Electromagnetism
Christopher R Prior
Fellow and Tutor in Mathematics
ASTeC Intense Beams Group
Trinity College, Oxford
Rutherford Appleton Laboratory
Contents
‰
Maxwell’s equations and Lorentz Force Law
‰
Motion of a charged particle under constant
Electromagnetic fields
‰
Relativistic transformations of fields
‰
Electromagnetic waves
‰
Waves in a uniform conducting guide
ƒ
ƒ
ƒ
ƒ
Simple example TE01 mode
Propagation constant, cut-off frequency
Group velocity, phase velocity
Illustrations
Reading
‰
J.D. Jackson: Classical Electrodynamics
‰
H.D. Young and R.A. Freedman: University Physics
(with Modern Physics)
‰
P.C. Clemmow: Electromagnetic Theory
‰
Feynmann Lectures on Physics
‰
W.K.H. Panofsky and M.N. Phillips: Classical
Electricity and Magnetism
‰
G.L. Pollack and D.R. Stump: Electromagnetism
What is electromagnetism?
‰
The study of Maxwell’s equations, devised in 1863 to represent
the relationships between electric and magnetic fields in the
presence of electric charges and currents, whether steady or
rapidly fluctuating, in a vacuum or in matter.
‰
The equations represent one of the most elegant and concise
way to describe the fundamentals of electricity and magnetism.
They pull together in a consistent way earlier results known
from the work of Gauss, Faraday, Ampère, Biot, Savart and
others.
‰
Remarkably, Maxwell’s equations are perfectly consistent with
the transformations of special relativity.
Maxwell’s Equations
‰
Relate Electric and Magnetic fields generated by charge
and current distributions.
E = electric field
D = electric displacement
H = magnetic field
B = magnetic flux density
ρ= charge density
j = current density
µ0 (permeability of free space) = 4π 10-7
ε0 (permittivity of free space) = 8.854 10-12
c (speed of light) = 2.99792458 108 m/s
r
∇⋅D = ρ
r
∇⋅B = 0
r
r
∂B
∇∧E = −
∂t
r
r r ∂D
∇∧H = j +
∂t
r
r r
r
In vacuum D = ε 0 E , B = µ 0 H , ε 0 µ 0 c 2 = 1
Maxwell’s 1st Equation
Equivalent to Gauss’ Flux Theorem:
r ρ
r r 1
∇⋅E =
⇔ ∫∫ E ⋅ dS =
ε0
ε0
S
∫∫∫ ρ dV =
V
r ρ
∇⋅E =
ε0
Q
ε0
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
r
E=
r r
∫∫ E ⋅ dS =
sphere
q
4πε 0 r 3
r
r
q
dS q
∫∫ r 2 = ε 0
4πε 0 sphere
Area integral gives a measure of the net charge
enclosed; divergence of the electric field gives the density
of the sources.
Maxwell’s 2nd Equation
r
∇⋅B = 0 ⇔
r
∇⋅ B = 0
r r
∫∫ B ⋅ dS = 0
Gauss’ law for magnetism:
The net magnetic flux out of any
closed surface is zero. Surround a
magnetic dipole with a closed surface.
The magnetic flux directed inward
towards the south pole will equal the
flux outward from the north pole.
If there were a magnetic monopole
source, this would give a non-zero
integral.
Gauss’ law for magnetism is then a statement that
There are no magnetic monopoles
Maxwell’s 3rd Equation
r
r
∂B
∇∧E = −
∂t
Equivalent to Faraday’s Law of Induction:
r
r r
∂B r
∫∫S ∇ ∧ E ⋅ dS = −∫∫S ∂t ⋅ dS
r
r r
dΦ
d
⇔
⋅
=
−
⋅
=
−
B
d
S
E
d
l
(for a fixed circuit C)
∫C
dt
dt ∫∫
S
The electromotive
force round a
r r
circuit ε = ∫ E ⋅ dl is proportional to
the rate of change of flux of magnetic
r r
field, Φ = ∫ B ⋅ dl through the circuit.
N
S
Faraday’s Law is the basis for electric generators. It also forms the
basis for inductors and transformers.
