A bar magnet of magnetic moment 6 `J//T` is aligned at `60^(@)` with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Explanation

Work done = mB `( cos theta _ 1 -cos theta _2) `
( i) ` theta _ 1 =60^(@) , theta _2 = 90^(@) `
`therefoe ` work done ` ( cos 60^(@) -0^(@)) `
`= mB ( (1)/(2) - 0) =(1)/(2) mB `
`= (1)/(2) xx 6xx 0.44J= 1.32J`
( ii) ` theta _1 =60^(@) , theta _2 =180^(@) `
` therefore ` work done `=mB ( cos 60^(@) -cos 180^(@) ) `
`mB ((1)/(2) - (-1) ) =(3)/(2) mB `
`= (3)/(2) xx 6xx 0.44J =3.96J`
[Also sccept calculations done through changes in potential energy.]