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Drift velocity from Hall experiment

Short answer

Drift velocity v_d = V_H/(B w) = 2.4×10⁻³ m/s. Carrier density n = I/(e v_d w t) ≈ 7.4×10²⁸ m⁻³.

In a Hall Effect experiment, a current of 3.2A lengthwise in a conductor 1.2 cm wide, 4.0 cm long and 9.5μm thick produces a transverse Hall voltage (across the width) of 40μV when a magnetic field of 1.4T is passed perpendicularly through the thin conductor. From this data, find (a) the drift velocity of the charge carriers and (b) the number density of charge carriers.

Formulas: the Hall field balances the magnetic force, E_H = v_d B, and V_H = E_H w. The carrier density follows from I = n e v_d (w t).

Given: I = 3.2 A, w = 1.2 cm = 0.012 m, t = 9.5 μm = 9.5×10⁻⁶ m, V_H = 40 μV = 4×10⁻⁵ V, B = 1.4 T.

(a) Drift velocity

v_d = V_H / (B w) = 4×10⁻⁵ / (1.4 × 0.012)
    = 4×10⁻⁵ / 0.0168
    = 2.38×10⁻³ m/s

(b) Number density

n = I / (e v_d w t)
  = 3.2 / (1.602×10⁻¹⁹ × 2.38×10⁻³ × 0.012 × 9.5×10⁻⁶)
  = 3.2 / (4.35×10⁻²⁹)
  = 7.36×10²⁸ m⁻³

(check with n = IB/(e t V_H) = 3.2×1.4 / (1.602×10⁻¹⁹ × 9.5×10⁻⁶ × 4×10⁻⁵) = 7.36×10²⁸ m⁻³)

Answer: (a) v_d ≈ 2.4×10⁻³ m/s; (b) n ≈ 7.4×10²⁸ m⁻³ (the length 4.0 cm is not needed).

✓ Answer checked by Claude Opus 5.5 (high effort)

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