Animated Solution for Physics - Electromagnetic Induction: The total number of turns and cross-section area in a solenoid is fixed. However, its length $L$ is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to

Unraveling Self-Inductance

Imagine you are holding a slinky, but instead of a toy, it's a tightly wound copper coil—a solenoid. The problem presents a fascinating scenario: we are stretching and compressing this solenoid like an accordion.

However, there are strict rules to this game. The total number of turns ($N$) and the cross-sectional area ($A$) of the coil are absolutely fixed. The only variable we are allowed to change is the overall length ($L$) of the solenoid by adjusting the gap between the windings. Our mission is to discover how the self-inductance ($L_{sol}$) responds to this stretching.

To understand self-inductance, we must first look at the magnetic field generated inside the core of the solenoid. The magnetic field $B$ of an ideal solenoid is given by the classic formula:

Here, $n$ represents the number of turns per unit length. But remember our constraints! The total number of turns $N$ is fixed, not $n$. Therefore, we must express $n$ in terms of our fixed parameter $N$ and our variable length $L$:

Substituting this back into our magnetic field equation, we get:

Now, let's calculate the total magnetic flux linkage ($\Phi$) through the entire solenoid. The flux through a single loop is simply the magnetic field multiplied by the area ($B\cdot A$). Since there are $N$ loops in total, the total flux linkage is:

Substituting our expression for $B$, we arrive at the master equation for flux:

By the very definition of self-inductance, the total magnetic flux linkage is directly proportional to the current flowing through the circuit. The constant of proportionality is the self-inductance ($L_{sol}$):

By equating our two expressions for the total flux $\Phi$, we can easily isolate the self-inductance:

Canceling the current $I$ from both sides yields the beautiful, final formula for the self-inductance of our solenoid:

Let's analyze our final formula. The magnetic permeability of free space (${\mu }_0$), the total number of turns ($N$), and the cross-sectional area ($A$) are all strictly constant.

Because all the terms in the numerator are constant, it becomes crystal clear that the self-inductance is inversely proportional to the length of the solenoid:

The Trap: Many students blindly memorize the formula $L_{sol}={\mu }_0{n}^{2}AL$ and immediately conclude that inductance is directly proportional to length ($L_{sol}\propto L$). This is a fatal error! That direct proportionality is only true if the turn density ($n$) is kept constant while adding more wire to increase the length. In our problem, the total wire length (and thus $N$) is fixed, making the inverse relationship the correct physical reality.