Question:
A current of $1.5 \mathrm{~A}$ is flowing through a triangle, of side $9 \mathrm{~cm}$ each. The magnetic field at the centroid of the triangle is:
(Assume that the current is flowing in the clockwise direction.)
Correct Option: , 4
Solution:
$\mathrm{B}=3\left[\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)\right]$
$\tan 60^{\circ}=\frac{\ell / 2}{r}$
Where $\mathrm{r}=\frac{9 \times 10^{-2}}{2 \sqrt{3}} \mathrm{M}$
$\therefore B=3 \times 10^{-5} \mathrm{~T}$
Current is flowing in clockwise direction so, $\overrightarrow{\mathrm{B}}$ is inside plane of triangle by right hand rule.