofphysicstopredicttheoutcomeofanexperiment.Then,closethevideowindowand
answerthequestionatright.Youcanwatchthevideoagainatanypoint.

PartA
Electricrailcarsoftenusemagneticbraking.Thebrakeconsistsofasetofelectromagnetsthatareheldjustabovetherails.Tobrakethetrain,the
electromagnetsareswitchedon,creatingamagneticfieldthatinduceseddycurrentsinthemetalrailspassingbeneaththem.
Inthefigure,whichofthechoicescorrectlyrepresentstheeddycurrentsinducedintherails?Thediagramsrepresentaviewfromabove,lookingdownat
therailthroughtheelectromagnet.Theelectromagnetmovestotheright,andthemagneticfieldpointsintothescreen.

Hint1.Howtoapproachtheproblem.
ThisproblemasksyoutouseLenzslaw.
First,thinkaboutthemagneticfluxthroughagivenportionontherailastheelectromagnetpassesoverthatportion.Isthefluxthesameunderthe
leadingandtrailingedgesoftheelectromagnet?Arethereplaceswherethefluxisincreasingordecreasing?
Theeddycurrentsinducedintherailbythechangingexternalfluxwillthemselvesinduceamagneticfield.AccordingtoLenzslaw,thisinduced
magneticfieldpointsinthedirectionthatopposesthechangeintheexternalmagneticflux.
Whichwaymusttheinducedmagneticfieldpointinthesegmentsofrailundertheleadingandtrailingedgesofthemovingelectromagnet?
TherighthandrulewilltellyouthedirectionagiveneddycurrentmusthavetoinducethemagneticfieldpredictedbyLenzslaw.
ANSWER:

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A
B
C
D

Correct
Themagneticfluxincreasesundertheleadingedgeoftheelectromagnetanddecreasesunderitstrailingedge.Therefore,byLenzslaw,theinduced
magneticfieldwillpointoutofthescreenundertheleadingedgeandintoitunderthetrailingedge.TheeddycurrentsinchoiceBhavetheright
directionstoinducesuchmagneticfields,accordingtotherighthandrule.

Exercise29.8
Aflat,circular,steelloopofradius75cmisatrestinauniformmagneticfield,asshowninanedgeonviewinthefigure.Thefieldischangingwithtime,
accordingtoB(t)

(0.057s

= (1.4T)e

)t

PartA
Findtheemfinducedintheloopasafunctionoftime(assumetisinseconds).
Expressyouranswerintermsofthevariablet.
ANSWER:
E

0.057t

0.122e

Correct

PartB
Whenistheinducedemfequalto

1
20

ofitsinitialvalue?

ANSWER:
t

= 52.6 s

Correct

PartC
Findthedirectionofthecurrentinducedintheloop,asviewedfromabovetheloop.
ANSWER:

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Counterclockwise
Clockwise

Correct

Exercise29.28
A1.05mlongmetalbarispulledtotherightatasteady4.3m/sperpendiculartoauniform,0.745Tmagneticfield.Thebarridesonparallelmetalrails
connectedthroughR=24.8,asshowninthefigure,sotheapparatusmakesacompletecircuit.
Youcanignoretheresistanceofthebarandtherails.

PartA
Calculatethemagnitudeoftheemfinducedinthecircuit.
Expressyouranswerusingtwosignificantfigures.
ANSWER:
E

= 3.4 V

Correct

PartB
Findthedirectionofthecurrentinducedinthecircuit.
ANSWER:
clockwise
counterclockwise

Correct

PartC
Calculatethecurrentthroughtheresistor.
Expressyouranswerusingtwosignificantfigures.
ANSWER:
I

= 0.14 A

Correct
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Exercise29.36
Along,thinsolenoidhas930turnspermeterandradius2.60cm.Thecurrentinthesolenoidisincreasingatauniformrateof60.0A/s.

PartA
Whatisthemagnitudeoftheinducedelectricfieldatapoint0.470cmfromtheaxisofthesolenoid?
ANSWER:
E1

= 1.65104 V/m

Correct

PartB
Whatisthemagnitudeoftheinducedelectricfieldatapoint1.00cmfromtheaxisofthesolenoid?
ANSWER:
E2

= 3.51104 V/m

Correct

Exercise29.42
Aparallelplateairfilledcapacitorisbeingchargedasinthefigure.Thecircularplateshaveradius4.03
cm,andataparticularinstant,theconductioncurrentinthewiresis0.278 A .

