Physics - Magnetism Question with Solution | TestHub

The magnetic induction at the centre of a circular coil of radius $R$ is given by

$B_c=\frac{{\mu }_{0}I}{2R}$

The magnetic induction at a distance $x$ on the axis of the coil is given by

$B_a=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

Given that $B_a=\frac{1}{27}{B}_c$, we have

$\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{1}{27}\left(\frac{{\mu }_{0}I}{2R}\right)$

$\frac{R^2}{(R^2+x^2{)}^{3/2}}=\frac{1}{27R}$

$\frac{R^3}{(R^2+x^2{)}^{3/2}}=\frac{1}{27}$

Taking the cube root on both sides, we get

$\frac{R}{\sqrt{R^2+x^2}}=\frac{1}{3}$

Squaring both sides, we get

$\frac{R^2}{R^2+x^2}=\frac{1}{9}$

$9R^2=R^2+x^2$

$x^2=8R^2$

$x=2\sqrt{2}R$

Answer: $2\sqrt{2}\mathrm{R}$