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February 20, 2016 CH 28 MagneticFields
I.
WhatProducesMagneticField?
A. Onewaythatmagneticfieldsareproducedistousemoving
electricallychargedparticles,suchasacurrentinawire,tomakean
electromagnet.Thecurrentproducesamagneticfieldthatis
utilizable.
B. Theotherwaytoproduceamagneticfieldisbymeansof
elementaryparticlessuchaselectrons,becausetheseparticleshave
anintrinsicmagneticfieldaroundthem.
1.
Themagneticfieldsoftheelectronsincertainmaterialsaddtogether
togiveanetmagneticfieldaroundthematerial.Suchadditionisthereason
whyapermanentmagnet,hasapermanentmagneticfield.
2.
Inothermaterials,themagneticfieldsoftheelectronscancelout,
givingnonetmagneticfieldsurroundingthematerial.
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February 20, 2016 II.
TheDefinitionofBfield:
A. Wecandefineamagneticfield,B,byfiringachargedparticle
throughthepointatwhichistobedefined,usingvariousdirections
andspeedsfortheparticleanddeterminingtheforcethatactsonthe
particleatthatpoint.Bisthendefinedtobeavectorquantitythatis
directedalongthezero‐forceaxis.
B. Themagneticforceonthechargedparticle,FB,isdefinedtobe:
Hereqisthechargeoftheparticle,visitsvelocity,andBthemagnetic
fieldintheregion.
C. Themagnitudeofthisforceisthen:
HereisthesmallestanglebetweenvectorsvandB.
D. FindingtheMagneticForceonaParticle:
1.
TheforceFBactingonachargedparticlemovingwithvelocityv
throughamagneticFieldBisALWAYSperpendiculartobothvandB.
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February 20, 2016 E. TheSIunitforBthatfollowsisnewtonpercoulomb‐meterper
second.Forconvenience,thisiscalledthetesla(T):
1.
Anearlier(non‐SI)unitforBisthegauss(G),and
2.
Table
F. Sampleproblem:
1.
Analphaparticletravelsatavelocity
ofmagnitude550m/sthrough
auniformmagneticfield ofmagnitude0.045T.(Analphaparticlehasa
chargeof+3.2×10‐19Candamassof6.6×10‐27kg.)Theanglebetween and
is52°.Whatisthemagnitudeof(a)theforce
actingontheparticle
duetothefieldand(b)theaccelerationoftheparticledueto
?(c)Does
thespeedoftheparticleincrease,decrease,orremainthesame?
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February 20, 2016 III.
MagneticFieldLines:
A. Thedirectionofthetangenttoamagneticfieldlineatanypoint
givesthedirectionofBatthatpoint.
B. ThespacingofthelinesrepresentsthemagnitudeofB—the
magneticfieldisstrongerwherethelinesareclosertogether,and
conversely.
C. Oppositemagneticpolesattracteachother,andlikemagnetic
polesrepeleachother.
D. NosuchthingasaMagneticMonopoleEveryNorthPolehasan
associatedSouthPole!
E. FieldlinesemanatefromtheNorthPoleandterminateinthe
SouthPole.
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February 20, 2016 IV.
CrossedFields,DiscoveryofanElectron:
A. Figure
B. WhenthetwofieldsinFig.28‐7areadjustedsothatthetwo
deflectingforcesactingonthechargedparticlecancel,wehave
C. Thus,thecrossedfieldsallowustomeasurethespeedofthe
chargedparticlespassingthroughthem.
D. Thedeflectionofachargedparticle,movingthroughanelectric
field,E,betweentwoplates,atthefarendoftheplates(inthe
previousproblem)is
Here,vistheparticle’sspeed,mitsmass,qitscharge,andListhe
lengthoftheplates.
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February 20, 2016 V.
CrossedFields,TheHallEffect:
A. AHallpotentialdifferenceVisassociatedwiththeelectricfield
acrossstripwidthd,andthemagnitudeofthatpotentialdifferenceis
V=Ed.Whentheelectricandmagneticforcesareinbalance(Fig.28‐
8b),
wherevdisthedriftspeed.But,
1.
WhereJisthecurrentdensity,Athecross‐sectionalarea,ethe
electroniccharge,andnthenumberofchargesperunitvolume.
