Set 5 — Self, Mutual Inductance and Induced EMF
A magnetic circuit becomes a circuit element the moment it is given terminals. Flux linkage per ampere is inductance, shared flux linkage per ampere is mutual inductance, and the rate of change of either is a voltage. This set works both quantities from the geometry, measures the second one the way a laboratory actually does — by comparing series aiding with series opposing — and puts numbers to the energy a coupled pair stores.
Alongside runs the other half of Faraday's law. An emf appears whether the flux changes around a stationary coil or a conductor moves through a steady field, and Problem 4 shows that \(Blv\) and \(N\,d\Phi/dt\) are the same statement with different factors varying. Problem 6 closes the set by showing that a coupling coefficient can fall well below unity without a single line of flux leaving the iron.
Self-inductance is the magnetic circuit seen from the terminals. Flux linkage per ampere, and therefore turns squared over reluctance:
\[ L = \frac{\lambda}{i} = \frac{N\Phi}{i} = \frac{N^2}{\mathcal{R}} \]Every reduction technique of Sets 1 to 3 feeds this one formula. Find \(\mathcal{R}\) by whatever series–parallel argument the core requires, then divide.
Mutual inductance is the same idea with two windings. The flux one coil produces that also links the other, per ampere:
\[ M = \frac{N_2\Phi_{21}}{i_1} = \frac{N_1\Phi_{12}}{i_2}, \qquad M = \frac{N_1N_2}{\mathcal{R}}\ \text{when all the flux is shared} \]The coupling coefficient measures what is shared. \(k = 1\) only if every line of flux links both coils:
\[ k = \frac{M}{\sqrt{L_1L_2}} \le 1 \]It falls below one for two reasons that look alike and are not: flux escaping into air, and flux dividing between iron branches — Problem 6.
Series aiding and series opposing measure \(M\) without a flux meter:
\[ L_{aid} = L_1+L_2+2M, \quad L_{opp} = L_1+L_2-2M \;\Longrightarrow\; M = \frac{L_{aid}-L_{opp}}{4} \]One law of induction, two ways of reading it. An emf appears whenever the flux linkage changes, whether the circuit stands still and the flux varies, or the flux stands still and the circuit moves:
\[ e = \frac{d\lambda}{dt} = N\frac{d\Phi}{dt} \quad\text{(statically induced)}, \qquad e = Blv \quad\text{(dynamically induced)} \]Energy stored in a coupled pair depends on the sign of the mutual term, which is the sign of the dot convention:
\[ W = \tfrac12 L_1 i_1^2 + \tfrac12 L_2 i_2^2 \pm M i_1 i_2 \]Requiring this to be non-negative for every possible pair of currents is what proves \(M \le \sqrt{L_1L_2}\) — Problem 5.
A coil of 500 turns is wound uniformly on an iron ring.
| Quantity | Value |
|---|---|
| Mean diameter of the ring | 25 cm |
| Cross-sectional area | 6 cm2 |
| Relative permeability | 1200 |
| Turns | 500 |
| Coil current | 0.5 A |
Find the reluctance of the ring, the inductance of the coil, the flux and flux density at 0.5 A, the energy stored, the emf induced if the current is switched off uniformly in 20 ms, and the number of turns that would be needed for an inductance of 0.5 H.
The mean path of a ring is its mean circumference:
Reluctance, then inductance. With \(\mu_0\mu_r = (4\pi\times10^{-7})(1200) = 1.508\times10^{-3}\):
Note that the current has not been used. Inductance is a property of the winding and the magnetic circuit, and while \(\mu_r\) stays constant it is a property of the current too.
Flux, linkage and density at 0.5 A, as a check on the inductance by the other route:
0.48 T is comfortably below the knee, which is what justifies treating \(\mu_r = 1200\) as a constant in the first place. At three times this current the assumption would fail and \(L\) would fall.
Stored energy:
The emf on switch-off, with the current falling linearly to zero in 20 ms:
Modest here. Interrupt the same coil in 200 µs instead and the emf is 720 V — the reason a field winding is never opened without a discharge path.
