Force and Torque in Magnetic Field Systems

Chapters 15 to 17 built the energy method one case at a time. This chapter states the results in their general form and then turns them into something a designer can use.

The two force expressions established in Section 16-2 hold for any magnetic field system whatever — any number of windings, any geometry, linear or saturating:

What did assume linearity was the convenient form \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\), which followed from writing \(W' = \tfrac{1}{2}Li^{2}\). Section 18-3 examines what happens when that step is not available.

With \(n\) windings, the coenergy of a linear system generalises the two-winding expression of Chapter 17 in the obvious way:

where \(L_{jj}\) is the self-inductance of winding \(j\) and \(L_{jk} = L_{kj}\) is the mutual inductance between windings \(j\) and \(k\). Differentiating gives the torque:

Setting \(n = 1\) recovers Chapter 16 and \(n = 2\) recovers Chapter 17, which is the check worth doing on any general formula. A three-phase machine has \(n = 4\) when the field winding is counted, giving four reluctance terms and six mutual terms — which is why machine analysis moves to the transformed coordinates of Part 5 rather than working with all ten.

Chapter 15 warned that energy and coenergy diverge once the core saturates. The consequence for torque calculation is severe and is worth stating plainly.

The correct procedure is unchanged in principle and only slightly more work in practice:

1. Obtain \(\lambda(i,\theta)\) for the system, from a magnetisation curve or a field solution.
2. Integrate to find the coenergy at the operating angle: \(W'(i,\theta) = \int_0^{i}\lambda(i',\theta)\,\mathrm{d}i'\).
3. Differentiate with respect to \(\theta\), holding \(i\) constant.

Where \(\lambda\) is known only numerically — the usual case in finite-element design work — steps 2 and 3 are done numerically, evaluating the coenergy at two nearby angles and taking the difference.

The energy method gives a total force but says nothing about where it acts. A complementary picture treats the magnetic field as exerting a stress on the surfaces it touches — which turns out to be the most useful way to think about a rotating machine.

At an iron surface the field exerts two stresses, and they do quite different jobs:

Once the torque is seen as a shear stress on a cylindrical surface, computing it is elementary mechanics.

Let the rotor have diameter \(D\) and axial length \(l\), so that its curved surface area is \(\pi D l\). If a mean tangential stress \(\sigma\) acts over that surface at a radius \(D/2\), the torque is

The quantity \(\sigma\) is called the air-gap shear stress or specific tangential force, and it is the single most useful figure of merit in machine design. Typical values are remarkably consistent across machine types:

Rearranging the shear-stress result gives the equation with which every machine design begins:

and since mechanical power is \(P = \omega T\),

The second point deserves emphasis. Halving the size of a motor while keeping its power means doubling its speed and fitting a 2:1 reduction gear. Whether that is worth doing is a question of cost, noise and reliability rather than electromagnetics — but the trade is always available, and it is why traction and aerospace drives run at speeds that would have seemed absurd in 1950.

The shear-stress picture is physical but requires knowing the field. A third formulation, equivalent to both, expresses torque in terms of the two mmf distributions that produce it — and this is the form Part 5 uses throughout.

A machine's stator and rotor each set up an mmf wave around the air gap, of peak values \(F_1\) and \(F_2\), separated by an angle \(\delta\). Working the coenergy through the gap geometry gives

Three features of this expression matter more than its exact constant:

• Both mmfs appear as a product. Weakening either one weakens the torque proportionally — which is why field weakening reduces torque in Chapter 37.

• The torque depends on the angle between them, not on their positions. Both waves may be rotating rapidly; only their relative displacement matters. This is what makes steady torque possible in an AC machine.

• It is inversely proportional to the gap. A smaller gap gives more torque for the same mmf, which is why machine gaps are made as small as mechanical tolerance permits — often under a millimetre.

Chapter 17 showed that a salient machine develops both kinds of torque. It is worth setting out, for each machine family, which component does the work.

• The general results \(f = -\partial W/\partial x|_{\lambda}\) and \(f = +\partial W'/\partial x|_{i}\), with their rotational counterparts, are exact for any magnetic field system.

• For \(n\) windings, \(T = \tfrac{1}{2}\sum_j i_j^{2}\,\mathrm{d}L_{jj}/\mathrm{d}\theta + \sum_{j\lt k}i_ji_k\,\mathrm{d}L_{jk}/\mathrm{d}\theta\) — self terms halved, mutual terms not.

