By the end of this chapter you should be able to:
Recognise a singly excited system and identify the coordinate on which its inductance depends.
Derive \(f = +\partial W'/\partial x|_i\) and \(f = -\partial W/\partial x|_{\lambda}\) from the energy balance.
Apply the linear result \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\).
Express the same force through reluctance as \(f = -\tfrac{1}{2}\Phi^{2}\,\mathrm{d}S/\mathrm{d}x\) and state the direction rule that follows.
Derive and use the magnetic pull \(f = B^{2}A/2\mu_0\) and the idea of magnetic pressure.
Explain why a gap-type actuator has a force rising as \(1/x^{2}\) while a proportional actuator can have constant force.
Compute reluctance torque from \(T = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}\theta\) and identify where it is zero and maximum.
Explain why a relay's pull-in current greatly exceeds its holding current.
What Is a Singly Excited System?
A singly excited magnetic system has exactly one electrical winding acting on a magnetic circuit that contains a movable member. There is no second winding, no permanent magnet, and no other source of mmf.
Its defining property is simple to state and easy to overlook: the inductance of the winding depends on the position of the movable member. As the member moves, it changes the reluctance of the flux path, and hence the inductance seen at the terminals:
Chapter 15 ended by noting that force follows from the derivative of coenergy with respect to position. In a singly excited system that derivative is available directly, because the coenergy is \(\tfrac{1}{2}L(x)i^{2}\) and only \(L\) depends on \(x\).
| Device | Moving member | Coordinate | Purpose |
|---|---|---|---|
| Relay, contactor | Armature | Gap length \(x\) | Switching |
| Solenoid actuator | Plunger | Insertion \(x\) | Linear motion |
| Lifting magnet | The load itself | Gap \(x\) | Holding, lifting |
| Moving-iron instrument | Iron vane | Angle \(\theta\) | Measurement |
| Reluctance motor | Salient rotor | Angle \(\theta\) | Continuous rotation |
| Magnetic bearing | Shaft | Clearance \(x\) | Support without contact |
Deriving the Force
Start from the energy balance of Chapter 15, with the mechanical work now written explicitly as force times displacement:
Route 1 — field energy at constant flux linkage. The field energy is a function of two independent variables, \(\lambda\) and \(x\), so its total differential is
Substituting and collecting terms:
Since \(\lambda\) and \(x\) are independent, this can hold for all \(\mathrm{d}\lambda\) and \(\mathrm{d}x\) only if both brackets vanish separately:
Route 2 — coenergy at constant current. Using the identity \(W' = \lambda i - W\) from Section 15-4:
where the balance equation was used to replace \(i\,\mathrm{d}\lambda - \mathrm{d}W_{\text{fld}}\) by \(f\,\mathrm{d}x\). Reading off the coefficients:
Note carefully what is held constant in each, and that the signs differ. At constant \(\lambda\) no electrical energy enters, so work must be taken out of the store — hence the minus. At constant \(i\) the supply contributes, and the plus sign follows.
Both give the same numerical force. They must: the device does not know which variable an analyst has chosen to hold fixed.
The Linear Result
If the magnetic circuit is linear, the coenergy is \(W' = \tfrac{1}{2}L(x)i^{2}\). Differentiating with \(i\) held constant leaves only the inductance to differentiate:
and for rotation,
Force exists wherever inductance varies with position. If \(L\) is constant there is no force at all, however large the current.
Three consequences follow immediately and are worth stating separately:
The force is proportional to \(i^{2}\). Doubling the current quadruples the pull, and reversing it changes nothing.
The force acts in the direction of increasing \(L\). Where \(\mathrm{d}L/\mathrm{d}x\) is positive the force is positive, so the member is drawn towards greater inductance.
At a maximum or minimum of \(L\) the force is zero. These are the equilibrium positions — the aligned and unaligned positions of a reluctance machine.
Problem. A relay has 800 turns and a single working air gap of area 6 cm² and length 2 mm. The iron reluctance is negligible. Find the inductance, the force on the armature at a current of 1.5 A, and verify the result independently.
