A small bar magnet having a magnetic moment of `9xx10^(-9)`Wb-m is suspended at its centre of gravity by a light torsionless string at distance of `10^(-2)`m vertically above a long straight horizontal wire carrying a current of 1.0 A. Find the frequency of oscillation of the magnet about its equilibrium position assuming that the motion is undamped. The moment of inertia of the magnet is `6 xx10^(-9)"kg m"^(2)`.

Explanation

The torque acting on a small magnet in external magnetic field
`=-MB sin theta =-MB theta`
[For small oscillations, `theta` is small, so sin `theta=theta`]
Here, `B=(mu_(0))/(4pi) .(2i)/(r )`
If `(d^(2)theta)/(dt^(2))` is the angular acceleration produced and I is M.I. about axis of oscillation,
`I xx (d^(2)theta)/(dt^(2)) = -MB theta = -M. (mu_(0))/(4pi).(2i)/(r ) theta`
or `(d^(2)theta)/(dt^(2)) = (mu_(0))/(4pi) .(2piM)/(Ir)theta=-omega^(2) theta`
So motion is SHM with frequency of oscillation,
`f=(omega)/( 2pi) =(1)/(2pi) sqrt((mu_(0))/(4pi) .(2piM)/(Ir)) = (1)/(2pi) sqrt((10^(-7)xx 2xx1 xx 9xx10^(-9))/(6xx10^(-9)xx10^(-2)))=8.7 xx10^(-4)Hz`