By the end of this chapter you should be able to:
Explain why an air gap is unavoidable in a rotating machine and deliberate in a reactor.
Correct a gap area for fringing and quantify the resulting reduction in gap reluctance.
Compute Carter's coefficient for a slotted machine surface and convert an actual gap into an effective gap.
Distinguish leakage flux from fringing flux, and apply a leakage coefficient correctly in a design calculation.
Explain how a gap linearises a magnetic circuit by swamping the iron's variable reluctance, and quantify the effect.
Compute the energy stored in a gap and show that it dominates the energy stored in the iron.
Design a gapped reactor: choose turns from the saturation limit and gap length from the required inductance.
State which corrections make a calculation optimistic and which make it conservative, and why that asymmetry matters.
Introduction
Chapters 2 and 3 built a working method for magnetic circuits on three quiet assumptions: that the gap is a neat prism of the same cross-section as the core, that all the flux goes where it is told, and that the iron's permeability is a fixed number. All three are false, and the first two fail hardest exactly where machines do their work — at the air gap.
This chapter pays off two of those debts. Fringing means the flux bulges outward as it crosses the gap, so the effective area is larger than the core's and the gap is easier to cross than the simple formula predicts. Leakage means some flux never reaches the gap at all, closing instead through the air around the winding, so the useful flux is less than the coil produces. The two effects are frequently confused, they act in opposite directions, and only one of them is dangerous to neglect.
Along the way the chapter makes a case for the air gap that may be surprising. It is usually introduced as a nuisance — the thing that eats 80 % of your ampere-turns. It is also the thing that stores essentially all the useful energy, that makes the flux almost independent of the iron's temperamental permeability, and that lets a reactor carry DC without saturating. The gap is not a defect to be minimised but a design variable to be chosen.
Why Machines Have Air Gaps
Air gaps arise for two quite different reasons, and it is worth separating them because they lead to opposite design pressures.
In any rotating machine the rotor must be free to turn inside the stator. The gap is a mechanical necessity, set by bearing tolerance, shaft deflection, thermal expansion and manufacturing runout.
Here the designer wants the gap as small as possible. Typical values: 0.2–0.5 mm for a small induction motor, 1–3 mm for a large one, 10–30 mm for a turbo-alternator whose rotor is metres across.
In a reactor, choke, relay or current transformer the gap is cut on purpose, to raise reluctance, lower inductance, and above all to allow a large DC current without saturating the core.
Here the designer wants the gap exactly right. It is the parameter that sets the inductance, and it is chosen last, once turns and core area are fixed.
Two further consequences of the gap are worth flagging now, because they are developed later in this chapter and relied on throughout the book. First, the gap is where essentially all of the field energy resides (Section 4-8), and therefore where torque is produced. Second, because the gap's reluctance is constant — air does not saturate — a dominant gap makes the whole circuit behave almost linearly (Section 4-7), which is what allows the equivalent circuits of Parts 3 to 5 to be linear circuits at all.
The MMF of the Air Gap
The starting point is the result already established. For a gap of length \(l_g\) and area \(A_g\),
Two numerical facts are worth committing to memory, because they let you estimate a gap mmf in your head and catch errors instantly:
At 1 T, one millimetre of air costs about 800 ampere-turns. At 0.8 T — a typical machine gap density — it costs about 640 AT per millimetre. If your calculated gap mmf is not close to \(800\,B\,l_g\) with \(l_g\) in millimetres, you have made an arithmetic error.
The second point is that \(A_g\) in the reluctance formula is not in general the core area. It is the area over which the flux actually crosses the gap, and that is larger than the core face because of fringing. Sections 4-4 and 4-5 give the two standard corrections: fringing for an isolated gap between pole faces, and Carter's coefficient for the slotted surfaces of a rotating machine.
Fringing at the Gap Edges
When the useful flux crosses the air gap it tends to bulge outwards at the edges, because magnetic lines set up in the same direction repel one another and are no longer constrained by iron. This spreading is called fringing.
Its consequences follow directly:
The effective area of the gap increases beyond the core cross-section.
Because the same flux now crosses a larger area, the flux density in the gap decreases.
