Magnetic Energy Stored Inside Small Cube at Center of Circular Loop

To find the magnetic energy stored inside the small cube placed at the center of the circular loop, we proceed step-by-step:

1. Calculate the Magnetic Field at the Center of the Circular Loop: The magnitude of the magnetic field $B$ at the center of a circular loop of radius $R$ carrying current $I$ is given by: $B = \frac{\mu_0 I}{2R}$

   Given:

   • Current, $I = 2\text{ A}$
   • Radius of the loop, $R = 10\text{ cm} = 0.1\text{ m} = 10^{-1}\text{ m}$
   • Permeability of free space, $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$

   Substituting the given values into the formula: $B = \frac{(4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}) \times 2\text{ A}}{2 \times 0.1\text{ m}} = 4\pi \times 10^{-6}\text{ T}$

2. Calculate the Magnetic Energy Density: The magnetic energy density $u_B$ (magnetic energy per unit volume) in a magnetic field $B$ is given by: $u_B = \frac{B^2}{2\mu_0}$

   Substituting the values of $B$ and $\mu_0$: $u_B = \frac{(4\pi \times 10^{-6}\text{ T})^2}{2 \times (4\pi \times 10^{-7}\text{ T}\cdot\text{m/A})} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5}\text{ J/m}^3$

3. Calculate the Total Magnetic Energy Stored in the Cube: Since the cube is very small (side $a = 1\text{ mm} = 10^{-3}\text{ m}$), the magnetic field inside the cube can be assumed to be uniform and equal to $B$.

   The volume $V$ of the cube is: $V = a^3 = (10^{-3}\text{ m})^3 = 10^{-9}\text{ m}^3$

   The total magnetic energy $U$ stored inside the cube is: $U = u_B \times V$ $U = (2\pi \times 10^{-5}\text{ J/m}^3) \times (10^{-9}\text{ m}^3) = 2\pi \times 10^{-14}\text{ J}$

   Using $\pi = 3.14$: $U = 2 \times 3.14 \times 10^{-14}\text{ J} = 6.28 \times 10^{-14}\text{ J}$

4. Determine the Value of $\alpha$: Given that $U = \alpha \times 10^{-14}\text{ J}$, we get: $\alpha = 6.28$

Correct Option: A