Explanation

(a) For a circular coil of radius $a$ carrying a current $i,$ the magnetic field at point $P,$ distance $x$ from coil is given by
[image:4_image109.emf]$B = \frac{\mu_{0}ia^{2}}{2\left( a^{2} + x^{2} \right)^{3/2}}\ {NA}^{- 1}\ m^{- 1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
At the centre of coil $x = 0$
$\therefore B^{''} = \frac{\mu_{0}i}{2a}\ {NA}^{- 1}\ m^{- 1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Given, $B = \frac{1}{8}\ B''$
$\therefore\ \frac{\mu_{0}ia^{2}}{2\left( a^{2} + x^{2} \right)^{3/2}} = \frac{1}{8}\left( \frac{\mu_{0}i}{2a} \right)$
$\Rightarrow \frac{a^{2}}{\left( a^{2} + x^{2} \right)^{3/2}} = \frac{1}{8a}$
$\Rightarrow 8a^{3} = \left( a^{2} + x^{2} \right)^{3/2}$
$\Rightarrow a^{2} + x^{2} = 4a^{2}$
$\Rightarrow x = \sqrt{3.a}$
Given, $a = R$
$\therefore x = \sqrt{3}R$