A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
2n2B
nB
n2B
2nB
Explanation:
$B = \dfrac{\mu_0i}{2R} = \dfrac{\mu_0i(2\pi)}{2l} = \dfrac{\mu_0\pi i}{2l}$
$B = \dfrac{\mu_0ni}{2r} = \dfrac{\mu_0ni}{2\left(\dfrac{l}{2n\pi}\right)} = \dfrac{n^2\mu_0\pi i}{2l} = n^2B$