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Q. A current carrying circular loop of radius $R$ is placed in the $x - y$ plane with centre at the origin. Half of the loop with $x > 0$ is now bent so that it now lies in the $y - z$ plane

Moving Charges and Magnetism

Solution:

For a circular loop of radius $R$ carrying current $I$ placed in $x-y$ plane, the magnetic moment $M = I\times \pi R^2$. It acts perpendicular to the loop i.e., along $z$-direction. When half of the current loop is bent in $y-z$ plane, then magnetic moment due to half current loop in $x-y$ plane, $M_1 = I(\pi R^2/2)$ acting along $z$-direction. Magnetic moment due to half current loop in $y - z$ plane, $M_2 = I(\pi R^2/2)$ along $x$ -direction.
Effective magnetic moment due to entire bent current loop,
$M' = \sqrt{M_{1}^{2}+M_{2}^{2}} $
$ = \sqrt{\left(I\pi R^{2} /2\right)^{2} + \left(I\pi R^{2} /2\right)^{2}} $
$ = \frac{I\pi R^{2}}{2} \sqrt{2} < M$
i.e., magnetic moment diminishes.