Step-by-Step Derivation:

1. Understanding the Physics: The problem asks for the angular momentum gained by the coil. According to the angular impulse-momentum theorem, the change in angular momentum ($ΔL$) of an object is equal to the angular impulse ($J_τ$) applied to it. Since the coil starts from rest, the angular momentum gained is equal to the total angular impulse. $ΔL = L_{final} - L_{initial} = L_{final} - 0 = J_τ$ The angular impulse is the integral of the torque ($τ$) over time. $J_τ = \int τ(t) dt$

2. Calculating the Torque on the Coil: A current-carrying coil in a magnetic field experiences a torque. The magnetic dipole moment ($μ$) of the coil with $N$ turns, area $A$, and current $i(t)$ is given by: $μ(t) = N i(t) A$ For a circular coil of radius $R$, the area is $A = πR^2$. So, $μ(t) = N i(t) (πR^2)$ The torque ($τ$) on this magnetic dipole in a uniform magnetic field $B_0$ is given by: $τ = μ × B_0$ The magnitude of the torque is $τ = μ B_0 \sin(θ)$, where $θ$ is the angle between the magnetic moment vector $μ$ and the magnetic field vector $B_0$.

3. Determining the Angle θ:

   • The magnetic field $B_0$ is horizontal and parallel to the plane of the coil.
   • The magnetic moment vector $μ$ is, by definition, perpendicular to the plane of the coil.
   • Therefore, the angle $θ$ between $μ$ and $B_0$ is $90°$ or $π/2$ radians.
   • This means $\sin(θ) = \sin(90°) = 1$, and the torque has its maximum magnitude.

4. Expressing the Torque Magnitude: Substituting the values into the torque magnitude equation: $τ(t) = μ(t) B_0 \sin(90°) = (N i(t) πR^2) B_0 (1)$ $τ(t) = N π R^2 B_0 i(t)$

5. Calculating the Angular Impulse: The capacitor discharges over a very short time, say from $t=0$ to $t=T$. The angular impulse is the integral of the torque over this period. The problem states that the discharge time is so short that the coil has hardly rotated. This justifies keeping the angle $θ$ constant at $90°$ throughout the discharge. $J_τ = \int_0^T τ(t) dt = \int_0^T (N π R^2 B_0 i(t)) dt$ Since $N$, $π$, $R$, and $B_0$ are constants, we can take them out of the integral: $J_τ = N π R^2 B_0 \int_0^T i(t) dt$

6. Relating Current Integral to Charge: The current $i(t)$ is the rate of flow of charge. The integral of the current over the discharge time gives the total charge that has passed through the coil. When the capacitor with initial charge $Q$ discharges fully, the total charge that flows through the coil is $Q$. $Q = \int_0^T i(t) dt$

7. Final Calculation of Angular Momentum: Substitute the result from step 6 into the equation for angular impulse from step 5: $J_τ = N π R^2 B_0 Q$ Since the angular momentum gained ($L$) is equal to the angular impulse ($J_τ$): $L = N Q B_0 π R^2$ Rearranging the terms to match the options: $L = π N Q B_0 R^2$

8. Comparing with Options: The calculated angular momentum is $π N Q B_0 R^2$. This matches option B.

   • A: ${\pi \over 2}NQ{B_0}{R^2}$ - Incorrect.
   • B: $\pi NQ{B_0}{R^2}$ - Correct.
   • C: $2\pi NQ{B_0}{R^2}$ - Incorrect.
   • D: $4\pi NQ{B_0}{R^2}$ - Incorrect.