Question

Consider the resonant cavity produced by closing off the two ends of a rectangular wave guide, at $\mathbf{z}\mathbf{=}\mathbf{0}$ and at $\mathbf{z}\mathbf{=}\mathbf{d}$, making a perfectly conducting empty box. Show that the resonant frequencies for both TE and TM modes are given by

${\mathbf{\omega }}_{\mathbf{lmn}}\mathbf{=}\mathbf{c\pi }\sqrt{{\left(\frac{l}{d}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{m}{a}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{n}{b}\right)}^{\mathbf{2}}}$ (9.204)

For integers l, m, and n. Find the associated electric and magneticfields.

Short answer

The resonant frequencies for both TE and TM modes are ${\mathrm{\omega }}_{\mathrm{lmn}}=\mathrm{c\pi }\sqrt{{\left(\frac{\mathrm{l}}{\mathrm{d}}\right)}^2+{\left(\frac{\mathrm{m}}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{n}}{\mathrm{b}}\right)}^2}$, the associated electric field is ${\mathrm{E}}_{\mathrm{x}}=\left[\begin{array}{c}\mathrm{Bcos}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{x}}+\mathrm{Dsin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{y}}+\\ \mathrm{Fsin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{z}}\end{array}\right]$, and the associated magnetic field is ${\mathrm{B}}_{\mathrm{x}}=\left[\begin{array}{c}\frac{-\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Fk}}_{\mathrm{y}}-{\mathrm{Dk}}_{\mathrm{z}}\right)\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{x}}-\\ \frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Bk}}_{\mathrm{z}}-{\mathrm{Fk}}_{\mathrm{x}}\right)\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{y}}-\\ \frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Dk}}_{\mathrm{x}}-{\mathrm{Bk}}_{\mathrm{y}}\right)\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{z}}\end{array}\right]$.

Step-by-step solution

Expression for the resonant frequencies for both TE and TM mode:

Write the expression for the resonant frequencies for both TE and TM mode.

${\mathbf{\omega }}^{\mathbf{2}}\mathbf{=}{\mathbf{c}}^{\mathbf{2}}\left({k_x}^2+{k_y}^2+{k_z}^2\right)$ …… (1)

Here, c is the speed of light and k is the wave number.

Here, the value of data-custom-editor="chemistry" ${\mathrm{k}}_{\mathrm{x}}$,data-custom-editor="chemistry" ${\mathrm{k}}_{\mathrm{y}}$ anddata-custom-editor="chemistry" ${\mathrm{k}}_{\mathrm{z}}$ are given as:

$$\begin{gathered} {\mathrm{k}}_{\mathrm{x}}=\frac{\mathrm{m\pi }}{\mathrm{a}} \\ {\mathrm{k}}_{\mathrm{y}}=\frac{\mathrm{n\pi }}{\mathrm{b}} \\ {\mathrm{k}}_{\mathrm{z}}=\frac{\mathrm{l\pi }}{\mathrm{d}} \end{gathered}$$

Prove the expression for resonant frequencies for both TE and TM mode:

Substitute ${\mathrm{k}}_{\mathrm{x}}=\frac{\mathrm{m\pi }}{\mathrm{a}}$,${\mathrm{k}}_{\mathrm{y}}=\frac{\mathrm{n\pi }}{\mathrm{b}}$ and${\mathrm{k}}_{\mathrm{z}}=\frac{\mathrm{l\pi }}{\mathrm{d}}$ in equation (1).

$$\begin{gathered} {\mathrm{\omega }}^2={\mathrm{c}}^2\left[{\left(\frac{\mathrm{m\pi }}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{n\pi }}{\mathrm{b}}\right)}^2+{\left(\frac{\mathrm{l\pi }}{\mathrm{d}}\right)}^2\right] \\ \mathrm{\omega }=\mathrm{c\pi }\sqrt{\left[{\left(\frac{\mathrm{m}}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{n}}{\mathrm{b}}\right)}^2+{\left(\frac{\mathrm{l}}{\mathrm{d}}\right)}^2\right]} \end{gathered}$$

Determine the associated electric field:

Write the expression for the x, y, and z components of an electric field.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left[\left(\mathrm{Asin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)+\mathrm{Bcos}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\right)\right]\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right) \\ {\mathrm{E}}_{\mathrm{y}}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\left[\left(\mathrm{Csin}\left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)+\mathrm{Dcos}\left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\right)\right]\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right) \\ {\mathrm{E}}_{\mathrm{z}}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\left[\left(\mathrm{Esin}\left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)+\mathrm{Fcos}\left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\right)\right] \end{gathered}$$.

