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Q.

A short bar magnet is placed in a uniform magnetic field. If the work done to turn the magnet from the direction of field to normal to field is W then potential energy of the magnet when it is in stable equilibrium position is

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a

zero

b

W

c

–W/2

d

–W

answer is B.

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Detailed Solution

W=MBcosθ1−cosθ2

=MBcos0−cos90=M.B

U=−M¯.B¯=−MBcosθ

⇒U=−W

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