Question

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 is
A. `15sqrt3N`
B. `75sqrt3N`
C. `3.75sqrt3N`
D. `25sqrt3N`

Answer 1

Correct Answer - C
Anticlockwise torque=clockwise torque.
`F=0.12=MB sin theta`
`F=(3.6xx025xxsqrt3//2)/(0.12)=3.75sqrt3N`