If ' N ' is the number of turns in a circular coil, the value of...

The magnetic field at the centre of a circular coil of radius $R$ having $N$ turns carrying current $I$ is given by

$B = \dfrac{\mu_0 N I}{2R}$

The magnetic flux linked with a single turn of the coil is

$\phi_1 = B A = \left(\dfrac{\mu_0 N I}{2R}\right) (\pi R^2) = \dfrac{\mu_0 \pi N I R}{2}$

The total magnetic flux linked with the entire coil of $N$ turns is

$\phi = N \phi_1 = N \left(\dfrac{\mu_0 \pi N I R}{2}\right) = \dfrac{\mu_0 \pi N^2 I R}{2}$

The self-inductance $L$ of the coil is defined by $\phi = L I$. Therefore,

$L = \dfrac{\phi}{I} = \dfrac{\mu_0 \pi N^2 R}{2}$

From this expression, it is clear that the self-inductance $L$ is directly proportional to the square of the number of turns.

$L \propto N^2$

Answer: $N^{2}$