Self induction and self inductance
A changing current in a coil induces an opposing emf in the same coil. Self inductance L gives NΦ = LI and ε = −L dI/dt.
What self induction is
When the current in a coil changes, the flux through the same coil changes. So an emf is induced in the coil itself. This is self induction. By Lenz's law, the emf opposes the change in current, so it is called a back emf. The flux linkage is proportional to the current: NΦ = LI. The constant L is the self inductance (coefficient of self induction). The induced emf is then ε = −L dI/dt. L depends only on the coil's shape, size, number of turns and core material, just as capacitance depends only on geometry. L acts like inertia in a circuit: it opposes any change in current, as mass opposes any change in velocity.
Self inductance of a long solenoid
- 1Take a long solenoid of length l, cross-sectional area A and n turns per unit length, carrying a current I. Total turns N = nl.
- 2Field inside, ignoring edge effects: B = μ₀nI.
- 3Flux through each turn: Φ = BA = μ₀nIA.
- 4Total flux linkage: NΦ = (nl)(μ₀nIA) = μ₀n²AlI.
- 5Compare with NΦ = LI: L = μ₀n²Al.
- 6★So, L = μ₀n²Al = μ₀N²A/l. With a core of relative permeability μr, L = μrμ₀n²Al.
★ key step
Energy stored in an inductor
- 1While the current grows, the source does work against the back emf, of magnitude ε = L dI/dt.
- 2Rate of doing work: dW/dt = εI = LI dI/dt.
- 3Work done in a small time: dW = LI dI.
- 4Total work to raise the current from 0 to I: W = ∫ LI dI from 0 to I = ½LI².
- 5★So, the energy stored in the magnetic field of the inductor is U = ½LI².
★ key step
Worked example: back emf and energy
The current in a coil of self inductance 0.2 H falls steadily from 5 A to 1 A in 0.1 s. Find (a) the self-induced emf and (b) the energy stored in the coil when the current is 5 A.
- (a) ε = L ΔI/Δt = 0.2 H × (5 − 1) A ÷ 0.1 s = 0.2 × 40 = 8 V.
- The emf tries to keep the current flowing, opposing its fall.
- (b) U = ½LI² = ½ × 0.2 H × (5 A)² = 0.1 × 25 = 2.5 J.
(a) 8 V; (b) 2.5 J
Worked example: inductance of a solenoid
An air-cored solenoid is 0.5 m long, has 500 turns and a cross-sectional area of 4 × 10⁻⁴ m². Find its self inductance. (μ₀ = 4π × 10⁻⁷ T m A⁻¹)
- Formula: L = μ₀N²A/l.
- N² = 500² = 2.5 × 10⁵.
- L = 4π × 10⁻⁷ × 2.5 × 10⁵ × 4 × 10⁻⁴ ÷ 0.5.
- 2.5 × 10⁵ × 4 × 10⁻⁴ = 100, and 100 ÷ 0.5 = 200.
- L = 4π × 10⁻⁷ × 200 = 8π × 10⁻⁵ H.
- Taking π = 3.14: L ≈ 25.1 × 10⁻⁵ H ≈ 2.5 × 10⁻⁴ H = 0.25 mH.
L = 8π × 10⁻⁵ H ≈ 2.5 × 10⁻⁴ H (0.25 mH)
Inductance as electrical inertia
| Mechanics | Electricity |
|---|---|
| Mass m opposes a change in velocity | Inductance L opposes a change in current |
| Velocity v | Current I |
| Force = m dv/dt | Back emf = L dI/dt |
| Kinetic energy ½mv² | Magnetic energy ½LI² |
Switching off a ceiling fan or a motor sometimes gives a small spark at the switch: the sudden fall in current induces a large self-induced emf. Older tube lights used a choke, an inductor, whose large self-induced emf helped start the tube.
2 marks: “Define self inductance and give its SI unit. On what factors does it depend?” L = NΦ/I, or the emf induced per unit rate of change of current; henry (H). Depends on the number of turns, area, length and core permeability, not on the current.
3 marks: “Derive an expression for the self inductance of a long air-cored solenoid.” B = μ₀nI; Φ = μ₀nIA per turn; NΦ = (nl)μ₀nIA = μ₀n²AlI; so L = μ₀n²Al.
2 marks: “Find the energy stored in a 40 mH inductor carrying a current of 3 A.” U = ½LI² = ½ × 0.04 H × (3 A)² = 0.02 × 9 = 0.18 J.
L is lazy: it resists any change in current, and it stores ½LI² just as a moving mass stores ½mv².