[SHIVOK SP212]
February 20, 2016 CH 30 InductionandInductance
I.
Faraday’sExperiments
A. Let’sexaminetwosimpleexperimentstoprepareforour
discussionaboutFaraday’sLaw:
B. FirstExperiment:
1.
MoveamagnetthroughaloopinducinganEMFintothatloop.
2.
Acurrentappearsonlyifthereisrelativemotionbetweentheloopand
themagnet(onemustmoverelativetotheother);thecurrentdisappears
whentherelativemotionbetweenthemceases.
3.
Fastermotionproducesagreatercurrent.
4.
Ifmovingthemagnet’snorthpoletowardtheloopcauses,say,
clockwisecurrent,thenmovingthenorthpoleawaycausescounterclockwise
current.Movingthesouthpoletowardorawayfromtheloopalsocauses
currents,butinthereverseddirections.
5.
Thecurrentthusproducedintheloopiscalledinducedcurrent.
Page1
[SHIVOK SP212]
February 20, 2016 C. SecondExperiment:
1.
ForthisexperimentweusetheapparatusofFig.30‐2,withthetwo
conductingloopsclosetoeachotherbutnottouching.
2.
IfwecloseswitchS,toturnonacurrentintheright‐handloop,the
metersuddenlyandbrieflyregistersacurrent—aninducedcurrent—inthe
left‐handloop.
3.
Ifwethenopentheswitch,anothersuddenandbriefinducedcurrent
appearsinthelefthandloop,butintheoppositedirection.
4.
Wegetaninducedcurrent(andthusaninducedemf)onlywhenthe
currentintheright‐handloopischanging(eitherturningonorturningoff)
andnotwhenitisconstant(evenifitislarge).
II.
Faraday’sLawofInduction:
A. AnEMFisinducedintheloopattheleftinFigures30‐1and30‐2
whenthenumberofmagneticfieldlinesthatpassthroughtheloopis
changing.
B. ThemagnitudeoftheEMF(ScriptE)inducedinaconductingloop
isequaltotherateatwhichthemagneticfluxBthroughthatloop
changeswithtime.
C. SupposealoopenclosinganareaAisplacedinamagneticfieldB.
Thenthemagneticfluxthroughtheloopis
Page2
[SHIVOK SP212]
February 20, 2016 D. Iftheloopliesinaplaneandthemagneticfieldisperpendicular
totheplaneoftheloop,andIfthemagneticfieldisconstant,then
E. TheSIunitformagneticfluxisthetesla–squaremeter,whichis
calledtheweber(abbreviatedWb):
F. Finally,Faraday’sLawofInduction:
1.
IfwechangethemagneticfluxthroughacoilofNturns,aninducedemf
appearsineveryturnandthetotalemfinducedinthecoilisthesumofthese
individualinducedemfs.Ifthecoilistightlywound(closelypacked),sothat
thesamemagneticfluxBpassesthroughalltheturns,thetotalemfinduced
inthecoilis
2.
Herearethegeneralmeansbywhichwecanchangethemagneticflux
throughacoil:
a)
ChangethemagnitudeBofthemagneticfieldwithinthecoil.
b)
Changeeitherthetotalareaofthecoilortheportionofthatarea
thatlieswithinthemagneticfield(forexample,byexpandingthecoilor
slidingitintooroutofthefield).
c)
ChangetheanglebetweenthedirectionofthemagneticfieldB
andtheplaneofthecoil(forexample,byrotatingthecoilsothatthe
fieldBisfirstperpendiculartotheplaneofthecoil,andthenalongthat
plane).
Page3
[SHIVOK SP212]
February 20, 2016 G. SampleProblem:
1.
InFig.below,a120‐turncoilofradius1.8cmandresistance5.3Ωis
coaxialwithasolenoidof220turns/cmanddiameter3.2cm.Thesolenoid
currentdropsfrom1.5AtozerointimeintervalΔt=25ms.Whatcurrentis
inducedinthecoilduringΔt?
a)
Solution:
Page4
[SHIVOK SP212]
February 20, 2016 III.
