A bar magnet when placed at an angle of 30∘to the direction of magnetic field induction of5×10−2T, experiences a moment of couple 25×10−6N−m. If the length of the magnet is 5 cm its pole strength is
A
2×10−2A−m
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B
5×10−2A−m
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C
2A−m
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D
5A−m
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Solution
The correct option is A2×10−2A−m τ=MBsinθ⇒τ=(mL)Bsinθ ⇒25×10−6=(m×5×10−2)×5×10−2×sin30 ⇒m=2×10−2A−m