Q. A circular loop carrying current i 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. Then new magnetic moment of the loop will be

A

$I\piR^2 \sqrt{2}$

B

$\frac {I\piR^2}{2}$

C

$2\pi IR^2$

D

$\frac {I\piR^2}{\sqrt {2}}$

Solution:

$M_{eq} = \left(\frac{I\pi R^{2}}{2}\right)\sqrt{2}$

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