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(C) The magnetic field at the center of a circular arc of radius $r$ carrying current $i$ subtending an angle $	heta$ at the center is given by $B =

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$1$

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(C) The magnetic field at the center of a circular arc of radius $r$ carrying current $i$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 i 	heta}{4\pi r}$.For the two arcs $ABC$ and $ADC$,the magnetic fields are $B_1 = \frac{\mu_0 i_1 	heta_1}{4\pi r}$ and $B_2 = \frac{\mu_0 i_2 	heta_2}{4\pi r}$,where $	heta_1 = 300^{\circ}$ and $	heta_2 = 60^{\circ}$.Since the two arcs are connected in parallel,the potential difference $V$ across them is the same. Thus,$V = i_1 R_1 = i_2 R_2$,where $R_1$ and $R_2$ are the resistances of the arcs $ABC$ and $ADC$ respectively.The resistance is proportional to the length of the arc,so $R \propto 	heta$. Therefore,$i_1 	heta_1 = i_2 	heta_2$,which implies $\frac{i_1}{i_2} = \frac{	heta_2}{	heta_1}$.Substituting this into the ratio of the magnetic fields: $\frac{B_1}{B_2} = \frac{i_1 	heta_1}{i_2 	heta_2} = \left(\frac{	heta_2}{	heta_1}ight) 	imes \left(\frac{	heta_1}{	heta_2}ight) = 1$.

$A$ cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of centre $O$ with angle $AOC = 60^{\circ}$. If $B_1$ and $B_2$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC$ and $ADC$ respectively,the ratio $\frac{B_1}{B_2}$ is

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