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Chapter2ParticleMotion inElectricandMagneticFieldsConsideringEandBtobegiven,westudythetrajectoryofparticlesundertheinfluenceofLorentzforce
F=q(E+vB) (2.1)2.1 ElectricFieldAlone
dvm =qE (2.2)
dtOrbitdependsonlyonratioq/m. UniformE uniformacceleration. Inone-dimensionz,Ez trivial. Inmultipledimensionsdirectlyanalogoustoparticlemovingunder influenceof
qgravity. AccelerationgravitygmE. Orbitsareparabolas. Energy isconservedtaking
z
x
__m
qE
Figure2.1: Inauniformelectricfield,orbitsareparabolic,analogoustogravity.intoaccountpotentialenergy
P.E.=q electricpotential (2.3)
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[Proofifneeded,regardlessofEspatialvariation,dv d
ddt
mdt.v
12
mv2
==
q.v=qd
dt(q)
dt (2.4)(2.5)
i.e. 1mv2 +q2 =const.]
Aparticlegainskineticenergyqwhenfallingthroughapotentialdrop-. Soconsidertheaccelerationandsubsequentanalysisofparticleselectrostatically: Howmuchdeflection
=0 a+ /2
E
L
Electrostatic AnalyserParticleSource
DetectingScreen
s
d
z
x
z0
a /2
Figure2.2: Schematicofelectrostaticaccelerationandanalysis.willtherebe? AfteraccelerationstageKE= 1mv2 =qs2 x
vx = 2qs . (2.6)m
SupposingEa,fieldofanalyser,tobepurelyz,thisvelocityissubsequentlyconstant. Withintheanalyser
dvz q q xm =qEa vz = Eat= Ea . (2.7)
dt m m vxSo
q t2 q 1x2z= vzdt= Ea = Ea . (2.8)
m 2 m 2v2xHenceheightatoutputofanalyseris
q 1L2 q 1L2 mzo = Ea = Ea
m 2v2 m 2 (2qs)x=
4
1Es
aL2 =+
4
1
a
sL
d2
(2.9)
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using Ea = a/d. Notice this is independent of q and m! We could see this directly byeliminatingthetimefromourfundamentalequationsnoting
dt
d=v
d
d(=v.) with v= 2q(
ms)
or v=
2m
qs +E
qs(2.10)
ifthereisinitialenergyEs. Soequationofmotionis mq
2q(s)m
dd 2q(s)m ddx=2
s d
d
sdxd =Ea = , (2.11)
whichisindependentofqandm. Trajectoryofparticleinpurelyelectrostaticfielddependsonlyonthefield(andinitialparticlekineticenergy/q). Ifinitialenergyiszero,cantdeduceanythingaboutq, m.
2.2 ElectrostaticAccelerationandFocussingAcceleratedchargedparticlebeamsarewidelyusedinscienceandineverydayapplications.Examples:X-raygenerationfrome-beams(Medical,Industrial)ElectronmicroscopesWelding. (e-beam)SurfaceionimplantationNuclearactivation(ion-beams)NeutrongenerationTelevisionand(CRT)MonitorsForapplicationsrequiring
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Clearly getting the required energy is simple. Ensure the potential difference is right andparticles are singly charged: Energy (eV) Potential V. More interesting question: Howtofocusthebeam? Whatdowemeanbyfocussing?
Object Image
Focal Point
(a) Optical Focussing
Lens
(b) Particle beam Focussing
Parallel Rays
Figure2.4: Analogybetweenopticalandparticle-beamfocussing.What is required of the Lens? To focus at a single spot we require the ray (particle
path) deviation from a thin lens to be systematic. Specifically, all initially parallel raysconvergetoapointifthelensdeviatestheirdirectionbysuchthat
r
Deviation Angle
Distance from axis
Parallel Rays
Figure2.5: Requirementforfocussingisthattheangulardeviationofthepathshouldbealinearfunctionofthedistancefromtheaxis.
r=ftan (2.12)andforsmallangles,,r=f . Thislineardependence(=r/f))ofthedeviation,ondistancefromtheaxis,r,isthekeyproperty. ElectrostaticLenswouldliketohave(e.g.)
