Magnetic fields at two points on the axis of a circular coil at a distance of $0.05 \mathrm{~m}$ and $0.2 \mathrm{~m}$ from the centre are in the ratio $8: 1$. The radius of coil is_________.
Correct Option: , 2
We know, the magnetic field on the axis of a current carrying circular ring is given by
$\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{NIA}}{\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$
$\therefore \frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}}=\frac{8}{1}=\left[\frac{\mathrm{R}^{2}+(0.2)^{2}}{\mathrm{R}^{2}+(0.05)^{2}}\right]^{3 / 2}$
$4\left[R^{2}+(0.05)^{2}\right]=\left[R^{2}+(0.2)^{2}\right]$
$4 R^{2}-R^{2}=(0.2)^{2}-4 \times(0.05)^{2}$
$4 R^{2}-R^{2}=(0.2)^{2}-(0.1)^{2}$
$3 \mathrm{R}^{2}=0.3 \times 0.1$
$\mathrm{R}^{2}=(0.1)^{2} \Rightarrow \mathrm{R}=0.1$