A current $\mathrm{I}$ is flowing through the sides of an , equilateral triangle of side a. The magnitude of the magnetic field at the centroid of the triangle is

Refer to figure let, $\mathrm{AB}=\mathrm{BC}=\mathrm{AC}=\mathrm{a}$, Let $\mathrm{OD}=\mathrm{r}$.

The magnetic field at the centroid $\mathrm{O}$ due to current $\mathrm{I}$ flowing inside $\mathrm{AB}$ of the triangle is given by 

$B_{AB}=\frac{{\mu }_{0}I}{4\mathrm{\pi r}}\left(\sin \alpha +\sin \beta \right)$

It is clear that $\alpha =\beta =60°$

 $$\begin{gathered} \mathrm{OD}=\mathrm{r}=\frac{\mathrm{AD}}{\mathrm{tan\alpha }} \\ =\frac{\mathrm{a}/2}{\tan 60°} \\ =\frac{\mathrm{a}/2}{\sqrt{3}} \\ =\frac{\mathrm{a}}{2\sqrt{3}} \end{gathered}$$

$\therefore {\mathrm{B}}_{\mathrm{AB}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\times \frac{2\sqrt{3}}{\mathrm{a}}\left(\sin 60°+\sin 60°\right)$

$=\frac{3{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi a}}$

By symmetry, the magnetic fields due to current in sides $\mathrm{BC}$ and  $\mathrm{AC}=$ that due to side $\mathrm{AB}$. Hence, the magnetic field at $\mathrm{O}$ due to the current in the three sides of triangle $\mathrm{ABC}$ is

              ${\mathrm{B}}_{\mathrm{AB}}+{\mathrm{B}}_{\mathrm{BC}}+{\mathrm{B}}_{\mathrm{CA}}=3{\mathrm{B}}_{\mathrm{AB}}=\frac{9{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi a}}$