Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'R' is placed inside a large square loop of wire of side L (L ≫ R). The loops are coplanar and their centres coincide :

Solution

Physics sir · 4 steps

Send current through the big square, find B at its centre, and treat it as uniform over the small loop.

1. Choose the loop. M is the same whichever loop carries the current. Send current I through the square, because its field at the centre is easy to find. Since L ≫ R, that field is nearly uniform over the small circle.
2. One side. Each side is a finite wire at perpendicular distance L/2 from the centre. The lines to its two ends each make 45° with that perpendicular, 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, so B = 4B₁ = (2√2 μ₀ I)/(π L). The flux through the circle is φ = B × π R².

   M = φ/I = (2√2 μ₀ I)/(π L) × (π R²)/I = (2√2 μ₀ R²)/L

4. So. The mutual inductance is M = (2√2 μ₀ R²)/L, which is option (c).

The idea to keep. For mutual inductance, put the current in the loop whose field you can find easily. A tiny loop at the centre of a big one just sees the centre field times its own area.

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