Question

The number of turns in a coil of wire of fixed radius is 600 and its self inductance is 108 mH. The self inductance of a coil of 500 turns will be (in mH)

Answer 1

$L\propto N^2$
$\frac{108}{L}={\left(\frac{600}{500}\right)}^2$
$L=108\times \frac{25}{36}=75\text{mH}$