What is the magnetic field at centre O of an equilateral triangle of side 2 cm with current flowing as shown below ? (Resistance of part ABC is 2Ω, and resistance of part ADC is 4Ω )

Let the total potential difference of the arrangement be V.
The equivalent resistance of the circuit is
$\begin{array}{l}\frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\\ \Rightarrow {\mathrm{R}}_{\mathrm{P}}=\frac{4}{3}\end{array}$

By ohm's law we know that V = IR

so, the potential difference of the total circuit is $V=10\times \frac{4}{3}=\frac{40}{3}$

Current in branch ABC will be

${\mathrm{I}}_1=\frac{\mathrm{V}}{{\mathrm{R}}_1}=\frac{40/3}{2}=\frac{20}{3}$

Current in branch ADC will be

${\mathrm{I}}_2=\frac{\mathrm{V}}{{\mathrm{R}}_2}=\frac{40/3}{4}=\frac{10}{3}$

Since current in branch ABC is greater than current in Branch ADC. the net magnetic field will be in the direction of the magnetic field due to branch ABC.

$\begin{array}{l}\text{ i.e }{\mathrm{B}}_{\mathrm{net}}={\mathrm{B}}_{\mathrm{ABC}}-{\mathrm{B}}_{\mathrm{ADC}}\\ =\frac{9}{4}\frac{{\mathrm{\mu }}_0{\mathrm{I}}_1}{\mathrm{\pi a}}-\frac{9}{4}\frac{{\mathrm{\mu }}_0{\mathrm{I}}_2}{\mathrm{\pi a}}\\ =\frac{9}{4}\frac{{\mathrm{\mu }}_0}{\mathrm{\pi a}}\left({\mathrm{I}}_1-{\mathrm{I}}_2\right)\odot \\ =\frac{9}{4}\times \frac{4\mathrm{\pi }\times {10}^{-7}}{\mathrm{\pi }\times 2\times {10}^{-2}}\left(\frac{20}{3}-\frac{10}{3}\right)\odot \\ =1.5\times {10}^{-4}\mathrm{T}\end{array}$