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Hall voltage and mobility in silver

Short answer

Using n = 5.86×10²⁸ m⁻³, Hall voltage V_H = IB/(n e t) = 3.73×10⁻⁵ V (≈ 37 μV). Electron mobility μ = σ/(n e) ≈ 7.2×10⁻³ m²V⁻¹s⁻¹.

In a Hall experiment, a current of 25A is passed through a long foil of silver, which is 0.1 mm thick and 3 m long. Calculate the Hall voltage produced across the width by a flux of 1.4 Wb/m². If the conduction of silver is 6.8×10⁷ mho/m, estimate the mobility of the electrons in silver.

Formulas: V_H = IB/(n e t) and, for mobility, μ = σ R_H = σ/(ne).

Assumption: the question does not give n, so I use the standard value for silver, one free electron per atom: n = 5.86×10²⁸ m⁻³. The foil length (3 m) is not needed. Here t = 0.1 mm, I = 25 A, B = 1.4 T, σ = 6.8×10⁷ S/m.

Hall voltage

n e t = 5.86×10²⁸ × 1.602×10⁻¹⁹ × 1×10⁻⁴ = 9.388×10⁵
V_H   = (25 × 1.4) / 9.388×10⁵
      = 35 / 9.388×10⁵
      = 3.73×10⁻⁵ V

Mobility

R_H = 1/(ne) = 1 / (5.86×10²⁸ × 1.602×10⁻¹⁹) = 1.065×10⁻¹⁰ m³/C
μ   = σ R_H  = 6.8×10⁷ × 1.065×10⁻¹⁰
    = 7.24×10⁻³ m²/(V·s)

Answer: V_H ≈ 37 μV; electron mobility ≈ 7.2×10⁻³ m²V⁻¹s⁻¹ (about 72 cm²/V·s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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