A “tenuous” plasma consists of free electric charges of
mass m and charge –e (where e 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** =
**E**_{0}exp(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** = **E**_{0}exp(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: md^{2}**r**/dt^{2}= -e**E**= -e**E**_{0}exp(i(kx - ωt)).

**r**(t) = (e**E**_{0}/(mω^{2}))exp(i(kx - ωt)).**v**(t) = (e**E**_{0}/(imω))exp(i(kx - ωt)) = (e/(imω))**E**(t) = -ne

j**v**(t) =**-**(ne^{2}/(imω))**E**(t).**j**= σ**E**. σ = (ine^{2}/(mω)).(b)

**∇**×**B**= μ_{0}σ**E**+ μ_{0}ε_{0}∂**E**/∂t.

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= μ_{0}σ(**∇**×**E**) + μ_{0}ε_{0}∂(**∇**×**E**)/∂t.

**∇**^{2}**B**- μ_{0}σ∂**B**/∂t - μ_{0}ε_{0}∂^{2}**B**/∂t^{2}= 0.

Similarly:**∇**^{2}**E**- μ_{0}σ∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}= 0.

Plane wave solutions of the form**E**=**E**_{0}exp(i(kx - ωt)) have to satisfy

-k^{2}+ iωμ_{0}σ + ω^{2}/c^{2}= 0, k^{2}= ω^{2}/c^{2}- (μ_{0}ne^{2}/m ) = (1/c^{2})(ω^{2}– ω_{p}^{2}),

k = (1/c)( ω^{2}– ω_{p}^{2})^{½},

where ω_{p}= (μ_{0}ne^{2}c^{2}/m)= (ne^{2}/(ε_{0}m)).

c/n = ω/k. n = ck/ω = (1 – ω_{p}^{2}/ω^{2})^{½}. For ω_{p}> ω the index of refraction n is imaginary, the wave is absorbed.

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 ρ_{f }= 0 and**j**= σ_{f }_{c}**E**Maxwell's equations can be used to show that both**E**and**B**satisfy the damped wave equation. - Details of the calculation:

Assume ε = ε_{0}, μ = μ_{0}, ρ_{f }= 0 and**j**= σ_{f }_{c}**E**in the conductor. Then Maxwell's equations become

**∇∙E**= 0,**∇**×**E**= -∂**B**/∂t,**∇∙B**= 0,**∇**×**B**= μ_{0}σ**E**+ μ_{0}ε_{0}∂**E**/∂t.

**∇**×(**∇**×**E**) =**∇**(**∇∙E**) -**∇**^{2}**E**= -∂(**∇**×**B**)/∂t = -μ_{0}∂**j**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}.

**∇**^{2}**E**= μ_{0}σ_{c}∂**E**/∂t + μ_{0}ε_{0}∂^{2}**E**/∂t^{2}, or

**∇**^{2}**E**- μ_{0}σ_{c}∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}= 0.

This is the damped wave equation for**E**.

Similarly,

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= μ_{0}σ(**∇**×**E**) + μ_{0}ε_{0}∂(**∇**×**E**)/∂t.

**∇**^{2}**B**- μ_{0}σ∂**B**/∂t - μ_{0}ε_{0}∂^{2}**B**/∂t^{2}= 0.

This is the damped wave equation for**B**.

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 45^{o}.

Solution:

- Concepts:

Maxwell's equations - Reasoning:

In regions with ρ_{f }= 0 and**j**= σ_{f }_{c}**E**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/ε_{0},**∇**×**E**= -∂**B**/∂t**, ∇∙B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t.Assume ε = ε

_{0}, μ = μ_{0}, ρ_{f }= 0 and**j**= σ_{f }_{c}**E**in the conductor. Then

**∇**×**B**= μ_{0}σ_{c}**E**+ μ_{0}ε_{0}∂**E**/∂t.

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= μ_{0}σ_{c}(**∇**×**E**) + μ_{0}ε_{0}∂(**∇**×**E**)/∂t.

**∇**^{2}**B**- μ_{0}σ_{c}∂**B**/∂t - μ_{0}ε_{0}∂^{2}**B**/∂t^{2}= 0.

Similarly:**∇**^{2}**E**- μ_{0}σ_{c}∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}= 0.

Both**E**and**B**satisfy the damped wave equation.

Try solutions of the form**E**(**r**,t) =**E**_{0 }exp(i(**k∙r**- ωt)).

Then k^{2}= iμ_{0}σ_{c}ω + μ_{0}ε_{0}ω^{2}= μ_{0}ε_{0}ω^{2}(1 + iσ_{c}/(ε_{0}ω)).

k^{2}= μ_{0}ε_{0}ω^{2}(1 + σ_{c}^{2}/(ε_{0}ω)^{2})^{½ }e^{iφ}. tanφ = σ_{c}/(ε_{0}ω).

k = (μ_{0}ε_{0})^{½}ω(1 + σ_{c}^{2}/(ε_{0}ω)^{2})^{¼ }e^{iφ/2}= |k|e^{iφ/2}.