Inductance. With only the gap contributing, \(S = g/\mu_0A\) and \(L = N^{2}/S\):
Rate of change with gap. Since \(L \propto 1/g\),
Force.
The negative sign means the force acts in the direction of decreasing \(g\) — it closes the gap, as expected. Its magnitude is 135.7 N.
Independent check. Working out the flux density and using the magnetic-pull formula of Section 16-5:
Comment. Note the \(1/g^{2}\) dependence. Halve the gap to 1 mm and the force quadruples to 543 N at the same current. This is why a relay slams shut rather than closing gently, and why the same device needs far less current to hold than to attract.
Force in Terms of Reluctance
The force can be written in a second way that makes its direction obvious. Since \(L = N^{2}/S\),
Substituting into \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\) and recognising \(Ni/S = \Phi\):
The minus sign is the whole point. Wherever moving in the \(+x\) direction would increase reluctance, the force is negative and pushes the other way. A magnetic system always tries to rearrange itself into the configuration of least reluctance — equivalently, of greatest inductance.
This single rule predicts the direction of every force in this chapter without any calculation:
An armature is attracted across a gap, because closing the gap reduces reluctance.
A plunger is drawn into a solenoid, because more iron in the path reduces reluctance.
A salient rotor aligns with the stator field, because the aligned position offers the shortest iron path.
An iron vane rotates into the strongest part of the field in a moving-iron instrument.
Magnetic Pull and Magnetic Pressure
For the commonest case of all — a plane air gap of area \(A\) and length \(g\) — the reluctance form collapses to a strikingly simple result. The gap reluctance is \(S = g/\mu_0A\), so
and therefore, with \(\Phi = BA\),
Dividing by the area gives the magnetic pressure — a force per unit area that depends on nothing but the flux density:
Note that this is numerically identical to the energy density \(B^{2}/2\mu_0\) of Chapter 1. That is not a coincidence: pressure and energy density have the same units, and the pull is the rate at which gap energy changes as the gap closes.
| \(B\) (T) | Pressure (kN/m²) | Equivalent | Where met |
|---|---|---|---|
| 0.8 | 255 | 2.5 bar | Lightly worked relay |
| 1.0 | 398 | 4.0 bar | Typical machine gap |
| 1.2 | 573 | 5.7 bar | Lifting magnet |
| 1.5 | 895 | 8.8 bar | Hard-worked pole face |
| 1.8 | 1289 | 12.7 bar | Near saturation — the practical ceiling |
Problem. A lifting magnet has two pole faces, each of area 80 cm², working at a flux density of 1.2 T. Find the magnetic pressure, the total lifting force, and the mass that can be lifted. Take \(g = 9.81\) m/s².
Magnetic pressure.
Force per pole.
Total force. Both poles pull on the load, so
Mass lifted.
Comment. Just under a tonne from 160 cm² of pole face — about the area of a sheet of A5 paper. Note the factor of two for the second pole, which is the commonest source of error in these problems: a horseshoe geometry has two working gaps in series magnetically but two faces pulling in parallel mechanically.
In practice the rated lift would be set well below 934 kg. The formula assumes perfect contact and no leakage; a rusty or uneven load surface introduces an unintended air gap, and since \(f \propto 1/g^{2}\) at fixed mmf, even a fraction of a millimetre matters.
Shaping the Force–Stroke Curve
The two examples so far behave quite differently, and the difference is entirely a matter of how \(L\) varies with position.
Weak at large gaps, violently strong as it closes. Excellent for relays and contactors, where a decisive snap is wanted. Poor where controlled motion is needed.
The same force at every point of the stroke. Achieved by shaping the plunger so that iron enters the flux path at a uniform rate. This is the design goal for solenoid valves and proportional actuators.
Problem. A solenoid actuator is designed so that its inductance varies linearly with plunger position: \(L(x) = 0.050 + 25x\) H, with \(x\) in metres over a stroke of 0 to 20 mm. The current is 3.0 A. Find the force, the inductance at each end of the stroke, and the work done over the full stroke.
Force.
Constant over the whole stroke, because \(\mathrm{d}L/\mathrm{d}x\) is constant.
