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Question

If X amount of work is required to rotate a bar magnet in a magnetic field by 600, from a position parallel to the field, What is the torque required to maintain it in new position.

A
35X
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B
3X
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C
32X
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D
33X
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Solution

The correct option is B 3X
work done in rotating a magnet from 0 to 60 will be,
X=M×B×(1cosθ)
=M×B×(0.5)
2X=M×B taking θ=60
torqur required to maintain in that position will be ,
τ=M×B×Sinθ
τ=2X×(3)2 taking θ=60
=3X

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