(a) Beginning with the free Maxwell equations in unbounded space, show that
the propagation of an electromagnetic wave is transverse. That is, show
that the electric and magnetic fields oscillate in a plane perpendicular to the
direction of propagation and perpendicular to each other.

(b) Now, consider the reflection of a plane electromagnetic wave from a flat
perfect conductor. Note that a perfect conductor develops a charge and
current distribution on the surface so that external fields do not penetrate it.
Find the charge and current densities on the conductor for the case of
polarization in the plane of incidence.

Solution:

- Concepts:

Maxwell's equations, boundary conditions - Reasoning:

Maxwell's equations in free space yield the wave equation for both**E**and**B**. They can also be used to show that**E**⊥**B**.**E**⊥**k**.**B**⊥**k**.**B**= (μ_{0}ε_{0})^{½}(**k**/k)×**E**. - 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 ρ

_{f}and**j**_{f}are zero in the medium. Then

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

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

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

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

Each Cartesian component of**E**and**B**satisfies the 3-dimensional, homogeneous wave equation.

Plane wave solutions**E**=**E**((**k**/k)∙**r**- ct).**B**=**B**((**k**/k)∙**r**- ct) exist. (**k**/k denotes the direction of propagation.)

Let u = (**k**/k)**∙r**- ct.

**∇∙E**= ∂E_{x}/∂x + ∂E_{y}/∂y +∂E_{z}/∂z

= (∂E_{x}/∂u)(k_{x}/k) + (∂E_{y}/∂u)(k_{y}/k) +(∂E_{z}/∂u)(k_{z}/k) = (∂/∂u)(**E∙**(**k**/k)) = 0.

**E∙k**= constant = 0 if we are not interested in static fields.

**∇∙E**= 0 requires that**E∙k**= 0 for radiation fields, i.e. that**E**is perpendicular to**k**.

Similarly,**∇∙B**= 0 requires that**B∙k**= 0 for radiation fields.

**∇**×**E**= -∂**B**/∂t requires that ((**k**/k)×**E**= c**B**. i.e.**B**is perpendicular to**E**and**k**.

[(**∇**×**E**)_{x}= ∂E_{z}/∂y - ∂E_{y}/∂z = (∂E_{u}/∂u)(k_{y}/k) - (∂E_{y}/∂u)(k_{z}/k)

= (∂/∂u)((**k**/k)×**E**)_{x}= -∂B_{x}/∂t = (∂B_{x}/∂u)c.

If we are not interested in static fields then ((**k**/k)×**E**)_{x}= cB_{x}.

Repeating this derivation for the y- and z- components we have ((**k**/k)×**E**= c**B**.]

(b) The tangential component of**E**is continuous at z = 0. This implies E_{i}cos(θ) = E_{r}cos(θ) at z = 0, E_{i}= E_{r}.

The normal component of**E**then is E_{i}sin(θ) + E_{r}sin(θ) = 2E_{i}sin(θ).

The surface charge density is σ = ε_{0}2E_{i}sin(θ). [(**E**_{2}-**E**_{1})**∙n**_{2}= σ/ε_{0}]

The tangential component of B at the z < 0 side of the surface is B_{r}+ B_{i}= 2B_{i}.

Therefore 2B_{i}/μ_{0}= k. (Here**k**= surface current density.) [(**B**_{2}-**B**_{1})**∙t**= μ_{0}**k∙n**]

Consider a rectangular conducting
waveguide of width a (along the x axis) and height b (along the y axis).
An EM wave of wavelength λ travels into the guide
along the z axis; the amplitude of the wave has the form E_{y }= E_{0}sink_{x}x.
Derive the form of the guided wavelength and discuss the meaning of the critical
(cutoff) wavelength.