PartA
WhatisthedisplacementcurrentdensityjD intheairspacebetweentheplates?
ANSWER:
j

= 54.5 A/m2

Correct

PartB
Whatistherateatwhichtheelectricfieldbetweentheplatesischanging?
ANSWER:
dE
dt

= 6.161012 V/m s

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Correct

PartC
Whatistheinducedmagneticfieldbetweentheplatesatadistanceof1.91cmfromtheaxis?
ANSWER:
B1

= 6.54107 T

Correct

PartD
Whatistheinducedmagneticfieldbetweentheplatesatadistanceof1.06cmfromtheaxis?
ANSWER:
B2

= 3.63107 T

Correct

Problem29.69
ArectangularloopwithwidthLandaslidewirewithmassmareasshowninthefigure.Auniform
magneticfieldBisdirectedperpendiculartotheplaneoftheloopintotheplaneofthefigure.The
slidewireisgivenaninitialspeedofv 0 andthenreleased.Thereisnofrictionbetweentheslidewire
andtheloop,andtheresistanceoftheloopisnegligibleincomparisontotheresistanceRofthe
slidewire.

PartA
ObtainanexpressionforF ,themagnitudeoftheforceexertedonthewirewhileitismovingatspeedv .
ExpressyouranswerintermsofthevariablesB,v ,R,L,andm.
ANSWER:

B L

2 v
R

Correct

PartB
Findthedistancexthatthewiremovesbeforecomingtorest.
ExpressyouranswerintermsofthevariablesB,v 0 ,R,L,andm.
ANSWER:

mRv0
2

B L

Correct

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Exercise29.35
Arectangularcircuitismovedataconstantvelocityof3.00m/sinto,through,andthenoutofauniform1.25Tmagneticfield,asshowninthefigure.The
magneticfieldregionisconsiderablywiderthan50.0cm.

PartA
Findthedirection(clockwiseorcounterclockwise)ofthecurrentinducedinthecircuitasitisgoingintothemagneticfield(thefirstcase),totallywithin
themagneticfieldbutstillmoving(thesecondcase),andmovingoutofthefield(thethirdcase).
ANSWER:
Thecurrentiszerointhefirstcase,counterclockwiseinthesecondcaseandclockwiseinthethirdcase.
Thecurrentiscounterclockwiseinthefirstcase,zerointhesecondcaseandclockwiseinthethirdcase.
Thecurrentisclockwiseinthefirstcase,counterclockwiseinthesecondcaseandzerointhethirdcase.
Thecurrentisclockwiseinthefirstcase,zerointhesecondcaseandcounterclockwiseinthethirdcase.
Thecurrentiscounterclockwiseinthefirstcase,clockwiseinthesecondcaseandzerointhethirdcase.

Correct

PartB
Findthemagnitudeofthecurrentinducedinthecircuitasitisgoingintothemagneticfield.
ANSWER:
I1

= 0.225 A

Correct

PartC
Findthemagnitudeofthecurrentinducedinthecircuitasitistotallywithinthemagneticfieldbutstillmoving.
ANSWER:
I2

= 0 A

Correct

PartD
Findthemagnitudeofthecurrentinducedinthecircuitasitismovingoutofthefield.
ANSWER:

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= 0.225 A

Correct

ASimpleWaytoMeasureMagneticFields
AloopofwireisattheedgeofaregionofspacecontainingauniformmagneticfieldB.Theplaneoftheloopisperpendiculartothemagneticfield.Nowthe
loopispulledoutofthisregioninsuchawaythattheareaAofthecoilinsidethemagneticfieldregionisdecreasingattheconstantratec.Thatis,
dA
dt

= c,with c > 0 .

PartA
TheinducedemfintheloopismeasuredtobeV .WhatisthemagnitudeBofthemagneticfieldthattheloopwasin?
ExpressyouranswerintermsofsomeorallofthevariablesA,c,andV .

Hint1.Theformulaforthemagneticfluxthroughaloop
Theformulaforthemagneticfluxthroughawireloopis

= B A = |B||A | cos()

where isthemagneticfield,Aistheareavectorassociatedwiththeloop,andistheanglebetweenthemagneticfieldandtheareavector.
Notethattheareavectorisnormaltotheplaneofthewireloopandcanbechosentobeparallelorantiparalleltothemagneticfieldinthiscase.
Thisambiguityastowhichareavectortochoosedoesnotaffecttheproblemaslongasthevectorchosenstaysthesamethroughoutthe
problem.

Hint2.Howtotakethederivativeoftheproductoftwofunctions
Leta(x)andb(x)betwofunctionsofx.Whichofthefollowinggivesthepropervaluefor
d
dx

[a(x)b(x)] ?