Because reluctance is inversely proportional to area, the gap reluctance decreases — the gap is easier to cross than the naive calculation suggests.
The effect is directly related to the length of the gap: a long gap fringes badly, a very short one hardly at all.
The standard empirical correction adds one gap length to each transverse dimension of the core face:
This rule is empirical, not derived. It assumes the gap is short compared with the face dimensions — say \(l_g < 0.1\,\min(a,b)\). Beyond that, the fringing flux ceases to be a small perturbation and only a field solution will do.
Note also the direction of the error if you ignore fringing entirely: you compute too small an area, hence too large a reluctance, hence too large a required mmf. Neglecting fringing is therefore a conservative error — the machine will need less current than you predicted. Contrast this with leakage in Section 4-6, where the error runs the other way.
Problem. A machine pole has a rectangular face measuring 60 mm × 100 mm and faces the armature across a 3 mm gap. The pole carries a flux of 7.2 mWb. Find the flux density in the pole and in the gap, and the gap mmf with and without the fringing correction.
Areas.
an increase of 8.15 %.
Flux densities. The same flux crosses both, so the gap runs at the lower density:
Gap mmf without fringing.
Gap mmf with fringing.
Check against the rule of thumb. \(800 \times B_g \times l_g = 800(1.110)(3) = 2664\) AT — within 1 % of the full calculation \(\checkmark\)
Comment. Ignoring fringing overstates the gap mmf by 8.2 %, matching the area increase exactly — as it must, since reluctance is inversely proportional to area at fixed length. For a machine needing perhaps 3500 AT per pole in total, 216 AT is a real quantity of copper, so the correction is worth making.
Slotting and Carter's Coefficient
A rotating machine's gap is not bounded by smooth iron. The stator (and usually the rotor) is slotted to hold the windings, so the flux sees an alternation of tooth and slot. Over a tooth the path is short and easy; over a slot opening the flux must travel further through air, spreading as it goes.
The effect is that a slotted gap behaves like a larger smooth gap. The correction factor is Carter's coefficient \(k_C\), defined so that
where \(g\) is the physical clearance and \(g_e\) the effective gap to be used in all reluctance and mmf calculations. Carter's expression is
with \(\tau_s\) the slot pitch, and the parameter \(\gamma\) depending on the ratio of slot opening \(b_o\) to gap:
That expression is awkward to evaluate by hand, and a simpler approximation is widely used and accurate to about 1 % over the normal range:
Problem. An induction motor stator has a slot pitch of 20 mm, a slot opening of 4 mm and an air gap of 1 mm. Find Carter's coefficient by both the exact and the approximate expressions, and the effective gap. If the gap flux density is 0.85 T, find the extra mmf per pole caused by slotting.
Approximate expression.
Exact expression. With \(b_o/2g = 4/2 = 2\):
The two agree to three figures, which is why the approximate form is used in practice.
Effective gap and extra mmf.
against \(676.4\) AT for a smooth gap — an increase of 67.0 AT, or 9.9 %.
Comment. Ten percent more magnetising ampere-turns, purely because the stator has slots in it. Since the magnetising current of an induction motor is already 25–40 % of full-load current, this is the difference between a good no-load power factor and a mediocre one — and it is why designers keep slot openings narrow, using semi-closed or closed slots wherever the winding can still be inserted.
Leakage Flux
Flux which does not follow the intended path in a magnetic circuit is called leakage flux. It exists because, as Chapter 2 established, there is no magnetic insulator — air is only a few thousand times worse than iron, not \(10^{20}\) times as a good electrical insulator is.
The mechanism is easiest to see in a solenoid wound on a gapped core. When a current is passed through the coil, magnetic flux is produced. Most of it is set up in the core and passes through the air gap along the intended path; this is the useful flux \(\Phi_u\). But some flux is set up around the coil itself, closing through the surrounding air without ever reaching the gap, and is not used for any work. This is the leakage flux \(\Phi_l\). The total flux produced by the solenoid is therefore
The coefficient is always greater than one, since some leakage is unavoidable. Typical values are 1.15 to 1.25 for a rotating machine with an appreciable gap, and much closer to unity — often below 1.05 — for a well-designed transformer whose windings are interleaved on a closed core.
