= Solution
In a <Hall effect> measurement, a current flows along $x$ and a <magnetic field> is applied along $z$. The transverse <Lorentz force> initially deflects the mobile charges, building a transverse <electric field> until the net transverse current vanishes. Defining $R_H=E_y/(j_xB_z)$, the one-carrier <Drude model> gives $R_H=1/(nq)$. Thus electron-like carriers give $R_H=-1/(ne)$ and hole-like carriers give $R_H=+1/(pe)$.
A nearly filled band can conduct through <holes>: its missing electrons behave as positive mobile charges. Metals can have both electron-like and hole-like <Fermi surface> sheets. In the weak-field two-carrier approximation the <Hall coefficient> is
$$
\boxed{R_H=\frac{p\mu_h^2-n\mu_e^2}{e(p\mu_h+n\mu_e)^2}.}
$$
The sign depends on the carrier densities and squared mobilities; it need not identify the most numerous carrier. Experimentally the <Hall voltage> changes sign when the field or current reverses, allowing subtraction of unwanted longitudinal-voltage pickup. Anisotropic bands and different scattering rates complicate a literal conversion of $R_H$ into a carrier count.
Back to article page