Maxwell’s 4th Equation
r
r
r 1 ∂E
∇ ∧ B = µ0 j + 2
c ∂t
r
r
∇ ∧ B = µ0 j
Originates from Ampère’s (Circuital) Law :
r r
r r
r r
∫ B ⋅ dl = ∫∫ ∇ ∧ B ⋅ dS = µ0 ∫∫ j ⋅ dS = µ0 I
C
Ampère
S
S
Satisfied by the field for a steady line current (Biot-Savart Law,
1820):
r µ0 I
B=
4π
Biot
∫
r r
dl ∧ r
r3
For a straight line current
µ0 I
Bθ =
2πr
Need for displacement current
‰
Faraday: vary B-field, generate E-field
Maxwell: varying E-field should then produce a B-field, but not covered by Ampère’s
Law.
‰ Apply Ampère to surface 1 (flat disk): line
integral of B = µ0I
Surface 2
Surface 1
‰
‰
Current I
‰
Closed loop
‰
Applied to surface 2, line integral is zero
since no current penetrates the deformed
surface.
Q
dQ
dE
In capacitor, E =
, so I =
= ε0 A
ε0 A
dt
dt
r
r
∂E
ε
j
=
Displacement current density is d
0
∂t
r
r
r r
r
∂E
∇ ∧ B = µ 0 ( j + jd ) = µ 0 j + µ 0ε 0
∂t
Consistency with charge conservation
‰
Charge conservation: Total
current flowing out of a region equals
the rate of decrease of charge within
‰
From Maxwell’s equations:
Take divergence of (modified)
Ampère’s equation
the volume.
⇔
⇔
r r
d
∫∫ j ⋅ d S = − dt ∫∫∫ ρ dV
r
∂ρ
∫∫∫ ∇ ⋅ j dV = − ∫∫∫ ∂ t dV
r ∂ρ
∇⋅ j +
=0
∂t
(
r
r 1 ∂
∇ ⋅ ∇ ∧ B = µ 0∇ ⋅ j + 2
∇⋅E
c ∂t
r
∂⎛ρ⎞
⇒ 0 = µ 0∇ ⋅ j + ε 0 µ 0 ⎜⎜ ⎟⎟
∂t ⎝ ε 0 ⎠
r ∂ρ
⇒ 0 = ∇⋅ j +
∂t
)
Charge conservation is implicit in Maxwell’s Equations
Maxwell’s Equations in Vacuo
‰
In vacuum
‰
r
r r
r
1
D = ε 0 E , B = µ0 H , ε 0 µ0 = 2
c
‰
‰
Source-free equations:
r
∇⋅B = 0
v
r ∂B
∇∧E+
=0
∂t
Source equations
r ρ
∇⋅E =
ε0
r
v 1 ∂E
r
∇∧B− 2
= µ0 j
c ∂t
Equivalent integral forms
(sometimes useful for
simple geometries)
r v 1
∫∫ E ⋅ dS =
ε0
∫∫∫ ρ dV
r r
∫∫ B ⋅ dS = 0
r r
r r
d
dΦ
⋅
=
−
⋅
=
−
E
d
l
B
d
S
∫
dt ∫∫
dt
r r
r r
r r 1 d
∫ B ⋅ dl = µ0 ∫∫ j ⋅dS + c 2 dt ∫∫ E ⋅ dS
Example: Calculate E from B
r r
d r
∫ E ⋅ dl = − dt ∫∫ B ⋅ dS
z
r
⎧ B0 cos ω t
Bz = ⎨
0
⎩
Also from
r < r0
r > r0
r
r
∂B
∇∧E = −
∂t
r < r0
2π rEθ = −π r 2 B0ω cos ω t
⇒
B0ω r
Eθ = −
cos ω t
2
r > r0
2π rEθ = −π r02 B0ω cos ω t
⇒
Eθ = −
ω r02 B0
2r
cos ω t
r
r
r 1 ∂E then gives current density necessary
∇ ∧ B = µ0 j + 2
c dt to sustain the fields
Lorentz force law
‰
Supplement to Maxwell’s equations, gives force on a charged
particle moving in an electromagnetic field:
r
r r r
f = q E+v ∧B
r
r r r
For continuous distributions, have a force density f d = ρ E + j ∧ B
‰
Relativistic equation of motion
‰
(
)
r r
⎛v⋅ f
dP
F=
⇒ γ ⎜⎜
,
dτ
⎝ c
ƒ 4-vector form:
r
r⎞
⎛ 1 dE dp ⎞
f ⎟⎟ = γ ⎜
, ⎟
⎝ c dt dt ⎠
⎠
ƒ 3-vector component:
(
r
r r r
r
d
(m0γ v ) = f = q E + v ∧ B
dt
)
Motion of charged particles in constant
electromagnetic fields
(
r
r r r
r
d
(m0γ v ) = f = q E + v ∧ B
dt
‰
‰
Constant E-field gives uniform
acceleration in straight line
q
d r
(
)
γ
v
=
E
Solution of dt
m0
2
⎤
⎡
⎛ qEt ⎞
m0 c 2 ⎢
⎟⎟ − 1⎥
x=
1 + ⎜⎜
⎥
qE ⎢
m0 c ⎠
⎝
⎦
⎣
1 qE 2
≈
t for qE << m0 c
2 m0
‰
Energy gain = qEx
‰
)
Constant magnetic field gives
uniform spiral about B with
constant energy.
