The resultant magnetic field at point O, due to current I in the conductor shown in figure is

$\vec{B}=\frac{\mu_0l}{4r}\left[\frac{2}{\pi}+1\right]$

The correct answer is option (2) : $\vec{B}=\frac{\mu_0l}{4r}\left[\frac{2}{\pi}+1\right]$

$\vec{B}_1=\frac{\mu_0l}{4\pi \gamma }(-\hat{k})$

$\vec{B}_2=\frac{\mu_0l}{4\pi }(\hat{k})$

$\vec{B}_3=\frac{\mu_0l}{4\pi }(-\hat{k})$

$\vec{B}=\vec{B}_1+\vec{B}_2+\vec{B}_3$

$=\left(\frac{\mu_0l}{4\pi \gamma }+\frac{\mu_0l}{4 \gamma }+\frac{\mu_0l}{4\pi \gamma }\right)=\frac{\mu_0l}{4\gamma }\left(\frac{2}{\pi}+1\right)(-\hat{k})$