The resistance of the part $\mathrm{ABC}$ will be: $\mathrm{R}_{1}=\rho l_{1}$

The resistance of the part $\mathrm{ADC}$ will be: $\mathrm{R}_{2}=\rho l_{2}$

Potential difference across

$\mathrm{AC}=\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}$

or $I_{1} \rho l_{1}=I_{2} \rho l_{2}$

or $\mathrm{I}_{1} l_{1}=\mathrm{I}_{2} l_{2}.........(i)$

Magnetic field at $O$ due to a current $I_{1}$ flowing in $ABC$ is given by:

$\mathrm{B}_{1}=\frac{\mu_{0} I_{1} \theta_{1}}{4 \pi R}=\frac{\mu_{0} I_{1} l_{1}}{4 \pi R^{2}} \otimes \quad\left(\text { As } l_{1}=\theta_{1} R\right)$

Magnetic field at $O$ due to current $I_2$, flowing in $\mathrm{ADC}$ is given by :-

$\mathrm{B}_{2}=\frac{\mu_{0} I_{2} \theta_{2}}{4 \pi R}=\frac{\mu_{0} I_{2} l_{2}}{4 \pi R^{2}} \odot \quad\left(\text { As } l_{2}=\theta_{2} R\right)$

or $\left.\quad \mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{1} l_{1}}{4 \pi \mathrm{R}^{2}} \odot \quad \text { [Using eqn. (i) }\right]$

The resultant magnetic field induction at $\mathrm{O}$ is, $\mathrm{B}=\mathrm{B}_{2}-\mathrm{B}_{1}=0$

$\left(\mathrm{As} \,\,\,\mathrm{B}_{2}=\mathrm{B}_{1}\right)$