Turns for 0.5 H. Since \(L \propto N^2\) and the magnetic circuit is unchanged:
Equivalently \(500\sqrt{0.5/0.288} = 659\). A 32% increase in turns for a 74% increase in inductance — and a 32% increase in resistance with it.
Two coils are wound on the same iron core. Coil 1 has 600 turns and coil 2 has 400 turns. When coil 1 carries 2 A it produces a total flux of 1.0 mWb, of which 0.85 mWb links coil 2; the remainder leaks through the air. The same fraction is found to link in the reverse direction when coil 2 is excited. Find:
- the self-inductance of each coil;
- the mutual inductance;
- the coefficient of coupling;
- the emf induced in coil 2 when the current in coil 1 is reduced to zero in 10 ms.
Self-inductance of coil 1 uses its own total flux — leakage included, because leakage flux still links the coil that made it:
Mutual inductance uses only the shared part:
The distinction between \(\Phi_1\) and \(\Phi_{21}\) is the whole of the leakage story, and it is the same distinction that becomes the leakage reactance of a transformer in Part 3.
Coil 2 follows from reciprocity. Mutual inductance is a single number, \(M_{12} = M_{21}\), so the flux that coil 2 must send into coil 1 per ampere is already fixed:
With the same coupling fraction of 0.85 in this direction, the total flux coil 2 produces is
Check against the turns ratio: \(L_1/L_2 = 0.300/0.1333 = 2.25 = (600/400)^2\). Two coils on the same magnetic circuit have inductances in the ratio of the squares of their turns, whatever the reluctance is.
Coupling coefficient:
Equal to the flux fraction, because the fraction was the same both ways. In general \(k = \sqrt{k_1k_2}\) with a separate fraction for each direction, so the symmetric case is the one where \(k\) and the flux fraction coincide.
The reluctance of the main path, as an independent check on \(L_1\):
Induced emf in coil 2:
Coil 2 is not connected to anything, carries no current, and still develops 34 V. That is the transformer, stripped of everything but the essential.
Two coils on a common core are connected in series and their combined inductance is measured twice, once in each relative polarity. The readings are 3.0 H and 1.0 H. A separate measurement on the first coil alone gives 1.6 H. Find:
- the mutual inductance;
- the self-inductance of the second coil;
- the coefficient of coupling;
- the energy stored in each connection when the series current is 2 A;
- how the two readings identify which connection is which, given that the terminal markings are missing.
Write the two connections. In series the same current flows in both coils, so each coil links its own flux and its neighbour's; the mutual term is added twice, once for each coil, and its sign is the sign of the winding sense:
Subtract and add. The difference isolates \(M\) and the sum isolates the two self-inductances:
The factor of 4, not 2, is where this calculation is most often lost. The two readings differ by \(4M\) because the mutual term flips sign and appears twice.
Coupling coefficient:
Comfortably below one, so the coils share rather less than two-thirds of their flux — two windings on separate limbs, not interleaved on the same one.
Energy at 2 A in each connection. A series pair is a single inductor of the measured value:
Same coils, same current, three times the stored energy. The mutual flux either reinforces or cancels, and 4 J of the 6 in the aiding case is the mutual contribution \(2Mi^2\).
Identifying the connections without the markings. Since \(M > 0\) by construction, \(L_{aid} > L_{opp}\) always. The larger reading is therefore the aiding connection, and no dot convention is needed to say so. The test also gives a free consistency check:
Here \(0.5 \le 1.0\), satisfied. A negative opposing reading would mean a measurement error, since stored energy cannot be negative.
Why the test is the practical one. It needs only an inductance bridge and a pair of links: measure both series combinations, then one coil alone. No flux has to be measured, no core dimension known, no permeability assumed. Everything in Problem 2 required knowing what the flux was doing inside the iron; this requires nothing at all.
Three situations, all governed by the same law of induction.
- A straight conductor 0.30 m long moves at 25 m/s through a uniform field of 0.9 T. Find the emf when the motion is perpendicular to the field, and when it is at 60° to it.
