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) $6.28$
Concept. The field at the centre of the loop, $\dfrac{\mu_0I}{2R}$, is uniform over the tiny cube. Energy $=u\times$volume with $u=\dfrac{B^2}{2\mu_0}$.
Step 1. Field. $B=\dfrac{4\pi\times10^{-7}\times2}{2\times0.1}=1.256\times10^{-5}$ T.
Step 2. Energy density. $u=\dfrac{(1.256\times10^{-5})^2}{2\times4\pi\times10^{-7}}=6.28\times10^{-5}$ J/m$^3$.
Step 3. Energy. $6.28\times10^{-5}\times10^{-9}=6.28\times10^{-14}$ J.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
