Animated Solution for Physics - Electromagnetic Induction: There are two long coaxial solenoids of same length $l$. The inner and outer coils have radii $r_1$ and $r_2$ and number of turns per unit length $n_1$ and $n_2$, respectively. The ratio of mutual inductance to the self-inductance of the inner coil is

Imagine you are looking at a cross-section of two perfectly aligned, infinitely long coaxial solenoids. The inner solenoid is tightly wound with a radius $r_1$ and has $n_1$ turns per unit length. Wrapping around it is the outer solenoid, with a larger radius $r_2$ and $n_2$ turns per unit length. Both solenoids share the exact same length $l$.

Our mission is to find the ratio of their mutual inductance $M$ to the self-inductance $L$ of the inner coil. To do this, we need to carefully construct the equations for both $M$ and $L$ from first principles.

Mutual inductance $M$ measures how effectively a changing current in one coil induces an electromotive force (EMF) in another. Thanks to the beautiful symmetry of physics (specifically, the Reciprocity Theorem), $M_{12}=M_{21}$. This means we can choose whichever coil makes our math easier to act as the "source" of the magnetic field.

Let's pass a current $I_2$ through the outer solenoid. This creates a uniform magnetic field everywhere inside it, given by the standard solenoid formula:

Now, we need to find the magnetic flux ${\varphi }_{12}$ that this field links with the inner solenoid. Here is the crucial conceptual trap: even though the magnetic field $B_2$ fills the entire volume of the outer solenoid (area $\pi r_2^2$), the inner solenoid can only "catch" the flux that passes through its own loops. Therefore, the relevant area is the cross-sectional area of the inner coil, $A_1=\pi r_1^2$.

The total number of turns in the inner solenoid is $N_1=n_{1}l$. The total flux linkage is the product of the number of turns, the magnetic field, and the area:

By definition, mutual inductance is the flux linkage per unit current ($M=\frac{{\varphi }_{12}}{I_2}$), which gives us:

Now let's shift our focus entirely to the inner coil to find its self-inductance $L$. Self-inductance measures how much a coil opposes a change in its own current.

If we pass a current $I_1$ through the inner coil, it generates its own magnetic field:

This field links with the inner coil's own turns. The total flux ${\varphi }_{11}$ linked with the inner coil itself is:

Dividing by the current $I_1$, we get the self-inductance $L$:

We have successfully derived both $M$ and $L$. The question asks for the ratio $\frac{M}{L}$. Let's set up the fraction and watch the magic of algebra happen:

Notice how beautifully the geometric factors cancel out! The permeability ${\mu }_0$, the area $\pi r_1^2$, and the length $l$ all vanish. Even one factor of $n_1$ cancels out, leaving us with a remarkably simple, elegant result:

This tells us that for coaxial solenoids of the same length, the ratio of mutual inductance to the inner coil's self-inductance depends purely on the ratio of their turn densities. Physics is deeply satisfying when complex setups reduce to such clean relationships!