Self-inductance
A coil that opposes itself
A current in a coil sends magnetic flux through the coil itself. So if that current changes, the coil’s own flux changes, and an emf is induced in the same coil. This is self-induction.
The flux linkage is proportional to the current: N Φ_B equals L I. The constant L is the self-inductance of the coil, measured, like M, in henry.
Differentiate, and the induced emf is ε equals minus L dI by dt. By Lenz’s law, it opposes the change of current. So it is called the back emf.
Let us find L for a long solenoid: length l, cross-sectional area A, n turns per unit length. A current I makes a field B equals μ₀ n I inside it.
The flux through each turn is μ₀ n I A, and there are n l turns. So N Φ_B equals μ₀ n squared A l times I, and L equals μ₀ n squared A l.
L grows with the square of the number of turns. Fill the solenoid with iron of relative permeability μr, and L becomes μr μ₀ n squared A l. Like M, it depends only on geometry and material.
An example. An air-cored solenoid is 30 centimetres long, with a cross-section of 25 centimetre squared and 500 turns. Using N squared A over l, L equals 4π × 10⁻⁷ × 500 squared × 25 × 10⁻⁴ over 0.3: about 2.6 × 10⁻³ henry.
It carries 2.5 amperes, and the current is switched off in 10⁻³ second. The average back emf is L ΔI over Δt: 2.6 × 10⁻³ × 2.5 over 10⁻³, about 6.5 volts, much more than you might expect. This is why switches can spark.
Self-inductance acts like inertia. Mass resists a change in velocity; inductance resists a change in current. It is electrical inertia.
Energy stored in an inductor
Connect a coil and a resistor to a battery. Without the coil, the current would jump at once to E over R. With the coil, it rises gradually.
At the start, the current is changing fastest, so the back emf is largest and holds the current back. As the current approaches E over R, it changes more slowly, and the back emf fades away.
Once the current is steady, dI by dt is zero, and the coil behaves just like a piece of wire. The final current is E over R, with or without the coil.
While the current grows, the source must push charge against the back emf. The rate of doing work is the back emf times the current: dW by dt equals L I dI by dt.
Add up this work as the current grows from zero to I. The total is W equals ½ L I squared.
This energy is stored in the magnetic field of the coil. Compare it with kinetic energy, ½ m v squared: L plays the part of mass, and I the part of velocity.
For a coil of 4 henry carrying 1.5 amperes, the stored energy is ½ × 4 × 1.5 squared, which is 4.5 joules.
Take the battery out of the circuit, and the current does not stop at once. The stored energy keeps it flowing through the resistor for a while, and it all turns into heat.
Our solenoid from before, 2.6 millihenry carrying 2.5 amperes, stores ½ × 2.6 × 10⁻³ × 2.5 squared, about 8.2 × 10⁻³ joule. Breaking the circuit quickly releases this energy as a spark across the switch.
Try it: a coil in a circuit
- Choose “Battery in”. The current rises gradually to E/R = 1.5 A, and the back emf (bottom graph) starts at 12 V and dies away.
- Watch the energy bar fill to ½LI² = ½ × 4 × 1.5² = 4.5 J.
- Now choose “Battery out”. The current keeps flowing for a while, and the heat produced in R adds up to the stored energy.
- Change L from 1 H to 10 H. The final current stays E/R; only the time it takes changes.
Switching off a solenoid
An air-cored solenoid of length 30 cm, cross-sectional area 25 cm² and 500 turns carries a current of 2.5 A. The current is switched off in 10⁻³ s. What is the average back emf?
- 6.5 × 10⁻³ V
- 13 mV
- 3.3 V
- 6.5 V
Answer
D. L = μ₀N²A/l = 4π × 10⁻⁷ × (500)² × 25 × 10⁻⁴ / 0.30 = 2.6 × 10⁻³ H. Then |ε| = L ΔI/Δt = 2.6 × 10⁻³ × 2.5 / 10⁻³ ≈ 6.5 V. Forgetting to divide by Δt gives 6.5 × 10⁻³ V; using N instead of N² gives about 13 mV.
Energy in a coil
A coil of self-inductance 4 H carries a steady current of 1.5 A. How much energy is stored in its magnetic field?
- 3 J
- 4.5 J
- 6 J
- 9 J
Answer
B. U = ½LI² = ½ × 4 × (1.5)² = ½ × 4 × 2.25 = 4.5 J. LI = 6 and ½LI = 3 forget to square the current; LI² = 9 forgets the ½.
Summary
- Definition:
NΦ_B = L I - Back emf:
ε = −L dI/dt - Long solenoid:
L = μr μ₀ n² A l - Energy stored:
U = ½ L I²
- A changing current in a coil induces an emf in the same coil: the back emf, which opposes the change of current.
- L depends on the coil’s geometry (it grows as n²) and on the core material (μr).
- Inductance is electrical inertia: the current grows and dies gradually, but its final value E/R is unchanged.
- The work done against the back emf is stored as ½LI² in the magnetic field, and released when the current falls (heat, or a spark at a switch).