A magnet of magnetic dipole moment `M` is released in a uniform magnetic field of induction `B` from the position shown in the figure.Find
(i) Its kinetic energy at `theta=90^(@)`
(ii)its maximum kinetic energy during the motion.

Explanation

(i) Apply energy conservation at `theta=120^(@)` and `theta=90^(@)`
`-MB cos 120^(@)+0`
`=-MB cos 90^(@)+(K.E)`
`KE=(MB)/2`
(ii) `K.E.` will be maximum where `P.E.` is minimum. `P.E` is minimum at `theta=0^(@)`.Now apply energy conservation between `theta=120^(@)` and `theta=0^(@)`
`-mB cos 120^(@)+0`
`=-mB cos 0^(@)+(K.E)_(max)`
`(KE)_(max)=3/2MB`
The `K.E` is max at `theta=0^(@)` can also be proved by torque method.From `theta=120^(@)` to `theta=0^(@)` the torque always acts on the dipole in the same direction (here it is clockwise) so its `K.E.` keeps on increase till `theta=0^(@)`, Beyond that reverses its direction and then `K.E.` starts decreasing
`:. theta=0^(@)` is the orientation of `M` to here the maximum `K.E.`
(iii) Since `theta` is not small.
`:.`the motion is not `S.H.M` but it is oscillatory and periodic amplitude is `120^(@)`