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

role="math" localid="1657446745988" ${\mathit{\omega }}_{\mathbf{l}\mathbf{m}\mathbf{n}}\mathbf{=}\mathit{c}\mathit{\pi }\sqrt{{\mathbf{\left(}\frac{\mathbf{l}}{\mathbf{d}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\frac{\mathbf{m}}{\mathbf{a}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\frac{\mathbf{n}}{\mathbf{b}}\mathbf{\right)}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{204}\mathbf{)}$

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

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

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

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

Here, the value of $k_x$, $k_y$and $k_z$are given as:

role="math" localid="1657449389970" $$\begin{gathered} k_x=\frac{m\pi }{a} \\ {k}_y=\frac{{\displaystyle n\pi }}{{\displaystyle b}} \\ {k}_z=\frac{{\displaystyle I\pi }}{{\displaystyle d}} \end{gathered}$$

Substitute$k_x=\frac{m\pi }{a},k_y=\frac{n\pi }{b}$and $k_z=\frac{l\pi }{d}$in equation (1).

localid="1657450941973" ${\omega }^2=c^2\left[{\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2+{\left(\frac{l\pi }{d}\right)}^2\right]$

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

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

$$\begin{gathered} E_x(x,y,z)=\left[A\sin \left(k_{x}x\right)+B\cos \left(k_{x}x\right)\right]\sin \left(k_{x}y\right)\sin \left(k_{z}z\right) \\ {E}_y(x,y,z)=\sin \left(k_{x}x\right)\left[C\sin \left(k_{y}y\right)+D\cos \left(k_{y}y\right)\right]\sin \left(k_{z}z\right) \\ {E}_z(x,y,z)=\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\left[E\sin \left(k_{z}z\right)+F\cos \left(k_{z}z\right)\right] \end{gathered}$$

Hence, the associated electric field will be,

$E_x=\left[\begin{array}{l}B\cos \left(k_{x}x\right)\sin \left(k_{z}z\right)\hat{x}+D\sin \left(k_{x}x\right)\cos \left(k_{y}y\right)\sin \left(k_{z}z\right)\hat{y}+\\ F\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\cos \left(k_{z}z\right)\hat{z}\end{array}\right]$

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

$B_x=-\frac{i}{\omega }\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\mid$

Substitute the value of and in the above expression.

$B_x=-\frac{i}{\omega }\left(F{k}_y\sin \left(k_{x}x\right)\cos \left(k_{y}y\right)\cos \left(k_{z}z\right)-D{k}_z\sin \left(k_{x}x\right)\cos \left(k_{y}y\right)\cos \left(k_{z}z\right)\right)$

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

$B_y=-\frac{i}{\omega }\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)$

Substitute the value of $E_x$and $E_z$in the above expression.

$B_y=-\frac{i}{\omega }\left(B{k}_z\cos \left(k_{x}x\right)\sin \left(k_{y}y\right)\cos \left(k_{z}z\right)-F{k}_x\cos \left(k_{x}x\right)\sin \left(k_{y}y\right)\cos \left(k_{z}z\right)\right)$

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

$B_z=-\frac{i}{\omega }\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)$

Substitute the value of $E_y$and $E_z$in the above expression.

$B_z=-\frac{i}{\omega }\left(D{k}_x\cos \left(k_{x}x\right)\cos \left(k_{y}y\right)\sin \left(k_{z}z\right)-B{k}_y\cos \left(k_{x}x\right)\cos \left(k_{y}y\right)\sin \left(k_{z}z\right)\right)$

Hence, the associated magnetic field will be,

$B=\left[\begin{array}{l}-\frac{i}{\omega }\left(F{k}_y-D{k}_z\right)\sin \left(k_{x}x\right)\cos \left(k_{y}y\right)\cos \left(k_{z}z\right)\hat{x}-\\ \frac{i}{\omega }\left(B{k}_z-F{k}_x\right)\cos \left(k_{x}x\right)\sin \left(k_{y}y\right)\cos \left(k_{z}z\right)\hat{y}-\\ \frac{i}{\omega }\left(D{k}_x-B{k}_y\right)\cos \left(k_{x}x\right)\cos \left(k_{y}y\right)\sin \left(k_{z}z\right)\hat{z}\end{array}\right]$

Therefore, the resonant frequencies for both TE and TM modes are

$\omega =c\pi \sqrt{\left[{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2+{\left(\frac{l}{d}\right)}^2\right]}$,the associated electric field is

$E_x=\left[\begin{array}{l}B\cos \left(k_{x}x\right)\sin \left(k_{z}z\right)\hat{x}+D\sin \left(k_{x}x\right)\cos \left(k_{y}y\right)\sin \left(k_{z}z\right)\hat{y}+\\ F\sin \left(k_{x}x\right)\sin \left(k_{y}y\right)\cos \left(k_{z}z\right)\hat{z}\end{array}\right]$,and the associated magnetic field is$B=\left[\begin{array}{l}-\frac{i}{\omega }\left(F{k}_y-D{k}_z\right)\sin \left(k_{x}x\right)\cos \left(k_{y}y\right)\cos \left(k_{z}z\right)\hat{x}-\\ \frac{i}{\omega }\left(B{k}_z-F{k}_x\right)\cos \left(k_{x}x\right)\sin \left(k_{y}y\right)\cos \left(k_{z}z\right)\hat{y}-\\ \frac{i}{\omega }\left(D{k}_x-B{k}_y\right)\cos \left(k_{x}x\right)\cos \left(k_{y}y\right)\sin \left(k_{z}z\right)\hat{z}\end{array}\right]$