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(C) For wire $I$: The magnetic field at center $P$ is the sum of the fields due to the two straight segments and the semicircular arc. The field due t

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$\frac{2+\pi}{1-\pi}$

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(C) For wire $I$: The magnetic field at center $P$ is the sum of the fields due to the two straight segments and the semicircular arc. The field due to a semi-infinite wire at distance $r$ is $B_{straight} = \frac{\mu_0 I}{4\pi r}$. Since there are two such segments, their contribution is $2 	imes \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{2\pi r}$. The field due to the semicircular arc is $B_{arc} = \frac{\mu_0 I}{4r}$. Thus, $B_1 = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} (\frac{2}{\pi} + 1) = \frac{\mu_0 I}{4r} (\frac{2+\pi}{\pi})$.For wire $II$: The magnetic field at center $Q$ is the difference between the field due to the semicircular arc and the field due to the straight segments. The straight segments contribute $B_{straight} = \frac{\mu_0 I}{4\pi r}$ each. However, based on the geometry, the net field is $B_2 = \frac{\mu_0 I}{4r} - \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4r} (1 - \frac{1}{\pi}) = \frac{\mu_0 I}{4r} (\frac{\pi-1}{\pi})$.Taking the ratio $\frac{B_1}{B_2} = \frac{\frac{2+\pi}{\pi}}{\frac{\pi-1}{\pi}} = \frac{2+\pi}{\pi-1}$.

Two identical long current-carrying wires are bent into the shapes shown in the following figures. If the magnitudes of the magnetic fields at the centers $P$ and $Q$ of the semicircular arcs are $B_1$ and $B_2$ respectively, then the ratio $\frac{B_1}{B_2}$ is . . . . . . .

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