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Hall voltage and mobility, copper strip

Short answer

Hall voltage V_H = IB/(net) ≈ 9.75 μV; Hall mobility μ_H = 1/(neρ) ≈ 4.32×10⁻³ m²V⁻¹s⁻¹.

A copper strip 3.0 cm wide and 2.0 mm thick is placed in a magnetic field 1.75T. If a current of 150 A is setup in the strip, calculate (i) Hall voltage and (ii) Hall mobility if the number of electrons per unit volume is 8.4×10²⁸ m⁻³ and resistivity is 1.72×10⁻⁸ ohm-m.

Formulas: the Hall field balances the magnetic force, eE_H = e v_d B. For a strip of width w and thickness t, v_d = I/(n e w t), so

  • V_H = IB/(n e t) (t is the thickness along B)
  • Hall coefficient R_H = 1/(ne) (magnitude)
  • Hall mobility μ_H = |R_H|/ρ = 1/(neρ)

Given: w = 3.0 cm, t = 2.0 mm, B = 1.75 T, I = 150 A, n = 8.4×10²⁸ m⁻³, ρ = 1.72×10⁻⁸ Ω·m, e = 1.602×10⁻¹⁹ C.

(i) Hall voltage

n e t = 8.4×10²⁸ × 1.602×10⁻¹⁹ × 2×10⁻³ = 2.691×10⁷
V_H   = (150 × 1.75) / 2.691×10⁷
      = 262.5 / 2.691×10⁷
      = 9.75×10⁻⁶ 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 ≈ 9.75 μV; μ_H ≈ 4.32×10⁻³ m²V⁻¹s⁻¹

✓ Answer checked by Claude Opus 5.5 (high effort)

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