A bar magnet of magnetic moment `6J.T^(-1)` is aligned at `60^(@)` with a uniform external magnetic field of `0.44T` . Calculate
the work done in turning the magnet to align ist magnetic moment (i) normal to the magnetic field , (ii) opposite to the magnetic field

Answer

3

Explanation

Here, `m=6JT^(-1),theta_(1)=60^(@),B=0.44T`
Work done in turning the magnet ,
`W=-mB(costheta_(2)-costheta_(1))`
(i) `thereforeW=-6xx0.44(cos90^(@)-cos60^(@))`
`=-6xx0.44(0-(1)/(2))=1.32J`
(ii) `theta_(1)=60^(@),theta_(2)=180^(@)` `thereforeW=-6xx0.44(cos180^(@)-cos60^(@))`
`=-6xx0.44(0-(1)/(2))=3.96J`