Field On The Axis Of A Circular Loop

From NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), the axial field of a circular loop at distance $x$ from its centre is $B_{axial}=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$, while the central field ($x=0$) is $B_{centre}=\frac{{\mu }_{0}I}{2R}$. Taking the ratio, $\frac{B_{axial}}{B_{centre}}=\frac{R^3}{(R^2+x^2{)}^{3/2}}$. At $x=R$, the denominator becomes $(R^2+R^2{)}^{3/2}=(2R^2{)}^{3/2}=2\sqrt{2}{R}^3$, so the ratio is $\frac{R^3}{2\sqrt{2}{R}^3}=\frac{1}{2\sqrt{2}}\approx 0.354$. The '$1/2$' option assumes a simple inverse-square drop, which is wrong. The 'equal' option ignores the distance dependence. The '$1/4$' option misuses the exponent. Plausibility check: the axial field must be smaller than the central field but not negligibly so at a distance equal to the radius, and $\frac{1}{2\sqrt{2}}\approx 0.35$ lies sensibly between 0 and 1. As a further test of the formula, at very large distances $x\gg R$ the expression reduces to $B\propto 1/x^3$, the characteristic dipole fall-off, which is exactly what a current loop should behave like when viewed from far away, so the general axial formula used here is consistent with the known dipole limit and the intermediate value at $x=R$ is trustworthy.