Worked example: A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) of length 50 m has resistance 0.85 Ω. (a) Find its cross-sectional area and diameter. (b) What length of nichrome wire (ρ = 1.1 × 10⁻⁶ Ω·m) of the same diameter gives the same resistance? Step 1: A = ρL/R = (1.7 × 10⁻⁸)(50)/0.85 = 1.0 × 10⁻⁶ m² = 1.0 mm². Diameter d = √(4A/π) = √(1.273 × 10⁻⁶) ≈ 1.13 × 10⁻³ m ≈ 1.1 mm. Step 2: For the same R and A, L = RA/ρ = (0.85)(10⁻⁶)/(1.1 × 10⁻⁶) ≈ 0.77 m. Step 3: Nichrome needs only 0.77 m versus copper's 50 m for the same resistance — 65× shorter — which is exactly why heater elements use nichrome: compact coils, big resistance. Conductivity check: σ_Cu = 1/ρ ≈ 5.9 × 10⁷ S/m.