Question

A circular coil has radius $R$, the distance from the centre of the coil on the axis where the magnetic induction will be ${\frac{1}{8}}^{th}$of its value at the centre of the coil is

• A. $\frac{R}{\sqrt{3}}$
• B. $R\sqrt{3}$
• C. $2\sqrt{3}R$
• D. $\frac{2}{\sqrt{3}}R$

Answer 1

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

The magnetic field due to a circular loop at the centre is given by,

$B_{\text{centre}}=\frac{{\mu }_{0}I}{2R}..........\left(1\right)$

Magnetic field due to a circular loop on the axis is,

$B_{\text{axis}}=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{3/2}}........\left(2\right)$

Now, Dividing $\left(1\right)$ and $(2$) we get,

$\frac{B_{\text{centre}}}{B_{\text{axis}}}=\frac{\frac{{\mu }_{0}I}{2R}}{\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{3/2}}}$

$\frac{B_{\text{centre}}}{B_{\text{axis}}}=\frac{(x^2+R^2{)}^{3/2}}{R^3}$

$\frac{B_{\text{centre}}}{B_{\text{axis}}}=\frac{(x^2+R^2{)}^{3/2}}{(R^2{)}^{3/2}}$

$\frac{B_{\text{centre}}}{B_{\text{axis}}}={(\frac{x^2}{R^2}+1)}^{3/2}$

As per the question,

$B_{\text{axis}}=\frac{1}{8}{B}_{\text{centre}}$

i.e. $\frac{B_{\text{centre}}}{B_{\text{axis}}}=8$

Therefore, ${(\frac{x^2}{R^2}+1)}^{\frac{3}{2}}=8$

$\frac{x^2}{R^2}+1=(8{)}^{\frac{2}{3}}=4$

$\frac{x^2}{R^2}+1=4$

$\frac{x^2}{R^2}=3\Rightarrow x^2=3R^2$

$\therefore x=\pm \sqrt{3}R$

Hence, correct option is $\left(b\right)$.

Why this Question ? Ratio of magnetic field due to circular loop at the centre to the magnetic field on the axis is, $\frac{B_{\text{centre}}}{B_{\text{axis}}}={(1+\frac{x^2}{R^2})}^{3/2}$