Lenz’sLaw:
A. Aninducedcurrenthasadirectionsuchthatthemagneticfield
dudetothecurrentopposesthechangeinthemagneticfluxthat
inducesthecurrent.
B. OppositiontoPoleMovement.Theapproachofthemagnet’snorth
poleinFig.30‐4increasesthemagneticfluxthroughtheloop,
inducingacurrentintheloop.Toopposethemagneticfluxincrease
beingcausedbytheapproachingmagnet,theloop’snorthpole(and
themagneticmoment)mustfacetowardtheapproachingnorthpole
soastorepelit.Thecurrentinducedintheloopmustbe
counterclockwiseinFig.30‐4.Ifwenextpullthemagnetawayfrom
theloop,acurrentwillagainbeinducedintheloop.Now,theloopwill
haveasouthpolefacingtheretreatingnorthpoleofthemagnet,soas
toopposetheretreat.Thus,theinducedcurrentwillbeclockwise.
Page5
[SHIVOK SP212]
February 20, 2016 C. Fig.30‐5Thedirectionofthecurrentiinducedinaloopissuch
thatthecurrent’smagneticfieldBindopposesthechangeinthe
magneticfieldinducingi.Thefieldisalwaysdirectedoppositean
increasingfield(a)andinthesamedirection(b)asadecreasingfield
B.Thecurled–straightright‐handrulegivesthedirectionofthe
inducedcurrentbasedonthedirectionoftheinducedfield.
D. Ifthenorthpoleofamagnetnearsaclosedconductingloopwith
itsmagneticfielddirecteddownward,thefluxthroughtheloop
increases.Toopposethisincreaseinflux,theinducedcurrentimust
setupitsownfieldBinddirectedupwardinsidetheloop,asshownin
Fig.30‐5a;thentheupwardfluxofthefieldBindopposestheincreasing
downwardfluxoffield.Thecurled–straightright‐handrulethentells
usthatimustbecounterclockwiseinFig.30‐5a.
Page6
[SHIVOK SP212]
February 20, 2016 E. SampleProblem(Lenz’sLaw):
1.
Thefollowingtwosituationsareseparate.Ontheleft,asquareloopof
wireispenetratedbyamagneticfieldoutofthepagethatisincreasingin
strength.Ontheright,thenorthpoleofamagnetismovingawayfromacoilof
wire.Therowthatcorrectlygivesthedirectionoftheinducedcurrentthrough
eachresistoris
Page7
[SHIVOK SP212]
February 20, 2016 IV.
InductionandEnergyTransfers:
A. Considerthepullingaconductingloopoutofamagneticfieldas
shownbelow:
B. Iftheloopispulledataconstantvelocityv,onemustapplya
constantforceFtotheloopsinceanequalandoppositemagnetic
forceactsonthelooptoopposeit.ThepowerisP=Fv.
C. Astheloopispulled,theportionofitsareawithinthemagnetic
field,andthereforethemagneticflux,decrease.Accordingto
Faraday’slaw,acurrentisproducedintheloop.Themagnitudeofthe
fluxthroughtheloopisB=BA=BLx.
D. Therefore,
E. Theinducedcurrentistherefore
F. Thenetdeflectingforceis:
G. Thepoweristherefore
Page8
[SHIVOK SP212]
February 20, 2016 H. SampleProblems:
1.
InFig.belowametalrodisforcedtomovewithconstantvelocity
alongtwoparallelmetalrails,connectedwithastripofmetalatoneend.A
magneticfieldofmagnitudeB=0.350Tpointsoutofthepage.(a)Iftherails
areseparatedbyL=25.0cmandthespeedoftherodis55.0cm/s,whatemfis
generated?(b)Iftherodhasaresistanceof18.0Ωandtherailsandconnector
havenegligibleresistance,whatisthecurrentintherod?(c)Atwhatrateis
energybeingtransferredtothermalenergy?
a)
Solution:
Page9
[SHIVOK SP212]
February 20, 2016 2.
