= 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> $R_H=E_y/(j_xB_z)$, 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 is
$$
\boxed{R_H=\frac{p\mu_h^2-n\mu_e^2}{e(p\mu_h+n\mu_e)^2},\qquad
\sigma=e(n\mu_e+p\mu_h).}
$$
The <two-carrier Hall coefficient> weights each species by mobility squared. Its sign changes when
$$
\boxed{p\mu_h^2=n\mu_e^2.}
$$
<Temperature> 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.
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