# 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 at an angle of 60° with respect to a uniform magnetic field of induction 0.25 Wb/m2, by applying a force F at right angles to its axis at a point 12 cm from pivot Calculate F. What will happened if the force F is removed?

Navjyot Kalra
8 years ago
Hello Student,
2ℓ = 0.25 m
Also, m x 2ℓ = 14.4
⇒ m = 14.4 / 0.25 = 57.6 A-m2
Torque due to magnetic field
= pm x B x sin 60°
= 14.4 x 0.25 x √3/2
The torque due to the force
= F x 0.12
For equilibrium F x 0.12 = 14.4 x 0.25 x √3/2
⇒ F = 25.98 N
If the force F is removed, the torque due to magnetic field will move the bar magnet. It will start oscillating about the mean position. Where the angle between$\underset{P}{\rightarrow}$m and $\underset{B}{\rightarrow}$is 0.
Thanks
Navjot Kalra