r
r
v// = constant
dv
q r r
=
v∧B
r
dt m0γ
x ⊥ = constant
Relativistic Transformations of E and B
‰
r
r
According to observer O in frame F, particle has velocity v, fields are E and B
and Lorentz force is
(
r)
r
f ′ = q′E ′
r
r r r
f = q E+v ∧B
‰
In Frame F′, particle is at rest and force is
‰
Assume measurements give same charge and force, so
r
r r
q = q′ and E ′ = E + v ∧ B
‰
r
Point charge q at rest in F: E =
‰
See a current in F′, giving a field
‰
r r 1 r r
B′ = B − 2 v ∧ E
c
Suggests
r
r
r
, B=0
3
4πε 0 r
q
Exact:
r r
r
µ0 q v ∧ r
1 r r
=− 2 v∧E
B=−
3
4π r
c
(
)
r
r
r
r r
E⊥′ = γ E⊥ + v ∧ B , E//′ = E//
r
r
r
r
⎛
v∧E⎞ r
B⊥′ = γ ⎜⎜ B⊥ − 2 ⎟⎟ , B//′ = B//
c ⎠
⎝
Electromagnetic waves
‰
Maxwell’s equations predict the existence of electromagnetic waves, later discovered
by Hertz.
‰
No charges, no currents:
r
r ∂D
∇∧H =
∂t
r
∇⋅D = 0
r
r
∂B
∇∧E = −
∂t
r
∇⋅B = 0
r
r
∂B
∇ ∧ ∇ ∧ E = −∇ ∧
∂t
r
r
r
2
r
∂
∇ ∧ ∇ ∧ E = ∇ ∇⋅E −∇ E
=−
∇∧B
r
2
∂t
= −∇ E
r
r
∂2D
∂2E
= −µ
= − µε
2
∂ t 2 3D wave equation :
∂t
(
)
(
)
(
)
(
)
r
r
r
r
2
2
2
2
r
∂ E ∂ E ∂ E
∂ E
∇ 2 E = 2 + 2 + 2 = µε 2
∂x
∂y
∂z
∂t
Nature of electromagnetic waves
‰
‰
‰
A general plane wave
r with angular frequency ω travelling in the direction of
the wave vector k has the form
r r
r r
r r
r r
E = E 0 exp[ j (ω t − k ⋅ x )] B = B0 exp[ j (ω t − k ⋅ x )]
r r
Phase ω t − k ⋅ x = 2π × number of waves and so is a Lorentz invariant.
Apply Maxwell’s equations
r
r
r
∇ ↔ − jk
∇⋅E = 0 = ∇⋅B
∂
r
r&
↔ jω
∇ ∧ E = −B
∂t
↔
↔
r r
r r
k ⋅E = 0 = k ⋅B
r r
r
k ∧ E =ωB
Waves are transverse to the direction of propagation, and
r
k are mutually perpendicular
s r
E, B
and
Plane electromagnetic wave
Plane Electromagnetic Waves
r
r 1 ∂E
∇∧B = 2
c ∂t
r r
ω r
↔ k ∧B=− 2E
c
r r
r
Combined with k ∧ E = ω B
r
E ω kc 2
deduce that r = =
ω
B k
2π
Wavelength λ = r
k
ω
Frequency ν =
2π
⇒
speed of wave in vacuum is
ω
r =c
k
r r
The fact that ω t − k ⋅ x is an invariant tells
us that
⎛ω r⎞
Λ = ⎜ ,k ⎟
⎝c ⎠
is a Lorentz 4-vector, the 4-Frequency vector.
