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12. The Maxwell Equations
a. Farady’s law and the Electromagnetic “force” (EMF).
Consider a charge, q, moved around a loop, c. The work done on it, per unit
charge is the electromagnetic force, ε , is
ε =∫
c
r r
F ⋅ dl
q
r
The force on a charge moving with a velocity v is the Lorentz force,
r r
r
r
⎛
v×B⎞
⎟
F = q⎜⎜ E +
⎟
c
⎝
⎠
but we know that (Stoke’s theorem) for a time-independent electric field
r r
r r
∫ E ⋅ d l = ∫ ∇ × E ⋅ nˆda = 0
c
However,
(
)
(
)
(
)
1 r r r 1 r r r
1d r r r
1 dΦ
l
ε = ∫ v × B ⋅ dl = ∫ B ⋅ dl × v = −
⋅
×
=
−
B
d
x
d
c
c
c dt ∫
c dt
On the other hand we define
1
r r
r r
ε = ∫ E ⋅ d l = ∫ ∇ × E ⋅ nˆ da
c
to obtain
r
r r 1 ∂B
∇× E +
=0
c ∂t
Collecting all our results, the Maxwell’s equations in vacuum are
r r
r r
∇ ⋅ E = 4πρ
∇ ⋅ B =r 0
r
r r 1 ∂B
r r 1 ∂E 4π r
J
= 0 ∇× B −
=
∇× E +
c ∂t
c ∂t
c
Implicit in those are the continuity equation,
r r ∂ρ
∇⋅J +
=0
∂t
and the Lorentz force
r r
r
r
⎛
v×B⎞
⎟⎟
F = q⎜⎜ E +
c
⎝
⎠
In the macroscopic media, the Maxwell’s equations are
2
r r
r r
∇ ⋅ D = 4πρ f
∇⋅B = 0
r
r
r r 1 ∂B
r r 1 ∂D 4π r
∇× E +
=
= 0 ∇× H −
Jf
c ∂t
c ∂t
c
with
r r
r
D = E + 4πP
r r
r
H = B − 4πM
b. Boundary conditions.
Let the fields in two neighboring media be denoted by 1 and 2. Let the surface
charge density on the surface between the two media be σ and let the surface
r
current density there be K . Then the boundary conditions are
r
r
( D2 − D1 ) ⋅ nˆ = 4πσ
r
r
( E2 − E1 ) × nˆ = 0
r
r
( B2 − B1 ) ⋅ nˆ = 0
r
r
4π r
nˆ × ( H 2 − H 1 ) =
K
c
c. Scalar and vector potentials
r r r
r
Let us introduce the vector potential A such that B = ∇ × A . This ensures that
r r
∇ ⋅ B = 0 . Then, from the Maxwell’s equations, we have
r
r ⎛ r 1 ∂A ⎞
⎟=0
∇ × ⎜⎜ E +
c ∂t ⎟⎠
⎝
3
and therefore we choose
r
r
⎛ r 1 ∂A ⎞
⎜E +
⎟ = −∇ Φ
⎜
c ∂t ⎟⎠
⎝
These choices satisfy the two homogeneous Maxwell’s equations. The other two
equations determine the scalar and the vector potentials,
1 ∂ r r
∇ Φ+
∇ ⋅ A = −4πρ
c
∂
t
r
2
r
r r r 1 ∂Φ
1
4π r
∂
A
2
)=−
∇ A − 2 2 − ∇ (∇ ⋅ A +
J
c ∂t
c ∂t
c
2
We note that both the electric and the magnetic field are unchanged by the
transformation
r
r r r
A → A' = A + ∇Λ
1 ∂Λ
Φ → Φ' = Φ −
c ∂t
and therefore we can choose the gauge
r r 1 ∂Φ
∇⋅ A+
=0
c ∂t
to obtain the Maxwell’s equations in the form
4
1 ∂ 2Φ
∇ Φ − 2 2 = −4πρ
c ∂tr
r 1 ∂2 A
4π r
2
∇ A− 2 2 = −
J
c ∂t
c
2
(Lorentz gauge)
r r
∇
⋅ A = 0 , and then to obtain
But of course, we could have stayed with
∇r2Φ = −4πρ
r 1 ∂ 2 A r 1 ∂Φ
4π r
2
∇ A − 2 2 − ∇(
)=−
J
c ∂t
c ∂t
c
(Coulomb gauge)
d. Poynting theorem.
