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A short bar magnet placed with its axis ...

A short bar magnet placed with its axis at `30^@` to a uniform magnetic field of `0*2T` experiences a torque of `0*06N-m`. Calculate magnetic moment of the magnet. What orientation of magnet corresponds to its stable equilibrium in the magnetic field?

Text Solution

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Here, `theta=30^@, B=0*2T`,
`tau=0*06N-m, M=?`
From `tau=Mbsintheta`,
`M=(tau)/(B sin theta)=(0*06)/(0*2 sin 30^@)=0*6Am^2`
Potential energy of magnetic dipole
`U=-MB cos theta`
In stable equilibrium, P.E. is minimum
`:. cos theta=1` or `theta=0^@`
i.e., bar magnet will be in stable equilibrium.
when its magnetic moment `vecM` is parallel to magnetic field `vecB`.
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