Energy, Coenergy, Force and Torque

Get \(L(x)\) first; everything else is a derivative of it. With perfect iron only the two gaps count:

Operating point. Flux, density, linkage and stored energy:

Route 1 — constant flux linkage. Express the stored energy in terms of \(\lambda\) and \(x\), which is the correct pair of independent variables for this derivative:

The minus sign says the force acts to reduce \(x\) — it pulls the armature closed. At constant \(\lambda\) no emf can appear, so no energy enters from the source, and the mechanical work comes entirely out of the field: the stored energy falls as the armature moves.

Route 2 — constant current. Now the coenergy is the right function, and for a linear device it equals the energy:

Identical, as it must be. At constant current the source supplies \(i\,d\lambda\), half of which goes into the growing field and half into mechanical work — the fact Problem 3 exploits.

Route 3 — the pole-face check. Each gap face pulls with \(B_g^2A_g/2\mu_0\), and there are two of them:

Three derivations, one number. The pole-face form is the quickest when the gap density is already known, and it makes the physics visible: force per unit area of gap is \(B^2/2\mu_0\), which at 0.63 T is 157 kPa — about 1.6 atmospheres.

Closing to 1.0 mm at the same current:

Four times the pull for half the gap. This is why relays snap shut rather than closing smoothly, and why the sealed force of a contactor is far above its pick-up force.

Work backwards from the force. The load is shared by two pole faces:

A working density, well below the knee, so the iron will not be the limitation. Note how little density is needed: \(B^2/2\mu_0\) at 0.70 T is 196 kPa, and the two pole faces together, 0.04 m², then carry \((1.962\times10^{5})(0.04) = 7848\) N — the 800 kg required.

Ampere-turns for the two gaps:

Ampere-turns for the iron, which here is not negligible because \(\mu_r\) is only 2000 and the path is 0.80 m long:

The iron takes 29% of the ampere-turns. Two 0.5 mm gaps are worth \(\mu_r \times 1.0\ \text{mm} = 2.0\) m of this iron, and the path is only 0.80 m long, which is why the split is not the lopsided one of Set 3.

Safety factor of 2. Force goes as \(B^2\) and, in this linear circuit, \(B\) goes as \(i\), so the current scales as \(\sqrt{2}\):

Doubling the pull costs only 41% more current, but 100% more \(I^2R\) loss in the coil. A lifting magnet is thermally limited, not magnetically.

If the residual gap doubles. At constant current the mmf is fixed, and with the iron drop rescaled the gap density falls. Working with reluctances, the gap term doubles:

The pull collapses to 2671 N — 34% of the original, and far below the 7848 N needed. Half a millimetre of scale, paint or rust is the difference between holding the load and dropping it, which is why lifting magnets are rated against a stated surface condition.

The inductance at the two positions. Two gaps of length \(x\) in series:

Stage 1 — energising at fixed position. Nothing moves, so every joule goes into the field:

Stage 2 — the stroke, at constant current. The source delivers \(i\,d\lambda\), and with \(i\) constant that integrates trivially:

Exactly half the electrical input became mechanical work and exactly half went into the field. That 50:50 split is a property of every magnetically linear device moved at constant current, and it is worth remembering as a check rather than rederiving each time.

Stages 3 and 4 — recovery and reset. At fixed position and falling current the stored energy returns to the supply (or is burned in the discharge resistor, which does not change the balance sheet):

Stage 4 moves the armature with no flux and no current, so all three quantities are zero.

The balance sheet for the cycle:

The field returns to its starting state, so its column must total zero — and it does. The net electrical input over the cycle equals the mechanical work delivered, which is what "lossless coupling field" means. Copper loss and the loss in the discharge path sit outside this accounting.

Average against instantaneous force. The average over the 4 mm stroke follows from the work:

The instantaneous value comes from \(f = \tfrac12 i^2 dL/dx\) with \(L = C/2x\) and \(C = \mu_0N^2A = 3.217\times10^{-4}\):

The integral reproduces the mechanical work exactly, which closes the argument. Note how badly the average represents the stroke: 7.2 N at pick-up, 181 N when sealed. The relay's return spring must be beaten at the open position, where the magnet is weakest.

Operating point. At \(x = 0.010\) m, \(k = 0.012/0.010 = 1.2\), so

The characteristic is visibly saturating: at 1 A it gives 0.80 Wb-t, so quadrupling the current has only doubled the linkage.

Coenergy is the integral under the curve, taken along \(i\):

The partial-fraction step \(i/(1+0.5i) = 2 - 4/(i+2)\) is worth checking by multiplying back out; it turns an awkward integral into two standard ones.

