In the figure shown,a coil of single turn is | vedclass

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(B) For the sphere to be in equilibrium,the net torque about the point of contact $O$ must be zero.The gravitational force $mg$ acts downwards through

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$\frac{mg}{\pi i R}$

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(B) For the sphere to be in equilibrium,the net torque about the point of contact $O$ must be zero.The gravitational force $mg$ acts downwards through the center of the sphere. The lever arm for the gravitational force about the point of contact $O$ is $R \sin 	heta$.Therefore,the gravitational torque is $	au_{g} = mg R \sin 	heta$.The magnetic moment of the coil is $\mu = i A = i (\pi R^2)$.The magnetic torque on the coil in a uniform magnetic field $B$ is $	au_{m} = \mu B \sin \phi$,where $\phi$ is the angle between the magnetic moment vector and the magnetic field vector. Since the coil plane is parallel to the incline,the magnetic moment vector is perpendicular to the incline. The magnetic field $B$ is vertical. The angle between the vertical and the normal to the incline is $	heta$. Thus,$\phi = 	heta$.Equating the torques: $	au_{m} = 	au_{g}$$i \pi R^2 B \sin 	heta = mg R \sin 	heta$Solving for $B$: $B = \frac{mg}{\pi i R}$.

In the figure shown,a coil of single turn is wound on a sphere of radius $R$ and mass $m$. The plane of the coil is parallel to the inclined plane and lies in the equatorial plane of the sphere. The current in the coil is $i$. If the sphere is in equilibrium,the value of the magnetic field $B$ is:

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