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%TCIDATA{Created=Fri Sep 27 09:26:49 1996}
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\markright{{\sf EM2 test, 26 Sept 95}}
\begin{document}


\begin{center}
\textbf{Phys 743 - EM II}

\textbf{In-class test 1 - 26 Sept. 1995}
\end{center}

\noindent \textbf{1.}

\textbf{(a) }What is the general behavior of $\epsilon (\omega )$ for $%
\omega \rightarrow \infty $? A full answer is needed for the following.

\textbf{(b)} Show that the sum rule 
\[
\frac 2\pi \int_0^\infty \omega \func{Im}\epsilon (\omega )\,d\omega =\frac{%
4\pi Ne^2}m 
\]
follows from the dispersion relation 
\[
\func{Re}\epsilon (\omega )=1+\frac 2\pi \,\Bbb{P}\int_0^\infty \frac{\omega
^{\prime }\func{Im}\epsilon (\omega ^{\prime })}{\omega ^{\prime \,2}-\omega
^2}\,d\omega ^{\prime } 
\]
and from the behavior of $\epsilon (\omega )$ for $\omega \rightarrow \infty
.$ Supply the meaning of $N$ and $m.$

\textbf{(c) }The intensity of a plane wave propagating in the $z$ direction
with frequency $\omega $ is observed to be proportional to $e^{-\alpha z}.$
Relate the attenuation constant $\alpha $ to the imaginary part of the index
of diffraction in the medium.

\textbf{(d) }What is the relation between $\epsilon (\omega )$ and the
complex index of diffraction $n(\omega )$? Can one write dispersion
relations for $n(\omega )?$

\textbf{(e)} By an argument similar to that in \textbf{(b)}, or otherwise,
evaluate 
\[
\frac 2\pi \int_0^\infty \alpha (\omega )\,d\omega =\cdots 
\]
for a non-magnetic medium ($\mu =1).$

\smallskip

\noindent \textbf{2.}

Consider a cylindrical cavity (not necessarily a circular cylinder) with the
axis in the $z$ direction. Assume that the walls are perfectly conducting.

\textbf{(a) }What are the boundary conditions for the \textbf{E }and \textbf{%
H }fields on the side walls?

\textbf{(b) }What are the boundary conditions for \textbf{E }and \textbf{H }
on the end walls?\textbf{\ }

\textbf{(c) }Describe qualitatively the lowest TM mode. You may use a cavity
with a rectangular or circular cross section as an example.

\textbf{(d) }Describe qualitatively the lowest TE mode. You may use an
example as above.

\textbf{(e) }Which mode has lowest frequency for a long, thin cavity? For a
short cavity?

\smallskip\ \ \ 

\noindent \textbf{3.}

\textbf{(a) }Write down the expression for the skin depth $\delta $ in a
medium of d.c. conductivity $\sigma $ and magnetic permeability $\mu ,$ at
frequency $\omega .$ Describe conditions under which this expression is
valid.

\textbf{(b) }Under the above conditions, how is the surface current $\mathbf{%
K}$ related to the $\mathbf{E}$ field at the surface? How is it related to
the $\mathbf{H}$ field at the surface?

\textbf{(c)} What is the power dissipated per unit surface area?

\begin{center}
\newpage\ 

\textbf{SOLUTIONS }
\end{center}

\noindent \textbf{1.}

\textbf{(a) }$\epsilon (\omega )\rightarrow 1-\dfrac{4\pi Ne^2}{m\omega ^2}$
\thinspace where $N$ is the electron density, $m$ the electron mass. This
formula also implies that $\func{Im}\epsilon \rightarrow 0$ faster than $%
\omega ^{-2}$ .

\smallskip\ 

\textbf{(b)} Compare $\epsilon (\omega )\rightarrow 1-\dfrac{4\pi Ne^2}{%
m\omega ^2}$ with $\func{Re}\epsilon (\omega )\rightarrow 1-\dfrac 2\pi \,%
\Bbb{P}\dint_0^\infty \dfrac{\omega ^{\prime }\func{Im}\epsilon (\omega
^{\prime })}{\omega ^2}\,d\omega ^{\prime }$.

\smallskip\ 

\textbf{(c)} $\dfrac \alpha 2=\dfrac \omega c\func{Im}n\quad \quad \quad
\quad \quad $ \textbf{(d) }$n=\sqrt{\epsilon \mu }.$ One can, because $%
\epsilon $ and $\mu $ have no zeros for $\func{Im}\omega >0$.