The stored energy is the complementary area, and the rectangle identity gives it without a second integration:

This is the practical reason for keeping both quantities in play. Integrating \(\int i\,d\lambda\) directly would require inverting \(\lambda(i)\) first; subtracting from the rectangle does not.

What a linear model would have said. An engineer who took \(L = \lambda/I = 1.60/4 = 0.40\) H and applied \(\tfrac12 Li^2\) would get 3.20 J for both quantities:

The sum is right because it is fixed by the operating point alone; the split is wrong in both directions. Since force follows the coenergy, the linear model would understate the pull by 26% — and it errs on the unsafe side for an energy-storage question and the safe side for a force question, which is exactly the sort of asymmetry worth knowing about.

Force, from the coenergy at constant current. The current is held while \(x\) varies, so only \(k(x)\) is differentiated:

Negative, so the plunger is pulled in — the direction that raises \(k\) and therefore the inductance, exactly as in the linear case. The magnitude is 433 N.

Write the general torque and see which terms survive. Every inductance that depends on \(\theta\) contributes:

\(L_{22}\) is constant — a cylindrical rotor sees the same magnetic circuit at every angle — so the middle term vanishes and

At \(\theta = 45^\circ\), where \(\sin2\theta = 1\) and \(\sin\theta = 0.7071\):

Both components are negative, so both act to reduce \(\theta\) — to drag the rotor back towards alignment. The reluctance term is more than twice the excitation term here, because the stator current is large and enters squared.

At \(\theta = 90^\circ\): \(\sin 180^\circ = 0\), so the reluctance term disappears altogether while the excitation term is at its largest:

The two mechanisms peak at different angles — the reluctance torque at 45° and 135°, the excitation torque at 90° — because one varies as \(\sin2\theta\) and the other as \(\sin\theta\). That offset is why a salient-pole synchronous machine has the two-term power-angle characteristic of Set 37.

Rotor unexcited. With \(i_2 = 0\) only the reluctance term remains:

The machine still develops torque with one winding dead. This is the reluctance motor, and it is also why a synchronous machine with a failed field can still pull into step at reduced torque.

Equilibrium positions. Setting \(T = 0\):

Stability requires the torque to oppose a small displacement, that is \(dT/d\theta < 0\):

Two stable rest positions per revolution, not one. The saliency repeats every 180° and by itself would make 0° and 180° equally good; the excitation torque, which repeats only every 360°, biases the pair without destroying either. Writing \(T = -a\sin2\theta - b\sin\theta\), the position at 180° survives as long as \(b < 2a\) — here \(4 < 12\). Raise the rotor current until \(b > 2a\) and 180° becomes unstable, leaving the single aligned rest position expected of a machine whose torque is dominated by excitation.

Direct current first. With one winding only:

Zero at \(\theta = 0\) and 90° — the aligned and the fully unaligned positions, where the inductance is stationary. Maximum halfway between, where it changes fastest.

Note what the torque does not depend on. The current enters as \(i^2\), so reversing it changes nothing. A reluctance machine therefore runs just as well on alternating current — which is the whole of part (b).

Alternating current at synchronous speed. Put \(i = \sqrt2\,I\sin\omega t\) and \(\theta = \omega t + \delta\):

Two products to average. The first, \(\sin(2\omega t+2\delta)\), averages to zero over a cycle. The second is the one that survives.

Take the average, using \(\left\langle\cos A\,\sin(A+B)\right\rangle = \tfrac12\sin B\):

The factor of one half is the point of the problem. Substituting the rms value into the dc expression would give 2.88 N·m and be wrong by a factor of two, because the torque follows \(i^2\), which pulsates, and only the component of that pulsation at \(2\omega\) beats against the rotor position to leave a steady term.

Maximum average torque and the load angle:

This is the pull-out torque. Load the machine beyond it and no value of \(\delta\) can supply the demand, the rotor slips a pole and synchronism is lost — the same mechanism, and the same 45°, as the reluctance component of the salient-pole synchronous machine.

Power, and why it will not start. A two-pole machine at 50 Hz runs at 3000 rpm:

At any other speed the derivation fails. If \(\theta = \omega_m t + \delta\) with \(\omega_m \ne \omega\), the term \(\cos2\omega t\,\sin(2\omega_m t + 2\delta)\) is a product of two different frequencies and averages to zero. The machine develops a pulsating torque with no mean value and does not turn.

A practical reluctance motor is therefore fitted with a cage winding and started as an induction motor, pulling into synchronism near full speed — or, in the modern form, fed from an inverter that supplies the stator at whatever frequency the rotor happens to be running at.