The magnetic field at the centre of an equilateral triangular loop of side $2\,L$ and carrying a current $i$ is
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Magnetic field at $\mathrm{O}$ is
$\mathrm{B}=3\left[\frac{\mu_{0}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)\right]$
$\tan {30^\circ } = \frac{{\text{r}}}{{\text{L}}}\quad \Rightarrow {\text{r}} = \frac{{\text{L}}}{{\sqrt 3 }}\quad $
$ = 3\left( {\frac{{{\mu _0}}}{{4\pi }}} \right)\frac{{({\text{i}})}}{{\text{L}}}\left( {\frac{{\sqrt 3 }}{2}} \right)(\sqrt 3 ) \times 2$
$ = \frac{{9{\mu _0}{\text{i}}}}{{4\pi {\text{L}}}}$
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