ANSWER:
d(ab)
dx

d(ab)
dx

d(ab)
dx

d(ab)
dx

da
dx

db

=a

dx

da
dx
1
b

b+a

da
dx

db

dx

ba

db
dx

Hint3.Theformulafortheemfinducedinaloop(Faraday'slaw)
TheformulafortheemfE inducedinawireloop(Faraday'slaw)is
E=

d
dt

whereisthemagneticfluxthroughthewireloop.
ANSWER:
B

V
c

Correct
Soyouseethatingeneral,therecanbecontributionstotheinducedemfinawireloopbothfromachangingmagneticfieldthroughtheloop(about
whichyoumayhavestudiedearlier)andfromthechangeintheareaoftheloop(withinthemagneticfieldregion),asinthisproblem.

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PartB
ForthecaseofasquareloopofsidelengthLbeingpulledoutofthemagneticfieldwithconstantspeedv (seethefigure),whatistherateofchangeof
areac =

dA
dt

ExpressyouranswerintermsofLandv .

Hint1.Howtoapproachtheproblem
Thinkaboutthefollowingquestions:
Whatistheshapeoftheregionofthecoilinthemagneticfield?
Whatistheformulafortheareaofsuchashape?
Forthepartofthecoilthatisinthemagneticfield,isthelengthchanging?Isthewidthchanging?
ANSWER:
c

Lv

Correct
Later,youwilllearn,ifyouhavenotalready,thatthe"motionalemf"E associatedwitharodoflengthLmovingthroughauniformmagneticfieldof
magnitudeBwithspeedv isgivenby
E = vLB

or,equivalently,
B=

E
vL

Thisisanotherwayofthinkingabouttheresultderivedabove.Ifyouhavealreadystudiedthis,canyouseewhichsidesofthesquareloop
contributetothemotionalemfandwhichdonot,andwhy?

Exercise29.5
Acircularloopofwirewitharadiusof13.0cmandorientedinthehorizontalxyplaneislocatedinaregionofuniformmagneticfield.Afieldof1.5Tis
directedalongthepositivezdirection,whichisupward.

PartA
Iftheloopisremovedfromthefieldregioninatimeintervalof2.7ms,findtheaverageemfthatwillbeinducedinthewireloopduringtheextraction
process.
Expressyouranswerusingtwosignificantfigures.
ANSWER:
E

= 29 V

Correct

PartB
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Ifthecoilisviewedlookingdownonitfromabove,istheinducedcurrentintheloopclockwiseorcounterclockwise?
ANSWER:
Clockwise
Counterclockwise

Correct

Exercise29.8
Aflat,circular,steelloopofradius75cmisatrestinauniformmagneticfield,asshowninanedgeonviewinthefigure.Thefieldischangingwithtime,
accordingtoB(t)

(0.057s

= (1.4T)e

)t

PartA
Findtheemfinducedintheloopasafunctionoftime(assumetisinseconds).
Expressyouranswerintermsofthevariablet.
ANSWER:
E

0.057t

0.122e

Correct

PartB
1

Whenistheinducedemfequalto 18 ofitsinitialvalue?
ANSWER:
t

= 50.7 s

Correct

PartC
Findthedirectionofthecurrentinducedintheloop,asviewedfromabovetheloop.
ANSWER:
Counterclockwise
Clockwise

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PSS29.1:Faraday'sLaw
LearningGoal:
TopracticeProblemSolvingStrategy29.1:Faraday'sLaw.
Ametaldetectorusesachangingmagneticfieldtodetectmetallicobjects.Supposeametaldetectorthatgeneratesauniformmagneticfieldperpendicularto
itssurfaceisheldstationaryatanangleof15.0 totheground,whilejustbelowthesurfacethereliesasilverbraceletconsistingof6circularloopsofradius
5.00cmwiththeplaneoftheloopsparalleltotheground.Ifthemagneticfieldincreasesataconstantrateof0.0250T/s,whatistheinducedemfE ?Take
themagneticfluxthroughanareatobepositivewhenBcrossestheareafromtoptobottom.
ProblemSolvingStrategy29.1:Faraday'sLaw
IDENTIFYtherelevantconcepts:
Faradayslawapplieswhenthereisachangingmagneticflux.Tousethelaw,identifyanareathroughwhichthereisafluxofmagneticfield.
SETUPtheproblemusingthefollowingsteps:
1.Determinewhatismakingthemagneticfluxchange.Istheconductormovingorchangingorientation?Isthemagneticfieldchanging?

2.ChooseadirectionfortheareavectorAordAthatisperpendiculartotheplaneofthearea.Whateverdirectionyouchoose,useitconsistently
throughouttheproblem.
EXECUTEthesolutionasfollows:
1.CalculatethemagneticfluxusingB = B A= BA cos ifBisuniformovertheareaand B
uniform,beingmindfulofthedirectionyouchosefortheareavector.
2.CalculatetheinducedemfusingE = N

d B
dt

= B dA = BdA cos ifitisn't

foraconductorthathasN turnsinacoil.Rememberthesignforthepositivedirectionofemf

anduseitconsistently.
3.IfthecircuitresistanceRisknown,youcancalculatethemagnitudeoftheinducedcurrentI usingE

= IR

EVALUATEyouranswer:
Doesyouranswermakephysicalsense?Checkoveryourunits,andbesurethatyou'vecorrectlyusedthesignrules.

