A bar magnet with poles 25.0 cm apart and of pole strength 14.4 Am rests with its centre on a friction less point. It is held in equilibrium at 600 to a uniform magnetic field of induction 0.25 T by applying a force F at right angle to the axis, 12 cm from its pivot. The magnitude of the force is
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a
153 N
b
753 N
c
3.753 N
d
253N
answer is C.
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Detailed Solution
Anticlockwise torque = clockwise torque.F x 0.12 = MB sin θF=3.6×0.25×3/20.12=3.753 N