Q13 of 26 Page 1

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) Given


Magnetic moment m = 6J/T


Magnetic field B = 0.44T


Initial angle θ1 = 60°


Now work done in turning the magnet is given as


W = -mB (cosθ2 -cosθ1)


i) For first case θ2 is 90°


Therefore


W = -mB (cos90 -cos60)


As Cos90° = 0 and cos60° = 1/2


We get,


W = -6× 0.44× (0-1/2)


W = 3× 0.44 = 1.32J


ii) For the second case θ2 is 180°


Therefore


W = -mB (cos180 -cos60)


Cos 180° = -1 and cos60° = 1/2




b) Torques is given as


τ = m× B = mBsinθ


where θ is the angle between m and B.


In the second orientation magnetic moment and magnetic field are anti parallel i.e. θ = 180° .


Now sin180° = 0.


Therefore, torque will be zero.


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