Using the coefficient in a calculation is straightforward, but the direction of the correction must be right. You specify the flux you need in the gap; the coil must produce more than that:
- Start from the required useful flux \(\Phi_u\) in the gap, fixed by the machine's EMF or torque requirement.
- Scale up to the total flux the pole core must carry: \(\Phi = \lambda\Phi_u\).
- Compute the mmf of the gap and armature using \(\Phi_u\) — the leakage never gets there.
- Compute the mmf of the pole core and yoke using \(\Phi\) — those sections carry the leakage too.
- Add the drops to get the total excitation.
| Leakage | Fringing | |
|---|---|---|
| Where the flux goes | Never reaches the gap | Crosses the gap, but spread out |
| Effect on useful flux | Reduces it | None — same flux crosses |
| Effect on gap reluctance | None | Reduces it |
| Effect on gap \(B\) | Reduces it | Reduces it |
| Handled by | Coefficient \(\lambda > 1\) on flux | Enlarged area \(A_g\) |
| If neglected, your answer is | Optimistic — dangerous | Conservative — safe |
The last row is the one to remember. Neglect fringing and you build a machine that needs slightly less current than you predicted — no harm done. Neglect leakage and you build a machine whose gap flux, and therefore whose EMF or torque, falls short of specification. That is why leakage estimates are treated seriously in design offices while fringing is often absorbed into a safety margin.
Looking forward. Leakage flux does not merely reduce the useful flux — because it links the winding, it contributes inductance. That inductance is the leakage reactance which dominates transformer voltage regulation (Chapter 45), limits transformer short-circuit current (Chapter 46), and sets the breakdown torque of an induction motor (Chapter 62). What appears here as a nuisance becomes, later, one of the most important parameters in the entire subject.
Problem. A machine pole must deliver a useful flux of 8.0 mWb across a 3 mm gap. The leakage coefficient is 1.18. The pole core has a cross-section of 80 cm² and a length of 100 mm with \(\mu_r = 1500\); the effective gap area, allowing for fringing, is 95 cm². Find the flux density in the pole core and in the gap, and the ampere-turns required by these two sections.
Total flux in the pole core. The core must carry the leakage as well as the useful flux:
Pole core.
Air gap. Only the useful flux crosses it:
Combined.
Comment. Two lessons sit in these numbers. First, the gap takes 97 % of the ampere-turns of these two sections — the familiar pattern. Second, and less obvious, the leakage does not change the gap mmf at all, because leakage flux never crosses the gap. What it does is drive the pole core 18 % harder, at 1.18 T instead of 1.00 T. In a machine running nearer saturation that extra 18 % could push the pole core over the knee, where its permeability collapses and its mmf demand rises steeply. Leakage bites hardest not through the gap but through the iron that carries it.
The Gap as a Linearising Element
So far the gap has looked like a pure liability. Here is the compensating virtue, and it is a large one.
Iron's reluctance is unreliable: \(\mu_r\) can change by a factor of ten across the operating range, so \(S_{\text{iron}}\) changes with it. Air's reluctance is perfectly constant — air does not saturate, and \(\mu_0\) is a constant of nature. When the two are in series and the gap dominates, the total reluctance is mostly a constant:
The \(\Phi\)-versus-mmf characteristic of a gapped circuit is very nearly a straight line, even though the iron's own characteristic is strongly curved. The gap does not remove saturation — the iron still saturates at the same \(B\) — but it makes the circuit behave linearly over a much wider range, and it pushes the knee to a far higher excitation.
Problem. A ring has a mean length of 0.50 m and a cross-section of 5 cm². As the core is driven harder its relative permeability falls from 3000 to 1000. Compare the change in total reluctance (a) with no gap, and (b) with a 1 mm gap. Take the flux at a fixed excitation of 1000 AT in each case.
(a) No gap.
The reluctance triples, exactly as \(\mu_r\) falls by three. Flux at 1000 AT falls from 3.77 mWb to 1.26 mWb.