- A coil of 250 turns is wound on a core of cross-section 40 cm2. The flux density in the core is raised uniformly from 0.2 T to 1.0 T in 0.05 s. Find the induced emf.
- A rectangular coil of 50 turns, 0.20 m by 0.15 m, rotates at 1500 rpm about an axis perpendicular to a uniform field of 0.8 T. Find the frequency, the peak emf and the rms emf.
Then show that the first and second are the same equation.
Part 1 — dynamically induced emf. The circuit moves, the field does not change:
Only the component of velocity perpendicular to the field sweeps flux, so at \(\theta\) to the field:
At \(\theta = 0\) the conductor slides along the flux lines, cuts none, and generates nothing — which is why the conductors of a machine lie parallel to the shaft and the flux crosses the gap radially.
Part 2 — statically induced emf. The circuit stands still, the flux changes. The area is fixed, so it is the density that is differentiated:
Part 3 — the rotating coil, which is both at once. A two-pole field gives one cycle per revolution:
The linkage is \(\lambda = NBA\cos\omega t\), so
Cross-check by the moving-conductor route: the two active sides each move at \(v = \omega r = (157.1)(0.075) = 11.78\) m/s, and at the instant of maximum cutting \(E_{max} = 2NBlv = 2(50)(0.8)(0.20)(11.78) = 188.5\) V. The same number by the other reading of the same law.
Showing that parts 1 and 2 are one equation. Put the moving conductor of part 1 on a pair of rails a distance \(l\) apart, closed at one end, and let \(x\) be the distance from the closed end. The circuit encloses a flux
Nothing was assumed about whether \(B\) or the geometry was doing the changing. There is one law, \(e = d\lambda/dt\); "dynamically induced" and "statically induced" name which factor in \(\lambda\) varies, not two different physics.
The coils of Problem 3 — \(L_1 = 1.6\) H, \(L_2 = 0.4\) H, \(M = 0.5\) H — are now supplied separately. Coil 1 carries 3 A and coil 2 carries 2 A. Find:
- the energy stored when both currents enter the dotted terminals;
- the energy stored when the current in coil 2 is reversed;
- the value of \(i_2\) that minimises the stored energy with \(i_1\) held at 3 A, and the value of that minimum;
- hence the condition on \(M\) that the stored energy can never be negative.
The energy expression, and where its three terms come from. Building the currents up from zero, each coil stores its own \(\tfrac12 Li^2\), and the work done against the emf each induces in the other supplies the third term:
The mutual term appears once, not twice, although the coupling acts both ways — the standard result of establishing the currents one after the other and adding the work.
Both currents into the dots — the fluxes aid:
Coil 2 reversed — the mutual term changes sign:
The self terms are untouched, since they carry \(i^2\). Reversing one current halves the stored energy here, and the 6 J difference is \(2Mi_1i_2\).
The minimum over \(i_2\), with \(i_1 = 3\) A fixed. Differentiate and set to zero:
A current of 3.75 A opposing in coil 2 is the most complete cancellation the pair allows, and even then 4.39 J remains stored.
Write the minimum in closed form and read the condition off it. Substituting \(i_2 = -Mi_1/L_2\) back:
Stored magnetic energy cannot be negative for any pair of currents, so the bracket must be non-negative:
So \(k \le 1\) is not a definition or a convention. It is an energy statement: a value of \(M\) above \(\sqrt{L_1L_2}\) would let a pair of currents extract energy from a passive pair of coils.
Check against Problem 3. There the coils were in series, so \(i_1 = i_2 = i\) and the general expression collapses:
At \(i = 2\) A this gives \(\tfrac12(3.0)(4) = 6.0\) J, the value found there. The series result is a special case of this one.
