Question

In the given 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 due to downward magnetic field $B$, the value of magnetic field $B$ is (Current in the coil is $i$ and acceleration due to gravity $=g$)

• A. $\frac{mg}{\pi ir}$
• B. $\frac{mg\sin \theta }{\pi i}$
• C. $\frac{2mg\sin \theta }{\pi i}$
• D. none of these

Answer 1

The correct option is A $\frac{mg}{\pi ir}$

The gravitational torque must be counter balanced by the magnetic torque about O, for equilibrium of the sphere.
The magnetic torque

$\overrightarrow{{\tau }_m}=\overrightarrow{\mu }\times \overrightarrow{B}=\left(i\overrightarrow{A}\right)\times \overrightarrow{B}$

$|\overrightarrow{{\tau }_m}|=i\left(\pi r^2\right)B\sin \theta$

$|\overrightarrow{{\tau }_{gr}}|=mg\left(r\sin \theta \right)$

$|\overrightarrow{{\tau }_m}|=|\overrightarrow{{\tau }_{gr}}|$

$\Rightarrow i\left(\pi r^2\right)B\sin \theta =mg\left(r\sin \theta \right)$

$\Rightarrow B=\frac{mg}{\pi ir}$