Skip to content

Question

A copper wire carries \( 4.5 \, \text{A} \) with a drift speed of \( 1.8 \times 10^{-4} \, \text{m/s}
\). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what
is the cross-sectional area?

Options

Choose one · Correct answer highlighted

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) . Rearrange: A = (I/n e v_d) . Substitute: A = (4.5/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.8 × 10⁻⁴) . Calculate: A = (4.5/2.448 × 10⁵) ≈ 1.84 × 10⁻⁵ m² . 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

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.