Question

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 B0 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. $\frac{\pi}{2} NQB _{o} R ^{2}$
• B. $\pi NQB _{o} R ^{2}$
• C. $2 \pi NQB _{o} R ^{2}$
• D. $4 \pi NQB _{o} R ^{2}$

Answer 1

Answer: (B)

$\pi NQB _{o} R ^{2}$

Answer 2

The torque on a magnetic dipole $\vec{\mu}$ in a magnetic field $\vec{B}_0$ is given by $\vec{\tau} = \vec{\mu} \times \vec{B}_0$. The magnitude of the torque is $\tau = \mu B_0 \sin\theta$. The magnetic dipole moment of the coil is $\vec{\mu} = NI\vec{A}$, where $N$ is the number of turns, $I$ is the current, and $\vec{A}$ is the area vector. The magnitude of the area vector is $A = \pi R^2$. Since the magnetic field $B_0$ is horizontal and parallel to the plane of the coil (which is vertical), the magnetic field is perpendicular to the normal vector of the coil's plane. Thus, $\theta = 90^\circ$, and $\sin\theta = 1$. The torque magnitude is $\tau = (NIA)B_0 = NI(\pi R^2)B_0$. The change in angular momentum $\Delta L$ is the impulse of the torque: $\Delta L = \int \tau dt$. Since the discharge time is very short and the coil hardly rotates, the angle between $\vec{\mu}$ and $\vec{B}_0$ remains $90^\circ$, so $\tau$ is proportional to $I$. $\Delta L = \int NI(\pi R^2)B_0 dt = N(\pi R^2)B_0 \int I dt$. The integral of current $I$ over the discharge time is the total charge $Q$ that flows from the capacitor. Therefore, $\Delta L = N(\pi R^2)B_0 Q = \pi N Q B_0 R^2$.