InFig.below,alongrectangularconductingloop,ofwidthL,resistance
R,andmassm,ishunginahorizontal,uniformmagneticfield thatis
directedintothepageandthatexistsonlyabovelineaa.Theloopisthen
dropped;duringitsfall,itacceleratesuntilitreachesacertainterminalspeed
vt.Ignoringairdrag,findanexpressionforvt.
a)
Solution:
Page
10
[SHIVOK SP212]
February 20, 2016 I.
EddyCurrents
V.
InducedElectricField:
A. Achangingmagneticfieldproducesanelectricfield.
Page
11
[SHIVOK SP212]
February 20, 2016 B. InducedElectricFields,ReformulationofFaraday’sLaw:
1.
Consideraparticleofchargeq movingaroundthecircularpath.The
0
workWdoneonitinonerevolutionbytheinducedelectricfieldisW=Eq ,
0
whereEistheinducedemf.
2.
Fromanotherpointofview,theworkis
Herewhereq Eisthemagnitudeoftheforceactingonthetestcharge
3.
0
and2risthedistanceoverwhichthatforceacts.
4.
Ingeneral,
C. InducedElectricFields,ANewLookatElectricPotential:
1.
ElectricPotentialhasmeaningonlyforelectricfieldsthatareproduced
bystaticcharges;ithasnomeaningforelectricfieldsthatareproducedby
induction.
2.
Whenachangingmagneticfluxispresent,theintegralisnotzerobutis
d /dt.
B
3.
Thus,assigningelectricpotentialtoaninducedelectricfieldleadsusto
concludethatelectricpotentialhasnomeaningforelectricfieldsassociated
withinduction.
Page
12
[SHIVOK SP212]
February 20, 2016 D. SampleProblem:
1.
Alongsolenoidhasadiameterof12.0cm.Whenacurrentiexistsinits
windings,auniformmagneticfieldofmagnitudeB=30.0mTisproducedinits
interior.Bydecreasingi,thefieldiscausedtodecreaseattherateof6.50
mT/s.Calculatethemagnitudeoftheinducedelectricfield(a)2.20cmand(b)
8.20cmfromtheaxisofthesolenoid.
a)
Solution:
Page
13
[SHIVOK SP212]
February 20, 2016 VI.
InductorsandInductance:
A. Aninductor(symbol)canbeusedtoproduceadesired
magneticfield.
B. Ifweestablishacurrentiinthewindings(turns)ofthesolenoid
whichcanbetreatedasourinductor,thecurrentproducesamagnetic
fluxBthroughthecentralregionoftheinductor.
C. Theinductanceoftheinductoristhen
D. TheSIunitofinductanceisthetesla–squaremeterperampere(T
2
m /A).
Wecallthisthehenry(H),afterAmericanphysicistJosephHenry.
Page
14
[SHIVOK SP212]
February 20, 2016 E. InductanceofaSolenoid:
1.
Consideralongsolenoidofcross‐sectionalareaA,withnumberof
turnsN,andoflengthl.Thefluxis
Herenisthenumberofturnsperunitlength.
2.
ThemagnitudeofBisgivenby:
3.
Therefore,
4.
Theinductanceperunitlengthnearthecenteristherefore:
Here,
F. Self‐Induction:
1.
Aninducedemfappearsinanycoilinwhichcurrentischanging.
2.
Thisprocess(seefigurebelow)iscalledself‐induction,andtheemfthat
appearsiscalledself‐inducedemf.ItobeysFaraday’slawofinductionjustas
otheremfsdo.
Page
15
[SHIVOK SP212]
February 20, 2016 3.
,butremember
4.
G. Sampleproblems:
1.
Theinductanceofacloselypackedcoilof400turnsis8.0mH.Calculate
themagneticfluxthroughthecoilwhenthecurrentis5.0mA.
a)
Solution:
2.