Inductance at the ends.
Work done over the stroke.
Check by coenergy. From Chapter 15, the mechanical work at constant current equals the change in coenergy:
Comment. Contrast this with Example 16.1, where the force rose as \(1/g^{2}\). Here it is flat, which is exactly what a valve or a positioning actuator needs. The designer's task is not to make the force large but to make \(\mathrm{d}L/\mathrm{d}x\) the right shape — usually by tapering the plunger nose so that iron enters the gap at a controlled rate rather than all at once.
Reluctance Torque
Replace linear motion by rotation and the force expression becomes a torque:
Consider a two-pole stator winding and a salient rotor — a plain bar of iron with no winding at all. Its inductance is greatest when the bar is aligned with the stator poles and least when it lies across them, and by symmetry the variation repeats twice per revolution:
Differentiating gives the reluctance torque:
The torque is zero at \(\theta = 0\) and \(\theta = 90^{\circ}\), and greatest in magnitude at \(45^{\circ}\). It varies at twice the mechanical frequency, and it always acts to pull the rotor back towards alignment.
The ratio \(L_{\max}/L_{\min}\) is called the saliency ratio, and it governs how much torque the machine can produce. Since \(L_{\max} = L_0 + L_2\) and \(L_{\min} = L_0 - L_2\), a large \(L_2\) means both a high saliency ratio and a high torque. Modern synchronous reluctance rotors achieve ratios of 6 to 10 by cutting multiple flux barriers into the lamination.
With an alternating current \(i = I_m\sin\omega t\) the torque contains \(i^{2} \propto \sin^{2}\omega t\), which has a steady component and a component at \(2\omega\). Averaged over a revolution at an arbitrary speed the torque comes to zero.
A useful average appears only when the rotor turns in step with the field, so that \(\theta\) and the current maintain a fixed phase relationship. A reluctance motor is therefore inherently a synchronous machine — it must be brought up to speed by some other means, or fed from a controlled inverter as in the switched reluctance drives of Chapter 91.
Problem. A singly excited rotating system has \(L(\theta) = 0.40 + 0.15\cos 2\theta\) H and carries 5.0 A. Find the maximum and minimum inductance, the saliency ratio, the torque as a function of angle, and the maximum torque.
Inductance extremes.
Saliency ratio.
Torque.
Maximum torque. At \(2\theta = 90^{\circ}\), that is \(\theta = 45^{\circ}\):
Comment. The negative sign means the torque acts to reduce \(\theta\), pulling the rotor back towards the aligned position at \(\theta = 0\) — which is where inductance is greatest and reluctance least, exactly as Section 16-4 requires.
Note that \(L_0\) plays no part in the torque at all. Only the varying part \(L_2\) matters, since a constant inductance has zero derivative. A designer wanting more torque must increase the saliency, not the average inductance — which is why reluctance rotors are cut with flux barriers rather than made of more iron.
Problem. A relay must exert 50 N on its armature. The pole face area is 4 cm², the coil has 600 turns, and the iron reluctance is negligible. Find the required flux density and the current when the gap is (a) 0.5 mm, the closed position, and (b) 5 mm, the open position.
Required flux density. From the magnetic pull formula, rearranged:
The same \(B\) is needed at either gap, since force depends only on flux density and area.
Field strength in the gap.
(a) Closed, \(g = 0.5\) mm.
(b) Open, \(g = 5\) mm.
Since the required mmf is proportional to gap length at constant flux density, the current ratio is simply the gap ratio: \(5.0/0.5 = 10\).
Every relay and contactor faces this. The coil must be sized for the pull-in current but then spends its working life carrying far less, so large contactors use an economy circuit — a series resistor or a change of tapping switched in once the armature has closed, cutting the holding current and the coil heating with it.
Comment. This is the practical face of the \(1/g^{2}\) law noted in Example 16.1. It also explains hysteresis in relay operation: a relay picks up at one voltage and drops out at a much lower one, because the force at the closed position is so much greater. The gap between pull-in and drop-out voltages is not a defect — it is what stops the contacts chattering.
