Mutual induction and mutual inductance
A changing current in one coil induces an emf in a nearby coil. N₁Φ₁ = MI₂ and ε₁ = −M dI₂/dt, where M is the mutual inductance.
What mutual induction is
Place two coils close together. When the current in one coil changes, its changing field changes the flux through the other coil. So an emf is induced in the second coil. This is mutual induction. The flux linkage with coil 1 is proportional to the current in coil 2: N₁Φ₁ = M₁₂I₂. The constant M₁₂ is the mutual inductance of coil 1 with respect to coil 2. The induced emf is ε₁ = −M₁₂ dI₂/dt. Its SI unit is also the henry. M depends on the number of turns, the sizes, the separation and the relative orientation of the coils, and on the core between them.
Mutual inductance of two long coaxial solenoids
- 1Take two long solenoids of the same length l, one inside the other. Inner solenoid S₁: radius r₁, n₁ turns per unit length. Outer solenoid S₂: n₂ turns per unit length.
- 2Let a current I₂ flow in S₂. The field inside it is B₂ = μ₀n₂I₂.
- 3This field passes through each turn of S₁ over the area πr₁². Flux linkage with S₁: N₁Φ₁ = (n₁l)(πr₁²)(μ₀n₂I₂).
- 4Compare with N₁Φ₁ = M₁₂I₂: M₁₂ = μ₀n₁n₂πr₁²l.
- 5Now let a current I₁ flow in S₁. Its field μ₀n₁I₁ exists only inside S₁, so it links S₂ over the area πr₁² only: N₂Φ₂ = (n₂l)(πr₁²)(μ₀n₁I₁), giving M₂₁ = μ₀n₁n₂πr₁²l.
- 6★So, M₁₂ = M₂₁ = M = μ₀n₁n₂πr₁²l. With a core of relative permeability μr, M = μrμ₀n₁n₂πr₁²l.
★ key step
Self induction vs mutual induction
Both are measured in henry, both oppose the change in current (Lenz's law), and both depend on geometry and core, not on the current.
Worked example: emf from mutual induction
Two coils have a mutual inductance of 0.5 H. The current in the first coil rises steadily from 0 to 8 A in 0.2 s. Find (a) the emf induced in the second coil and (b) the change in flux linkage with the second coil.
- (a) ε₂ = M ΔI₁/Δt = 0.5 H × 8 A ÷ 0.2 s = 0.5 × 40 = 20 V.
- (b) Change in flux linkage = M ΔI₁ = 0.5 H × 8 A = 4 Wb.
(a) 20 V; (b) 4 Wb
Worked example: coaxial solenoids
A long solenoid of length 0.5 m has 500 turns per metre. Inside it, on the same axis, is a solenoid of radius 2 cm with 1000 turns per metre and the same length. Find their mutual inductance. (μ₀ = 4π × 10⁻⁷ T m A⁻¹; take π² ≈ 10)
- Formula: M = μ₀n₁n₂πr₁²l, with r₁ the radius of the inner solenoid.
- n₁ = 1000 m⁻¹, n₂ = 500 m⁻¹, r₁ = 0.02 m so r₁² = 4 × 10⁻⁴ m², and l = 0.5 m.
- n₁n₂r₁²l = 1000 × 500 × 4 × 10⁻⁴ × 0.5 = 100.
- M = 4π × 10⁻⁷ × π × 100 = 4π² × 10⁻⁵ H.
- Taking π² ≈ 10: M ≈ 40 × 10⁻⁵ H = 4 × 10⁻⁴ H = 0.4 mH.
M ≈ 4 × 10⁻⁴ H (0.4 mH)
What changes M
- Closer coils sharing the same axis have a larger M.
- A soft iron core through both coils increases M by the factor μr.
- More turns in either coil increase M.
- Coils with their planes at right angles share almost no flux, so M is nearly zero.
- M does not depend on the current in either coil.
The distribution transformer near your home and a wireless phone charger both use mutual induction: a changing current in one coil induces an emf in another coil that it does not touch.
2 marks: “Define mutual inductance. Write any two factors on which it depends.” M = N₁Φ₁/I₂, or the emf induced in one coil per unit rate of change of current in the other; unit henry. Depends on the turns, sizes, separation, orientation and core (any two).
5 marks: “Derive an expression for the mutual inductance of two long coaxial solenoids and show that M₁₂ = M₂₁.” B₂ = μ₀n₂I₂; flux linkage with S₁ = n₁l × πr₁² × μ₀n₂I₂, so M₁₂ = μ₀n₁n₂πr₁²l; a current I₁ in S₁ gives the same expression for M₂₁, since its field is confined inside S₁.
1 marks: “Two coils are placed with their planes perpendicular to each other. What can you say about their mutual inductance?” It is nearly zero, as almost no flux of one coil links the other.
M needs two coils: change in one, emf in the other. Coaxial solenoids: μ₀n₁n₂πr₁²l, always with the inner radius.