JEE Main 2026 · 4 Apr · Shift 1 · Physics Q41

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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 $\alpha\times10^{-2}$ A.m². The value of $\alpha$ is ________.

4.4

2.64

3.25

1.2

Answer and worked solution

Official answer: B

Turns per unit radius $\frac N{r_2-r_1}$ . A ring of width $dr$ has moment $I\pi r^2\frac{N\,dr}{r_2-r_1}$ :
$m=\frac{NI\pi(r_2^3-r_1^3)}{3(r_2-r_1)}=\frac{200(0.02)\pi(216-27)\times10^{-6}}{0.09}=2.64\times10^{-2}\text{ A m}^2.$

(B)

Topics:Torque on a current loop

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JEE Main 2026 · 4 Apr · Shift 1, Physics question 41