Deduce frequency transforms as
(
)
r r
c−v
ω′ = γ ω − v ⋅ k = ω
c+v
Waves in a conducting medium
‰
r
r
For a medium of conductivity σ, j = σ E
‰
r r
r&
r
r&
Modified Maxwell: ∇ ∧ H = j + ε E = σ E + ε E
‰
Put
r r
r r
E = E 0 exp[ j (ω t − k ⋅ x )]
r r
r r
B = B0 exp[ j (ω t − k ⋅ x )]
r r
r
r
− j k ∧ H = σ E + jω ε E
Dissipation factor
σ
D=
ωε
conduction
current
displacement
current
Copper : σ = 5.8 ×107 , ε = ε 0
⇒ D = 1012
Teflon : σ = 3 ×10-8 , ε = 2.1ε 0
⇒ D = 2.57 ×10 − 4
Attenuation in a Good Conductor
r r
r
r
− j k ∧ H = σ E + jω ε E
r r
r
r&
r
Combine with ∇ ∧ E = − B ⇒ k ∧ E = ω µ H
r r r
r r
r
⇒ k ∧ k ∧ E = ω µ k ∧ H = −ω µ (− jσ + ω ε )E
(
)
k 2 = ω µ (− jσ + ω ε )
⇒
σ >> ω ε
For a good conductor D >> 1,
Wave form is
where
δ =
k 2 ≈ − jω µσ
⎡ ⎛
x ⎞⎤
⎛ x⎞
exp ⎢ j ⎜ ω t − ⎟ ⎥ exp ⎜ − ⎟
δ ⎠⎦
⎝ δ⎠
⎣ ⎝
2
ω µσ
is the skin - depth
copper.mov
water.mov
⇒ k≈
ω µσ
2
(1 − j )
Maxwell’s Equations in a uniform perfectly
conducting guide
z
Hollow metallic cylinder with perfectly
conducting boundary surfaces
Maxwell’s equations with time dependence exp(jωt)are:
r
r
r
r
r
r
2
∂B
∇ E = ∇ ∇⋅E −∇ ∧ ∇ ∧ E
∇∧E = −
= − jω µ H
r
∂t
⇒
= jω µ ∇ ∧ H
r
r ∂D
r
r
2
∇∧H =
= jω ε E
= −ω εµ E
t
∂
144444444444244
4444444443
r
⎧
⎫
⎛
⎞⎪ E ⎪
2
⎪
⎜ 2
⎟⎪
⎜⎜ ∇ + ω µε ⎟⎟ ⎨ r ⎬ = 0
⎝
⎠ ⎪⎪ H ⎪⎪
⎩
⎭
(
x
y
r
r
Assume E ( x, y, z , t ) = E ( x, y )e ( jω t −γ z )
r
r
H ( x, y, z , t ) = H ( x, y )e ( jω t −γ z )
γ is the propagation constant
)
(
r
⎧E ⎫
Then ∇ t2 + (ω 2εµ + γ 2 ) ⎨ r ⎬ = 0
⎩H ⎭
[
]
Can solve for the fields completely
in terms of Ez and Hz
)
Special cases
‰
Transverse magnetic (TM modes):
ƒ Hz=0 everywhere, Ez=0 on cylindrical boundary
‰
Transverse electric (TE modes):
ƒ Ez=0 everywhere,
‰
∂H z
= 0 on cylindrical boundary
∂n
Transverse electromagnetic (TEM modes):
ƒ Ez=Hz=0 everywhere
ƒ requires γ 2 + ω 2εµ = 0 or γ = ± jω εµ
A simple model with Ez=0
Transport between two infinite parallel conducting plates:
r
E = (0,1,0) E ( x) e ( jω t −γ z )
where E ( x) satisfies
2
d
E
∇ 2t E = 2 = − K 2 E , K 2 = ω 2εµ + γ 2
dx
⎧ sin ⎫
i.e. E = A⎨ ⎬ Kx
⎩cos ⎭
y
x
To satisfy boundary conditions, E=0 on x=0 and x=a, so
z
E = A sin Kx, K = Kn =
a
x=
0
x=
nπ
, n integer
a
Propagation constant is γ = K n2 − ω 2εµ
=
nπ
a
⎛ω ⎞
1 − ⎜⎜ ⎟⎟
⎝ ωc ⎠
2
where ω c =
Kn
εµ
Cut-off frequency, ωc
2
γ=
‰
⎛ω⎞
nπ
nπ
nπx jω t −γ z
e
1 − ⎜⎜ ⎟⎟ , E = A sin
, ωc =
a
a
a εµ
⎝ ωc ⎠
ω<ωc gives real solution for γ, so
attenuation only. No wave propagates:
cut-off modes.