For a plane wave propagating into the z-direction we have

**E**(**r**,t) =**E**_{0 }exp(ikz)exp(-iωt).

**∇**×**E**= i**k**×**E**= -∂**B**/∂t.**B**(**r**,t) =**B**(**r**)_{ }exp(-iωt). Therefore i**k**×**E**= iω**B**. ik(**z**/z)×**E**= iω**B**.

Choose the coordinate system so that**E**= E(**x**/x), then**B**(**r**) = (**y**/y)(k/ω)ε_{0}exp(ikz).**B**(**r**) = (**y**/y)(|k|/ω)ε_{0}e^{ikz}e^{iφ/2}= (**y**/y)B_{0}e^{ikz}e^{iφ/2},

where B_{0}= (|k|/ω)ε_{0}.**B**_{0}= (**y**/y)B_{0}.

**E**(**r**,t) =**E**_{0 }exp(i(**k∙r**- ωt)).**B**(**r**,t) =**B**_{0 }exp(i(**k∙r**- ωt + φ/2)).

For a good conductor σ_{c}/(ε_{0}ω) >> 1. tanφ --> ∞. φ -- > 90^{o}. φ/2 --> 45^{o}.

[Example: copper

σ_{c}= 5.98*10^{7}(Ωm)^{-1}, ε_{0}= 8.85*10^{-12}C^{2}/(Nm^{2}), σ_{c}/(ε_{0}ω) = 6.75*10^{18}s^{-1}/ω.]

An electromagnetic wave with
frequency of 10^{6} Hz "travels" along the z-axis in an aluminum medium
located at z ≥ 0. The conductivity of aluminum is
38.2∙ 10^{6} (Ωm)^{-1}
and its relative permeability is k_{m }= 1. Just inside the
conductor at z = +0, the electric field amplitude is E_{0}** 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 ρ_{f }= 0 and**j**= σ_{f }_{c}**E**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/ε_{0},**∇**×**E**= -∂**B**/∂t**, ∇∙B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t.

Assume ε = ε_{0}, μ = μ_{0}, ρ_{f }= 0 and**j**= σ_{f }_{c}**E**in the conductor. Then

**∇**×**B**= μ_{0}σ_{c}**E**+ μ_{0}ε_{0}∂**E**/∂t.

**∇**×(**∇**×**B**) =**∇**(**∇∙B**) -**∇**^{2}**B**= μ_{0}σ_{c}(**∇**×**E**) + μ_{0}ε_{0}∂(**∇**×**E**)/∂t.

**∇**^{2}**B**- μ_{0}σ_{c}∂**B**/∂t - μ_{0}ε_{0}∂^{2}**B**/∂t^{2}= 0.

Similarly:**∇**^{2}**E**- μ_{0}σ_{c}∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}= 0.

Both**E**and**B**satisfy the damped wave equation.

Plane wave solutions are of the form**E**(**r**,t) =**E**_{0 }exp(i(**k∙r**- ωt)).

k^{2}= iμ_{0}σ_{c}ω + μ_{0}ε_{0}ω^{2}= μ_{0}ε_{0}ω^{2}(1 + iσ_{c}/(ε_{0}ω)) = μ_{0}ε_{0}ω^{2}(1 + σ_{c}^{2}/(ε_{0}ω)^{2})^{½ }e^{iφ}. tanφ = σ_{c}/(ε_{0}ω).

k is a complex number, k = β + iα/2.

k = (μ_{0}ε_{0})^{½}ω(1 + σ_{c}^{2}/(ε_{0}ω)^{2})^{¼ }e^{iφ/2}= |k|e^{iφ/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 + tan^{2}φ)^{-½}in the first quadrant.

Therefore cos(φ/2) = 2^{-½}(1 + (1 + (σ_{c}/(ε_{0}ω))^{2})^{-½})^{½}.

sin(φ/2) = 2^{-½}(1 - (1 + (σ_{c}/(ε_{0}ω))^{2})^{-½})^{½}.

k = (μ_{0}ε_{0}/2)^{½}ω[(1 + (σ_{c}/(ε_{0}ω))^{2})^{½}+ 1]^{½}+ i(μ_{0}ε_{0}/2)^{½}ω[(1 + (σ_{c}/(ε_{0}ω))^{2})^{-½}- 1]^{½}.

β = (μ_{0}ε_{0}/2)^{½}ω[(1 + (σ_{c}/(ε_{0}ω))^{2})^{½}+ 1]^{½}. α/2 = (μ_{0}ε_{0}/2)^{½}ω[(1 + (σ_{c}/(ε_{0}ω))^{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. σ_{c}/(ε_{0}ω) = (38.2*10^{6}/Ωm)/(8.85*10^{-12}C^{2}/(Nm^{2}))/(2π*10^{6}/s) = 6.87*10^{11}.

