= Solution
Take a two-dimensional conducting sheet in the $xy$ plane with <magnetic flux density> $\mathbf B=B\hat{\mathbf z}$ and <electric current density> along $x$. Moving carriers feel the transverse <Lorentz force> $q\mathbf v_d\times\mathbf B$. They accumulate at opposite edges, producing a transverse <electric field> until its force balances the magnetic deflection:
$$
q(\mathbf E_H+\mathbf v_d\times\mathbf B)=0.
$$
For the transverse-voltage convention $E_y/j_x$, this gives $E_y=v_xB$ and
$$
\frac{E_y}{j_x}=\frac{B}{nq}.
$$
The <Hall effect> therefore measures carrier sign as well as density in the single-carrier approximation: electrons give the opposite transverse-voltage sign from positive carriers. Some tensor conventions call $E_y/j_x$ $\rho_{yx}$ rather than $\rho_{xy}$; the physical sign is fixed by the stated lead and field directions.
The <Hall voltage> is the integral of this transverse <electric field> across the sample width. In a three-dimensional slab of thickness $t$, its magnitude is $|V_H|=IB/(n|q|t)$; for a sheet, use the sheet number density $n_s$ and omit $t$.
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