A cell is connected between the points A and C of a circular conductor ABCD of centre O, angle AOC = 60°. If B1 and B2 are the magnitude of the magnetic fields at O due to the currents in ABC and ADC respectively, the ratio of B1/B2 is 
Answer & explanation
Correct answer: option 4
Let the lengths of ABC and ADC parts be l1 and l2 respectively. Then
$\frac{l_1}{l_2} = \frac{360 - 60}{60} = 5$
Further $\frac{i_1}{i_2} = \frac{l_2}{l_1} = \frac{1}{5} \Rightarrow i_2 = 5i_1$
$B_1 = \frac{5}{6} \frac{\mu_0 i_1}{2r}$
$B_2 = \frac{1}{6} \frac{\mu_0 i_2}{2r} = \frac{1}{6} \frac{\mu_0 5 i_1}{2r} = \frac{5}{6} \frac{\mu_0 i_1}{2r} = B_1$