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ElectromagneticTheory
G.Franchetti,GSICERNAccelerator– SchoolBudapest,2-14/10/2016
3/10/16 G.Franchetti 1
MathematicsofEM
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Fieldsare3dimensionalvectorsdependentoftheirspatialposition(anddependingontime)
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Products
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Scalarproduct
~A
~B
~A · ~B = AB cos ✓
Vectorproduct
~A
~B
~A⇥ ~B = AB sin ✓v
~A⇥~B
✓ ✓
Thegradientoperator
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~r = (@x
, @y
, @z
)
Isanoperatorthattransformspacedependentscalarinvector
Example:givenf(x, y, z)
~rf(x, y, z) = (@x
f, @
y
f, @
z
f)
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Divergence/Curlofavectorfield
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~
A(x, y, z)
Divergenceofvectorfield
~
A(x, y, z)
(@y
A
z
� @
z
A
y
)x+
(@z
A
x
� @
x
A
z
)y+
(@x
A
y
� @
y
A
x
)z
~r⇥ ~
A(x, y, z) =
~r · ~A(x, y, z) =
@x
Ax
+ @y
Ay
+ @z
Az
Curlofvectorfield
Relations
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~A⇥ ( ~B ⇥ ~C) = �( ~A · ~B) ~C + ~B( ~A · ~C)
~r⇥ (~r⇥ ~C) = �(~r · ~r) ~C + ~r(~r · ~C)
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FluxConcept
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Examplewithwater
ab
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Volumepersecond
dV
dt= ava
a
or
L
θ
dV
dt
= Lbvcos✓
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Flux
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θ~E
d ~A
A
d�( ~E) = ~E · d ~A
Fluxthroughasurface
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�( ~E) =
Z
S
~E · d ~A
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Fluxthroughaclosedsurface:Gausstheorem
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Z
S
~E · d ~A =
Z
Vr · ~E dV
Anyvolumecanbedecomposedinsmallcubes
Fluxthroughaclosedsurface
Stokestheorem
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x
y
Ey(D,0,0)
Ey(0,0,0)
I
�
~E · d~l = Ex
(0, 0, 0)�+ Ey
(�, 0, 0)�
� Ex
(0,�, 0)�� Ey
(0, 0, 0)�
I
�
~
E · d~l =✓�@E
x
@y
+@E
y
@x
◆�2
I
�z
~E · d~l = (~r⇥ ~E)z�2
�
�
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foranarbitrarysurface
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I
�
~E · d~l =Z
S(~r⇥ ~E) · d ~A
Howitworks
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ElectricChargesandForces
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+ -Twocharges
Experimentalfacts
+ ++ -
--
Coulomblaw
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q1
q2r
~F
~F =1
4⇡✏0
q1q2r2
r
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Units
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SystemSI
~Fq1r
~F =1
4⇡✏0
q1q2r2
r
Newton
Coulomb
Meters
✏0 = 8.8541⇥ 10�12
C2N-1 m-2
permettivity offreespace
Superpositionprinciple
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q1
q2r
~F
~F =1
4⇡✏0
q1q2r2
r
q3
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ElectricField
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q1
q2r
~Fq3
~E =~F
q1
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~F
~F = q1 ~E
~E
q1
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Byknowingtheelectricfieldtheforceonachargeiscompletelyknown
Workdonealongapath
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W = q
Z
�
~E · d~l�
workdonebythecharge
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Electricpotential
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V (P ) = �Z P
1~E · d~l
ForconservativefieldV(P)doesnotdependonthepath!
