Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.3
NCERT Solutions for Class 12 Physics Chapter 5 Question 3
A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet?
Share
Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T = 4.5 x 10-2 J
Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0
Therefore , M = T/Bsin 0
= (4.5 x 10-2 )/(0.25 x sin 30º) =0.36 J T⁻¹
Hence, the magnetic moment of the magnet is 0.36 IT-1.