JEE Main 2026 Physics question
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 . (π=3.14)
Answer: 314
Solution
τ = NIAB sinθ = 125×1×π(0.02)²×0.4×sin30° = 0.0314 N·m = 314×10⁻⁴ N·m.