A long conducting wire having a current $$I$$ flowing through it, is bent into a circular coil of $$N$$ turns. Then it is bent into a circular coil of $$n$$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:

A long conducting wire of total length $$L$$ carrying current $$I$$ is bent into a circular coil of $$N$$ turns, and then into a coil of $$n$$ turns. We need the ratio of magnetic fields at the centres.

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

where $$N$$ is the number of turns and $$R$$ is the radius.

For $$N$$ turns: $$L = N \times 2\pi R_1$$, so $$R_1 = \frac{L}{2\pi N}$$

For $$n$$ turns: $$L = n \times 2\pi R_2$$, so $$R_2 = \frac{L}{2\pi n}$$

$$ B_1 = \frac{\mu_0 N I}{2R_1} = \frac{\mu_0 N I}{2} \times \frac{2\pi N}{L} = \frac{\mu_0 \pi N^2 I}{L} $$

$$ B_2 = \frac{\mu_0 n I}{2R_2} = \frac{\mu_0 n I}{2} \times \frac{2\pi n}{L} = \frac{\mu_0 \pi n^2 I}{L} $$

$$ \frac{B_1}{B_2} = \frac{N^2}{n^2} $$

The ratio of magnetic field in the first case to the second case is Option C: $$N^2 : n^2$$.