Inductance and Self-Induction: Henry, RL Circuits, Energy

A Coil That Resists Change

Switch off an electric motor or a fluorescent tube's choke and you may see a spark jump across the switch. Car ignition coils turn a 12 V battery into the thousands of volts needed to fire a spark plug. Both effects come from self-induction: a coil opposes any change in the current flowing through it by inducing an e.m.f. in itself. This tutorial covers inductance for WAEC, JAMB and first-year university physics: self-induction and Lenz's law, the definition of self-inductance and the henry, the inductance of a solenoid, energy stored in an inductor, growth and decay of current in an RL circuit, mutual inductance, and uses and hazards, with worked examples. It builds on the tutorial on electromagnetic induction and Faraday's law.

What Is Self-Induction?

When a current flows in a coil, it produces a magnetic field, and the coil's own turns are linked by that field's flux. If the current changes, the flux through the coil changes, and by Faraday's law an e.m.f. is induced in the coil itself. This is self-induction.

By Lenz's law, the induced e.m.f. always opposes the change that produces it:

When the current is switched on and rising, the induced e.m.f. opposes the rise, so the current grows gradually instead of instantly. This opposing e.m.f. is called a back e.m.f.
When the current is switched off and falling, the induced e.m.f. tries to keep the current flowing. Because the current falls very quickly, this e.m.f. can be very large, which causes the spark at the switch.

An inductor acts like electrical inertia: just as mass resists changes in velocity, inductance resists changes in current.

Self-Inductance and the Henry

The induced e.m.f. is proportional to the rate of change of current:

E = −L (dI/dt)

The constant L is the self-inductance (or simply inductance) of the coil, and the minus sign expresses Lenz's law. The unit of inductance is the henry (H).

A coil has an inductance of 1 henry if an e.m.f. of 1 volt is induced in it when the current changes at a rate of 1 ampere per second.

Inductance can also be defined through flux linkage: if a current I produces flux Φ through each of N turns, then NΦ = LI, so L = NΦ/I.

Worked example 1

The current in a coil of inductance 0.5 H falls steadily from 4 A to zero in 0.02 s. Find the size of the induced e.m.f.

E = L × ΔI/Δt = 0.5 × 4 ÷ 0.02 = 100 V. A modest coil and current produce 100 V, because the change is so fast.

Worked example 2

A current of 2 A in a coil of 200 turns produces a flux of 3 × 10⁻⁴ Wb through each turn. Find the inductance.

L = NΦ/I = 200 × 3 × 10⁻⁴ ÷ 2 = 0.03 H (30 mH).

Inductance of a Solenoid

For a long solenoid of N turns, length l and cross-sectional area A, the field inside is B = μ0NI/l. The flux linkage is NBA, so:

L = μ0N²A ÷ l

The inductance depends only on the coil's shape and on what is inside it:

It is proportional to N²: doubling the number of turns (in the same length) quadruples L.
A larger cross-sectional area gives a larger L.
Inserting a soft iron core multiplies L by the relative permeability of the iron, which can be hundreds or thousands. This is why chokes and transformers have iron cores.

Worked example 3

An air-cored solenoid has 500 turns, a length of 25 cm and a radius of 1.0 cm. Find its inductance. (μ0 = 4π × 10⁻⁷ H/m)

A = π × 0.01² = 3.14 × 10⁻⁴ m².

L = 4π × 10⁻⁷ × 500² × 3.14 × 10⁻⁴ ÷ 0.25 ≈ 3.9 × 10⁻⁴ H (0.39 mH).

Energy Stored in an Inductor

Work must be done against the back e.m.f. to build up a current in an inductor. This work is stored as energy in the magnetic field:

W = ½LI²

Compare this with ½CV² for a capacitor (energy in an electric field) and ½mv² for a moving mass. When the current is switched off, this stored energy must go somewhere, which is why a spark appears at the switch.

Worked example 4

Find the energy stored in a 2 H inductor carrying 3 A.

W = ½ × 2 × 3² = 9 J.