IDENTIFYtherelevantconcepts
Thereisachangingmagneticfluxthroughtheareaenclosedbytheloopsofthebraceletduetothemagneticfieldproducedbythemetaldetector.Therefore,
Faraday'slawappliestothisproblem.

SETUPtheproblemusingthefollowingsteps
PartA
Whichofthefollowingisthereasonwhythemagneticfluxthroughthebraceletischanging?
ANSWER:
Thebraceletismovinginthemagneticfield.
Themagnitudeofthemagneticfieldischanging.
Themagneticfieldischangingdirectionwithrespecttothebracelet.

Correct

PartB
Thediagrambelowshowsasideviewofthesituationdescribedinthisproblem.Thebraceletisdepictedasasingleloopparalleltotheground.
DrawtheareavectorAforthebracelet.UsetheblackdotasthestartingpointofA,thatis,makecertaintodrawyourvectorsothatitstailisonthedot.
Onlytheorientationofyourvectorwillbegraded,notitslength.
ANSWER:

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WhilethedirectionofAisfixed(bydefinition,Aisperpendiculartotheplaneofthearea),itsorientationisarbitrary.Inthisspecificproblem,Acan
beupwardordownward.However,sincetheproblemintroductionstatesthatthemagneticfluxthroughanareaistobetakenpositivewhenB
crossestheareafromtoptobottom,weneedtotakeAtobedownward,asshowninthediagrambelow,sothattheanglebetweenBandAisless
than90 anditscosineispositive.

UsethischoiceofdirectionforAtoworkthroughthenextpartoftheproblem.

EXECUTEthesolutionasfollows
PartC
BasedonthedefinitionofAgiveninthepreviouspart,whatisthetotalinducedemfE inthebracelet?
Expressyouranswerinvoltstothreesignificantfigures.
Youdidnotopenhintsforthispart.
ANSWER:
E

1.8910

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EVALUATEyouranswer
PartD
Thisquestionwillbeshownafteryoucompletepreviousquestion(s).

Exercise29.17
TwoclosedloopsAandC areclosetoalongwirecarryingacurrentI .

PartA
Findthedirection(clockwiseorcounterclockwise)ofthecurrentinducedinloopAifI issteadilyincreasing.
ANSWER:
ThecurrentinloopAiscounterclockwise.
ThecurrentinloopAisclockwise.
ThecurrentinloopAiszero.

Correct

PartB
Findthedirection(clockwiseorcounterclockwise)ofthecurrentinducedintheloopBifI issteadilyincreasing.
ANSWER:
ThecurrentinloopC iscounterclockwise.
ThecurrentinloopC isclockwise.
ThecurrentinloopC iszero.

Correct

PartC
WhileI isincreasing,whatisthedirectionofthenetforcethatthewireexertsonloopA?
ANSWER:

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ThewireexertsanupwardforceonloopA.
ThewireexertsadownwardforceonloopA.
ThewireexertsnoforceonloopA.

Correct

PartD
WhileI isincreasing,whatisthedirectionofthenetforcethatthewireexertsonloopC ?
ANSWER:
ThewireexertsanupwardforceonloopC .
ThewireexertsadownwardforceonloopC .
ThewireexertsnoforceonloopC .

Correct

PartE
Explainhowyouobtainyouranswer.
Dragthetermsonthelefttotheappropriateblanksontherighttocompletethesentences.
ANSWER:

Reset

Help

decreasing

Whenthecurrentinthewireisincreasingthe

increasing

magnitudeofthemagneticfluxintheloopsis

oppose
support

increasing .
Inducedcurrentsintheloopsandtheforces
exertedonthemaredirectedsothey oppose
thischange.

Correct

Exercise29.25
InthefigureaconductingrodoflengthL=35.0cmmovesinamagneticfieldBofmagnitude0.420Tdirectedintotheplaneofthefigure.Therodmoves
withspeedv =5.90m/sinthedirectionshown.

