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Free electron density from Hall emf

Short answer

Use Hall voltage V_H = IB/(n e t) → n = IB/(e t V_H). With I=150 A, B=2.0 T, thickness t=1.5×10⁻³ m, V_H=1.79×10⁻⁵ V, n ≈ 7.0×10²⁸ m⁻³.

A flat silver strip of width 1.5cm and thickness 1.5mm carries a current of 150A. A magnetic field of 2.0 Tesla is applied perpendicular to the flat face of the strip. The emf developed across the width of the strip is measured to be 17.9μV. Estimate the number density of free electrons in the metal.

From the Hall effect, V_H = IB/(n e t), where t is the thickness along the field. So

n = IB/(e t V_H)

Given: I = 150 A, B = 2.0 T, t = 1.5 mm = 1.5×10⁻³ m, V_H = 17.9 μV = 1.79×10⁻⁵ V, e = 1.602×10⁻¹⁹ C. (The width 1.5 cm is not needed.)

n = (150 × 2.0) / (1.602×10⁻¹⁹ × 1.5×10⁻³ × 1.79×10⁻⁵)
  = 300 / (4.301×10⁻²⁷)
  = 6.97×10²⁸ m⁻³

Answer: n ≈ 7.0×10²⁸ electrons/m³, close to one free electron per silver atom (5.86×10²⁸ m⁻³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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