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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a
2π x 10-7 Nm
b
2π x 10-5 Nm
c
0.5 Nm
d
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