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PartA
Whatisthepotentialdifferencebetweentheendsoftherod?
ANSWER:
V

= 0.867 V

Correct

PartB
Whichpoint,aorb,isathigherpotential?
ANSWER:
a
b

Correct

PartC
Whenthechargesintherodareinequilibrium,whatisthemagnitudeoftheelectricfieldwithintherod?
ANSWER:
E

= 2.48 V/m

Correct

PartD
Whatisthedirectionoftheelectricfieldwithintherod?
ANSWER:
Fromatob
Frombtoa

Correct

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PartE
Whenthechargesintherodareinequilibrium,whichpoint,aorb,hasanexcessofpositivecharge?
ANSWER:
a
b

Correct

PartF
Whatisthepotentialdifferenceacrosstherodifitmovesparalleltoab?
ANSWER:
V

= 0 V

Correct

PartG
Whatisthepotentialdifferenceacrosstherodifitmovesdirectlyoutofthepage?
ANSWER:
V

= 0 V

Correct

MagneticLightSwitchRankingTask
Sixidenticalverticalmetalbarsstartatthepositionsshownbelowandmoveatconstantvelocitiesthroughidenticalmagneticfields.Thebarsmakeelectrical
contactwithandmovealongfrictionlessmetalrodsattachedtolightbulbs.

PartA
Attheinstantshown,rankthesesixscenariosonthebasisofthemagnitudeofthecurrentinthelightbulb.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.

Hint1.Faraday'slawofinduction
Themagnitudeofthecurrentinthelightbulbisproportionaltothemagnitudeoftheemfinducedineachscenario.Theemfinducedisequaltothe
timerateofchangeofthemagneticfluxthroughtheclosedloopformedbytheslidingbar.

Hint2.Changeinmagneticflux
Magneticfluxistheproductofthemagneticfieldperpendiculartoagivenareaandtheareaitself.Sincethemagneticfieldisconstantinthis
case,anychangeinfluxmustbeduesolelytoachangeinarea.

Hint3.Changeinarea
Theareaoftheclosedloopformedbytheslidingbarchangesataratethatisproportionaltothevelocityoftheslidingbar.
ANSWER:

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Reset

largest

Help

smallest

Thecorrectrankingcannotbedetermined.

Correct

Exercise29.30
A0.650mlongmetalbarispulledtotherightatasteady6.0 m/sperpendiculartoauniform,0.550Tmagneticfield.Thebarridesonparallelmetalrails
connectedthrougha25,resistor,sotheapparatusmakesacompletecircuit.Ignorethe
resistanceofthebarandtherails.

PartA
Calculatethemagnitudeoftheemfinducedinthecircuit.
Expressyouranswerwiththeappropriateunits.
ANSWER:
E

= 2.1V

Correct

PartB
Findthedirectionofthecurrentinducedinthecircuit.

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ANSWER:
clockwise
counterclockwise

Correct

PartC
Calculatethecurrentthroughtheresistor.
Expressyouranswerwiththeappropriateunits.
ANSWER:
I

= 8.6102A

Correct

RailGun
Thisproblemexploreshowacurrentcarryingwirecanbeacceleratedbyamagneticfield.YouwillusetheideasofmagneticfluxandtheEMFduetochange
offluxthroughaloop.NotethatthereisaninvolvedfollowuppartthatwillbeshownonceyouhavefoundtheanswertoPartB.

PartA
Aconductingrodisfreetoslideontwoparallelrailswithnegligiblefriction.Attherightendoftherails,avoltagesourceofstrengthV inserieswitha
resistorofresistanceRmakesaclosedcircuittogetherwiththerailsandtherod.Therailsandtherodaretakentobeperfectconductors.Therails
extendtoinfinityontheleft.Thearrangementisshowninthefigure.
ThereisauniformmagneticfieldofmagnitudeB,pervadingallspace,perpendiculartothe
planeofrodandrails.Therodisreleasedfromrest,anditisobservedthatitacceleratestothe
left.Inwhatdirectiondoesthemagneticfieldpoint?

Hint1.Theforceonaconductingrodduetoamagneticfield
Thereisaforceontherodbecauseacurrentisflowingthroughit.Hencechargesaremovingperpendiculartothemagneticfield.Therod

experiencesaforceF ,whichisgivenby
F =

IL B

Hint2.Thedirectionofthemagneticfield
Usetherighthandrulewiththecrossproduct:Takeyourrighthandandpointwithyourindexfingerinthedirectionoftherail'smotion,andpoint
yourmiddlefingerinthedirectionofL.Yourthumbwillthenpointinthedirectionofthemagneticfield.
ANSWER:
intotheplaneofthefigure
outoftheplaneofthefigure

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PartB
Assumingthattherailshavenoresistance,whatisthemostaccuratequalitativedescriptionofthemotionoftherod?

Hint1.Lenz'slaw
AnEMFisinducedinthecircuitduetothechangeinmagneticfluxthroughit.ButwillthisEMFincreasethecurrentthroughtheloopordecrease
itbyopposingthevoltagecomingfromthesource?Lenz'slawstatesthatinducedcurrentswillalwaysflowinsuchawaythattheyopposethe
changeinfluxthatcausedthem.

