5, Magnetism
Permanentmagnets
Earthsmagneticfield
Magneticforce
Motionofchargedparticlesinmagneticfields
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March 2012
University Physics, Chapter 27
1
Permanent Magnets
Examplesofpermanentmagnetsincluderefrigeratormagnetsandmagnetic
doorlatches
Theyareallmadeofcompoundsofiron,nickel,orcobalt
Ifyoutouchanironneedletoapieceofmagneticlodestone,theironneedle
willbemagnetized
Ifyouthenfloatthisironneedleinwater,theneedlewillpointtowardthe
northpoleoftheEarth(approximately!)
Wecalltheendofthemagnetthatpointsnorththenorthpoleofthemagnet
andtheotherendthesouthpoleofthemagnet
March 5, 2012
University Physics, Chapter 27
2
Permanent Magnets - Poles
Magnetsexertforcesononeanother
attractiveorrepulsivedependingon
orientation.
Ifwebringtogethertwopermanent
magnetssuchthatthetwonorthpoles
aretogetherortwosouthpolesare
together,themagnetswillrepeleach
other
Ifwebringtogetheranorthpoleanda
southpole,themagnetswillattract
eachother
March 5, 2012
University Physics, Chapter 27
3
Magnetic Field Lines (1)
Permanentmagnetsinteractwitheachotheratadistance,withouttouching
Inanalogywiththeelectricfield,wedefineamagneticfieldtodescribethe
magneticforce
Aswedidfortheelectricfield,wemayrepresentthemagneticfieldusing
magneticfieldlines
Themagneticfielddirectionisalwaystangenttothemagneticfieldlines
March 5, 2012
University Physics, Chapter 27
4
Magnetic Field Lines (2)
Themagneticfieldlinesfromapermanentbarmagnetareshownbelow
Twodimensionalcomputercalculation
March 5, 2012
University Physics, Chapter 27
Threedimensionalreallife
5
Broken Permanent Magnet
Ifwebreakapermanent
magnetinhalf,wedonot
getaseparatenorthpole
andsouthpole
Whenwebreakabar
magnetinhalf,wealways
gettwonewmagnets,each
withitsownnorthandsouthpole
Unlikeelectricchargethatexistsaspositive(proton)andnegative(electron)
separately,therearenoseparatemagneticmonopoles(anisolatednorthpoleoran
isolatedsouthpole)
Scientistshavecarriedoutextensivesearchesformagneticmonopoles;allresultsare
negative
Magnetismisnotcausedbymagneticparticles!Magnetismiscausedbyelectric
currents
March 5, 2012
University Physics, Chapter 27
6
Magnetic Field Lines
Fortheelectricfield,theelectricforcepointsinthe
samedirectionastheelectricfieldandtheelectricforcewasdefinedintermsof
apositivetestparticle
However,becausethereisnomagneticmonopole,wemustemployothermeansto
definethemagneticforce
Wecandefinethedirectionofthemagneticfieldintermsofthedirectiona
compassneedlewouldpoint
Acompassneedle,withanorthpoleandasouthpole,willorientitselfin
equilibriumsuchthatitsnorthpolepointsinthedirectionofthemagneticfield
Thusthedirectionofthefieldcanbemeasuredatanypointbymovinga
compassneedlearoundinamagneticfieldandnotingthedirectionthatthe
compassneedlepoints
March 5, 2012
University Physics, Chapter 27
7
The Earths Magnetic Field
TheEarthitselfisamagnet
Ithasamagneticfieldsortoflikeabar
magnet(butnotreallylikeabarmagnet)
ThepolesoftheEarths