(b) With a 1 mm gap. The gap adds a fixed \(S_g\), and the iron path shortens to 0.499 m:
Flux at 1000 AT falls only from 0.539 mWb to 0.419 mWb.
| No gap | 1 mm gap | |
|---|---|---|
| Flux at 1000 AT, \(\mu_r = 3000\) | 3.770 mWb | 0.539 mWb |
| Flux at 1000 AT, \(\mu_r = 1000\) | 1.257 mWb | 0.419 mWb |
| Change in flux | −66.7 % | −22.3 % |
| Change in total reluctance | 3.00 × | 1.285 × |
Comment. A three-to-one collapse in the iron's permeability produces only a 1.29-to-one change in the gapped circuit's reluctance. The gap has converted a wildly variable circuit into a nearly predictable one — at the cost of about seven-eighths of the flux. That is the bargain, and it is one designers take willingly whenever predictability matters more than magnetising current: in reactors, in current transformers that must not saturate on fault current, and in any inductor carrying DC bias.
Energy Storage in the Gap
Chapter 1 asserted that essentially all of a machine's field energy sits in the air gap. The magnetic-circuit machinery now available lets that be proved in one line.
The energy stored in a region of volume \(V\) at uniform flux density is \(W = B^{2}V/2\mu\). For two sections in series carrying the same flux, the ratio of stored energies is therefore
which is precisely the reluctance ratio of Section 3-4. Equivalently, in terms of reluctance,
Whatever fraction of the reluctance a section holds, it holds the same fraction of the stored energy. Since the gap typically holds 80 % of the reluctance, it holds 80 % of the energy — and since force is the rate of change of stored energy with displacement (Chapter 18), that is exactly why the gap is where torque is produced.
This is also the practical basis of reactor design. An inductor's job is to store energy, and iron is bad at storing it — the same iron that makes flux easy to establish makes energy hard to accumulate. A gapped core is therefore not a compromised inductor but the correct design: the iron guides the flux, and the gap stores the energy.
Problem. Design a reactor of 5 mH to carry 10 A peak without exceeding 1.2 T. The core has a cross-section of 6 cm², a mean iron path of 0.25 m and \(\mu_r = 2500\). Find the number of turns and the gap length, and check how the stored energy divides between iron and gap.
Turns from the saturation limit. At peak current the flux linkage is \(\lambda = LI = N\Phi = NB_{\max}A\), so
Required total reluctance. Since \(L = N^{2}/S\):
Iron contribution.
Gap. The gap must supply the balance:
Energy check. With \(\Phi = B_{\max}A = (1.2)(6\times10^{-4}) = 7.2\times10^{-4}\) Wb:
Cross-check against \(\tfrac{1}{2}LI^{2} = \tfrac{1}{2}(0.005)(10)^{2} = 0.250\) J — agreeing to within the rounding of \(N\) from 69.4 to 69 \(\checkmark\)
Comment. The gap holds \(0.2124/0.2468 = 86\,\%\) of the energy while occupying 0.25 % of the magnetic path length. Note also how little gap is needed: 0.6 mm of air does the work of about 1.5 m of this iron. In a reactor the gap is not a defect — it is the component. Fringing at a gap this short would add a further few percent to the effective area, so a real design would iterate once with the corrected area.
Applications
The gap is made as small as bearings permit, then corrected by Carter's coefficient for slotting on both sides. Magnetising current, and hence no-load power factor, follows almost entirely from \(g_e\). This is the single most consequential dimension in the machine.
A gapped core carries large DC bias without saturating, because the gap dominates the reluctance and stores the energy. Ungapped, the same core would saturate at a fraction of the current and the inductance would collapse when it is most needed.
Protection CTs must stay linear during fault currents many times rated. A small deliberate gap linearises the characteristic (Section 4-7) and prevents saturation, at the cost of a larger magnetising current and some phase error.
Force comes from the rate of change of gap energy with position. Since almost all energy is in the gap, the force is largest when the gap is smallest — which is why relays snap closed and why holding current is far below pull-in current.
The design goal is a high, uniform \(B\) across a narrow annular gap holding the voice coil. Fringing is deliberately controlled by the pole-piece geometry, because non-uniform gap flux produces distortion.
