CH 28 Structured Notes 2015 - USNA 28 Structured Notes with...electrically charged particles, such as a current in a wire, to make an electromagnet. The current produces a magnetic

[SHIVOK SP212] February 20, 2016

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CH 28

MagneticFields

I. WhatProducesMagneticField?

A. Onewaythatmagneticfieldsareproducedistousemovingelectricallychargedparticles,suchasacurrentinawire,tomakeanelectromagnet.Thecurrentproducesamagneticfieldthatisutilizable.

B. Theotherwaytoproduceamagneticfieldisbymeansofelementaryparticlessuchaselectrons,becausetheseparticleshaveanintrinsicmagneticfieldaroundthem.

1. Themagneticfieldsoftheelectronsincertainmaterialsaddtogethertogiveanetmagneticfieldaroundthematerial.Suchadditionisthereasonwhyapermanentmagnet,hasapermanentmagneticfield.

2. Inothermaterials,themagneticfieldsoftheelectronscancelout,givingnonetmagneticfieldsurroundingthematerial.


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II. TheDefinitionofBfield:

A. Wecandefineamagneticfield,B,byfiringachargedparticlethroughthepointatwhichistobedefined,usingvariousdirectionsandspeedsfortheparticleanddeterminingtheforcethatactsontheparticleatthatpoint.Bisthendefinedtobeavectorquantitythatisdirectedalongthezero‐forceaxis.

B. Themagneticforceonthechargedparticle,FB,isdefinedtobe:

Hereqisthechargeoftheparticle,visitsvelocity,andBthemagneticfieldintheregion.

C. Themagnitudeofthisforceisthen:

HereisthesmallestanglebetweenvectorsvandB.

D. FindingtheMagneticForceonaParticle:

1. TheforceFBactingonachargedparticlemovingwithvelocityvthroughamagneticFieldBisALWAYSperpendiculartobothvandB.


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E. TheSIunitforBthatfollowsisnewtonpercoulomb‐meterpersecond.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)Doesthespeedoftheparticleincrease,decrease,orremainthesame?


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Sample problem continued:


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III. MagneticFieldLines:

A. ThedirectionofthetangenttoamagneticfieldlineatanypointgivesthedirectionofBatthatpoint.

B. ThespacingofthelinesrepresentsthemagnitudeofB—themagneticfieldisstrongerwherethelinesareclosertogether,andconversely.

C. Oppositemagneticpolesattracteachother,andlikemagneticpolesrepeleachother.

D. NosuchthingasaMagneticMonopoleEveryNorthPolehasanassociatedSouthPole!

E. FieldlinesemanatefromtheNorthPoleandterminateintheSouthPole.


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IV. CrossedFields,DiscoveryofanElectron:

A. Figure

B. WhenthetwofieldsinFig.28‐7areadjustedsothatthetwodeflectingforcesactingonthechargedparticlecancel,wehave

C. Thus,thecrossedfieldsallowustomeasurethespeedofthechargedparticlespassingthroughthem.

D. Thedeflectionofachargedparticle,movingthroughanelectricfield,E,betweentwoplates,atthefarendoftheplates(inthepreviousproblem)is

Here,vistheparticle’sspeed,mitsmass,qitscharge,andListhelengthoftheplates.


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V. CrossedFields,TheHallEffect:

A. AHallpotentialdifferenceVisassociatedwiththeelectricfieldacrossstripwidthd,andthemagnitudeofthatpotentialdifferenceisV=Ed.Whentheelectricandmagneticforcesareinbalance(Fig.28‐

8b), wherevdisthedriftspeed.But,

1. WhereJisthecurrentdensity,Athecross‐sectionalarea,etheelectroniccharge,andnthenumberofchargesperunitvolume.

B. Therefore, here,l=(A/d),thethicknessofthestrip.

1. Fig.28‐8Astripofcoppercarryingacurrentiisimmersedinamagneticfield.(a)Thesituationimmediatelyafterthemagneticfieldisturnedon.Thecurvedpaththatwillthenbetakenbyanelectronisshown.(b)Thesituationatequilibrium,whichquicklyfollows.Notethatnegativechargespileupontherightsideofthestrip,leavinguncompensatedpositivechargesontheleft.Thus,theleftsideisatahigherpotentialthantherightside.(c)Forthesamecurrentdirection,ifthechargecarrierswerepositivelycharged,theywouldpileupontherightside,andtherightsidewouldbeatthehigherpotential.


