An insulated wire is wound so that it forms a flat coil with N = 200 turns. The radius of the innermost turn is r₁ = 3 cm, and of the outermost turn r₂ = 6 cm. If 20 mA current flows in it then the magnetic moment will be α × 10⁻² A.m². The value of α is ______. (a) 4.4 (b) 2.64 (c) 3.25 (d) 1.2

Answer: (b). Turns per unit radius = (N)/(r₂ - r₁); a ring of width dr has dm = Iπ r²(N)/(r₂ - r₁)dr. m = (Iπ N)/(r₂ - r₁)·(r₂³ - r₁³)/(3) = (Iπ N)/(3)(r₂² + r₁r₂ + r₁²) m = (0.02 × π × 200)/(3)(36 + 18 + 9)×10⁻⁴ = 84π×10⁻⁴ m ≈ 2.64×10⁻² A m² So the answer is (b) 2.64.

An insulated wire is wound so that it forms a flat coil with N = 200… | JEE Main 2026 (4 Apr, Shift 1)

Q18JEE Main 2026 · 4 APR · SHIFT 1Difficult

Current Loop as a Magnetic Dipole

An insulated wire is wound so that it forms a flat coil with $N = 200$ turns. The radius of the innermost turn is $r_1 = 3\text{ cm}$ , and of the outermost turn $r_2 = 6\text{ cm}$ . 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 ______.

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Answer: (b)

Turns per unit radius $= \frac{N}{r_2 - r_1}$ ; a ring of width $dr$ has $dm = I\pi r^2\frac{N}{r_2 - r_1}dr$ .
$m = \frac{I\pi N}{r_2 - r_1}\cdot\frac{r_2^3 - r_1^3}{3} = \frac{I\pi N}{3}(r_2^2 + r_1r_2 + r_1^2)$
$m = \frac{0.02 \times \pi \times 200}{3}(36 + 18 + 9)\times10^{-4} = 84\pi\times10^{-4}$
$m \approx 2.64\times10^{-2}\text{ A m}^2$
So the answer is (b) 2.64.

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