
$\mathrm{B}=\mathrm{B}_{2}+\mathrm{B}_{3}\left(\mathrm{B}_{2}=\mathrm{B}_{3}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}} \odot\right)$
$\mathrm{B}=\frac{2 \mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \odot$
Reason : $I_1 = I_2$ implies that the fields due to the current $I_1$ and $I_2$ will be balanced.