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A bar magnet with poles 25.0 cm apart an...

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`

Text Solution

Verified by Experts

The correct Answer is:
C
Anticlockwise torque=clockwise torque.
`F=0.12=MB sin theta`
`F=(3.6xx025xxsqrt3//2)/(0.12)=3.75sqrt3N`
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