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

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$n^2B$

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(D) The magnetic field at the centre of a circular coil with $n$ turns is given by $B = \frac{\mu_0 n i}{2r}$,where $i$ is the current and $r$ is the radius of the loop.Let the length of the wire be $L$. For a single turn $(n_1 = 1)$,the circumference is $2\pi r_1 = L$,so $r_1 = L / (2\pi)$.The magnetic field is $B = \frac{\mu_0 (1) i}{2 r_1} = \frac{\mu_0 i}{2 (L / 2\pi)} = \frac{\mu_0 i \pi}{L}$.When the same wire is bent into $n$ turns $(n_2 = n)$,the new circumference is $2\pi r_2 = L/n$,so $r_2 = L / (2\pi n)$.The new magnetic field $B'$ is $B' = \frac{\mu_0 n i}{2 r_2} = \frac{\mu_0 n i}{2 (L / 2\pi n)} = \frac{\mu_0 n^2 i \pi}{L}$.Comparing the two,$B' = n^2 \left( \frac{\mu_0 i \pi}{L} \right) = n^2 B$.

$A$ 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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