5.12 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.

Magnetic moment of the bar magnet,  M=0.48 J T-1
(a) Distance, d = 10 cm 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=μ04π2Md3
Where, μ0 = Permeability of free space = =4π×10-7 Tm A-1
 
B=4π×107×2×0.484π×(0.1)3=0.96×104T=0.96G
The magnetic field is along the S-N direction.
(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on
equatorial line of the magnet is given as:
B=μ04πMd3B=4π×107×0.484π×(0.1)3=0.48G
The magnetic field is along the N - S direction.