Interleaving windings reduces leakage flux and hence leakage reactance; deliberately separating them, or inserting a magnetic shunt, increases it. Welding transformers and ballasts exploit exactly this to limit current.
Summary and Key Formulas
Air gaps are unavoidable in rotating machines (mechanical clearance) and deliberate in reactors, chokes and relays (to set inductance and prevent saturation).
Useful shortcut: \(\mathcal{F}_g \approx 800\,B_g\,l_g\) with \(l_g\) in millimetres. At 1 T, one millimetre of air costs about 800 AT.
Fringing makes the flux bulge outward, enlarging the effective gap area, lowering gap \(B\) and lowering gap reluctance. Correct with \(A_g = (a+l_g)(b+l_g)\). Neglecting it is conservative.
Carter's coefficient converts a slotted gap into an equivalent smooth one: \(g_e = k_C g\), typically 1.05–1.30. With both surfaces slotted the coefficients multiply.
Leakage flux never reaches the gap. \(\Phi = \Phi_u + \Phi_l\) and \(\lambda = \Phi/\Phi_u > 1\), typically 1.15–1.25 for machines. Neglecting it is optimistic and dangerous.
Leakage does not affect the gap mmf — it loads the pole core and yoke, driving them closer to saturation.
A dominant gap linearises the circuit, because air's reluctance is constant while iron's is not. A 3:1 change in \(\mu_r\) may produce only a 1.3:1 change in total reluctance.
Energy divides in proportion to reluctance. The gap holds most of the reluctance, therefore most of the energy — and therefore is where force and torque are produced.
| Quantity | Formula | Notes |
|---|---|---|
| Gap reluctance | \(S_g = \dfrac{l_g}{\mu_0 A_g}\) | \(A_g\) is the effective area |
| Gap mmf | \(\mathcal{F}_g = \dfrac{B_g l_g}{\mu_0}\) | \(\approx 800\,B_g\,l_g(\mathrm{mm})\) |
| Fringing, rectangular face | \(A_g \approx (a+l_g)(b+l_g)\) | short gaps only |
| Fringing, circular face | \(A_g \approx \dfrac{\pi}{4}(d+l_g)^{2}\) | short gaps only |
| Carter's coefficient | \(k_C = \dfrac{\tau_s}{\tau_s - \gamma g}\) | exact form |
| Carter, approximate | \(k_C \approx \dfrac{\tau_s}{\tau_s - \dfrac{b_o^{2}}{5g + b_o}}\) | within about 1 % |
| Effective gap | \(g_e = k_C g\) | both sides slotted: \(k_{C1}k_{C2}g\) |
| Total flux | \(\Phi = \Phi_u + \Phi_l\) | useful plus leakage |
| Leakage coefficient | \(\lambda = \dfrac{\Phi}{\Phi_u}\) | always \(> 1\) |
| Stored energy | \(W = \tfrac{1}{2}S\Phi^{2} = \tfrac{1}{2}LI^{2}\) | joules |
| Energy share | \(\dfrac{W_g}{W_{\text{tot}}} = \dfrac{S_g}{S_g + S_i}\) | same split as reluctance |
| Energy ratio | \(\dfrac{W_g}{W_i} = \mu_r\dfrac{l_g}{l_i}\) | equal areas |
| Reactor turns | \(N = \dfrac{LI}{B_{\max}A}\) | from the saturation limit |
| Reactor gap | \(l_g = \left(\dfrac{N^{2}}{L} - S_i\right)\mu_0 A\) | gap sets the inductance |
Common Mistakes
Confusing leakage with fringing. Leakage never reaches the gap and reduces useful flux; fringing crosses the gap and reduces its reluctance. They act on different quantities and in opposite directions.
Applying the leakage coefficient to the gap. The gap carries only \(\Phi_u\). Use \(\lambda\Phi_u\) for the pole core and yoke, and \(\Phi_u\) for the gap and armature.
Dividing by the leakage coefficient instead of multiplying. The coil must produce more flux than reaches the gap, so \(\Phi = \lambda\Phi_u\) with \(\lambda > 1\). If your total flux comes out smaller than the useful flux, you have inverted it.
