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(A) The current $i$ splits into $i_1$ and $i_2$ at the junction. Since the two arcs are in parallel,the potential difference across them is the same.

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(A) The current $i$ splits into $i_1$ and $i_2$ at the junction. Since the two arcs are in parallel,the potential difference across them is the same. Thus,$i_1 R_1 = i_2 R_2$,where $R_1$ and $R_2$ are the resistances of the arcs. Since $R \propto 	ext{length} \propto 	heta$,we have $i_1 	heta_1 = i_2 	heta_2$. Given $	heta_1 = 90^\circ = \pi/2$ and $	heta_2 = 270^\circ = 3\pi/2$,we get $i_1(\pi/2) = i_2(3\pi/2)$,which implies $i_2 = 3i_1$. Since $i_1 + i_2 = i$,we have $i_1 + 3i_1 = i$,so $i_1 = i/4$ and $i_2 = 3i/4$.The magnetic field at the center due to an arc of angle $	heta$ is $B = \frac{\mu_0 i 	heta}{4 \pi R}$.For the upper arc (angle $270^\circ = 3\pi/2$): $B_1 = \frac{\mu_0 i_1 (3\pi/2)}{4 \pi R} = \frac{3 \mu_0 i_1}{8 R} = \frac{3 \mu_0 (i/4)}{8 R} = \frac{3 \mu_0 i}{32 R}$ (inward).For the lower arc (angle $90^\circ = \pi/2$): $B_2 = \frac{\mu_0 i_2 (\pi/2)}{4 \pi R} = \frac{\mu_0 i_2}{8 R} = \frac{\mu_0 (3i/4)}{8 R} = \frac{3 \mu_0 i}{32 R}$ (outward).Since $B_1$ and $B_2$ are equal in magnitude and opposite in direction,the net magnetic field at the center $P$ is $B_{net} = B_1 - B_2 = 0$.

$A$ straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is

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