Physics - Magnetism Question with Solution | TestHub

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

$B_2=\frac{{\mu }_{0}I}{2R}$

The magnetic field at an axial point at a distance $x$ from the centre is given by

$B_1=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

Given that $B_1=\frac{B_2}{8}$, we have

$\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{1}{8}\left(\frac{{\mu }_{0}I}{2R}\right)$

$\frac{R^2}{(R^2+x^2{)}^{3/2}}=\frac{1}{8R}$

$(R^2+x^2{)}^{3/2}=8R^3$

Taking the cube root on both sides, we get

$(R^2+x^2{)}^{1/2}=2R$

Squaring both sides, we get

$R^2+x^2=4R^2$

$x^2=3R^2$

$x=R\sqrt{3}$

Answer: $\mathrm{R}\sqrt{3}$