Consider a bar magnet of magnetic dipole moment \(\vec M\), suspended by a light twistless fibre in a uniform magnetic field \(\vec B\) in such a way that it is free to rotate in a horizontal plane. In the rest position θ = 0, \(\vec M\) is is parallel to \(\vec B\).
If magnet is given a small angular displacement θ from its rest position and released, the magnet performs angular or torsional oscillations about the rest position.
Let I be the moment of inertia of the bar magnet about the axis of oscillation and α the angular acceleration. The deflecting torque (in magnitude) is
τd = Iα = I \(\cfrac{d^2\theta}{dt^2}\)………….. (1)
However, the restoring torque tries to bring back the oscillating bar magnet in the rest position. The restoring torque (in magnitude) is,
τr = – MB sin θ …………. (2)
The minus sign in Eq. (2) indicates that restoring torque is opposite in direction to the angular deflection.
In equilibrium, both the torques balance each other. From Eqs. (1) and (2),
I\(\cfrac{d^2\theta}{dt^2}\) = - MB sin \(\theta\)....(3)
For small \(\theta\), sin \(\theta\) = \(\theta\), Thus Eq, (3) can be written as
Eq. (4) represents angular simple harmonic motion.
Writing ω2 = \(\cfrac{MB}I\), the angular frequency ω of the motion is
ω = \(\sqrt{\cfrac{MB}I}\) ………….. (5)
The time period of oscillations of the bar magnet is
T = \(\cfrac{2\pi}\omega\) = 2π \(\sqrt{\cfrac I{MB}}\)…………… (6)
This is the required expression.
