Side length of larger square =L
Side length of smaller square = l
Let current flowing in larger square = I
current flowing in smaller square = i
Magnetic field at a distance L/2 from the centre of a current carrying wire of length L is B = μ×i [sinα + sinβ] / 4πd
i.e., B = μ×i × 2 sin 45 / 4π (L/2)
So due to 4 wires of length L, Magnetic field in the centre of the square
B₁ = μ 8√2 i / 4π L ( Note that the will just get added.)
Since the smaller loop is at the centre and l<<L, we can assume the the magnetic field in the whole smaller square is uniform and equal to B₁.
So, the flux linked with smaller loop, Ф₂ = B₁S₂ = μ 8√2 i l² / 4πL
and mutual inductance is given by M = Ф₂ / i = 2√2 μ l² / πL
