A small bar magnet having a magnitic moment of `9xx10^(-3)Am^(-2)` is suspended at its centre of gravity by a light torsionless string at a distance of `10^(-2)m` vertically above a long, straight horizontal wire carrying a current of 1.0A from east to west. Find the frequency of oscillation of the magnet about its equilibrium position. The moment of inertia of the magnet is `6xx10^(-9)kgm^(2).(H=3xx10^(-5)T).

Explanation

The magnetic moment of the bar magnet is,
` " " M=9xx10^(-9)A-m^(2)`
The magnitude of the magnetic field at the location of the magnet due to current carrying wire is:
` " " M=(mu_(0))/(2pi)(I)/(r)=((2xx10^(-7))(1.0))/(10^(-2))`
` " " 2=10^(-5)Wb//m^(2)`
The earth's horizontal magnetic field is ,
` " " H=3xx10^(-5)T=3xx10^(-5)Wb//m^(2)`
The frequency of oscillation will be,
` " " v=(1)/(2pi) sqrt((M(B+H))/I) ["here,I=moment of inerita"]` ` " " v=(1)/(2pi) sqrt((9xx10^(-9))(5xx10^(-5)))/(6xx10^(-9))`
` " " =1.38xx10^(-3)H`