A small circular loop of wire of radius r is placed inside a large equilateral triangular loop of wire of side a ( a r ). The two loops are coplanar and their centres (centroids) coincide. The mutual inductance of the arrangement is:

Step-by-step solution

Let a steady current I flow through the large equilateral triangular loop. The perpendicular distance from the centroid to any side of the equilateral triangle is d = a 2 3 . The magnetic field at the centroid due to one side of the triangle is given by the finite wire formula: B₁ = ₀ I 4 d ( 60^ + 60^ ) Substituting the values: B₁ = ₀ I 4 ( a 2 3 ) ( 3 2 + 3 2 ) = 2 3 ₀ I 4 a ( 3 ) = 6 ₀ I 4 a = 3 ₀ I 2 a Since the triangle has three identical sides, the total magnetic field at the centroid is: B = 3 B₁ = 9 ₀ I 2