Angular Momentum Gained by Coil Discharging Capacitor in Magnetic Field

To determine the magnitude of the angular momentum acquired by the coil, we calculate the magnetic torque exerted on the coil during the capacitor's discharge and integrate it over time.

1. Magnetic Dipole Moment of the Coil

The magnetic dipole moment $\vec{\mu}$ of a coil carrying an instantaneous current $I(t)$ with $N$ turns and radius $R$ is given by: $\vec{\mu}(t) = N I(t) \vec{A}$

where $\vec{A}$ is the area vector of the coil. The magnitude of the area vector is: $A = \pi R^2$

The vector $\vec{A}$ points normal to the vertical plane of the loop.

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2. Magnetic Torque on the Loop

The magnetic field $\vec{B}_o$ is horizontal and parallel to the plane of the loop. Therefore, the angle between the normal vector $\vec{A}$ (and hence $\vec{\mu}$) and the magnetic field $\vec{B}_o$ is $\theta = 90^\circ$.

The torque $\vec{\tau}(t)$ acting on the coil is given by: $\vec{\tau}(t) = \vec{\mu}(t) \times \vec{B}_o$

The magnitude of the torque at any instant $t$ is: $\tau(t) = |\vec{\mu}(t)| |\vec{B}_o| \sin(90^\circ) = N I(t) (\pi R^2) B_o$

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3. Angular Momentum Gained by the Coil

The angular momentum $L$ acquired by the coil is equal to the angular impulse of the magnetic torque over the discharge duration $\Delta t$: $L = \int_{0}^{\Delta t} \tau(t) \, dt$

Substituting the expression for $\tau(t)$: $L = \int_{0}^{\Delta t} N I(t) \pi R^2 B_o \, dt$

Since the discharge happens extremely fast and the coil hardly rotates during this time, the orientation of the loop remains unchanged during the integration. Thus, $N, \pi, R,$ and $B_o$ remain constant and can be taken out of the integral: $L = \pi N B_o R^2 \int_{0}^{\Delta t} I(t) \, dt$

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4. Total Charge Discharged

The integral of current over the discharge time represents the total charge $Q$ originally stored on the capacitor: $\int_{0}^{\Delta t} I(t) \, dt = Q$

Substituting this into the expression for angular momentum yields: $L = \pi N Q B_o R^2$

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Conclusion

Thus, the magnitude of the angular momentum gained by the coil is: $\mathbf{\pi N Q B_o R^2}$

This corresponds to Option (B).