Electrical Machines · Chapter 9

Permanent Magnets and Their Circuits

Part 1 · Principles of Energy Conversion — a magnet is a magnetic circuit that supplies its own mmf and never gets it back. Everything about designing with one comes down to where its load line crosses its demagnetisation curve.

Prof. Mithun Mondal Engineering Devotion Digital Textbook
i Learning Objectives

By the end of this chapter you should be able to:

  • Interpret the demagnetisation curve — the second quadrant of the hysteresis loop — and distinguish normal from intrinsic coercivity.

  • Derive the load line of a magnet circuit from Ampère's law and flux continuity, and compute the permeance coefficient.

  • Locate the operating point graphically and, for a linear magnet, algebraically using \(B_m/B_r = P_c/(P_c + \mu_{\text{rec}})\).

  • Show that the gap energy is \(\tfrac{1}{2}|B_mH_m|V_m\), and hence that the energy product \((BH)_{\max}\) sets the minimum magnet volume.

  • Design magnet dimensions to deliver a specified gap flux density with the least material.

  • Compare NdFeB, SmCo, Alnico and ferrite on remanence, energy product, temperature capability and cost.

  • Explain recoil, irreversible loss and the purpose of stabilisation and of a keeper.

  • Apply reversible temperature coefficients, and explain why ferrite is at risk when cold while NdFeB is at risk when hot.

Section 9-1

Introduction

Every magnetic circuit so far has needed a coil. Ampere-turns were supplied by a current, and when the current stopped, the flux stopped with it. A permanent magnet supplies its own magnetomotive force indefinitely, with no supply, no copper loss and no excitation winding at all.

That is an enormous practical advantage, and it comes with an equally distinctive analytical problem. Because there is no external mmf, the magnet must drive flux through the rest of the circuit and through itself, which means it operates with \(H\) opposing its own \(B\). In other words, a working permanent magnet always sits in the second quadrant of its hysteresis loop, partially demagnetising itself. How far into that quadrant it sits is determined entirely by the shape of the circuit around it — and that is the whole subject of this chapter.

The construction used to find the answer will be familiar. In Chapter 5 the air-gap line was superimposed on the B–H curve, and the intersection gave the operating point. Here a load line is superimposed on the demagnetisation curve, and the intersection does the same job. It is the same idea applied to a magnet instead of a coil, and it is worth noticing the parallel, because it is the last appearance of a construction that has now served three chapters.

Section 9-2

The Demagnetisation Curve

Chapter 5 traced the full hysteresis loop and identified two points on it: the remanence \(B_r\), where the flux density survives at zero field, and the coercivity \(H_c\), where a reverse field drives the flux density to zero. The arc joining them — running from \((0, B_r)\) to \((-H_c, 0)\) — is the demagnetisation curve, and for a permanent magnet it is the only part of the loop that matters.

Remanence \(B_r\)

The flux density remaining when the magnetising field is removed and the magnetic circuit is closed — that is, with no air gap at all. It is the largest flux density the magnet can ever provide, and no practical circuit achieves it.

Coercivity \(H_c\)

The reverse field that reduces the flux density in the magnet to zero. It measures resistance to demagnetisation and is the property that separates a magnet from a core material.

! Normal versus Intrinsic Coercivity

Two coercivities are quoted for modern magnets and they are not the same number.

  • \(H_{cB}\), the normal coercivity, is where the flux density \(B\) reaches zero. It is what the demagnetisation curve of this chapter shows.

  • \(H_{cJ}\), the intrinsic coercivity, is where the magnetisation of the material itself reaches zero — the point of genuine, permanent destruction.

Always \(H_{cJ} > H_{cB}\), and for NdFeB it may be two or three times greater. At \(H_{cB}\) the magnet has been temporarily cancelled but not damaged; at \(H_{cJ}\) it has been wiped. Grade suffixes such as M, H, SH and UH on NdFeB denote increasing intrinsic coercivity, and hence increasing tolerance of heat and of demagnetising fields.

Linear and non-linear magnets. For NdFeB, SmCo and ferrite the demagnetisation curve is very nearly a straight line over its whole useful length:

\[B_m = B_r + \mu_0\mu_{\text{rec}} H_m \qquad (H_m < 0)\]

where \(\mu_{\text{rec}}\) is the recoil permeability, typically 1.02 to 1.10 — a magnet is barely more permeable than air. Setting \(B_m = 0\) gives the useful relation

\[H_{cB} = \frac{B_r}{\mu_0\mu_{\text{rec}}}\]

Alnico is the exception: its curve bends sharply, with a pronounced knee at modest reverse field. That single geometric fact — not any deficiency of remanence, which is as high as NdFeB's — is why Alnico demagnetises so easily and why it was displaced.

Section 9-3

The Load Line and Operating Point

Consider the simplest magnet circuit: a magnet of length \(l_m\) and area \(A_m\), a soft-iron return path of negligible reluctance, and an air gap of length \(l_g\) and area \(A_g\). There is no winding anywhere.

Step 1 — Ampère's law. With no enclosed current, the line integral of \(H\) around the closed path is zero:

\[\oint \mathbf{H}\cdot\mathrm{d}\mathbf{l} = 0 \quad\Longrightarrow\quad H_m l_m + H_g l_g = 0 \quad\Longrightarrow\quad H_m l_m = -H_g l_g\]

Since \(H_g\) is positive, \(H_m\) must be negative. This is the algebraic statement of the fact announced in Section 9-1: a working magnet is always partly demagnetising itself.

