Question

A circular current loop of magnetic moment $M$ is in an arbitrary orientation in an external magnetic field $B$. The work done to rotate the loop by ${30}^{\circ }$ about an axis perpendicular to its plane is

• A. zero
• B. $MB$
• C. $\frac{MB}{2}$
• D. $\sqrt{3}\frac{MB}{2}$

Answer 1

The correct option is A zero

As per the diagram, the rotation of the loop by ${30}^{\circ }$ about an axis perpendicular to its plane makes no variation in the angle made by axis of the loop with the direction of magnetic field, i.e.


Therefore, the work done to rotate the loop is given by,

$W=MB(\cos {\theta }_1-\cos {\theta }_2)$

Hence,

$W=MB(\cos \alpha -\cos \alpha )=0$

Work done to rotate the loop is zero.

$\text{Final Answer}:$ (d)