Magnetic Circuits with Air Gaps

Count the gaps before doing anything else. The flux path leaves the rotor, crosses into the stator, travels round the stator yoke and re-enters the rotor — so it crosses two gaps in series, of total length \(2g\). By symmetry the flux density in each is the same.

Only the gaps are reluctant. With \(\mu_r = \infty\) the iron contributes nothing, so the entire magnetic circuit is two gaps in series:

Remember the conversions: \(2000\ \text{cm}^2 = 0.2\ \text{m}^2\) and \(2g = 0.02\ \text{m}\).

The flux is then the mmf divided by that reluctance:

The gap flux density is best taken from the unrounded flux, or better still written directly so that the area cancels:

Rounding \(\Phi\) to 0.13 Wb before dividing would give 0.65 T, an error of 3.5% introduced purely by the arithmetic. Carry full precision to the last step.

Sanity check on the number. Machine gap densities sit between about 0.5 and 1.0 T — high enough to give useful torque, low enough to keep the stator teeth out of saturation. 0.628 T is squarely in that range, and the 10 A field current producing it is likewise realistic for a winding of 1000 turns.

Draw the equivalent circuit. The winding is a source of mmf \(\mathcal{F} = Ni\); the iron is a perfect conductor of flux; the two gaps therefore hang directly across the source, in parallel:

The total reluctance is the parallel combination of the two gap reluctances, and the total flux follows:

The inductance is the flux linkage per ampere. Substituting the two reluctances gives a result in the physical dimensions alone:

The two terms are simply the inductances the gaps would have separately: parallel reluctances mean inductances in addition, \(L = L_1 + L_2\). Adding a second gap path therefore increases the inductance, whereas putting a gap in series with the first would reduce it.

The density in gap 1 alone. Because the branches are in parallel, gap 1 sees the full mmf \(Ni\) regardless of what the other branch is doing:

Note what has cancelled: \(B_1\) depends on \(g_1\) only — not on \(A_1\), and not on \(g_2\) or \(A_2\) at all. Halving \(g_1\) doubles \(B_1\); widening gap 2 changes it not at all.

Check the flux sums. Adding the branch fluxes must return the total:

and dividing by \(i\) after multiplying by \(N\) returns the inductance found above — the two parts of the problem are consistent.

Inductance from the total reluctance. The two reluctances are in series, and the flux linkage per ampere is:

The gap reluctance is 117 times the core's, so to a good approximation \(L \approx N^2/\mathcal{R}_g\). Close the gap and the same winding would measure about 66 H — and would be hopelessly nonlinear.

Stored energy at the stated operating point, using the current found in Set 1, Problem 5:

Since the reluctance — and hence the mmf — is 99% gap, so is the energy. An air gap is where a magnetic circuit stores what it holds, which is why every inductor meant to store energy has one.

The induced voltage, from Faraday's law. The area is constant, so only the density is differentiated:

Check against the transformer emf equation, which Part 3 uses throughout. The rms value of that voltage is:

The two agree because \(4.44 = 2\pi/\sqrt2\) — the same calculation with the differentiation already performed.

Change of method, and why. A reluctance \(l/\mu_0\mu_r A\) presupposes a single \(\mu_r\); a magnetisation curve says there is none. So work in mmf drops instead, using Ampère's law around the loop:

Read the core field strength off the curve at the required density. At \(B_c = 1.0\) T the figure gives:

The gap needs no curve. Air is linear, and with no fringing the gap carries the same density as the core, \(B_g = B_c = 1.0\) T:

Add the drops and divide by the turns:

The gap takes 99.2% of the ampere-turns. Setting \(\mathcal{F}_c = 0\) outright would have given 0.796 A — the same answer to two figures.

Reconcile with Problem 3. The curve reading implies an effective relative permeability at this point of

which is the 70,000 assumed in Problem 3, and the current is the same 0.80 A. The two methods are consistent because 1.0 T lies on the straight part of this material's curve; push to 1.8 T and \(H_c\) would rise by orders of magnitude while the gap drop only doubled.

