The magnetic induction field at the centroid of an equilateral triangle of side ‘l’ and carrying a current ‘i’ is

$\begin{array}{l}\tan {60}^{\circ }=\frac{\mathrm{AD}}{\mathrm{B\theta }}\\ \sqrt{3}=\frac{\mathrm{l}/2}{\mathrm{r}};\mathrm{r}=\frac{\mathrm{l}}{2\sqrt{3}}\\ {\mathrm{\theta }}_1={\mathrm{\theta }}_2={60}^{\circ }\end{array}$
Magnetic induction due to AB part is
$\begin{array}{l}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi r}}\left(\sin {60}^{\circ }+\sin {60}^{\circ }\right)\\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{i}\times 2\sqrt{3}}{\mathrm{l}}\left(\sin {60}^{\circ }+\sin {60}^{\circ }\right)\\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{i}}{\mathrm{l}}2\sqrt{3}\times \frac{\sqrt{3}}{2}\times 2\Rightarrow \mathrm{B}=\frac{6{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi l}}\end{array}$
Magnetic induction due to entire triangle
${\mathrm{B}}^1=3\mathrm{B}=\frac{18{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi l}}\Rightarrow {\mathrm{B}}^{\mathrm{\prime }}=\frac{9{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi l}}$