Applications
The archetype. Design centres on providing enough mmf to pull in across the open gap, then reducing it for economical holding.
The plunger is shaped so that \(\mathrm{d}L/\mathrm{d}x\) — and hence force — matches the load over the stroke, rather than rising steeply at the end.
Rated directly from \(B^{2}A/2\mu_0\), then derated heavily for surface irregularity, since even a small parasitic gap costs force rapidly.
Deflecting torque is \(\tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}\theta\), so the reading follows \(i^{2}\) and the instrument reads true RMS on AC or DC alike — the reason the scale is non-linear.
Torque from saliency alone, with no rotor winding, magnet or brushes. Robust and cheap, and increasingly attractive as rare-earth prices rise (Chapters 91 and 92).
Force grows as the gap closes, so the system is open-loop unstable and needs active control — a direct consequence of the \(1/g^{2}\) characteristic.
Summary and Key Formulas
A singly excited system has one winding acting on a magnetic circuit with a movable member, so that \(L = L(x)\) or \(L(\theta)\).
From the energy balance, \(f = -\partial W_{\text{fld}}/\partial x|_{\lambda}\) and \(f = +\partial W'_{\text{fld}}/\partial x|_{i}\). Both give the same force; the coenergy form is preferred because supplies fix current.
For a linear system, \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\) and \(T = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}\theta\). No variation of \(L\) means no force.
Force is proportional to \(i^{2}\), so it is independent of current direction. Singly excited systems attract only, never repel.
Equivalently \(f = -\tfrac{1}{2}\Phi^{2}\,\mathrm{d}S/\mathrm{d}x\): the force always acts to reduce reluctance, that is, towards maximum inductance.
Across a plane air gap the magnetic pull is \(f = B^{2}A/2\mu_0\), and the magnetic pressure \(B^{2}/2\mu_0\) is numerically the gap energy density.
Saturation caps the pressure at roughly 1.3 MN/m², which sets the size of every electrical machine.
A gap-type actuator gives \(f \propto 1/x^{2}\); a proportional actuator with \(L = L_0 + kx\) gives constant force.
With \(L = L_0 + L_2\cos2\theta\), the reluctance torque is \(T = -i^{2}L_2\sin2\theta\): zero at alignment, maximum at 45°, twice per revolution. Only \(L_2\) matters.
A relay's pull-in current exceeds its holding current in the ratio of the gaps, which is why economy circuits exist and why pick-up and drop-out voltages differ.
| Quantity | Formula | Notes |
|---|---|---|
| Force, constant \(\lambda\) | \(f = -\left.\dfrac{\partial W_{\text{fld}}}{\partial x}\right|_{\lambda}\) | note the minus |
| Force, constant \(i\) | \(f = +\left.\dfrac{\partial W'_{\text{fld}}}{\partial x}\right|_{i}\) | the practical form |
| Linear force | \(f = \tfrac{1}{2}i^{2}\dfrac{\mathrm{d}L}{\mathrm{d}x}\) | newtons |
| Linear torque | \(T = \tfrac{1}{2}i^{2}\dfrac{\mathrm{d}L}{\mathrm{d}\theta}\) | newton-metres |
| Reluctance form | \(f = -\tfrac{1}{2}\Phi^{2}\dfrac{\mathrm{d}S}{\mathrm{d}x}\) | acts to reduce \(S\) |
| Magnetic pull | \(f = \dfrac{B^{2}A}{2\mu_0}\) | per gap; double for two poles |
| Magnetic pressure | \(p = \dfrac{B^{2}}{2\mu_0}\) | N/m²; equals energy density |
| Required flux density | \(B = \sqrt{\dfrac{2\mu_0 f}{A}}\) | design form |
| Gap-type inductance | \(L(g) = \dfrac{N^{2}\mu_0 A}{g}\) | iron neglected |
| Gap-type force | \(f = \tfrac{1}{2}i^{2}\dfrac{L}{g} = \dfrac{i^{2}N^{2}\mu_0A}{2g^{2}}\) | \(\propto 1/g^{2}\) |
| Proportional actuator | \(L = L_0 + kx \Rightarrow f = \tfrac{1}{2}i^{2}k\) | constant force |
| Salient inductance | \(L(\theta) = L_0 + L_2\cos2\theta\) | two-pole rotor |
| Reluctance torque | \(T = -i^{2}L_2\sin2\theta\) | max \(= i^{2}L_2\) at 45° |
| Saliency ratio | \(\dfrac{L_{\max}}{L_{\min}} = \dfrac{L_0+L_2}{L_0-L_2}\) | governs torque capability |
| Pull-in current ratio | \(\dfrac{i_{\text{open}}}{i_{\text{closed}}} = \dfrac{g_{\text{open}}}{g_{\text{closed}}}\) | at equal force |
Common Mistakes
Omitting the factor of one half. It is \(f = \tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\), not \(i^{2}\,\mathrm{d}L/\mathrm{d}x\), because only half the electrical energy becomes work.