magneticfieldarenotaligned
withtheEarthsgeographic
polesdefinedastheendpoints
oftheaxisoftheEarthsrotation
TheEarthsmagneticfieldisnotasimpledipolefieldbecauseitis
distortedbythesolarwind
ProtonsfromtheSunmovingat400km/s
Themagneticfieldinside the Earthis ve ryc o m p le x
March 5, 2012
University Physics, Chapter 27
8
Earths Magnetic Field Strength
T h e s tre ng th o fth e Ea rth s m a g ne tic fie ld a tth e
s urfa c e o fEa rth va rie s b e twe e n 0 .2 5 G a n d 0 .6 5 G
Geomagneticfieldstrength
(NationalGeophysicalDataCenter)
C o m p a re to He a ling m a g ne ts (inshoes,asbracelets)
Fieldisatmost1Gatadistanceofd=1mmfromthe
magnetcenteranddecreasesas1/d3
Magnetholderthickness~5mm,skin~1mm
fieldis~0.003Ginmusclesandtissue
Andofcoursethehumanbodyisnonmagnetic
March 5, 2012
University Physics, Chapter 27
9
Earths Magnetic Poles
Thenorthandsouthmagneticpolesarenotexactlylocatedatthenorthand
southgeographicpoles
ThemagneticnorthpoleislocatedinCanada
ThemagneticsouthpoleislocatedontheedgeofAntarctica
Themagneticpolesmovearound,atarateof40kmperyear
Bytheyear2500themagneticnorthpolewillbe
locatedinSiberia
ThereareindicationsthattheEarthsmagneticfieldreverses(NS)onthetimescale
of1millionyearsorso.
March 5, 2012
University Physics, Chapter 27
10
Magnetic Force
Wedefinethemagneticfieldintermsofitseffectonanelectricallycharged
particle(q).
Recallthatanelectricfieldexertsaforceonaparticlewithcharge qgivenby
F = qE ( x )
Amagneticfieldexertsnoforceonastationarycharge
Butamagneticfielddoesexertaforceonachargethatmovesacrossthefield
Thedirectionoftheforceisperpendiculartoboththevelocityofthemoving
chargedparticleandthemagneticfield;themagneticforceissideways
T h e Lo re ntzfo rc e
F = qv B ( x )
March 5, 2012
University Physics, Chapter 27
11
Right Hand Rule (1)
is ive n b yth e rig h t
T h e d ire c tio no fth e c ro s s p ro d uc t() B
v g
h a n d rule
To a p p lyth e rig h th a n d ru le
Us e yo u rrig h th a nd !
Alig n th u m b in th e d ire c tio n o f v
Alig n yo urind e xfing e rwith th e m a g ne tic fie ld
Yo urm id d le fing e rwillp o intin th e d ire c tio n o fth e c ro s s
p ro d uc t vx B
March 5, 2012
University Physics, Chapter 27
12
Right Hand Rule (2)
Wh a ta b o utth e s ig no fth e c h a rg e ?
If qis p o s itive ,th e nis in th e s a m e
F
d ire c tio n a s
v B
If q is ne g a tive , Fis F
inth e o p p o s ite
d ire c tio n
March 5, 2012
University Physics, Chapter 27
13
Magnitude of Magnetic Force
T h e m a g nitu d e o fth e m a g n e tic fo rc e o n a m o ving c h a rg e is
FB = qvB sin
wh e re is th e a ng le b e twe e n th e ve lo c ityo fth e c h a rg e d
p a rtic le a nd th e m a g n e tic fie ld .
Do yo u s e e th a tth e re is no m a g n e tic fo rc e o n a c h a rg e d
p a rtic le m o vin g p a ra lle lto th e m a g ne tic fie ld ?
(Be c a u s e is ze ro a n d s in(0 ) =0 )
Do yo u s e e wh e nth e m a g ne tic fo rc e is m o s ts tro n g ?
