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(C) The magnetic field at the center of a circular arc is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$, where $	heta$ is the angle subtended by the

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$6:5$

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(C) The magnetic field at the center of a circular arc is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$, where $	heta$ is the angle subtended by the arc at the center in radians.For wire $A$: $I_A = 2 \, A$, $R_A = 2 \, cm$, and the angle subtended is $	heta_A = 360^\circ - 90^\circ = 270^\circ = \frac{3 \pi}{2} \, 	ext{radians}$.Thus, $B_A = \frac{\mu_0 (2) (3 \pi / 2)}{4 \pi (2)} = \frac{3 \mu_0}{8}$.For wire $B$: $I_B = 3 \, A$, $R_B = 4 \, cm$, and the angle subtended is $	heta_B = 360^\circ - 60^\circ = 300^\circ = \frac{5 \pi}{3} \, 	ext{radians}$.Thus, $B_B = \frac{\mu_0 (3) (5 \pi / 3)}{4 \pi (4)} = \frac{5 \mu_0}{16}$.The ratio is $\frac{B_A}{B_B} = \frac{3 \mu_0 / 8}{5 \mu_0 / 16} = \frac{3}{8} 	imes \frac{16}{5} = \frac{6}{5}$.

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