a)
$ TEM \to E_z = H_z = 0\\ \left\{ \begin{array}{ll} \bar{E} = E(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \gamma = \alpha + j\beta\\ \bar{H} = H(x,y)e^{-\gamma z}e^{j\omega t} \hspace{1cm} \beta = \omega\sqrt{\mu \epsilon} \end{array} \right. $ assume these solutions in region between conductors.
solve wave equations:
$ \left\{ \begin{array}{ll} \nabla^2\bar{E} + k^2 \bar{E} =0 \hspace{1cm} \text{with BC's to find}\\ \nabla^2\bar{H} + k^2 \bar{E} =0 \hspace{1cm} \bar{E} \text{ and } \bar{H} \end{array} \right. Z = \frac{|E|}{|H|} $
Alternative: from transmission line theory :
$ Z_0 = \sqrt{\frac{L}{C}} (lossless) $
find C by assuming same V on line (or Q)
find L by assuming same I on the line
$ \textbf{Note:}\\ TEM\\ Z_{TEM} = \frac{E_x}{E_y}\\ \bar{H} = \frac{1}{Z_{\text{TEM}}}(\hat{z}x\bar{E}) $
b)
$ find C: C= \frac{Q}{V} \oint \bar{D}\cdot d\bar{s} = Q \\ \int_0^L \int_0^{2\pi}\epsilon E_r(rd\phi dz) = Q\\ \epsilon E_r(2\pi r)L = Q\\ \bar{E} = \frac{Q}{2\pi r\epsilon(L)}\hat{r}\\ \begin{align*} V_2 - V_1 &= - \int_1^2 \bar{E}\cdot dl\\ &=-\int_b^a \frac{Q}{2\pi L\epsilon}\bigg(\frac{1}{r}\bigg)dr\\ &=\frac{Q}{2\pi L\epsilon}\ln\bigg(\frac{b}{a}\bigg) \end{align*} \[C= \frac{2\pi L\epsilon}{\ln\big(\frac{b}{a}\big)}\]\\ $
$ find L: L = \frac{\Phi}{NI}\\ \oint \bar{H}\cdot d\bar{l} = I_{enc}\\ \int_0^{2\pi} H_\phi(rd\phi) = I\\ \bar{H} = \frac{I}{2\pi r}\hat{\phi}\\ \begin{align*} \Phi &= \int \bar{B}\cdot d\bar{s}\\ &=\int_0^L\int_a^b \frac{\mu I}{2\pi r}drdz\\ &= \frac{\mu IL}{2\pi}\ln\bigg(\frac{b}{a}\bigg) \end{align*} dl = dr\hat{r} + rd\phi\hat{\phi} +dz\hat{z}\\ ds_{\phi} = drdz\\ L = \frac{\mu L \ln \big(b/a\big)}{2\pi}\\ Z_o = \sqrt{\frac{L}{C}} = \sqrt{\frac{\frac{\mu L\ln(b/a)}{2\pi}}{{\frac{2\pi L\epsilon}{\ln(b/a)}}}} = \sqrt{\frac{\mu}{\epsilon}\bigg(\frac{\ln(b/a)}{2\pi}\bigg)^2}=\frac{\ln(b/a)}{2\pi}\sqrt{\frac{\mu}{\epsilon}}\\ $
c) c) still supports a TEM mode; a uniformly lossy dielectric results in a non-zero attenuation constant $ \alpha $ and a conductance $ G $ such that Ez = 0 and Hz = 0. However, if the {\bf conductors} have a large but finite conductivity (no longer PEC), then the Poynting vector must have a component normal to the conductor surface, necessitating the presence of a small axial electric field Ez != 0 (see David K. Cheng, Field and Wave Electromagnetics, 2nd Ed. page 433).
