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

• A

  $\frac{\mu_{0}}{4 \pi}\left(\frac{4 L}{L}\right)$
• B

  $\frac{\mu_{0}}{2 \pi}\left(\frac{4 L}{L}\right)$
• C

  $\frac{\mu_{0}}{4 \pi}\left(\frac{2 L}{L}\right)$
• Correct answer
  D

  zero

Solution

The given situation can be shown as The magnetic field at centroid $O$ = Magnetic field due to left part + Magnetic field due to right part $\therefore$ $$ \mathrm{B}=\mathrm{B}_{1}+\mathrm{B}_{2}=0 $$ (: direction of $B$ due to both parts is different of $O$ )

Hence, the correct answer is option: 'D'.