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2014-12-30T10:25:13+05:30

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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 B
₁.

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

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