Question

A current carrying circular loop of radius r is placed in the x-y plane with center at the origin. Half of the loop with x > 0 is now bent so that it now lies in y - z plane. Magnitude of change in magnetic dipole momentum is [$I$ is current in loop]

• A. $I\pi r^2\frac{(\sqrt{2}-1)}{/sqrt2}$
• B. $\frac{I\pi r^2}{2}$
• C. $\frac{I\pi r^2}{\sqrt{2}}$
• D. $\frac{I\pi r^2\sqrt{2}}{(\sqrt{2}-1)}$

Answer 1

The correct option is C $\frac{I\pi r^2}{\sqrt{2}}$

${\overrightarrow{\mu }}_i=I\pi R^2\hat{k}$
assuming wire is bent towards negative z plane
${\overrightarrow{\mu }}_f=\frac{I\pi R^2}{2}\hat{k}+\frac{I\pi R^2}{2}\hat{i}$
$\Delta \overrightarrow{\mu }={\overrightarrow{\mu }}_f-{\overrightarrow{\mu }}_i=\frac{\pi r^2}{2}I(\hat{i}-\hat{k})$.
So the magnitude of the change is $\frac{\pi r^{2}I}{2}\times \sqrt{2}=\frac{\pi r^{2}I}{\sqrt{2}}$