StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A circular coil of radius R having N turns and negligible resistance hangs vertically (plane vertical) from two light wires. The wires connect the coil, through a switch, to a capacitor carrying charge Q. A uniform horizontal magnetic field B0 lies in the plane of the coil. On closing the switch the capacitor discharges completely through the coil in an extremely short time, during which the coil barely rotates. Find the magnitude of the angular momentum acquired by the coil.

  1. (A) (pi/2) N Q B0 R²
  2. (B) pi N Q B0 R²
  3. (C) 2 pi N Q B0 R²
  4. (D) 4 pi N Q B0 R²

Correct answer: (B) pi N Q B0 R²

Solution

The field lies in the plane of the coil, so the magnetic moment (normal to coil) is perpendicular to B0 and torque = N*I*A*B0. The angular momentum gained equals the angular impulse = integral(tau dt) = N*A*B0*integral(I dt) = N*A*B0*Q. With A = pi*R², L = pi N Q B0 R².

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →