Question

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

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

Answer 1

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

At the centre of the coil,
$B_c=\frac{{\mu }_{0}NI}{2R}$
At a point, on the axis of the coil,
$B_a=\frac{{\mu }_{0}NI{R}^2}{2(R^2+x^2{)}^{3/2}}$

As, $B_a=\frac{1}{8}{B}_c$

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

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

Taking cube root of both sides, we get,

$\frac{R}{(R^2+x^2{)}^{1/2}}=\frac{1}{2}$

Squaring both sides,

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

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

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

Hence, $\left(B\right)$ is the correct answer.