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 

1

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 $\frac{i_1}{i_2} = \frac{l_2}{l_1} = \frac{1}{5} \Rightarrow i_2 = 5i_1$

$B_1 = \frac{5}{6} \frac{\mu_0 i_1}{2r}$

$B_2 = \frac{1}{6} \frac{\mu_0 i_2}{2r} = \frac{1}{6} \frac{\mu_0 5 i_1}{2r} = \frac{5}{6} \frac{\mu_0 i_1}{2r} = B_1$