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Question: A bar magnet with it's poles 25 *cm* apart and of pole strength 24 *amp*×*m* rests with it'scentre o...

A bar magnet with it's poles 25 cm apart and of pole strength 24 amp×m rests with it'scentre on a frictionless pivot. A force F is applied on the magnet at a distance of 12 cm from the pivot so that it is held in equilibrium at an angle of 30° with respect to a magnetic field of induction 0.25 T. The value of force F is

A

5.62 N

B

2.56 N

C

6.52 N

D

6.25 N

Answer

6.25 N

Explanation

Solution

In equilibrium

Magnetic torque = Deflecting torque

⇒ MBsin⁡θ=F.dM B \sin \theta = F . d or F=mlBsin⁡θdF = \frac { m l B \sin \theta } { d }