Step 2 — flux continuity. Neglecting leakage,

\[B_m A_m = B_g A_g\]

Step 3 — combine. Substituting \(H_g = B_g/\mu_0 = B_m A_m/(\mu_0 A_g)\) into the first equation:

\[H_m l_m = -\frac{B_m A_m l_g}{\mu_0 A_g} \quad\Longrightarrow\quad B_m = -\mu_0\left(\frac{A_g}{A_m}\right)\left(\frac{l_m}{l_g}\right)H_m\]
📐
The Load Line
A straight line through the origin
\[B_m = -\mu_0 P_c H_m, \qquad P_c = \frac{A_g\, l_m}{A_m\, l_g}\]

\(P_c\) is the permeance coefficient (also called the load-line slope or operating slope). It is dimensionless and depends only on geometry — never on the magnet material. A long magnet facing a short gap gives a large \(P_c\) and a steep line; a short magnet facing a long gap gives a shallow one.

The operating point is where the load line meets the demagnetisation curve, because only there are both the material law and the circuit law satisfied. This is precisely the construction of Section 5-8, with two differences: the line now passes through the origin (there is no external mmf to displace it), and it lies in the second quadrant.

H (kA/m) — negative in the magnet B (T) −300−200−1000 0.10.20.30.4 demagnetisation curve B = B_r + μ₀μ_rec H B_r = 0.40 T −H_cB load line B = −μ₀ P_c H operating point P_c = A_g l_m / A_m l_g Long magnet, short gap → large P_c → steep line → operates near B_r Short magnet, long gap → small P_c → shallow line → operates near H_c Geometry alone sets the line; the material alone sets the curve.
The load-line construction. The intersection satisfies both the material law and the circuit law.

For a linear demagnetisation curve the intersection can be found algebraically rather than graphically. Equating the two expressions for \(B_m\):

\[-\mu_0 P_c H_m = B_r + \mu_0\mu_{\text{rec}}H_m \quad\Longrightarrow\quad H_m = \frac{-B_r}{\mu_0\left(P_c + \mu_{\text{rec}}\right)}\]
🎯
Operating Point of a Linear Magnet
A single line of algebra replaces the graph
\[\frac{B_m}{B_r} = \frac{P_c}{P_c + \mu_{\text{rec}}}\]

Since \(\mu_{\text{rec}} \approx 1\), the fraction of the remanence actually delivered is roughly \(P_c/(P_c+1)\). A permeance coefficient of 1 gives half of \(B_r\); a coefficient of 9 gives ninety percent of it. The magnet's own geometry decides how much of its potential it realises.

1 Worked Example 9.1 — Finding the Operating Point

Problem. A ferrite magnet has \(B_r = 0.40\) T and \(\mu_{\text{rec}} = 1.1\). It is 20 mm long with a cross-section of 800 mm², and drives flux across a 3 mm air gap of the same area through a soft-iron return path of negligible reluctance. Find the permeance coefficient, the operating point, and the gap flux density.

Permeance coefficient.

\[P_c = \frac{A_g l_m}{A_m l_g} = \frac{(800)(20)}{(800)(3)} = \frac{20}{3} = 6.667\]

Operating flux density.

\[\frac{B_m}{B_r} = \frac{P_c}{P_c + \mu_{\text{rec}}} = \frac{6.667}{6.667 + 1.1} = \frac{6.667}{7.767} = 0.8584\]
\[B_m = (0.8584)(0.40) = 0.3434~\mathrm{T}\]

Operating field. From the load line,

\[H_m = -\frac{B_m}{\mu_0 P_c} = -\frac{0.3434}{\left(4\pi\times10^{-7}\right)(6.667)} = -40\,984~\mathrm{A/m}\]

Check against the demagnetisation curve: \(B_r + \mu_0\mu_{\text{rec}}H_m = 0.40 + \left(1.382\times10^{-6}\right)(-40\,984) = 0.40 - 0.0566 = 0.3434\) T \(\checkmark\)

Gap flux density. The areas are equal, so \(B_g = B_m = 0.343\) T.

Comment. The magnet delivers 86 % of its remanence, which sounds excellent. Section 9-4 will show that it is in fact rather wasteful: the energy product at this point is

\[\left|B_mH_m\right| = (0.3434)(40\,984) = 14\,072~\mathrm{J/m^{3}}\]

against a maximum available of about 28 900 J/m³. Operating close to \(B_r\) is not the same as operating efficiently — a distinction that catches out most people the first time they meet it.

Section 9-4

The Energy Product

The figure of merit for a permanent magnet is not its remanence, nor its coercivity, but the product of the two quantities at its operating point. The reason emerges from a short energy calculation.

The energy stored in the air gap is its volume times the energy density of Chapter 1:

\[W_g = \frac{B_g^{2}}{2\mu_0}\left(A_g l_g\right) = \frac{1}{2}\left(B_g A_g\right)\left(H_g l_g\right)\]

Now substitute the two circuit relations of Section 9-3, \(B_g A_g = B_m A_m\) and \(H_g l_g = -H_m l_m\):

\[W_g = \frac{1}{2}\left(B_m A_m\right)\left(-H_m l_m\right) = \frac{1}{2}\left|B_m H_m\right| A_m l_m\]
Key Result
Gap energy per unit magnet volume is half the energy product
\[W_g = \tfrac{1}{2}\left|B_m H_m\right| V_m \quad\Longrightarrow\quad V_m = \frac{2 W_g}{\left|B_m H_m\right|}\]

To deliver a given energy to the gap, the magnet volume required is inversely proportional to the energy product at which it works. Operating at \((BH)_{\max}\) therefore uses the least material — which, for magnets costing what NdFeB costs, is the whole of the design problem.

