An equilateral triangle is formed by the wire ABC as shown in the...

Let the total current entering at vertex $A$ be $I$. The current leaves from the midpoint $M$ of the side $BC$.

The current divides into two parallel paths: path $A \rightarrow B \rightarrow M$ and path $A \rightarrow C \rightarrow M$.

Since the wire is uniform, the resistance of each path is proportional to its length. The length of path $ABM$ is $L + L/2 = 3L/2$, and the length of path $ACM$ is $L + L/2 = 3L/2$.

As the lengths and resistances of both paths are equal, the current splits equally. The current in branch $ABM$ is $I/2$ and the current in branch $ACM$ is $I/2$.

Let $d$ be the perpendicular distance from the centroid $O$ to each side of the equilateral triangle.

The magnetic field at $O$ due to the segment $AB$ is B_{AB} = \dfrac{\mu_0 (I/2)}{4 \pi d} (\sin 60^{}^{{}^{\circ}} + \sin 60^{}^{{}^{\circ}}). By the right-hand rule, its direction is into the plane.

The magnetic field at $O$ due to the segment $AC$ is B_{AC} = \dfrac{\mu_0 (I/2)}{4 \pi d} (\sin 60^{}^{{}^{\circ}} + \sin 60^{}^{{}^{\circ}}). Its direction is out of the plane.

Since $B_{AB}$ and $B_{AC}$ are equal in magnitude and opposite in direction, they cancel each other out.

The magnetic field at $O$ due to the segment $BM$ is B_{BM} = \dfrac{\mu_0 (I/2)}{4 \pi d} (\sin 60^{}^{{}^{\circ}} + \sin 0^{}^{{}^{\circ}}). Its direction is out of the plane.

The magnetic field at $O$ due to the segment $CM$ is B_{CM} = \dfrac{\mu_0 (I/2)}{4 \pi d} (\sin 60^{}^{{}^{\circ}} + \sin 0^{}^{{}^{\circ}}). Its direction is into the plane.

Similarly, $B_{BM}$ and $B_{CM}$ are equal in magnitude and opposite in direction, so they also cancel each other out.

Therefore, the net magnetic field at the center $O$ of the triangle is zero.

Answer: Zero