Question

A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be ${\left(\frac{1}{8}\right)}^{th}$ to its value at the centre of the coil is

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

Answer 1

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

Magnetic field on the axis of a circular coil is,

$B_1=\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$

Where, $R=\text{is radius of the coil and}x=\text{is the field point alog the axis of the coil from its centre}$

Magnetic field at the centre of the circular coil is,

$B_2=\frac{{\mu }_{0}i}{2R}$

Given, $B_1=\frac{B_2}{8}$

$\Rightarrow \frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{1}{8}\frac{{\mu }_{0}i}{2R}$

$8R^3=(R^2+x^2{)}^{3/2}$

$64R^6=(R^2+x^2{)}^3$

$(4R^2{)}^3=(R^2+x^2{)}^3$

$4R^2=R^2+x^2$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.