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,
(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:
$\mathrm{B}=\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\mathrm{d}}^{3}}$
Where, ${\mathrm{\mu }}_{0}$ = Permeability of free space =

$\begin{array}{rl}\therefore B& =\frac{4\pi ×{10}^{-7}×2×0.48}{4\pi ×\left(0.1{\right)}^{3}}\\ & =0.96×{10}^{-4}\mathrm{T}=0.96\mathrm{G}\end{array}$
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:
$\begin{array}{l}B=\frac{{\mu }_{0}}{4\pi }\frac{M}{{d}^{3}}\\ ⇒B=\frac{4\pi ×{10}^{-7}×0.48}{4\pi ×\left(0.1{\right)}^{3}}\\ =0.48G\end{array}$
The magnetic field is along the N - S direction.