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Hall voltage, coefficient and mobility

Short answer

V_H = IB/(n e t) = (23×0.65)/(8.5×10^28 × 1.60×10^-19 × 1.5×10^-4) ≈ 7.3 μV. Hall coefficient R_H = 1/(n e) ≈ −7.3×10^-11 m^3/C. Mobility μ_H = |R_H|/ρ ≈ 4.3×10^-3 m^2 V^-1 s^-1.

A copper strip 150μm thick is placed in a magnetic field of strength 0.65T perpendicular to the plane of the strip and current of 23Amp is set up in the strip. Calculate: (i) the Hall voltage (ii) Hall coefficient and (iii) Hall mobility, if the number of electrons per unit volume is 8.5×10²⁸/m³ and resistivity is 1.72×10⁻⁸ Ohm-m.

Formulas: V_H = IB/(n e t); R_H = 1/(ne) (negative for electrons, magnitude used); μ_H = |R_H|/ρ.

Given: t = 150 μm = 1.5×10⁻⁴ m, B = 0.65 T, I = 23 A, n = 8.5×10²⁸ m⁻³, ρ = 1.72×10⁻⁸ Ω·m, e = 1.602×10⁻¹⁹ C.

(i) Hall voltage

n e t = 8.5×10²⁸ × 1.602×10⁻¹⁹ × 1.5×10⁻⁴ = 2.0426×10⁶
V_H   = (23 × 0.65) / 2.0426×10⁶
      = 14.95 / 2.0426×10⁶
      = 7.32×10⁻⁶ V

(ii) Hall coefficient

R_H = 1/(ne) = 1/(8.5×10²⁸ × 1.602×10⁻¹⁹)
    = 7.34×10⁻¹¹ m³/C   (negative sign for electrons)

(iii) Hall mobility

μ_H = |R_H|/ρ = 7.34×10⁻¹¹ / 1.72×10⁻⁸
    = 4.27×10⁻³ m²/(V·s)

Answer: (i) V_H ≈ 7.3 μV; (ii) R_H ≈ −7.3×10⁻¹¹ m³/C; (iii) μ_H ≈ 4.3×10⁻³ m²V⁻¹s⁻¹

✓ Answer checked by Claude Opus 5.5 (high effort)

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