Hint2.AppylingLenz'slawtothisproblem
LetusapplyLenz'slawhere.Wewanttofindthedirectionoftheinducedcurrent.Thechangeinfluxthatinducedthiscurrentiscausedbythe
motionoftherod,whichinturniscausedbythecurrentflowingaroundthecircuitduetothevoltagesource.Theinducedcurrentworksagainst
thatsourcecurrent,andreducesit.Alternatively,onecouldsaythattheinducedEMFopposesthevoltagefromthesource.

Hint3.Thevelocityoftherod
Thehigherthevelocityoftherod,thehighertheinducedEMF,andthelowerthecurrentflowingthroughtheloop.Buttheforceacceleratingtherod
isproportionaltothecurrent.Hencetheaccelerationgoesdownasthevelocitygoesup.Thevelocitycannotincreasebeyondthepointatwhich
theinducedEMFisequalandoppositetoV .
ANSWER:
Therodwillacceleratebutthemagnitudeoftheaccelerationwilldecreasewithtimethevelocityoftherodwillapproachbutneverexceeda
certainterminalvelocity.
Undertheseidealizedconditionstherodwillexperienceconstantaccelerationandthevelocityoftherodwillincreaseindefinitely.
Therodwillaccelerateindefinitelywithaccelerationproportionaltoits(increasing)velocity.

Correct

PartC
Whatistheaccelerationar (t)oftherod?Takemtobethemassoftherod.
ExpressyouranswerasafunctionofV ,B,thevelocityoftherodv r (t),L,R,andthemassoftherodm.

Hint1.FindtheinducedEMF
Todeterminethecurrent,weneedtheEMFE(t) inducedbythechangeinmagneticflux(t)
Faraday'slawstatesthatE(t) =

d(t)
dt

= B A (t)

throughthecurrentloopofareaA(t).

.ThemotionoftherodwillchangeA(t)accordingto

d(t)
dt

dA(t)
dt

B .WhatistheEMF E(t) ?

ExpressE intermsofthevelocityv r (t),theseparationoftherailsL,andthemagneticfieldB.

BLv r (t)

Hint2.Findthecurrentintherod
WecanfindthecurrentthroughtherailbyusingKirchhoff'sruleforaclosedcircuit:I R
currentI (t)?

= V +E

,whereE istheinducedEMF.Whatisthe

Expressyouranswerintermsofv r (t)andothergivenquantities.
ANSWER:

I (t)

V BLvr (t)
R

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Hint3.Findtheaccelerationoftherod
Tofindtheaccelerationar (t),weneedtheforceontherod.Wehavealreadydetermineditsdirection.ThemagnitudeofthisforceisgiveninPart
A.i.Writedownanexpressionforar (t),usingNewton'ssecondlaw.
Expresyouranswerintermsofmassm,currentI (t),B,andL.
ANSWER:

ar (t)

I (t)BL
m

ANSWER:

ar (t)

(V BLvr (t))LB
Rm

Correct
Makingthesubstitutionar (t) =

dvr (t)
dt

,youobtainthedfferentialequation
dvr (t)
dt

BL
mR

(V BLvr (t)),

whichyoucansolvetofindthevelocityoftherodasafunctionoftime:
vr (t) =

V
BL

(1 e

2
2
B L
mR

).

Toachieveahighacceleration,whichisnecessaryforausefulgun,amagneticfieldoflargemagnitudeandahighvoltageareadvantageous.

PartD
Whatistheterminalvelocityv t reachedbytherod?

Hint1.Findanexpressionfortheterminalvelocity
Theterminalvelocityisreachedwhentheaccelerationiszero.Whatistheconditionforthis?
Writedownthiscondition,expressingtheappliedvoltageV asafunctionoftheterminalvelocityv t .
ANSWER:
V

BLv t

ANSWER:
vt

V
BL

Correct
Alargermagneticfieldincreasestheaccelerationoftherod,butlowerstheterminalvelocity:atradeoffforrailgunengineers!

Exercise29.43
Supposethattheparallelplatesinthefigurehaveanareaof3.00cm2 andareseparatedbya2.50mmthicksheetofdielectricthatcompletelyfillsthe
volumebetweentheplates.Thedielectrichasdielectricconstant4.70.(Youcanignorefringingeffects.)Atacertaininstant,thepotentialdifferencebetween
theplatesis120VandtheconductioncurrentiC equals6.00mA.

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PartA
Atthisinstant,whatisthechargeqoneachplate?
ANSWER:
|q|

= 5.991010 C

Correct

PartB
Atthisinstant,whatistherateofchangeofchargeontheplates?
ANSWER:
dq
dt

= 6.00103 A

Correct

PartC
Atthisinstant,whatisthedisplacementcurrentinthedielectric?
ANSWER:
id

= 6.00103 A

Correct

Exercise29.45
Attemperaturesnearabsolutezero,Bc approaches0.142Tforvanadium,atypeIsuperconductor.Thenormalphaseofvanadiumhasamagnetic
susceptibilityclosetozero.Consideralong,thinvanadiumcylinderwithitsaxisparalleltoanexternalmagneticfieldB0 inthe+xdirection.Atpointsfarfrom
theendsofthecylinder,bysymmetry,allthemagneticvectorsareparalleltothexaxis.

