Q67 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
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, $\pi = 3.14$)
- Correct answerA$6.28$
- B$6.28 \times 10^{-6}$
- C$628$
- D$6.28 \times 10^{-4}$
Solution
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: $6.28$
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: $6.28$
Hence, the correct answer is option: 'A'.