7.8.12 The TM₀₁₀ mode

Energy. With ${\int }_0^a{J}_0^2(x_{01}\rho /a)\rho d\rho =\frac{a^2}{2}{J}_1^2(x_{01})$, $W_e=\frac{\epsilon }{4}|E_0{|}^{2}d2\pi \frac{a^2}{2}{J}_1^2(x_{01})$, and $W=2W_e$ gives Eq. (7.8.33). Loss. The side wall carries $|H_{\varphi }(a)|=|E_0|J_1(x_{01})/\eta$ over the area $2\pi ad$. The two end plates carry $H_{\varphi }(\rho )$, and since ${\int }_0^a{J}_1^2(x_{01}\rho /a)\rho d\rho =\frac{a^2}{2}{J}_1^2(x_{01})$ as well (because $J_0(x_{01})=0$), each plate loses $\frac{R_s}{2}\frac{|E_0{|}^2}{{\eta }^2}\pi a^2{J}_1^2(x_{01})$. In total $P_L=\frac{R_s|E_0{|}^2}{{\eta }^2}{J}_1^2(x_{01})\pi a(d+a)$. Ratio. With $\omega \epsilon {\eta }^2=\omega \mu =k\eta$ and $k=x_{01}/a$, $Q=\omega W/P_L=\frac{k\eta d{a}^2}{2R_{s}a(d+a)}=\frac{\eta }{R_s}\frac{x_{01}}{2(1+a/d)}$. A longer cavity has a higher $Q$, since the loss of the two end plates is spread over more stored energy.