Series and Parallel Magnetic Circuits

One flux, one flux density. The three sections are in series, so each carries the whole 2.4 mWb; and because the area happens to be the same in all three, the density is the same as well:

The equality of the three densities is a consequence of the equal areas, not of the series connection. Problem 2 removes it.

Read \(H\) for each material at 1.2 T. Two materials at the same density need very different field strengths, which is the whole point of a composite core:

The cast-steel section is three times shorter and still absorbs almost as much mmf, because it is nearly three times harder to magnetise.

The gap takes no curve. Air is linear and, with fringing neglected, carries the same density as the iron:

Add the drops and divide by the turns. Series sections add their mmf, exactly as series resistors add their voltage drops:

Effective permeabilities, for comparison only. Each material has an apparent \(\mu_r\) at this density:

Both are numbers valid at 1.2 T and nowhere else. Quoting them as if they were material constants is the error Problem 5 punishes.

The flux is common, the density is not. Nothing leaves the core, so all three sections carry 2.0 mWb; the density follows the area:

This is the step with no electrical counterpart. A current is the same in every series resistor; a current density is not, and it is flux density, not flux, that the magnetisation curve is drawn against.

Field strengths from the table of Problem 1. The value at 1.25 T lies between the 1.2 T and 1.3 T entries and is interpolated linearly:

One mmf drop per section, then add:

Section A is 40% of the path and takes 62% of the mmf, purely because it was made narrow.

Reluctances follow by division, since every section carries the same flux and \(\mathcal{R}_k = \mathcal{F}_k/\Phi\):

Inductance at this point only:

Equivalently \(L = N\Phi/I = (400)(2.0\times10^{-3})/0.805 = 0.994\) H. Push the flux to 2.5 mWb and section A reaches 1.56 T, where \(H\) is about five times larger; the reluctance climbs and \(L\) falls. An iron-cored inductance is a number attached to an operating point.

Recognise the topology. The coil drives flux up the central limb; at the top yoke the flux has two identical routes back to the bottom of the central limb. So the circuit is one reluctance in series with two equal reluctances in parallel — the magnetic copy of a resistor feeding two equal resistors.

Compute the three reluctances. With \(\mu_0\mu_r = (4\pi\times10^{-7})(1200) = 1.508\times10^{-3}\):

Combine, remembering which rule applies where. The two outer paths sit across the same pair of yoke nodes, so they are in parallel; being identical, their combination is simply half of one:

Flux from the magnetic Ohm's law:

Symmetry does the division: each outer limb takes exactly half, \(\Phi_o = 1.034\) mWb. No divider formula is needed when the two branches are identical.

Flux densities — and a deliberate piece of design. The outer limbs have half the area and half the flux:

Equal densities throughout. Halving the area of a limb that carries half the flux is how a real three-limb core is proportioned, so that no part saturates before the rest.

How the 400 AT divides, and the check that it must sum:

\(114.3 + 285.7 = 400\) AT. The same drop of 285.7 AT appears across both outer paths, which is what "in parallel" means magnetically.

Inductance:

Set up the analogy explicitly, because the whole problem is a current divider in disguise:

Reluctance of each path. With \(\mu_0\mu_r = 1.885\times10^{-3}\) and \(A = 40\times10^{-4}\ \text{m}^2\), the iron denominator is \(\mu_0\mu_r A = 7.540\times10^{-6}\):

Path C is iron in series with air:

Half a millimetre of air is worth \(\mu_r g = 1500 \times 0.5\) mm, that is 0.75 metres, of this iron. That single line is why the flux will not divide equally.

Reduce and find the total flux:

Divide it — with the opposite branch on top:

Check: \(2.812+0.979 = 3.791\) mWb, and the ratio \(\Phi_B/\Phi_C = \mathcal{R}_C/\mathcal{R}_B = 2.87\). The gapped limb takes the smaller share, as the higher-resistance branch of a current divider does.