Using the wrong sign convention. Constant \(\lambda\) takes a minus and the energy; constant \(i\) takes a plus and the coenergy. Mixing them reverses the force.
Differentiating \(\tfrac{1}{2}Li^{2}\) with respect to \(x\) while letting \(i\) vary. The current is held constant in this derivative; only \(L\) depends on \(x\).
Expecting reversing the current to reverse the force. The force goes as \(i^{2}\) — an electromagnet attracts either way.
Forgetting the second pole face. A horseshoe magnet has two faces pulling in parallel, so the total force is twice \(B^{2}A/2\mu_0\).
Confusing gap area with pole-face area. Use the area the flux actually crosses, allowing for fringing where the gap is large (Chapter 4).
Including \(L_0\) in the torque. A constant inductance contributes nothing; only the varying part \(L_2\) produces torque.
Expecting a reluctance motor to start on AC. Averaged over an arbitrary speed the torque is zero — it must be brought into synchronism.
Assuming a lifting magnet's rating holds on a rough surface. Force goes as \(1/g^{2}\) at fixed mmf, so a fraction of a millimetre of parasitic gap is significant.
Treating pull-in and holding current as the same. They differ in the ratio of the gaps, often by a factor of ten.
Chapter Review
Check the direction of every force against the rule that it must reduce reluctance.
P16.1 An actuator has \(\mathrm{d}L/\mathrm{d}x = 40\) H/m and carries 4.0 A. Find the force.
Show answer
Positive, so the force acts in the direction of increasing \(x\) — towards greater inductance.\[f = \tfrac{1}{2}i^{2}\frac{\mathrm{d}L}{\mathrm{d}x} = \tfrac{1}{2}(16.0)(40) = 320~\mathrm{N}\]P16.2 An air gap of area 25 cm² carries a flux density of 1.0 T. Find the magnetic pressure and the pull.
Show answer
\[p = \frac{(1.0)^{2}}{2\left(4\pi\times10^{-7}\right)} = 3.979\times10^{5}~\mathrm{N/m^{2}}\]\[f = pA = \left(3.979\times10^{5}\right)\left(2.5\times10^{-3}\right) = 995~\mathrm{N}\]P16.3 A lifting magnet has two pole faces each of 50 cm² working at 1.1 T. Find the total force and the mass it can lift.
Show answer
\[f_{\text{pole}} = \frac{(1.1)^{2}\left(5.0\times10^{-3}\right)}{2\left(4\pi\times10^{-7}\right)} = 2407~\mathrm{N}\]\[f = 2(2407) = 4814~\mathrm{N}, \qquad m = \frac{4814}{9.81} = 491~\mathrm{kg}\]P16.4 A proportional solenoid has \(L = L_0 + 18x\) H and carries 2.5 A. Find the force and comment on how it varies over the stroke.
Show answer
The force is constant over the whole stroke, because \(\mathrm{d}L/\mathrm{d}x = 18\) H/m does not depend on \(x\). Note that \(L_0\) never enters — only the rate of change matters.\[f = \tfrac{1}{2}(6.25)(18) = 56.25~\mathrm{N}\]P16.5 A salient rotor system has \(L(\theta) = 0.30 + 0.10\cos2\theta\) H and carries 6.0 A. Find the saliency ratio and the maximum torque.
