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) $2.64$
Concept. A flat spiral has turns of every radius between $r_1$ and $r_2$; turns per unit radius $\dfrac{N}{r_2-r_1}$. Add $I\pi r^2$ over all turns.
Step 1. Integral. $M=\dfrac{NI\pi}{r_2-r_1}\displaystyle\int_{r_1}^{r_2}r^2dr=\dfrac{NI\pi(r_2^3-r_1^3)}{3(r_2-r_1)}$.
Step 2. Value. $\dfrac{200\times0.02\times\pi\times189\times10^{-6}}{0.09}=2.64\times10^{-2}$ A m$^2$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
