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(D) The magnetic field at the centre of a circular coil with $n$ turns carrying current $i$ and radius $r$ is given by $B = \frac{\mu_0 n i}{2 r}$.Let

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$n^2 B$

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(D) The magnetic field at the centre of a circular coil with $n$ turns carrying current $i$ and radius $r$ is given by $B = \frac{\mu_0 n i}{2 r}$.Let the length of the wire be $L$. For a single turn $(n_1 = 1)$, the circumference is $2 \pi r_1 = L$, so $r_1 = \frac{L}{2 \pi}$.The magnetic field is $B_1 = B = \frac{\mu_0 (1) i}{2 r_1} = \frac{\mu_0 i}{2 (L / 2 \pi)} = \frac{\mu_0 i \pi}{L}$.When the same wire is bent into $n$ turns $(n_2 = n)$, the new circumference of each turn is $2 \pi r_2 = \frac{L}{n}$, so $r_2 = \frac{L}{2 \pi n}$.The new magnetic field $B_2$ is $B_2 = \frac{\mu_0 n i}{2 r_2} = \frac{\mu_0 n i}{2 (L / 2 \pi n)} = \frac{\mu_0 n i \pi n}{L} = n^2 \left( \frac{\mu_0 i \pi}{L} \right)$.Substituting $B$ for the term in the bracket, we get $B_2 = n^2 B$.

$A$ steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns, the magnetic field at the centre of the coil is

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