Two conducting circular loops of radii a and b are placed in the same plane with their centre coinciding. If a >> b, the mutual inductance M between then will be directly proportional to:

$\frac{b^2}{a}$

The correct answer is Option (4) → $\frac{b^2}{a}$

The mutual inductance (M) is -

$M=\frac{μ_0}{2\pi}×\frac{ab}{\sqrt{a^2+b^2}}$

$⇒M∝\frac{b}{\sqrt{a}}∝\frac{b^2}{a}$