An electric current i is carried by a wire of length l which is...

Let the length of each side of the equilateral triangle be $a$. Since the total length of the wire is $l$, we have $a = \dfrac{l}{3}$.

The perpendicular distance $d$ from the centroid (center) of the equilateral triangle to any of its sides is given by:

$d = \dfrac{a}{2\sqrt{3}} = \dfrac{\dfrac{l}{3}}{2\sqrt{3}} = \dfrac{l}{6\sqrt{3}}$

The magnetic field at the center due to one side of the triangle can be found using the Biot-Savart law for a finite straight wire:

$B_1 = \dfrac{\mu_0 i}{4\pi d} (\sin \theta_1 + \sin \theta_2)$

For an equilateral triangle, the line segments joining the center to the vertices subtend an angle of $120^{\circ}$ at the center. The perpendicular bisects this angle, giving $\theta_1 = \theta_2 = 60^{\circ}$.

Substituting the values:

$B_1 = \dfrac{\mu_0 i}{4\pi \left(\dfrac{l}{6\sqrt{3}}\right)} (\sin 60^{\circ} + \sin 60^{\circ})$

$B_1 = \dfrac{6\sqrt{3} \mu_0 i}{4\pi l} \left(\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}\right)<br><br>$B_1 = $\dfrac{6\sqrt{3} \mu_0 i}{4\pi l} (\sqrt{3}) = \dfrac{18 \mu_0 i}{4\pi l} = \dfrac{9 \mu_0 i}{2\pi l}<br><br>Since the current flows in the same sense through all three sides, the magnetic field vectors due to each side point in the same direction at the center. The total magnetic field$B is the sum of the fields due to all three sides:

$B = 3B_1 = 3 \times$\dfrac{9 \mu_0 i}{2\pi l} = \dfrac{27 \mu_0 i}{2\pi l}