JEE Advanced 2020 · Paper 1 · Physics Q4

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PhysicsSingle correct+3 / -1Question 4 of 54

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 $B_o$ 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 circular coil hangs, plane vertical, from two parallel wires. The wires lead up to a circuit containing a capacitor (top) and an open switch (right). Horizontal arrows of a uniform magnetic field B_o pass across the coil, parallel to its plane.

\dfrac{\pi}{2}NQB_oR^2

\pi NQB_oR^2

2\pi NQB_oR^2

4\pi NQB_oR^2

Answer and worked solution

Official answer: B

The field $B_o$ lies in the plane of the coil, so the magnetic moment $NIA$ is perpendicular to $B_o$ and the torque is maximal:
$\tau=NI\pi R^2B_o.$

The coil barely turns while the capacitor discharges, so this stays the torque throughout. The angular momentum gained is the angular impulse:
$L=\int\tau\,dt=N\pi R^2B_o\int I\,dt=\pi NQB_oR^2.$
(B)

Topics:Torque on a current loop Angular momentum and its conservation

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JEE Advanced 2020 · Paper 1, Physics question 4