Work out the theory of TM modes for a rectangular wave guide. In particular, find the longitudinal electric field, the cutoff frequencies, and the wave and group velocities. Find the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency, for a given wave guide. [Caution: What is the lowest TM mode?]

Write the expression for the components of electric and magnetic fields along the z-axis in a rectangular wave.

$\begin{array}{c}[\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+{\left(\frac{\omega }{c}\right)}^2-k^2]E_z=0\\ [\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+{\left(\frac{\omega }{c}\right)}^2-k^2]B_z=0\end{array}$

Here, $E_z$is the longitudinal component of electric field and$B_z$ is the longitudinal component of a magnetic field,$\omega$ is the frequency of a wave, c is the speed of light, and k is the wavenumber.

For the TM wave, the value of the longitudinal component of the magnetic field is zero.

Write the boundary conditions at the wall.

$\begin{array}{c}{E}^{\parallel }=0\\ B^{\perp }=0\end{array}$

Let, $E_z(x,y)=X\left(x\right)Y\left(y\right)$.

Here, .$X\left(x\right)=A\sin \left(k_{x}x\right)+B\cos \left(K_{x}x\right)$

At walls $E_z=0$, then at$x=0$and $x=a$, the value of X and B will be,

$\begin{array}{c}X=0\\ B=0\end{array}$

Hence, it is known that:

$\begin{array}{c}{k}_x=\frac{m\pi }{a};m=1,2,\mathrm{3....}\\ k_y=\frac{n\pi }{a};n=1,2,\mathrm{3....}\end{array}$

So, the longitudinal electric field will be,

$E_z=E_0\sin \left(\frac{m\pi x}{a}\right)\sin \left(\frac{n\pi y}{a}\right)$

${\omega }_{mn}=c\pi \sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$Write the expression for wave number.

Hence, the cut off frequency will be,

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

Write the expression for the wave velocity,

$v=\frac{\omega }{k}$ …… (1)

Write the expression for the lowest cut-off frequency $\left({\omega }_{11}\right)$for mode${\text{TM}}_{11}$ .

${\omega }_{11}=c\pi \sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}$ …… (2)

Hence, the wavenumber in terms of the cut-off frequency will be,

$k=\frac{1}{c}\sqrt{{\omega }^2-{\omega }_{mn}^2}$