
\item {\bf  Skin depth. } \smallskip

\noindent (1) 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

\noindent (2) Using the table of appendix~B,
$$ \mu=10^3\,,~~~\omega = 2\pi\,10^{10}\,[rad/s]\,,~~~\sigma^c = 
   1/\rho^r = \frac{c_f^2\,10^9}{20}\,
   \left[\frac{statampere}{statvolt\,m}\right]\,,$$
$$ {\rm so\ that}~~~\delta = \frac{c}{\sqrt{4\pi}}\,
   \sqrt{\frac{2}{\mu\,\omega\,\sigma^c}} = 0.0712\,[cm]\ .$$
\bigskip