Using the actual gap instead of the effective gap in a slotted machine. Carter's coefficient is not optional; a 10 % error here is a 10 % error in magnetising current.
Forgetting that Carter's coefficient applies twice when both stator and rotor are slotted.
Applying the fringing rule to a long gap. The correction assumes \(l_g\) is small relative to the face dimensions. For a gap comparable with the face, it is meaningless.
Assuming the gap and the core run at the same flux density. With fringing they do not — the gap is always at the lower density, since the same flux crosses a larger area.
Believing a gap prevents saturation of the iron. It does not. The iron saturates at the same \(B\) as ever. What the gap does is require far more mmf to reach that \(B\), so a given winding is less likely to drive the core there.
Designing an inductor without a gap "to maximise inductance". You will maximise inductance at zero current and lose all of it at rated current. Energy storage needs the gap.
Treating fringing and leakage as interchangeable safety factors. One is conservative and one is optimistic. Lumping them together can silently cancel a real design shortfall.
Chapter Review
Use the rule of thumb \(\mathcal{F}_g \approx 800\,B_g\,l_g(\mathrm{mm})\) as a running check on every gap calculation.
P4.1 A pole face measures 50 mm × 80 mm with a 2.5 mm gap. Find the effective gap area and the percentage increase over the core area.
Show answer
\[A_{\text{core}} = (0.050)(0.080) = 4.000\times10^{-3}~\mathrm{m^{2}}\]An increase of 8.28 %. The gap reluctance is therefore 7.65 % lower than the uncorrected value.\[A_g = (0.0525)(0.0825) = 4.331\times10^{-3}~\mathrm{m^{2}}\]P4.2 A machine has a slot pitch of 25 mm, a slot opening of 5 mm and a gap of 1.2 mm. Find Carter's coefficient by the approximate formula and the effective gap.
Show answer
\[\frac{b_o^{2}}{5g+b_o} = \frac{25}{5(1.2)+5} = \frac{25}{11} = 2.273~\mathrm{mm}\]\[k_C = \frac{25}{25-2.273} = 1.100, \qquad g_e = (1.100)(1.2) = 1.320~\mathrm{mm}\]P4.3 Repeat P4.2 using the exact Carter expression, and compare.
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With \(b_o/2g = 5/2.4 = 2.083\):\[\gamma = \frac{4}{\pi}\left[(2.083)\arctan(2.083) - \ln\sqrt{1+2.083^{2}}\,\right] = 1.2732\left[2.340 - 0.838\right] = 1.913\]Against 1.100 from the approximation — a difference of 0.1 %, which is why the simple form is used.\[k_C = \frac{25}{25-(1.913)(1.2)} = \frac{25}{22.70} = 1.101\]P4.4 A machine produces a total flux of 7.2 mWb of which 6.0 mWb is useful. Find the leakage coefficient, the leakage flux, and the leakage as a percentage of the total.
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Leakage is \(1.2/7.2 = 16.7\,\%\) of the total. Note that 16.7 % of the total corresponds to a coefficient of 1.20, not 1.167 — the two percentages have different denominators, which is a common source of confusion.\[\lambda = \frac{7.2}{6.0} = 1.200, \qquad \Phi_l = 7.2 - 6.0 = 1.2~\mathrm{mWb}\]P4.5 A pole must deliver 5.0 mWb useful flux. The leakage coefficient is 1.22 and the pole core area is 50 cm². Find the flux density in the pole core.
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Without allowing for leakage the designer would have expected 1.000 T and a comfortable margin. At 1.22 T the pole is close to the knee for many steels.\[\Phi = (1.22)(5.0) = 6.10~\mathrm{mWb}, \qquad B_{\text{pole}} = \frac{6.10\times10^{-3}}{50\times10^{-4}} = 1.220~\mathrm{T}\]P4.6 A gap of 1.5 mm has a flux density of 0.90 T. Find the gap mmf using the rule of thumb and exactly, and comment.