EaEr = r (2.13)
abutthelenscanthavechargedsolidsinitsmiddlebecausethebeamsmustpassthroughso(initially)= 0 .E=0. ConsequentlypureEr isimpossible(0=.E= 1(rEr)/r=r
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2Ea/a Ea =0). Foran axisymmetric lens (/ =0)we musthave bothEr and Ez.Perhaps the simplest way to arrange appropriate Er is to have an aperture between two
E1
E2
Potential Contours
Figure2.6: Potentialvariationnearanaperturebetweentworegionsofdifferentelectricfieldgivesrisetofocussing.regions of unequal electric field. The potential contours bow out toward the lower fieldregion: givingEr.Calculating focal lengthofapertureRadialacceleration.
1 2
r
z
Figure2.7: Coordinatesnearanaperture.dvr q
= Er (2.14)dt m
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Sodvrdz =
1vz
dvrdt =
qm
Ervz (2.15)
But.E=0 1
r(rEr)
r +
Ezz
=0 (2.16)Neartheaxis,onlythelinearpartofEr isimportanti.e.
ErEr(r, z)r (2.17)
rr=0
r=0
So1 Er
rEr 2rr (2.18)r
andthusEr Ez
2 + =0 (2.19)r z
r=0andwemaywriteEr 21rEz/z. Then
dvr qr Ezdz =2mvz z , (2.20)
whichcanbeintegratedapproximatelyassumingthatvariationsinrandvz canbeneglectedinlenstoget
final qr 2vr = [vr]initial = [Ez]1 (2.21)2mvzTheangulardeviationistherefore
= vr = +qr [Ez2 Ez1] (2.22)vz 2mvz
2andthefocallengthisf =r/
f = 2mvz2 = 4E (2.23)q(Ez2 Ez1) q(Ez2 Ez1)
WhenE1 isanacceleratingregionandE2 iszeroorsmallthelensisdiverging. Thismeansthatjust depending on an extractor electrode to form an ion beam will give a divergingbeam. Needtodomorefocussingdownstream: moreelectrodes.2.2.1 ImmersionLensTwotubesatdifferentpotentialseparatedbygap InthiscasethegapregioncanbethoughtofasanaperturebutwiththeelectricfieldsE1, E2 thesame(zero)onbothsides. Previouseffectiszero. Howevertwoothereffects,neglectedpreviously,givefocussing:
1. vz isnotconstant.2. r isnotconstant.
Consideranacceleratinggap: q(2 1)
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ExtractionElectrode
DivergingSource
Figure2.8: Theextractionelectrodealonealwaysgivesadivergingbeam.
Region 1 Region 2
Tube
1 2
Figure2.9: AnImmersionLensconsistsofadjacentsectionsoftubeatdifferentpotentials.Effect (1) ions are converged in region 1, diverged in region 2. However because of z-acceleration, vz is higher in region 2. The diverging action lasts a shorter time. Henceoverallconverging.Effect (2) TheelectricfieldEr isweakeratsmallerr. Becauseofdeviation,r issmallerindivergingregion. Henceoverallconverging.
