A circular current carrying coil has a radius $R$. The distance from the center of the coil on the axis, where the magnetic induction will be $\dfrac{1}{8}$ th to its value at the center of the coil is-
A.$\dfrac{R}{\sqrt{3}}$
B.$R\sqrt{3}$
C.$2\sqrt{3}R$
D.$\dfrac{2}{\sqrt{3}} R$

Hint: By applying the formula for the magnitude of magnetic induction for a current carrying circular coil at a distance from its axis, this question can be resolved. We will then plug in the required value given in the question and solve the issue Equation to get the distance's value.
Formula used:
$B=\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{\left(x^{2}+R^{2}\right)^{\dfrac{3}{2}}}$
Where $I$ is the current in the coil, $\mathrm{R}$ is the radius of the coil, $x$ is the distance on the axis from its centre and $\mu_{0}$ is the
magnetic permeability of free space (vacuum) equal to $\mu_{0}=4 \pi \times 10^{-7} N . A^{-2}$

Complete answer:
A current carrying a circular coil behaves like a two-pole magnet. The magnetic field's value at a distance $x$ from
the centre on its axis is given by,
$B=\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{\left(x^{2}+R^{2}\right)^{\dfrac{3}{2}}}$
Where I is the current in the coil, $\mathrm{R}$ is the radius of the coil, $x$ is the distance on the axis from its centre and $\mu_{0}$ is the
magnetic permeability of free space (vacuum) equal to $\mu_{0}=4 \pi \times 10^{-7} N . A^{-2}$
Now, Let us analyse the problem. It is given that at a distance, the value of magnetic induction is $\left(\dfrac{1}{8}\right)^{t h}$ of its value at the center of the coil.
Let the distance be $x$.
Now, using (1)
For magnetic induction at the centre, $x=0$. Putting this in (1), we get,
$B(\text { centre })=\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{\left(0^{2}+R^{2}\right)^{\dfrac{3}{2}}}=\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{\left(R^{2}\right)^{\dfrac{3}{2}}}=\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{R^{3}}=\dfrac{\mu_{0} I}{2 R}$- (II)
where $B$ (centre) is the value of the magnetic induction at the centre.
Using (1), magnetic induction at the distance $x$ on the axis is given by,
$B($ at $\text{x})=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}$- (III)
Now, by the problem $\dfrac{B(\text { at } \mathrm{x})}{B(\text { at centre })}=\dfrac{1}{8}$
Therefore, putting (2) and (3) in (4), we get, $\dfrac{\mu_{0} I}{2} \dfrac{R^{2}}{3}$
$\left(x^{2}+R^{2}\right)^{\dfrac{3}{2}}$
$\Rightarrow \dfrac{{{\mu }_{0}}I}{2R}=\dfrac{1}{8}$
$\Rightarrow \dfrac{{{R}^{3}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}={{\left( \dfrac{1}{2} \right)}^{3}}$
Putting cube root on both sides, $\therefore \dfrac{R}{\left(x^{2}+R^{2}\right)^{\dfrac{1}{2}}}=\dfrac{1}{2}$
Squaring both sides, $\dfrac{R^{2}}{\left(x^{2}+R^{2}\right)}=\dfrac{1}{4}$
$4{{R}^{2}}={{x}^{2}}+{{R}^{2}}$
$3{{R}^{2}}={{x}^{2}}$
Putting square root on both sides,
$\therefore x=R \sqrt{3}$
Hence the required distance is $R \sqrt{3}$.

The correct option is (B) $R \sqrt{3}$.

Note:
To check whether or not a student has written formula (1) correctly, the value $x=0$ can be set to get the familiar magnetic induction equation at the centre, which is $\dfrac{\mu {0} I}{2 R}$. This is a good way to verify that in an exam you are not going wrong. A circular current carrying coil.