$A$ 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.

(N/A) Magnetic moment of the bar magnet,$M = 0.48 \; J \; T^{-1}$.
$(a)$ 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:
$B = \frac{\mu_{0}}{4 \pi} \frac{2M}{d^{3}}$
Where,$\mu_{0} = 4 \pi \times 10^{-7} \; T \; m \; A^{-1}$.
Substituting the values:
$B = \frac{10^{-7} \times 2 \times 0.48}{(0.1)^{3}} = \frac{0.96 \times 10^{-7}}{10^{-3}} = 0.96 \times 10^{-4} \; T = 0.96 \; G$.
The direction of the magnetic field is along the $S-N$ direction.
$(b)$ The magnetic field at a distance of $10 \; cm$ $(d = 0.1 \; m)$ on the equatorial line of the magnet is given by:
$B = \frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}$
Substituting the values:
$B = \frac{10^{-7} \times 0.48}{(0.1)^{3}} = \frac{0.48 \times 10^{-7}}{10^{-3}} = 0.48 \times 10^{-4} \; T = 0.48 \; G$.
The direction of the magnetic field is along the $N-S$ direction.