A small square loop of wire of side 1 mm is placed inside a large square loop of wire of side 10 m. The loops are coplanar and their centres coincide. Find the mutual inductance of the system. (μ₀=4π×10⁻⁷ SI units)

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(3) 8√2×10⁻¹⁴ H. Given: small loop side a=1 mm=10⁻³ m, big loop side L=10 m, coplanar and concentric. Idea: send a current I round the big loop and find the flux through the small one; then M=Φ/I. The small loop is tiny, so the field over it is the field at the centre. Each side is L/2 from the centre and subtends 45° on either side: B₁=(μ₀I)/(4π(L/2))(sin 45°+sin 45°)=(√2 μ₀I)/(2π L). Four sides: B=4B₁=(2√2 μ₀I)/(π L). Flux through the small loop: Φ=Ba², so M=(2√2 μ₀a²)/(π L). M=(2√2×4π×10⁻⁷×10⁻⁶)/(π×10). So M=8√2×10⁻¹⁴ H.

A small square loop of wire of side 1 mm is placed inside a large square loop of wire of side 10 m. The loops are coplanar and their centres coincide. Find the mutual inductance of the system. (μ₀=4π×10⁻⁷ SI units)

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