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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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(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 $B = \frac{\mu_0 I 	heta}{4 \pi R}$.Let $R$ be the radius of the circular conductor. The current $I$ from the cell splits into two parts $I_1$ and $I_2$ through the arcs $ABC$ and $ADC$ respectively.The resistance of an arc is proportional to its length,which is proportional to the angle subtended at the center. Let $	heta_1 = 300^o$ be the angle for arc $ABC$ and $	heta_2 = 60^o$ be the angle for arc $ADC$.Since the arcs are in parallel,the potential difference $V$ across them is the same: $V = I_1 R_1 = I_2 R_2$,where $R_1 \propto 	heta_1$ and $R_2 \propto 	heta_2$.Thus,$I_1 	heta_1 = I_2 	heta_2$,which implies $I_1 (300^o) = I_2 (60^o)$,or $I_1 = \frac{I_2}{5}$.The magnetic field due to arc $ABC$ is $B_1 = \frac{\mu_0 I_1 	heta_1}{4 \pi R}$ and due to arc $ADC$ is $B_2 = \frac{\mu_0 I_2 	heta_2}{4 \pi R}$.The ratio is $\frac{B_1}{B_2} = \frac{I_1 	heta_1}{I_2 	heta_2}$.Substituting $I_1 	heta_1 = I_2 	heta_2$,we get $\frac{B_1}{B_2} = \frac{I_2 	heta_2}{I_2 	heta_2} = 1$.

$A$ cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of center $O$ with angle $AOC = 60^o$. 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 $B_1/B_2$ is

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