A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is (Assume that, the current is flowing in the clockwise direction.)

Given, electric current, I = 1.5 A

Side of triangle, l = 9 cm = 9 × 10^-2 m

Let OD = OE = OF = r

       $\mathrm{AD}=\frac{1}{2}$

$\begin{array}{l}\ln \mathrm{\Delta AOD}, \frac{\mathrm{OD}}{\mathrm{AD}}=\tan {30}^{\circ }\\ \mathrm{OD}=\mathrm{AD} \tan {30}^{\circ }\\ \therefore \mathrm{r}=\mathrm{OD}=\frac{\mathrm{l}}{2}\times \frac{1}{\sqrt{3}}=\frac{\mathrm{l}}{2\sqrt{3}}\end{array}$

Magnetic field due to AB at O,

$\begin{array}{l} {\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi r}}\left[\sin {60}^{\circ }+\sin {60}^{\circ }\right]=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi r}}\times \frac{2\sqrt{3}}{2}\\ \Rightarrow {\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{I}\sqrt{3}}{4\mathrm{\pi r}} (\mathrm{perpendicular} \mathrm{inward})\end{array}$

Now, total magnetic field at O due to all sides,

               $\mathrm{B}=3{\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{l}}{4\mathrm{\pi r}}3\sqrt{3}$

By substituting the values, we get

             $\mathrm{B}=\frac{{10}^{-7}\times 1.5\times 3\sqrt{3}\times 2\sqrt{3}}{9\times {10}^{-2}}=3\times {10}^{-5} \mathrm{T} (\mathrm{inside} \mathrm{the} \mathrm{plane} \mathrm{of} \mathrm{triangle})$