Question

A small square loop of wire of side 'l' is placed inside a large square loop of wire of side L(>>l).
The loops are co-planar and their centers coincide. What is the mutual inductance of the system ?
Please do justify your answer with a good explanation !

Answer 1

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

Answer 2

Answer:

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