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 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 ————. $(\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}, \pi = 3.14)$
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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.
Updated On: Apr 9, 2026
- 6.28
- $6.28 \times 10^{-6}$
- 628
- $6.28 \times 10^{-4}$
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The Correct Option is A
Solution and Explanation
To find the magnetic energy stored in a volume, we use the magnetic energy density formula. The energy density $u_B$ (energy per unit volume) in a magnetic field $B$ is given by:
$$u_B = \frac{B^2}{2\mu_0}$$
First, let's find the magnetic field $B$ at the center of the circular loop. The formula for the magnetic field at the center of a loop of radius $R$ carrying current $I$ is:
$$B = \frac{\mu_0 I}{2R}$$
Given: $I = 2$ A, $R = 10$ cm $= 0.1$ m, and $\mu_0 = 4\pi \times 10^{-7}$ Tm/A.
$$B = \frac{(4\pi \times 10^{-7}) \times 2}{2 \times 0.1} = \frac{4\pi \times 10^{-7}}{0.1} = 40\pi \times 10^{-7} = 4\pi \times 10^{-6} \text{ T}$$
Next, calculate the energy density $u_B$:
$$u_B = \frac{(4\pi \times 10^{-6})^2}{2 \times (4\pi \times 10^{-7})} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5} \text{ J/m}^3$$
The total energy $U$ stored in the cube of side $a = 1$ mm $= 10^{-3}$ m is the product of energy density and the volume of the cube ($V = a^3$):
$$V = (10^{-3})^3 = 10^{-9} \text{ m}^3$$
$$U = u_B \times V = (2\pi \times 10^{-5}) \times 10^{-9} = 2\pi \times 10^{-14} \text{ J}$$
Using $\pi = 3.14$:
$$U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} \text{ J}$$
Comparing this with $\alpha \times 10^{-14}$, we find $\alpha = 6.28$.
$$u_B = \frac{B^2}{2\mu_0}$$
First, let's find the magnetic field $B$ at the center of the circular loop. The formula for the magnetic field at the center of a loop of radius $R$ carrying current $I$ is:
$$B = \frac{\mu_0 I}{2R}$$
Given: $I = 2$ A, $R = 10$ cm $= 0.1$ m, and $\mu_0 = 4\pi \times 10^{-7}$ Tm/A.
$$B = \frac{(4\pi \times 10^{-7}) \times 2}{2 \times 0.1} = \frac{4\pi \times 10^{-7}}{0.1} = 40\pi \times 10^{-7} = 4\pi \times 10^{-6} \text{ T}$$
Next, calculate the energy density $u_B$:
$$u_B = \frac{(4\pi \times 10^{-6})^2}{2 \times (4\pi \times 10^{-7})} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5} \text{ J/m}^3$$
The total energy $U$ stored in the cube of side $a = 1$ mm $= 10^{-3}$ m is the product of energy density and the volume of the cube ($V = a^3$):
$$V = (10^{-3})^3 = 10^{-9} \text{ m}^3$$
$$U = u_B \times V = (2\pi \times 10^{-5}) \times 10^{-9} = 2\pi \times 10^{-14} \text{ J}$$
Using $\pi = 3.14$:
$$U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} \text{ J}$$
Comparing this with $\alpha \times 10^{-14}$, we find $\alpha = 6.28$.
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