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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

$F=\frac{3{\mu}_{0}}{2\pi}\frac{{m}_{1}{m}_{2}}{{r}^{4}}$

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.

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