A current of 2 A is flowing in the sides of an equilateral triangle of side 9 cm . The magnetic field at the centroid of the triangle is

Step-by-step solution

The magnetic field at the centre O due to the current through side AB is B 1 = μ 0 I 4 π a sin ⁡ θ 1 + sin ⁡ θ 2 As the magnetic field due to each of the three sides is the same in magnitude and direction. So, the total magnetic field at O is sum of all the fields. i.e. B = 3 B 1 = 3 μ 0 I 4 π θ sin ⁡ θ 1 + sin ⁡ θ 2 Here, tan ⁡ θ 1 = A D O D ⇒     tan ⁡ 60 o = 1 2 a   ⇒       a = 1 2 3 =