6.7.2. Self-Inductance

In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction.

Self-inductance occurs when a changing current in a coil generates a changing magnetic field, which in turn induces an emf (electromotive force) in the same coil. This means that if the current flowing through a coil varies, it affects the magnetic flux linked to that coil, and this changing flux induces an emf that opposes the change in current. Thus, the behavior of the coil is influenced by its own magnetic field.

Think of self-induction like a person who begins to swing a heavy bat. As they swing it faster (changing how they use their energy), they feel a resistance (back emf) that pushes back against that energy. Just as it takes effort to speed up or slow down the swing, it takes energy to change the current through the coil.

In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as NF ∝ I_B. NF = LI_B where constant of proportionality L is called self-inductance of the coil.

The equation NF = LI_B indicates that the total magnetic flux (NF) linked with the coil is proportional to the current (I_B) flowing through it, with L being the self-inductance constant. This 'L' value varies depending on the coil’s physical characteristics and the material it’s made from, similar to how the strength of a spring depends on its material and construction.

Imagine using a rubber band. How much you stretch it (current) affects how much force it pushes back against you (self-inductance). The performance of a rubber band can vary based on how thick or thin it is, just like inductance can vary based on how the coil is built.

When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using Eq. (6.13), the induced emf is given by ( ) dNF e = – B / dt dI e = –L (6.14) dt. Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil.

The induced emf, represented mathematically as e = -L (dI/dt), illustrates that the self-induced emf always works against the change in current. If the current attempts to increase, the self-induced emf will act to decrease it, and vice versa. This is akin to inertia in mechanics, where an object tends to resist changes to its state of motion.

Think of riding a bicycle. When you try to speed up (increase current), the bike pushes back a bit in the form of resistance from air and inertia (self-induced emf). If you suddenly want to slow down, the same resistance pushes you forward, making it more challenging. This pushback effect is the self-inductance at work.

It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is B = µ₀ n I (neglecting edge effects, as before).

For a long solenoid, we can derive the formula for self-inductance. The total flux linked with the solenoid can be expressed as NF = µ₀ n² Al I, leading us to the conclusion that L (the self-inductance) = µ₀ n² Al, where A is the cross-sectional area, l is the length, and n is the number of turns per unit length. This clearly shows how self-inductance depends on the solenoid's characteristics.

Try visualizing a water pipe: the flow rate can be influenced by the pipe’s radius (cross-sectional area), length, and how many bends it has (turns), similar to how the inductance is affected by length, area, and turn density. A thicker pipe allows more water current (current flow) just as a highly inductive solenoid allows for greater magnetic behavior.

The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. … Therefore, using Faraday’s law, the total amount of work done in establishing the current I is W = (1/2) LI².

The work done against the back emf while establishing current in the coil gives rise to stored energy in the magnetic field around the coil. This work manifests as magnetic potential energy, showcasing the inherent energy associated with self-inductance. The expression W = (1/2) LI² indicates that as the current increases, the energy stored grows significantly, reflecting the efficiency of electromagnetic systems.

Consider a balloon. When you inflate it (establish current), you must work against the internal pressure of the air, similar to how you must work against back emf. The more air (current) you add, the greater the energy is stored in the stretched balloon, akin to how energy is stored in the magnetic field of the coil.

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