A semiconductor strip of thickness 0.500 mm carries 20.0 mA along +x in a 0.800 T field along +z. The measured Hall voltage across its width is −12.0 mV, with V(+y)−V(−y) = −12.0 mV. Determine the sign of the mobile carriers and their number density magnitude using V_H = BI/(nqt). Explain the transverse force balance, find the Hall coefficient, and identify one source of systematic error from contact placement.

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Carriers are positive; n = 1.67 × 10²² m⁻³. Hall coefficient R_H = +3.74 × 10⁻⁴ m³ C⁻¹. Magnetic and electric transverse forces balance in steady state. For positive carriers drifting along +x with B along +z, q(v_d × B) points toward −y. Positive charge accumulates there, so V(−y) exceeds V(+y); the measured negative voltage therefore identifies positive carriers. Using magnitudes, n = BI/(|q|t|V_H|) = (0.800)(0.0200)/[(1.602 × 10⁻¹⁹)(0.500 × 10⁻³)(0.0120)] = 1.67 × 10²² m⁻³. At steady state transverse electric force qE_H balances magnetic force qv_dB, so charge no longer accumulates at the edges. Hall coefficient is R_H = 1/(nq) including carrier sign, giving +1/[(1.67 × 10²²)(1.602 × 10⁻¹⁹)] = +3.74 × 10⁻⁴ m³ C⁻¹. Misaligned Hall contacts can include an ordinary longitudinal voltage. Reverse B and take half the difference of the two Hall readings; the Hall term reverses while a fixed contact-offset voltage largely cancels.

Use the Hall effect to identify charge carriers

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