Solution:

- Concepts:

Waveguides, boundary conditions - Reasoning:

In a rectangular channel with its axis in the z-direction TE_{0n}waves of the form**E**= E_{0}sink_{x}x exp(i(k_{z}z-ωt))**j**can propagate. They have to satisfy the wave equation and the boundary conditions**E∙t**= 0 and**B∙n**= 0 on the walls of the channel. These conditions can only be satisfied if ω < ω_{c}, where ω_{c}is called the cutoff frequency. - Details of the calculation:

A wave traveling into the z-direction that satisfies the boundary conditions**E∙t**= 0 on the walls if we have**E**= E_{0}sink_{x}x exp(i(k_{z}z-ωt))**j**. k_{x}a = nπ. The wave equation requires

∇^{2}E_{y}= (1/c^{2})∂^{2}E_{y}/∂t^{2}or k_{x}^{2}E_{y}+ k_{z}^{2}E_{y}- (ω^{2}/c^{2})E_{y}= 0.

k_{x}^{2}+ k_{z}^{2}= (ω^{2}/c^{2}). Since we know k_{x}we can solve for k_{z}.

k_{z}= ((ω^{2}/c^{2}) - (nπ/a)^{2})^{½}.

Check if**B**satisfies the boundary conditions**B∙n**= 0 at the walls:

**∇**×**E**= -(∂E_{y}/∂x)(**z**/z) + (∂E_{y}/∂z)(**x**/x) = iω**B**.

**B**= [(k_{z}/ω)E_{0}sink_{x}x(**x**/x) + (ik_{x}/ω))E_{0}cosk_{x}x(**z**/z)] exp(i(k_{z}z-ωt)).

**B∙n**= 0 on the walls.

The wave satisfies the wave equation and the boundary conditions.

The electric field is transverse to the channel, we have a TE wave. The mode is TE_{0n}.

k_{z}becomes imaginary and the wave does not propagate if

(ω^{2}/c^{2}) < (nπ/a)^{2}or ω < nπc/a.

λ = 2πc/ω. If λ > λ_{c}= 2a/n then the TE_{0n}mode cannot propagate.

If a > b then λ_{c}= 2a is the cutoff wavelength for all TE modes.

A rectangular waveguide of sides a = 7.21cm and b = 3.40cm is used in the
transverse magnetic (TM) mode. Assume that the walls are perfect
conductors.

(a) By calculating the lowest cut off frequency, determine whether TM radiation
of angular frequency ω = 6.1*10^{10 }s^{-1}
will propagate in the waveguide.

(b) What is the dispersion relation for this guide?

(c) Find the attenuation length for a frequency ω
that is half the cut off frequency.

Solution:

- Concepts:

Waveguides, boundary conditions - Reasoning:

A TM wave is a superposition of waves of the form= B

B_{0x}sink_{x}x exp(i(k_{z}z-ωt))**i**+ B_{0y}sink_{y}y exp(i(k_{z}z-ωt))**j**,

with k_{x}a = nπ, k_{y}b = mπ, with m and n nonzero, positive integers.

These waves is called the TM_{nm}modes of the waveguide.

TM_{10}or TM_{01}guided waves do not exist. - Details of the calculation:

(a) B_{x}= B_{0x}sink_{x}x exp(i(k_{z}z-ωt)), B_{y}= B_{0y}sink_{y}y exp(i(k_{z}z-ωt).

**B∙n**= 0 on the walls requires

k_{x}a = nπ, k_{y}b = mπ, with m and n nonzero, positive integers.

We need**∇∙B**= 0, ∂**B**/∂x + ∂_{x}**B**/∂y = 0._{y}

This yields k_{x}B_{0x}cosk_{x}x + k_{y}B_{0y}cosk_{y}y = 0,

B_{0x}= -B_{0y}(k_{y}sink_{y}y)/(k_{x}cosk_{x}x).

If we choose B_{0y}= k_{x}B_{0}cosk_{x}x and B_{0x}= -k_{y}B_{0}cosk_{y}y

then the equation is satisfied.

We have**B**= (-k_{y}cosk_{y}y sink_{x}x**i**+ k_{x}cosk_{x}x_{ }sink_{y}y**j**)B_{0}exp(i(k_{z}z-ωt))

The wave equation requires**∇**^{2}**B**- (1/c^{2})∂^{2}**B**/∂t^{2}= 0.

k_{x}^{2}**B**+ k_{y}^{2}**B**+ k_{z}^{2}**B**_{ }- (ω^{2}/c^{2})**B**= 0.

k_{x}^{2}+ k_{y}^{2}+ k_{z}^{2}= (ω^{2}/c^{2}). Since we know k_{x}and k_{y}we can solve for k_{z}.

k_{z}= ((ω^{2}/c^{2}) - (nπ/a)^{2}- (mπ/b)^{2})^{½}.

