Ratio of Magnetic Fields at Centre of Semicircular Arcs

To find the ratio of the magnitudes of the magnetic fields $B_1$ and $B_2$ at the centers $P$ and $Q$ of the semicircular arcs, we analyze the contribution of each segment of the wires using the Biot-Savart law.

---

1. Calculation of Magnetic Field $B_1$ at Point $P$ (Figure I)

Figure (I) consists of three current-carrying segments:

1. Top semi-infinite horizontal wire: Extends from $x = -\infty$ to $x = 0$ at a distance $r$ above point $P$. Current $I$ flows in the $+x$ direction. The magnetic field at $P$ due to this semi-infinite wire is: $B_1^{(1)} = \frac{\mu_0 I}{4\pi r} \quad \text{(pointing into the page, } \otimes\text{)}$

2. Semicircular arc: Has a radius $r$ and carries current $I$ in a clockwise direction. The magnetic field at the center $P$ due to the semicircular arc is: $B_1^{(2)} = \frac{\mu_0 I}{4r} = \frac{\mu_0 I \pi}{4\pi r} \quad \text{(pointing into the page, } \otimes\text{)}$

3. Bottom semi-infinite horizontal wire: Extends from $x = 0$ to $x = -\infty$ at a distance $r$ below point $P$. Current $I$ flows in the $-x$ direction. The magnetic field at $P$ due to this semi-infinite wire is: $B_1^{(3)} = \frac{\mu_0 I}{4\pi r} \quad \text{(pointing into the page, } \otimes\text{)}$

Since all three magnetic field components point into the page ($\otimes$), they add up algebraically: $B_1 = B_1^{(1)} + B_1^{(2)} + B_1^{(3)}$ $B_1 = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I \pi}{4\pi r} + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} (2 + \pi)$

---

2. Calculation of Magnetic Field $B_2$ at Point $Q$ (Figure II)

Figure (II) also consists of three current-carrying segments:

1. Top semi-infinite horizontal wire: Identical to the top wire in Figure (I). Its magnetic field contribution at $Q$ is: $B_2^{(1)} = \frac{\mu_0 I}{4\pi r} \quad \text{(pointing into the page, } \otimes\text{)}$

2. Semicircular arc: Identical to the semicircular arc in Figure (I). Its magnetic field contribution at $Q$ is: $B_2^{(2)} = \frac{\mu_0 I}{4r} = \frac{\mu_0 I \pi}{4\pi r} \quad \text{(pointing into the page, } \otimes\text{)}$

3. Vertical semi-infinite wire: Extends vertically downwards from the end of the semicircular arc along the line passing through point $Q$. Since the line along which this wire lies passes directly through point $Q$, the angle between the current element vector $d\vec{l}$ and the position vector $\vec{r}$ is $180^\circ$ ($\sin 180^\circ = 0$). Thus, its contribution to the magnetic field at $Q$ is zero: $B_2^{(3)} = 0$

Summing the non-zero magnetic field contributions at $Q$: $B_2 = B_2^{(1)} + B_2^{(2)} + B_2^{(3)}$ $B_2 = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I \pi}{4\pi r} + 0 = \frac{\mu_0 I}{4\pi r} (1 + \pi)$

---

3. Ratio of the Magnetic Fields

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

Thus, the correct option is A.