Class 12th Physics 2018 Set2 Delhi Board Paper Solution
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).
(a) (i)
M = 6 J/T
Q1 = 60º Q2 = 90º
B = 0.44T
W = MB(cosQ1 - cosQ2)
= 6 x 0.44(cos60º - cos90º)
= 6 x 0.44 x ½
= 1.32 J
(a) (ii)
Q1 = 60º Q2 = 180º (opposite to magnetic field)
W = MB(cosQ1 - cosQ2)
= 6 x 0.44(cos60º - cos90º)
= 6 x 0.44 x (½ - (-1))
= 6 x 0.44 x 3/2
= 3.96 J
(b) 
= MB sin Ø
= 6 x 0.44 x sin 180º
= 6 x 0.44 x 0
= 0
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