For a linear demagnetisation curve the maximum is easy to locate. Writing \(B = B_r + \mu_0\mu_{\text{rec}}H\) and maximising \(|BH|\) gives

\[\frac{\mathrm{d}}{\mathrm{d}H}\left(B_rH + \mu_0\mu_{\text{rec}}H^{2}\right) = 0 \quad\Longrightarrow\quad H = -\frac{B_r}{2\mu_0\mu_{\text{rec}}}, \qquad B = \frac{B_r}{2}\]
\[\boxed{(BH)_{\max} = \frac{B_r^{2}}{4\mu_0\mu_{\text{rec}}} \qquad\text{at}\qquad B_m = \frac{B_r}{2}}\]

Combining this with the operating-point relation of Section 9-3 gives a strikingly simple design rule. Setting \(B_m/B_r = 1/2\) in \(B_m/B_r = P_c/(P_c+\mu_{\text{rec}})\) requires

\[P_c = \mu_{\text{rec}} \approx 1\]
demagnetisation curve −H B B_r B = B_r / 2 B·H (kJ/m³) (BH)max The energy product peaks exactly halfway up the demagnetisation curve. (BH)max = B_r² / 4μ₀μ_rec , reached when the permeance coefficient P_c = μ_rec ≈ 1
Maximum energy product occurs at half the remanence — not near it.
The counter-intuitive conclusion. The most material-efficient operating point delivers only half the remanence. A designer who lengthens a magnet to push its operating point up towards \(B_r\) gets a higher gap flux density, but pays for it with disproportionately more magnet. Whether that is worth doing depends on whether the constraint is flux density or cost — and with rare-earth material it is almost always cost.

A note on units. Energy product is quoted in kJ/m³ in SI and in megagauss-oersteds (MGOe) in the older literature, with \(1~\mathrm{MGOe} = 7.958~\mathrm{kJ/m^{3}}\). The familiar NdFeB grade designations — N35, N42, N52 — are simply the energy product in MGOe.

2 Worked Example 9.2 — How Much Magnet Is Being Wasted?

Problem. For the ferrite circuit of Example 9.1, find the energy stored in the air gap, the maximum energy product of the material, and the minimum magnet volume that could in principle deliver the same gap energy.

Gap energy. With \(B_g = 0.3434\) T and \(V_g = (800~\mathrm{mm^{2}})(3~\mathrm{mm}) = 2.4\times10^{-6}~\mathrm{m^{3}}\):

\[W_g = \frac{B_g^{2}}{2\mu_0}V_g = \frac{(0.3434)^{2}}{2\left(4\pi\times10^{-7}\right)}\left(2.4\times10^{-6}\right) = (46\,906)\left(2.4\times10^{-6}\right) = 0.1126~\mathrm{J}\]

Cross-check with the magnet-side formula, using \(V_m = (800)(20) = 1.6\times10^{-5}~\mathrm{m^{3}}\):

\[W_g = \tfrac{1}{2}\left|B_mH_m\right|V_m = \tfrac{1}{2}(14\,072)\left(1.6\times10^{-5}\right) = 0.1126~\mathrm{J} \;\checkmark\]

Maximum energy product.

\[(BH)_{\max} = \frac{B_r^{2}}{4\mu_0\mu_{\text{rec}}} = \frac{(0.40)^{2}}{4\left(4\pi\times10^{-7}\right)(1.1)} = \frac{0.16}{5.529\times10^{-6}} = 28\,940~\mathrm{J/m^{3}}\]

Minimum magnet volume.

\[V_{m,\min} = \frac{2W_g}{(BH)_{\max}} = \frac{2(0.1126)}{28\,940} = 7.78\times10^{-6}~\mathrm{m^{3}} = 7781~\mathrm{mm^{3}}\]

Comment. The magnet actually fitted has a volume of 16 000 mm³ — 2.06 times the minimum. That ratio is exactly the inverse of the energy-product ratio computed in Example 9.1 (\(28\,940/14\,072 = 2.06\)), as it must be.

The same gap energy could be obtained from less than half the ferrite by making the magnet shorter and wider — reducing \(P_c\) from 6.67 towards 1.1, which moves the operating point down the curve to \(B_r/2\). The gap flux density would fall, so the gap would have to be reshaped too; the next section shows how to do both at once.

Section 9-5

Designing a Magnet Circuit

The design problem is usually stated the other way round: a gap of given dimensions must carry a given flux density — what magnet is needed? Working at \((BH)_{\max}\), where \(B_m = B_r/2\) and \(H_m = -B_r/2\mu_0\mu_{\text{rec}}\), both dimensions follow directly.

Area — from flux continuity, \(B_mA_m = B_gA_g\):

\[A_m = \frac{B_g A_g}{B_m} = \frac{2 B_g A_g}{B_r}\]

Length — from the mmf balance, \(|H_m| l_m = H_g l_g = B_g l_g/\mu_0\):

\[l_m = \frac{B_g l_g}{\mu_0 \left|H_m\right|} = \frac{B_g l_g}{\mu_0}\cdot\frac{2\mu_0\mu_{\text{rec}}}{B_r} = \frac{2\mu_{\text{rec}} B_g l_g}{B_r}\]
📏
Minimum-Volume Design
Two formulas size the magnet completely
\[A_m = \frac{2 B_g A_g}{B_r}, \qquad l_m = \frac{2\mu_{\text{rec}} B_g l_g}{B_r}\]

Both scale with the required gap flux density and inversely with the remanence — a better magnet is smaller in both dimensions. Multiplying them confirms the volume result of Section 9-4, and dividing them confirms that \(P_c = \mu_{\text{rec}}\) as required.

Two practical corrections. Real designs never use these figures unmodified:

  • Leakage. As Chapter 4 established, some flux never reaches the gap. Multiply \(A_m\) by a leakage factor of 1.1 to 1.5 — considerably worse than for a wound circuit, because a magnet's own permeability is close to that of air and gives leakage flux little to distinguish between.

