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 option is R₂²/R₁.

To find the mutual inductance between two concentric, coplanar circular loops of radii $R_1$ and $R_2$ with $R_1>>R_2$, we can assume a current $I_1$ flows through the outer loop of radius $R_1$.

The magnetic field $B_1$ at the center of the outer loop is given by:
$B_1=\frac{{\mu }_0{I}_1}{2R_1}$

Since the radius of the outer loop is much larger than the radius of the inner loop ($R_1>>R_2$), we can assume that this magnetic field $B_1$ is uniform over the entire area of the smaller inner loop.

The area of the smaller inner loop is:
$A_2=\pi R_2^2$

The magnetic flux ${\Phi }_2$ linked with the inner loop due to the current in the outer loop is:
${\Phi }_2=B_1{A}_2$
Substituting the value of $B_1$ and $A_2$:
${\Phi }_2=\left(\frac{{\mu }_0{I}_1}{2R_1}\right)·\left(\pi R_2^2\right)$
${\Phi }_2=\frac{{\mu }_0\pi R_2^2}{2R_1}{I}_1$

By definition of mutual inductance $M$:
${\Phi }_2=M{I}_1$

Comparing the two equations for flux, we get:
$M=\frac{{\mu }_0\pi R_2^2}{2R_1}$

From this equation, we can see that:
$M\propto \frac{R_2^2}{R_1}$

Therefore, the mutual inductance between the two loops is directly proportional to $\frac{R_2^2}{R_1}$.