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 _______. (\(\mu_o = 4\pi \times 10^{-7}\) Tm/A, \(\pi = 3.14\))

The magnetic field at the centre of the circular loop is given by \(B = \dfrac{\mu_0 I}{2R}\)

Substituting the given values, \(B = \dfrac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6}\) T

The magnetic energy density is \(u = \dfrac{B^2}{2\mu_0}\)

\(u = \dfrac{(4\pi \times 10^{-6})^2}{2 \times 4\pi \times 10^{-7}} = \dfrac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5}\) J/m\(^3\)

The volume of the small cube is \(V = (1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9}\) m\(^3\)

The magnetic energy stored inside the cube is \(U = u \times V\)

\(U = 2\pi \times 10^{-5} \times 10^{-9} = 2\pi \times 10^{-14}\) J

Using \(\pi = 3.14\), we get \(U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14}\) J

Comparing with \(\alpha \times 10^{-14}\) J, we get \(\alpha = 6.28\)

Answer: \(6.28\)