(Created page with "\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} \...")
 
 
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\item \[R_s =\frac{1}{\sigma\delta}\]
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a)
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<math>
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R_s =\frac{1}{\sigma\delta}\\
 
\begin{align*}
 
\begin{align*}
 
\bar{P} &= \frac{1}{2}R_s|J_s|^2 (A)\\
 
\bar{P} &= \frac{1}{2}R_s|J_s|^2 (A)\\
 
& = \frac{1}{2\sigma\delta}|H_y|^2(b)\\
 
& = \frac{1}{2\sigma\delta}|H_y|^2(b)\\
\bar{P} &= \frac{b|H_y|^2}{2\sigma\delta}
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\bar{P} &= \frac{b|H_y|^2}{2\sigma\delta}\\
 
\end{align*}
 
\end{align*}
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\\
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BC: H_{1t} - \cancelto{0}{H_{2t}} = J_s \\
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H_{it} = J_s \\
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|J_s| = |H_y| \text{ where: } \bar{H} = Hx\hat{x} + Hy\hat{y} \\
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</math>
  
\[BC: H_{1t} - \cancelto{0}{H_{2t}} = J_s\]
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b)
\[H_{it} = J_s\]
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<math>
\[|J_s| = |H_y| \text{where:} \bar{H} = Hx\hat{x} + Hy\hat{y}\]
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\alpha = \frac{1}{\delta} \\
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\delta = \frac{1}{\sqrt{\pi f \mu \sigma}} \text{ (conductor)} \\
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\alpha = \sqrt{\pi f \mu \sigma}\\
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</math>
  
\item \[\alpha = \frac{1}{\delta}\]
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c)
\[\delta = \frac{1}{\sqrt{\pi f \mu \sigma}} \text{ (conductor)}\]
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<math>
\[\alpha = \sqrt{\pi f \mu \sigma}\]
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BCs: E_{1t} = E_{2t} = 0\\
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assuming TEM: $E_z = 0 \\
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</math>
  
\item BCs: $E_{1t} = E_{2t} = 0$\\
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d)
assuming TEM: $E_z = 0$
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<math>
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assuming TEM: E_z = 0\\
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non-TEM: E_z \approx 0
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</math>
  
\item assuming TEM: $E_z = 0$\\
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*Aside:
non-TEM: $E_z \approx 0$
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<math>
 
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P = \frac{1}{2}IV = \frac{1}{2}I^2R \\
Aside: \\
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I = J_s(l)\\
\begin{itemize}
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P = \frac{1}{2}|J_s|^2(l^2)R_s\\
\item $P = \frac{1}{2}IV = \frac{1}{2}I^2R$\\
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P = \frac{1}{2}R_s|J_s|^2A\\
$I = J_s(l)$\\
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P = \frac{1}{2}\int \bar{E}\cdot\bar{J}dv\\
$P = \frac{1}{2}|J_s|^2(l^2)R_s$\\
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  = \frac{1}{2}\int E\cdot \big(\frac{J_s}{\delta}\big)ds\\
$P = \frac{1}{2}R_s|J_s|^2A$
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  = \frac{1}{2}\big(\frac{J_s}{\sigma}\big)\big(\frac{J_s}{\delta}\big)(A)\\
\item $P = \frac{1}{2}\int \bar{E}\cdot\bar{J}dv$\\
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  = \frac{1}{2}|J_S|^2\big(\frac{1}{\sigma\delta}\big)A\\
$ = \frac{1}{2}\int E\cdot \big(\frac{J_s}{\delta}\big)ds$\\
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  = \frac{1}{2}R_s|J_s|^2A\\
$ = \frac{1}{2}\big(\frac{J_s}{\sigma}\big)\big(\frac{J_s}{\delta}\big)(A)$\\
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</math>
$ = \frac{1}{2}|J_S|^2\big(\frac{1}{\sigma\delta}\big)A$\\
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$ = \frac{1}{2}R_s|J_s|^2A$
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\end{itemize}
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Latest revision as of 20:42, 3 June 2017

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