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Question

A copper wire of cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) carries a current of \( 1.5
\, \text{A} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \,
\text{C} \), what is the drift speed?

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Explanation

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.5/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1 × 10⁻⁶) . Calculate: v_d = (1.5/1.36 × 10⁴) ≈ 1.1 × 10⁻⁴ m/s . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε

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