Centralforcesareconservative
UNITS:Joule/Coulomb=Volt
Workdonealongapath
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�
workdonebythecharge
A
B
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ElectricFieldßà ElectricPotential
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E
x
= � @
@x
V (x, y, z)
Ey = � @
@y
V (x, y, z)
Ez = � @
@z
V (x, y, z)
Invectorialnotation
~E = �~rV
Electricpotentialbyonecharge
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Takeoneparticlelocatedattheorigin,then
V (~r) = �Z ~r
1
1
4⇡✏0
q
(r1 � r0)3(~r1 � ~r0) · d~l
V (~r) =1
4⇡✏0
q
r
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ElectricPotentialofanarbitrarydistribution
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setofNparticles
q1
q2
q3
~r1
~r2
~r3
~rorigin
V (~r)
V (~r) =1
4⇡✏0
X
i
qip(~r � ~ri)2
Electricpotentialofacontinuousdistribution
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Splitthecontinuousdistributioninagrid
qi = ⇢(~ri)dV
~ri
origin
V (~r) =1
4⇡✏0
X
i
qip(~r � ~ri)2
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Energyofachargedistribution
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itistheworknecessarytobringthechargedistributionfrominfinity
Moresimply U =X
j
qjX
i=1,j
1
4⇡✏0
qi|~rj � ~ri|
Moresimply
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Inintegralform
Using ~E = �~rV andthedivergencetheoremitcanbeprovedthat
✏0E2
2isthedensityofenergyoftheelectricfield
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Fluxoftheelectricfield
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θ~E
d ~A
A
d�( ~E) = ~E · d ~A
Fluxofelectricfieldthroughasurface
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�( ~E) =
Z
S
~E · d ~A
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ApplicationtoCoulomblaw
Z
S
~E · d ~A =q
✏0
Onasphere
Thisresultisgeneralandapplytoanyclosedsurface
Onanarbitraryclosedcurve
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q1
q2
Z
S
~E · d ~A =1
✏0(q1 + q2)
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FirstMaxwellLaw
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Z
S
~E · d ~A =Q
✏0integralform
forainfinitesimalsmallvolume differentialform
(trytoderiveit.Hint:usedGausstheorem)
~r · ~E =⇢
✏0
Physicalmeaning
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Ifthereisachargeinoneplace,theelectricfluxisdifferentthanzero
Onechargecreateanelectricflux.
�( ~E) =q
✏0
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PoissonandLaplaceEquations
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As ~E = �~rV and
combiningbothwefind
r2V = � ⇢
✏0
Invacuum: r2V = 0
~r · ~E =⇢
✏0
~r · ~rV = � ⇢
✏0
Poisson
Laplace
MagneticField
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Thereexistnotamagneticcharge!(FindamagneticmonopoleandyougettheNobelPrize)
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Ampere’sexperiment
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I1
I2
F = µ0I1I22⇡r
Ampere’sLaw
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I
~B =µ0
2⇡
I
rv
I2
F
allexperimentalresultsfitwiththislaw
~F = I2d~l ⇥ ~B
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Units
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~B =µ0
2⇡
I
rv
~F = dl~I2 ⇥ ~BFromN
Am= T [Tesla]
From
µ0 = 4⇡ ⇥ 10�7 N A�2
follows
Tohave1Tat10cmwithonecable I=5x105 Amperes!!
Biot-SavartLaw
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I
~B =µ0
2⇡
I
rv
~r
d~l
d ~B =µ0
4⇡
d~l ⇥ ~r
r3
Fromanalogywiththeelectricfield
dB
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Lorentzforce
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~F = q~v ⇥ ~B
Achargenotinmotiondoesnotexperienceaforce!
v
B
F
Noaccelerationusingmagneticfield!
+
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Fluxofmagneticfield
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Thereexistnotamagneticcharge!Nomatterwhatyoudo..
Themagneticfluxisalwayszero!Z
S
~B · d ~A = 0
SecondMaxwellLaw
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r · ~B = 0
Integralform
Differentialfrom
Z
S
~B · d ~A = 0
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ChangingthemagneticFlux...
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~E = ~v ⇥ ~B
h
L
Magneticflux
Followingthepath
�( ~B) = hLB
E =1
h
d�( ~B)
dt
Z
�
~E · d~l = �d�( ~B)
dt
Faraday’sLaw
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integralform
validinanywaythemagneticfluxischanged!!!
Z
�
~E · d~l = �d�( ~B)
dt
(Reallynotobvious!!)
