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Question 5.7: 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. (a) 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, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-201

Answer»

(a) work REQUIRED to turn the dipole is GIVEN by
W=MB[cos\theta_i-cos\theta_f]
here \theta_i is the initial ANGLE made by dipole with magnetic field and \theta is the FINAL angle made by dipole with magnetic field.

(i) magnetic moment is normal to the field direction.
so, \theta_i=0\:and\:\theta_f=90
W = 1.5 × 0.22 [ cos0° - cos90° ]
W = 0.33J

(ii) magnetic moment is opposite to the field direction.
so, \theta_i=0\:and\:\theta_f=180
now, W = 1.5 × 0.22 [ cos0° - cos180°]
= 0.33 [ 1 - (-1) ]
= 0.66 J

(b) torque when \theta=90°
\tau=MBsin\theta
=1.5 × 0.22 × sin90°
= 0.33 Nm

torque when \theta=180°
\tau=MBsin\theta
= 1.5 × 0.22 × SIN 180°
= 0.33 × 0 = 0 Nm


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