Question

A circular current carrying coil has a radius $R$. The distance of a point from the center of the coil on the axis where the magnetic induction will be $1/8^{\text{th}}$ of its value at the center of the coil -

$[$0.77 Mark$]$

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

Answer 1

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

We know that magnetic induction due to a current carrying coil at a point on its axis is given by,

$B=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{(R^2+x^2{)}^{3/2}}$

At its center, $x=0$

So, $B_O=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I}{R}$

Suppose, at $x=r$, from the center of the coil on the axis where the magnetic induction is $1/8^{\text{th}}$ of its value at the center of the coil.

$\therefore B_{x=r}=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{(R^2+r^2{)}^{3/2}}$

$\Rightarrow \frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{(R^2+r^2{)}^{3/2}}=\frac{1}{8}\times \frac{{\mu }_0}{4\pi }\times \frac{2\pi I}{R}\left[\text{Given}\right]$

$\Rightarrow 8R^3=(R^2+r^2{)}^{3/2}$

On solving this, we get,

$r=R\sqrt{3}$