Question

A circular coil of radius r is carrying current I. at what distance from the centre of loop on the axis magnetic field is one eight the magnetic field at the centre.

Answer 1

We know,
$\underset{B}{\to }$ at axial point of current carrying circular loop,
$B=\frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{3/2}}$
Where R= radius of circular loop.
x= distance of point from the centre.
according to the question,
$$\begin{gathered} B_{atx}=\frac{1}{8}{B}_{centre} \\ \Rightarrow \frac{{\mu }_{0}I{R}^2}{2{\left(x^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{8}\frac{{\mu }_{0}I{R}^2}{2{\left(0^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \Rightarrow \frac{1}{{\left(x^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{8}\frac{1}{{\left(R^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \Rightarrow \frac{1}{\left(x^2+R^2\right)}={\left(\frac{1}{8}\right)}^{\frac{2}{3}}\frac{1}{\left(R^2\right)} \end{gathered}$$
After calculating,
Hence required distance, x = $\sqrt[\pm ]{3}R$