### Plane waves in conductors, dielectric-conductor boundaries

#### Problem:

A "tenuous" plasma consists of free electric charges of mass m and charge -qe (where qe is positive).  There are n charges per unit volume.  Assume that the density is uniform and that the interactions between the charges may be neglected.  Also assume that the charges can be treated classically.
A linearly-polarized electromagnetic wave of frequency ω is incident on the plasma.
Let the electric field component of the plane wave be E = E0exp(i(kx - ωt)).
(a)  Solve the equation of motion for a single charge and find the current density j and the conductivity σ of the plasma as a function of ω.
(b)  Assume a plane wave of the form E = E0exp(i(kx - ωt)) propagate in the plasma with conductivity σ.  Find the dispersion relation (the relation between k and ω) for the electromagnetic wave in the plasma and the index of refraction as a function of ω.

Solution:

• Concepts:
Newton's 2nd law, charge and current density, plane waves in a conducting medium
• Reasoning:
The "tenuous" plasma is a conducting medium.  We can find its conductivity by solving for the motion of the free charges using Newton's second law.
• Details of the calculation:
(a)  Equation of motion for a single charge: md2r/dt2 = -qeE = -qeE0exp(i(kx - ωt)).
r(t) = (qeE0/(mω2))exp(i(kx - ωt)).  v(t) = (qeE0/(imω))exp(i(kx - ωt)) = (qe/(imω))E.
j
(t) = -nqev(t) = -(nqe2/(imω))E(t).  j = σE.  σ = (inqe2/(mω)).

(b)  ×B = μ0σE + μ0ε0E/∂t.
×(×B) = (∇∙B) - 2B = μ0σ(×E) + μ0ε0∂(×E)/∂t.
2B - μ0σ∂B/∂t - μ0ε02B/∂t2 = 0.
Similarly:  2E - μ0σ∂E/∂t - μ0ε02E/∂t2 = 0.
Plane wave solutions of the form E = E0exp(i(kx - ωt)) have to satisfy
-k2 + iωμ0σ + ω2/c2 = 0,  k2 = ω2/c2 - (μ0nqe2/m ) = (1/c2)(ω2 - ωp2),
k = (1/c)(ω2 - ωp2)½,
where ωp = μ0nqe2c2/m = nqe2/(ε0m).
c/n = ω/k.  n = ck/ω = (1 - ωp22)½.  For ωp > ω the index of refraction n is imaginary, the field decreases exponentially.

#### Problem:

In a medium with conductivity σ but no net charge, write down Maxwell's equations, and derive the wave equation for the electric field, E, in this medium.

Solution:

• Concepts:
Maxwell's equations
• Reasoning:
In regions with  ρf = 0  and  jf = σcE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
Assume ε = ε0, μ = μ0, ρf = 0 and jf = σcE in the conductor.  Then Maxwell's equations become
∇∙E = 0,  ×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = μ0σE + μ0ε0E/∂t.
×(×E) = (∇∙E) - 2E = -∂(×B)/∂t = -μ0j/∂t - μ0ε02E/∂t2.
2E = μ0σcE/∂t + μ0ε02E/∂t2, or
2E - μ0σcE/∂t - μ0ε02E/∂t2 = 0.
This is the damped wave equation for E.
Similarly,
×(×B) = (∇∙B) - 2B = μ0σ(×E) + μ0ε0∂(×E)/∂t.
2B - μ0σ∂B/∂t - μ0ε02B/∂t2 = 0.
This is the damped wave equation for B.

#### Problem:

Starting from Maxwell's equations, prove that a plane EM wave propagates in a good conductor such that the electric and magnetic fields are out of phase by 45o.

Solution:

• Concepts:
Maxwell's equations
• Reasoning:
In regions with  ρf = 0  and  jf = σcE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
Maxwell's equations:
∇∙E = ρ/ε0×E = -∂B/∂t,  ∇∙B = 0,  ×B = μ0j + (1/c2)∂E/∂t.
Assume ε = ε0, μ = μ0, ρf = 0 and jf = σcE in the conductor.  Then
×B = μ0σcE + μ0ε0E/∂t.
×(×B) = (∇∙B) - 2B = μ0σc(×E) + μ0ε0∂(×E)/∂t.
2B - μ0σcB/∂t - μ0ε02B/∂t2 = 0.
Similarly:  2E - μ0σcE/∂t - μ0ε02E/∂t2 = 0.
Both E and B satisfy the damped wave equation.
Try solutions of the form E(r,t) = E0 exp(i(k∙r - ωt)).
Then k2 = iμ0σcω + μ0ε0ω2 = μ0ε0ω2(1 + iσc/(ε0ω)).
k2 = μ0ε0ω2(1 + σc2/(ε0ω)2)½ e.  tanφ = σc/(ε0ω).
k = (μ0ε0)½ω(1 + σc2/(ε0ω)2)¼ eiφ/2 = |k|eiφ/2.
For a plane wave propagating into the z-direction we have
E(r,t) = E0 exp(ikz)exp(-iωt).
×E = ik×E = -∂B/∂t.
B(r,t) = B(r) exp(-iωt).  Therefore  ik×E  = iωB.  ik(z/z)×E  = iωB.
Choose the coordinate system so that E = E(x/x), then B(r) = (y/y)(k/ω)E0exp(ikz).
B(r) = (y/y)(|k|/ω)E0eikzeiφ/2 = (y/y)B0eikzeiφ/2,
where  B0 = (|k|/ω)E0B0 = (y/y)B0.
E(r,t) = E0 exp(i(k∙r - ωt)).  B(r,t) = B0 exp(i(k∙r - ωt + φ/2)).
For a good conductor σc/(ε0ω) >> 1.  tanφ --> ∞.  φ -- > 90o.  φ/2 --> 45o.
[Example:  copper
σc = 5.98*107 (Ωm)-1, ε0 = 8.85*10-12 C2/(Nm2), σc/(ε0ω) = 6.75*1018 s-1/ω.]

#### Problem:

An electromagnetic wave with frequency of 106 Hz "travels" along the z-axis in an aluminum medium located at z ≥ 0.  The conductivity of aluminum is 38.2∙ 106 (Ωm)-1 and its relative permeability is km = 1.  Just inside the conductor at z = +0, the electric field amplitude is E0 i

(a)  Write down an expression for the electric field inside the conductor.
(b)  Find the skin depth, wave velocity, and wavelength of the wave in aluminum.
(c)  Determine the corresponding magnetic field.
(d)  Find the phase difference between the electric and magnetic fields at each fixed location in aluminum.

Solution:

• Concepts:
Maxwell's equations, plane waves in conductors.
• Reasoning:
In regions with  ρf = 0  and  jf = σcE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
(a)  Maxwell's equations:
∇∙E = ρ/ε0×E = -∂B/∂t,  ∇∙B = 0,  ×B = μ0j + (1/c2)∂E/∂t.
Assume ε = ε0, μ = μ0, ρf = 0 and jf = σcE in the conductor.  Then
×B = μ0σcE + μ0ε0E/∂t.
×(×B) = (∇∙B) - 2B = μ0σc(×E) + μ0ε0∂(×E)/∂t.
2B - μ0σcB/∂t - μ0ε02B/∂t2 = 0.
Similarly:  2E - μ0σcE/∂t - μ0ε02E/∂t2 = 0.
Both E and B satisfy the damped wave equation.
Plane wave solutions are of the form E(r,t) = E0 exp(i(k∙r - ωt)).
k2 = iμ0σcω + μ0ε0ω2 = μ0ε0ω2(1 + iσc/(ε0ω)) = μ0ε0ω2(1 + σc2/(ε0ω)2)½ e.  tanφ = σc/(ε0ω).
k is a complex number, k = β + iα/2.
k = (μ0ε0)½ω(1 + σc2/(ε0ω)2)¼ eiφ/2 = |k|eiφ/2.
To find the real and imaginary parts we have to find cos(φ/2) and sin(φ/2) in terms of tanφ.
cos(φ/2) = ((1 + cosφ)/2)½.  sin(φ/2) = ((1 - cosφ)/2)½.  cosφ = (1 + tan2φ) in the first quadrant.
Therefore cos(φ/2) = 2(1 + (1 + (σc/(ε0ω))2))½
sin(φ/2) = 2(1 - (1 + (σc/(ε0ω))2))½.
k = (μ0ε0/2)½ω[(1 + (σc/(ε0ω))2)½ + 1]½ + i(μ0ε0/2)½ω[(1 + (σc/(ε0ω))2) - 1]½.
β = (μ0ε0/2)½ω[(1 + (σc/(ε0ω))2)½ + 1]½.  α/2 = (μ0ε0/2)½ω[(1 + (σc/(ε0ω))2)½ - 1]½.
(b)  The skin depth δ is the distance it takes to reduce the amplitude by a factor of 1/e.
δ = 2/α.  σc/(ε0ω) =  (38.2*106 /Ωm)/(8.85*10-12 C2/(Nm2))/(2π*106/s) = 6.87*1011.
σc/(ε0ω) >> 1, α/2 ≈ (μ0ε0/2)½ω(σc/(ε0ω))½ = (μ0σcω/2)½,
δ = (2/(μ0σcω))½ = (2/(4π*10-7)/(38.2*106)/(2π*106))½m = 8.14*10-5m.
β ≈ α/2 ≈ (μ0σcω/2)½ = 1/δ.
Wave velocity = ω/β = ω*δ = (2π*106)(8.14*10-5) m/s = 5.1*102 m/s.
Wavelength = 2π/β = 2π*δ = 5.1*10-4 m.
All numbers are for aluminum.
(c)  Assume the plane wave is propagating into the z-direction.
E(r,t) = E0 exp(ikz)exp(-iωt).
×E = ik×E = -∂B/∂t.
B(r,t) = B(r) exp(-iωt).  Therefore  ik×E  = iωB.  ik(z/z)×E  = iωB.
Choose the coordinate system so that E = E(x/x), then B(r) = (y/y)(k/ω)ε0exp(ikz).
B(r) = (y/y)(|k|/ω)E0eikzeiφ/2 = (y/y)B0eikzeiφ/2,
where  B0 = (|k|/ω)E0.  |k| = (μ0ε0)½ω(1 + σc2/(ε0ω)2)¼ ≈ (μ0σcω)½ = 2½.
B
0 = (y/y)B0.
E(r,t) = E0 exp(i(kz - ωt)) = E0 e-αz/2exp(i(βz - ωt)).
B
(r,t) = B0 exp(i(kz - ωt + φ/2)) = B0 e-αz/2exp(i(βz - ωt + φ/2)).
(d)  For aluminum σc/(ε0ω) >> 1.  tanφ --> ∞.  φ -- > 90o.  φ/2 --> 45o.
B0 e-αz/2exp(i(βz - (ωt - π/4))).
"B lags behind E" by φ/2 = π/4.