The cutoff frequency for the TM_{nm}mode is ω_{c}= c((nπ/a)^{2}+ (mπ/b)^{2})^{½}.

The cutoff frequency for the TM_{11}mode is ω_{c}= c((π/a)^{2}+ (π/b)^{2})^{½}= 3.1*10^{-10 }s^{-1}, so TM radiation of angular frequency ω = 6.1*10^{10 }s^{-1}will propagate in the waveguide. - (b) The dispersion relation gives ω as
a function of k
_{z}.

ω = c(k_{z}^{2}+ (nπ/a)^{2}+(mπ/b)^{2})^{½}.

(c) k_{z}= (1/c)(ω^{2}- ω_{c}^{2})^{½}. If ω = ω_{c}/2 then k_{z}= (i/c)(√3)ω_{c}/2.

We have: B proportional to exp(-k_{z}z) = exp(-(√3)ω_{c}z/(2c)).

Therefore: Power P proportional to B^{2}proportional to exp(-(√3)ω_{c}z/c).

We want: P(z)/P(0) = 1/e, z = c/(√3)ω_{c}= 3*10^{8}m/(3.1*10^{-10}*√3) = 5.6*10^{-3}m.

A rectangular waveguide made of perfectly conducting material has sides of length a and b as shown in the figure below.

The ends of a section of length l are covered with plates of conducting material, i.e. the waveguide is effectively a resonant cavity. If the electric field is given by the real part of

**E**(x,y,z,t) = E_{0}(x,z)e^{iωt}
**j**,

what is ω for the cavity mode with the lowest resonant frequency?

Solution:

- Concepts:

Maxwell's equations, waveguides, resonant cavities - Reasoning:

The electric field in the cavity must satisfy Maxwell's equations and the boundary conditions**E∙t**= 0 and**B∙n**= 0 on the walls of the cavity. - Details of the calculation:

Given:**E**(x,y,z,t) = E_{0}(x,z)e^{iωt}(**y**/y),**B**proportional to e^{iωt}.

The field has to satisfy Maxwell's equations. Does it?

**∇∙E**= 0, ∂E_{y}/∂y = 0. This equation is satisfied.

**∇**×**E**= -(∂E_{y}/∂z)(**x**/x) + (∂E_{y}/∂x)(**z**/z) = -∂**B**/∂t = -iω**B**.

This implies that**B**= (1/(iω))(∂E_{y}/∂z)(**x**/x) - (1/(iω))(∂E_{y}/∂x)(**z**/z).

**∇∙B**= (1/(iω))[∂^{2}E_{y}/∂x∂z - ∂^{2}E_{y}/∂z∂x] = 0. This equation is satisfied.

**B**= B_{x}(x,z)(**x**/x) + B_{z}(x,z)(**z**/z).

**∇**×**B**= [∂B_{x}/∂z) - (∂B_{z}/∂x)](**y**/y) = (1/c^{2})∂**E**/∂t = (iω/c^{2})E_{0}(x,z)e^{iωt}(**y**/y).

The given field satisfies Maxwell's equations.

If we choose E_{0}(x,z) = Asin(k_{x}x)sin(k_{z}z) with k_{x}a = nπ, k_{z}l = mπ, then**E∙t**= 0 is satisfied on all walls.

With this choice B_{x}= (1/(iω))k_{z}Asin(k_{x}x)sin(k_{z}z), B_{x}= 0 at x = 0 and x = a.

B_{z}= (-1/(iω))k_{x}Asin(k_{x}x)sin(k_{z}z), B_{z}= 0 at z = 0 and z = l.

**B∙n**= 0 is satisfied on all walls.

[∂B_{x}/∂z) - (∂B_{z}/∂x)] = (iω/c^{2})E_{0}(x,z)e^{iωt}requires -(1/(iω))k_{z}^{2}- (1/(iω))k_{x}^{2}= (iω/c^{2}).

-k_{z}^{2}- k_{x}^{2}+ ω^{2}/c^{2}= 0. ω^{2}/c^{2}= (mπ/l)^{2}+ (nπ/a)^{2}.

The lowest resonance frequency in the cavity is ω = c*[(π/l)^{2}+ (π/a)^{2}]^{½}.