Show answer
\[L_{\max} = 0.40~\mathrm{H}, \qquad L_{\min} = 0.20~\mathrm{H}, \qquad \frac{L_{\max}}{L_{\min}} = 2.0\]\[T = -i^{2}L_2\sin2\theta, \qquad T_{\max} = (36.0)(0.10) = 3.60~\mathrm{N\,m} \ \text{at } \theta = 45^{\circ}\]P16.6 What flux density is needed to produce 80 N across a pole face of 5 cm²?
Show answer
Comfortably within the capability of any steel, so the design would be limited by mmf rather than by saturation.\[B = \sqrt{\frac{2\mu_0 f}{A}} = \sqrt{\frac{2\left(4\pi\times10^{-7}\right)(80)}{5\times10^{-4}}} = \sqrt{0.4021} = 0.634~\mathrm{T}\]P16.7 A magnet has 500 turns and a gap of area 4 cm² and length 1.5 mm, carrying 2.0 A. Neglecting iron, find the inductance and the force by both methods.
Show answer
\[L = \frac{N^{2}\mu_0A}{g} = \frac{(250\,000)\left(4\pi\times10^{-7}\right)\left(4\times10^{-4}\right)}{1.5\times10^{-3}} = 0.0838~\mathrm{H}\]Check via flux density: \(S = 2.984\times10^{6}\), \(\Phi = 3.351\times10^{-4}\) Wb, \(B = 0.838\) T, and\[f = \tfrac{1}{2}i^{2}\frac{L}{g} = \tfrac{1}{2}(4.0)\frac{0.0838}{1.5\times10^{-3}} = 111.7~\mathrm{N}\]\[f = \frac{B^{2}A}{2\mu_0} = \frac{(0.7018)\left(4\times10^{-4}\right)}{2.513\times10^{-6}} = 111.7~\mathrm{N} \;\checkmark\]P16.8 A contactor's gap is 8 mm open and 0.4 mm closed. By what factor must the coil current be reduced after closing to keep the same force?
Show answer
Equal force means equal flux density, and the mmf needed is \(\mathcal{F} = Bg/\mu_0\), which is proportional to the gap. SoThe current can be cut to one twentieth once closed. In practice the coil is left at a somewhat higher current than the bare minimum, to give margin against vibration and voltage dips, but an economy circuit is essential in a device this size or the coil would burn out.\[\frac{i_{\text{open}}}{i_{\text{closed}}} = \frac{8.0}{0.4} = 20\]P16.9 For the actuator of P16.4, find the work done and the electrical energy supplied over a 15 mm stroke at constant current.
Show answer
\[W_{\text{mech}} = f\,\Delta x = (56.25)\left(15\times10^{-3}\right) = 0.844~\mathrm{J}\]Half becomes work and half increases the stored field energy, as the fifty-fifty rule of Chapter 15 requires: \(\Delta W_{\text{fld}} = 1.688 - 0.844 = 0.844\) J \(\checkmark\)\[\Delta L = (18)\left(15\times10^{-3}\right) = 0.270~\mathrm{H}, \qquad W_{\text{elec}} = i^{2}\Delta L = (6.25)(0.270) = 1.688~\mathrm{J}\]P16.10 A cylindrical iron plunger is partly inserted into a solenoid. State the direction of the force, and explain what happens if the current is reversed.
Show answer
The plunger is drawn further into the solenoid. Inserting more iron shortens the high-reluctance air path and lengthens the low-reluctance iron path, so reluctance falls and inductance rises — and by \(f = -\tfrac{1}{2}\Phi^{2}\,\mathrm{d}S/\mathrm{d}x\) the force acts towards smaller \(S\).Reversing the current changes nothing. The force depends on \(i^{2}\), so both the magnitude and the direction are unaffected. A singly excited system cannot be made to push; producing a bidirectional force requires a second excitation, which is the subject of Chapter 17.
