 Solution

The Hall effect is a transverse electric response to a magnetic force on moving charge carriers. Take conventional current density $\mathbf{J}=J_x\hat{\mathbf{x}}$ and magnetic field $\mathbf{B}=B_z\hat{\mathbf{z}}$ in a conducting slab. For one carrier type of number density $n$ and signed electric charge $q$, the drift velocity is ${\mathbf{v}}_d=\mathbf{J}/(nq)$. Carriers initially deflect sideways until a Hall electric field balances the magnetic Lorentz force:The Hall coefficient is thereforeFor Electrons it is $-1/(ne)$, whereas a simple hole-dominated conductor gives $+1/(pe)$. The sign reveals the effective carrier sign; the magnitude estimates their density in the one-carrier approximation.

For width $w$ along $y$ and thickness $t$ along $z$, $I=J_{x}wt$. Define the signed Hall voltage by $V_H=E_{y}w$, namely the electric potential at the lower-$y$ edge minus that at the higher-$y$ edge. ThenThis explicit contact convention fixes the voltage sign. A Hall bar with longitudinal and transverse contacts measures both electrical resistivity and the Hall response. Reverse the field and take the part of the transverse voltage odd in $B_z$ to suppress longitudinal-voltage pickup from misaligned contacts; reversing current provides another useful check.

A Hall measurement is useful for carrier identification, semiconductor doping measurements and magnetic-field sensing. Combining a one-carrier $R_H$ with electrical conductivity gives a mobility magnitude $\mu =\sigma |R_H|$. However, mixed Electron and hole conduction is not described by a single density: in weak fields,The response weights mobilities differently from ordinary conductivity. Complicated energy bands, anisotropic Fermi surfaces and multiple bands can likewise change the simple interpretation in metals; a positive Hall coefficient does not mean that the material contains positively charged fundamental Electrons.