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Electromagnetism
Christopher R Prior
ASTeC Intense Beams Group
Rutherford Appleton Laboratory
Fellow and Tutor in Mathematics
Trinity College, Oxford
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2
Contents
• Review of Maxwell’s equations and Lorentz Force Law• Motion of a charged particle under constant
Electromagnetic fields• Relativistic transformations of fields
• Electromagnetic waves – Waves in vacuo – Waves in conducting medium
• Waves in a uniform conducting guide – Simple example TE 01 mode
– Propagation constant, cut-off frequency – Group velocity, phase velocity – Illustrations
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3
Reading
• J.D. Jackson: Classical Electrodynamics (Wiley, 1998)
• H.D. Young, R.A. Freedman & L. Ford: University Physics(with Modern Physics) (Addison-Wesley,2007)
• P.C. Clemmow: Electromagnetic Theory (CUP, 1973)
• Feynmann Lectures on Physics (Basic Books, 2011)
• W.K.H. Panofsky & M.N. Phillips: Classical Electricity and
Magnetism (Addison-Wesley, 2005)• G.L. Pollack & D.R. Stump: Electromagnetism (Addison-
Wesley, 2001)
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Gradient is normal to
surface = constant.
For a scalar function (x,y,z , t ),
gradient: =∂∂ x
, ∂∂ y
, ∂∂ z
For a vector F = F 1 , F 2 , F 3 :
divergence: · F = ∂ F 1∂ x
+ ∂ F 2∂ y
+ ∂ F 3∂ z
curl: F =∂ F 3∂ y
− ∂ F 2
∂ z, ∂ F 1
∂ z−
∂ F 3∂ x
, ∂ F 2
∂ x−
∂ F 1∂ y
4
Vector Calculus
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· F G = G · F − F · G
φ = 0 , · F = 0
( F ) = ( · F ) − 2 F
Stokes’ Theorem
Basic Vector Calculus
5
Divergence or Gauss’Theorem
Closed surface S, volume V,outward pointing normal
S
F · d
S = C
F · d r
V
·
F d V = S
F · d
S
nd
S = n d S
Oriented
boundary C
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What is Electromagnetism?
• The study of Maxwell’s equations , devised in 1863 torepresent the relationships between electric and magneticfields in the presence of electric charges and currents,whether steady or rapidly fluctuating, in a vacuum or inmatter.
• The equations represent one of the most elegant andconcise way to describe the fundamentals of electricity andmagnetism. They pull together in a consistent way earlierresults known from the work of Gauss, Faraday, Ampère,Biot, Savart and others.
• Remarkably, Maxwell’s equations are perfectly consistentwith the transformations of special relativity.
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Maxwell’s EquationsRelate Electric and Magnetic fields generated bycharge and current distributions.
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E = electric eld D = electric displacement
H = magnetic eld B = magnetic ux density
ρ = electric charge density j = current density
µ 0 = permeability of free space, 4 π 10− 7
0 = permittivity of free space, 8 .854 10 − 12
c = speed of light, 2 .99792458 10 8
In vacuum:
D = 0 E,
B = µ 0
H, 0 µ 0 c
2
= 1
· D = ρ
· B = 0
E = −∂ B∂ t
H = j + ∂ D∂ t
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E = q 4 π 0
rr 3
= sphere
E · d S = q 4 π 0
sphere
d S r 2
= q
0
Equivalent to Gauss’ Flux Theorem:
The flux of electric field out of a closed region is proportional to the total
electric charge Q enclosed within the surface. A point charge q generates an electric field:
Maxwell’s 1 st Equation· E =
ρ0
· E = ρ
0 V
· E d V = S
E · d S = 1
0 V
ρ d V = Q
0
Area integral gives a measure of the net charge enclosed; divergence ofthe electric field gives the density of the sources. 8
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Gauss’ law for magnetism:
The net magnetic ux out of any
closed surface is zero. Surround amagnetic dipole with a closed surface.The magnetic ux directed inwardtowards the south pole will equal theux outward from the north pole.
If there were a magnetic monopolesource, this would give a non-zerointegral.
Maxwell’s 2 nd Equation
Gauss’ law for magnetism is then astatement that
There are no magnetic monopoles 9
·
B = 0
B · d
S = 0
·
B = 0
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Equivalent to Faraday’s Law of Induction:
(for a xed circuit C)
The electromotive force round a circuit
is proportional to the rate of change of ux of
magnetic eld through the circuit.
Maxwell’s 3 rd Equation
! "= l d E !!
#
!! "=# S d B!!