The rate of work done by the electromagnetic fields on a moving charge q is
For a system of current density, that rate will be
r r
qv ⋅ E .
r
r
r
r
r
r
r
⎛
r
r
1
∂D ⎞
⎜
∫ dr J ⋅ E = 4π ∫ dr ⎜⎝ cE ⋅ (∇ × H ) − E ⋅ ∂t ⎟⎟⎠
Using
r r r
r r r
r r r
∇ ⋅ ( E × H ) = H ⋅ (∇ × E ) − E ⋅ (∇ × H )
we find
r
r
r
r
r
r
r
r
r
r
r⎛
1
∂D
∂B ⎞
⎜
⎟⎟
(
)
d
r
J
⋅
E
=
−
d
r
c
∇
⋅
E
×
H
+
E
⋅
+
H
⋅
∫
∫
⎜
4π
∂t
∂t ⎠
⎝
5
Denoting the total energy density of the electromagnetic field by U,
(
1 r r r r
U=
E⋅D+ B⋅H
8π
)
we obtain the conservation law
r r
∂U r r
+ ∇ ⋅ S = −J ⋅ E
∂t
total energy rate
mechanical work
with the vector pointing
r
c r r
S=
E×H
4π
Example: A segment of a wire of length L and radius a, carrying the current I. At the
surface of the wire the magnetic field is
r 2I
B = ϕ̂
ca
and the electric field there is
6
r V
E = zˆ ,
L
where V is the voltage creating the current. The Poynting vector at the surface of
the wire is
r
c r r
I V
S=
E×H =
ρˆ ,
4π
2πa L
which is the power supplied by the battery,
the wire,
2πaL .
P = IV , divided by the surface area of
13. Electromagnetic waves
a. In the absence of sources (i.e., no free charges or free currents) the Maxwell’s
equations read
r r
r r
∇ ⋅ D = 0r
∇ ⋅ B = 0r
r r 1 ∂B
r r 1 ∂D
∇× E +
= 0 ∇× H −
=0
c ∂t
c ∂t
Let us describe the medium by its dielectric constant,
permeability,
r
r
B = µH . We then have
r
r
D = εE , and magnetic
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r r
r r
∇ ⋅ E = 0r
∇ ⋅ B = 0r
r r 1 ∂B
r r µε ∂E
∇× E +
= 0 ∇× B −
=0
c ∂t
c ∂t
We see that each of the filed components satisfy the wave equation
1 ∂ 2u
∇ u− 2 2 =0
v ∂t
2
with the velocity
c
v=
εµ
The solution is (in general)
r r
r r
r
ik ⋅ r − i ω t
ik ⋅ r − iω t
u ( r , t ) = Ae
+ Be
with the dispersion relation relating the wave-vector with the frequency
k=
ω
v
= εµ
ω
c
Let us now solve for the electromagnetic fields. We write
r iknˆ ⋅ rr − iωt
r r
E (r , t ) = E0e
r r
r iknˆ ⋅ rr − iωt
B (r , t ) = B0e
8
with the convention that the physical fields are the real parts of the above
expressions. From the homogeneous Maxwell’s equations we have that
r
E0 ⋅ nˆ = 0
r
B0 ⋅ nˆ = 0
and therefore both the electric and the magnetic fields are perpendicular to the
direction of the wave propagation. This wave is called transverse wave. From the
other two Maxwell’s equations we have
r
r
B0 = εµ nˆ × E0
The Poynting vector, averaged over a period, is given by the real part of
r
c r r*
〈S 〉 =
E×H
8π
Explanation:
r
c 1 r r* 1 r r *
c r r r* r * r r * r* r
〈S 〉 =
〈 ( E + E ) × ( H + H )〉 =
〈E × H + E × H + E × H + E × H 〉
4π 2
2
16π
c r r * r* r
=
〈E × H + E × H 〉
16π
In our case
9
r
c
〈S 〉 =
8π
ε r 2
| E0 | nˆ
µ
The energy density, averaged over a period is
1
〈u 〉 =
16π
⎛ r r* 1 r r* ⎞
⎜⎜ εE ⋅ E + B ⋅ B ⎟⎟
µ
⎠
⎝
and in our case
ε r 2
〈u 〉 =
| E0 |
8π
b. Electromagnetic waves in a conducting medium. We consider the
charge-free case.
We have
r r
r r
r r
∇ ⋅ D = ε∇ ⋅ E
∇ ⋅ B r= 0
r =0
r r 1 ∂B
r r 1 ∂D 4π r
∇× E +
= 0 ∇× H −
Jf
=
c ∂t
c ∂t
c
Using Ohm’s law, relating the current density to the electric field
Via the conductivity
r
r
J = σE
the fourth equation becomes
10
v
r
r
1
ε ∂E 4π r
∇× B −
=
σE
µ
c ∂t
c
Therefore we now find
r
r
r
2
⎛
⎛
⎞
⎛
⎞
εµ ∂ ⎜ E ⎞⎟
4πµσ ∂ E
2 E
⎜
⎟
⎜
⎟
r⎟− 2 2 ⎜ r⎟ = 0
∇ ⎜ r⎟−
2
⎜
∂
c
t
⎝ B ⎠ c ∂t ⎝ B ⎠
⎝B⎠
Using again a plane-wave solution, the dispersion law (relating the wave
vector to the frequency) is now
k = µε
2
ω2 ⎛
4πσ ⎞
+
i
1
⎜
⎟
ωε ⎠
c2 ⎝
and the wave is damped.
