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In an isotropic single-carrier Drude model, let the electric current density point along $x$, with zero transverse current and magnetic field along $z$. The Hall coefficient is $R_H=1/(nq)$ and the longitudinal electrical conductivity is $\sigma =n|q|\mu$, where the charge carrier mobility $\mu$ is positive. Consequently $E_y/E_x=R_H\sigma B=\mathrm{sgn}(q)\mu B$. This signed Hall angle convention describes the angle of the electric field relative to the current; a convention measuring current relative to field has the opposite sign.

For one isotropic carrier species in the Drude model, the longitudinal electrical resistivity at zero transverse electric current is ${\rho }_{xx}=1/(ne\mu )$, and the Hall electrical resistivity has magnitude $|{\rho }_{yx}|=B/(ne)$. Equivalently, the electric field balances the magnetic deflection through $E_y=v_{x}B$, while $|v_x|=\mu |E_x|$. Therefore the magnitude of the Hall angle obeysThe required ratio givesThe electron density cancels. With electric current along $+x$ and field along $+z$, electrons have negative Hall coefficient and hence $E_y<0$; the requested one-percent condition concerns the magnitude. This calculation assumes electrons dominate transport and the transverse contacts draw no electric current, so a Hall field can develop.

For a single carrier species with positive charge carrier mobility $\mu$, the Hall coefficient and longitudinal electrical conductivity areTake the electric current density along $x$, magnetic field along $z$, and no transverse current. Then $E_y=R_H{j}_{x}B$ and $j_x=\sigma E_x$, giving the Hall angle relationThe sign specifies the electron Hall polarity; the ratio of field magnitudes is $|E_H|/|E|=\mu |B|$. Converting the stated electron mobility to SI gives $\mu =900\times 10^{-4}=0.0900{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. ThereforeThis uses the homogeneous single-carrier Drude model; the carrier density cancels.