Question

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 ————. $(\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}, \pi = 3.14)$

• A. 6.28
• B. $6.28 \times 10^{-6}$
• C. 628
• D. $6.28 \times 10^{-4}$

Hint:

Calculate the magnetic field B at the center of the loop, then find the energy density (B^2 / 2*mu_0) and multiply by the volume of the cube.

Answer 1

The Correct Option is A

The magnetic energy $U$ stored in a small volume $V$ where the magnetic field is approximately uniform is:
$$U = \frac{B^2}{2\mu_0} V$$
For a circular loop of radius $R$ and current $I$, the field at the center is $B = \frac{\mu_0 I}{2R}$.
Substituting $B$ into the energy equation:
$$U = \frac{(\frac{\mu_0 I}{2R})^2}{2\mu_0} V = \frac{\mu_0^2 I^2}{4R^2 \cdot 2\mu_0} V = \frac{\mu_0 I^2 V}{8R^2}$$
Now, plug in the numerical values:
$\mu_0 = 4\pi \times 10^{-7}$ Tm/A
$I = 2$ A
$R = 0.1$ m
$V = (1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9} \text{ m}^3$

$$U = \frac{(4\pi \times 10^{-7}) \times (2)^2 \times 10^{-9}}{8 \times (0.1)^2}$$
$$U = \frac{4\pi \times 10^{-7} \times 4 \times 10^{-9}}{8 \times 0.01}$$
$$U = \frac{16\pi \times 10^{-16}}{0.08} = \frac{16\pi \times 10^{-16}}{8 \times 10^{-2}}$$
$$U = 2\pi \times 10^{-14} \text{ J}$$
Given $U = \alpha \times 10^{-14}$ J, we have $\alpha = 2\pi = 2 \times 3.14 = 6.28$.