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foranarbitrarysurface
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I
�
~E · d~l =Z
S(~r⇥ ~E) · d ~A
Faraday’sLawindifferentialform
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Z
S(~r⇥ ~E) · d ~A = � d
dt
Z
S
~B · d ~A
~r⇥ ~E = �@ ~B
@t
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SummaryFaraday’sLaw
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Integralform
differentialform ~r⇥ ~E = �@ ~B
@t
I
�
~E · d~l = �d�( ~B)
dt
Importantconsequence
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Currentcreatesmagneticfield
magneticfieldcreatemagneticflux
�(B) = LI
ChangingthemagneticfluxcreatesaninducedemfI
�
~E · d~l = �d�( ~B)
dt
L=inductance[Henry]
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I = 0
U =1
2LI2energynecessarytocreatethemagneticfield
h
r
✏emf = �d�(B)
dtdU = ✏emfIdt = �d�(B)
dtIdt
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B = µ0NI
�(B) = V olume
B
2
µ0I
Therefore
V olume
B
2
µ0I= LI
U =1
2LI
2 = V olume
B
2
2µ0
B2
2µ0
Energydensityofthemagneticfield
Fieldinsidethesolenoid
Magneticflux �(B) = ⇡r2BNh
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Ampere’sLaw
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I
B
I
�
~B · d~l = µ0I
DisplacementCurrent
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I
�
~B · d~l = µ0I
I
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DisplacementCurrent
3/10/16 G.Franchetti 57
I
�
~B · d~l = µ0IHerethereisavaryingelectricfieldbutnocurrent!
I
DisplacementCurrent
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StationarycurrentIà electricfieldchangeswithtime
I = ✏0@
@t
Z
S
~E · d ~A
I
ThisdisplacementcurrenthastobeaddedintheAmperelaw
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FinalformoftheAmperelaw
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I
�
~B · d~l = µ0
✓I + ✏0
@
@t
Z
S
~E · d ~A◆
~r⇥ ~B = µ0
✓~j + ✏0
@
@t~E
◆
integralform
differentialform
MaxwellEquationsinvacuum
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Integralform DifferentialformZ
S
~E · d ~A =Q
✏0r · ~E =
⇢
✏0Z
S
~B · d ~A = 0 r · ~B = 0
~r⇥ ~E = �@ ~B
@tI
�
~B · d~l = µ0
✓I + ✏0
@
@t
Z
S
~E · d ~A◆
~r⇥ ~B = µ0
✓~j + ✏0
@
@t~E
◆
I
�
~E · d~l = �d�( ~B)
dt
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Magneticpotential?
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Canwefinda“potential”suchthat ~B = �~rV
~r · ~B = �r2VMaxwellequation
~r · ~B = 0
r2V = 0
~r⇥ ~B = ~r⇥ ~rV = 0But itmeansthatwecannotincludecurrents!!
?
Example:2Dmultipoles
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For2Dstaticmagneticfieldinvacuum(onlyBx,By)
~B = �~rV~B = ~r⇥ ~A
�@x
V = @y
Az
@y
V = @x
Az
~B = (@y
Az
,�@x
Az
, 0)
~B = (�@x
V,�@y
V, 0)
ThesearetheCauchy-ReimannThatmakesthefunction
A+ iVanalytic
By
+ iBx
= �@x
(A+ iV )
By
+ iBx
= BX
n
(bn
+ ian
)zn
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VectorPotential
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Ingeneralwerequire ~B = ~r⇥ ~A
~r · ~A = 0 (thischoiceisalwayspossible)
Automatically ~r · ~B = 0~r⇥ ~B = µ0
~J r2 ~A = �µ0~J
Solution
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r2 ~A = �µ0~Jr2V = � ⇢
✏0
V (~r) =1
4⇡✏0
Z⇢(~ri)
|~r � ~ri|dV ~A =
µ0
4⇡
Z ~J(~r0)
|~r0 � ~r|dV
Electricpotential Magneticpotential
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Effectofmatter
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Electricfield Magneticfield
ConductorsDielectric
DiamagnetismParamagnetismFerrimagnetism
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Maxwellequationinvacuumarealwaysvalid,evenwhenweconsidertheeffectofmatter
Microscopicfield
Thatisthefieldis“local”betweenatomsandmovingcharges
Averagedfield
thisisafieldaveragedoveravolumethatcontainmanyatomsormolecules
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Conductors
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freeelectrons boundedtobeinsidetheconductor
Conductorsandelectricfield
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surfacedistributionofelectrons
boundedtobeinsidetheconductor
onthesurfacetheelectricfieldisalwaysperpendicular
~Eext
~E = 0
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�
� = ✏0EApplyingGausstheorem
�( ~E) = AE
Boundarycondition
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Thesurfacesofmetalsarealwaysequipotential
++
+
- -
--
++ +
+ +
-
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Ohm’sLaw
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freeelectrons
R =l
A⇢
l
A
⇢
[Ω]
resistivity
~E = ⇢ ~Jor
~J = � ~E
� =1
⇢conductivity
[Ωm]
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Whoiswho?