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Question

A copper wire of cross-sectional area \( 5 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1 \,
\text{A} \). 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

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 5 × 10⁻⁷) . Calculate: v_d = (1/6.8 × 10³) ≈ 1.47 × 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 - I r and P = I²R,

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