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If a bar magnet of magnetic moment 'm' is freely suspended in a uniform magnetic field B, the work done in rotating the magnet through an angle `thete` is

A

`mB(1- sin theta)`

B

`mB sin theta`

C

`mB cos B`

D

`mB(1-costheta)`

Text Solution

Verified by Experts

The correct Answer is:
D
Work done W = `mB[cos theta_1 - cos theta_2]`
As here `theta_1 = 0^@ and theta_2 = theta` , hence W = `mB[cos 0^@ - cos theta] = mB(1-cos theta)`
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