Question

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).

Answer 1

a) 

i) `W = vecMvecB (cos theta_1-cos theta_2)`

=  6 × 0.44×(cos 60 −cos 90)

=6 × 0.44×(0.5 - 0)

= 1.32 J

ii) `W = vecMvecB (cos theta_1 - cos theta_2)`

= 6 x 0.44 x (xos 60 - cos 180)

= 6 x 0.44 x [0.5 - (-1)]

= 3.96 J

b) r = `vecM xx vecB`

= 6 x 0.44 x sin 180°

= 0