The magnetic field at the center of current carrying circular loop is $$B_1$$. The magnetic field at a distance of $$\sqrt{3}$$ times radius of the given circular loop from the center on its axis is $$B_2$$. The value of $$\frac{B_1}{B_2}$$ will be

The magnetic field at the center of a current-carrying circular loop of radius $$R$$ is $$B_1 = \frac{\mu_0 I}{2R}$$.

The magnetic field on the axis of the loop at a distance $$d$$ from the center is given by $$B = \frac{\mu_0 I R^2}{2(R^2 + d^2)^{3/2}}$$.

Here, $$d = \sqrt{3}R$$, so $$R^2 + d^2 = R^2 + 3R^2 = 4R^2$$, and $$(R^2 + d^2)^{3/2} = (4R^2)^{3/2} = 8R^3$$.

So $$B_2 = \frac{\mu_0 I R^2}{2 \times 8R^3} = \frac{\mu_0 I}{16R}$$.

Now, $$\frac{B_1}{B_2} = \frac{\mu_0 I / 2R}{\mu_0 I / 16R} = \frac{16R}{2R} = 8$$.

So $$\frac{B_1}{B_2} = 8:1$$.

Hence, the correct answer is Option C.