Discussion Forum

A circular coil carrying current 'l' has radius ' R'  and magnetic  field at the  centre ' B'  . At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?

Answer:

Explanation:

According to the question, 

Magnetic field at the centre of circular coil is given by 

$B_{centre}=\frac{\mu_{0}Ni}{2R}$  ...........(i)

 $B_{axis}=\frac{\mu_{0}}{4\pi}\frac{2\pi Ni R^{2}}{(x^{2}+R^{2})^{3/2}}$

$\frac{B}{8}=\frac{\mu_{0} Ni R^{2}}{2(x^{2}+R^{2})^{3/2}}$

 From Eq.(i) , we get

 $\frac{\mu_{0} Ni}{8\times 2R}=\frac{\mu_{0} Ni R^{2}}{2(x^{2}+R^{2})^{3/2}}$

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

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

$8 R^{3}=(x^{2}+R^{2})^{3/2}$  (on taking cube root)

$2 R^{}=(x^{2}+R^{2})^{1/2}$  ( on taking square root)

$4 R^{2}=(x^{2}+R^{2})^{}$

 $4 R^{2}-R^{2}=x^{2}$

$3 R^{2}=x^{2}$

x= $R\sqrt{3}$