Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres P and Q of a semicircular arc are B$$_1$$ and B$$_2$$ respectively, then the ratio $$\frac{B_1}{B_2}$$ is __________.

We need the magnetic fields at the centres P and Q of semicircular arcs in two different wire configurations.

The magnetic field at the centre of a semicircular arc of radius $$r$$ carrying current $$I$$ is:

$$B_{semi} = \frac{\mu_0 I}{4r}$$

The magnetic field due to a semi-infinite straight wire at a perpendicular distance $$d$$ from its end is:

$$B_{wire} = \frac{\mu_0 I}{4\pi d}$$

Also, an important fact: if the observation point lies on the line of the wire (i.e., on the wire's axis extended), the magnetic field from that wire at the point is zero (since $$d\vec{l} \times \hat{r} = 0$$).

Figure (I) -- Field at P:

The wire comes from infinity on the left along the top, goes to the right, makes a semicircular arc of radius $$r$$ (centre P, opening to the right), and returns to the left along the bottom to infinity.

The two straight portions are horizontal, each at a perpendicular distance $$r$$ from P. Neither wire's axis passes through P, so both contribute a non-zero field.

Using the right-hand rule, the semicircular arc (current going clockwise from top to bottom) and both semi-infinite wires all produce a field at P directed into the page.

$$B_1 = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r}$$

$$= \frac{\mu_0 I}{4\pi r}(\pi + 1 + 1) = \frac{\mu_0 I}{4\pi r}(\pi + 2)$$

Figure (II) -- Field at Q:

The wire comes horizontally from the right, makes a semicircular arc of radius $$r$$ (centre Q, opening to the right), and then goes vertically downward to infinity.

The horizontal wire is at perpendicular distance $$r$$ from Q, so it contributes $$\frac{\mu_0 I}{4\pi r}$$ (into the page by right-hand rule).

The vertical wire starts from the bottom of the semicircle, which is directly below Q at distance $$r$$. When extended, this wire's axis passes through Q. Therefore, Q lies on the line of the wire, and the magnetic field at Q due to this wire is zero.

$$B_2 = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r} + 0$$

$$= \frac{\mu_0 I}{4\pi r}(\pi + 1)$$

Now we take the ratio:

$$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi r}(\pi + 2)}{\frac{\mu_0 I}{4\pi r}(\pi + 1)} = \frac{\pi + 2}{\pi + 1}$$

Hence, the correct answer is Option A.