A three-limb core carries two windings: coil 1 of 400 turns on the central limb and coil 2 of 200 turns on the right-hand outer limb. The two outer return paths are identical. The material has a constant relative permeability of 1200 and there is no air gap; leakage into the surrounding air is neglected entirely.
| Path | Mean length | Cross-section | Winding |
|---|---|---|---|
| Central limb | 0.20 m | 25 cm2 | coil 1, 400 turns |
| Left outer path | 0.50 m | 12.5 cm2 | — |
| Right outer path | 0.50 m | 12.5 cm2 | coil 2, 200 turns |
Find \(L_1\), \(L_2\), \(M\) and \(k\). Verify the reciprocity of \(M\) by computing it from both directions, and explain how \(k\) can be well below unity when no flux escapes into the air at all.
Reluctances first. With \(\mu_0\mu_r = 1.508\times10^{-3}\):
Conveniently \(\mathcal{R}_o = 5\mathcal{R}_c\) exactly, which will make the flux fractions come out as clean ratios.
Excite coil 1 alone. It sees the central limb in series with the two outer paths in parallel — the circuit of Set 3, Problem 3:
By symmetry exactly half the central flux goes down the right-hand limb, so coil 2 links half of it:
Now excite coil 2 alone — and note that it sees a different circuit. Flux driven round the right-hand limb returns partly through the central limb and partly through the left-hand limb, which are in parallel with each other:
The two coils do not see the same magnetic circuit, which is why \(L_1/L_2 = 6.67\) and not the \((N_1/N_2)^2 = 4\) of Problem 2. Turns-squared scaling holds only when both coils sit on the same path.
Reciprocity check. The flux coil 2 produces divides between the central and left-hand paths in inverse proportion to their reluctances, so the fraction reaching coil 1 is
The same 0.2154 H from a completely different route — a different reluctance, a different flux fraction, a different pair of turns. Reciprocity is a theorem, and reproducing it is the strongest available check on a two-winding calculation.
Coupling coefficient:
| Quantity | Circuit seen | Value |
|---|---|---|
| \(L_1\) | \(\mathcal{R}_c + \mathcal{R}_o/2\) | 0.8617 H |
| \(L_2\) | \(\mathcal{R}_o + \mathcal{R}_c\!\parallel\!\mathcal{R}_o\) | 0.1293 H |
| \(M\) | half of \(\Phi_1\); 83.3% of \(\Phi_2\) | 0.2154 H |
| \(k\) | \(\sqrt{(0.500)(0.8333)}\) | 0.645 |
The last row is the point of the problem: \(k = \sqrt{k_1k_2}\) with \(k_1 = 0.500\) and \(k_2 = 0.8333\), and \(\sqrt{0.4167} = 0.6455\).
Where the coupling went, given that nothing leaked. Not into the air — the problem forbade that. Half of coil 1's flux simply went down the left-hand limb, where coil 2 is not. From coil 2's side, one-sixth of its flux returned through the left-hand limb rather than through the centre. Both are perfectly good iron paths that happen not to pass through the other winding, and a magnetic circuit does not distinguish between flux lost to air and flux sent down a branch that misses the second coil.
This is exactly how a magnetic shunt is used to detune a coupled pair on purpose: add a third limb and the coupling falls without a single line of flux leaving the iron.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Self-inductance | \(L = \dfrac{\lambda}{i} = \dfrac{N\Phi}{i} = \dfrac{N^2}{\mathcal{R}}\) | Operating-point value — Problems 1, 6 |
| Turns for a target \(L\) | \(N = \sqrt{L\mathcal{R}}\) | Square law — Problem 1 |
| Mutual inductance | \(M = \dfrac{N_2\Phi_{21}}{i_1} = \dfrac{N_1\Phi_{12}}{i_2}\) | Shared flux only — Problems 2, 6 |
| Ideal coupling | \(M = \dfrac{N_1N_2}{\mathcal{R}}\) | All flux common — Problem 2 |
| Coupling coefficient | \(k = \dfrac{M}{\sqrt{L_1L_2}} = \sqrt{k_1k_2}\) | Never exceeds 1 — Problems 2, 3, 6 |