Approximately, we find
2π µ
σ
c
c ε
2πωµσ
k = (1 + i )
c
k = εµ
ω
+i
4πσ
<< 1
ωε
4πσ
>> 1
ωε
11
In a good conductor (high conductivity, law frequency), the electromagnetic
wave penetrates only over a length given by
δ=
c
2πµωσ
which is called `skin-depth’.
Example: A wave of frequency ω polarized in the x-direction, and propagates
along the z-direction in a good conductor occupying the z>0 space. Let us
assume that
ε , µ = 1.
The wave is then given by
r
E = E0 xˆe − iωt e − z (1− i )δ
The current density which is established in the conductor is
r
J = σE0 xˆe − iωt e − z (1− i ) / δ
Let us consider a slab of dimensions
in the slab (along the x-direction) is
l × h × dz . Then the current flowing
dI = σE0 e − iωt e − z (1− i ) / δ hdz
and the total current becomes
12
∞
I = ∫ σE0 e − iωt e − z (1− i ) / δ hdz = E0
0
σhδ
1− i
e − iωt
Taking the real part to obtain the `true’ current, we find
I = E0σhδ
cos(ωt +
π
4
)
2
We see that the current is not in=phase with the field.
The resistance of a slab of thickness
R=
δ
is
l
σhδ
Writing
V = E0 l
for the potential difference on the slab, we find
IR = V
cos(ωt +
π
4
)
2
13
14. Waveguides
Consider a rectangular waveguide of cross section of dimensions a and b and axis
along the z-direction. In the absence of charge and current distributions, the
electric and magnetic fields satisfy
r
r
2
⎛E⎞ 1 ∂ ⎛E⎞
∇ 2 ⎜⎜ r ⎟⎟ − 2 2 ⎜⎜ r ⎟⎟ = 0
⎝ H ⎠ c ∂t ⎝ H ⎠
When the wave propagates along the z-direction, and the electric field is only in the
x-y plane, then
r
E = xˆE x + yˆ E y
r
H = xˆH x + yˆ H y + zˆH z
and all the z-dependence of the fields is in the form
e
ik g z
. Denoting
ω = ck
we have
∂2H z ∂2H z
2
2
+
+
(
k
−
k
g )H z = 0
2
2
∂x
∂y
14
r
r r 1 ∂B
= 0 we then find
From the Maxwell’s equation ∇ × E +
c ∂t
k
k
H y , Ey = − H x
Ex =
kg
kg
and from his other equation
ikE x +
r
r r 1 ∂D
∇× H −
=0
c ∂t
we have
∂H z
∂H z
− ik g H y = 0, ikE y −
+ ik g H x = 0
∂y
∂x
Therefore,
⎛ ⎡ k ⎤2 ⎞
∂H z
ik g ⎜1 − ⎢ ⎥ ⎟ H x =
⎜ ⎢k ⎥ ⎟
∂x
⎝ ⎣ g⎦ ⎠
⎛ ⎡ k ⎤2 ⎞
∂H z
ik g ⎜1 − ⎢ ⎥ ⎟ H y =
⎜ ⎢k ⎥ ⎟
∂y
⎝ ⎣ g⎦ ⎠
and it is sufficient to find H z to find all fields. Now we solve for that component
by variables separation. Writing
H z ( x, y ) = h1 ( x ) h2 ( y )
we have
15
d 2 h1
2
+
k
h1 = 0,
x
2
dx
d 2 h2
2
+
k
h2 = 0
y
2
dy
with
k x2 + k y2 − k 2 + k g2 = 0
We now have the following solution
Hz = e
E x = −i
Ey = i
kk y
k
2
g
ik g z
( A1 sin k x x sin k y y + A2 sin k x x cos k y y + A3 cos k x x sin k y y + A4 cos k x x cos k y y )
(1 − [
k 2 −1 ik g z
] ) e ( A1 sin k x x cos k y y − A2 sin k x x sin k y y + A3 cos k x x cos k y y − A4 cos k x x sin k y y )
kg
kk x
k
ik z
(1 − [ ]2 ) −1 e g ( A1 cos k x x sin k y y + A2 cos k x x cos k y y − A3 sin k x x sin k y y − A4 sin k x x cos k y y )
2
kg
kg
Now we need to apply boundary conditions. Taking the surface of the waveguide as
a conductor then
Ey = 0
at x=0 and x=a gives
A1 = A2 = 0 and
k x = mπ / a
16
where m is an integer. Taking
Ex = 0
at y=0 and y=b gives
A1 = A3 = 0
and
k y = nπ / b
where n is an integer. Therefore the final solution is
Hz = e
ik g z
∑A
mn cos(
nm
mπx
nπy
) cos(
)
a
b
with
⎛ mπ ⎞ ⎛ nπ ⎞
k g2 = k 2 − ⎜
⎟ −⎜
⎟
a
b
⎝
⎠ ⎝
⎠
2
2
limiting the allowed values of m and n!
17