The energy stored in the magnetic field at a solenoid 50 cm long and 5 cm diameter wound with 1000 turns of wire carrying a current at 10 A is
- A. 0.025 joules
- B. 0.25 joules
- C. 0.5 joules
- D. 1.25 joules
Answer: B. 0.25 joules
Explanation: Inductance L = µ0 N² A / l = (4π×10^-7 × 1000² × π×0.025²) / 0.5 ≈ 4.935 mH. Energy E = 1/2 L I² = 0.5 × 4.935×10^-3 × 10² ≈ 0.25 J.