(d) $\ \ \ \ \ \ \ F = \frac{\mu_{0}}{4\pi}.\frac{m_{1}m_{2}}{r^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
When pole strength of each pole become double.
$\therefore\ F^{'} = \frac{\mu_{0}}{4\pi}.\frac{\left( 2m_{1} \right)\left( 2m_{2} \right)}{(2r)^{2}} = F$