A small square loop of wire of side l is placed inside a large square loop of wire L L ≫ l . Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is

Step-by-step solution

Let a current, I pass through the square loop of side, L . The magnetic field at the centre O , B = 4 × Magnetic   field   due   to   each   side ⇒ B = 4 × μ 0 4 π ⋅ I L / 2 sin   45 ∘ + sin   45 ∘ ⇒ B = 2 μ 0 I π L 1 2 + 1 2 = 2 2 μ 0 I π L Magnetic flux linked with the small square loop, ϕ = B A = B l 2 = 2 2 μ 0 I l 2 π L Mutual inductance of the coil, M = ϕ I = 2 2 μ 0 l 2 π L .