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