Home
Class 12
PHYSICS
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

Promotional Banner

Similar Questions

Explore conceptually related problems

A bar magnet with poles 25cm apart and pole strength 14.4 Am rests with its center on a frictionless pivot. If it is held in equilibrium at 60^(0) to a uniform magnetic field on induction 0.25 T by applying a force F at right angles to its axis 10cm from the pivolt, the value of F in newton is (nearly)

A bar magnetic with poles 25cm apart and pole strength 14*4A.m rests with its centre on a frictionless pivot. It is held in equilibrium at 60^@ to a uniform magnetic field of induction 0*25T by applying a force F at right angles to its axis, 10cm from the pivot. Calculate the value of F. What will happen if the force is removed?

A bar magnet with poles sqrt3cm apart and pole strength 14*4Am rests with its centre on a frictionless pivot. The magnet is held in equilibrium at an angle of 60^@ to a uniform magnetic field of intensity 0*20T by applying a force F at right angles to its axis, 0*1m from its pivot. What is the value of force F?

A bar magnet with poles 25 cm apart and of strength 14.4 amp - m rests with centre on a frictionless pivot. It is held in equilibrium aat an angle of 60@ with respect to a uniform magnetic field of induction 0.25 Wb//m^(2) , by applying a force F at right angles to its axis at a point 12 cm from pivot. Calculate F . What will happen if the force F is removed?

A bar magnet with poles 25 cm apart and of strength 14.4 amp - m rests with centre on a frictionless pivot. It is held in equilibrium aat an angle of 60^@ with respect to a uniform magnetic field of induction 0.25 Wb//m^(2) , by applying a force F at right angles to its axis at a point 10 cm from pivot. Calculate F.

A bar magnet with its poles 25 cm apart and of pole strength 24.0 A-m rests with its centre 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