A circular coil of radius R and N turns carries a current I. The magnetic field at its center is B. At what distance from the center along the axis o…

The magnetic field on the axis of a circular coil at distance $x$ is $B_x = \frac{\mu_0 NIR^2}{2(R^2 + x^2)^{3/2}}$. At the center ($x=0$), $B = \frac{\mu_0 NI}{2R}$. Given $B_x = B/8$, so $\frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8} \implies \frac{R}{\sqrt{R^2 + x^2}} = \frac{1}{2} \implies R^2 + x^2 = 4R^2 \implies x = \sqrt{3}R$.