Hence, the associated electric field will be,

role="math" localid="1657688765767" ${\mathrm{E}}_{\mathrm{x}}=\left[\begin{array}{c}\mathrm{Bcos}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{x}}+\mathrm{Dsin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{y}}+\\ \mathrm{Fsin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{z}}\end{array}\right]$

Determine the associated magnetic field:

Write the expression for the x component of a magnetic field.

${\mathrm{B}}_{\mathrm{x}}=-\frac{\mathrm{i}}{\mathrm{\omega }}\left(\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{y}}-\frac{\partial {\mathrm{E}}_{\mathrm{y}}}{\partial \mathrm{z}}\right)$

Substitute the value of${\mathrm{E}}_{\mathrm{z}}$ and${\mathrm{E}}_{\mathrm{y}}$ in the above expression.

${\mathrm{B}}_{\mathrm{x}}=-\frac{\mathrm{i}}{\mathrm{\omega }}\left[{\mathrm{Fk}}_{\mathrm{y}}\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)-{\mathrm{Dk}}_{\mathrm{z}}\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\right]$

Write the expression for the y component of a magnetic field.

${\mathrm{B}}_{\mathrm{y}}=-\frac{\mathrm{i}}{\mathrm{\omega }}\left(\frac{\partial {\mathrm{E}}_{\mathrm{x}}}{\partial \mathrm{z}}-\frac{\partial {\mathrm{E}}_{\mathrm{z}}}{\partial \mathrm{x}}\right)$

Substitute the value ofdata-custom-editor="chemistry" ${\mathrm{E}}_{\mathrm{x}}$ anddata-custom-editor="chemistry" ${\mathrm{E}}_{\mathrm{z}}$ in the above expression.

data-custom-editor="chemistry" ${\mathrm{B}}_{\mathrm{y}}=-\frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Bk}}_{\mathrm{z}}\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)-{\mathrm{Fk}}_{\mathrm{x}}\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\right)$

Write the expression for the z component of a magnetic field.

data-custom-editor="chemistry" ${\mathrm{B}}_{\mathrm{z}}=-\frac{\mathrm{i}}{\mathrm{\omega }}\left(\frac{\partial {\mathrm{E}}_{}}{\partial \mathrm{x}}-\frac{\partial {\mathrm{E}}_{\mathrm{x}}}{\partial \mathrm{y}}\right)$

Substitute the value ofdata-custom-editor="chemistry" ${\mathrm{E}}_{\mathrm{y}}$ anddata-custom-editor="chemistry" ${\mathrm{E}}_{\mathrm{x}}$ in the above expression.

data-custom-editor="chemistry" ${\mathrm{B}}_{\mathrm{z}}=-\frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Dk}}_{\mathrm{x}}\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)-{\mathrm{Bk}}_{\mathrm{y}}\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\right)$

Hence, the associated magnetic field will be,

data-custom-editor="chemistry" $\mathrm{B}=\left[\begin{array}{c}-\frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Fk}}_{\mathrm{y}}-{\mathrm{Dk}}_{\mathrm{z}}\right)\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{x}}-\\ \frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Bk}}_{\mathrm{z}}-{\mathrm{Fk}}_{\mathrm{x}}\right)\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{y}}-\\ \frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Dk}}_{\mathrm{x}}-{\mathrm{Bk}}_{\mathrm{y}}\right)\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{z}}\end{array}\right]$

Therefore, the resonant frequencies for both TE and TM modes are $\omega =c\mathrm{\pi }\sqrt{\left[{\left(\frac{\mathrm{m}}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{n}}{\mathrm{b}}\right)}^2+{\left(\frac{\mathrm{l}}{\mathrm{d}}\right)}^2\right]}$, the associated electric field is ${\mathrm{E}}_{\mathrm{x}}=\left[\begin{array}{c}\mathrm{Bcos}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{x}}+\mathrm{Dsin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{y}}+\\ \mathrm{Fsin}\left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{z}}\end{array}\right]$, and the associated magnetic field is data-custom-editor="chemistry" $\mathrm{B}=\left[\begin{array}{c}-\frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Fk}}_{\mathrm{y}}-{\mathrm{Dk}}_{\mathrm{z}}\right)\sin \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{x}}-\\ \frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Bk}}_{\mathrm{z}}-{\mathrm{Fk}}_{\mathrm{x}}\right)\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\sin \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\cos \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{y}}-\\ \frac{\mathrm{i}}{\mathrm{\omega }}\left({\mathrm{Dk}}_{\mathrm{x}}-{\mathrm{Bk}}_{\mathrm{y}}\right)\cos \left({\mathrm{k}}_{\mathrm{x}}\mathrm{x}\right)\cos \left({\mathrm{k}}_{\mathrm{y}}\mathrm{y}\right)\sin \left({\mathrm{k}}_{\mathrm{z}}\mathrm{z}\right)\hat{\mathrm{z}}\end{array}\right]$.