A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the direction of the magnetic field. The torque acting on the coil is α × 10⁻⁴ N.m. The value of α is 2em0.4pt. (π = 3.14) Numerical answer type. Work it out, then open the answer.

Answer: 314. τ = NIπ r² Bsin θ τ = 125 × 1 × 3.14 × (0.02)² × 0.4 × sin 30° τ = 0.157 × 0.4 × 0.5 = 3.14×10⁻² = 314×10⁻⁴ N m So α = 314.

A circular coil of radius 2 cm and 125 turns carries a current of 1… | JEE Main 2026 (4 Apr, Shift 2)

Q16JEE Main 2026 · 4 APR · SHIFT 2Medium

Current Loop as a Magnetic Dipole

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of $30^\circ$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times 10^{-4}\text{ N.m}$ . The value of $\alpha$ is $\rule{2em}{0.4pt}.\ (\pi = 3.14)$

Numerical answer type. Work it out, then open the answer.

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Answer: 314

$\tau = NI\pi r^2 B\sin\theta$
$\tau = 125 \times 1 \times 3.14 \times (0.02)^2 \times 0.4 \times \sin30^\circ$
$\tau = 0.157 \times 0.4 \times 0.5 = 3.14\times10^{-2} = 314\times10^{-4}\text{ N m}$
So $\alpha = 314$ .

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