A bar magnet having a magnetic moment of 2.0 x 10 5 JT -1 , is placed along the direction of uniform magnetic field of magnitude B= 14 x 10 -5 T. The work done in rotating the magnet slowly through 60° from the direction of field is:
Text Solution
Verified by ExpertsThe correct answer is:
A
Work done = MB (cos 
- cos 
)
= 2x 10 5 x 14 x 10 -5 (1-1/2)
= 14J
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