Question

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

• A. $\frac{{\mu }_0}{4\pi }\left(\begin{array}{c}\frac{4I}{L}\end{array}\right)$
• B. $\frac{{\mu }_0}{2\pi }\left(\begin{array}{c}\frac{4I}{L}\end{array}\right)$
• C. $\frac{{\mu }_0}{4\pi }\left(\begin{array}{c}\frac{2I}{L}\end{array}\right)$
• D. Zero

Answer 1

Answer: (D)

Zero

Equal current flows in part 1 and 2 of triangular loop as shown in the figure.

As both part of the triangular loop is symmetric about point O, so magnetic field at O due to both parts is equal and opposite to each other. Thus magnitude of magnitude of magnetic field at O due to part 1 cancel out the magnitude of magnetic field at O due to part 2. Hence net magnetic field at O is zero.