• Under saturation the coenergy derivative still holds but \(\tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}\theta\) does not. Using an apparent \(L = \lambda/i\) gives a materially wrong answer.

• Maxwell stress has a normal component \(B_n^{2}/2\mu_0\) which pulls radially and produces no torque, and a tangential component \(B_nB_t/\mu_0\) which produces all of it.

• Torque as shear: \(T = \sigma\pi D^{2}l/2\), so torque is proportional to rotor volume.

• The sizing equation \(D^{2}l = 2T/\sigma\pi = 2P/\sigma\pi\omega\) means torque needs volume and power needs volume times speed.

• Achievable \(\sigma\) is 4 to 100 kN/m², fifty to a hundred times below the magnetic pressure, because it is limited by cooling rather than by saturation.

• In terms of mmf waves, \(T = -kF_1F_2\sin\delta\) with \(k \propto Dl/g\) — proportional to both mmfs, dependent only on their relative angle, and inversely proportional to the gap.

• Machine families differ in which component they use: DC, induction and round-rotor synchronous are pure alignment; reluctance machines are pure reluctance; salient and interior-magnet machines use both.

• Using \(\tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}\theta\) on a saturated machine. It is a linear result. Example 18.2 showed a 25 % error.

• Putting a factor of one half on the mutual terms. Self terms carry \(\tfrac{1}{2}\); mutual terms do not, because each pair appears twice in the double sum.

• Thinking the normal stress produces torque. It acts radially and contributes nothing but bearing load and noise.

• Confusing magnetic pressure with air-gap shear stress. The first reaches 1300 kN/m²; the second rarely exceeds 100. They differ by the ratio \(B_t/B_n\), which is a few percent.

• Expecting torque to depend on the winding details. At the sizing stage it depends only on rotor volume and shear stress.

• Confusing the torque and power sizing rules. \(D^{2}l\) is set by torque; power brings speed into it.

• Scaling a machine by \(D\) alone. Torque goes as \(D^{2}l\), so uniform scaling gives \(s^{3}\) — a 26 % size increase for double torque, not 41 %.

• Believing a larger air gap is safer. Torque varies as \(1/g\), so gaps are made as small as mechanical tolerance allows.

• Assuming all machines use alignment torque. Reluctance and switched-reluctance machines use none at all.

• Adding alignment and reluctance peaks directly. They peak at different angles, so the total peak is less than the sum and occurs somewhere between.

Where a machine is described by its dimensions, check the implied shear stress against Table 18.1 — an answer far outside that range signals an error.

1. P18.1 A machine develops 150 N·m from a rotor 180 mm in diameter and 220 mm long. Find the air-gap shear stress and comment.

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   \[\sigma = \frac{2T}{\pi D^{2}l} = \frac{2(150)}{\pi(0.180)^{2}(0.220)} = \frac{300}{0.02240} = 13\,400~\mathrm{N/m^{2}} = 13.4~\mathrm{kN/m^{2}}\]

   Squarely in the medium-industrial band of Table 18.1, so the design is plausible.

2. P18.2 A 22 kW motor runs at 2900 rev/min with \(\sigma = 20\) kN/m². Find the required \(D^{2}l\), and the dimensions if \(l = 1.2D\).

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   \[\omega = \frac{2\pi(2900)}{60} = 303.7~\mathrm{rad/s}, \qquad T = \frac{22\,000}{303.7} = 72.44~\mathrm{N\,m}\]

   \[D^{2}l = \frac{2T}{\sigma\pi} = \frac{2(72.44)}{\left(20\,000\right)\pi} = 2.306\times10^{-3}~\mathrm{m^{3}}\]

   With \(l = 1.2D\), \(1.2D^{3} = 2.306\times10^{-3}\), so

   \[D = \left(1.922\times10^{-3}\right)^{1/3} = 0.1244~\mathrm{m} \approx 124~\mathrm{mm}, \qquad l = 149~\mathrm{mm}\]

3. P18.3 In a machine gap \(B_n = 0.90\) T and \(B_t = 0.025\) T. Find the normal and tangential stresses and their ratio.

