91Ó°ÊÓ

$\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{z}}}{\mathbf{∂}\mathbf{y}}\mathbf{-}\mathit{i}\mathit{K}{\mathit{E}}_{\mathbf{y}}\mathbf{=}\mathit{i}\mathit{Ó\neg }{\mathit{B}}_{\mathbf{x}}$ …… (1)

For a rectangular waveguide, as $\frac{\mathrm{Ó\neg }}{c}=k\text{and}{E}_z=0$then, equation (1) becomes,

localid="1657514438209" $$\begin{gathered} \frac{∂\left(0\right)}{∂y}-i\frac{\mathrm{Ó\neg }}{c}{E}_y=i\mathrm{Ó\neg }{B}_x \\ {E}_y=-c{B}_x \end{gathered}$$

Substitute $k=\frac{\mathrm{Ó\neg }}{c}\text{and}\frac{∂E_z}{∂x}=0$in the equation (2).

role="math" localid="1657513845935" $$\begin{gathered} i\frac{\mathrm{Ó\neg }}{C}{E}_x-0=i\mathrm{Ó\neg }{B}_y \\ {E}_x=c{B}_y \end{gathered}$$

Substitute $E_x=c{B}_y$in the equation (3).

$$\begin{gathered} \frac{∂B_z}{∂y}-ik{B}_y=-i\frac{\mathrm{Ó\neg }}{c^2}\left(c{B}_y\right) \\ \frac{∂B_z}{∂y}=ik{B}_y-ik{B}_yâˆ\pounds \\ \frac{∂B_z}{∂y}=0 \end{gathered}$$

Substitute $E_y=c{B}_x$in the equation (4).

$ik{B}_x-\frac{∂B_z}{∂x}=-\frac{i\mathrm{Ó\neg }}{c^2}\left(-c{B}_x\right)$

$ik{B}_x-\frac{∂B_z}{∂x}=ik{B}_x$

$\frac{∂B_z}{∂x}=0$

Hence,

$\frac{∂B_z}{∂x}=\frac{∂B_z}{∂y}=0$

If the boundary is just inside the metal, the value of E will be zero. So, the value of B will also be zero.

Hence, this is a TEM mode, and $T{E}_{00}$mode cannot occur in a rectangular waveguide.

Therefore, the role="math" localid="1657514122509" $T{E}_{00}$mode cannot occur in a rectangular waveguide.