PartA
Attemperaturesnearabsolutezero,whatisthemagnitudeoftheresultantmagneticfieldBinsidethecylinderforB0

0.130T)^i ?

= (

ANSWER:
B

= 0 T

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Correct

PartB
WhatisthedirectionoftheresultantmagneticfieldBinsidethecylinderforthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
thefieldiszero

Correct

PartC
WhatisthemagnitudeoftheresultantmagneticfieldBoutsidethecylinder(farfromtheends)forthiscase?
ANSWER:
B

= 0.130 T

Correct

PartD
WhatisthedirectionoftheresultantmagneticfieldBoutsidethecylinder(farfromtheends)forthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
thefieldiszero

Correct

PartE
WhatisthemagnitudeofthemagnetizationMinsidethecylinderforthiscase?
ANSWER:
M

= 1.03105 A/m

Correct

PartF
WhatisthedirectionofthemagnetizationMinsidethecylinderforthiscase?
ANSWER:

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inthe+xdirection
inthexdirection
perpendiculartothexaxis
themagnetizationiszero

Correct

PartG
WhatisthemagnitudeofthemagnetizationMoutside(farfromtheends)thecylinderforthiscase?
ANSWER:
M

= 0 A/m

Correct

PartH
WhatisthedirectionofthemagnetizationMoutsidethecylinder(farfromtheends)forthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
themagnetizationiszero

Correct

PartI
Attemperaturesnearabsolutezero,whatisthemagnitudeoftheresultantmagneticfieldBinsidethecylinderforB0

^
= (0.260T) i

ANSWER:
B

= 0.260 T

Correct

PartJ
WhatisthedirectionoftheresultantmagneticfieldBinsidethecylinderforthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
thefieldiszero

Correct

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PartK
WhatisthemagnitudeoftheresultantmagneticfieldBoutsidethecylinder(farfromtheends)forthiscase?
ANSWER:
B

= 0.260 T

Correct

PartL
WhatisthedirectionoftheresultantmagneticfieldBoutsidethecylinder(farfromtheends)forthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
thefieldiszero

Correct

PartM
WhatisthemagnitudeofthemagnetizationMinsidethecylinderforthiscase?
ANSWER:
M

= 0 A/m

Correct

PartN
WhatisthedirectionofthemagnetizationMinsidethecylinderforthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
themagnetizationiszero

Correct

PartO
WhatisthemagnitudeofthemagnetizationMoutsidethecylinderforthiscase?
ANSWER:
M

= 0 A/m

Correct

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PartP
WhatisthedirectionofthemagnetizationMoutsidethecylinderforthiscase?
ANSWER:
inthe+xdirection
inthexdirection
perpendiculartothexaxis
themagnetizationiszero

Correct

Exercise29.60
Ametalrodwithalengthof23.0cmliesinthexy planeandmakesanangleof38.7 withthepositivexaxisandanangleof51.3 withthepositiveyaxis.
Therodismovinginthe+xdirectionwithaspeedof6.80m/s.TherodisinauniformmagneticfieldB =

^
^
^
(0.160T) i (0.200T) j (0.0200T )k

PartA
Whatisthemagnitudeoftheemfinducedintherod?
ANSWER:
E

0.0467

IncorrectTryAgain22attemptsremaining

Problem29.49
Averylong,straightsolenoidwithacrosssectionalareaof1.84cm2 iswoundwith86.4turnsofwirepercentimeter.Startingatt=0,thecurrentinthe
solenoidisincreasingaccordingtoi(t) =(0.165A/s2 )t2 .Asecondarywindingof5turnsencirclesthesolenoidatitscenter,suchthatthesecondary
windinghasthesamecrosssectionalareaasthesolenoid.

PartA
Whatisthemagnitudeoftheemfinducedinthesecondarywindingattheinstantthatthecurrentinthesolenoidis3.2A?
Expressyouranswerwiththeappropriateunits.
ANSWER:
= 1.45105V

|E|

Correct

ASuperconductingCylinder
Attemperaturesnearabsolutezero,thecriticalfieldBc forvanadium,atypeIsuperconductor,approaches0.142tesla.Thenormalphaseofvanadiumhasa
magneticsusceptibilityclosetozero.Consideralong,thin,solidvanadiumcylinderwithitsaxisparalleltoanexternalmagneticfieldB0 inthe+xdirection.
Atpointsfarfromtheendsofthecylinder,bysymmetry,allthemagneticvectorsareparalleltothexaxis.
First,consideranappliedmagneticfieldB0

^
= (0.130 T) i

PartA
Attemperaturesnearabsolutezero,whatisthemagnitudeoftheresultantmagneticfieldBinsidethecylinder(farfromtheends)?
Expressyouranswerinteslas.