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VI. ACirculatingChargedParticle:

A. Consideraparticleofchargemagnitude|q|andmassmmovingperpendiculartoauniformmagneticfieldB,atspeedv.

B. Themagneticforcecontinuouslydeflectstheparticle,andsinceBandvarealwaysperpendiculartoeachother,thisdeflectioncausestheparticletofollowacircularpath.

C. Themagneticforceactingontheparticlehasamagnitudeof|q|vB.

D. Foruniformcircularmotion


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E. Diagram

1. Fig.28‐10Electronscirculatinginachambercontaininggasatlowpressure(theirpathistheglowingcircle).Auniformmagneticfield,B,pointingdirectlyoutoftheplaneofthepage,fillsthechamber.NotetheradiallydirectedmagneticforceF

B;forcircularmotiontooccur,F

Bmustpoint

towardthecenterofthecircle,(CourtesyJohnLeP.Webb,SussexUniversity,England)

VII. HelicalPaths:

A. Fig.28‐11(a)Achargedparticlemovesinauniformmagneticfield,theparticle’svelocityvmakingananglefwiththefielddirection.(b)Theparticlefollowsahelicalpathofradiusrandpitchp.(c)Achargedparticlespiralinginanonuniformmagneticfield.(Theparticlecanbecometrapped,spiralingbackandforthbetweenthestrongfieldregionsateitherend.)Notethatthemagneticforcevectorsattheleftandrightsideshaveacomponentpointingtowardthecenterofthefigure.


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B. Thevelocityvector,v,ofsuchaparticleresolvedintotwocomponents,oneparalleltoandoneperpendiculartoit:

C. Theparallelcomponentdeterminesthepitchpofthehelix(thedistancebetweenadjacentturns(Fig.28‐11b)).Theperpendicularcomponentdeterminestheradiusofthehelix.

Pitch = and Radius =

D. Themorecloselyspacedfieldlinesattheleftandrightsidesindicatethatthemagneticfieldisstrongerthere.Whenthefieldatanendisstrongenough,theparticle“reflects”fromthatend.Iftheparticlereflectsfrombothends,itissaidtobetrappedinamagneticbottle.


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VIII. SampleproblemsofChargedParticlesinMagneticFields

A. CrossedFields:

1. Anelectricfieldof1.50kV/mandaperpendicularmagneticfieldof0.400Tactonamovingelectrontoproducenonetforce.Whatistheelectron'sspeed?

B. TheHallEffect:

1. Astripofcopper150μmthickand4.5mmwideisplacedinauniform

magneticfield ofmagnitude0.65T,with perpendiculartothestrip.Acurrenti=23AisthensentthroughthestripsuchthataHallpotentialdifferenceVappearsacrossthewidthofthestrip.CalculateV.(Thenumberofchargecarriersperunitvolumeforcopperis8.47×1028electrons/m3.)


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C. ACirculatingChargedParticle:

1. AnelectronisacceleratedfromrestthroughpotentialdifferenceVandthenentersaregionofuniformmagneticfield,whereitundergoesuniformcircularmotion.Figure28‐37givestheradiusrofthatmotionversusV1/2.Theverticalaxis

scaleissetbyrs=3.0mm,andthehorizontalaxisscaleissetby .Whatisthemagnitudeofthemagneticfield?


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IX. MagneticForceonaCurrent‐CarryingWire:

A. AforceactsonacurrentcarryingwireinaBfield:

B. ConsideralengthLofthewireinthefigure.Alltheconductionelectronsinthissectionofwirewilldriftpastplanexxinatimet=L/vd.


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C. Thus,inthattimeachargewillpassthroughthatplanethatisgivenby

HereLisalengthvectorthathasmagnitudeLandisdirectedalongthewiresegmentinthedirectionofthe(conventional)current.

