Question

A circular coil of radius R carries a current I . The magnetic field as it's centre is B. At what distance from the centre ,on the axis of the coil ,the magnetic field will be B/8:-

Answer 1

As you know that magnetic field at point on the axis of current carrying ring is
$\bold{B_x =\frac{\mu_0NiR^2}{2(R^2+x^2)^{\frac{3}{2}}}}$
where x is the point on the axis of ring, R is the radius of ring , i is the current carrying on ring and N is the number of turns .

at centre magnetic field , $\bold{B_c =\frac{\mu_0Ni}{2R}}$

Now, A/C to question,
magnetic field at x = magnetic field at centre/8
$\frac{\mu_0NiR^2}{2(R^2+x^2)^{3/2}}=\frac{\mu_0Ni}{8\times2R}\\\\\frac{R^2}{2(R^2+x^2)^{3/2}}=\frac{1}{16R}\\\\8R^3=(R^2+x^2)^{3/2}$

This is possible only when x = ±√3R
Hence, √3R distance from the centre magnetic field is equal to magnetic field at centre .

Answer 2

As you know that magnetic field at point on the axis of current carrying ring is

where x is the point on the axis of ring, R is the radius of ring , i is the current carrying on ring and N is the number of turns .

at centre magnetic field ,  

Now, A/C to question,

magnetic field at x = magnetic field at centre/8

This is possible only when x = ±√3R

Hence, √3R distance from the centre magnetic field is equal to magnetic field at centre .