Consider a point P located on the axial line of a short bar magnet of magnetic length 21 and strength m. Let us find B at a point P which is at a distance z from the centre of magnet. 
Magnetic field at point P due to N-pole,

$\mathbf{B}=\frac{{\mu }_0}{4\pi }\frac{2M}{z^3}\hat{\mathbf{M}}$

$B_1=\frac{{\mu }_0}{4\pi }\left[\frac{m}{(z-l{)}^2}\right]\text{ along }NP$

Similarly, magnetic field at point P due to S-pole

$B_2=\frac{{\mu }_0}{4\pi }\left[\frac{m}{(z+l{)}^2}\right]\text{ along PS}$

Net magnetic field at P,

$B=B_1-B_2=\frac{{\mu }_0}{4\pi }\left[\frac{m}{(z-l{)}^2}-\frac{m}{(z+l{)}^2}\right]$

$=\frac{{\mu }_0}{4\pi }\left[\frac{4mlz}{{\left(z^2-l^2\right)}^2}\right]$

$=\frac{{\mu }_0}{4\pi }\left[\frac{m\times 2}{{\left(z^2-l^2\right)}^2}2z\right]$

$B=\frac{{\mu }_0}{4\pi }\frac{2Mz}{{\left(z^2-l^2\right)}^2}$   $[\therefore M=m(2l\left)\right]$

$B=\frac{{\mu }_0}{4\pi }·\frac{2Mz}{z^4}$   $(\because z>>l)$

Hence, in vector form, magnetic field at point P due to N-pole can be written as

$\mathbf{B}=\frac{{\mu }_0}{4\pi }\frac{2M}{z^3}\hat{\mathbf{M}}$

where, $\hat{\mathrm{M}}$ = unit vector along magnetic dipole moment.