# 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
${F}_{new}=\frac{3{\mu }_{0}}{2\pi }\frac{{m}_{1}{m}_{2}}{{\left(\frac{r}{2}\right)}^{4}}=16\frac{3{\mu }_{0}}{2\pi }\frac{{m}_{1}{m}_{2}}{{r}^{4}}=16F$
So new force will be 16 times of the old force.

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