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:

$B=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}$

Where, $\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \,T\,m\,A ^{-1}$

$\therefore B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}}$

$=0.96 \times 10^{-4} T =0.96 \,G$

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 the equatorial of the magnet is given as:

$B=\frac{\mu_{0} \times M}{4 \pi \times d^{3}}$

$=\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi(0.1)^{3}}$

$=0.48 \,G$

The magnetic field is along the $N-S$ direction.