A small bar magnet having magnetic moment of `9xx10^9Wb-m` is suspended at its centre of gravity by a light torsionless string at a distance of `1m` vertically above a long straight horizontally wire carrying a current of `1A`. 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 `6xx10^-9kgm^2`.

Explanation

Here, `M=9xx10^-9Wb-m`
`r=1cm=10^-2m`, `i=1A v=?`
`I=6xx10^-9kgm^2`
Magnetic field strength at `1cm` from the wire carrying current
`B=(mu_0)/(4pi)(2i)/(r)`
The torque acting on the small magnet in this external magnetic field.
`tau=-MB sin theta=-MB theta`, when `theta` is small.
If `alpha` is angular acceleration produced by the torque, then
`tau=lalpha=-MB theta=-M((mu_0)/(4pi)(2i)/(r))theta`
`alpha=-(mu_0)/(4pi)(2iM)/(Ir)theta=-omega^2theta`
where `omega^2=(mu_0)/(4pi)(2iM)/(Ir)`
`:.` Motion of magnet is simple harmonic and its frequency is
`n=(omega)/(2pi)=(1)/(2pi)sqrt((mu_0)/(4pi)(2iM)/(Ir))`
`=(1)/(2xx3.14)sqrt((10^-7xx2xx1.0xx9xx10^-9)/(6xx10^-9xx10^-2))`
`n=8*7xx10^-4Hz`