D. Ifawireisnotstraightorthefieldisnotuniform,wecanimaginethewirebrokenupintosmallstraightsegments.Theforceonthewireasawholeisthenthevectorsumofalltheforcesonthesegmentsthatmakeitup.Inthedifferentiallimit,wecanwrite

andwecanfindtheresultantforceonanygivenarrangementofcurrentsbyintegratingEq.28‐28overthatarrangement.

E. Sampleproblem:

1. Awire1.80mlongcarriesacurrentof13.0Aandmakesanangleof35.0°withauniformmagneticfieldofmagnitudeB=1.50T.Calculatethemagneticforceonthewire.


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X. TorqueonaCurrentLoop:

A. Considernow,aloopofwirewithcurrentflowingthroughit:

B. ThetwomagneticforcesFand–Fproduceatorqueontheloop,tendingtorotateitaboutitscentralaxis.

C. Todefinetheorientationoftheloopinthemagneticfield,weuseanormalvectornthatisperpendiculartotheplaneoftheloop.Figure28‐19bshowsaright‐handruleforfindingthedirectionofn.InFig.28‐19c,thenormalvectoroftheloopisshownatanarbitraryangletothedirectionofthemagneticfield.

RHR for µ


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D. Forside2themagnitudeoftheforceactingonthissideisF2=ibB

sin(90°‐)=ibBcos=F4.F

2andF

4canceloutexactly.

E. ForcesF1andF

3havethecommonmagnitudeiaB.AsFig.28‐19c

shows,thesetwoforcesdonotsharethesamelineofaction;sotheyproduceanettorque.

F. ForNloops,whenA=ab,theareaoftheloop,thetotaltorqueis:

G. SampleProblem

1. Anelectronmovesinacircleofradiusr=5.29×10‐11mwithspeed2.19×106m/s.Treatthecircularpathasacurrentloopwithaconstantcurrentequaltotheratiooftheelectron'schargemagnitudetotheperiodofthemotion.IfthecircleliesinauniformmagneticfieldofmagnitudeB=7.10mT,whatisthemaximumpossiblemagnitudeofthetorqueproducedontheloopbythefield?


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XI. TheMagneticDipoleMoment,:

A. Definition:

Here,Nisthenumberofturnsinthecoil,iisthecurrentthroughthecoil,andAistheareaenclosedbyeachturnofthecoil.

B. Direction:Thedirectionofisthatofthenormalvectortotheplaneofthecoil.

C. Thedefinitionoftorquecanberewrittenas:

D. Justasintheelectriccase,themagneticdipoleinanexternalmagneticfieldhasanenergythatdependsonthedipole’sorientationinthefield:

E. Amagneticdipolehasitslowestenergy(‐Bcos0=‐B)whenitsdipolemomentislinedupwiththemagneticfield.Ithasitshighestenergy(‐Bcos180°=+B)whenisdirectedoppositethefield.

F. Fromthebelowequations,onecanseethattheunitofcanbethejoulepertesla(J/T),ortheampere–squaremeter.


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G. SampleProblem:

1. Amagneticdipolewithadipolemomentofmagnitude0.020J/Tisreleasedfromrestinauniformmagneticfieldofmagnitude52mT.Therotationofthedipoleduetothemagneticforceonitisunimpeded.Whenthedipolerotatesthroughtheorientationwhereitsdipolemomentisalignedwiththemagneticfield,itskineticenergyis0.80mJ.(a)Whatistheinitialanglebetweenthedipolemomentandthemagneticfield?(b)Whatistheanglewhenthedipoleisnext(momentarily)atrest?

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CH 28 Structured Notes 2015 - USNA 28 Structured Notes with...electrically charged particles, such as a current in a wire, to make an electromagnet. The current produces a magnetic