A solenoid has 500 turns, length 0.50 m, and cross-sectional area 0.010 m². (a) Find its self-inductance. (b) If the current rises from 0 to 2.0 A in 0.020 s, find the induced emf. (c) Find the energy stored at 2.0 A. Step 1: L = μ₀N²A/ℓ = (4π×10⁻⁷)(500)²(0.010)/0.50. (500)² = 250,000; × 0.010 = 2500; /0.50 = 5000; × 4π×10⁻⁷ ≈ 6.28×10⁻³ H = 6.28 mH. Step 2: dI/dt = 2.0/0.020 = 100 A/s. ε = −L(dI/dt) = −(6.28×10⁻³)(100) = −0.628 V. The negative sign means it opposes the rise of current (it acts like a 0.628 V battery fighting the increase). Step 3: Stored energy U = ½LI² = 0.5 × 6.28×10⁻³ × 4 = 0.0126 J ≈ 12.6 mJ. This energy sits in the magnetic field and is returned to the circuit when the current falls.