Compute the skin depth for copper (σ = 5.8×10⁷ S/m, μ ≈ μ₀ = 4π×10⁻⁷ H/m) at (a) 50 Hz, (b) 1.0 MHz. (c) At 1 MHz, what fraction of the surface field remains at 0.20 mm depth? Step 1: δ = √(2/(μσω)). At 50 Hz: μσω = (4π×10⁻⁷)(5.8×10⁷)(2π×50) = (4π×10⁻⁷)(5.8×10⁷)(314) ≈ 2.29×10⁴. δ = √(2/2.29×10⁴) = √(8.73×10⁻⁵) ≈ 9.3×10⁻³ m ≈ 9.3 mm. Step 2: At 1 MHz, ω is 20,000× larger, so δ shrinks by √20000 ≈ 141: δ ≈ 9.3 mm/141 ≈ 66 μm. Step 3: Field ratio = e^(−z/δ) = e^(−200/66) = e^(−3.03) ≈ 0.048 — only ~5% of the field survives 0.2 mm in; 95% of the current flows in the outer 0.2 mm skin. This is why RF conductors are tubes or plated: the middle is dead weight.