Growth and Decay of Current in an RL Circuit

Connect an inductor L in series with a resistor R across a battery of e.m.f. V. The back e.m.f. stops the current from jumping to its final value. Instead it grows exponentially:

I = I0 (1 − e^(−Rt/L)), where I0 = V/R

The time constant of an RL circuit is:

τ = L ÷ R

After one time constant, the current reaches about 63% of its final value. When the battery is removed and the coil is connected across the resistor, the current decays as I = I0 e^(−Rt/L), falling to 37% after one time constant. These are the same shapes as the voltage curves in the tutorial on RC circuits, with L/R playing the role of RC.

Worked example 5

A coil of inductance 0.4 H and resistance 20 Ω is connected to a 12 V battery. Find (a) the time constant, (b) the final current, (c) the current after one time constant, and (d) the time to reach 90% of the final current.

(a) τ = L/R = 0.4 ÷ 20 = 0.02 s.

(b) I0 = 12 ÷ 20 = 0.6 A.

(d) 0.9 = 1 − e^(−t/0.02), so e^(−t/0.02) = 0.1 and t = 0.02 × ln 10 ≈ 0.046 s.

Once the current is steady, dI/dt = 0, so there is no back e.m.f. In steady d.c., an inductor behaves like an ordinary wire with only its resistance.

Mutual Inductance

When two coils are close together, a changing current in one (the primary) changes the flux through the other (the secondary), inducing an e.m.f. in it. This is mutual induction:

E2 = −M (dI1/dt)

where M is the mutual inductance of the pair, also measured in henrys. M is larger when the coils have many turns, are close together and share an iron core. Mutual induction is the principle of the transformer and of the induction coil.

Worked example 6

Two coils have a mutual inductance of 0.15 H. The current in the primary rises from 0 to 2 A in 0.01 s. Find the e.m.f. induced in the secondary.

E2 = M × ΔI1/Δt = 0.15 × 2 ÷ 0.01 = 30 V.

Inductors in A.C. Circuits

With alternating current, the current is always changing, so an inductor continuously produces a back e.m.f. This gives it an opposition to a.c. called inductive reactance, XL = 2πfL, which increases with frequency. So an inductor passes d.c. and low frequencies easily but blocks high frequencies. The tutorial on alternating current circuits covers reactance, impedance and resonance.

Uses and Hazards of Self-Induction

Chokes limit the current in fluorescent lamps and help start them with a high-voltage pulse.
Ignition and induction coils use a rapid collapse of current to produce very high voltages for spark plugs and gas lighters.
Filters and tuned circuits combine inductors with capacitors to select radio frequencies and smooth power supplies.
Hazard: breaking the current in a large inductor (motors, relays, electromagnets) produces a high back e.m.f. that can cause sparks, damage switches and destroy transistors. A diode connected across the coil (a flyback diode) gives the current a safe path to die away.
Non-inductive resistors, used in precise measurement, have their wire wound back on itself so the magnetic fields of the two halves cancel.

Inductance: Practice Questions

Define the henry.
The current in a 0.2 H coil changes at 50 A/s. Find the induced e.m.f.
Find the energy stored in that coil when it carries 5 A.
How does the inductance of a solenoid change if the number of turns is tripled, with the length unchanged?
A coil has L = 0.6 H and R = 30 Ω. Find its time constant.

Answers: 1. A coil has an inductance of 1 H if an e.m.f. of 1 V is induced in it when the current through it changes at 1 A/s. 2. E = 0.2 × 50 = 10 V. 3. W = ½ × 0.2 × 25 = 2.5 J. 4. L is proportional to N², so it becomes 9 times larger. 5. τ = 0.6 ÷ 30 = 0.02 s.

Conclusion

Self-induction is a coil's opposition to changes in its own current. The induced e.m.f. is E = −L dI/dt, where the inductance L is measured in henrys and equals μ0N²A/l for a solenoid. An inductor stores energy ½LI² in its magnetic field, and in an RL circuit the current grows and decays exponentially with time constant L/R. Mutual inductance links two coils and underlies the transformer, while the large e.m.f. produced when current is cut off is both useful, in ignition coils, and hazardous, in switches and electronics.