  • Reluctance drop in the iron. The return path was assumed lossless. Multiply \(l_m\) by a reluctance factor of 1.05 to 1.3 to supply the mmf the steel actually consumes.

3 Worked Example 9.3 — Sizing a Magnet

Problem. A gap of area 600 mm² and length 2 mm must carry a flux density of 0.55 T. The available magnet material is NdFeB with \(B_r = 1.25\) T and \(\mu_{\text{rec}} = 1.05\). Find the minimum-volume magnet dimensions, and verify that the design does operate at \((BH)_{\max}\).

Magnet area.

\[A_m = \frac{2B_gA_g}{B_r} = \frac{2(0.55)(600)}{1.25} = \frac{660}{1.25} = 528~\mathrm{mm^{2}}\]

Magnet length.

\[l_m = \frac{2\mu_{\text{rec}}B_g l_g}{B_r} = \frac{2(1.05)(0.55)(2)}{1.25} = \frac{2.31}{1.25} = 1.848~\mathrm{mm}\]
\[V_m = A_m l_m = (528)(1.848) = 976~\mathrm{mm^{3}}\]

Verification 1 — the permeance coefficient.

\[P_c = \frac{A_g l_m}{A_m l_g} = \frac{(600)(1.848)}{(528)(2)} = \frac{1108.8}{1056} = 1.050 = \mu_{\text{rec}} \;\checkmark\]

Verification 2 — the energy balance.

\[W_g = \frac{B_g^{2}}{2\mu_0}\left(A_gl_g\right) = \frac{(0.55)^{2}}{2\left(4\pi\times10^{-7}\right)}\left(1.2\times10^{-6}\right) = 0.1444~\mathrm{J}\]
\[(BH)_{\max} = \frac{(1.25)^{2}}{4\left(4\pi\times10^{-7}\right)(1.05)} = 296\,000~\mathrm{J/m^{3}}\]
\[V_{m,\min} = \frac{2(0.1444)}{296\,000} = 9.76\times10^{-7}~\mathrm{m^{3}} = 976~\mathrm{mm^{3}} \;\checkmark\]

Comment. Under a cubic centimetre of NdFeB produces 0.55 T across a 2 mm gap of 6 cm² — a striking demonstration of what rare-earth material can do. Note also the aspect ratio: the magnet is broad and thin, 528 mm² in area but under 2 mm thick. Minimum-volume magnets are almost always disc-like rather than rod-like, which is exactly how magnets in loudspeakers and BLDC rotors are shaped.

Applying practical corrections — say a leakage factor of 1.2 on the area and a reluctance factor of 1.1 on the length — would give roughly 634 mm² by 2.03 mm, or about 1290 mm³.

Section 9-6

Permanent Magnet Materials

Four families dominate, and they differ by more than an order of magnitude in energy product, price and temperature capability.

Table 9.1 — Permanent magnet materials compared. Values are typical; consult manufacturer's data for design.
Material\(B_r\) (T)\(H_{cB}\) (kA/m)\((BH)_{\max}\) (kJ/m³)Max temp (°C)Relative cost
NdFeB (sintered)1.20–1.45900–1000280–40080–200High
SmCo0.95–1.10700–800160–240300–350Very high
Alnico 51.20–1.3050–6040–50500+Moderate
Ferrite (ceramic)0.35–0.43200–29026–34250Very low
−H (kA/m) B (T) 1000800600 4002000 0.51.0 NdFeB SmCo ferrite Alnico Alnico reaches almost the same B_r as NdFeB — but collapses under the slightest reverse field. Its knee is what disqualified it, not its remanence.
Same axes, four materials. Area under the curve, not height, is what counts.
Why Alnico lost. Alnico's remanence matches NdFeB's, so on the vertical axis it looks competitive. But its coercivity is roughly one-twentieth, and its curve knees over sharply. A load line of even modest slope intersects it below the knee, where recoil (Section 9-7) causes permanent loss. Alnico must be used as a long, thin rod to keep \(P_c\) high — which is why old magnets are horseshoe-shaped or long cylinders, while modern ones are thin discs. The shape of a magnet tells you what it is made of.
4 Worked Example 9.4 — Choosing a Material

Problem. A BLDC rotor requires a certain gap energy. Using the representative energy products \((BH)_{\max}\) of 320, 200, 44 and 29 kJ/m³ for NdFeB, SmCo, Alnico and ferrite, compare the magnet volume each would need, and comment on the choice.

Solution. From Section 9-4, \(V_m = 2W_g/(BH)_{\max}\), so the volume is inversely proportional to the energy product:

Table 9.2 — Relative magnet volume for the same gap energy.
Material\((BH)_{\max}\) (kJ/m³)Relative volumePractical verdict
NdFeB3201.00 ×Smallest and lightest; temperature-limited and costly
SmCo2001.60 ×Chosen when 200 °C or more is required
Alnico447.27 ×Bulky and easily demagnetised; now rare in motors
Ferrite2911.0 ×Eleven times the volume — but a small fraction of the price

Comment. The volume ratios are decisive for anything that must rotate: eleven times the magnet volume means a far larger rotor, more inertia, and a bigger machine to house it. That is why traction motors for electric vehicles use NdFeB almost universally despite its cost and its supply-chain risk.

But the ranking is not a ranking of merit. Ferrite dominates by unit volume shipped, because in a fan motor, a loudspeaker or a door catch the extra volume is free and the price difference is not. Selection is about the binding constraint — volume, temperature, or cost — and only in the first case does NdFeB win automatically.

Section 9-7

Recoil and Stabilisation

So far the operating point has been treated as fixed. In a real machine it moves — when the rotor turns, when armature reaction opposes the magnet, when the magnet is removed from its circuit for assembly. What happens then depends on how far it moves.