| Series aiding / opposing | \(L_{eq} = L_1+L_2\pm 2M\) | Same current in both — Problem 3 |
| Measuring \(M\) | \(M = \dfrac{L_{aid}-L_{opp}}{4}\) | Factor 4, not 2 — Problem 3 |
| Self-induced emf | \(e = L\dfrac{di}{dt}\) | Switch-off transient — Problem 1 |
| Mutually induced emf | \(e_2 = M\dfrac{di_1}{dt}\) | Open secondary still develops it — Problem 2 |
| Statically induced emf | \(e = N A_c \dfrac{dB}{dt}\) | Fixed area — Problem 4 |
| Dynamically induced emf | \(e = Blv\sin\theta\) | Perpendicular component only — Problem 4 |
| Rotating coil | \(E_{max} = NBA\omega,\quad E_{rms} = E_{max}/\sqrt2\) | Ancestor of every machine emf — Problem 4 |
| Energy, single coil | \(W = \tfrac12 Li^2\) | Problems 1, 3 |
| Energy, coupled pair | \(W = \tfrac12L_1i_1^2+\tfrac12L_2i_2^2\pm Mi_1i_2\) | Sign set by the dots — Problem 5 |
| Limit on \(M\) | \(M \le \sqrt{L_1L_2}\) | From \(W \ge 0\) — Problem 5 |
Common Mistakes
Using the linking flux for self-inductance. \(L_1\) counts all the flux coil 1 produces, leakage included; only \(M\) is restricted to the shared part. Using 0.85 mWb for both would give \(k = 1\) and hide the leakage entirely — Problem 2.
Writing \(M = (L_{aid}-L_{opp})/2\). The mutual term appears twice in each connection and reverses between them, so the two readings differ by \(4M\) — Problem 3.
Assuming \(L_1/L_2 = (N_1/N_2)^2\) always. True only when both coils see the same magnetic circuit. On a three-limb core the ratio is 6.67 where the turns ratio would predict 4 — Problem 6.
Forgetting the mutual term in the stored energy. \(\tfrac12L_1i_1^2 + \tfrac12L_2i_2^2\) gives 8.0 J where the true answer is 11.0 J aiding or 5.0 J opposing — Problem 5.
Counting the mutual energy twice. It is \(Mi_1i_2\), not \(2Mi_1i_2\) — the factor 2 belongs to the series connection, where \(i_1 = i_2\) — Problems 3 and 5.
Dropping \(\sin\theta\) from \(Blv\). Only the velocity component perpendicular to the field cuts flux; at 60° the emf is 5.85 V, not 6.75 V — Problem 4.
Confusing peak with rms for a rotating coil. \(NBA\omega\) is the peak; the meter reads \(188.5/\sqrt2 = 133.3\) V — Problem 4.
Using rpm where rad/s is required. \(\omega = 2\pi N/60\); substituting 1500 for 157.1 inflates the emf by a factor of nearly ten — Problem 4.
Treating \(k\) as a measure of air leakage only. Flux that stays in the iron but takes a limb the other coil is not on reduces \(k\) just as effectively — Problem 6.
Quoting an iron-cored inductance without its operating point. \(L = N^2/\mathcal{R}\) is only as constant as \(\mu_r\) is, and \(\mu_r\) falls sharply above the knee — Problems 1 and 6.
The magnetic circuit now has terminals. \(L = N^2/\mathcal{R}\) converts every reluctance calculation of Sets 1 to 3 into a quantity a bridge can measure; \(M\) and \(k\) do the same for two windings; and \(e = d\lambda/dt\) connects both to a voltage. Problem 6 made the point that matters most for what follows: coupling is decided by where the flux goes, and a limb that misses the second winding costs exactly as much as a path through air.
What remains is the mechanical terminal. So far the flux has produced voltages; it also produces force. The bridge between them is energy: what the source delivers must equal what the field stores plus what the shaft or the armature receives, and differentiating the stored energy with respect to position is what converts a magnetic circuit into a machine.
Next: Set 6 — Energy, Coenergy, Force and Torque, where the force on an armature is found from \(\partial W/\partial x\) at constant flux linkage and from \(\partial W'/\partial i\) at constant current, coenergy is separated from energy for a saturating core, and torque is obtained by differentiating inductances with respect to rotor angle.