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 `vec(B)` is
`|vec(tau)| = p_(m) B"sin" theta`
The magnitude of restoring torque acting on the bar magnet can be written as
`|vec(tau)| = I (d^(2) theta)/(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_(m)" B sin " theta`
the negative sign implies that both are in opposite 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 use small angle approximation. i.e., sin `theta approx theta` , we get
`(d^(2) theta)/(dt^(2)) = - (p_(m) B)/(I) theta`
this inear second order homogeneous differential equation is a simple Harmonic differential equation. therefore,
comparing with simple Harmonic Motion (SHM) differential equation `(d^(2)x)/(dt^(2)) = -omega^(2) x ` where `omega` is the angular frequency of the osxillation .
`omega^(2) = (p_(m)B)/(I) rArr omega = sqrt((p_(m)B)/(I))`
T = `2pi sqrt((I)/(p_(m)B))`
T = `2pi sqrt((I)/(p_(m)B_(H)))` in second
where, `B_(H)` is the horizontal component of Earth.s magnetic field.