Mutual Inductance & Transformers

Run a current i₁ through coil 1 and it fills space with a magnetic field, including the space inside coil 2. Call Φ₂₁ the flux through one turn of coil 2 from that field; the flux linkage of coil 2 is N₂Φ₂₁. In a linear medium the field is proportional to the current that makes it, so the linkage is proportional too, and the constant of proportionality is the mutual inductance: N₂Φ₂₁ = M i₁. Put numbers on it. Take a long solenoid, ℓ = 0.50 m with N₁ = 1200 turns and cross-section A = 8.0 cm², and wind a short 300-turn coil tightly round its middle. Inside the solenoid B = μ₀(N₁/ℓ)i₁ and outside it the field is negligible, so every turn of the outer coil catches the same Φ₂₁ = BA. Then M = μ₀N₁N₂A/ℓ = (4π × 10⁻⁷)(1200)(300)(8.0 × 10⁻⁴)/0.50 = 7.2 × 10⁻⁴ H. Notice what is missing from that answer: i₁. M is geometry and material only, which is exactly why it stays usable when the current refuses to hold still.

Now reverse the roles and ask what flux the 300-turn coil sends through the solenoid. That calculation is horrible: the small coil's field spreads like a dipole's, weakening and tilting along the solenoid's length, with no symmetry to exploit. Yet the answer is known before you start, because mutual inductance is reciprocal: M₁₂ = M₂₁, and we simply write M. The reason is energetic. The work needed to establish a given pair of final currents cannot depend on the order in which you raise them, and writing that work out both ways leaves M₁₂ = M₂₁ as the only possibility. So the 0.72 mH computed the easy way is the answer in both directions: ramp the 300-turn coil at 500 A s⁻¹ and the solenoid's terminals show 7.24 × 10⁻⁴ × 500 = 0.36 V. The working rule is to calculate whichever direction has the symmetry, then read the result backwards.

Faraday and Lenz together give ε₂ = −M di₁/dt: the induced emf opposes the change that produced it. But opposing means nothing until you have chosen which terminal of coil 2 you call positive, and a coil can be wound either handedness onto the same core without changing a single magnitude. The dot convention settles this by marking one terminal on each coil, chosen so that currents entering the dotted terminals produce fluxes that add. The rule is then mechanical: if i₁ enters the dotted terminal of coil 1, the dotted terminal of coil 2 is the positive one for v₂ = +M di₁/dt. That plus sign and the minus in ε₂ = −M di₁/dt are the same physics seen twice — one is the terminal voltage a meter reads, the other the emf opposing the change. Reverse the secondary winding and the dot moves to the other end, inverting the output while every magnitude stays put. When both coils carry changing currents the two effects add: v₁ = L₁ di₁/dt ± M di₂/dt, with the sign taken from the dots.

You rarely know a real geometry well enough to compute M, so you measure it instead. Wire the coils in series so their fluxes aid, and an inductance meter reads L_aid = L₁ + L₂ + 2M; swap the connections to one coil so the fluxes oppose, and it reads Lₒₚₚ = L₁ + L₂ − 2M. The difference is 4M, so M = (L_aid − Lₒₚₚ)/4, while the sum is a free check on the two self-inductances. Then express M as a fraction of the most the pair could manage: k = M/√(L₁L₂). The ceiling k ≤ 1 is not a convention but an energy statement — if M exceeded √(L₁L₂), some pair of currents would give the coupled coils negative stored energy. Real values run the whole range. Two air-cored coils side by side on a bench manage k ≈ 0.01 to 0.3; wind both onto one closed laminated core and k climbs past 0.99. What k < 1 measures is leakage: flux that one coil makes and the other never sees.

Drive a closed ferromagnetic core and k ≈ 1, so the two windings stop being two circuits with a shared fringe and become two counts of a single flux Φ(t). Faraday applies turn by turn: v₁ = N₁ dΦ/dt and v₂ = N₂ dΦ/dt, so v₂/v₁ = N₂/N₁ at every instant, for any waveform at all. Idealise once more — no winding resistance, no core loss, no leakage, and a permeability high enough that magnetising the core costs no current — and energy conservation adds I₁V₁ = I₂V₂, hence I₁/I₂ = N₂/N₁. For a sinusoidal flux Φ = Φₘₐₓ sin ωt, Faraday gives the design equation Vᵣₘₛ = 2πfNΦₘₐₓ/√2 = 4.44 f N Bₘₐₓ A. Two consequences are worth keeping. Everything here rests on dΦ/dt, so a transformer does nothing on DC: 240 V DC across an 8 Ω primary gives 30 A and 7.2 kW of smoke, not a stepped voltage. And Bₘₐₓ goes as 1/f, which is why a 50 Hz transformer is content at 400 Hz while a 400 Hz one saturates at 50 Hz.

Hang a resistance R on the secondary and the primary does not merely supply it, it sees it, rescaled. With V₂ = (N₂/N₁)V₁ and I₂ = V₂/R, the current ratio gives I₁ = (N₂/N₁)I₂, so V₁/I₁ = (N₁/N₂)²R: impedance transfers as the square of the turns ratio. That is the entire point of an output transformer, which makes an 8.0 Ω loudspeaker look like 800 Ω to an amplifier through a 10:1 winding. Now subtract reality. Winding resistance dissipates I²R in both coils, so copper loss climbs with the square of the load current. Hysteresis costs the area of the B–H loop once per cycle, so it scales with f. Eddy currents circulating in the core scale as f²B²t² for lamination thickness t, which is why cores are stacks of thin insulated sheets. Leakage flux, k < 1, acts as a series reactance that makes the output droop under load. Small transformers reach 95% to 97%; grid transformers pass 99.5%.