Above the knee, excursions are reversible: the operating point slides up and down the demagnetisation curve and returns to where it started.

Below the knee, they are not. If a demagnetising field drives the magnet past the knee and is then removed, the magnet does not return along the original curve. It follows a recoil line of slope \(\mu_0\mu_{\text{rec}}\) — parallel to the straight part of the original curve — arriving at a new, lower effective remanence. The magnet has been permanently weakened.

Design Rule
Keep the operating point above the knee under all conditions

"All conditions" includes the worst case: maximum armature reaction, at the highest expected temperature, during a short-circuit fault. Since the knee rises with temperature for NdFeB (Section 9-8), the critical condition is usually a fault at maximum operating temperature — not at rated load, and not when cold.

Stabilisation turns this liability into a controlled quantity. The magnet is deliberately exposed once to a demagnetising field slightly worse than anything it will meet in service. It loses a few percent of its flux, and thereafter operates reversibly along its recoil line. A stabilised magnet is slightly weaker but entirely predictable — a trade every instrument designer accepts.

Why magnets used to come with keepers. Removing a magnet from its circuit leaves it with only air as a return path, so \(P_c\) collapses and the operating point slides far down towards \(H_c\). For Alnico, whose knee is high, this alone could ruin the magnet — hence the soft-iron keeper bridging the poles of every horseshoe magnet in an old laboratory. Modern NdFeB and ferrite have knees below the origin at room temperature and need no keeper, which is why keepers have quietly disappeared.
Section 9-8

Temperature Effects

Every magnetic property is temperature-dependent, and for permanent magnets the dependence is strong enough to govern the design. Two coefficients are quoted, and they behave differently.

Table 9.3 — Reversible temperature coefficients and Curie temperatures.
Material\(B_r\) coefficient (%/°C)\(H_{cJ}\) coefficient (%/°C)Curie temp (°C)
NdFeB−0.11 to −0.12−0.55 to −0.65310–400
SmCo−0.03 to −0.04−0.20 to −0.30700–800
Alnico−0.02−0.02860
Ferrite−0.20+0.30 to +0.40450
! Ferrite Demagnetises When Cold, Not When Hot

Read the ferrite row carefully. Its coercivity coefficient is positive: heating a ferrite magnet makes it harder to demagnetise, and cooling it makes it easier. The knee moves up as the temperature falls.

The consequence is that a ferrite-magnet motor is at greatest risk of irreversible demagnetisation on a cold start — a winter morning, or an aircraft at altitude — precisely when a designer's instinct says the magnet is safest. NdFeB behaves the opposite way, with its risk at maximum temperature. The worst case must be checked at both ends of the temperature range, and which end is critical depends on the material.

5 Worked Example 9.5 — Temperature Derating

Problem. An NdFeB magnet has \(B_r = 1.25\) T and \(H_{cJ} = 1200\) kA/m at 20 °C, with reversible coefficients of \(-0.11\) %/°C for \(B_r\) and \(-0.60\) %/°C for \(H_{cJ}\). Find both at 120 °C, and compare with a ferrite magnet of \(B_r = 0.40\) T at the same temperature. Comment on the design implications.

NdFeB at 120 °C. The rise is \(\Delta T = 100\) °C:

\[B_r' = (1.25)\left[1 - (0.0011)(100)\right] = (1.25)(0.89) = 1.113~\mathrm{T}\]
\[H_{cJ}' = (1200)\left[1 - (0.0060)(100)\right] = (1200)(0.40) = 480~\mathrm{kA/m}\]

The remanence has fallen 11 %, which is manageable. The intrinsic coercivity has fallen 60 %, which is not — the knee has risen dramatically, and an operating point that was comfortably safe at room temperature may now sit below it.

Ferrite at 120 °C.

\[B_r' = (0.40)\left[1 - (0.0020)(100)\right] = (0.40)(0.80) = 0.320~\mathrm{T}\]

A larger proportional loss of remanence — 20 % against 11 % — but its coercivity has improved, so it is in no danger of permanent damage.

Comment. The two failure modes are quite different. NdFeB loses little flux but becomes vulnerable to permanent demagnetisation; ferrite loses more flux but stays safe. For NdFeB the temperature limit is set by coercivity, not by remanence — which is exactly why the grade suffixes (M, H, SH, UH, EH) denote coercivity and why a high-temperature grade costs more for no gain in strength.

Check the cold end too. At \(-20\) °C the same ferrite magnet's coercivity would fall by about \((0.35)(40) = 14\,\%\), moving its knee upward and creating the cold-start risk described above.

Section 9-9

Applications

PMSM and BLDC Rotors

Surface-mounted or interior NdFeB magnets replace the field winding entirely, eliminating rotor copper loss and slip rings. This is why permanent-magnet machines dominate electric traction. Chapters 89 and 90 treat them fully.

Loudspeakers

A magnet, pole piece and top plate form a circuit with a narrow annular gap holding the voice coil. The design goal — high uniform \(B_g\) in the smallest magnet — is exactly the minimum-volume problem of Section 9-5.

Magnetos and Small Generators

A permanent-magnet field needs no excitation supply, so a magneto generates from the first revolution with no battery. This self-sufficiency is why they persist in small engines and aircraft ignition.

Moving-Coil Instruments

The PMMC movement of Chapter 7's damping discussion relies on a stabilised magnet, so that its field — and hence the instrument's calibration — does not drift after an overload.

Magnetic Couplings and Bearings

Torque transmitted through a sealed wall, with no shaft penetration. Here the load line is deliberately shallow and varies as the coupling slips, so demagnetisation margin must be checked at the worst relative position.

