Explanation

(b) The magnetic field at the centre of circular coil is
$B = \ \frac{\mu_{0}i}{2r}$
Where, $r =$ radius of circle $= \frac{l}{2\pi}\ \ \ \ \ \ \ \ \ \ \ (\because l = 2\pi r)$
$\therefore B = \frac{\mu_{0}i}{2} \times \frac{2\pi}{l}$
$\ \ \ \ \ \ \ \ = \frac{\mu_{0}i\pi}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
When wire of length $l$ bents into a circular loops of $n$ turns, then
$\ \ \ \ \ \ \ \ l = n\ \times 2\pi r'$
$\Rightarrow r^{'} = \frac{1}{n \times 2\pi}$
Thus, new magnetic field
$\ \ \ \ \ B^{'} = \ \frac{\mu_{0}ni}{2r'} = \frac{\mu_{0}ni}{2} \times \frac{n \times 2\pi}{l}$
$\ \ \ \ \ \ \ \ \ \ \ = \frac{\mu_{0}i\pi}{l} \times n^{2}$
$\ \ \ \ \ \ \ \ \ \ \ = n^{2}B\ \ \ \ $ [From Eq. (i)]