Show the time period of oscillation when a bar magnet is kept in a uniform magnetic field is T = `2pi sqrt((l)/(p_(m)B))` . In second, where I represents moment of inertia of the bar magnet, `p_(m)` is the magnetic moment and is the magnetic field.

Explanation

The magnitude of deflecting torque ( the torque which makes the object rotate ) acting on the bar magnet which will tend to align the bar magnet parallel to the direction of the uniform magnetic field `vecB` is
`|vectau|=p_m, B sin theta`
The magnitude of restoring torque acting on the bar magnet can be written as
`|vectau|=I(d^2theta)/(dt^2)`
Under equilibrium conditions, both magnitude of deflecting torque and restoring torque will be equal but act in the opposite directions, which means
`I(d^2 theta)/(dt^2)=-p_mB sin theta`
The negative sign implies that both are in oppsite directions. the above equation can be written as
`(d^2 theta)/(dt^2)=-(p_m B)/I sin theta`
This is non-linear second order homogeneous differential equation. In order to make it linear, we used small angle approximation i.e .,` sin theta ~~ theta`, we get
`(d^2theta)/(dt^2)=-(p_mB)/I theta`
This linear second order homogeneous differential equation is a simple Harmonic differential equation. Therefore,
Comparing with Simple Harmonic motion (SHM) differential equation `(d^2x)/(dt^2)=-omega^2x`
where `omega` is the angular frequency of the oscillation.
`omega^2=(p_mB)/I rArr omega=sqrt((p_mB)/I)`
`T=2pisqrt(I/(p_mB))`
`T=2pi sqrt(I/(p_mB_H)) ` in second
where `B_H` is the horizontal component of Earth.s magnetic field.