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Hall voltage and mobility, wide copper strip

Short answer

V_H = IB/(n e t) = (200×1.5)/(8.4×10^28 × 1.60×10^-19 × 0.01) ≈ 2.2×10^-6 V. Hall mobility μ_H = 1/(n e ρ) = 1/(8.4×10^28 × 1.60×10^-19 × 1.72×10^-8) ≈ 4.3×10^-3 m^2 V^-1 s^-1.

A copper strip 2 cm wide and 10mm thick is placed in a magnetic field of 1.5T. If a current of 200A is set up in the strip, calculate (i) Hall voltage (ii) Hall mobility if the no. of electrons per unit volume is 8.4×10²⁸/m³ and resistivity is 1.72×10⁻⁸ Ohm-m.

Formulas: V_H = IB/(n e t) with t the thickness along B; Hall mobility μ_H = |R_H|/ρ = 1/(neρ).

Given: w = 2 cm, t = 10 mm = 1×10⁻² m (as printed), B = 1.5 T, I = 200 A, n = 8.4×10²⁸ m⁻³, ρ = 1.72×10⁻⁸ Ω·m.

(i) Hall voltage

n e t = 8.4×10²⁸ × 1.602×10⁻¹⁹ × 1×10⁻² = 1.3457×10⁸
V_H   = (200 × 1.5) / 1.3457×10⁸
      = 300 / 1.3457×10⁸
      = 2.23×10⁻⁶ V

(ii) Hall mobility

R_H = 1/(ne) = 1/(8.4×10²⁸ × 1.602×10⁻¹⁹) = 7.43×10⁻¹¹ m³/C
μ_H = R_H/ρ = 7.43×10⁻¹¹ / 1.72×10⁻⁸ = 4.32×10⁻³ m²/(V·s)

Answer: V_H ≈ 2.2 μV; μ_H ≈ 4.3×10⁻³ m²V⁻¹s⁻¹

✓ Answer checked by Claude Opus 5.5 (high effort)

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