10
S
E · d S = − S
∂ B
∂ t · d S
C
E · d l = −d
d t S
B · d S = −d Φ
d t
E = −∂ B
∂ t
Faraday’s Law is the basis for electric generators. It also
forms the basis for inductors and transformers.
Michael Faraday
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B = µ0 I
4 π d l rr 3
Maxwell’s 4 th Equation
Ampère
11
Biot
B = µ 0 j +
1
c2
∂ E ∂ t
Originates from Ampère’s (Circuital) Law :
Satised by the eld for a steady line current(Biot-Savart Law, 1820):
B = µ 0
j
C
B · d
l = S
B · d
S = µ0 S
j · d
S = µ0 I
B =µ 0 I 2 π r
For a straight line current
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• pply mpere to sur ace 1 a atdisk): the line integral of B =µ0 I .
• Applied to surface 2, line integralis zero since no current penetratesthe deformed surface.
• In a capacitor,
E = Q
0 A and I =
dQdt
= 0 AdE dt
,
so there is a current density j d = 0
∂ E ∂ t
.
12
Displacement Current
Surface 1 Surface 2
Closed loop
Current I
• Faraday : vary B-field, generate E-field
• Maxwell : varying E-field should then produce a B-field, but not covered by Ampère’s Law.
B = µ0 ( j + j d ) = µ0 j + 0 µ 0
∂ E ∂ t
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Consistency with Charge Conservation
Charge conservation is implicit in Maxwell’s Equations
j · d S = −
d
d t ρ d V
· j d V = − ∂ρ∂ t
d V
· j + ∂ρ∂ t = 0
B = µ0 j +1
c2
∂ E ∂ t
= · B = µ0 · j + 1c2 ∂
∂ t· E
= 0 = · j + 0 µ 0∂ ∂ t
ρ0
= 0 = · j +∂ρ
∂ t
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In vacuum:
D = 0 E, B = µ 0 H, 0 µ 0 c2 = 1
Source-free equations:
· B = 0
E + ∂ B
∂ t = 0
Source equations:
· E = ρ0
B − 1c2
∂ B∂ t
= µ0 j
Equivalent integral form (useful forsimple geometries):
E · d S = 1
0 ρ dV
B · d S = 0
E · d l = − ddt B · d S = − dΦ
dt
B
· d
l = µ0
j d
S +
1c2
ddt
E
· d
S
Maxwell’s Equations in Vacuum
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Also from E = − ∂ B∂ t
B = µ0 j +
1c2
∂ E ∂ t
then gives current density necessary tosustain the elds
Example: Calculate E from B
E · d
l = −d
d t
B · d S
!"#
>
<=
0
00
0
sin
r r
r r t B B
z
$
r
z
15
2 π rE θ = −d
d tπ r 2 B 0 sin ω t = − π r 2 B 0 ω cos ω t
= E θ = −1
2B 0 ω r cos ω t
r < r 0
2 π rE θ = −d
d tπ r 2
0 B 0 sin ω t = − π r 20 B 0 ω cos ω t
= E θ = −ω r 2
0 B 0
2 r cos ω t
r > r 0
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E · d l = −d t B · d S
= 2 π rE θ = −d Φ
d t
The BetatronMagnetic
flux, !
Generated E -field
r
Particles accelerated by the rotational electriceld generated by a time-varying magnetic eld
16
−mv 2
r = evB = B = −
per
=∂
∂ t B (r, t ) = −
1er
d pdt = −
E r =
12π r 2
dΦ
dt
= B (r, t ) = 1
21
π r 2 B dS
B-field on orbit needs to be one half the average B over the circle. This imposes a limit on the energythat can be achieved. Nevertheless the constant radius principle is attractive for high energy circularaccelerators.
For circular motion at a constant radius:
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Lorentz Force Law
• Thought of as a supplement to Maxwell’s equations but actuallyimplicit in relativistic formulation, gives force on a charged particlemoving in an electromagnetic field:
• For continuous distributions, use force density:
• Relativistic equation of motion
– 4-vector form:
– 3-vector component: Energy component:
F =dP dτ
= γ v · f
c , f = γ
1
cdE dt
, d pdt
d
d tm 0 γ v = f = q E + v B v · f =
d E d t
= m 0 c2 d γ d t
17
f = q
E +
v
B
f d = ρ E + j B
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d t m 0 γ v = f = q E + v B = q v Bd
d tm 0 γ c2 = v · f = q v · v B = 0
Motion in Constant Magnetic Fields
• From energy equation, ! is constant
• From momentum equation,
18
No acceleration with a magnetic eld
|
v | constant and |
v | constant
= |
v | also constant
= |
v | is constant
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Motion in Constant Magnetic Field
Constant magnetic eld givesuniform spiral about B with
constant energy.
