A bar magnet of magnetic moment 3.0A−m2is placed in a uniform magnetic induction field of2×10−5T. If each pole of the magnet experiences a force of6×104N, the length of the magnet is
A
0.5 m
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B
0.3 m
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C
0.2 m
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D
0.1 m
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Solution
The correct option is D 0.1 m F=mB⇒F=ML×B ⇒6×10−4=3L×2×10−5⇒L=0.1m