(Ma x.fo rc e is fo r =9 0 d e g re e s ;th e n F=qvB)
March 5, 2012
University Physics, Chapter 27
14
Units of Magnetic Field Strength
Themagneticfieldstrengthhasreceiveditsownnamedunit,thetesla(T),
namedinhonorofthephysicistandinventorNikolaTesla(18561943)
Check unit
consistency:
F=qvB
N = C (m/s) T
Ateslaisaratherlargeunitofthemagneticfieldstrength
Sometimesyouwillfindmagneticfieldstrengthstatedinunitsofgauss(G),(not
anofficialSIunit)
Ns
N
1T=1
=1
Cm
Am
-4
1 G = 10 T
March 5, 2012
10 k G = 1 T
University Physics, Chapter 27
15
Example: Proton in B Field
Consideraregionofuniformmagneticfield
(greendots)comingoutofthepagewith
magnitudeB=1.2mT.Aprotonwithkinetic
energyK=8.48 1013Jentersthefield,
movingverticallyfromthebottomtothetop.
Question:Calculatethemagneticdeflectingforceontheproton.
Answer:UseF=qvBsin.First,figureoutthevelocityv.
12
2K
K = mv v =
2
m
v=
March 5, 2012
(
2 8.48 13 J
10
1.67
10
27
kg
) = 3.2 10
University Physics, Chapter 27
7
m/s
16
Example: Proton in B Field (2)
AnglebetweenBfielddirectionand
velocityoftheproton: = 9 0 deg
F = qvB sin
F = ( 1.60 19 C ) ( 3.2 7 m/s ) ( 1.2 3 T ) sin 90
10
10
10
F = 6.1 15 N
10
F
6.1 15 N
10
a= =
= 3.7 12 m/s 2
10
27
m 1.67
10 kg
Smallforcebutlargeaccelerationforlight
particle
DirectionofF:UseRightHandRule
March 5, 2012
University Physics, Chapter 27
17
Van Allen Radiation Belts
March 5, 2012
University Physics, Chapter 27
18
Example: Cathode Ray Tube (1)
Consideracathoderaytube.
Inthistubeelectronsformanelectron
beamwhenacceleratedhorizontallybya
voltageof136Vinanelectrongun.
Themassofanelectronis9.1094 1031k gwhiletheelementarychargeis
1.6022 1019C.
Question:Calculatethevelocityoftheelectronsinthebeamafterleavingthe
electrongun.
Answer:
K U = = qV
1
2
March 5, 2012
mv 2 = eV
implies
v = 6.92 106 m/s
University Physics, Chapter 27
19
Example: Cathode Ray Tube (2)
Question:Ifthetubeisplacedinauniformmagneticfield,inwhatdirectionisthe
electronbeamdeflected?
Answer:
downward
Question:Calculatethemagnitudeofaccelerationofanelectronifthefield
strengthis3.65104T.
Answer:
F = ma = qvB
1.6022 19 C 6.92 6 m/s 3.65 4 T
10
10
10
qvB
a=
=
= 4.44 14 m/s 2
10
31
m
9.1094
10 kg
(
March 5, 2012
)(
)(
University Physics, Chapter 27
)
20
Particle Orbits in a Uniform B (1)
Particle
(1)
Tieastringtoarockandtwirlitatconstantspeedinacircleoveryourhead.
Thetensionofthestringprovidesthecentripetalforcethatkeepstherock
movinginacircle.
Thetensiononthestringalwayspointstothecenterofthecircleandcreatesa
centripetalacceleration.
Aparticlewithchargeqandmassmmoveswithvelocityvperpendiculartoa
uniformmagneticfieldB.
Theparticlewillmoveinacirclewithaconstantspeedvandthemagneticforce
F=qvBwillkeeptheparticlemovinginacircle.
March 5, 2012
University Physics, Chapter 27
21
Particle Orbits in Uniform B (2)
Particle
Recallcentripetala c c e le ra tio n
Ne wto n s s e c o nd la w
v2
a=
r
F = ma
S o ,fo ra c h a rg e d p a rtic le qinc irc ula rm o tio n ina m a g n e tic fie ld
B
2
v
m = qvB
r
O r
March 5, 2012
v
m = qB
r
p
= qB
r
University Physics, Chapter 27
22
Example: Mass Spectrometer (1)
Themagneticfieldinamassspectrometer
is80mT,andthepotentialdifferenceis
1000V.Achargedion(1.60221019C)
entersthechamberandstrikesthe
detectoratadistancex=1.6254m.
