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 is1RP=12+14=34⇒RP=43By ohm's law we know that V = IRso, the potential difference of the total circuit is V=10×43=403Current in branch ABC will beI1=VR1=40/32=203Current in branch ADC will beI2=VR2=40/34=103Since 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. i.e Bnet=BABC−BADC=94μ0I1πa−94μ0I2πa=94μ0πaI1−I2⊙=94×4π×10−7π×2×10−2203−103⊙=1.5×10−4T