Derive an expression for potential energy of a bar magnet when placed in an external magnetic field.

Explanation

Potential of a bar magnet. The torque `vectau` acting on a magnetic dipole placed in a uniform magnetic field `vecB` is given by :
`vectau=vecMxxvecB`
or `tau=MBsintheta" "...(1)`

This torque tends to align the magnetic dipole in the direction of `vecB`. If the dipole is rotated against the action of this torque, work has to be done, which is stored in the form of potential energy of the dipole.
Work done in rotating the dipole through an angle `d theta` against the torque is given by
`dW=taud theta=MBsinthetad theta`
If the magnetic dipole is rotated from `theta_(1)andtheta_(2)`, total work done will be
`W=intdW=MBint_(theta_(1))^(theta_(2))sinthetad theta`
= `-MB(costheta_(2)-costheta_(1))`
This work done is the energy stored U
`therefore" "U=-MB(costheta_(2)-costheta_(1))`
If the dipole is rotated from
`theta_(1)=90^(@)" to "theta_(2)=theta` then
`U=-MB(costheta-cos90^(@))`
`U=-MBcostheta`
or `U=-vecM*vecB`.