Arrange the following in ascending order of magnetic induction at a point on axis of circular loop of radius(r) at a distance (x) from its Centre (for the same currents in the loops)

a) r=R; x=$\sqrt{3}R$

b)   r=2R; x=$\sqrt{5}R$

c) r=$\frac{\mathrm{R}}{2}$; x=2$\sqrt{2}R$

Given Data:

Arrange the following in ascending order of magnetic induction at a point on axis of circular loop of radius(r) at a distance (x).

Formula Used:

$B=\frac{{\mu }_{0}i{r}^2}{2{\left(r^2+x^2\right)}^{3/2}}$

Detailed Solution:

a)

$$\begin{gathered} B_a=\frac{{\mu }_{0}i{R}^2}{2{\left(R^2+3R^2\right)}^{3/2}} \\ =\frac{{\mu }_{0}i{R}^2}{2R^3\left(2^3\right)R}=\frac{{\mu }_{0}i}{8\left(2R\right)} \end{gathered}$$

b) $r=2 R, x=\sqrt{5R}$

$$\begin{gathered} B_b=\frac{{\mu }_{0}i{R}^2}{2{\left(4R^2+5R^2\right)}^{3/2}} \\ =\frac{4{\mu }_{0}i}{2R^3\left(3^3\right)R}=\frac{4{\mu }_{0}i}{27\left(2R\right)} \end{gathered}$$

$c)r=\frac{R}{2},x=2\sqrt{2}R$

$$\begin{gathered} B_c=\frac{{\mu }_{0}i{R}^2}{4\times 2{\left({\displaystyle \frac{R^2}{4}}+8R^2\right)}^{3/2}} \\ =\frac{{\mu }_{0}i}{4\times 2{\left(\frac{1}{4}+8\right)}^{3/2}}=\frac{{\mu }_{0}i}{8\left(2R\right)}\times \frac{1}{4{\left({\displaystyle \frac{33}{4}}\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$$B_a:B_b:B_c=\frac{{\mu }_{0}i}{8\left(2R\right)}:\frac{4{\mu }_{0}i}{27\left(2R\right)}:\frac{{\mu }_{0}i}{\left(2R\right)}\times \frac{1}{4{\left({\displaystyle \frac{33}{4}}\right)}^{{\displaystyle \frac{3}{2}}}}$Hence, the correct option is D.