Physics - Magnetism Question with Solution | TestHub

The magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is given by:

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

The magnetic field on the axis of the circular loop at a distance $x$ from the center is given by:

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

Substituting $x=R$, we get:

$B_{axis}=\frac{{\mu }_{0}I{R}^2}{2(R^2+R^2{)}^{3/2}}=\frac{{\mu }_{0}I{R}^2}{2(2R^2{)}^{3/2}}=\frac{{\mu }_{0}I{R}^2}{2(2\sqrt{2}{R}^3)}=\frac{{\mu }_{0}I}{4\sqrt{2}R}$

The ratio of the magnetic fields is:

$\frac{B_{center}}{B_{axis}}=\frac{\frac{{\mu }_{0}I}{2R}}{\frac{{\mu }_{0}I}{4\sqrt{2}R}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}=\sqrt{8}$

The required ratio is $\sqrt{8}:1$.