‰
ω>ωc gives purely imaginary solution for
γ, and a wave propagates without
attenuation.
γ = jk , k = εµ (ω − ω
2
‰
2
c
)
1
2
⎛ ω
= ω εµ ⎜⎜1 −
⎝ ω
ω only a finite
number of modes can propagate.
nπ
aω
ω > ωc =
⇒ n<
εµ
π
a εµ
2
c
2
⎞
⎟⎟
⎠
1
2
For a given frequency
For given frequency,
convenient to choose a s.t.
only n=1 mode occurs.
Propagatedr electromagnetic fields
‰
From
r
∂B
∇∧E =−
,
∂t
(assuming A is real)
⎧
Ak
⎛ nπ x ⎞
H
=
−
sin
⎟ cos(ω t − kz )
⎜
⎪
x
ωµ ⎝ a ⎠
r
r
⎪
j
H=
Hy = 0
∇∧E ⇒ ⎨
ωµ
⎪
A nπ
⎛ nπ x ⎞
cos⎜
⎪H z = −
⎟ sin (ω t − kz )
ωµ a
⎝ a ⎠
⎩
x
z
Phase and group velocities
∞
Plane wave exp j(ωt-kx) has
constant phase ωt-kx at peaks
⇔
ω ∆ t − k∆ x = 0
ω
∆x
vp =
∆t
=
k
j [ω ( k ) t − kx ]
A
(
k
)
e
dk
∫
−∞
Superposition of plane waves. While
shape is relatively undistorted,
pulse travels with the group velocity
dω
vg =
dk
Wave packet structure
‰
Phase velocities of individual plane waves making up
the wave packet are different,
‰
The wave packet will then disperse with time
Phase and group velocities in the simple
wave guide
‰
(
k = εµ ω − ω
Wave number is
so wavelength in guide λ =
‰ Phase velocity is v p =
‰
ω
k
>
2
2
c
)
1
2
< ω εµ
2π
2π
>
, the
k ω εµ
1
εµ
free-space wavelength
, larger than free-space velocity
Group velocity is less than infinite space value
(
k = εµ ω − ω
2
2
2
c
)
⇒
dω
k
vg =
=
<
dk ωεµ
1
εµ
Calculation of wave properties
‰
If a=3 cm, cut-off frequency of lowest order mode is
ωc
1
=
≅ 5 GHz
2π 2a εµ
fc =
‰
At 7 GHz, only the n=1 mode propagates and
(
k = εµ ω − ω
λ=
2
c
)
1
2
≈ 103 m −1
2π
≈ 6 cm
k
vp =
vg =
2
ω
k
≈ 4.3 ×108 ms −1
k
ωεµ
= 2.1× 108 ms −1
Waveguide animations
‰
TE1 mode above cut-off
ppwg_1-1.mov
‰
TE1 mode, smaller ω
ppwg_1-2.mov
‰
TE1 mode at cut-off
ppwg_1-3.mov
‰
TE1 mode below cut-off
ppwg_1-4.mov
‰
TE1 mode, variable ω
ppwg_1_vf.mov
‰
TE2 mode above cut-off
ppwg_2-1.mov
‰
TE2 mode, smaller
ppwg_2-2.mov
‰
TE2 mode at cut-off
ppwg_2-3.mov
‰
TE2 mode below cut-off
ppwg_2-4.mov