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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 11} \bigskip

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\noindent
{\bf 30.\ Skin depth (E.95). } \smallskip

\begin{enumerate}

\item We define a good conductor by the validity of the 
approximation $\sigma^c >> \omega \epsilon^c$ so that we have
$$  \vec{E}^c = {c\over 4\pi\,\sigma^c}\,\nabla\times\vec{H}^c\,.$$
Using the homogeneous Maxwell equation for the electric field together 
with the $e^{-i\,\omega\,t}$ time dependence of the magnetic field one 
finds
$$ \nabla\times\vec{E}^c = -\,{\mu^c\over c}\, 
  {\partial\vec{H}^c\over\partial t\ } = {i\,\omega\mu^c\over c}\,
  \vec{H}^c ~~{\rm or}~~ \vec{H}^c = - {i\,c\over \mu^c\,\omega}\,
  \nabla \times \vec{E}^c\, . $$
We assume smooth variations of the fields parallel to the interface 
and approximate the nabla operator by a derivative perpendicular to 
the interface
$$\nabla\ \approx\ \hat{n}\, {\partial\ \over \partial n}\, .$$
Here $\hat{n}$ is the unit normal vector pointing {\it into} the
conductor and $n$ is the corresponding coordinate. In this
approximation 
$$  \vec{E}^c = {c\over 4\pi\,\sigma^c}\,
    \hat{n} \times {\partial\vec{H}^c \over \partial n} =
    {c\over 4\pi\,\sigma^c}\,
    \hat{n} \times {\partial\vec{H}^c_{\|} \over \partial n}
    = \vec{E}^c_{\|}\,,  $$
because $\hat{n}\times\vec{H}^c_{\perp} = 0$ holds. Further,
\begin{eqnarray} \nonumber 
  \vec{H}^c \approx \vec{H}^c_{\|} &=& - {i\,c\over \omega\,\mu^c}\,
  \hat{n} \times {\partial \vec{E}^c_{\|} \over \partial n}\, =\,
  - {i\,c^2\over 4\pi\,\omega\,\mu^c\,\sigma^c}\, \hat{n}\times
  \left( \hat{n} \times {\partial^2 \vec{H}^c_{\|} \over 
  \partial\,n^2} \right) \\ \label{Hc2}
  &=& {i\,c^2\over 4\pi\,\omega\,\mu^c\,\sigma^c}\, 
  {\partial^2 \vec{H}^c_{\|} \over \partial\,n^2}\,,
\end{eqnarray}
where the previous relation between $\vec{E}^c_{\|}$ and $\vec{H}^c_{\|}$
has been used. The resulting differential equation is elementary:
$$ {\partial^2\vec{H}^c_{\|} \over \partial n^2}  = -
 {2\,i \over \delta^2 }\, \vec{H}^c_{\|}~~~{\rm where}~~~
 \delta = {c\over \sqrt{4\pi} }\,
 {\sqrt{2\over \omega\,\mu^c\,\sigma^c} } $$
is the skin depth.  Its interpretation follows from the solution 
$$ \vec{H}^c_{\|}=\vec{H}^0_{\|}\,e^{-n/\delta}\,e^{+i\,n/\delta}\,.$$
Proof of this equation: The ansatz $\exp(\alpha\,n)$ gives $\alpha^2=
-2i/\delta^2$, i.e., $\alpha=\pm\sqrt{2}\,\sqrt{-i}/\delta$. The square 
root of $-i$ is taken by writing $-i=\exp(-i\,\pi/2)$ $\Rightarrow\ 
\sqrt{-i}=\exp(-i\,\pi/4)=(1-i)/\sqrt{2}$. The result follows after 
discarding the plus sign solution as unphysical. It remains 
$\alpha=-(1-i)/\delta$. \medskip

\item (2) Using the table of appendix~B,
$$ \mu=10^3\,,~~~\omega = 2\pi\,10^{10}\,[rad/s]\,,~~~\sigma^c = 
   1/\rho^c = \frac{c_f^2\,10^9}{20}\,
   \left[\frac{statampere}{statvolt\,m}\right]\,,$$
$$ {\rm so\ that}~~~\delta = \frac{c}{\sqrt{4\pi}}\,
   \sqrt{\frac{2}{\mu\,\omega\,\sigma^c}} = 0.0712\,[cm]\ .$$
However, with $\epsilon =3$ we find
$$ \frac{\epsilon\,\omega}{c} \approx 629 > \frac{4\pi}{c\,\rho^c} 
   \approx 19 $$
so that our approximation for the skin depth is invalid. This comes
because $\rho^2=20\,[\Omega\,m]$ is an unrealistically large 
resistivity. For instance iron has a resistivity $\rho^c=10^{-7}
\,[\Omega\,m]$. With this we obtain
$$ \delta = 0.0523\,[\mu m]\,, $$
and the approximation is valid because
$$ \frac{\epsilon\,\omega}{c} \approx 629 \ll \frac{4\pi}{c\,\rho^c} 
   \approx 3.5\times 10^9 $$
holds.
\end{enumerate}
\bigskip