Holding and Separation

Door catches, chucks, lifting magnets and the eddy-current separators of Chapter 7. Ferrite dominates these by volume, since the constraint is price rather than size.

Section 9-10

Summary and Key Formulas

  • A permanent magnet supplies its own mmf and therefore always operates in the second quadrant, with \(H_m\) negative — partly demagnetising itself.

  • The demagnetisation curve runs from \(B_r\) to \(-H_{cB}\). For NdFeB, SmCo and ferrite it is essentially straight, with slope \(\mu_0\mu_{\text{rec}}\) and \(\mu_{\text{rec}} \approx 1.05\).

  • Intrinsic coercivity \(H_{cJ}\) exceeds normal coercivity \(H_{cB}\) and is what governs permanent damage and temperature capability.

  • The load line \(B_m = -\mu_0 P_c H_m\) passes through the origin, with the permeance coefficient \(P_c = A_gl_m/A_ml_g\) depending on geometry alone.

  • For a linear magnet, \(B_m/B_r = P_c/(P_c+\mu_{\text{rec}})\) — no graph is needed.

  • Gap energy is \(\tfrac{1}{2}|B_mH_m|V_m\), so minimum magnet volume requires operating at \((BH)_{\max}\), which for a linear magnet occurs at \(B_m = B_r/2\) and requires \(P_c = \mu_{\text{rec}}\).

  • Design formulas: \(A_m = 2B_gA_g/B_r\) and \(l_m = 2\mu_{\text{rec}}B_gl_g/B_r\), then corrected for leakage and iron reluctance.

  • Driving the operating point below the knee causes irreversible loss along a recoil line. Stabilisation makes subsequent behaviour reversible.

  • NdFeB is at risk when hot; ferrite is at risk when cold, because its coercivity coefficient is positive.

Table 9.4 — Formulas introduced in this chapter.
QuantityFormulaNotes
Demagnetisation curve\(B_m = B_r + \mu_0\mu_{\text{rec}}H_m\)linear magnets; \(H_m < 0\)
Normal coercivity\(H_{cB} = \dfrac{B_r}{\mu_0\mu_{\text{rec}}}\)where \(B\) reaches zero
Ampère's law, no current\(H_ml_m + H_gl_g = 0\)forces \(H_m\) negative
Flux continuity\(B_mA_m = B_gA_g\)neglecting leakage
Load line\(B_m = -\mu_0 P_c H_m\)through the origin
Permeance coefficient\(P_c = \dfrac{A_g l_m}{A_m l_g}\)geometry only
Operating point\(\dfrac{B_m}{B_r} = \dfrac{P_c}{P_c + \mu_{\text{rec}}}\)linear magnets
Gap energy\(W_g = \tfrac{1}{2}\left|B_mH_m\right|V_m\)joules
Magnet volume\(V_m = \dfrac{2W_g}{\left|B_mH_m\right|}\)minimised at \((BH)_{\max}\)
Maximum energy product\((BH)_{\max} = \dfrac{B_r^{2}}{4\mu_0\mu_{\text{rec}}}\)at \(B_m = B_r/2\)
Optimum geometry\(P_c = \mu_{\text{rec}}\)condition for \((BH)_{\max}\)
Magnet area\(A_m = \dfrac{2B_gA_g}{B_r}\)minimum-volume design
Magnet length\(l_m = \dfrac{2\mu_{\text{rec}}B_gl_g}{B_r}\)minimum-volume design
Temperature derating\(B_r(T) = B_r\left[1 + \alpha\,\Delta T\right]\)\(\alpha\) negative; see Table 9.3
Unit conversion1 MGOe = 7.958 kJ/m³NdFeB grade N\(xx\) = \(xx\) MGOe
Section 9-11

Common Mistakes

  • Expecting the gap flux density to equal \(B_r\). Remanence is achieved only in a fully closed circuit with no gap. Any real circuit operates below it, and the optimum design operates at half of it.

  • Assuming a higher operating point is a better design. Operating near \(B_r\) wastes magnet. Maximum energy product — minimum material — occurs at \(B_r/2\).

  • Getting the sign of \(H_m\) wrong. With no external mmf, Ampère's law forces \(H_ml_m = -H_gl_g\), so \(H_m\) is negative. A positive value means the load line has been drawn in the wrong quadrant.

  • Confusing \(H_{cB}\) with \(H_{cJ}\). Reaching \(H_{cB}\) cancels the flux temporarily; reaching \(H_{cJ}\) destroys the magnet.

  • Thinking \(P_c\) depends on the material. It is purely geometric — areas and lengths only. The material enters through the curve, not the line.

  • Inverting the permeance coefficient. \(P_c = A_gl_m/A_ml_g\): magnet length over gap length, gap area over magnet area. Getting it upside down inverts the whole design.

  • Ignoring leakage. A magnet's own permeability is close to that of air, so leakage is far worse than in a wound iron circuit — factors of 1.1 to 1.5 are normal.

  • Checking demagnetisation only at rated load and room temperature. The worst case is a fault at the temperature extreme, and which extreme depends on the material.

  • Assuming all magnets are at risk when hot. Ferrite's coercivity coefficient is positive; its danger is cold.

  • Applying the linear formulas to Alnico. Its curve is strongly knee'd, so \((BH)_{\max} = B_r^{2}/4\mu_0\mu_{\text{rec}}\) badly overestimates. Alnico must be handled graphically.

Section 9-12

Chapter Review

Practice Problems

Take \(\mu_0 = 4\pi\times10^{-7}\) H/m throughout, and assume linear demagnetisation curves unless told otherwise.

  1. P9.1 A magnet 8 mm long with a cross-section of 500 mm² faces a 1.5 mm gap of the same area. Find the permeance coefficient.