Foradeceleratinggapyoucaneasilyconvinceyourselfthatbotheffectsarestillconverging.[Timereversalsymmetryrequiresthis.] Onecanestimatethefocallengthas
1 3f
16
2q2 E
2z
dz (forweakfocussing) (2.24)r=0
but numerical calculations give the values in figure 2.10 where 1 = E/q. Here E is theenergy inregion1. Effect(2)above,thatthe focussingordefocussingdeviation isweakerat points closer to the axis, means that it is a general principle that alternating lenses ofequal converging and diverging power give a net converging effect. This principle can beconsideredtobethebasisfor
40
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Figure 2.10: Focal length of Electrostatic Immersion Lenses. Dependence on energy perunitcharge()inthetworegions,fromS.Humphries19862.2.2 AlternatingGradientFocussingIdeaistoabandonthecylindricallysymmetricgeometrysoastoobtainstrongerfocussing.ConsideranelectrostaticconfigurationwithEz =0and
dEx dExEx = x with =const. (2.25)dx dxSince.E=0,wemusthave
dEx dEy dEy dExdx + dx = 0 dy =const Ey = dx y (2.26)
Thissituationarisesfromapotential 1dEx
= x2 2 (2.27)y2 dx
so equipotentials arehyperbolas x2 y2 =const. If qdEx/dx is negative, thenthisfield isconverginginthex-direction,butdEy/dy=dEx/dx,soitis,atthesametime,diverginginthey-direction. Byusingalternatingsectionsof+veand-vedEx/dxanetconvergingfocuscanbeobtained inboththexandy directions. Thisalternatinggradientapproach isveryimportant for high energy particle accelerators, but generally magnetic, not electrostatic,fieldsareused. Sowellgointoitmorelater.
41
Image removed due to copyright restrictions.
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2.3 UniformMagneticfielddv
m =q(vB) (2.28)dt
TakeB in z-direction. Neveranyforcein z-dir. vz =constant. Perpendiculardynamicsareseparate.
v B
zy
B
v
x
Figure2.11: Orbitofaparticleinauniformmagneticfield.
2.3.1 Brute forcesolution:vx = q vyB vy = qvxB (2.29)
m m 2 2qB qBvx =
m vx vy =
m vy (2.30)
SolutionqB qB
vx =vsin t vy =vcos tmm qB m mqB , (2.31)x=v cos t+x0 y=v sin t+y0
qB m qB mthe equation of a circle. Center (x0, y0) and radius (vm/qB) are determined by initialconditions.2.3.2 PhysicsSolution
1. Magneticfield forcedoesnoworkonparticlebecauseFv. Consequentlytotal |v|isconstant.
2. Forceisthusconstant,tov. Givesrisetoacircularorbit.v2 Force vB mv3. Centripetalaccelerationgivesr = mass =qm i.e. r= qB. Thisradius iscalledthe
Larmor(orgyro)Radius.4. Frequency of rotation v = qB is called the Cyclotron frequency (angular fre
r mquency,s1,notcycles/sec,Hz).
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Whenweaddtheconstantvz wegetahelicalorbit. Cyclotronfrequency=qB/mdependsonlyonparticlecharacterq, mandB-strengthnotv(nonrelativistically,seeaside). LarmorRadiusr=mv/qBdependsonparticlemomentummv. All(non-relativistic)particleswithsame q/m have same . Different energy particles have different r. This variation can beusedtomakemomentumspectrometers.2.3.3 RelativisticAsideRelativisticdynamicscanbewritten
ddtp=q([E+]vB) (2.32)
whererelativisticmomentumisp=mv= m0v
1v2
c2. (2.33)
Massmisincreasedbyfactor 12v2
= 1c2 (2.34)
relative to restmass m0. Since forE=0the velocity |v| = const, isalso constant, andso ism. Thereforedynamicsofaparticle inapurelymagneticfieldcanbecalculatedasifitwere non-relativistic: m dv/dt=q(vB), exceptthattheparticle hasmass greater byfactor thanitsrestmass.2.3.4 MomentumSpectrometersParticlespassingverticallythroughslittakedifferentpathsdependingonmv/q. Bymea-
x
v
Different
Momenta
Unform
B
Slit Detection Plane
Figure2.12: Differentmomentumparticlesstrikethedetectionplaneatdifferentpositions.suringwhereaparticlehitsthedetectionplanewemeasureitsmomentum/q :
mv mv Bx2
qB =x : q = 2 . (2.35)43
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Whymakethedetectionplaneadiameter? Becausedetectionpositionis leastsensitivetovelocitydirection. Thisisaformofmagneticfocussing. Ofcoursewedontneedtomakethefull360,soanalysercanbereducedinsize.
