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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 `60^(@)` to a uniform magnetic field of induction `0.25 T` by applying a formce `F` at right angle to the axis, `12 cm` from its pivot. The magnitude of the force isA. `15sqrt3N`B. `75sqrt3N`C. `3.75sqrt3N`D. `25sqrt3N` |
Answer» Correct Answer - C Anticlockwise torque=clockwise torque. `F=0.12=MB sin theta` `F=(3.6xx025xxsqrt3//2)/(0.12)=3.75sqrt3N` |
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