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Magnetism And Matter

Question
CBSEENPH12037811

A short bar magnet placed with its axis inclined at 30° to the external magnetic field of 800 G acting horizontally experiences a torque of 0.016 Nm. Calculate:
(i) the magnetic moment of the magnet,
(ii) the work done by an external force in moving it from most stable to most unstable position,
(iii) What is the work done by the force due to the external magnetic field in the process mentioned in (ii)?

Solution

(i) Given,
Magnetic field, B = 800 G 
θ = 30o 
Torqua acting, τ = 0.016 Nm

Since, Torque acting on magnet, τ = MB sin θ

               M = τB sin θ
              M = 0.016800 × 10-4 × sin 30° 

                  M = 0.40 Am2        [ 1G = 10-4T]
which, is the required magnetic moment. 

(ii) Work done be an external force to move the magnet from stable to unstable position,W = -MBcos θ2- cos θ1 

              W = -MB (cos 180 ° - cos 0°) 

                 = -MB (-1-1) = 2 MB 

              W = 2 × 0.40 × 800 × 10-4     = 0.064 J. 

(iii) The displacement and the torque due to the magnetic field are in opposite direction.
So work done by the force due to the external magnetic field is W
B = – 0.064 J.



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