\noindent
{\bf 31.\ TM waves in a rectangular wave guide (E.96).}
\smallskip

\begin{enumerate}

\item 
$$ (\nabla_t^2 + \gamma^2)\, E^z = 0\,,\qquad E^z|_S = 0\,, $$
$$ E^z = E^0\, \sin\left(\frac{m\pi x}{a}\right)\,
               \sin\left(\frac{n\pi y}{b}\right)\,
               e^{i\,k\,z}\,e^{-i\,\omega\,t}\,,            $$
$$ \gamma^2_{n\,m} = \pi^2\,\left[\left(\frac{m}{a}\right)^2 +
                            \left(\frac{n}{b}\right)^2 \right]\,,
                   ~~~m=1,2,\dots\,,~~ n=1,2,\dots\,,          $$
\item
$$ \gamma^2_{1\,1} = \pi^2\,\left(a^{-2}+b^{-2}\right)\,,~~~~~
   \omega_{1\,1} = \frac{c\,\pi}{\sqrt{\mu\,\epsilon}}\
   \sqrt{a^{-2}+b^{-2}}                                        $$ 
\item
$$ \vec{E}_t = \frac{i\,k}{\gamma^2}\,\nabla_t\,E^z\,,~~~~~~
   {\rm lowest\ mode:} $$
\begin{eqnarray} \nonumber
  E^x &=& E_0\,\frac{i\,k}{\pi\,a\,(a^{-2}+b^{-2})}\,
          \cos\left(\frac{\pi x}{a}\right)\,
          \sin\left(\frac{\pi y}{b}\right)\,
          e^{i\,k\,z}\,e^{-i\,\omega\,t}\,, \\ 
  E^y &=& E_0 \,\frac{i\,k}{\pi\,b\,(a^{-2}+b^{-2})}\,
          \sin\left(\frac{\pi x}{a}\right)\,
          \cos\left(\frac{\pi y}{b}\right)\,
          e^{i\,k\,z}\,e^{-i\,\omega\,t}\,.\nonumber
\end{eqnarray}
\item
$$ \vec{H}_t = \frac{\epsilon\,\omega}{c\,k}\,\hat{z}\times\vec{E}_t\,,
   ~~~~~~{\rm lowest\ mode:} $$
\begin{eqnarray} \nonumber
  H^x&=&-E_0\,\frac{i}{b\,(a^{-2}+b^{-2})}\,\sqrt{\frac{\epsilon}{\mu}}
  \, \sin\left(\frac{\pi x}{a}\right)\,
     \cos\left(\frac{\pi y}{b}\right)\,
          e^{i\,k\,z}\,e^{-i\,\omega\,t}\,, \\ 
  H^y&=&+E_0\,\frac{i}{a\,(a^{-2}+b^{-2})}\,\sqrt{\frac{\epsilon}{\mu}}
  \, \cos\left(\frac{\pi x}{a}\right)\,
     \sin\left(\frac{\pi y}{b}\right)\,
          e^{i\,k\,z}\,e^{-i\,\omega\,t}\,.\nonumber
\end{eqnarray}
\item
$$ \vec{K} = \frac{c}{4\pi}\,\hat{n}\times\vec{H} $$
Let us choose the convention that $\hat{n}$ points out of the conductor
(into the wave guide). Then,
$$ {\rm For\ the}\ (x\!-\!z)\!=\!{\rm plane}\ y=0,b:
~~~\hat{n}=\pm\hat{y}\,,~~~\frac{4\pi}{c}\,\vec{K}=\mp\hat{z}\,H^x\,,$$
$$ {\rm For\ the}\ (y\!-\!z)\!=\!{\rm plane}\ x=0,a:
~~~\hat{n}=\pm\hat{x}\,,~~~\frac{4\pi}{c}\,\vec{K}=\pm\hat{z}\,H^y\,.$$

\end{enumerate}
\bigskip


\noindent
{\bf 32.\ Cubic cavity oscillator (E.99).}
\smallskip

$$ \vec{E}\ =\ {\hat x}\, E_0\, \sin \left( {\pi y\over a} \right)\,
   \sin \left( {\pi z\over a} \right)\, e^{-i\omega t}\, .        $$

\begin{enumerate}
\item 
\begin{eqnarray} \nonumber
  \vec{H} &=& -\frac{i\,c}{\omega\,\mu}\, \nabla\times\vec{E}\ =
  \\ \nonumber 
  &-&\frac{i\,c}{\omega\,\mu}\,E_0\,\frac{\pi}{a}\,\left[
  - \hat{z}\,\cos\left(\frac{\pi y}{a}\right)\,
             \sin\left(\frac{\pi z}{a}\right)\,
  + \hat{y}\,\sin\left(\frac{\pi y}{a}\right)\,
             \cos\left(\frac{\pi z}{a}\right)\,\right]\,
