Question 9.31: Work out the theory of TM modes for a rectangular wave guide...

Following Sect. 9.5.2, the problem is to solve Eq. 9.181 with E_{z} \neq 0, B_{z}=0 , subject to the boundary conditions 9.175. Let E_{z}(x, y)=X(x) Y(y) ; as before, we obtain X(x)=A \sin \left(k_{x} x\right)+B \cos \left(k_{x} x\right) . But the boundary condition requires E_{z}=0 (and hence X = 0) when x = 0 and x = a, so B = 0 and k_{x}=m \pi / a . But this time m = 1, 2, 3, . . . , but not zero, since m = 0 would kill X entirely. The same goes for Y (y). Thus E_{z}=E_{0} \sin \left(\frac{m \pi x}{a}\right) \sin \left(\frac{n \pi y}{b}\right) with n, m = 1, 2, 3, . . . . 

\left. \begin{matrix} \text { (i) }\left[\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+(\omega / c)^{2}-k^{2}\right] E_{z}=0 \text {, } \\ \text { (ii) }\left[\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+(\omega / c)^{2}-k^{2}\right] B_{z}=0 \text {. } \end{matrix} \right\}                                 (9.181)

The rest is the same as for TE waves:  \omega_{m n}=c \pi \sqrt{(m / a)^{2}+(n / b)^{2}} is the cutoff frequency, the wave velocity is v=c / \sqrt{1-\left(\omega_{m n} / \omega\right)^{2}} , and the group velocity is v_{g}=c \sqrt{1\left(\omega_{m n} / \omega\right)^{2}} . The lowest TM mode is 11, with cutoff frequency \omega_{11}=c \pi \sqrt{(1 / a)^{2}+(1 / b)^{2}} . So the ratio of the lowest TM frequency to the lowest TE frequency is  \frac{c \pi \sqrt{(1 / a)^{2}+(1 / b)^{2}}}{(c \pi / a)}=\sqrt{1+(a / b)^{2}}.