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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`. |
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Answer» 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` |
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