 Solution

A current-carrying sample in a transverse magnetic field develops a transverse electric field because the Lorentz force deflects its charge carriers. In steady state this Hall voltage balances the transverse deflection. With Hall coefficient $R_H=E_y/(j_x{B}_z)$, one isotropic electron band gives $R_H=-1/(ne)$ and one hole band gives $R_H=+1/(pe)$.

When Electrons and electron holes both conduct, the weak-field two-carrier result isThe two-carrier Hall coefficient weights each species by mobility squared. Its sign changes whenTemperature changes dopant ionization, intrinsic electron-hole generation, and the two charge carrier mobilities, so it can move the system across this condition. For example, a p-type semiconductor can be hole-dominated in its extrinsic regime but become negative in its intrinsic regime if $n\simeq p$ and ${\mu }_e>{\mu }_h$. A sign change is possible, not mandatory: in many n-type semiconductors Electrons dominate at both temperatures. The density majority alone does not determine the Hall sign in a two-carrier material.