A bar magnet with poles 25 cm apart and of strengt

Question

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 · Accepted Answer

Hello Student,

Please find the answer to your question

2ℓ = 0.25 m

Also, m x 2ℓ = 14.4

⇒ m = 14.4 / 0.25 = 57.6 A-m²

Torque due to magnetic field

= p_m 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
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