Question

The wire $ABC$ shown in figure forms an equilateral triangle. Find the magnetic field $B$ at the centre $O$ of the triangle assuming the wire to be uniform.

Answer 1

For $AB$ $B$ is along $\odot$ $B=\frac{{\mu }_{0}i}{4\pi r}(\sin {60}^{\circ }+\sin {60}^{\circ })$

For $AC$ $B$ $\otimes$ $B=\frac{{\mu }_{0}i}{4\pi r}(\sin {60}^{\circ }+\sin {60}^{\circ })$

For $BD$ $B$ $\odot$ $B=\frac{{\mu }_{0}i}{4\pi r}\left(\sin {60}^{\circ }\right)$

For $DC$ $B$ $\otimes$ $B=\frac{{\mu }_{0}i}{4\pi r}\left(\sin {60}^{\circ }\right)$

$\therefore$ Net $B=0$