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Hint1.Whatisthecriticalfield?
ChoosethecorrectdefinitionofthecriticialfieldBc .
ANSWER:
themaximummagnitudeofmagneticfieldintowhichanobjectcanbeplacedandstillexhibitsuperconductivity(atagiventemperature
belowthecriticaltemperature)
themagneticfieldduetocoulomb(orconservative)forces.

ANSWER:

|B|

= 0 T

Correct

PartB
WhatisthemagnitudeoftheresultantmagneticfieldBoutsidethecylinder(farfromtheends)forthesituationdescribedintheproblemintroduction?
Expressyouranswerinteslas.

Hint1.Findthemagneticfieldduetothecylinder
ThinkofthemagneticfieldoutsidethecylinderasanappliedfieldB0 plusamagneticfieldBcyl duetothecylinder.WhatisthemagnitudeofBcyl
?

Hint1.Asolenoidanalogy
Ifyougivesomethoughttothisproblem,youmaybeabletosolveitwithoutperforminganycalculations.Weknowthatthemagneticfield
insidethecylinderiszero.Weknowthattheobjectisbehavingasthoughtherewereacurrentflowingonitssurfacecreatingamagnetic
fieldthatexactlycancelstheappliedmagneticfield.Becausetheappliedmagneticfieldisparalleltothexaxis,theinducedmagneticfield
mustalsobeparalleltothexaxis.
Let'srecap:Wehaveacylindricalobjectbehavingasthoughacurrentwereflowingalongitssurfaceinsuchawaythatthereisamagnetic
fieldinsideitparalleltoitsaxis.Doesthissoundfamiliar?Thecylinderisactinglikeasolenoid.Currentflowingaroundthecircumference
producesamagneticfieldinsidethatisparalleltothexaxis(whichisalsotheaxisofthecylinder).Doyourecallwhatthemagneticfieldis
outsideofasolenoid?
ANSWER:

Bcyl

= 0

ANSWER:

|B|

= 0.130 T

Correct

PartC
WhatisthemagnitudeofthemagnetizationMin insidethecylinder?WhatisthemagnitudeofthemagnetizationMout outsidethecylinder?
Expressyouranswersinamperespermeterseparatedbyacomma.

Hint1.Howtoapproachtheproblem

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MagnetizationMisdefinedbytheequationB = B0 + 0 M,whereBisthetotalmagneticfieldandB0 istheappliedmagneticfield.Allofthe

Hint2.Valueof0
Recallthat0

= 4 10

Tm
A

instandardSIunits.

ANSWER:

M in

,Mout = 1.03105,0 A/m,A/m

Correct
Recallthatthereneedstobeamaterialpresenttoinducemagnetization.Inotherwords,themagnetizationoffreespaceisalwayszero.

NowconsideranappliedmagneticfieldB0

^
= (0.260 T) i

PartD
Attemperaturesnearabsolutezero,whatisthemagnitudeoftheresultantmagneticfieldBinsidethecylinder(farfromtheends)?

Hint1.Howtoapproachtheproblem
RecallthatwhenB0 > Bc ,anobjectwillnotexhibitsuperconductivity.

Hint2.Moreonsuperconductivity
Whenanobjectisnotinthesuperconductingphase,themagneticdipolesofwhichitismadecannotalignsoastocanceltheappliedmagnetic
field.Whenthecriticalfieldisexceeded,makingthematerialnolongersuperconducting,themagnetizationdependsonthemagneticsusceptibility
ofthematerialinitsnormalphase.
ANSWER:

|B|

= 0.260 T

Correct

PartE
WhatisthemagnitudeoftheresultantmagneticfieldBoutsidethecylinder(farfromtheends)?

Hint1.Howtoapproachtheproblem
ThispartisconceptuallysimilartoPartE,inthatanymagneticfieldcausedbythecylinderwillbebasedonitsnormalphasemagnetic
susceptibility.
ANSWER:

|B|

= 0.260 T

Correct

PartF
WhatisthemagnitudeofthemagnetizationMin insidethecylinder?WhatisthemagnitudeofthemagnetizationM_outoutsidethecylinder?
Expressyouranswersinamperespermeterseparatedbyacomma.

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ANSWER:

M in

,Mout = 0,0 A/m,A/m

Correct
ScoreSummary:
Yourscoreonthisassignmentis91.6%.
Youreceived22.9outofapossibletotalof25points.

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