Exercise 5.5.8: Potentials of a guided wave between two plates

About the further exercises of Section 5.5

Ten further exercises in the spirit of Problems 5.5.1–5.5.6, the home exercises and the worked examples, graded from easy to harder. Exercises 5.5.1–5.5.2 are about travelling pulses, Exercises 5.5.3–5.5.5 about the wave equation itself, Exercises 5.5.6–5.5.8 about potentials and gauges, and Exercises 5.5.9–5.5.10 about retarded potentials. They use only the equations listed at the head of this subsection. All media are nonmagnetic.

Exercise 5.5.8

Potentials of a guided wave between two plates

In free space between perfectly conducting plates at $x = 0$ and $x = a$ , $a = 2 cm$ , a vector potential is

A = A_{0} sin (\frac{π x}{a}) cos (ω t - β z) a_{z}

at $f = 12.5 GHz$ . (a) Find $β$ such that $A$ satisfies the homogeneous wave equation. (b) Find $V$ from the Lorenz gauge (no static part). (c) Find $E$ and $B$ , and the amplitudes of $E_{x}$ , $E_{z}$ and $c B_{y}$ in units of $ω A_{0}$ . (d) Show that the tangential $E$ vanishes on the plates. Why is this field not zero, unlike the longitudinal potential of Problem 5.5.5?

\vec E at t=0 over one wavelength 2\pi/\beta. — $E$ at $t = 0$ over one wavelength $2 π / β$ .

Try it yourself first.

Strategy

Problem 5.5.3 for the potential instead of the field, then Problem 5.5.5 with a potential that varies across the guide. The Lorenz gauge gives $V$ , and Eq. (5.5.37) gives the fields.

$A_{z}$ is a Cartesian component. Its second derivatives give $- (π / a)^{2} A_{z}$ , $- β^{2} A_{z}$ and $- ω^{2} A_{z}$ , so the wave equation requires

β^{2} \frac{ω}{c} \frac{π}{a} = \frac{ω ^{2}}{c ^{2}} - (\frac{π}{a})^{2}, = \frac{2 π \times 1.25 \times 1 0 ^{10}}{3 \times 1 0 ^{8}} = \frac{250 π}{3}, = 50 π = \frac{150 π}{3},

β = \frac{200 π}{3} = 209.4 rad/m .

The three numbers form a $3$ : $4$ : $5$ triangle: $π / a = \frac{3}{5} ω / c$ and $β = \frac{4}{5} ω / c$ .

$\nabla \cdot A = \partial A_{z} / \partial z = β A_{0} sin (π x / a) sin (ω t - β z)$ , and $\partial V / \partial t = - c^{2} \nabla \cdot A$ gives

V \frac{c ^{2} β}{ω} = \frac{c ^{2} β}{ω} A_{0} sin (\frac{π x}{a}) cos (ω t - β z), = \frac{4}{5} c = 2.4 \times 1 0^{8} m/s .

$E = - \nabla V - \partial A / \partial t$ and $B = \nabla \times A = - a_{y} \partial A_{z} / \partial x$ :

E_{x} E_{z} B_{y} = - \frac{\partial V}{\partial x} = - \frac{c ^{2} β}{ω} \frac{π}{a} A_{0} cos (\frac{π x}{a}) cos (ω t - β z), = - \frac{\partial V}{\partial z} - \frac{\partial A _{z}}{\partial t} = (ω - \frac{c ^{2} β ^{2}}{ω}) A_{0} sin (\frac{π x}{a}) sin (ω t - β z) = \frac{c ^{2}}{ω} (\frac{π}{a})^{2} A_{0} sin (\frac{π x}{a}) sin (ω t - β z), = - \frac{π}{a} A_{0} cos (\frac{π x}{a}) cos (ω t - β z) .

In units of $ω A_{0}$ , with $(c / ω) (π / a) = \frac{3}{5}$ and $c β / ω = \frac{4}{5}$ , the amplitudes are $\frac{4}{5} \cdot \frac{3}{5}$ , $(\frac{3}{5})^{2}$ and $\frac{3}{5}$ :

Answer

∣ E_{x} ∣_{m a x} = 0.48 ω A_{0}, ∣ E_{z} ∣_{m a x} = 0.36 ω A_{0}, c ∣ B_{y} ∣_{m a x} = 0.60 ω A_{0}

Check

$\partial E_{x} / \partial x$ and $\partial E_{z} / \partial z$ are both $\pm \frac{c ^{2} β}{ω} (\frac{π}{a})^{2} A_{0} sin (π x / a) cos (ω t - β z)$ with opposite signs, so $\nabla \cdot E = 0$ . The $x$ component of $\nabla \times B = c^{- 2} \partial E / \partial t$ holds: $- \partial B_{y} / \partial z$ and $c^{- 2} \partial E_{x} / \partial t$ both equal $\frac{π β}{a} A_{0} cos (π x / a) sin (ω t - β z)$ .

On the plates the tangential components are $E_{y} = 0$ and $E_{z} \propto sin (π x / a) = 0$ at $x = 0$ and $x = a$ . In Problem 5.5.5 the two contributions to $E_{z}$ , $- \partial V / \partial z$ and $- \partial A_{z} / \partial t$ , cancel because $β = ω / c$ . Here the variation across the guide makes $β < ω / c$ , the cancellation is incomplete, and what is left is $(c^{2} / ω) (π / a)^{2}$ : a real field with an $E_{z}$ along the direction of travel.

Remark

This is the TM $_{1}$ mode of the parallel-plate waveguide (Section 7.2), here obtained from a single potential component. Its cutoff is at $β = 0$ , $f_{c} = c / (2 a) = 7.5 GHz$ , and the ratio $E_{x} / B_{y} = c^{2} β / ω = 2.4 \times 1 0^{8} m/s$ is $c^{2}$ divided by the phase velocity $ω / β = 3.75 \times 1 0^{8} m/s$ .

cf. Problem 5.5.5