Question

The magnetic field at a distance x on the axis of a circular coil of radius R is $\dfrac{1}{8}th$ of that of the centre. The value of x is
A) $\dfrac{R}{\sqrt{3}}$
B)$\dfrac{2R}{\sqrt{3}}$
C)$R\sqrt{3}$
D)$R\sqrt{2}$

Answer 1

Hint: In the above question the magnetic field along the axis of the coil is at its centre is given to us. The magnetic field along the axis is governed by a particular relation. Hence comparing the magnetic field at the centre of the coil and at distance x such that the magnetic field is $\dfrac{1}{8}th$ of that of the centre, will enable us to determine the required value of x.

Formula used:
$B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}$

Complete step-by-step answer:
Let us say we have a circular coil of radius R carrying current ‘i’ as shown in the figure below.

The magnetic field at a distance x from the centre of the coil is given by,
$B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}$ where ${{\mu }_{\circ }}$ is the permeability of free space.
At x = 0, i.e. the centre of the coil the magnetic field is equal to,
$\begin{align}
  & B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}} \\
 & \Rightarrow B=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{\left( 0 \right)}^{2}}+{{R}^{2}})}^{3/2}}} \\
 & \therefore B=\dfrac{{{\mu }_{\circ }}i}{2R}.....(1) \\
\end{align}$
Similarly the magnetic field at a distance x from the centre is equal to,
${{B}_{1}}=\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}.....(2)$
In the question it is given that the magnetic field at a distance x on the axis of a circular coil of radius R is $\dfrac{1}{8}th$ of that of the centre. Therefore taking the ratio of equation 1 and 2 we get,
$\begin{align}
  & \dfrac{{{B}_{1}}}{B}=\dfrac{\dfrac{{{\mu }_{\circ }}i{{R}^{2}}}{2{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}}{\dfrac{{{\mu }_{\circ }}i}{2R}} \\
 & \Rightarrow \dfrac{{{B}_{1}}}{B}=\dfrac{{{R}^{3}}}{{{({{x}^{2}}+{{R}^{2}})}^{3/2}}} \\
 & \because {{B}_{1}}=\dfrac{1}{8}B \\
 & \Rightarrow \dfrac{{{B}_{1}}}{8{{B}_{1}}}=\dfrac{{{R}^{3}}}{{{({{x}^{2}}+{{R}^{2}})}^{3/2}}} \\
 & \Rightarrow 8{{R}^{3}}={{({{x}^{2}}+{{R}^{2}})}^{3/2}} \\
 & \Rightarrow 2R={{({{x}^{2}}+{{R}^{2}})}^{1/2}} \\
 & \Rightarrow 4{{R}^{2}}={{x}^{2}}+{{R}^{2}} \\
 & \Rightarrow {{x}^{2}}=3{{R}^{2}} \\
 & \therefore x=\sqrt{3}R \\
\end{align}$

So, the correct answer is “Option C”.

Note: It is to be noted that we have taken the cube root on both the sides of the equation so that the power 3 on both the sides becomes 1. Further squaring on both the sides will help to get rid of the square root in the subsequent step. In the above question we have assumed the coil to be of single loop as the value of magnetic field in case of comparison does not matter.