Question:Whatisthemassoftheion?
Answer:
Drawthepicture,addthepathofthe
particle
Thismustbeaboutthetrajectoryofachargedparticle
inamagneticfield,andhowtheiongetsacceleratedonthe
straightsegmentofitspath.
March 5, 2012
University Physics, Chapter 27
23
Example: Mass Spectrometer (2)
Answer:
Separatetheproblemintotwoparts:
Accelerationinelectricfield,ionacquiresenergy
energyconservation
Bendinginmagneticfieldcirculartrajectory
Whichpartismassdependent?
March 5, 2012
University Physics, Chapter 27
24
Example: Mass Spectrometer (3)
Electricfield,energyconservation
12
K = mv
2
U = qV
K + U = 0
12
mv qV = 0
2
mv
=
Themagneticfieldisgivenbyandrr=0.5x
qB
March 5, 2012
University Physics, Chapter 27
25
Example: Mass Spectrometer (4)
12
2qV
mv qV = 0 v =
2
m
mv m 2qV 1 2mV x
r=
=
=
=
qB qB m
B
q
2
B 2 qx 2
m=
=
8V
10
( 0.08 T ) ( 1.6022 19 C ) ( 1.6254 m )
2
m=
2
8 ( 1000 V )
m = 3.3863 25 kg
10
March 5, 2012
University Physics, Chapter 27
26
Example: Mass Spectrometer (5)
Canthisbetrue?
M=3.41025k g
atomicmassunit:
1u=1.71027k g
thisionis~200u,
anditmightbemercury
March 5, 2012
University Physics, Chapter 27
27
Example: Charged Particle Tracks
Inelementaryparticlephysicsandnuclearphysics,particlesareproducedinthe
collisionsofprotonsorofothernuclei
Examples:
protonantiprotonattheTevatronatFermilab,
protonprotonattheLHCatCERN ,
AuAuatRHICatBrookhaven
Eachcollisionproducesmanyparticlesperpendiculartothebeamdirection
Themomentumofeachparticleismeasured
bytracingthepathofeachparticleinamagneticfield
March 5, 2012
University Physics, Chapter 27
28
Orbits in a Constant Magnetic Field
Ifaparticleperformsacompletecircularorbitinsideaconstantmagneticfield,
thentheperiodofrevolutionoftheparticleisjustthecircumferenceofthecircle
dividedbythespeed
2 r 2 m
T=
=
v
qB
Fromtheperiodwecangetthefrequencyandangularfrequency
1
qB
qB
f= =
= 2 f =
Thefrequencyoftherotationisindependentofthespeedoftheparticle.
T 2 m
m
Isochronousorbits
Basisforthecyclotron
March 5, 2012
University Physics, Chapter 27
29
Force on a Current Carrying Wire (1)
Consideralong,straightwire
carryingacurrenti inaconstant
magneticfieldB
Themagneticfieldwillexert
aforceonthemovingcharges
inthewire
Thechargeqflowinginthewire
inagiventimet inalengthLofwireisgivenby
L
q = ti = i
v
wherevisthedriftvelocityoftheelectrons
March 5, 2012
University Physics, Chapter 27
30
Force on a Current Carrying Wire (2)
Themagnitudeofthemagneticforceisthen
L
F = qvB sin = i vB sin = iLB sin
v
istheanglebetweenthecurrentandthemagneticfield
Thedirectionoftheforceisperpendiculartoboththecurrentandthemagnetic
fieldandisgivenbytherighthandrule
Thisequationcanbeexpressedasavectorcrossproduct
F = iL B F = iLB for L B
iLrepresentsthecurrentinalengthLofwire
March 5, 2012
University Physics, Chapter 27
31
Torque on a Current-Carrying Loop (1)
Electricmotorsrelyonthemagneticforceexertedonacurrentcarryingwire
Thisforceisusedtocreateatorquethatturnsashaft
Asimpleelectricmotorisdepictedbelowconsistingofasingleloopcarrying
