\documentclass[12pt]{article}
\usepackage{epsfig}
\topmargin=-0.5cm
\oddsidemargin=-0.5cm
\evensidemargin=-0.5cm
\textheight=23cm
\textwidth=17cm
\begin{document}
\newcommand{\bold}[1]{\mbox{\boldmath $#1$}}
\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

\noindent
V{\aa}r 2007

\end{bf}

\begin{center}
\begin{large}
L{\o}sningsforslag til {\o}ving 10
\end{large}
\end{center}

\noindent
Veiledning uke 11


\vspace{0.5cm}
\noindent
{\em Oppgave 1}

\vspace{0.2cm}
\noindent
The conductor may be divided into sylindracal tubes with inner radius
$r$, outer radius $r+dr$, and hence a cross section with area
$$dA = 2 \pi r \; dr$$
The current running in such a tube is
$$dI = j \cdot dA = j_0\, e^{-r/R} \cdot 2\pi r \; dr$$
Total current $I$ is then found by integrating $dI$ over the
cross section of the conductor, i.e., by letting $r$ vary from 0 to $R$:
\begin{eqnarray}
I &=& \int dI \nonumber \\
&=& \int_0^R j_0\, e^{-r/R} \cdot 2\pi r \; dr \nonumber \\
&=& 2\pi j_0 \, R^2 \int_0^1 x\,e^{-x}\,dx \nonumber \\
&=& 2\pi j_0 \, R^2 \left( 1 - \frac{2}{e} \right) \nonumber
\end{eqnarray}
Here, we have substituted $x=r/R$ so that $dr = R\, dx$ and $r = R\, x$.
The integral is easily solved with integration by parts.

\end{document}