\smallskip\ 

\textbf{(e) }Compare $n(\omega )\rightarrow 1-\dfrac{2\pi Ne^{2}}{m\omega
^{2}}$ \ with\ \ $\func{Re}n(\omega )\rightarrow 1-\dfrac{2}{\pi }\,\Bbb{P}%
\dint_{0}^{\infty }\dfrac{\omega ^{\prime }\func{Im}n(\omega ^{\prime })}{%
\omega ^{2}}\,d\omega ^{\prime }$. \smallskip \noindent Get$\;\dfrac{2\pi
Ne^{2}}{m}=\dfrac{2}{\pi }\,\dint_{0}^{\infty }\omega ^{\prime }\func{Im}%
n(\omega ^{\prime })\,d\omega ^{\prime }$\ or\ \ $\dfrac{4\pi Ne^{2}}{mc}=%
\dfrac{2}{\pi }\,\dint_{0}^{\infty }\alpha (\omega ^{\prime })\,d\omega
^{\prime }$ 

\smallskip \ 

\noindent \textbf{2.}

\textbf{(a) }and \textbf{(b) } If we introduce components parallel and
perpendicular to the wall, such as $\mathbf{E}_{\parallel }$ and $E_n,$ the
conditions on any wall are $\mathbf{E}_{\parallel }=0,$ $\partial _nE_n=0$
and $\partial _n\mathbf{H}_{\parallel }=0,$ $H_n=0.$ In particular on the
end walls $E_x=0,$ $E_y=0,$ $\partial _zE_z=0\;$ and $\;\partial
_zH_x=0,\;\partial _zH_y=0,\;H_z=0.$

\textbf{(c) }In the lowest TM mode $E_z$ does not depend on $z$, hence the
frequency does not depend on the length of the cavity $L_z$ and is equal to $%
\omega _c,$ the lowest TM cutoff for a waveguide with the same cross section.

\textbf{\thinspace (d)} In the lowest TE mode $H_z\propto \sin (\pi z/L_z)$
and the frequency $\omega $ depends on $L_z$ according to 
\begin{equation}
\omega =\sqrt{\omega _c^{\prime \;2}+(\pi c/L_z)^2}.  \label{o1}
\end{equation}
where $\omega _c^{\prime }$ is the lowest TE cutoff for a waveguide with the
same cross section.

\textbf{(e)} Since $\omega _{c}^{\prime }<\omega _{c}$ (see the Examples),
the lowest mode is TM for a short cavity, TE for a long cavity.

\smallskip \ 

\noindent \textbf{Example 1. }For a box of sides $L_{x},L_{y},L_{z}$ with $%
L_{x}\geq L_{y},$ the lowest mode is TM$_{110}$ for $L_{z}<L_{x}$ (short
cavity) and TE$_{101}$ for $L_{z}>L_{x}$ (long cavity). Explicitly, with $%
k_{x}=\pi /L_{x}$ and $k_{y}=\pi /L_{y}$:

\begin{itemize}
\item  The TM$_{110}$ mode has fields 
\begin{eqnarray*}
E_{x} &=&0 \\
E_{y} &=&0 \\
E_{z} &=&E\sin k_{x}x\sin k_{y}y
\end{eqnarray*}
\begin{eqnarray*}
H_{x} &=&-i(k_{y}c/\omega )E\sin k_{x}x\cos k_{y}y \\
H_{y} &=&i(k_{x}c/\omega )E\cos k_{x}x\sin k_{y}y \\
H_{z} &=&0
\end{eqnarray*}
where $\omega =c\sqrt{k_{x}^{2}+k_{y}^{2}}=c\sqrt{\left( \pi /L_{x}\right)
^{2}+(\pi /L_{y})^{2}}$ is the frequency. We see that $\omega $ does not
depend on $L_{z}$ and is equal to the TM threshold $\omega _{c}=$ $\omega
_{11}$ (in the usual notation). Restoring the time-dependence and taking the
real part, the non-zero field components are

\begin{eqnarray*}
E_{z} &=&E\sin k_{x}x\sin k_{y}y\,\cos \omega t \\
H_{x} &=&-(k_{y}c/\omega )E\sin k_{x}x\,\cos k_{y}y\,\sin \omega t \\
H_{y} &=&(k_{x}c/\omega )E\cos k_{x}x\,\sin k_{y}y\,\sin \omega t
\end{eqnarray*}

\item  The TE$_{101}$ mode has fields 
\begin{eqnarray*}
E_{x} &=&0 \\
E_{y} &=&i(\omega /k_{x}c)H\sin k_{x}x\sin k_{z}z \\
E_{z} &=&0
\end{eqnarray*}
\begin{eqnarray*}
H_{x} &=&-(k_{z}/k_{x})H\sin k_{x}x\cos k_{z}z \\
H_{y} &=&0 \\
H_{z} &=&H\cos k_{x}x\sin k_{z}z
\end{eqnarray*}
where $\omega =c\sqrt{k_{x}^{2}+k_{z}^{2}}=c\sqrt{\left( \pi /L_{x}\right)
^{2}+(\pi /L_{z})^{2}}$ is the frequency. We see that $\omega $ is of the
form (\ref{o1}) with $\omega _{c}^{\prime \;}=c\pi /L_{x}=$ $\omega _{10}$
(in the usual notation).
\end{itemize}

