Perfect conductor boundaries, Waveguides

Problem:

(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 EB.  Ek.  Bk.  B = (μ0ε0)½(k/k)×E.
• Details of the calculation:
(a)  Maxwell's equations:
∇∙E = ρ/ε0×E = -∂B/∂t,  ∇∙B = 0,  ×B = μ0j + (1/c2)∂E/∂t.
Assume ρf and jf are zero in the medium.  Then
×(×E) = (∇∙E) - 2E = -(∂/∂t)(×B) = -μ0ε02E/∂t2.
2E = μ0ε02E/∂t2.
×(×B) = (∇∙B) - 2B = μ0ε0(∂/∂t)(×E) = -μ0ε02B/∂t2.
2B = μ0ε02B/∂t2.
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 = ∂Ex/∂x + ∂Ey/∂y +∂Ez/∂z
= (∂Ex/∂u)(kx/k) + (∂Ey/∂u)(ky/k) +(∂Ez/∂u)(kz/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 = cB.  i.e.  B is perpendicular to E and k.
[(×E)x = ∂Ez/∂y - ∂Ey/∂z = (∂Eu/∂u)(ky/k) - (∂Ey/∂u)(kz/k)
= (∂/∂u)((k/k)×E)x = -∂Bx/∂t = (∂Bx/∂u)c.
If we are not interested in static fields then (k/k)×E)x = cBx.
Repeating this derivation for the y- and z- components we have ((k/k)×E = cB.]

(b)  The tangential component of E is continuous at z = 0.  This implies Eicos(θ) = Ercos(θ) at z = 0, Ei = Er.
The normal component of E then is Eisin(θ) + Ersin(θ) = 2Eisin(θ).
The surface charge density is σ = ε02Eisin(θ).  [(E2 - E1)∙n2 = σ/ε0 ]
The tangential component of B at the z < 0 side of the surface is Br + Bi = 2Bi.
Therefore 2Bi0 = k.  (Here k = surface current density.)  [(B2 - B1)∙t = μ0k∙n]

Problem:

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 Ey = E0sinkxx.  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 TE0n waves of the form E = E0sinkxx exp(i(kzz-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 < ωc, where  ωc 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 = E0sinkxx exp(i(kzz-ωt)) j.  kxa = nπ.  The wave equation requires
2Ey = (1/c2)∂2Ey/∂t2 or kx2Ey + kz2Ey - (ω2/c2)Ey = 0.
kx2 + kz2 = (ω2/c2).  Since we know kx we can solve for kz.
kz = ((ω2/c2) - (nπ/a)2)½.
Check if B satisfies the boundary conditions B∙n = 0 at the walls:
×E = -(∂Ey/∂x)(z/z) + (∂Ey/∂z)(x/x) = iωB.
B = [(kz/ω)E0sinkxx(x/x) + (ikx/ω))E0coskxx(z/z)] exp(i(kzz-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 TE0n.
kz becomes imaginary and the wave does not propagate if
2/c2) < (nπ/a)2 or ω < nπc/a.
λ = 2πc/ω.  If λ > λc = 2a/n then the TE0n mode cannot propagate.
If a > b then λc = 2a is the cutoff wavelength for all TE modes.

Problem:

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*1010 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
= B0xsinkxx exp(i(kzz-ωt)) i + B0ysinkyy exp(i(kzz-ωt)) j,
with  kxa = nπ,  kyb = mπ, with m and n nonzero, positive integers.
These waves is called the TMnm modes of the waveguide.
TM10 or TM01 guided waves do not exist.
• Details of the calculation:
(a)  Bx = B0xsinkxx exp(i(kzz-ωt)),  By = B0ysinkyy exp(i(kzz-ωt).
B∙n = 0 on the walls requires
kxa = nπ,  kyb = mπ, with m and n nonzero, positive integers.
We need ∇∙B = 0, ∂Bx/∂x + ∂By/∂y = 0.
This yields kxB0xcoskxx + kyB0ycoskyy = 0,
B0x = -B0y(kysinkyy)/(kxcoskxx).
If we choose B0y = kxB0coskxx and B0x = -kyB0coskyy
then the equation is satisfied.
We have B = (-kycoskyy sinkxx i + kxcoskxx sinkyy j)B0exp(i(kzz-ωt))
The wave equation requires  2B - (1/c2)∂2B/∂t2 = 0.
kx2B + ky2B + kz2B - (ω2/c2)B = 0.
kx2 + ky2 + kz2 = (ω2/c2).  Since we know kx and ky we can solve for kz.
kz = ((ω2/c2) - (nπ/a)2 - (mπ/b)2)½.
The cutoff frequency for the TMnm mode is ωc = c((nπ/a)2 + (mπ/b)2)½.
The cutoff frequency for the TM11 mode is ωc = c((π/a)2 + (π/b)2)½ = 3.1*10-10 s-1, so TM radiation of angular frequency ω = 6.1*1010 s-1 will propagate in the waveguide.
• (b)  The dispersion relation gives ω as a function of kz.
ω = c(kz2 + (nπ/a)2 +(mπ/b)2)½.
(c)  kz = (1/c)(ω2 - ωc2)½.  If ω = ωc/2 then kz = (i/c)(√3)ωc/2.
We have:  B proportional to exp(-kzz) = exp(-(√3)ωcz/(2c)).
Therefore:  Power P proportional to B2 proportional to exp(-(√3)ωcz/c).
We want:  P(z)/P(0) = 1/e,  z = c/(√3)ωc = 3*108 m/(3.1*10-10*√3) = 5.6*10-3 m.

Problem:

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) = E0(x,z)eiωt 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) = E0(x,z)eiωt (y/y),  B proportional to eiωt.
The field has to satisfy Maxwell's equations.  Does it?
∇∙E = 0, ∂Ey/∂y = 0.  This equation is satisfied.
×E = -(∂Ey/∂z)(x/x) + (∂Ey/∂x)(z/z) = -∂B/∂t = -iωB.
This implies that B = (1/(iω))(∂Ey/∂z)(x/x) - (1/(iω))(∂Ey/∂x)(z/z).
∇∙B = (1/(iω))[∂2Ey/∂x∂z - ∂2Ey/∂z∂x] = 0.  This equation is satisfied.
B = Bx(x,z)(x/x) + Bz(x,z)(z/z).
×B = [∂Bx/∂z) - (∂Bz/∂x)](y/y) = (1/c2)∂E/∂t = (iω/c2)E0(x,z)eiωt (y/y).
The given field satisfies Maxwell's equations.
If we choose E0(x,z) = Asin(kxx)sin(kzz) with kxa = nπ, kzl = mπ, then E∙t = 0 is satisfied on all walls.
With this choice Bx = (1/(iω))kzAsin(kxx)sin(kzz), Bx = 0 at x = 0 and x = a.
Bz = (-1/(iω))kxAsin(kxx)sin(kzz), Bz = 0 at z = 0 and z = l.
B∙n = 0 is satisfied on all walls.
[∂Bx/∂z) - (∂Bz/∂x)] = (iω/c2)E0(x,z)eiωt requires -(1/(iω))kz2 - (1/(iω))kx2 = (iω/c2).
-kz2 - kx2 + ω2/c2 = 0.  ω2/c2 = (mπ/l)2 + (nπ/a)2.
The lowest resonance frequency in the cavity is ω = c*[(π/l)2 + (π/a)2]½.