A circular current carrying coil has a radius .R. The distance from the centre of the coil, on the axis where the magnetic induction will be 1/8th to its value at the centre of the coil is

The magnetic field on the axis of a current i carrying coil of turns n, radius R and at a distance x from the centre of coil isB=μ04π×2πiR2R2+x23/2At centre Bcentre =μ04π×2πiR (∵x=0)In the given problemμ04π×2πiR2R2+x23/2=18μ04π×2πiRThis givesR2+x23/2=8R3Solving, we get x=3⋅R