Sample width is fixed at w = 10 mm; elementary charge q = 1.602e-19 C. The animation shows carrier drift and deflection; in steady state the Hall field balances the Lorentz force.
Carriers drift along current I, the field B (out of page) exerts a Lorentz force that deflects them, they pile up on one face, the Hall field builds and the paths straighten at steady state. Toggle the carrier type and the accumulation face and V_H sign flip.
X = B [T], Y = V_H. Slope = I/(nqt); the sign flips with carrier type. The yellow dot marks the current operating point.
In steady state on a Hall bar, the Lorentz force on the carriers is balanced by the electric field from the charge that has piled up at the sides. This gives the Hall voltage $V_H$ in terms of the current $I$, the magnetic field $B$, the carrier density $n$, and the sample thickness $t$.
Hall voltage:
$$V_H = \frac{I\,B}{n\,q\,t}$$Hall coefficient and Hall field:
$$R_H = \frac{1}{n\,q},\qquad E_H = \frac{V_H}{w}$$Current density:
$$j = \frac{I}{w\,t}$$Steady-state balance condition:
$$q\,E_H = q\,v_d\,B,\qquad v_d=\frac{I}{nqwt}$$Here $q = 1.602\times10^{-19}$ C is the elementary charge, $w$ the sample width, and $t$ the sample thickness. The sign of $R_H$ indicates the carrier type — negative for electrons, positive for holes — and the animation flips the accumulation face and the V_H sign accordingly.
Validation: $I=10$ mA, $B=0.5$ T, $n=10^{23}$/m³, $t=100$ μm → $V_H\approx 3.12$ mV, $R_H\approx 6.24\times10^{-5}$ m³/C. For copper ($n=8.5\times10^{28}$) $V_H\approx 0.37$ μV (tiny), so lower-density semiconductors give a much larger $V_H$.