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(A) Let the lengths of the two arcs be $l_1$ (arc $ABC$) and $l_2$ (arc $ADC$). The resistances of these arcs are $R_1 = \rho \frac{l_1}{A_{cross}}$ a

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(A) Let the lengths of the two arcs be $l_1$ (arc $ABC$) and $l_2$ (arc $ADC$). The resistances of these arcs are $R_1 = ho \frac{l_1}{A_{cross}}$ and $R_2 = ho \frac{l_2}{A_{cross}}$.Since the arcs are in parallel,the potential difference across them is the same: $V = I_1 R_1 = I_2 R_2$.This implies $I_1 l_1 = I_2 l_2$,or $\frac{I_1}{I_2} = \frac{l_2}{l_1}$.The magnetic field at the center due to an arc of length $l$ carrying current $i$ is $B = \frac{\mu_0 i 	heta}{4\pi R} = \frac{\mu_0 i l}{4\pi R^2}$.For the two arcs,the magnetic fields are $B_1 = \frac{\mu_0 I_1 l_1}{4\pi R^2}$ and $B_2 = \frac{\mu_0 I_2 l_2}{4\pi R^2}$.Since $I_1 l_1 = I_2 l_2$,we have $B_1 = B_2$.The currents flow in opposite directions around the center,so the magnetic fields produced by the two arcs are in opposite directions.Therefore,the resultant magnetic field at the center is $B_{net} = |B_1 - B_2| = 0$.

$A$ uniform wire is bent in the form of a circle of radius $R$. $A$ current $I$ enters at $A$ and leaves at $C$ as shown in the figure. If the length $ABC$ is half of the length $ADC$,the magnetic field at the centre $O$ will be

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