-?. The Hall effect and the sign of charge carriers. Consider a rectangular conductor of width $w$ and thickness $t$ carrying a current $I$ in a uniform magnetic field $\mathbf{B}$ perpendicular to the plane of the conductor, as shown in Fig. $14.45 .$ The charge carriers, with charge $q$ and drift velocity $\mathbf{v}$, experience a Lorentz force $\mathbf{F}=q \mathbf{v} \times \mathbf{B}$, which tends to push them to one side of the conductor. The resulting steady-state charge buildup causes a voltage difference $V_{\mathrm{H}}$, called the Hall voltage, between the sides of the conductor.
(a) Show that the Hall voltage is given by $$
V_{\mathrm{H}}=\frac{B I}{n q t}
$$
where $n$ is the concentration of charge carriers. [Hint:
In steady state, there is no net force on the charge carriers transverse to the current direction, so the force due to the transverse $E$ field must be canceled by the force from the $B$ field.] (b) Show, with an appropriate diagram, that the sign of the Hall voltage gives the sign of the charge carriers. Thus, the Hall voltage allows one to determine whether the carriers in a doped semiconductor are electrons or holes. (c) Compute the magnitude of the Hall voltage in an $n$ -type semiconductor sample at room temperature with donor concentration $$
\begin{aligned}
&N_{\mathrm{d}}=10^{22} \mathrm{~m}^{-3} \text { and the following parameters: } t=1 \mathrm{~mm} \text { , }\\
&I=0.01 \mathrm{~A}, B=1.0 \mathrm{~T}
\end{aligned}
$$