Discussion Forum

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 $\frac{1}{8}$ th to its value at the centre of the coil is 

Answer:

Explanation:

 Here the ratio, $\frac{B_{centre}}{B_{axis}}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3/2}$

Also, $B_{axis}=\frac{1}{8}B_{centre}$

$\frac{8}{1}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3/2} \Rightarrow   4=1+\frac{x^{2}}{R^{2}}$

$3=\frac{x^{2}}{R^{2}} \Rightarrow X^{2}=3R^{2}$

or ,$x= \sqrt{3}R$