A straight copper wire of radius 1.0 mm carries 10 A DC; the electric field along the wire is E = 5.41×10⁻² V/m (driving the current against resistance). (a) Find B at the wire surface. (b) Find the Poynting vector magnitude and direction there. (c) Show the total power flowing in equals I²R per unit length (resistivity ρ = 1.7×10⁻⁸ Ω·m). Step 1: B = μ₀I/(2πr) = (4π×10⁻⁷ × 10)/(2π × 10⁻³) = 2.0×10⁻³ T, circling the wire. Step 2: S = EB/μ₀ = (5.41×10⁻²)(2.0×10⁻³)/(4π×10⁻⁷) ≈ 86.1 W/m², pointing radially inward (E along the wire × B around the wire = inward). Step 3: Power per unit length = S × (surface area per unit length) = 86.1 × 2πr = 86.1 × 6.283×10⁻³ ≈ 0.541 W/m. Check against I²R: resistance per metre R = ρ/(πr²) = 1.7×10⁻⁸/(π×10⁻⁶) ≈ 5.41×10⁻³ Ω/m; I²R = 100 × 5.41×10⁻³ = 0.541 W/m ✓. Energy flows in from the surrounding field and becomes heat — it does not travel through the wire's interior.