A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field Bo parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)-
(A)2πNQB0R2
(B)πNQB0R2✓ Correct
(C)2πNQB0R2
(D)4πNQB0R2
Answer: (B)
Solution
The torque on the coil is τ = M × B, where M is the magnetic moment. At the initial instant shown, the angle between M and B is π/2, so τ = iπR²NB₀. Integrating the angular impulse: ∫τ dt = ΔL = ∫iπR²NB₀ dt = πR²NB₀∫i dt = πR²NB₀Q. Therefore, the angular momentum gained is πNQB₀R².
JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Single correct option