Question

A coil has moment of inertia 0.8 kg/m^2 released in uniform magnetic field 4T when there is 60° angle between magnetic field and magnetic moment of coil. magnetic moment of coil is 20 A-m^2 find the angular speed of coil when it passes through stable equilibrium​

Answer 1

It has given that, moment of inertia of coil, I = 0.8 kgm² , magnetic field , B = 4T angle between magnetic field and magnetic moment, θ = 60° , magnetic moment, M = 20 Am²

To find : the angular speed of coil when it passes through stable equilibrium.

solution : magnet will be in stable equilibrium when angle between magnetic field and magnetic moment equals 0°.

from conservation of energy,

rotational kinetic energy of coil = -change in potential energy

⇒1/2 Iω² = -(Uf - Ui)

⇒1/2 × 0.8 × ω² = Ui - Uf = -MBcos60° - (-MBcos0°)

⇒0.4 × ω² = MB(1 - 1/2) = MB/2

⇒ω² = (20 × 4)/(2 × 0.4) = 100

⇒ω = 10 rad/s

Therefore the angular speed of coil is 10 rad/s