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A bar magnet of magnetic moment 3.0Am2 is placed in a uniform induction field of 2×10−5T . If each pole of the magnet experiences a force of 3×10−4N , the length of the magnet is

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a

0.5

b

0.3

c

0.2

d

0.1

answer is C.

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Detailed Solution

F=mB But m=M2l ⇒F=M2lB ⇒2l=MBF=3×2×10−53×10−4=0.2 m

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