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(C) Let the length of the wire be $L$. For a single turn of radius $R_1$, the circumference is $2\pi R_1 = L$, so $R_1 = L / (2\pi)$. The magnetic fie

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$1 : n^2$

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(C) Let the length of the wire be $L$. For a single turn of radius $R_1$, the circumference is $2\pi R_1 = L$, so $R_1 = L / (2\pi)$. The magnetic field at the centre is $B_1 = \frac{\mu_0 I}{2 R_1} = \frac{\mu_0 I}{2 (L / 2\pi)} = \frac{\mu_0 I \pi}{L}$. For $n$ turns of radius $R_n$, the total length of wire is $n(2\pi R_n) = L$, so $R_n = L / (2\pi n)$. The magnetic field at the centre is $B_n = \frac{n \mu_0 I}{2 R_n} = \frac{n \mu_0 I}{2 (L / 2\pi n)} = \frac{n^2 \mu_0 I \pi}{L}$. The ratio of the magnetic field for one turn to that of $n$ turns is $B_1 / B_n = (\frac{\mu_0 I \pi}{L}) / (\frac{n^2 \mu_0 I \pi}{L}) = 1 / n^2$.

$A$ long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having $n$ turns. If the same current is passed in both the cases, the ratio of magnetic field produced at the centre for one turn to that of $n$ turns is

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