A bar magnet of length 10 cm and having the pole strength equal to 10-3Wb is kept in a magnetic field having magnetic induction (B) equal to 4π x 10-3T. It makes an angle of 30o with the direction of magnetic induction. The value of the torque acting on the magnet is

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2π x 10-7 Nm

2π x 10-5 Nm

0.5 Nm

0.5 x 102 Nm

answer is A.

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Detailed Solution

Torque, τ = MB sinθ= 0.1 x l0-3 x 4π x10-3 x sin30o =10-7 x4π x 12= 2π x l0-7 N m

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