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 = (\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 = (a)/(2\sqrt3)`

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

`B = (\mu_0 i*2\sqrt3)/(4\pi a)(\sqrt3)`

`B = (3\mu_0 i)/(2\pi a)`

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

`B = (9\mu_0 i)/(2\pi a)`