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A bar magnet having a magnetic moment of...

A bar magnet having a magnetic moment of `2xx10^(4) JT^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B=6xx10^(-4)T` exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction `60^(@)` from the field is

A

2J

B

0.6J

C

12J

D

6J

Text Solution

Verified by Experts

The correct Answer is:
D
`W=MB(cos theta_(1)-cos theta _(2))`
`= 2 xx 10^(4) xx 6xx 10^(-4)( cos theta - cos 60^(@))`
`=12xx(1)/(2)=6J`
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