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Rule of thumb: \(800(0.90)(1.5) = 1080\) AT.The shortcut is high by 0.5 %, because \(1/\mu_0 = 7.958\times10^{5}\) rather than exactly \(8\times10^{5}\). Ample for checking.\[\mathcal{F}_g = \frac{(0.90)(1.5\times10^{-3})}{4\pi\times10^{-7}} = \frac{1.35\times10^{-3}}{1.2566\times10^{-6}} = 1074.3~\mathrm{AT}\]P4.7 A ring of iron path 0.40 m and area 4 cm² has \(\mu_r\) falling from 2000 to 800. Find the ratio of total reluctances (a) without a gap and (b) with a 0.8 mm gap.
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(a) No gap: the ratio is simply \(2000/800 = 2.50\).
(b) With gap: iron path 0.3992 m, area \(4\times10^{-4}\) m².\[S_g = \frac{0.0008}{(4\pi\times10^{-7})(4\times10^{-4})} = 1.5915\times10^{6}\]\[S_i(2000) = 3.9721\times10^{5}, \qquad S_i(800) = 9.9302\times10^{5}\]A 2.5:1 swing reduced to 1.30:1.\[\text{ratio} = \frac{9.9302\times10^{5}+1.5915\times10^{6}}{3.9721\times10^{5}+1.5915\times10^{6}} = \frac{2.5845\times10^{6}}{1.9887\times10^{6}} = 1.300\]P4.8 Find the energy stored in an air gap 2 mm long and 8 cm² in area at a flux density of 0.90 T.
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\[W = \frac{B^{2}}{2\mu_0}\,A l_g = \frac{(0.90)^{2}}{2(4\pi\times10^{-7})}\left(8\times10^{-4}\right)\left(2\times10^{-3}\right)\]\[= \left(3.2228\times10^{5}\right)\left(1.6\times10^{-6}\right) = 0.5157~\mathrm{J}\]P4.9 Design a 8 mH reactor to carry 12 A peak at no more than 1.1 T. The core has an area of 8 cm², a mean iron path of 0.30 m and \(\mu_r = 3000\). Find the turns and the gap.
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\[N = \frac{LI}{B_{\max}A} = \frac{(0.008)(12)}{(1.1)(8\times10^{-4})} = \frac{0.096}{8.8\times10^{-4}} = 109.1 \ \to\ N = 109\]\[S = \frac{N^{2}}{L} = \frac{11\,881}{0.008} = 1.4851\times10^{6}~\mathrm{AT/Wb}\]\[S_i = \frac{0.30}{(4\pi\times10^{-7})(3000)(8\times10^{-4})} = 9.947\times10^{4}\]\[S_g = 1.4851\times10^{6} - 9.947\times10^{4} = 1.3857\times10^{6}\]The gap supplies 93 % of the reluctance — and therefore stores 93 % of the energy.\[l_g = S_g\mu_0 A = (1.3857\times10^{6})(1.2566\times10^{-6})(8\times10^{-4}) = 1.393~\mathrm{mm}\]P4.10 A gap between faces of 40 mm × 60 mm is doubled from 1 mm to 2 mm. Find the gap reluctance in each case with fringing included, and show that it does not quite double.