B. Therefore,
here,l=(A/d),thethicknessofthestrip.
1.
Fig.28‐8Astripofcoppercarryingacurrentiisimmersedina
magneticfield.(a)Thesituationimmediatelyafterthemagneticfieldisturned
on.Thecurvedpaththatwillthenbetakenbyanelectronisshown.(b)The
situationatequilibrium,whichquicklyfollows.Notethatnegativechargespile
upontherightsideofthestrip,leavinguncompensatedpositivechargeson
theleft.Thus,theleftsideisatahigherpotentialthantherightside.(c)For
thesamecurrentdirection,ifthechargecarrierswerepositivelycharged,they
wouldpileupontherightside,andtherightsidewouldbeatthehigher
potential.
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February 20, 2016 VI.
ACirculatingChargedParticle:
A. Consideraparticleofchargemagnitude|q|andmassmmoving
perpendiculartoauniformmagneticfieldB,atspeedv.
B. Themagneticforcecontinuouslydeflectstheparticle,andsinceB
andvarealwaysperpendiculartoeachother,thisdeflectioncauses
theparticletofollowacircularpath.
C. Themagneticforceactingontheparticlehasamagnitudeof|q|vB.
D. Foruniformcircularmotion
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February 20, 2016 E. Diagram
1.
Fig.28‐10Electronscirculatinginachambercontaininggasatlow
pressure(theirpathistheglowingcircle).Auniformmagneticfield,B,
pointingdirectlyoutoftheplaneofthepage,fillsthechamber.Notethe
radiallydirectedmagneticforceFB;forcircularmotiontooccur,FBmustpoint
towardthecenterofthecircle,(CourtesyJohnLeP.Webb,SussexUniversity,
England)
VII.
HelicalPaths:
A. Fig.28‐11(a)Achargedparticlemovesinauniformmagnetic
field,theparticle’svelocityvmakingananglefwiththefield
direction.(b)Theparticlefollowsahelicalpathofradiusrandpitch
p.(c)Achargedparticlespiralinginanonuniformmagneticfield.
(Theparticlecanbecometrapped,spiralingbackandforthbetween
thestrongfieldregionsateitherend.)Notethatthemagneticforce
vectorsattheleftandrightsideshaveacomponentpointingtoward
thecenterofthefigure.
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February 20, 2016 B. Thevelocityvector,v,ofsuchaparticleresolvedintotwo
components,oneparalleltoandoneperpendiculartoit:
C. Theparallelcomponentdeterminesthepitchpofthehelix(the
distancebetweenadjacentturns(Fig.28‐11b)).Theperpendicular
componentdeterminestheradiusofthehelix.
Pitch = and Radius = D. Themorecloselyspacedfieldlinesattheleftandrightsides
indicatethatthemagneticfieldisstrongerthere.Whenthefieldatan
endisstrongenough,theparticle“reflects”fromthatend.Ifthe
particlereflectsfrombothends,itissaidtobetrappedinamagnetic
bottle.
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February 20, 2016 VIII. SampleproblemsofChargedParticlesinMagneticFields
A. CrossedFields:
1.
Anelectricfieldof1.50kV/mandaperpendicularmagneticfieldof
0.400Tactonamovingelectrontoproducenonetforce.Whatisthe
electron'sspeed?
B. TheHallEffect:
1.
Astripofcopper150μmthickand4.5mmwideisplacedinauniform
magneticfield ofmagnitude0.65T,with perpendiculartothestrip.A
currenti=23AisthensentthroughthestripsuchthataHallpotential
differenceVappearsacrossthewidthofthestrip.CalculateV.(Thenumberof
chargecarriersperunitvolumeforcopperis8.47×1028electrons/m3.)
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February 20, 2016 C. ACirculatingChargedParticle:
1.
AnelectronisacceleratedfromrestthroughpotentialdifferenceVandthen
entersaregionofuniformmagneticfield,whereitundergoesuniformcircular
motion.Figure28‐37givestheradiusrofthatmotionversusV1/2.Theverticalaxis
scaleissetbyrs=3.0mm,andthehorizontalaxisscaleissetby
Whatisthemagnitudeofthemagneticfield?
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February 20, 2016 IX.
