Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.12
NCERT Solutions for Class 12 Physics Chapter 5 Question 12
A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
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Magnetic moment of the bar magnet, M = 0.48 J T-1
Ans (a).
Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:
B = μ0 /4π x (2M/d³)
Where, μ0 = Permeability of free space =4π x 10-7 T m A-1
Therefore ,B = (4π x 10-7 x 2 x 0.48)/[4π x (0.1)³]
The magnetic field is along the S – N direction.
Ans (b).
The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:
B = μ0 /4π x (M/d³)
=> B = (4π x 10-7 x 0.48)/[4π x (0.1)³]
= 0.48 G
The magnetic field is along the N – S direction.