magnetic field due to a finite straight conductor is given by
[tex]B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)[/tex]
here since it forms an equilateral triangle so we will have
[tex]\theta_1 = \theta_2 = 60 degre[/tex]
also the perpendicular distance of the point from the wire is
[tex]r = \frac{a}{2\sqrt3}[/tex]
now from the above equation magnetic field due to one wire is given by
[tex]B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)[/tex]
[tex]B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)[/tex]
[tex]B = \frac{3\mu_0 i}{2\pi a}[/tex]
now since in equilateral triangle there are three such wires so net magnetic field will be
[tex]B = \frac{9\mu_0 i}{2\pi a}[/tex]
