$$B_X$$ and $$B_Y$$ are the magnetic field at the centre of two coils $$X$$ and $$Y$$ respectively, each carrying equal current. If coil $$X$$ has $$200$$ turns and $$20 \text{ cm}$$ radius and coil $$Y$$ has $$400$$ turns and $$20 \text{ cm}$$ radius, the ratio of $$B_X$$ and $$B_Y$$ is

We need to find the ratio $$B_X : B_Y$$ for two coils carrying equal current. Coil $$X$$ has $$N_X = 200$$ turns and radius $$R_X = 20 \text{ cm}$$. Coil $$Y$$ has $$N_Y = 400$$ turns and radius $$R_Y = 20 \text{ cm}$$.

The magnetic field at the centre of a circular coil of $$N$$ turns, radius $$R$$, carrying current $$I$$ is:

$$B = \frac{\mu_0 N I}{2R}$$

For coil $$X$$:

$$B_X = \frac{\mu_0 N_X I}{2R_X} = \frac{\mu_0 \times 200 \times I}{2 \times 20}$$

For coil $$Y$$:

$$B_Y = \frac{\mu_0 N_Y I}{2R_Y} = \frac{\mu_0 \times 400 \times I}{2 \times 20}$$

Since both coils have the same radius ($$R_X = R_Y = 20 \text{ cm}$$) and carry the same current:

$$\frac{B_X}{B_Y} = \frac{N_X}{N_Y} = \frac{200}{400} = \frac{1}{2}$$

Therefore, $$B_X : B_Y = 1 : 2$$.

The correct answer is Option B: $$1 : 2$$.