    e^{-i\omega t}\,.
\end{eqnarray}
\item 
$$ \nabla\cdot\vec{E} = 0~~~{\rm as}\ \vec{E}\ {\rm does\ not\
   depend\ on}\ x\ {\rm and}\ \vec{E}\sim\hat{x}\,, $$
$$ \nabla\cdot\vec{H} \sim \left[ 
   - \cos\left(\frac{\pi y}{a}\right)\,
     \cos\left(\frac{\pi z}{a}\right)\,
  +  \cos\left(\frac{\pi y}{a}\right)\,
     \cos\left(\frac{\pi z}{a}\right)\,\right]\,
    e^{-i\omega t} = 0\,.                                           $$
\begin{eqnarray} \nonumber
  \nabla\times\vec{H} &=& - \frac{i\,c}{\omega\,\mu}\,E_0\,\frac{\pi^2}{a^2}
  \,\left[ \hat{x}\,\sin\left(\frac{\pi y}{a}\right)\,
           \sin\left(\frac{\pi z}{a}\right)\,
  + \hat{x}\,\sin\left(\frac{\pi y}{a}\right)\,
             \sin\left(\frac{\pi z}{a}\right)\,\right]\,e^{-i\omega t}
\\ \nonumber &=& - \frac{i\,c}{\omega\,\mu}\,\frac{2\,\pi^2}{a^2}\,\vec{E}
            \,=\, - \frac{i\,\omega}{c}\,\epsilon\,\vec{E}
\end{eqnarray}
where the result of (3) below was used in the last step. BCs: From
$\sin(\pi a/a)=\sin(\pi)=0$ we see that $\vec{E}|_S=0$ holds on the 
boundaries parallel to $\vec{E}$ and on the boundaries where 
$\vec{H}$ has a perpendicular component $\vec{H}_{\perp}|_S=0$ holds.
\bigskip

\item 
$$ 0 = \left( \nabla^2-\frac{\mu\epsilon}{c^2}\,
   \frac{\partial^2~}{\partial t^2}\right)\,\vec{E} = 
   \left( -\frac{\pi^2}{a^2} -\frac{\pi^2}{a^2} +
   \frac{\mu\epsilon}{c^2}\,\omega^2 \right)\,\vec{E}\
   \Rightarrow\ \omega = 
   \sqrt{\frac{2}{\mu\epsilon}}\,\frac{c\,\pi}{a}\,. $$
\bigskip

\item Surface currents:
$$ \vec{K} = \frac{c}{4\pi}\,\hat{n}\times\vec{H}\,. $$
Top, bottom $(x\!-\!y)\!-\!{\rm plane},\ z=0,a$:
$$ \vec{K} = \pm\frac{c}{4\pi}\,\hat{z}\times\vec{H} =
             \pm\hat{x}\,\frac{i\,c^2}{4\omega\mu\epsilon a}\,
             \sin\left(\frac{\pi y}{a}\right)\,e^{-i\omega t}\,, $$
Front, back $(z\!-\!x)\!-\!{\rm plane},\ y=0,a$:
$$ \vec{K} = \pm\frac{c}{4\pi}\,\hat{y}\times\vec{H} =
             \mp\hat{x}\,\frac{i\,c^2}{4\omega\mu\epsilon a}\,
             \sin\left(\frac{\pi y}{a}\right)\,e^{-i\omega t}\,, $$
Left, right $(y\!-\!z)\!-\!{\rm plane},\ x=0,a$:
\begin{eqnarray} \nonumber
 &\vec{K}& =\,\pm\frac{c}{4\pi}\,\hat{x}\times\vec{H}\,=\\ \nonumber
 &\mp & \frac{i\,c^2}{4\omega\mu\epsilon a}\,\left[
   \hat{y}\,\cos\left(\frac{\pi y}{a}\right)\,
            \sin\left(\frac{\pi z}{a}\right) +
   \hat{z}\,\sin\left(\frac{\pi y}{a}\right)\,
            \cos\left(\frac{\pi z}{a}\right)\,
            \right]\, e^{-i\omega t}\,.
\end{eqnarray}
Surface charge density in the $(y\!-\!z)\!-\!{\rm plane}:$
$$ \sigma = \frac{1}{4\pi}\,\frac{2\pi}{\omega^2\mu}\,E_0\,
   \sin\left(\frac{\pi y}{a}\right)\,
   \sin\left(\frac{\pi z}{a}\right)\,e^{-i\omega t}\ .$$
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\end{enumerate}

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