Densities and the mmf check. All three limbs have the same area, so the densities are in the same ratio as the fluxes:

The mmf drops confirm the reduction:

And the same 149.2 AT drives both branches: \(149.2/5.305\times10^{4} = 2.81\) mWb and \(149.2/1.525\times10^{5} = 0.979\) mWb. Either route — divider formula or common mmf — gives the same answer, and computing both is the cheapest available check.

Why the direction matters. Problems 1 and 2 gave the flux and asked for the ampere-turns, so every step ran forward: \(\Phi \to B \to H \to Hl\). Here the ampere-turns are given and the flux is wanted, which needs \(H\) inverted — and \(H(B)\) is a table, not a formula. So the equation to be satisfied is written first and then solved numerically:

The gap coefficient is \(g/\mu_0 = 0.8\times10^{-3}/4\pi\times10^{-7} = 636.6\) AT per tesla — a straight line. Only the first term bends.

Test the perfect-iron shortcut first, because when it works there is no need to iterate at all. Setting \(H_c = 0\):

No electrical steel reaches 2.51 T. The shortcut has failed, and it has failed in the informative way: the answer it returns is above saturation, which is precisely the condition under which the iron cannot be ignored.

First trial — guess a density near the knee. Take \(B = 1.40\) T, for which the table gives \(H_c = 1000\) A/m:

More than the 1600 AT available, so 1.40 T is too high.

Second trial — drop to \(B = 1.35\) T. Interpolating the table between 1.3 T (600 A/m) and 1.4 T (1000 A/m) gives \(H_c = 800\) A/m:

Too low. The answer is bracketed between 1.35 T and 1.40 T.

Interpolate between the two trials, treating \(\mathcal{F}(B)\) as locally straight:

Verify — never skip this step. At \(B = 1.389\) T the table interpolation gives \(H_c = 600 + 0.89(400) = 956\) A/m:

The trial reproduces 1601 AT against the 1600 AT available — an error of 0.06%, far inside the accuracy with which the curve itself is known. The flux follows:

Read the split, and see how the balance has shifted. The iron now takes \(717/1600 = 44.8\%\) of the ampere-turns, against 32.5% in Problem 1 and under 1% in the lightly worked cores of Set 2. The reason is the shape of the curve: from 1.2 T to 1.39 T the gap requirement rises by 16% while the iron requirement rises by 139%.

Beyond the knee the iron stops being a detail and starts being the limit. Raising the current by another 25% here would gain under 5% more flux.

Separate the two fluxes before doing anything else. The leakage coefficient is defined at the coil, and it multiplies the useful flux to give the flux the iron under the winding must actually carry:

The leakage flux takes a short cut through the air between the two limbs. It never reaches the gaps, so it does no useful work — but it does magnetise the iron it passes through.

Flux density part by part. Same area everywhere, two different fluxes:

Twenty per cent more flux in the yoke, and the cast-steel curve turns that into 53% more field strength — the effect of leakage is amplified by the nonlinearity.

The two gaps. Air is linear, and there are two of them in series on the path:

The two iron parts, each at its own density. From the cast-steel row of the Problem 1 table, \(H = 720\) A/m at 1.0 T and 1100 A/m at 1.2 T:

Total and current:

What ignoring leakage would have cost. Setting \(\lambda_L = 1\) puts the yoke at 1.0 T, where \(H = 720\) A/m:

An 8.3% underestimate of the current, and therefore of the copper loss and the temperature rise. Built to that figure, the magnet would not develop its rated pull.

A note on what the model assumes. Applying a single coefficient of 1.20 to the whole yoke-and-limb path treats the leakage as though it all left at the pole tips. In reality it leaves progressively along the limbs, so the flux — and the mmf drop — is largest near the coil and smallest near the poles. The lumped coefficient therefore slightly overstates the iron drop, which is the safe direction to err. Values between 1.15 and 1.25 are usual for lifting magnets and dc-machine poles; the figure comes from measurement or from a field solution, never from this calculation.