Write the equation of torque on the needle placed in a uniform magnetic field and obtain the equation of its periodic time `T = 2 pi sqrt((I)/( mB))`.

Explanation

A magnetic needle placed in a uniform magnetic field `overset(to) (B) `. Its dipole moment is `overset(to) (M)`.

The forces acting on both the poles N and S, constitute a couple of force and hence torque acting on it.
The magnetic field at distance `r` from pole strength `q_m` in magnetism is
`overset(to) (B) = (q_m)/( 4pi r^2)` and in electricity electric field at distance `r` from changes `q` is `overset(to) (E) = (q)/( 4 pi r^2)`.
In electricity the torque acting on a electric dipole `overset(to) (tau) = overset(to) (p) xx overset(to) (epsilon)`, similarly torque in magnetism `overset(to) (tau) = overset(to) (m) xx overset(to) (B) ` (Where `overset(to) (m) = ` dipole moment of magnetic needle).
`therefore tau = m B sin theta" "...(1)`
Here `tau` is restoring torque and `theta` is the angle between `overset(to) (m) and overset(to) (B) `. Therefore, in equilibrium,
`I alpha = m B sin theta ( because tau = I alpha)`
`therefore (Id^(2 ) theta)/( dt^(2) ) = - m B sin theta" "[ because alpha = (d^(2) theta)/( dt^(2) )]`
Negative sign with mBsine indicates that restoring torque is in the opposite direction to the depleting torque. For small value of `theta, sin theta ~~ theta,` so
`(Id^(2) theta)/( dt^(2) ) = - m B theta ` or `(d^(2) theta)/( dt^(2) ) = - (m B )/( I) theta`
This equation represents a simple harmonic motion. Comparing with it to general equation, `(d^(2) x)/( dt^(2) ) = - omega^(2) x`
Here, `omega^(2) = (m B)/( I)`
`therefore omega = sqrt((mB)/( I))`
`therefore (2pi)/( T) = sqrt((mB)/( I))`
`therefore T= 2pi sqrt((I)/( m B) ) .... (1)` or
`B= (4pi^(2) I)/( m T^(2) ) .... (2)`
Where I = moment of inertia of bar magnet `= (1)/(12) ML^(2)`
(M = Mass of bar magnet, L = magnetic length of bar magnet)