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Question

A copper wire carries \( 4 \, \text{A} \) with a drift speed of \( 1.0 \times 10^{-4} \, \text{m/s} \).
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?

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Explanation

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁴) . Calculate: A = (4/1.36 × 10⁵) ≈ 2.94 × 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 - I r and P = I²R, evaluation

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