UnformB
v
Different
Momenta
Unform
B
Slit Detection Plane
(b)
v
Slit Focus
Different Angles
(a)
Entrance
Figure2.13: (a)Focussingisobtainedfordifferentinputanglesbyusing180degreesoforbit.(b)Theotherhalfoftheorbitisredundant.Even
so,
it
may
be
inconvenient
to
produce
uniform
B
of
sufficient
intensity
over
sufficiently
largeareaifparticlemomentumislarge.2.3.5 HistoricalDayDream(J.J.Thomson1897)Cathoderays: howtotelltheirchargeandmass?ElectrostaticDeflectionTells only their energy/q = E/q and we have no independent way to measure E since thesamequantityE/qjustequalsacceleratingpotential,whichisthethingwemeasure.MagneticDeflectionTheradiusofcurvatureis
mvr= (2.36)
qBSocombinationofelectrostaticandelectromagneticgivesus
1 2mv mv2 =M1 and =M2 (2.37)
q qHence
2M1 qM22 = m. (2.38)Wecanmeasurethecharge/massratio. Inordertocompletethejobanindependentmeasureofq(orm)wasneeded. Millikan(1911-13). [ActuallyTownsendinJ.J.Thomsonslabhadanexperimenttomeasureqwhichwaswithinfactor2correct.]
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2.3.6 PracticalSpectrometerIn fusion research fast ion spectrum is often obtained by simultaneous electrostatic andelectromagnetic analysis E
1 parallel to B. This allows determination of E/q and q/m velocity of particle [E = mv2]. Thus e.g. deuterons and protons can be distinguished.2
y
x
B
E
Particle y
x
Figure2.14: EparalleltoBanalyserproducesparabolicoutputlocusasafunctionofinputvelocity. Thelociaredifferentfordifferentq/m.
qHowever,He4 andD2 havethesamem soonecantdistinguishtheirspectraonthebasis
ofionorbits.2.4 DynamicAcceleratorsInadditiontotheelectrostaticaccelerators,thereareseveraldifferenttypesofacceleratorsbasedontime-varyingfields. WiththeexceptionoftheBetatron,theseareallbasedonthegeneralprincipleofarrangingforaresonancebetweentheparticleandtheoscillatingfieldssuchthatenergyiscontinuallygiventotheparticle. Simpleexample
1+ 2+ 2- 3+1- 3-
1 2 3
Figure2.15: Sequenceofdynamicallyvaryingelectrodepotentialsproducescontinuousacceleration. Valuesat3timesareindicated.Particle isacceleratedthroughsequenceofelectrodes3attimes(1)(2)(3). Thepotentialofelectrodeisraisedfromnegativeto+vewhileparticleisinsideelectrode. SoateachgapitseesanacceleratingEz. Canbethoughtofasasuccessivemovingpotentialhill:Waveofpotentialpropagatesatsamespeedasparticleso it iscontinuouslyaccelerated.Historicallyearliestwidespreadacceleratorbasedonthisprinciplewasthecyclotron.
