1. Magnetic field at the center of a circular coil

For a circular coil of radius $R$, having $N$ turns and carrying current $I$, the magnetic field at the center is

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

2. Use the fact that the same wire is used in both cases

Let the total length of the wire be $L$.

• When it is bent into a coil of $N$ turns and radius $R_1$:

$L = N(2\pi R_1)$

So,

$R_1 = \frac{L}{2\pi N}$

Hence the magnetic field in the first case is

$B_1 = \frac{\mu_0 N I}{2R_1} = \frac{\mu_0 N I}{2\left(\frac{L}{2\pi N}\right)}$

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

Thus,

$B_1 \propto N^2$

3. Second case: coil of $n$ turns

Similarly, if the same wire is bent into a coil of $n$ turns and radius $R_2$,

$L = n(2\pi R_2)$

So,

$R_2 = \frac{L}{2\pi n}$

Now the magnetic field at the center is

$B_2 = \frac{\mu_0 n I}{2R_2} = \frac{\mu_0 n I}{2\left(\frac{L}{2\pi n}\right)}$

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

Thus,

$B_2 \propto n^2$

4. Take the ratio

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

So the required ratio is

$N^2 : n^2$

5. Option check

• A: $N^2 : n^2$ ✅
• B: $N : n$ ❌
• C: $n : N$ ❌
• D: $n^2 : N^2$ ❌

Therefore, the correct option is A.