MagneticForceonaCurrent‐CarryingWire:
A. AforceactsonacurrentcarryingwireinaBfield:
B. ConsideralengthLofthewireinthefigure.Alltheconduction
electronsinthissectionofwirewilldriftpastplanexxinatimet
=L/vd.
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February 20, 2016 C. Thus,inthattimeachargewillpassthroughthatplanethatis
givenby
HereLisalengthvectorthathasmagnitudeLandisdirectedalong
thewiresegmentinthedirectionofthe(conventional)current.
D. Ifawireisnotstraightorthefieldisnotuniform,wecanimagine
thewirebrokenupintosmallstraightsegments.Theforceonthe
wireasawholeisthenthevectorsumofalltheforcesonthe
segmentsthatmakeitup.Inthedifferentiallimit,wecanwrite
andwecanfindtheresultantforceonanygivenarrangementof
currentsbyintegratingEq.28‐28overthatarrangement.
E. Sampleproblem:
1.
Awire1.80mlongcarriesacurrentof13.0Aandmakesanangleof
35.0°withauniformmagneticfieldofmagnitudeB=1.50T.Calculatethe
magneticforceonthewire.
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February 20, 2016 X.
TorqueonaCurrentLoop:
A. Considernow,aloopofwirewithcurrentflowingthroughit:
B. ThetwomagneticforcesFand–Fproduceatorqueontheloop,
tendingtorotateitaboutitscentralaxis.
C. Todefinetheorientationoftheloopinthemagneticfield,weusea
normalvectornthatisperpendiculartotheplaneoftheloop.Figure
28‐19bshowsaright‐handruleforfindingthedirectionofn.InFig.
28‐19c,thenormalvectoroftheloopisshownatanarbitraryangle
tothedirectionofthemagneticfield.
RHR for µ  Page
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February 20, 2016 D. Forside2themagnitudeoftheforceactingonthissideisF2=ibB
sin(90°‐)=ibBcos=F4.F2andF4canceloutexactly.
E. ForcesF1andF3havethecommonmagnitudeiaB.AsFig.28‐19c
shows,thesetwoforcesdonotsharethesamelineofaction;sothey
produceanettorque.
F. ForNloops,whenA=ab,theareaoftheloop,thetotaltorqueis:
G. SampleProblem
1.
Anelectronmovesinacircleofradiusr=5.29×10‐11mwithspeed
2.19×106m/s.Treatthecircularpathasacurrentloopwithaconstant
currentequaltotheratiooftheelectron'schargemagnitudetotheperiodof
themotion.Ifthecircleliesinauniformmagneticfieldofmagnitude
B=7.10mT,whatisthemaximumpossiblemagnitudeofthetorqueproduced
ontheloopbythefield?
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February 20, 2016 XI.
TheMagneticDipoleMoment,:
A. Definition:
Here,Nisthenumberofturnsinthecoil,iisthecurrentthroughthe
coil,andAistheareaenclosedbyeachturnofthecoil.
B. Direction:Thedirectionofisthatofthenormalvectortothe
planeofthecoil.
C. Thedefinitionoftorquecanberewrittenas:
D. Justasintheelectriccase,themagneticdipoleinanexternal
magneticfieldhasanenergythatdependsonthedipole’sorientation
inthefield:
E. Amagneticdipolehasitslowestenergy(‐Bcos0=‐B)whenits
dipolemomentislinedupwiththemagneticfield.Ithasitshighest
energy(‐Bcos180°=+B)whenisdirectedoppositethefield.
F. Fromthebelowequations,onecanseethattheunitofcanbethe
joulepertesla(J/T),ortheampere–squaremeter.
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February 20, 2016 G. SampleProblem:
1.
Amagneticdipolewithadipolemomentofmagnitude0.020J/Tis
releasedfromrestinauniformmagneticfieldofmagnitude52mT.The
rotationofthedipoleduetothemagneticforceonitisunimpeded.Whenthe
dipolerotatesthroughtheorientationwhereitsdipolemomentisaligned
withthemagneticfield,itskineticenergyis0.80mJ.(a)Whatistheinitial
anglebetweenthedipolemomentandthemagneticfield?(b)Whatisthe
anglewhenthedipoleisnext(momentarily)atrest?
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