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(B) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $\vec{	au} = \vec{M} 	imes \vec{B}_0$,where $\vec{M}$ i

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$\pi N Q B_0 R^2$

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(B) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $\vec{	au} = \vec{M} 	imes \vec{B}_0$,where $\vec{M}$ is the magnetic moment of the coil.The magnitude of the torque is $	au = M B_0 \sin(	heta)$. Since the plane of the coil is vertical and the magnetic field $B_0$ is horizontal and parallel to the plane of the coil,the angle between the area vector (normal to the plane) and the magnetic field is $	heta = 90^\circ$. Thus,$\sin(90^\circ) = 1$.The magnetic moment of the coil is $M = N i A = N i (\pi R^2)$.Therefore,the torque is $	au = N i \pi R^2 B_0$.From the angular impulse-momentum theorem,the change in angular momentum $L$ is given by $\Delta L = \int 	au dt$.Substituting the expression for torque: $\Delta L = \int (N i \pi R^2 B_0) dt = N \pi R^2 B_0 \int i dt$.Since the total charge $Q$ discharged through the coil is $Q = \int i dt$,we get:$L = N \pi R^2 B_0 Q$.

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