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Q.

A bar magnet with a magnetic moment 5.0Am2 is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _________ 

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a

zero

b

2J

c

1J

d

4J

answer is B.

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Detailed Solution

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The potential energy stored in a bar magnet is given by 

U=−M→⋅B→

Therefore, work done to alter the position of the bar magnet can be calculated by w=ΔU=Uf−Ui=−mBcos⁡π−(−MBcos⁡0)

⇒w = 2mB = 2×5×0.4 = 4J

 

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