A circular coil of radius 2 cm has 125 turns and carries a current of 1 A. It is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the magnetic field. If the torque is α×10⁻⁴ N·m, take π = 3.14 and find α.
Physics · Magnetic effect of current · mabnetic moment and torque on current carring loop in magnetic field
JEE Mains 2026 4 April Evening Shift

Question

A circular coil of radius 2 cm has 125 turns and carries a current of 1 A. It is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the magnetic field. If the torque is α×10⁻⁴ N·m, take π = 3.14 and find α.

Physics · Magnetic effect of current · mabnetic moment and torque on current carring loop in magnetic field

JEE Mains 2026 4 April Evening Shift

G1 Physics · Accepted Answer

The magnetic dipole moment of the coil is M = NIA, where A = πr². The torque magnitude is τ = MB sinθ = NIπr²B sinθ. Substitute: N = 125, I = 1 A, r = 2 cm = 2×10⁻² m, B = 0.4 T, θ = 30°. Therefore, τ = 125×1×3.14×(2×10⁻²)²×0.4×(1 ÷ 2) = 3.14×10⁻² N·m = 314×10⁻⁴ N·m. Hence α = 314.