If ' \(N\) ' is the number of turns in a circular coil, the value of its self-inductance varies as

The correct option is (C)
Concept: The flux associated with the loop is \(\phi=\mathrm{BA}\) and the coefficient of self-inductance \(\mathrm{L}\) is defined as \(\mathrm{L}=\frac{\phi}{\mathrm{I}}\), where the I is the current flowing through the coil.
The magnetic magnetic field \(B\) at the center of a coil carrying current \(\mathrm{i}\), with radius \(\mathrm{r}\) is given by,
\(B=\frac{\mu_0 I}{2 r}\)
So for a coil with \(\mathrm{n}\) no. of turns: \(\mathrm{B}_{\mathrm{N}}=\mathrm{N}\left(\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\right)\)
The self flux associated with one coil is : \(\phi 1=\mathrm{B}_{\mathrm{N}} \mathrm{A}\)
Now, the self flux associated with coil of \(\mathrm{N}\) turns is: \(\phi \mathrm{N}=\mathrm{N} \phi_1\)
Therefore,
\(\phi N=N\left\{N\left(\frac{\mu_0 I}{2 r}\right) A\right\}_1\)
Now using the definition of self-inductance: \(L=\frac{\theta}{I}\), we get, \(\mathrm{L} \propto \mathrm{N}^2\)
So, self-inductance is proportional to square of number of turns in the coil.