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   \[\sigma_n = \frac{(0.90)^{2}}{2\left(4\pi\times10^{-7}\right)} = 3.223\times10^{5}~\mathrm{N/m^{2}} = 322~\mathrm{kN/m^{2}}\]

   \[\sigma_t = \frac{(0.90)(0.025)}{4\pi\times10^{-7}} = 1.790\times10^{4}~\mathrm{N/m^{2}} = 17.9~\mathrm{kN/m^{2}}\]

   \[\frac{\sigma_n}{\sigma_t} = \frac{322}{17.9} = 18.0\]

   The radial pull is eighteen times the useful shear — and does no work at all.

4. P18.4 A four-winding system has all self-inductances constant. How many torque terms are there, and of what kind?

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   With constant self-inductances, all four reluctance terms vanish. The mutual terms number

   \[\frac{n(n-1)}{2} = \frac{(4)(3)}{2} = 6\]

   so there are six alignment terms — one for each pair of windings — and no reluctance terms.

5. P18.5 A saturating system has \(\lambda = k(\theta)\sqrt{i}\). Show that the coenergy is twice the field energy, whatever \(k\) and \(i\).

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   \[W' = \int_0^{i}k\sqrt{i'}\,\mathrm{d}i' = \tfrac{2}{3}k\,i^{3/2}\]

   \[W = \lambda i - W' = k\sqrt{i}\cdot i - \tfrac{2}{3}k i^{3/2} = k i^{3/2}\left(1 - \tfrac{2}{3}\right) = \tfrac{1}{3}k i^{3/2}\]

   \[\frac{W'}{W} = \frac{2/3}{1/3} = 2\]

   The ratio depends only on the exponent of the characteristic, not on the operating point.

6. P18.6 A motor is redesigned with every linear dimension increased by 15 %. By what factor does its torque rise, at constant shear stress?

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   \[\frac{T'}{T} = s^{3} = (1.15)^{3} = 1.521\]

   A 52 % increase in torque for a 15 % increase in size — the strong reward for volume that makes machines scale so favourably.

7. P18.7 A salient machine has \(T = 200\sin\delta + 50\sin2\delta\) N·m. Find the maximum torque and the share from each component.

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   Setting \(200\cos\delta + 100\cos2\delta = 0\):

   \[200\cos^{2}\delta + 200\cos\delta - 100 = 0 \quad\Longrightarrow\quad \cos\delta = 0.3660\]

   \[\delta_{\max} = 68.5^{\circ}, \qquad T_{\text{align}} = 200(0.9306) = 186.1, \qquad T_{\text{rel}} = 50(0.6812) = 34.1\]

   \[T_{\max} = 220.2~\mathrm{N\,m} \quad (84.5\,\% \text{ alignment}, \ 15.5\,\% \text{ reluctance})\]

8. P18.8 Two machines have the same rotor volume, but one runs at 1500 rev/min and the other at 3000. Compare their power ratings.

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   Equal volume at equal shear stress means equal torque. Since \(P = \omega T\) and the speed is doubled, the faster machine delivers twice the power from the same amount of active material.

   This is why high-speed machines are attractive wherever a gearbox is acceptable, and why direct-drive low-speed machines — wind turbines, ship propulsion — are so large.

9. P18.9 Explain why an air-gap shear stress of 20 kN/m² is respectable, when Chapter 16 showed the gap can support 1300 kN/m² of magnetic pressure.

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   The two are different components of the same stress. Magnetic pressure uses \(B_n\), which saturation limits to about 2 T; shear uses the product \(B_nB_t\), and \(B_t\) is produced by the winding current.

   That current is limited by heat removal, not by magnetics, so \(B_t\) stays at a few percent of \(B_n\) and the shear stress at a few percent of the pressure. Machine torque density is a thermal problem wearing a magnetic disguise, which is why liquid cooling buys so much.

10. P18.10 A designer proposes doubling a motor's air gap to ease manufacturing tolerances. What happens to the torque at constant mmf, and what would have to change to restore it?

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    Since \(k \propto 1/g\), doubling the gap halves the torque at the same mmfs.

    To restore it, the product \(F_1F_2\) must double — for example by increasing both mmfs by \(\sqrt{2} = 1.41\), which raises the copper loss by a factor of two and worsens the thermal problem that already limits \(\sigma\). This is why machine air gaps are made as small as the bearings and shaft stiffness permit, often well under a millimetre in small machines.