Two long coaxial solenoids of the same length L, have inner and outer coils with diameters $D_1$ and $D_2$ and number of turns $N_1$ and $N_2$, respectively. What will be the ratio of the self-inductance of the inner solenoid to the mutual inductance of both the solenoids?

$N_1\,{N_2}^{-1}$

The correct answer is Option (1) → $N_1\,{N_2}^{-1}$

Given:

Inner coil turns = $N_1$, outer coil turns = $N_2$, length = $L$, inner coil area = $A_1 = \pi\left(\frac{D_1}{2}\right)^2$

Self-inductance of inner solenoid:

$L_{\text{self}} = \mu_0 \frac{N_1^2 A_1}{L}$

Mutual inductance (flux through inner area):

$M = \mu_0 \frac{N_1 N_2 A_1}{L}$

Ratio:

$\frac{L_{\text{self}}}{M} = \frac{\mu_0 \frac{N_1^2 A_1}{L}}{\mu_0 \frac{N_1 N_2 A_1}{L}} = \frac{N_1}{N_2}$

Ratio of self-inductance of inner solenoid to mutual inductance = $\frac{N_1}{N_2}$