An equilateral triangle is made by uniform wi | vedclass

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(D) The current $I$ enters at $A$ and splits into two paths: one through $AB$ and the other through $AC$. Since the wires are uniform, the resistance

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zero

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(D) The current $I$ enters at $A$ and splits into two paths: one through $AB$ and the other through $AC$. Since the wires are uniform, the resistance of path $AB$ is $R$ and the resistance of path $AC$ is $R$. At the midpoint of $BC$ (let's call it $M$), the current from $B$ to $M$ and $C$ to $M$ recombines to exit. Due to the symmetry of the equilateral triangle and the current distribution, the magnetic field produced by the current in segment $AB$ and $BM$ at the centroid $O$ is equal and opposite to the magnetic field produced by the current in segment $AC$ and $CM$ at the centroid $O$. Specifically, the magnetic field at the centroid $O$ due to the current flowing through the left half of the triangle $(A-B-M)$ is equal in magnitude but opposite in direction to the magnetic field due to the current flowing through the right half of the triangle $(A-C-M)$. Therefore, the net magnetic field at the centroid $O$ is $B_{net} = B_{left} + B_{right} = 0$.

An equilateral triangle is made by uniform wires $AB, BC, CA$. A current $I$ enters at $A$ and leaves from the midpoint of $BC$. If the length of each side of the triangle is $L$, the magnetic field $B$ at the centroid $O$ of the triangle is:

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