A circular coil of radius 2R is carrying current ${}^{\text{'}}i^{\text{'}}$. The ratio of magnetic fields at the centre of the coil and at a point at a distance 6R from the centre of the coil on the axis of the coil is

Given data:

radius 2R, current I and distance 6 R

Concept used:  moving charges and magnetism 

Detailed solution:

$B_1=\frac{{\mu }_0\mathrm{i}}{2\times 2\mathrm{R}}=\frac{{\mu }_0\mathrm{i}}{4\mathrm{R}}$at the center of coil

${\mathrm{B}}_2=\frac{{\mu }_0\mathrm{i}(2\mathrm{R}{)}^2}{2{\left(4{\mathrm{R}}^2+{\mathrm{x}}^2\right)}^{3/2}}$ on the axis of the coil.

Put $x=6 R$

$\Rightarrow {\mathrm{B}}_2=\frac{{\mu }_0\mathrm{i}}{2\mathrm{R}}\times \frac{4}{{40}^{3/2}}=\frac{{\mu }_0\mathrm{i}}{40\sqrt{10\mathrm{R}}}$

$\frac{{\mathrm{B}}_1}{{ \mathrm{B}}_2}=10\sqrt{10}$

Hence, option (B) is the solution.