two short magnets apply a force F on each other, if the separation between them is halved then what's the new force between them approximately?

ans.. 16F

A bar magnet can be considered as a magnetic dipole. Now the force between two magnetic dipole is given by
$F=\frac{3{\mu }_0}{2\pi }\frac{m_1{m}_2}{r^4}$
Where r is the distance between them. Now if the distance is halved the new force will be
$F_{new}=\frac{3{\mu }_0}{2\pi }\frac{m_1{m}_2}{{\left({\displaystyle \frac{r}{2}}\right)}^4}=16\frac{3{\mu }_0}{2\pi }\frac{m_1{m}_2}{r^4}=16F$
So new force will be 16 times of the old force.