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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.

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.

Choose conventional surface current density along $+x$, magnetic field along $+z$, and negative electron charge. An isotropic in-plane response has the following sheet conductivity and sheet resistivity tensors for $s\ge 0$, $k>0$:These are matrix inverses, rather than componentwise reciprocals. The sheet conductivity has units siemens, the sheet resistivity ohms. With $K_y=0$, the sign is $E_y/K_x<0$, agreeing with the negative Hall coefficient of a single electron species. Contact polarity can reverse the measured voltage sign, so quote the Hall magnitude. For a homogeneous strip of width $W$ and longitudinal probe separation $L$, $I=W{K}_x$ givesAt $s=0$, $k=N$, these reduce to $R_K/N$ and zero. A nonzero transverse sheet resistivity creates no local Joule heating: the antisymmetric contribution cancels in $\mathbf{E}\cdot \mathbf{K}=(e^2/h)s|\mathbf{E}{|}^2$.

For an original nonideal example, choose $k=2$, $s=0.01$ and $L/W=2$. Then $V_{xx}/I\simeq 129.061\Omega$, while the Hall magnitude is below $R_K/2$ by the relative amount $s^2/(k^2+s^2)\simeq 25.0$ parts per million. An unwanted admixture $0.02V_{xx}$ in the Hall readout would instead introduce approximately $2.58\Omega$. The chosen calculation shows why a small intrinsic tensor correction does not by itself guarantee an accurate contact measurement.

For isotropic electrons and electron holes with positive mobilities, adding their weak-field electrical conductivity tensors and imposing zero transverse total current givesThe Hall coefficient changes sign at $p{\mu }_h^2=n{\mu }_e^2$. Carrier abundance alone does not fix the sign; mobilities enter quadratically. The formula assumes ${\mu }_{e}B\ll 1$, ${\mu }_{h}B\ll 1$, independent semiclassical carriers, and no anomalous Hall contribution.