A bar magnet having a magnetic moment of 2.0 × 105 JT-1, is placed along the direction of uniform magnetic field of magnitude B = 14 × 10-5 T. The work done in rotating the magnet slowly through 60° from the direction of field is:

Option 1 -

14 J

Option 2 -

8.4 J

Option 3 -

4 J

Option 4 -

1.4 J

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 1


    Detailed Solution:

     u=MBcosθ2 (MBcosθ1)

    =MB (cosθ2cosθ1)=2×105×14×105 (cos60°cos0°)

    =28 (121)

    =282=14J

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