Biot-Savart Law & Circular Coils question from JEE Main 2026 (6 Apr, Shift 1)

Answer: (a)

1. Field at the centre: $B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 2}{0.2} = 4\pi \times 10^{-6}\text{ T}$.
2. Energy density: $u = \frac{B^2}{2\mu_0} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5}\text{ J/m}^3$.
3. Cube volume $= (10^{-3})^3 = 10^{-9}\text{ m}^3$, so $U = 2\pi \times 10^{-14} = 6.28 \times 10^{-14}\text{ J}$.
4. So the answer is (a) $\alpha = 6.28$.