a) $ 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} \\ $

b) $ \alpha = \frac{1}{\delta} \\ \delta = \frac{1}{\sqrt{\pi f \mu \sigma}} \text{ (conductor)} \\ \alpha = \sqrt{\pi f \mu \sigma}\\ $

c) $ BCs: E_{1t} = E_{2t} = 0\\ assuming TEM: $E_z = 0 \\ $

d) $ assuming TEM: E_z = 0\\ non-TEM: E_z \approx 0 $

  • Aside:

$ 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\\ 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\\ $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn