JEE Main 2026 — Magnetic Effects of Current Question with Solution

Let the number of turns per unit radial width be $n=\frac{N}{r_2-r_1}$.

Consider an elemental ring of radius $r$ and width $dr$. The number of turns in this element is $dN=ndr$.

The magnetic moment of this elemental ring is $dM=IdNA=I\left(\frac{N}{r_2-r_1}dr\right)(\pi r^2)$.

Integrating from $r_1$ to $r_2$ to find the total magnetic moment $M$:

$M={\int }_{r_1}^{r_2}\frac{I\pi N}{r_2-r_1}{r}^{2}dr=\frac{I\pi N}{3(r_2-r_1)}(r_2^3-r_1^3)$

Substituting the given values: $I=20\times 10^{-3}$ A, $N=200$, $r_1=0.03$ m, and $r_2=0.06$ m:

$M=\frac{20\times 10^{-3}\times \pi \times 200}{3(0.06-0.03)}\left((0.06{)}^3-(0.03{)}^3\right)$

$M=\frac{4\pi }{0.09}(216\times 10^{-6}-27\times 10^{-6})$

$M=\frac{4\pi }{0.09}\times 189\times 10^{-6}=4\pi \times 2100\times 10^{-6}$

$M=84\pi \times 10^{-4}\approx 84\times 3.14\times 10^{-4}=263.76\times 10^{-4}\approx 2.64\times 10^{-2}$ A.m${}^2$

Comparing with $\alpha \times 10^{-2}$ A.m${}^2$, we get $\alpha =2.64$.

Answer: $2.64$