Question

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 (1/8)th of its value at the center of the coil, is :

• A. $R/\sqrt{3}$
• B. $R\sqrt{3}$
• C. $2R\sqrt{3}$
• D. $\left(2\sqrt{3}\right)R$

Answer 1

Answer: (B)

$R\sqrt{3}$

$B_{axis}=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{{(R^2+x^2)}^{3/2}}$
At centre, $B_{center}=\frac{{\mu }_0}{4\pi }\times \frac{2\pi l}{R}$
In the given problem,
$\frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{{(R^2+x^2)}^{3/2}}=\frac{1}{8}[\frac{{\mu }_0}{4\pi }\times \frac{2\pi l}{R}]$
or ${(R^2+x^2)}^{3/2}=8R^3$
solving we get $x=R\sqrt{3}$