The magnetic induction at the centre of a circular coil of radius 10 cm is $5\sqrt{5}$ times the magnetic induction at a point on its axis. The distance of the point from the centre of the coil, in cm is:

Given data: 

radius$10 \mathrm{cm}$ and magnetic induction $5\sqrt{5}$

Concept used: moving charges and magnetism 

Detailed solution:

The magnetic field in the center of an R-radius circular loop that is carrying current is  $B_{\text{center }}=\frac{{\mu }_0\mathrm{i}}{2R}$

The magnetic field on the axis is :${\mathrm{B}}_{\mathrm{axis}}=\frac{{\mu }_{0}i{\mathrm{R}}^2}{2{\left({\mathrm{R}}^2+{\mathrm{x}}^2\right)}^{3/2}} \mathrm{where} \mathrm{R} \mathrm{is} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{loop}$carrying the current, and X is the distance from the loop’s center .

$\text{ Given }{B}_{\text{center }}=5\sqrt{5}{B}_{\text{axis }}$

$\therefore \frac{{\mu }_0\mathrm{i}}{2R}=5\sqrt{5}\frac{{\mu }_0{\mathrm{R}}^2}{2{\left({\mathrm{R}}^2+{\mathrm{x}}^2\right)}^{3/2}}$

$\Rightarrow \frac{{\left({\mathrm{R}}^2+{\mathrm{x}}^2\right)}^3}{{\mathrm{R}}^6}=(5\sqrt{5}{)}^2=125$

$\Rightarrow {\left(\frac{{\mathrm{R}}^2+{\mathrm{x}}^2}{{\mathrm{R}}^2}\right)}^3=125$

$\Rightarrow \frac{{\mathrm{R}}^2+{\mathrm{x}}^2}{{\mathrm{R}}^2}=5$

$\Rightarrow 1+\frac{x^2}{R^2}=5$

$\frac{x^2}{R^2}=4$

 $\mathrm{x}=2\mathrm{R}=2\times 10=20 \mathrm{cm}$