45
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z
Figure2.16: Oscillatingpotentialsgiverisetoapropagatingwave.+ -
Poles
Electric
MagneticPoles
Orbit
Plan
Elevation
B
Figure2.17: SchematicofaCyclotronaccelerator.2.4.1 Cyclotron
qBTake advantage of the orbit frequency in a uniform B-field =m . Apply oscillating
potentialtoelectricpoles,atthisfrequency. Eachtimeparticlecrossesthegap(twice/turn)itseesanacceleratingelectricfield. Resonantfrequency
f = = qB = 1.52107B Hz (2.39)2 m2
15.2MHz/Tforprotons. Ifmagnetradius isRparticle leavesacceleratorwhen itsLarmorradiusisequaltoR
mv=R 1mv2 =1q2B2R2 (2.40)
qB 2 2mIf iron is used for magnetic pole pieces then B
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2.4.2 LimitationsofCyclotronAcceleration:RelativityMass increase (1v2/c2)21 breaks resonance, restricting maximum energy to 25MeV(protons). Improvement: sweep oscillator frequency (downward). Synchrocyclotron allowed energy up to 500MeV but reduced flux. Alternatively: Increase B with radius.Leads to orbit divergence parallel to B. Compensate with azimuthally varying field forfocussingAVF-cyclotron. Advantagecontinuousbeam.2.4.3 SynchrotronVaryboth frequencyandfield intimetokeepbeam inresonanceatconstantradius. Highenergyphysics(to800GeV).2.4.4 LinearAcceleratorsAvoid limitations of electron synchrotron radiation. Come in 2 main types. (1) Induction(2) RF (linacs) with different pros and cons. (RF for highest energy electrons). Electronacceleration: v=cdifferentproblemsfromion.2.5 MagneticQuadrupoleFocussing(AlternatingGra-
dient)Magneticfocussingispreferredathighparticleenergy. Why? Itsforceisstronger.MagneticforceonarelativisticparticleqcB.ElectricforceonarelativisticparticleqE.E.g. B = 2T cB = 6108 same force as an electric field of magnitude6108V /m =0.6MV/mm! Howevermagnetic force isperpendiculartoBsoanaxisymmetric lenswould
liketohavepurelyazimuthalB field B=B. Howeverthiswouldrequireacurrentright
vB
v B^
Figure2.18:
Impossible
ideal
for
magnetic
focussing:
purely
azimuthal
magnetic
field.
wherethebeamis:
B.d=oI. (2.41)Axisymmetricmagneticlensisimpossible. Howeverwecanfocusinonecartesiandirection(x, y)atatime. Thenusethefactthatsuccessivecombinedfocus-defocushasanetfocus.
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2.5.1 PreliminaryMathematicsConsider
z =0 purely transverse field (approx)Bx, By. This can berepresented byB= AwithA=zAso A= (zA) =z A(sincez=0).Inthevacuumregionj=0(nocurrent)so
0 = B= (z A) =z2A+ ( (2.42)z.)A=0
i.e. 2A = 0. A satisfies Laplaces equation. Notice then that solutions of electrostaticproblems, 2=0 are also solutions of (2-d) vacuum magnetostatic problems. The samesolutiontechniqueswork.2.5.2 MultipoleExpansionPotentialcanbeexpandedaboutsomepointinspaceinakindofTaylorexpansion. Chooseoriginatpointofexpansionandusecoordinates(r, ),x=rcos,y=rsin.
1 A 1 2A2A=
rrrr +r 2 =0 (2.43)2LookforsolutionsintheformA=u(r).w().Theserequire
d2wd2 =const.w (2.44)
andd du
r r =const.u. (2.45)dr dr
Hencewsolutionsaresinesandcosinesw=cosn or sinn (2.46)
where n2 is the constant in the previous equation and n integral to satisfy periodicity.Correspondingly
u=rn or lnr , rn (2.47)Thesesolutionsarecalledcylindricalharmonicsor(cylindrical)multipoles:
1 lnrrncosn rncosn (2.48)r
nsin
n
rn
sinn
Ifourpointofexpansionhasnosourceatit(nocurrent)thentheright-handcolumnisruledoutbecausenosingularityatr=0ispermitted. Theremainingmultipolesare
1 constantirrelevanttoapotentialrcos(=x) uniformfield,Axr2cos2=r2(cos2sin2) =x2 y2 non-uniformfieldHigherorders neglected.
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Thesecondordersolution,x2y2 iscalledaquadrupolefield(althoughthisissomethingof a misnomer). [Similarly r3cos3 hexapole, r4cos octupole.] We already dealtwiththispotentialintheelectriccase.
