5.7 a) A bar magnet of magnetic moment 1.5 J T ^{-1}  lies aligned with the direction of a uniform magnetic field of 0.22 T.
 What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i)  normal to the field direction 

Answers (1)

Given.

Magnetic moment, M= 1.5 J T^{-1}

Magnetic field strength, B= 0.22 T

Now,

The initial angle between the axis and the magnetic field, \theta_1= 0°

Final angle, \theta_2= 90°

We know, The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = - MB ( cos \theta_2 - cos \theta_1)

\implies W = - (1.5)(0.22) ( cos 90^{\circ} - cos 0^{\circ}) = -0.33(0-1)
  = 0.33 J

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