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Class 12th Physics 2018 Set1 Delhi Board Paper Solution

Question 13

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

(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) \tau   = M X B

         = MB sin Ø

         = 6 x 0.44 x sin 180º

         = 6 x 0.44 x 0

        = 0
  

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