Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.7
NCERT Solutions for Class 12 Physics Chapter 5 Question 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)?
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Ans (a).
Magnetic moment, M = 1.5 J T-1 ,
Magnetic field strength, B = 0.22 T
(i) Initial angle between the axis and the magnetic field, 0₁ = 0°
Final angle between the axis and the magnetic field, 02 = 90°
The work required to make the magnetic moment normal to the direction of magnetic field is given as:
W = -MB(cos02 – cos 0₁)
= -1.5 x 0.22(cos 90° – cos 0°)
= -0.33(0-1)
= 0.33 J
(ii) Initial angle between the axis and the magnetic field, 0₁= 0°
Final angle between the axis and the magnetic field, 02 = 180°
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
W = -MB(cos02 – cos 0₁)
= -1.5 x 0.22(cos 180 – cos 0°)
= —0.33(—1 -1)
= 0.66 J
Ans (b).
For case (i): 0 = 02= 900
Therefore ,Torque, τ = MBsin 6
= 1.5×0.22 sin 90°
= 0.33 J
For case (ii): 0 = 02= 180°
Therefore Torque, τ = MB sin 0
= MBsin 180° = 0 J