A= x2 y2 =2xx2yy . (2.49)So
z A=2xy2yx (2.50)Forceonlongitudinallymovingcharge:
F = qvB=qv( A) (2.51)= qv(z A) =q(v.z)A qvzA (2.52)
Magneticquadrupoleforceisidenticaltoelectricquadrupoleforcereplacing Avz (2.53)
Consequentlyfocussinginx-direction defocussinginy-directionbutalternatinggradientsgivenetfocussing. Thisisbasisofallstrongfocussing.2.6 ForceondistributedcurrentdensityWehaveregardedtheLorentzforcelaw
F =q(E+vB) (2.54)as
fundamental.
However
forces
are
generally
measured
in
engineering
systems
via
the
interaction of wires or conducting bars with B-fields. Historically, of course, electricity and
magnetismwerebasedonthesemeasurements. Acurrent(I)isaflowofcharge: Coulombs/sAmp. Acurrentdensityj isaflowofchargeperunitareaA/m2. Thecharge iscarriedbyparticles:
j= niviqi (2.55)species i
HencetotalforceoncurrentcarriersperunitvolumeisF= niqi(vi B) =jB (2.56)
iAlso,forafinewirecarryingcurrentI, if itsarea is,thecurrentdensityaveragedacrossthesectionis
Ij= (2.57)
Volumeperunitlengthis. Andtheforce/unitlength=jB. =IBperpendiculartothewire.
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2.6.1 ForcesondipolesWe saw that the field of a localized current distribution, far from the currents, could beapproximatedasadipole. Similarlytheforcesonalocalizedcurrentbyanexternalmagneticfieldthatvariesslowlyintheregionofcurrentcanbeexpressedintermsofmagneticdipole.[Sameistrueinelectrostaticswithanelectricdipole].Total force
F= jBd3x (2.58)whereBisanexternalfieldthatisslowlyvaryingandsocanbeapproximatedas
B(x) =B0 + (x.)B (2.59)wherethetensorB(Bj/xi)issimplyaconstant(matrix). Hence
F
=j
Bo +j(x.B)d3x
= jd3x Bo + j(x.B)d3x (2.60)Thefirsttermintegraliszeroandthesecondistransformedbyourpreviousidentity,whichcanbewrittenas
x (x j)d3x = 2x. jxd3x =2x. xjd3x (2.61)foranyx. UsethequantityBforx(i.e. xi Bj)givingxi
12 (x
j)d3x B=m B= j(x.)Bd3x (2.62)
This tensor identity is then contracted by an internal cross-product [ijk Tjk ] to give thevectoridentity
(m )B= j[(x.)B]d3x (2.63)Thus
F= (m )B=(m.B)m(.B) (2.64)(remembertheoperatesonlyonBnotm). Thisistheforceonadipole:
F=(m.B) . (2.65)TotalTorque(Momentof force)is
M = x (jB)d3x (2.66)= j(x.B)B(x.j)d3x (2.67)
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Bhereis(tolowestorder)independentofx:B0 sosecondtermiszerosince 1 x.jd3x =
2 . |x|2j |x|2.j d3x = 0. (2.68)Thefirsttermisofthestandardformofouridentity.
1M=B. xjd3x =
2B (x j)d3x (2.69)M=mB Momentonadipole. (2.70)
y
x
zdx
dy I
Figure2.19: Elementarycircuitforcalculatingmagneticforce.
2.6.2 ForceonanElementaryMagneticMomentCircuitConsideraplanerectangularcircuitcarryingcurrentI havingelementaryareadxdy=dA.Regardthisasavectorpointing inthez directiondA. Theforceonthiscurrent inafieldB(r)isFsuchthat
BzFx = Idy[Bz(x+dx)Bz(x)]=Idydx (2.71)
xBz
Fy = Idx[Bz(y+dy)Bz(y)]=Idydx (2.72)y
Fz = Idx[By(y+dy)By(y)]Idy[Bx(x+dx)Bx(x)]Bx By Bz
= Idxdyx + y =Idydx z (2.73)
(using .B = 0). Hence, summarizing: F = IdydxBz. Now definem = IdA = Idydxzandtakeitconstant. Thenclearlytheforcecanbewritten
F=(B.m) (2.74)orstrictly(B).m.