MCQ 1. In a singly excited system the force is given by:
(a) \(i^{2}\,\mathrm{d}L/\mathrm{d}x\) (b) \(\tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\) (c) \(\tfrac{1}{2}iL\) (d) \(Li\,\mathrm{d}i/\mathrm{d}x\)Show answer
(b) \(\tfrac{1}{2}i^{2}\,\mathrm{d}L/\mathrm{d}x\) — the coenergy derivative at constant current.MCQ 2. Reversing the current in an electromagnet causes the force to:
(a) reverse (b) halve (c) stay the same (d) become zeroShow answer
(c) stay the same, since it depends on \(i^{2}\).MCQ 3. The force in a singly excited system always acts to:
(a) increase reluctance (b) reduce reluctance (c) reduce inductance (d) reduce fluxShow answer
(b) reduce reluctance — equivalently, towards maximum inductance.MCQ 4. The pull across an air gap of area \(A\) is:
(a) \(BA/\mu_0\) (b) \(B^{2}A/\mu_0\) (c) \(B^{2}A/2\mu_0\) (d) \(B^{2}/2\mu_0 A\)Show answer
(c) \(B^{2}A/2\mu_0\), per gap.MCQ 5. Magnetic pressure is numerically equal to:
(a) the flux density (b) the gap energy density (c) the reluctance (d) the permeabilityShow answer
(b) the gap energy density. Both are \(B^{2}/2\mu_0\) and share the same units.MCQ 6. For a gap-type actuator the force varies with gap as:
(a) \(g\) (b) \(1/g\) (c) \(1/g^{2}\) (d) independent of \(g\)Show answer
(c) \(1/g^{2}\) at constant current, which is why relays snap shut.MCQ 7. In \(L(\theta) = L_0 + L_2\cos2\theta\), the torque depends on:
(a) \(L_0\) only (b) \(L_2\) only (c) both equally (d) neitherShow answer
(b) \(L_2\) only. A constant inductance has zero derivative and produces no torque.MCQ 8. Reluctance torque is maximum when the rotor is at:
(a) 0° (b) 45° (c) 90° (d) 180°Show answer
(b) 45°, where \(\sin2\theta = 1\) and the inductance is changing fastest.MCQ 9. A relay needs more current to pull in than to hold because:
(a) the coil heats up (b) the gap is larger when open (c) the flux reverses (d) the contacts add resistanceShow answer
(b) the gap is larger when open, so more mmf is needed for the same flux density.MCQ 10. The practical ceiling on magnetic pressure in a machine is set by:
(a) the supply voltage (b) the number of turns (c) saturation of the iron (d) the winding resistanceShow answer
(c) saturation of the iron, limiting \(B\) to about 2 T and the pressure to roughly 1.3 MN/m².
Derive both force expressions from the energy balance, explaining why the signs differ and why each holds a different variable constant.
Explain why a singly excited system can only attract and never repel, and what would be needed to change this.
Show that \(f = -\tfrac{1}{2}\Phi^{2}\,\mathrm{d}S/\mathrm{d}x\) follows from the coenergy result, and use it to predict the direction of force in three different devices.
Derive the magnetic pull formula from the reluctance form, and explain why magnetic pressure equals the gap energy density.
Explain how saturation of the iron sets a limit on the size of electrical machines, and why this limit cannot be evaded by better control.
Contrast a gap-type actuator with a proportional one, and explain how a designer achieves constant force over a stroke.
Explain why a reluctance motor produces no average torque unless it runs in synchronism, and what this implies for starting it.
A singly excited system can pull but not push, and its force depends on \(i^{2}\) alone. Chapter 17 adds a second winding. In a doubly excited system the coenergy gains a mutual term \(M(\theta)i_1i_2\), and with it a torque component proportional to the product of the two currents — which can reverse, and which does not require saliency at all.
That mutual torque is the mechanism of every DC and synchronous machine in this book, and the reluctance torque of this chapter survives alongside it as a second component wherever the rotor is salient. Chapter 18 assembles both into the general force and torque expressions and shows how the two divide in a real machine. The orphaned material on torque by field alignment, which your project already contains, belongs there. From Chapter 19 the treatment turns to windings distributed in slots, and Part 2 begins.