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# 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 about an axis perpendicular to its plane is:MB√3MB2zeroMB2

A
3MB2
B
zero
C
MB
D
MB2
Solution
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#### The rotation of the loop by 30° about an axis perpendicular to its plane make no change in the angle made by axis of the loop with the direction of magnetic field, therefore, the work done to rotate the loop is zero.Important point: The work done to rotate the loop the loop in magnetic field W=MB(cos θ1−cos θ2), where signs are as usual.

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