    Show answer
    \[P_c = \frac{A_gl_m}{A_ml_g} = \frac{(500)(8)}{(500)(1.5)} = \frac{8}{1.5} = 5.333\]
    Since the areas are equal, \(P_c\) reduces to the simple length ratio \(l_m/l_g\).
  2. P9.2 The magnet of P9.1 is ferrite with \(B_r = 0.38\) T and \(\mu_{\text{rec}} = 1.1\). Find its operating point.

    Show answer
    \[\frac{B_m}{B_r} = \frac{5.333}{5.333+1.1} = \frac{5.333}{6.433} = 0.8290 \quad\Longrightarrow\quad B_m = 0.315~\mathrm{T}\]
    \[H_m = -\frac{B_m}{\mu_0 P_c} = -\frac{0.315}{\left(4\pi\times10^{-7}\right)(5.333)} = -47\,000~\mathrm{A/m}\]
  3. P9.3 An NdFeB grade has \(B_r = 1.35\) T and \(\mu_{\text{rec}} = 1.05\). Find \((BH)_{\max}\) in kJ/m³.

    Show answer
    \[(BH)_{\max} = \frac{(1.35)^{2}}{4\left(4\pi\times10^{-7}\right)(1.05)} = \frac{1.8225}{5.278\times10^{-6}} = 345\,300~\mathrm{J/m^{3}} = 345~\mathrm{kJ/m^{3}}\]
  4. P9.4 Express the answer to P9.3 in MGOe and identify the likely grade designation.

    Show answer
    \[\frac{345.3}{7.958} = 43.4~\mathrm{MGOe}\]
    This corresponds to roughly grade N42 (grades are named for the energy product in MGOe, rounded down to the standard step).
  5. P9.5 A gap of area 400 mm² and length 2 mm must carry 0.60 T. If the magnet material has \((BH)_{\max} = 300\) kJ/m³, find the minimum magnet volume.

    Show answer
    \[V_g = \left(400\times10^{-6}\right)\left(2\times10^{-3}\right) = 8.0\times10^{-7}~\mathrm{m^{3}}\]
    \[W_g = \frac{(0.60)^{2}}{2\left(4\pi\times10^{-7}\right)}\left(8.0\times10^{-7}\right) = (143\,239)\left(8.0\times10^{-7}\right) = 0.1146~\mathrm{J}\]
    \[V_{m,\min} = \frac{2(0.1146)}{300\,000} = 7.64\times10^{-7}~\mathrm{m^{3}} = 764~\mathrm{mm^{3}}\]
  6. P9.6 A gap of area 500 mm² and length 1.8 mm must carry 0.50 T, using a magnet with \(B_r = 1.20\) T and \(\mu_{\text{rec}} = 1.05\). Find the minimum-volume magnet dimensions.

    Show answer
    \[A_m = \frac{2B_gA_g}{B_r} = \frac{2(0.50)(500)}{1.20} = 416.7~\mathrm{mm^{2}}\]
    \[l_m = \frac{2\mu_{\text{rec}}B_gl_g}{B_r} = \frac{2(1.05)(0.50)(1.8)}{1.20} = 1.575~\mathrm{mm}\]
    \[V_m = (416.7)(1.575) = 656~\mathrm{mm^{3}}\]
    Check: \(P_c = (500)(1.575)/\left[(416.7)(1.8)\right] = 1.05 = \mu_{\text{rec}}\) \(\checkmark\)
  7. P9.7 An NdFeB magnet has \(B_r = 1.30\) T at 20 °C with a coefficient of \(-0.12\) %/°C. Find \(B_r\) at 100 °C.

    Show answer
    \[B_r' = (1.30)\left[1 - (0.0012)(80)\right] = (1.30)(0.904) = 1.175~\mathrm{T}\]
    A loss of 9.6 %, which is fully recovered on cooling — this coefficient describes a reversible change.
  8. P9.8 A ferrite-magnet motor operates safely at 25 °C. Explain, with a rough calculation, why it may fail at \(-25\) °C.

    Show answer
    Ferrite's coercivity coefficient is positive, about \(+0.35\) %/°C. Cooling by 50 °C therefore reduces the coercivity by
    \[(0.35)(50) = 17.5\,\%\]
    The knee of the demagnetisation curve moves upward by a comparable amount. An operating point that sat safely above the knee at 25 °C may now lie below it, so the demagnetising field of a starting current or a stalled rotor could push the magnet past the knee and cause irreversible loss. Ferrite must be checked at its minimum operating temperature, not its maximum.
  9. P9.9 A magnet is removed from its iron circuit and left in free air. Explain what happens to its operating point and why a keeper was traditionally fitted.

    Show answer
    In free air the return path has enormous reluctance, so the effective gap is very long and \(P_c = A_gl_m/A_ml_g\) becomes small. The load line flattens and the operating point slides down the demagnetisation curve towards \(H_{cB}\).

    For a material with a high knee — Alnico especially — this alone drives the magnet below the knee, causing irreversible loss along a recoil line. A soft-iron keeper bridging the poles restores a low-reluctance return path, raising \(P_c\) and holding the operating point near \(B_r\). Modern NdFeB and ferrite have knees below the origin at room temperature and tolerate open-circuit storage, which is why keepers are no longer supplied.

  10. P9.10 A designer doubles the length of a magnet, keeping its area and the gap unchanged. What happens to \(P_c\), to \(B_m\), and to the energy product at which it operates? Take \(B_r = 1.20\) T, \(\mu_{\text{rec}} = 1.05\), and an original \(P_c\) of 1.05.

