Question

In the figure shown a coil of single turn is wound on a sphere of radius rand mass $m$ . The plane of the coil is parallel to the inclined plane and lies in the equatorial plane of the sphere. If sphere is in rotational equilibrium, the value of B external is (current in the coil is I):-

• A. $\frac{mg}{\pi Ir}$
• B. $\frac{mg\sin \theta }{\pi I}$
• C. $\frac{mgr\sin \theta }{nI}$
• D. none of these

Answer 1

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

$\begin{array}{c}\text{For the rotational equilibrium we can say -}\\ \text{that sum of all the torque must be zero}\\ \Rightarrow \text{toqgue due to friction = Torque due to coil}\end{array}$

$\begin{array}{c}\text{Torque due to friction}{\tau }_f=f\cdot r=\text{mgresintheta}\\ \text{torque due to coil}{\tau }_c=MB\sin \theta M\to \text{magnetic moment.}\\ {\tau }_c=n\cdot \pi r^2\cdot I\cdot B\sin \theta n\to \text{turns.}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow \text{from}& {\tau }_f={\tau }_c\\ & mg{r}_r\sin \theta =n\pi r^{2}IB\sin \theta \\ \Rightarrow & B=\frac{mg}{n\pi Ir}\end{array}\\ & \text{for single twen}n=I\\ & B=\frac{mg}{\pi Ir}\end{array}$