Q 12-04-043JEE MainJEE Main 2026 (4 Apr, Shift 1)Hard
An insulated wire is wound so that it forms a flat coil with $N = 200$ turns. The radius of the innermost turn is $r_1 = 3$ cm, and of the outermost turn $r_2 = 6$ cm. If $20$ mA current flows in it then the magnetic moment will be $\alpha \times 10^{-2}\ \text{A.m}^2$. The value of $\alpha$ is ______.
Answer: (B) $2.64$
The turns are spread evenly between $r_1$ and $r_2$: $dN = \dfrac{N}{r_2 - r_1}dr$.
$$M = \int I\pi r^2\,dN = \frac{NI\pi}{r_2 - r_1}\cdot\frac{r_2^3 - r_1^3}{3} = \frac{NI\pi}{3}(r_1^2 + r_1r_2 + r_2^2)$$
$= \dfrac{200 \times 0.02 \times \pi}{3}(9 + 18 + 36) \times 10^{-4} = 84\pi \times 10^{-4} \approx 2.64 \times 10^{-2}\ \text{A m}^2$.
Solution by Sreeraj P, M.Sc Physics