Mutual Inductance
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Direct answer
Mutual inductance M measures how much flux a current in one coil threads through another: M = N2Φ21/i1, with the induced emf in the second coil being −M di1/dt. Its value depends only on geometry and medium — for a long inner solenoid of radius r nested coaxially inside an outer one sharing length l, M = μ0 n1 n2 π r² l, using the smaller solenoid's radius because the field (and hence flux) exists only inside the inner coil. Coupled coils obey M = k√(L1L2) with coupling 0 ≤ k ≤ 1, k approaching 1 only when the flux of one links nearly every turn of the other. The reciprocity theorem M12 = M21 holds regardless of which coil drives, and connecting two coils in series gives a measured L of L1 + L2 + 2M (aiding) or L1 + L2 − 2M (opposing) — the laboratory backdoor into M.
What you must remember
- Definition: M = N2Φ21/i1; emf induced in coil 2 is e2 = −M di1/dt; the unit is the henry, numerically weber-turn per ampere.
- Coaxial solenoid formula: M = μ0 n1 n2 π r² l, with r the radius of the smaller (inner) solenoid — flux produced by the outer coil's return field outside itself is negligible, so the inner coil's area is what counts.
- Coupling coefficient: M = k√(L1L2), 0 ≤ k ≤ 1; k ≈ 1 for a transformer's tightly interleaved windings on a shared iron core, small for well-separated coils.
- Reciprocity: M12 = M21 always — driving coil 1 and reading coil 2 gives the same M as the reverse, a genuinely non-obvious theorem worth quoting.
- Series-aiding and series-opposing: L_aid = L1 + L2 + 2M and L_opp = L1 + L2 − 2M, so connecting a pair both ways and measuring each equivalent inductance yields M = (L_aid − L_opp)/4.
- Iron-core multiplication: inserting a soft-iron core multiplies M by the relative permeability (hundreds to thousands), the entire operating principle of the transformer.
- Sign conventions: the minus sign belongs to Lenz — induced emf opposes the change of current that creates it.
Two experiments, one M
Build the standard pair: an inner solenoid with 500 turns over 50 cm and radius 2 cm inside an outer solenoid with 1000 turns over the same length. Turn densities are n1 = 1000 per metre and n2 = 2000 per metre, so M = μ0 n1 n2 π r² l = 4π × 10⁻⁷ × 2 × 10⁶ × π × 4 × 10⁻⁴ × 0.5 ≈ 1.6 × 10⁻³ H. Ramping the inner current at 2 A per second induces e = M di/dt = 3.2 mV in the outer coil — small, which is exactly why practical mutual inductance lives on iron cores.
The series-coil measurement is the JEE-loved inverse problem. Two coils connected in series read 1.05 H with their fields aiding and 0.45 H with one coil flipped. The difference of the two readings is 4M, so M = (1.05 − 0.45)/4 = 0.15 H, and their average must recover the sum of the self-inductances: (1.05 + 0.45)/2 = 0.75 H, which is exactly what the coils measure alone (0.30 H and 0.45 H). This cross-check — averaging the two series measurements to rebuild L1 + L2 — is precisely how examiners test whether the ±2M is understood rather than memorised.
Where students slip
The radius choice in the solenoid formula is the most common error: students multiply by the outer coil's cross-section, forgetting the inner solenoid's field is confined to its own bore, so only πr_inner² carries flux. Second is the coupling inequality: M ≤ √(L1L2) with equality only at perfect coupling, so any computed M exceeding the geometric mean flags a mistake. In the series-coil problems, the trap is halving instead of quartering — the difference L_aid − L_opp equals 4M because both +2M and −2M shift the same base L1 + L2. And a conceptual favourite in JEE Advanced assertion–reason: mutual inductance stays constant while currents change (it is geometry), but the induced emf changes with di/dt; students who conflate the two quantities answer the wrong question. Finally, remember M carries no dependence on the current itself — a coil pair has the same M at 1 ampere as at 100.
Frequently asked questions
What is the mutual inductance of two coaxial solenoids?
M = μ0 n1 n2 π r² l, where r is the inner solenoid's radius, since only that cross-section carries the flux that links the outer coil.
How is M related to the self-inductances of the coils?
Through M = k√(L1L2) with coupling factor 0 ≤ k ≤ 1; perfect flux linkage (k = 1) is approached only with shared iron cores or interleaved windings.
How can M be measured with just an LCR bridge?
Connect the coils in series both ways: the readings L1 + L2 ± 2M differ by 4M, so M is one quarter of the difference between aiding and opposing connections.
Does mutual inductance depend on the current in the coils?
No — M is fixed by geometry, turns and the magnetic medium; the current sets the flux and the rate of change sets the induced emf, but not M itself.
Why does the reciprocity M12 = M21 hold?
It is a theorem of the magnetic field equations: the flux linking geometry from coil 1 to coil 2 equals that from 2 to 1, independent of which carries the current.