Question

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 plane and lies in the equitorial plane of sphere. Current in the coil is $i$. Then find the value of $B$ if sphere is in its equilibrium

• A. $\frac{mg\cos \theta }{\pi iR}$
• B. $\frac{mg}{\pi iR}$
• C. $\frac{mg\tan \theta }{\pi iR}$
• D. $\frac{mg\sin \theta }{\pi iR}$

Answer 1

The correct option is B $\frac{mg}{\pi iR}$

The current carrying coil will experience a torque due to magnetic field.

Hence for sphere to be in equilibrium, this torque must be balanced by some other torque.
${\tau }_{mg}=0$ as $mg$ passes through centre.

Thus there must be presence of friction which can give the required torque and also balance the $mg\sin \theta$ component of weight of the sphere.


For translational equilibrium,

$f=mg\sin \theta ........\left(i\right)$

The torque due to friction will tend to rotate sphere in clockwise sense. Hence the direction of current in coil should be such that it produces torque which tend to rotate in anticlockwise sense.


Thus we can infer that a clock wise sense of current will lead to direction of magnetic moment $\overrightarrow{\mu }$ such that $\overrightarrow{\mu }\times \overrightarrow{B}=\overrightarrow{\tau }$ will give the torque in the desired direction.

Let angle between $\overrightarrow{\mu }$ and $\overrightarrow{B}$ is $\theta$, as shown in figure.

Now for rotational equilibrium ;

${\tau }_B={\tau }_f$

$\mu B\sin \left(\theta \right)=fR$

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

$B=\frac{mg\sin \theta R}{\lambda R^{2}i\sin \theta }$

$\therefore B=\frac{mg}{\pi iR}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (b) is the correct answer.