    Show answer
    \(P_c \propto l_m\), so it doubles from 1.05 to 2.10.
    \[\frac{B_m}{B_r}: \ \frac{1.05}{2.10} = 0.500 \ \to\ \frac{2.10}{3.15} = 0.667\]
    \[B_m: \ 0.600 \ \to\ 0.800~\mathrm{T}\]
    \(|H_m| = B_m/\mu_0P_c\): originally \(0.600/(\mu_0)(1.05) = 4.547\times10^{5}\); now \(0.800/(\mu_0)(2.10) = 3.032\times10^{5}\) A/m.
    \[|BH|: \ 272\,800 \ \to\ 242\,500~\mathrm{J/m^{3}}\]
    So twice the magnet buys 33 % more flux density and an 11 % worse energy product. The original design was already optimal, and lengthening it moves away from \((BH)_{\max}\) — an expensive way to gain flux.
Multiple-Choice Questions
  1. MCQ 1. A working permanent magnet operates in the:
    (a) first quadrant   (b) second quadrant   (c) third quadrant   (d) fourth quadrant

    Show answer
    (b) the second quadrant, with \(B\) positive and \(H\) negative, because Ampère's law with no current forces \(H_ml_m = -H_gl_g\).
  2. MCQ 2. The permeance coefficient \(P_c\) depends on:
    (a) the magnet material   (b) the geometry only   (c) the temperature   (d) the remanence

    Show answer
    (b) the geometry only — areas and lengths. The material enters through the demagnetisation curve, not the load line.
  3. MCQ 3. Maximum energy product for a linear magnet occurs at:
    (a) \(B = B_r\)   (b) \(B = B_r/2\)   (c) \(B = 0\)   (d) \(H = 0\)

    Show answer
    (b) \(B = B_r/2\), which requires \(P_c = \mu_{\text{rec}} \approx 1\).
  4. MCQ 4. To deliver a given gap energy, the required magnet volume is:
    (a) proportional to \((BH)_{\max}\)   (b) inversely proportional to \((BH)_{\max}\)   (c) independent of it   (d) proportional to \(B_r\)

    Show answer
    (b) inversely proportional, from \(V_m = 2W_g/(BH)\).
  5. MCQ 5. Intrinsic coercivity \(H_{cJ}\) compared with normal coercivity \(H_{cB}\) is:
    (a) smaller   (b) equal   (c) larger   (d) unrelated

    Show answer
    (c) larger, always. \(H_{cB}\) cancels the flux density; \(H_{cJ}\) destroys the magnetisation itself.
  6. MCQ 6. Alnico was displaced by NdFeB mainly because Alnico has:
    (a) low remanence   (b) low coercivity and a sharp knee   (c) poor temperature stability   (d) high cost

    Show answer
    (b) low coercivity and a sharp knee. Its remanence is comparable to NdFeB's and its temperature stability is excellent — the knee is the problem.
  7. MCQ 7. Driving a magnet below the knee of its demagnetisation curve causes it to:
    (a) recover fully   (b) follow a recoil line to a lower remanence   (c) reverse polarity   (d) heat up

    Show answer
    (b) follow a recoil line to a lower remanence — an irreversible loss.
  8. MCQ 8. A ferrite magnet is most at risk of irreversible demagnetisation when:
    (a) hot   (b) cold   (c) at room temperature   (d) never

    Show answer
    (b) cold. Its coercivity coefficient is positive, so coercivity falls as the temperature drops and the knee rises.
  9. MCQ 9. For NdFeB, the maximum operating temperature is set principally by:
    (a) remanence   (b) intrinsic coercivity   (c) the Curie temperature   (d) mechanical strength

    Show answer
    (b) intrinsic coercivity, which falls at about \(-0.6\) %/°C — five times faster than remanence. The Curie temperature is far higher and is not the binding limit.
  10. MCQ 10. A magnet operating at \(P_c = 9\) with \(\mu_{\text{rec}} = 1\) delivers a flux density of about:
    (a) \(0.5B_r\)   (b) \(0.7B_r\)   (c) \(0.9B_r\)   (d) \(B_r\)

    Show answer
    (c) \(0.9B_r\). From \(B_m/B_r = P_c/(P_c+\mu_{\text{rec}}) = 9/10\). High flux density, but well away from \((BH)_{\max}\) and therefore wasteful of material.
Conceptual Questions
  1. Explain from Ampère's law why a permanent magnet must always operate with \(H\) opposing its own \(B\), and what that implies about the second quadrant.

  2. Compare the load-line construction of this chapter with the air-gap-line construction of Chapter 5. What is the same, what differs, and why does the line here pass through the origin?

  3. Operating a magnet near its remanence gives more gap flux but is described as wasteful. Reconcile these two statements.

  4. Explain why the permeance coefficient is purely geometric, and what that implies about swapping one magnet material for another in an existing design.

  5. Alnico has almost the same remanence as NdFeB yet is far inferior in practice. Identify the property responsible and explain how it constrains the shape of an Alnico magnet.

  6. Explain why the worst-case demagnetisation check occurs at high temperature for NdFeB but at low temperature for ferrite.

  7. What is the purpose of stabilising a magnet, and why is a deliberately weakened magnet preferable in an instrument?

Looking Ahead

Part 1's treatment of the magnetic circuit is complete. Chapters 2 to 5 established how flux is set up and what governs it, Chapters 6 and 7 accounted for what it costs in heat, Chapter 8 set it alternating, and this chapter has shown how a magnet can supply the mmf with no supply at all.

From Chapter 10 the subject changes character. The next five chapters take up electromagnetic induction in its own right — Faraday's laws, Lenz's law and the right-hand rule, statically and dynamically induced EMF, and self and mutual inductance. The EMF equation of Chapter 8 will turn out to be one special case of a much more general result, and inductance will be revealed as nothing more than turns squared times permeance, as Chapter 2 hinted. From Chapter 15, energy and coenergy finally produce force, and the machine begins to move.