The ratio of magnetic field at the centre of a current carrying coil of radius $$r$$ to the magnetic field at distance $$r$$ from the centre of coil on its axis is $$\sqrt{x}$$ : 1. The value of $$x$$ is _____.

The magnetic field at the centre of a current-carrying coil of radius $$r$$ is:

$$B_{\text{centre}} = \frac{\mu_0 I}{2r}$$

The magnetic field at a distance $$r$$ from the centre on the axis of the coil is:

$$B_{\text{axis}} = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2 \cdot 2\sqrt{2} \cdot r^3} = \frac{\mu_0 I}{4\sqrt{2} \cdot r}$$

Now taking the ratio:

$$\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4\sqrt{2} r}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} = \sqrt{8}$$

We are given that the ratio is $$\sqrt{x} : 1$$, so:

$$\sqrt{x} = 2\sqrt{2} = \sqrt{8}$$

$$x = 8$$

Hence, the answer is $$8$$.