The current passing through a conducting loop in the form of equilateral triangle of side 4√3 cm is 2 A . The magnetic field at its centroid is α× 10⁻⁵ T. The value of α is ____ .
(Given : μₒ=4π× 10⁻⁷ SI units)

Solution

Physics sir · 4 steps

Each side is a finite straight wire. Find the field of one side at the centroid, then multiply by three.

1. Distance to each side. The centroid sits at the inradius from every side, r = a/2√3. With a = 4√3 cm, this gives r = 2 cm = 0.02 m.
2. Angles at the centroid. From the centroid, the foot of the perpendicular is the midpoint of the side. Each half of the side subtends 60°, so both angles in the finite-wire formula are 60°.

   B_(side) = (μ₀ I)/(4π r)(sin 60° + sin 60°) = (10⁻⁷ × 2)/0.02 × √3 = √3 × 10⁻⁵ T

3. Add all three sides. The current flows the same way around the loop, so all three sides give fields in the same direction at the centroid. The total is B = 3 × √3 × 10⁻⁵ = 3√3 × 10⁻⁵ T.
4. So. α = 3√3, which is option (a).

The idea to keep. For any regular polygon loop, the centre is the same distance from every side. Find the field of one finite side using (μ₀ I)/(4π r)(sin θ₁ + sin θ₂), then multiply by the number of sides.

Answer from the NTA final answer key. Solution by PrepWiser.