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%TCIDATA{Created=Mon Sep 30 15:59:03 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 - Oct. 1, 1996}
\end{center}

\noindent \textbf{1.}

Give three examples of physical quantities that obey Kramers-Kronig
relations.

\smallskip

\noindent \textbf{2.}

\textbf{(a) }Find the lowest electromagnetic frequency for a cubic cavity of
side $L,$ assuming that the walls are perfectly conducting and the cavity is
empty.

\textbf{(b)} Describe three mutually orthogonal modes that have this
frequency. For the present purpose, use the simple criterion that two modes
are orthogonal if their $\mathbf{E}$\textbf{\ }fields are mutually
perpendicular.

\textbf{(c) }Estimate the $Q$ of this cavity\textbf{\ }in its lowest mode if
the walls have a finite d.c. conductivity $\sigma .$

\smallskip

\noindent \textbf{3.}

\textbf{(a)} Derive the following ``equations of telegraphy'' from Maxwell's
equations for a homogeneous conducting medium: 
\[
\nabla ^{2}\mathbf{H}-\frac{\mu \varepsilon }{c^{2}}\frac{\partial ^{2}%
\mathbf{H}}{\partial t^{2}}-\frac{4\pi \mu \sigma }{c^{2}}\frac{\partial 
\mathbf{H}}{\partial t}=0 
\]
\[
\nabla ^{2}\mathbf{E}-\frac{\mu \varepsilon }{c^{2}}\frac{\partial ^{2}%
\mathbf{E}}{\partial t^{2}}-\frac{4\pi \mu \sigma }{c^{2}}\frac{\partial 
\mathbf{E}}{\partial t}=0 
\]
Show that there are monochromatic plane wave solutions and find the
dispersion relation for these waves. (Or, if you prefer, find the plane
waves first and then fourier-transform back to obtain the equations of
telegraphy)

\textbf{(b) }Find the attenuation constant of a wave as a function of its
frequency and discuss the two limits of $\sigma $ very small and very large
(compared to what?)

\smallskip

\noindent \textbf{4.}

A current loop of area $\mathcal{A}$ rotates with angular frequency $\omega $
around its diameter, while the current $I$ is kept constant. Choose the axis
of rotation as the $z$ axis. To leading order in $\omega ^{2}\mathcal{A}%
/c^{2},$

\textbf{(a) }Find the angular distribution of the radiated power

\textbf{(b) }Find the total radiated power

\textbf{(c) }Find the angular distribution of $z$-polarized radiation.

It may help to recall the a small current loop is equivalent to a magnetic
dipole. By convention, the direction of polarization is determined by the
electric field.

\newpage

\begin{center}
\textbf{Answer 1}
\end{center}

Standard KK relations are satisfied, for instance, by $\sigma (\omega ),$
and, for insulators, by $\varepsilon (\omega )-1$ and $n(\omega )-1.$
Another example that was discussed in class is $\dfrac{1}{\varepsilon
(\omega )}-1.$

\begin{center}
\textbf{Answer 2}
\end{center}

The lowest frequency is $\omega =c\sqrt{2}\,\pi /L$ and is triply
degenerate. A complete orthogonal set of eigenmodes with this frequency is TM%
$_{110},$ TE$_{011},$ TE$_{101}.$ Equivalently, these modes can be described
as TM$_{110}$ and the two modes obtained from it by cyclically permuting $%
xyz.$ The fields are given below, with  $H=-iE/\sqrt{2}\,,$ or $E=iH\sqrt{2}%
\,$

\begin{itemize}
\item  TM$_{110}$

\begin{center}
\begin{tabular}{lll}
$E_{x}=0$ &  & $H_{x}=H\sin (\pi x/L)\cos (\pi y/L)$ \\ 
$E_{y}=0$ &  & $H_{y}=-H\cos (\pi x/L)\sin (\pi y/L)$ \\ 
$E_{z}=E\sin (\pi x/L)\sin (\pi y/L)$ &  & $H_{z}=0$%
\end{tabular}
\end{center}

\item  TE$_{011}$ (can be obtained from TM$_{110}$ by letting $%
xyz\rightarrow yzx$)

\begin{center}
\begin{tabular}{lll}
$E_{x}=E\sin (\pi y/L)\sin (\pi z/L)$ &  & $H_{x}=0$ \\ 
$E_{y}=0$ &  & $H_{y}=H\sin (\pi y/L)\cos (\pi z/L)$ \\ 
$E_{z}=0$ &  & $H_{z}=-H\cos (\pi y/L)\sin (\pi z/L)$%
\end{tabular}
.
\end{center}

\item  TE$_{101}$ (can be obtained from TM$_{110}$ by letting $%
xyz\rightarrow zxy$)

\begin{center}
\begin{tabular}{lll}
$E_{x}=0$ &  & $H_{x}=-H\cos (\pi z/L)\sin (\pi x/L)$ \\ 
$E_{y}=E\sin (\pi z/L)\sin (\pi x/L)$ &  & $H_{y}=0$ \\ 
$E_{z}=0$ &  & $H_{z}=H\sin (\pi z/L)\cos (\pi x/L)$%
\end{tabular}
\end{center}
\end{itemize}

It can be seen by inspection of these fields that modes with two indices
equal to 0 cannot exist because they have $\mathbf{E}=0.$ See also Jackson,
just above eq. (8.44) and footnote on page 346.
\newpage
\begin{center}
\textbf{Answer 3}
\end{center}

The key here is to start from Maxwell's equations with $\mathbf{J}=\sigma 
\mathbf{E},\;\mathbf{D}=\varepsilon \mathbf{E,}$ $\mathbf{B}=\mu \mathbf{H},$
namely $\mathbf{\nabla \cdot E}=0,\mathbf{\nabla \cdot H}=0$ and 
\begin{eqnarray*}
\mathbf{\nabla \times H} &=&\frac{4\pi \sigma }{c}\,\mathbf{E}+\frac{%
\varepsilon }{c}\,\frac{\partial \mathbf{E}}{\partial t} \\
\mathbf{\nabla \times E} &=&-\frac{\mu }{c}\,\frac{\partial \mathbf{H}}{%
\partial t}
\end{eqnarray*}
The dispersion relation for waves of the form $\exp (i\mathbf{k\cdot x}%
-\omega t)$ is 
\[
k^{2}=\frac{\varepsilon \mu }{c^{2}}\,\omega ^{2}+\frac{4\pi \mu \sigma }{%
c^{2}}\,\,i\omega 
\]
If the first term on the r.h.s. is dominant, 
\[
k\simeq \frac{\sqrt{\varepsilon \mu }}{c}\,\omega \left( 1+\frac{2\pi
i\sigma }{\varepsilon \omega }\right) =\frac{\varepsilon \mu }{c}\,\omega +%
\sqrt{\frac{\mu }{\varepsilon }}\frac{2\pi i\sigma }{c}
\]
If the second term is dominant, 
\[
k\simeq \sqrt{2\pi \mu \sigma \omega }\,\frac{1+i}{c}
\]
In this case it is customary to put $k=(1+i)/\delta ,$where $\delta $ is the
skin depth. The attenuation constant is $\alpha =2\func{Im}k.$ Thus we get $%
\alpha =\sqrt{\dfrac{\mu }{\varepsilon }}\dfrac{4\pi \sigma }{c}$ for small $%
\sigma /\omega \sqrt{\varepsilon }$ and $\alpha =\dfrac{2}{\delta }=\dfrac{%
\sqrt{8\pi \mu \sigma \omega }}{c}$ for large $\sigma /\omega \sqrt{%
\varepsilon }.$

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