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 :

Solution

Physics sir · 5 steps

Put the current in the big loop: its field is nearly uniform over the tiny loop, so flux is B times area.

1. Choose the loop. Send a current I through the large loop. Because L ≫ l, its field over the small loop is almost constant and equal to the field at the centre O. That makes the flux easy to find.
2. Field of one side. Each side is a finite wire at distance d = L/2 from O, and its ends subtend 45° on either side. So B₁ = (μ₀ I)/(4π (L/2))(sin 45° + sin 45°) = (√2 μ₀ I)/(2π L).
3. All four sides. All four sides give fields in the same direction at O, so the total field is four times B₁.

   B_O = 4 × (√2 μ₀ I)/(2π L) = (2√2 μ₀ I)/(π L)

4. Flux and M. The flux through the small loop is φ = B_O l² = (2√2 μ₀ I l²)/(π L). Then M = φ / I.
5. So. M = (2√2 μ₀ l²)/(π L), which is option (c).

The idea to keep. Mutual inductance is the same whichever coil carries the current. Pick the coil whose field is easy to find over the other coil, usually the big one, whose field is nearly uniform over the small one.

Answer from the NTA final answer key. Solution by PrepWiser.