σ_{c}/(ε_{0}ω) >> 1, α/2 ≈ (μ_{0}ε_{0}/2)^{½}ω(σ_{c}/(ε_{0}ω))^{½}= (μ_{0}σ_{c}ω/2)^{½},

δ = (2/(μ_{0}σ_{c}ω))^{½}= (2/(4π*10^{-7})/(38.2*10^{6})/(2π*10^{6}))^{½}m = 8.14*10^{-5}m.

β ≈ α/2 ≈ (μ_{0}σ_{c}ω/2)^{½}= 1/δ.

Wave velocity = ω/β = ω*δ = (2π*10^{6})(8.14*10^{-5}) m/s = 5.1*10^{2 }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) =**E**_{0 }exp(ikz)exp(-iωt).

**∇**×**E**= i**k**×**E**= -∂**B**/∂t.**B**(**r**,t) =**B**(**r**)_{ }exp(-iωt). Therefore i**k**×**E**= iω**B**. ik(**z**/z)×**E**= iω**B**.

Choose the coordinate system so that**E**= E(**x**/x), then**B**(**r**) = (**y**/y)(k/ω)ε_{0}exp(ikz).**B**(**r**) = (**y**/y)(|k|/ω)E_{0}e^{ikz}e^{iφ/2}= (**y**/y)B_{0}e^{ikz}e^{iφ/2},

where B_{0}= (|k|/ω)E_{0}. |k| = (μ_{0}ε_{0})^{½}ω(1 + σ_{c}^{2}/(ε_{0}ω)^{2})^{¼}≈ (μ_{0}σ_{c}ω)^{½}= 2^{½}/δ**.**

B_{0}= (**y**/y)B_{0}.

**E**(**r**,t) =**E**_{0 }exp(i(kz - ωt)) =**E**_{0 }e^{-αz/2}exp(i(βz - ωt)).(

B**r**,t) =**B**_{0 }exp(i(kz - ωt + φ/2)) =**B**_{0 }e^{-αz/2}exp(i(βz - ωt + φ/2)).

(d) For aluminum σ_{c}/(ε_{0}ω) >> 1. tanφ --> ∞. φ -- > 90^{o}. φ/2 --> 45^{o}.**B**_{0 }e^{-αz/2}exp(i(βz - (ωt - π/4))).

"B lags behind E" by φ/2 = π/4.

An electric field **E** = **i **E_{0}exp(-iωt)is applied at the
interface of a vacuum (z < 0) and a conductor (z > 0) of real average
conductivity σ_{c}.
Assume ε = ε_{0}, μ = μ_{0},
in the conductor.

(a) For σ_{c}
>> ε_{0}ω, 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∙E**dV,
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 ρ_{f }= 0 and**j**= σ_{f }_{c}**E**Maxwell's equations can be used to show that both**E**and**B**satisfy the damped wave equation. - Details of the calculation:

Assume**j**= σ_{f }_{c}**E**in the conductor. Then

**∇**×**B**= μ_{0}σ_{c}**E**+ μ_{0}ε_{0}∂**E**/∂t.

**∇**×(**∇**×**E**) =**∇**(**∇∙E**) -**∇**^{2}**E**= -∂(**∇**×**B**)/∂t. = -μ_{0}σ_{c}∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t.

Therefore**∇**^{2}**E**- μ_{0}σ_{c}∂**E**/∂t - μ_{0}ε_{0}∂^{2}**E**/∂t^{2}= 0.

The tangential component of**E**is continuous across the boundary, therefore**E**(z = 0) =**i**E_{0}exp(-iωt) just inside the conductor.

∂^{2}E_{x}/∂z^{2}+ iμ_{0}ωσ_{c}E_{x}+ μ_{0}ε_{0}ω^{2}E_{x}= 0 inside the conductor.

Since σ_{c}>> ε_{0}ω, we can write ∂^{2}E_{x}/∂z^{2}+ iμ_{0}ω∂E_{x}= 0.

Try a solutions of the form E_{x}(z,t) = E_{0 }exp(i(kz - ωt)).

Then k^{2}= iμ_{0}ωσ_{c}, k = (μ_{0}ωσ_{c})^{½}exp(iπ/4) = 2^{-½}(1 + i)(μ_{0}ωσ_{c})^{½}.

Im(k) = Re(k) = (μ_{0}ωσ_{c}/2)^{½}.

E_{x}(z,t) = E_{0 }exp(-Im(k)z) exp(i(Re(k)z - ωt)).

Skin depth: E_{0 }exp(-Im(k)d) = 1/e. d = (2/(μ_{0}ωσ_{c}))^{½}.

(b) dW/dt =**j∙E**dV = 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**= σ_{c}E_{0}^{2}_{ }exp(-2Im(k)z) cos^{2}(Re(k)z - ωt). <**j∙E**> = σ_{c}½E_{0}^{2}_{ }exp(-2Im(k)z).

dW/dt = σ_{c}E_{0}^{2}_{ }exp(-2Im(k)z) cos^{2}(Re(k)z - ωt)dV.

<dW/dt> = σ_{c}½E_{0}^{2}_{ }exp(-2Im(k)z)dV.