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Drift speed in copper wire, 17 mA

Short answer

Electron density n = (density/M) N_A ≈ 8.49×10^28 m^-3. Cross-section A = π(9×10^-4)^2 ≈ 2.55×10^-6 m^2. Drift speed v_d = I/(nAe) ≈ 4.9×10^-7 m/s.

What is the drift speed of the conduction electrons in a copper wire (molecular mass = 63.54 g/mol, density 8.96 g/cm³) with radius 900 μm when it has a uniform current 17mA flowing in the wire?

The free-electron density is found from the density and molar mass, taking one free electron per copper atom. Then v_d = I/(nAe).

Given: M = 63.54 g/mol, density = 8.96 g/cm³ = 8.96×10³ kg/m³, r = 900 μm = 9×10⁻⁴ m, I = 17 mA = 1.7×10⁻² A, N_A = 6.022×10²³ mol⁻¹.

n = (density / M) × N_A
  = (8.96×10⁶ g/m³ / 63.54 g/mol) × 6.022×10²³
  = 1.410×10⁵ mol/m³ × 6.022×10²³
  = 8.49×10²⁸ m⁻³

A = π r² = π (9×10⁻⁴)² = 2.545×10⁻⁶ m²

v_d = I / (n A e)
    = 1.7×10⁻² / (8.49×10²⁸ × 2.545×10⁻⁶ × 1.602×10⁻¹⁹)
    = 1.7×10⁻² / 3.462×10⁴
    = 4.91×10⁻⁷ m/s

Answer: v_d ≈ 4.9×10⁻⁷ m/s (about 0.5 μm per second).

✓ Answer checked by Claude Opus 5.5 (high effort)

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