Q 12-06-031JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
A small cube of side $1$ mm is placed at the centre of a circular loop of radius $10$ cm carrying a current of $2$ A. The magnetic energy stored inside the cube is $\alpha \times 10^{-14}$ J. The value of $\alpha$ is ______. ($\mu_0 = 4\pi \times 10^{-7}$ Tm/A, $\pi = 3.14$)
Answer: (A) $6.28$
Field at the centre: $B = \dfrac{\mu_0I}{2R} = \dfrac{4\pi \times 10^{-7} \times 2}{0.2} = 4\pi \times 10^{-6}$ T (uniform over the tiny cube).
Energy density: $u = \dfrac{B^2}{2\mu_0} = \dfrac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5}$ J/m³.
Energy in the cube: $u \times (10^{-3})^3 = 6.28 \times 10^{-14}$ J, so $\alpha = 6.28$.
Solution by Sreeraj P, M.Sc Physics