Question

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 1

(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

Answer 2

Answer:

0.66J and 0Nm

Explanation:

Hope it is useful..