 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 $R_H=E_y/(j_x{B}_z)$, 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 isThe 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.