In Fig. 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]

Options

A. `mgcos theta//(piIR)`
B. `mg//piIR`
C. `mgtantheta//piIR`
D. `mgsin theta//piIR`

Answer

B

Explanation

Magnetic dipole moment of the coil carrying current, `M=IpiR^2`. Its direction is perpendicular to the plane upwards. Resolving `vecB` into two rectangular components we have, `Bcostheta` acts perpendicular to the plane downwards and `Bsin theta` acts along the plane downwards.

For rotational equilibrium of sphere about point P, `(mg sin theta)R=M(Bsin theta)`
or `mgRsin theta=IpiR^2Bsin theta` or `B=(mg)/(piIR)`