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(C) The magnetic field at the center due to a circular arc of radius $R$ and angle $	heta$ is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$. The stra

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$B_3 > B_1 > B_2$

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(C) The magnetic field at the center due to a circular arc of radius $R$ and angle $	heta$ is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$. The straight radial segments do not contribute to the magnetic field at the center.For circuit $1$: The field is due to a large arc of radius $3r$ and a small arc of radius $r$. Since the current flows in opposite directions in these two arcs relative to the center, the fields subtract: $B_1 = \frac{\mu_0 I}{4 \pi} (\frac{\pi}{3r} - \frac{\pi}{r}) = \frac{\mu_0 I}{4} (\frac{1}{r} - \frac{1}{3r}) = \frac{\mu_0 I}{6r}$.For circuit $2$: The field is due to two arcs of radius $r$ and $3r$ covering an angle of $\pi/2$. The fields subtract: $B_2 = \frac{\mu_0 I}{4 \pi} (\frac{\pi/2}{r} - \frac{\pi/2}{3r}) = \frac{\mu_0 I}{8} (\frac{2}{3r}) = \frac{\mu_0 I}{12r}$.For circuit $3$: The field is due to two arcs of radius $r$ and $3r$ covering an angle of $3\pi/2$. The fields subtract: $B_3 = \frac{\mu_0 I}{4 \pi} (\frac{3\pi/2}{r} - \frac{3\pi/2}{3r}) = \frac{3\mu_0 I}{8} (\frac{2}{3r}) = \frac{\mu_0 I}{4r}$.Comparing the magnitudes: $B_3 = 0.25 \frac{\mu_0 I}{r}$, $B_1 = 0.166 \frac{\mu_0 I}{r}$, $B_2 = 0.083 \frac{\mu_0 I}{r}$.Thus, $B_3 > B_1 > B_2$.

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