Given two long coaxial solenoids of equal length $l$:

• Inner solenoid: radius $r_1$, turns per unit length $n_1$
• Outer solenoid: radius $r_2$, turns per unit length $n_2$

We need: $\frac{M}{L_1}$ where $M$ is the mutual inductance and $L_1$ is the self-inductance of the inner solenoid.

Self-inductance of the inner solenoid

For a long solenoid, $L = \mu_0 n^2 A l$ where $A$ is cross-sectional area.

For the inner solenoid, $A_1 = \pi r_1^2$ so $L_1 = \mu_0 n_1^2 \pi r_1^2 l$

Mutual inductance between the two solenoids

Magnetic field produced by the inner solenoid is $B_1 = \mu_0 n_1 I_1$

Since the solenoids are coaxial and the inner one is smaller, the field of the inner solenoid effectively links the turns of the outer solenoid only through the area of the inner solenoid, i.e. $\pi r_1^2$.

Flux through each turn of outer solenoid due to current $I_1$ in inner solenoid: $\phi = B_1 A_1 = \mu_0 n_1 I_1 \pi r_1^2$

Total number of turns in outer solenoid: $N_2 = n_2 l$

Total flux linkage with outer solenoid: $N_2 \phi = n_2 l \cdot \mu_0 n_1 I_1 \pi r_1^2$

Therefore, $M = \frac{N_2 \phi}{I_1} = \mu_0 n_1 n_2 \pi r_1^2 l$

Take the ratio

$\frac{M}{L_1} = \frac{\mu_0 n_1 n_2 \pi r_1^2 l}{\mu_0 n_1^2 \pi r_1^2 l} = \frac{n_2}{n_1}$

Match with options

$\frac{M}{L_1} = \frac{n_2}{n_1}$

So the correct option is: B.