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Core area \(= 2.400\times10^{-3}\) m².\[A_g(1\,\mathrm{mm}) = (0.041)(0.061) = 2.501\times10^{-3}, \qquad S_g = \frac{0.001}{(4\pi\times10^{-7})(2.501\times10^{-3})} = 3.182\times10^{5}\]\[A_g(2\,\mathrm{mm}) = (0.042)(0.062) = 2.604\times10^{-3}, \qquad S_g = \frac{0.002}{(4\pi\times10^{-7})(2.604\times10^{-3})} = 6.112\times10^{5}\]The shortfall is because fringing grows with gap length: the longer gap gains 8.5 % of area against the shorter one's 4.2 %. Doubling a gap always gives slightly less than double the reluctance.\[\frac{6.112}{3.182} = 1.921 \quad \text{rather than } 2.000\]
MCQ 1. Fringing at an air gap causes the effective area to:
(a) decrease (b) increase (c) stay the same (d) become zeroShow answer
(b) increase. Lines in the same direction repel and spread once iron no longer confines them.MCQ 2. The fringing effect is:
(a) independent of gap length (b) greater for a longer gap (c) greater for a shorter gap (d) present only in slotted machinesShow answer
(b) greater for a longer gap. The correction adds one gap length to each dimension, so it scales directly with \(l_g\).MCQ 3. Leakage flux is flux that:
(a) crosses the gap but spreads (b) never reaches the intended path (c) flows in the copper (d) reverses directionShow answer
(b) never reaches the intended path, closing instead through the air around the coil.MCQ 4. The leakage coefficient of a practical machine is typically:
(a) 0.8 (b) 1.0 (c) 1.2 (d) 5.0Show answer
(c) about 1.2. It must exceed 1, and 1.15–1.25 is the usual range for a machine with an appreciable gap.MCQ 5. Carter's coefficient accounts for:
(a) leakage (b) hysteresis (c) the effect of slotting on the gap (d) eddy currentsShow answer
(c) the effect of slotting. It converts a slotted gap into the equivalent smooth gap, \(g_e = k_C g\).MCQ 6. Approximately how many ampere-turns does 1 mm of air gap absorb at 1 T?
(a) 80 (b) 800 (c) 8000 (d) 80 000Show answer
(b) about 800 AT, since \(1/\mu_0 \approx 8\times10^{5}\) AT/m per tesla.MCQ 7. An air gap is introduced into an inductor's core mainly to:
(a) increase inductance (b) allow more current before saturation (c) reduce copper loss (d) reduce costShow answer
(b) allow more current before saturation — and to store the energy, which iron does badly.MCQ 8. If a gap holds 80 % of a circuit's reluctance, the fraction of stored energy it holds is:
(a) 20 % (b) 50 % (c) 80 % (d) 100 %Show answer
(c) 80 %. Since \(W = \tfrac{1}{2}S\Phi^2\) and the flux is common, energy divides exactly as reluctance does.MCQ 9. Neglecting leakage in a design calculation makes the predicted useful flux:
(a) too high (b) too low (c) unchanged (d) negativeShow answer
(a) too high — the optimistic and therefore dangerous direction. Neglecting fringing errs the other, safer way.MCQ 10. Adding an air gap to a saturating core:
(a) raises the \(B\) at which the iron saturates (b) leaves that \(B\) unchanged but requires much more mmf to reach it (c) eliminates saturation entirely (d) lowers the saturation \(B\)Show answer
(b). Saturation is a property of the material and is untouched by the gap. The gap simply makes it far harder for a given winding to drive the iron there.
Fringing and leakage both reduce the flux density in the gap, yet only one of them reduces the useful flux. Explain the distinction and why it matters to a designer.
Explain why neglecting fringing is a conservative error while neglecting leakage is an optimistic one, and what that implies about which correction deserves more care.
A colleague argues that since the gap absorbs 80 % of the ampere-turns, machines should be built with the smallest gap physically possible and reactors with no gap at all. Evaluate both halves of that claim.
Explain, without algebra, why a gapped circuit's flux is far less sensitive to the iron's permeability than an ungapped one's.
Leakage flux does no useful work, yet it produces leakage reactance, which is one of the most important machine parameters. Reconcile these two statements.
Why does Carter's coefficient apply twice when both stator and rotor are slotted, and what does the product represent physically?
Energy divides between iron and gap exactly as reluctance does. Use this to explain why the force on a relay armature is greatest just before the armature closes.
Two of the three debts from Chapter 2 are now settled. Fringing and leakage have been given quantitative corrections, and the air gap has been promoted from nuisance to design variable. The third debt — the fiction of a constant \(\mu_r\) — is the subject of Chapter 5.
There the measured \(B\)–\(H\) curve replaces the single quoted permeability, and the graphical construction promised in Section 3-7 finally becomes available: plot the core's characteristic, superimpose the air-gap line of Section 4-7, and read the operating point off the intersection. Chapters 6 and 7 then show what all this flux costs in heat, through hysteresis and eddy currents — and Chapter 9 applies the same intersection construction to permanent magnets, where the gap sets the operating point of the magnet itself.