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 R₂² / R₁.

Step-by-step Derivation:

Let the outer circular loop have radius R₁ and carry a current I₁.
The magnetic field B₁ at the center of the outer loop is given by:

$B_1=\frac{{\mu }_0{I}_1}{2R_1}$

Since R₁ >> R₂, the magnetic field B₁ can be assumed to be approximately uniform over the area of the smaller, inner loop of radius R₂.
The magnetic flux Φ₂ passing through the inner loop is:

${\Phi }_2=B_1\cdot A_2$

Where A₂ is the area of the inner loop, A₂ = π R₂².
Substituting the value of B₁ into the flux equation:

${\Phi }_2=\left(\frac{{\mu }_0{I}_1}{2R_1}\right)\cdot \left(\pi R_2^2\right)$

Simplifying the expression:

${\Phi }_2=\frac{{\mu }_0\pi R_2^2}{2R_1}{I}_1$

By definition, the magnetic flux through the second loop is related to the current in the first loop by the mutual inductance M:

${\Phi }_2=M{I}_1$

Comparing the equations, we find the mutual inductance M:

From this expression, it is clear that: