Question

A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field intensity at the center of the triangle.

Answer 1

magnetic field due to a finite straight conductor is given by

$B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)$

here since it forms an equilateral triangle so we will have

$\theta_1 = \theta_2 = 60 degre$

also the perpendicular distance of the point from the wire is

$r = \frac{a}{2\sqrt3}$

now from the above equation magnetic field due to one wire is given by

$B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)$

$B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)$

$B = \frac{3\mu_0 i}{2\pi a}$

now since in equilateral triangle there are three such wires so net magnetic field will be

$B = \frac{9\mu_0 i}{2\pi a}$