19
dt (m 0 γ v) = q v B
=d vdt
= q m 0 γ
v B
=v 2
ρ = q
m 0 γ v B
= circular motion with radius ρ = m 0 γ v
qB
at an angular frequency ω = v
ρ =
qBm 0 γ
= qB
m
Magnetic Rigidity
B ρ =
m 0 γ vq
=
pq
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d
d tm 0 γ v = f = q E + v B =
d
d tm 0 γ v = q E
20
Motion in Constant Electric Field
Constant E-eld gives uniform acceleration instraight line
Solution is γ v = q E m 0
t
Then γ 2 = 1 +γ vc
2
= γ = 1 +q Etm 0 c
2
If E = ( E, 0, 0), dx
dt =
(γ v)γ
= x = x 0 + m 0 c2
qE 1 +qEtm 0 c
2
− 1
≈ x 0 + 1
2
qE
m 0t 2 for qE m 0 c
Energy gain is m 0 c2 (γ − 1) = qE (x − x 0 )
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• According to observer O in frame F, particle has velocity , fields are andand Lorentz force is
• In Frame F ’ , particle is at rest and force is
• Assume measurements give same charge and force, so
• Point charge q at rest in F:
• See a current in F’
, giving a field
• Suggests
E = q 4 π 0
rr 3 , B = 0
v
E
B
Relativistic Transformations of E and B
f = q
E
q = q and
E =
E +
v ×
B
B = −
µ 0 q 4 π
v × rr 3 = −
1
c2 v × E
B =
B −1
c 2
v ×
E
21
f = q
E +
v
B
R o u g h i d
e a
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h d l
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ω ∆ t − k ∆ x = 0
v p
=∆ x
∆ t
=ω
k
Superposition of plane waves. Whileshape is relatively undistorted, pulsetravels with the Group Velocity
Phase and Group Velocities
[ ]! "
"#
# dk ek A kxt k i )()( $
Plane wave has constantphase at peaks
sin(ω t − kx )ω t − kx = 1
2π
vg
=
d ω
dk
23
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24
Wave Packet Structure
• Phase velocities of individual plane waves makingup the wave packet are different,
• The wave packet will then disperse with time
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N f El i W
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Nature of Electromagnetic Waves• A general plane wave with angular frequency ! travelling in the
direction of the wave vector has the form
• Phase number of waves and so is a Lorentzinvariant.
• Apply Maxwell’s equations:
• Waves are transverse to the direction of propagation; and aremutually perpendicular
k
E =
E 0e i ( ω t − k · x ) ,
B =
B 0 e i ( ω t − k · x )
↔ − i k∂
∂ t ↔ i ω
26
ω t − k · x = 2 π ×
E,
B
k
· E = 0 = · B ←→ k · E = 0 = k · B
E = −
∂ B
∂ t ←→
k E = ω
B
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Plane Electromagnetic Wa e
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Plane Electromagnetic Wave
27Thursday, 25 October 2012
Plane Electromagnetic Waves
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= speed of electromagnetic waves in vacuum is ω
k = c
Plane Electromagnetic Waves
!
" #
! $
2Frequency
k
2Wavelength
=
= !
Reminder: The fact that is aninvariant tells us that
is a Lorentz 4-vector, the 4-Frequency vector.Deduce frequency transforms as
Λ =
ω
c ,
k
ω t − k · x
ω = γ (ω − v · k ) = ω c − v
c + v
B =
1
c 2
∂ E
∂ t
k B = −
ω
c 2
E
Combined with k E = ω B =| E || B |
= ω
k =
kc 2
ω
28
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Waves in a Conducting Medium
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Waves in a Conducting Medium
• (Ohm’s Law) For a medium of conductivity ! ,
• Modified Maxwell:
• Putconduction
currentdisplacement
current
Dissipationfactor
j = σ E
H = j +∂ E ∂ t
= σ E +∂ E ∂ t
− i k H = σ E + i ω E
D =
σ
ω
E =
E 0e i (ω t − k · x ) ,
B =
B 0e i ( ω t − k · x )
29
40
8-
120
7
1057.21.2,103:Teflon
10,108.5:Copper !
