Question

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

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

Answer 1

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


The field on the axis of a current carrying at distance "x" from centre is given by
$B=\frac{{\mu }_{0}Ni{R}^2}{2(R^2+x^2{)}^{\frac{3}{2}}}$
At centre, x = 0
$\therefore B_{centre}=\frac{{\mu }_{0}Ni}{2R}$
$\frac{B_{centre}}{B_{axis}}={(1+\frac{x^2}{R^2})}^{\frac{3}{2}}$
also $B_{axis}=\frac{1}{8}{B}_{centre}$
$\Rightarrow \frac{8}{1}={(1+\frac{x^2}{R^2})}^{\frac{3}{2}}\Rightarrow 2={(1+\frac{x^2}{R^2})}^{\frac{1}{2}}$
$\Rightarrow 4=1+\frac{x^2}{R^2}\Rightarrow 3=\frac{x^2}{R^2}\Rightarrow x^2=3R^2\Rightarrow x=\sqrt{3}R$