currentiinaconstantmagneticfieldB
Thetwomagneticforces,FandF,showninthefigureareofequalmagnitude
andoppositedirection
Theseforcescreateatorquethattendstorotatethelooparounditsaxis
March 5, 2012
University Physics, Chapter 27
32
Torque on a Current-Carrying Loop (2)
Asthecoilturnsinthefield,theforcesonthesidesoftheloopperpendicularto
themagneticfieldwillchange
Theforcesactonthesidesofthesquareloopbelow,whereistheanglebetween
anormalvector,n,andthemagneticfieldB
Thenormalvectorisperpendiculartotheplaneofthewireloopandpointsina
directiongivenbytherighthandrulebasedonthecurrentflowingintheloop
March 5, 2012
University Physics, Chapter 27
33
Torque on a Current-Carrying Loop (3)
Herethecurrentisflowingupwardinthetop
segmentanddownwardinthelowersegment
asillustratedbythearrowfeathersand
arrowhead
Theforceofeachoftheverticalsegmentsis
F = iaB (remember F = iLB )
Theforceontheothertwosidesisparallelorantiparalleltotheaxisofrotation
andcannotcauseatorque
March 5, 2012
University Physics, Chapter 27
34
Torque on a Current-Carrying Loop (4)
Thesumofthetorqueontheuppersideplusthetorqueonthelowersidegives
thetorqueexertedonthecoilaboutthecenteroftheloop
=rF
a
a
1 = iaB sin + iaB sin = ia 2 B sin = iAB sin
2
2
whereA=a2
()
March 5, 2012
()
University Physics, Chapter 27
35
Magnetic Dipole Moment (1)
IfwereplacethisloopwithNloopswoundclose
togetherwecanwrite
= N 1 = NiAB sin
Althoughwederivedthisexpressionforasquareloop,thisexpressionappliesto
circularloopsaswell,aslongasthemagneticfieldisuniform(anytypeloops
elliptical,triangle,etc.)
Wecandescribethiscoilwithoneparameterconsistingofinformationaboutthe
coilonly,combinedwithinformationaboutthemagneticfield
Wedefinethemagnitudeofthemagneticdipolemomentofthecoilabovetobe
= Ni A
March 5, 2012
University Physics, Chapter 27
36
Magnetic Dipole Moment (2)
Thedirectionofthemagneticdipolemoment,,isgivenbytherighthandrule
andpointsinthedirectionofthesurfacenormal vector
n
Wecanrewriteourexpressionforthetorqueas
= ( NiA ) B sin = B sin
whichwecangeneralizeto
= B
Thetorquewillalwaysbeperpendicular
tothemagneticdipolemomentandthe
magneticfield
March 5, 2012
University Physics, Chapter 27
37
Example: Current Carrying Loop (1)
Asingleturncurrentloop,carryingacurrent
of4A,isintheshapeofatrianglewithsides
50,120,and130cm.Theloopisinauniform
magneticfieldofstrengthB=75mT(direction
paralleltothe130cmsideoftheloop).
Question:Whatisthemagnitudeofthemagnetic
forceonthe130cmside?
Answer:
For the 130 cm side:
F = iL B
iL B F = 0 N
March 5, 2012
University Physics, Chapter 27
38
Example: Current Carrying Loop (2)
Question:Whatisthemagnitudeofthemagnetic
forceonthe120cmside?
Answer:Lookatthexandycomponentsofthe
magneticfield!
Bx = B cos
By = B sin
= tan 1 ( 50 cm /120 cm ) = 22.6
There is no contribution from Bx on iL120 since Bx PiLx
Force due to By is iL120 By
F = iLx By = ( 4 A ) ( 1.20 m ) ( 0.075 T ) sin ( 22.6 ) = 0.138 N
March 5, 2012
University Physics, Chapter 27
39