An equilateral triangular loop of wire of side length 0.5 m carries a steady current of 10 A. The magnetic field at the centroid of the triangle is X 10⁻⁶ T. The value of X is _________.

Step-by-step solution

The magnetic field at the centroid of an equilateral triangle due to one of its sides is given by the Biot-Savart law for a finite straight wire: B₁ = ₀ I 4 d ( ₁ + ₂) For an equilateral triangle of side a , the perpendicular distance d from the centroid to any side is the inradius: d = a 2 3 = 0.5 2 3 = 1 4 3 m The angles subtended by the half-segments of the side at the centroid are ₁ = ₂ = 60^ . Substituting the values into the formula for one side: B₁ = 4 10⁻⁷ 10 4 ( 1 4 3 ) ( 60^ + 60^ ) B₁ = 10⁻⁶ 4 3 ( 3 2 +