\item \[R_s =\frac{1}{\sigma\delta}\] \begin{align*} \bar{P} &= \frac{1}{2}R_s|J_s|^2 (A)\\ & = \frac{1}{2\sigma\delta}|H_y|^2(b)\\ \bar{P} &= \frac{b|H_y|^2}{2\sigma\delta} \end{align*}
\[BC: H_{1t} - \cancelto{0}{H_{2t}} = J_s\] \[H_{it} = J_s\] \[|J_s| = |H_y| \text{where:} \bar{H} = Hx\hat{x} + Hy\hat{y}\]
\item \[\alpha = \frac{1}{\delta}\] \[\delta = \frac{1}{\sqrt{\pi f \mu \sigma}} \text{ (conductor)}\] \[\alpha = \sqrt{\pi f \mu \sigma}\]
\item BCs: $E_{1t} = E_{2t} = 0$\\ assuming TEM: $E_z = 0$
\item assuming TEM: $E_z = 0$\\ non-TEM: $E_z \approx 0$
Aside: \\ \begin{itemize} \item $P = \frac{1}{2}IV = \frac{1}{2}I^2R$\\ $I = J_s(l)$\\ $P = \frac{1}{2}|J_s|^2(l^2)R_s$\\ $P = \frac{1}{2}R_s|J_s|^2A$ \item $P = \frac{1}{2}\int \bar{E}\cdot\bar{J}dv$\\ $ = \frac{1}{2}\int E\cdot \big(\frac{J_s}{\delta}\big)ds$\\ $ = \frac{1}{2}\big(\frac{J_s}{\sigma}\big)\big(\frac{J_s}{\delta}\big)(A)$\\ $ = \frac{1}{2}|J_S|^2\big(\frac{1}{\sigma\delta}\big)A$\\ $ = \frac{1}{2}R_s|J_s|^2A$ \end{itemize}