Atagiveninstantthecurrentandself‐inducedemfinaninductorare
directedasindicatedinFig.below.(a)Isthecurrentincreasingordecreasing?
(b)Theinducedemfis17V,andtherateofchangeofthecurrentis25kA/s;
findtheinductance.
a)
Solution:
Page
16
[SHIVOK SP212]
February 20, 2016 VII.
RLCircuits:
A. Initially,aninductoractstoopposechangesinthecurrent
throughit.Alongtimelater,itactslikeanordinaryconnectingwire.
B. Thecircuit:
1.
Currenteqn:
2.
Riseofcurrent:
3.
Thusthetimeconstant:
4.
Ifwesuddenlyremovetheemffromthissamecircuit,thechargedoes
notimmediatelyfalltozerobutapproacheszeroinanexponentialfashion:
Page
17
[SHIVOK SP212]
February 20, 2016 5.
Graphs:
C. Sampleproblem:
1.
Asolenoidhavinganinductanceof6.30μHisconnectedinserieswith
a1.20kΩresistor.(a)Ifa14.0Vbatteryisconnectedacrossthepair,how
longwillittakeforthecurrentthroughtheresistortoreach80.0%ofitsfinal
value?(b)Whatisthecurrentthroughtheresistorattimet=1.0τL?
a)
Solution:
Page
18
[SHIVOK SP212]
February 20, 2016 VIII. EnergyStoredinaMagneticField:
A. Thecircuit:
B. KVL:
C. Power:
D. ThisistherateatwhichmagneticpotentialenergyUBisstoredinthe
magneticfield.
E. Thus
F. Finally,thetotalenergystoredbyaninductorLcarryingacurrent
iis:
Page
19
[SHIVOK SP212]
February 20, 2016 G. Sampleproblem:
1.
ForthecircuitofFig.below,assumethat
=10.0V,R=6.70Ω,andL
=5.50H.Theidealbatteryisconnectedattimet=0.(a)Howmuchenergyis
deliveredbythebatteryduringthefirst2.00s?(b)Howmuchofthisenergyis
storedinthemagneticfieldoftheinductor?(c)Howmuchofthisenergyis
dissipatedintheresistor?
Page
20
[SHIVOK SP212]
February 20, 2016 H. EnergyDensityofaMagneticField:
1.
Consideralengthlnearthemiddleofalongsolenoidofcross‐sectional
areaAcarryingcurrenti;thevolumeassociatedwiththislengthisAl.
2.
TheenergyU storedbythelengthlofthesolenoidmustlieentirely
B
withinthisvolumebecausethemagneticfieldoutsidesuchasolenoidis
approximatelyzero.Also,thestoredenergymustbeuniformlydistributed
withinthesolenoidbecausethemagneticfieldis(approximately)uniform
everywhereinside.
3.
Thus,theenergystoredperunitvolumeofthefieldis
4.
I.
Sampleproblem:
1.
Asolenoidthatis85.0cmlonghasacross‐sectionalareaof17.0cm2.
Thereare950turnsofwirecarryingacurrentof6.60A.(a)Calculatethe
energydensityofthemagneticfieldinsidethesolenoid.(b)Findthetotal
energystoredinthemagneticfieldthere(neglectendeffects).
Page
21
[SHIVOK SP212]
February 20, 2016 IX.
MutualInduction:
A. Diagram
B. ThemutualinductanceM21ofcoil2withrespecttocoil1is
definedas
1.
2.
Therightsideofthisequationis,accordingtoFaraday’slaw,justthe
magnitudeoftheemfE appearingincoil2duetothechangingcurrentincoil
2
1.
3.
Similarly,
C. Sampleproblem:Twocoilsareatfixedlocations.Whencoil1has
nocurrentandthecurrentincoil2increasesattherate15.0A/s,the
emfincoil1is25.0mV.(a)Whatistheirmutualinductance?(b)
Whencoil2hasnocurrentandcoil1hasacurrentof3.60A,whatis
thefluxlinkageincoil2?
Page
22