"=#="=
=#="=
D
D
$ $ %
$ $ %
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Attenuation in a Good Conductor
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− i k H = σ E + i ω E k H = i σ E − ω E = ( i σ − ω ) E
ombine with E = −∂ B∂ t
= k E = ω µ H
= k ( k E ) = ω µ k H = ω µ(iσ − ω ) E
= ( k · E ) k − k2
E = ω µ(iσ − ω ) E
= k2 = ω µ(− iσ + ω ) since k · E = 0
Attenuation in a Good Conductor
For a good conductor, D 1, σ
ω
, k2
≈ − iω
µσ
= k ≈ ω µσ
2 (1 − i) =
1δ
(1 − i) where δ = 2ω µσ
is the skin-depth
Wave-form is: e i (ω t − kx ) = e i ( ω t − (1 − i )x ) = e− x e i ( ω t − x )
30
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Charge Density in a Conducting Material
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• Inside a conductor (Ohm’s law)
• Continuity equation is
• Solution is
• Charge density decays exponentially with time. For a verygood conductor, charge flows instantly to the surface toform a surface current density and (for time varying fields)a surface current. Inside a perfect conductor:
Charge Density in a Conducting Material j = σ E
ρ = ρ 0 e− σ t/
32
∂ρ∂ t
+ · j = 0 ∂ρ
∂ t + σ · E = 0
∂ρ∂ t
+ σ ρ = 0 .
(σ → ∞ )
E = H = 0
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A Uniform Perfectly Conducting Guide
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E = −∂ B∂ t
= − iω µ H
H = ∂ D
∂ t = iω E
=
2 E = ( · E ) − E
= iω µ H
= − ω 2 µ E
2 + ω 2 µ E
H = 0
A Uniform Perfectly Conducting Guide
Hollow metallic cylinder with perfectly conducting boundarysurfaces
Maxwell’s equations with time dependence e i! t are:
Assume )(
)(
),(),,,(
),(),,,( z t i
z t i
e y x H t z y x H
e y x E t z y x E ! "
! "
#
#
=
=
!!
!!
! is the propagation constant
Can solve for the fields completelyin terms of E
z and H
z
zx
y
33
Then 2t + ω2 µ + γ 2
E H
= 0
Helmholtz Equation
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A simple model: “Parallel Plate Waveguide”
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To satisfy oun ary con itions: E = 0 on x = 0 an x = a.
= E = A sin Kx, with K = K n ≡n π
a , n integer
E = (0 , 1, 0)E (x)e i ω t − γ z where E satises
2t E =
d2 E dx 2 = − K 2 E, K 2 = ω2 µ + γ 2 .
with solution E = A cos Kx or A sin Kx
Propagation constant is
γ = K 2n − ω
2 µ = n π
a 1 − ω
ω c
2
, ω c = K n√ µ
34
A simple model: Parallel Plate Waveguide
Transport between two infinite conducting plates (TE 01 mode):
z
x
y
x = 0 x = a
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Cut-off Frequency "
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γ =
n π
a 1
− ω
ω c
2
, E = sin
n π x
a ei ω t − γ z
, ω c =
n π
a√ µ
35
Cut off Frequency, c
! " <" c gives real solution for # , so attenuationonly. No wave propagates: cut-off modes.
! " >" c gives purely imaginary solution for # ,and a wave propagates without attenuation.
! For a given frequency " only a finite number of
modes can propagate .
For given frequency, convenient tochoose a s.t. only n=1 modeoccurs.
γ = ik, k = √ µ ω2
− ω2c
12 = ω √ µ 1
−ω
2c
ω2
12
ω > ω c =
π
a √ µ= n <
ω
π√ µ
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Phase and Group Velocities
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k = √ µ ω 2
−ω 2
c
1
2 < ω √ µ
λ =2 π
k >
2 π
ω √ µ,
k 2 = µ ω2
−ω
2
c
= vg =
ω
d k =
kω µ
< 1
√ µ
37
Phase and Group Velocities
v p =
ω
k >
1
√ µ
# free-space wavelength
# larger than free-spacevelocity
# smaller than free-space velocity
• Wave number
• Wavelength
• Phase velocity
• Group velocity
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Calculation of Wave Properties
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Calculation of Wave Properties• If cm, cut-off frequency of lowest order mode is
• At 7 GHz, only the n=1 mode propagates and
a = 3
f c = ω c
2 π=
1
2 a√ µ ≈3 × 10 8
2 × 0 .03 ≈ 5 GHz ω c = n π
a√ µ
c38
k = √ µ ω 2
− ω 2c
12
≈ 2π (72
− 52 ) 12 ×109 / 3 × 108 = 103 m − 1
λ = 2π
k ≈ 6 cm
v p = ω
k = 4.3 × 108 ms− 1 > c
vg = kω µ
= 2.1 × 108 ms− 1 < c
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