Question: Q. 4. 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) $M B$. (b) $\frac{\sqrt{3} M B}{2}$. (c) $\frac{M B}{2}$. (d) zero.
[NCERT Exemplar]
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Solution:
Ans. Correct option : (d)
Explanation : The work done to rotate the loop in magnetic field $W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)$.
When current carrying coil is rotated then there will be no change in angle between magnetic moment and magnetic field.
Here, $\theta_{1}=\theta_{2}=\alpha$
$\Rightarrow \quad W=M B(\cos \alpha-\cos \alpha)=0$.
