Skip to content

Question

A conductor has a resistivity of \( 6 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times
10^{-3} \, ^\circ\text{C}^{-1} \) at \( 20^\circ \text{C} \). What is its resistivity at \( 90^\circ
\text{C} \)?

Options

Choose one · Correct answer highlighted

Explanation

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 6 × 10⁻⁸ [1 + 4 × 10⁻³ (90 - 20)] . Calculate: rho_t = 6 × 10⁻⁸ [1 + 0.28] = 6 × 10⁻⁸ × 1.28 = 7.68 × 10⁻⁸ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t

Discussion

Comments

0 comments

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