The two reluctances. With \(\mu \to \infty\) the iron vanishes from the problem, and with no fringing the gap area is the core area:

The whole magnetic circuit is now this one number.

Flux, flux linkage and inductance at \(i = 1.5\) A:

Confirmed independently by \(L = N^2/\mathcal{R}_g = 83^2/1.017\times10^{6} = 6.78\ \text{mH}\), which also shows that \(L\) does not depend on the current — the point of gapping the core.

Turns for a specified inductance. Since \(L = N^2/\mathcal{R}_g\) and the reluctance is unaffected by the winding:

Equivalently \(83\sqrt{12/6.78} = 110\): raising the inductance by 77% needs only 33% more turns, because of the square law.

The current that saturates the core to 1.0 T. With no fringing \(B_g = B_c\), so the flux is fixed by the core area, and the current follows from the inductance:

This is the rating of the inductor: beyond about 16.5 A the real core would leave its linear region and \(L\) would collapse. (Rounding 110.5 down to 110 turns actually gives \(L = 11.9\) mH and a saturation current of 16.6 A; the design value 16.5 A quoted at the nominal 12 mH is the conservative figure.)

The driving mmf, common to both outer branches:

aWith \(\mu_r = \infty\), only the gaps count. The iron carries no mmf drop, so each outer limb sees the whole 500 AT across its own gap alone:

Each outer limb independently:

Half the gap, twice the flux: the branches are in parallel across a fixed mmf, so the shorter gap takes proportionally more.

The central limb carries the sum, by continuity of flux at the junction:

bWith \(\mu_r = 4500\) the iron must be added as reluctance in series within each branch. The analogous electrical circuit is shown below; the iron path lengths come from the figure's dimensions:

The iron reluctances:

Each is roughly an eighth of the gap it sits with — small, but no longer negligible.

Reduce the network. Each branch is a series pair; the two branches are in parallel; the central limb is in series with the combination:

Total flux and central-limb density:

Divide the flux between the branches, putting the opposite branch reluctance on top:

Check: \(0.653+1.168 = 1.821\ \text{mWb} = \Phi\) ✓

What the iron cost, side by side:

The short-gap branch suffers most, because its iron reluctance is a larger fraction of its total. Assuming perfect iron overstates the flux by nearly a third here — far more than in Problem 1, where the gap was proportionally much longer.

The gap reluctance, with \(A_g = A_c = 2\times2 = 4\ \text{cm}^2 = 4\times10^{-4}\ \text{m}^2\):

The iron of the centre limb, of mean length 51.5 cm excluding the gap, with \(\mu_c = \mu_0\mu_r\):

Half a centimetre of air is twelve times more reluctant than half a metre of this iron. From symmetry the two outer paths are identical, \(\mathcal{R}_{bafe} = \mathcal{R}_{bcde} = 2.58\times10^{6}\ \text{At/Wb}\), corresponding to a mean length of about 156 cm each.

Write the two loop equations. Each loop contains its own outer path plus the shared centre limb, and because the two fluxes reinforce in that limb, the shared reluctance appears in both equations with the same sign:

This is mesh analysis with flux for current, mmf for voltage and reluctance for resistance — the magnetic circuit analogy carried as far as it goes.

Substitute the numbers, in At/Wb and A·t:

Symmetry solves the pair by inspection. The coefficient matrix is symmetric and the two right-hand sides are equal, so \(\Phi_1 = \Phi_2\) and the equations collapse to one:

The gap carries both fluxes:

Check the mmf balance round one loop, which must return the 5000 A·t the winding supplies:

The 5 mm gap absorbs 82% of the available ampere-turns even though it is under 1% of the flux path. Note also that \(H_g\) is 1200 times the field intensity in the iron at the same density — the whole story of this set in one ratio.