In figure. 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. If the sphere is in rotational equilibrium, the value of B is [Current in the coil is i]

A non-conducting non-magnetic rod having circular cross section of radius R is suspended from a rigid support as shown in the figure. A light and small coil of 300 turns is wrapped tightly at the left end of the rod where uniform magnetic field B exists in vertically downward direction.Air of density $\rho$ hits the half of the right part of the rod with velocity V as shown in the figure. What should be current in clockwise direction (as seen from O) in the coil so that rod remains horizontal? Give answer in m A. Given

$\frac{2}{Lv}\sqrt{\frac{\pi RB}{\rho }}=\frac{1}{\sqrt{5}}{ \mathrm{A}}^{-1/2}$