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A bar magnet having a magnetic moment of 2 x 104 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10-4 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:

Option: 1

12 J


Option: 2

6 J


Option: 3

2 J


Option: 4

0.6 J


Answers (1)

best_answer

 work done = u_{f}-u_{i}

= -MB .\cos 60^{\circ}-\left ( -MB \cos 0^{\circ} \right )

= \frac{MB}{2}= \frac{2\times 10^{4}\times 6\times 10^{-4}}{2}J

= 6J

Posted by

vishal kumar

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