and a current density $j_x$ is flowing in that wire.
There is a magnetic field $B$ pointing in the positive $z$-direction.
Electrons are deflected in the negative $y$-direction
-due to the Lorentz force $F_L=-evB$
+due to the Lorentz force $F_L=-ev\times B$
until they run against the sides of the wire.
An electric field $E_y$ builds up opposing the Lorentz force
and thus preventing further electron accumulation at the sides.
\end{itemize}
In this tutorial the treatment of the Hall problem is based on a simple
Drude model analysis.
-\\\\
+
First of all the effect of individual electron collisions can be expressed
-by a frictional damping term into the equation of motion for the momentum
+by a frictional damping term in the equation of motion for the momentum
per electron.
+By inserting the forces acting on the elecron into the equation of motion
+the expression for the Hall coefficient $R_H=-\frac{1}{ne}$ can be found.
\begin{enumerate}
\item Recall the Drude model.
To find an expression for the Hall coefficient use the second equation
and the fact that there must not be transverse current $j_y$
while determining the Hall field.
+ \item Calculate the Hall field for a rectangular wire
+ (width: $l=15\, cm$, thickness: $d=4\, mm$) made out of silver
+ if there is a current $I=200\, A$ flowing
+ and a magnetic field $B=1.5\, Vs/m^2$ applied
+ perpendicular to the current.
+ Silver has the relative atomic mass $A_r=107.87$ and
+ density $\rho=10.5\, g/cm^3$.
+ Assume that there is one valence electron per atom.
+ The atomic mass unit is $u=1.6605 \cdot 10^{-27} \, kg$
+ and $e=1.602 \cdot 10^{-19} \, As$.
+ {\bf Hint:} Start by calculating the Hall coefficient of silver first.
\end{enumerate}
\end{document}