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+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics I}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ WS 2007/08\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 2 - proposed solutions}
+\end{center}
+
+\section{Drude theory of metallic conduction}
+\begin{enumerate}
+ \item $U=IR \Rightarrow EL=jA\rho\frac{L}{A}
+ \Rightarrow E=j\rho$
+ \item distance: $v\,dt$\\
+ number of electrons crossing $A$: $n(v\,dt)A$\\
+ $\Rightarrow$ $j=\frac{I}{A}=\frac{dQ/dt}{A}=\frac{-e\,n(v\,dt)A/dt}{A}
+ =-nev$
+ \item \begin{itemize}
+ \item In the absence of an electric field, electrons are as likely
+ to be moving in any one direction as in any other.
+ The velocity averages to zero.
+ As expected, according to the above equation, there is no
+ net electric current density.
+ \item Since electrons emerge in a random direction
+ there will be no contribution from the thermal velocity
+ to the average electronic velocity.
+ \item $v_{average}=at=\frac{F}{m}\tau=-\frac{eE}{m}\tau$
+ \end{itemize}
+ \item \begin{itemize}
+ \item $j=\left(\frac{ne^2\tau}{m}\right)E$\\
+ \item $j=\sigma E \Rightarrow \sigma=\frac{ne^2\tau}{m}$
+ \end{itemize}
+ \item Energy transfer: $\frac{m}{2}v_{drift}^2$,
+ $\qquad v_{drift}$:
+ end drift velocity of the accelerated electron\\
+ $v_{drift} \ne v_{average}$
+
+
+\end{enumerate}
+
+\end{document}