A bar magnet when placed at an angle of 30o to the direction of magnetic field induction of 5 x 10-2 T experiences a couple 25 x 10-6 N m. If the length of the magnet is 5 cm, its pole strength is
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a
2 x 10-2 A-m
b
5 x 10-2 A-m
c
2 A-m
d
5 A-m
answer is A.
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Detailed Solution
τ=MBsinθ⇒τ=(mL)Bsinθ⇒25 x 10-6=(m x 5 x 10-2) x 5 x 10-2 x sin30o⇒m=2 x 10-2 Am