A short bar magnet placed with its axis at `30^@` with a uniform external magnetic field of `0*16T` experiences a torque of magnitude `0*032J`. Estimate the magnetic moment of the magnet. If the bar were free to rotate, which orientations would correspond to its (i) , stable and (ii), unstable equilibrium? What is the potential energy of the magnet in the two case?

Answer

`0*4Am^2` (i) `theta=0^@` (ii) `theta=180^@`; `-0*064J`; `+0*064J`

Explanation

Here, `theta=30^@`, `B=0*16T`, `tau=0*032J`, `M=?`
As `tau=MBsintheta`
`:. M=(tau)/(B sin theta)=(0*032)/(0*16sin 30^@)=(0*032)/(0*16xx1//2)`
`M=0*4Am^2`, `vecM` is directed from S to N.
The bar will be in stable equilibrium, when `vecM` is parallel to `vecB`, i.e. `theta=0^@`
Potential energy in this position is minimum.
i.e. `PE_1=-MB=-0*4xx0*16=-0*064J`
The bar will be in unstable equilibrium, when `vecM` is antiparallel to `vecB`, i.e. `theta=180^@`
`:. P.E_2=-MB cos 180^@=+MB=+0*064J`