Magnetic fields at two points on the axis of a circular coil at a...

We know, the magnetic field on the axis of a current carrying circular ring is given by

$B = \dfrac{{\mu}_{0}}{4\pi}\dfrac{2NIA}{{\left({R}^{2} + {x}^{2}\right)}^{3 / 2}}$

$∴ \dfrac{{B}_{1}}{{B}_{2}} = \dfrac{8}{1} = {\left[\dfrac{{R}^{2} + {\left(0 . 2\right)}^{2}}{{R}^{2} + {\left(0 . 05\right)}^{2}}\right]}^{3 / 2}$

$4\left[{R}^{2} + {\left(0 . 05\right)}^{2}\right] = \left[{R}^{2} + {\left(0 . 2\right)}^{2}\right]$

$4{R}^{2} - {R}^{2} = {\left(0 . 2\right)}^{2} - 4 \times {\left(0 . 05\right)}^{2}$

$4{R}^{2} - {R}^{2} = {\left(0 . 2\right)}^{2} - {\left(0 . 1\right)}^{2}$

$3{R}^{2} = 0 . 3 \times 0 . 1$

${R}^{2} = {\left(0 . 1\right)}^{2} ⇒ R = 0 . 1$