#### Problem:

An electric field E = i E0exp(-iωt)is applied at the interface of a vacuum (z < 0) and a conductor (z > 0) of real average conductivity σc.  Assume ε = ε0, μ = μ0, in the conductor.
(a)  For σc >> ε0ω, calculate how deeply the electric field penetrates into the conductor, i.e. calculate the depth at which the electric field amplitude has decreased to 1/e of its value at the surface.
(b)  Calculate dW/dt =  j∙EdV, the rate at which work is done by the field on the charges in a volume dV, as a function of z.

Solution:

• Concepts:
Maxwell's equations
• Reasoning:
In regions with ρf = 0 and jf = σcE  Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
• Details of the calculation:
Assume jf = σcE in the conductor.  Then
×B = μ0σcE + μ0ε0E/∂t.
×(×E) = (∇∙E) - 2E = -∂(×B)/∂t. = -μ0σcE/∂t - μ0ε02E/∂t.
Therefore  2E - μ0σcE/∂t - μ0ε02E/∂t2 = 0.
The tangential component of E is continuous across the boundary, therefore E(z = 0) = i E0exp(-iωt) just inside the conductor.
2Ex/∂z2 + iμ0ωσcEx + μ0ε0ω2Ex = 0 inside the conductor.
Since σc >> ε0ω, we can write ∂2Ex/∂z2 + iμ0ω∂Ex = 0.
Try a solutions of the form Ex(z,t) = E0 exp(i(kz - ωt)).
Then k2 = iμ0ωσc, k = (μ0ωσc)½exp(iπ/4) = 2(1 + i)(μ0ωσc)½.
Im(k) = Re(k) = (μ0ωσc/2)½.
Ex(z,t) = E0 exp(-Im(k)z) exp(i(Re(k)z - ωt)).
Skin depth:  E0 exp(-Im(k)d) = 1/e.  d = (2/(μ0ωσc))½.
(b)  dW/dt =  j∙EdV = rate at which work is done by the field on the charges in dV.
When evaluating products, use the real parts of the complex expressions.
j∙E = σcE02 exp(-2Im(k)z) cos2(Re(k)z - ωt).  <j∙E> = σc½E02 exp(-2Im(k)z).
dW/dt = σcE02 exp(-2Im(k)z) cos2(Re(k)z - ωt)dV.
<dW/dt> = σc½E02 exp(-2Im(k)z)dV.