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(C) The magnetic field at the center of a circular arc carrying current $i$ is given by $B = \frac{\mu_0 i 	heta}{4 \pi r}$,where $	heta$ is the ang

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$1$

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(C) The magnetic field at the center of a circular arc carrying current $i$ 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 the two arcs,the magnetic fields are $B_1 = \frac{\mu_0 i_1 	heta_1}{4 \pi r}$ and $B_2 = \frac{\mu_0 i_2 	heta_2}{4 \pi r}$.Since the arcs are connected in parallel to the same cell,the potential difference across them is the same. Thus,$i_1 R_1 = i_2 R_2$,where $R$ is the resistance of the arc. Since $R \propto 	ext{length} \propto 	heta$,we have $i_1 	heta_1 = i_2 	heta_2$,which implies $\frac{i_1}{i_2} = \frac{	heta_2}{	heta_1}$.Substituting this into the ratio of the magnetic fields:$\frac{B_1}{B_2} = \frac{i_1 	heta_1}{i_2 	heta_2} = \left( \frac{	heta_2}{	heta_1} ight) \left( \frac{	heta_1}{	heta_2} ight) = 1$.

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