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S
N
m
M
B
Figure2.20: Momentonabarmagnetinauniformfield.
S N
N Sm
m
Net Force:
B
|B|
Figure2.21: Amagneticmomentintheformofabarmagnetisattractedorrepelledtowardthestrongerfieldregion,dependingonitsorientation.2.6.3 ExampleSmallbarmagnet: archetypeofdipole. InuniformBfeelsjustatorquealigningitwithB.Inauniformfield,nonetforce.
Non-uniformfield: Ifmagnettakesitsnaturalrestingdirection,mparalleltoB,forceisF=m|B| (2.75)
Abarmagnetisattractedtohighfield. AlternativelyifmparalleltominusBthemagnetpointsotherway
F=m|B| repelledfromhigh|B|. (2.76)Samewouldbetrueforanelementarycircuitdipole. It isattracted/repelledaccordingtowhether itactsto increaseordecreaseB locally. Achargedparticlemoving in itsLarmororbitisalwaysdiamagnetic: repelledfromhigh|B|.
2.6.4 IntuitionThereissomethingslightlynon-intuitiveaboutthenaturalbehaviorofanelementarywirecircuitandaparticleorbitconsideredassimilartothiselementarycircuit. Theircurrentsflow in opposite directions when the wire is in its stable orientation. The reason is thatthestrengthofthewiresustainsitagainsttheoutwardmagneticexpansionforce,whiletheparticleneedsaninwardforcetocausethecentripetalacceleration.
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Paramagnetic Attracted
Diamagnetic Repelled
B
|B|
Net Force:
Figure 2.22: Elementary circuit acting as a dipole experiences a force in a non-uniformmagneticfield.
I
Orbiting ParticleWire Loop
j B j B^ outwards. ^ inwards.
To balance accelerationStable Orientation
qv
Figure2.23: Differencebetweenawireloopandaparticleorbitintheirnaturalorientation.2.6.5 AngularMomentumIfthe localcurrent ismadeupofparticleshavingaconstantratioofchargetomass: q/Msay (Notational accidentm is magnetic moment). Then the angular momentum isL =
iMixi vandmagneticmomentism= 1 qixi vi. So2qm= L. Classical (2.77)
2MThiswouldalsobetrueforacontinuousbodywithconstant(chargedensity)/(massdensity)(/m). Elementaryparticles,e.g. electronsetc.,have spinwithmomentsm,L.Howevertheydonotobeytheaboveequation. Instead
qm=g L (2.78)
2MwiththeLandeg-factor(2 forelectrons). This isattributedtoquantumandrelativisticeffects. However the classical value might not occur if /m were not constant. So weshouldnotbesurprisedthatg isnotexactly1forparticlesspin.2.6.6 PrecessionofaMagneticDipole(formedfromchargedpar-
ticle)The result of a torquemB is a change in angular momentum. Sincem=gLq/2M wehave
dL q=mB=g (LB)) (2.79)
dt 2M53
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m B^
m B^
mL
Figure2.24: PrecessionofanangularmomentumLandalignedmagneticmomentmaboutthemagneticfield.ThisistheequationofacirclearoundB. [Comparewithorbitequation dv = qvB]. The
dt mdirectionofLprecesseslikeatiltedtoparounddirectionofB withafrequency
qB=g (2.80)
2MForan electron (g =2) this is equaltothe cyclotron frequency. Forprotonsg = 22.79[Writtenlikethisbecausespinis 1]. Forneutronsg= 2(1.93).
2Precessionfrequencyisthus
electronf = = (28GHz)(B/Tesla) (2.81)
2proton
= (43MHz)(B/Tesla) (2.82)2
This isthe(classical)basisofNuclearMagneticResonancebutofcoursethatreallyneedsQM.
54