Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ >> R₂ , the mutual inductance M between them will be directly proportional to :

The correct answer is that the mutual inductance $\frac{R_2^2}{R_1}$ is directly proportional to the square of the smaller loop’s radius divided by the larger loop’s radius.

When two circular loops are coaxial and lie in the same plane, the magnetic flux through the smaller loop due to a current I₁ in the larger loop is given by

where $B$₁ is the magnetic field produced by the larger loop at the location of the smaller loop, and $A$₂ = π R₂² is the area of the smaller loop.

For a single-turn loop of radius R₁ carrying current I��, the on‑axis magnetic field at a distance z from its centre is

Since the loops are in the same plane, the smaller loop lies at z =���0, so

Here we used the approximation R₁ ≫ R₂, which allows us to treat the field as uniform over the area of the small loop.

The mutual inductance M is defined by Φ₂ = M I₁, therefore

All constant factors (μ₀, π, 1/2) are the same for any given material and geometry, so the dependence of M on the radii is captured by the term

Thus, when R₁ ≫ R₂, the mutual inductance is directly proportional to the square of the smaller radius divided by the larger radius, i.e., M ∝ R₂² / R₁.