\noindent \textbf{Example 2. }For a circular cylinder of radius $R$ and
length $L_z,$ the lowest mode is TM$_{010}$ for $L_z<2.03R$ (short cavity)
and TE$_{111}$ for $L_z>2.03R$ (long cavity). Explicitly, with $k_\rho
=2.405/R,$ $k_\rho ^{\prime }=1.841/R$ and $k_z=\pi /L_z$:

\begin{itemize}
\item  The TM$_{010}$ mode has fields 
\begin{eqnarray*}
E_{\rho } &=&0 \\
E_{\phi } &=&0 \\
E_{z} &=&EJ_{0}(k_{\rho }\rho )
\end{eqnarray*}
\begin{eqnarray*}
H_{r} &=&0 \\
H_{\phi } &=&-iEJ_{1}(k_{\rho }\rho ) \\
H_{z} &=&0
\end{eqnarray*}
The frequency is $\omega =ck_{\rho }=2.405c/R$. We see that $\omega $ does
not depend on $L_{z}$ and is equal to the TM threshold $\omega _{c}=$ $%
\omega _{01}$ (in the usual notation).

\item  The TE$_{111}$ mode is doubly degenerate ($m=\pm 1$). For comparison
with the fields in a box it is good to use the linear combination 
\begin{eqnarray*}
E_{\rho } &=&-i(\omega /ck_{\rho }^{\prime \,2}\rho )HJ_{1}(k_{\rho
}^{\prime }\rho )\sin \phi \sin k_{z}z \\
E_{\phi } &=&-i(\omega /ck_{\rho }^{\prime })HJ_{1}^{\,\prime }(k_{\rho
}^{\prime }\rho )\cos \phi \sin k_{z}z \\
E_{z} &=&0
\end{eqnarray*}
\begin{eqnarray*}
H_{\rho } &=&(k_{z}/k_{\rho }^{\prime })HJ_{1}^{\,\prime }(k_{\rho }^{\prime
}\rho )\cos \phi \cos k_{z}z \\
H_{\phi } &=&-(k_{z}/\rho k_{\rho }^{\prime \,2})HJ_{1}(k_{\rho }^{\prime
}\rho )\sin \phi \cos k_{z}z \\
H_{z} &=&HJ_{1}(k_{\rho }^{\prime }\rho )\cos \phi \sin k_{z}z
\end{eqnarray*}
where $\omega =c\sqrt{k_{\rho }^{\prime \,2}+k_{z}^{2}}=c\sqrt{\left(
1.841/R\right) ^{2}+(\pi /L_{z})^{2}}$ is the frequency. We see that $\omega 
$ is of the form (\ref{o1}) with $\omega _{c}^{\prime \;}=1.841c/R=$ $\omega
_{11}^{\prime }$ (in the usual notation).
\end{itemize}

\noindent \textbf{Comment: }Jackson uses $k$ for $k_z.$ For the box he uses $%
\gamma ^2=k_x^2+k_y^2$ and for the circular cylinder he uses $\gamma $ for $%
k_\rho .$

\smallskip

\noindent \textbf{3.}

\textbf{(a) }$\delta =\sqrt{\dfrac c{2\pi \mu \omega \sigma }}$ \ is valid
when $\func{Im}\epsilon >>\func{Re}\epsilon $. One can always write $%
\epsilon =\func{Re}\epsilon +4\pi i\sigma /\omega $ and the condition is
then $4\pi \sigma >>\omega \func{Re}\epsilon .$ In many good conductors $%
\sigma $ and $\func{Re}\epsilon $ depend weakly on $\omega $ up to
frequencies in the infrared. The surface must also be nearly flat on the
scale of $\delta .$

\smallskip\ 

\textbf{(b)} $\mathbf{K}=\sigma \delta \,\dfrac{1+i}2\mathbf{E}_{\parallel }$
\quad and \quad $\mathbf{K}=\dfrac c{4\pi }\,\mathbf{n\times H}$

\smallskip\ 

\textbf{(c) }$\dfrac{dP}{da}=\dfrac 12\,\sigma \delta \left| \mathbf{E}%
_{\parallel }\right| ^2=\dfrac 1{2\sigma \delta }\,\left| \mathbf{K}\right|
^2$

\end{document}
