B-H curve and magnetic losses

Plot the magnetic flux density $B$ inside an iron core against the field intensity $H$ that drives it, and you get a curve, not a straight line — the resistor's tidy $V$ - $I$ proportionality has no counterpart here. At low $H$ the curve climbs steeply; as $H$ grows the material runs out of "room" to magnetise further and the curve bends over into saturation. Drive the core around a full AC cycle and $B$ traces a closed loop instead of retracing its own path — the material remembers where it has been. That memory costs energy every cycle, and a second, independent loss mechanism (circulating eddy currents) adds to it. Together these two losses are why a transformer sitting on no load, doing no useful work at all, still hums and gets warm. This article builds the B-H curve, both loss laws, and the reasoning that makes lamination the standard fix for one of them but not the other.

The core insight

The non-linear, history-dependent B-H curve makes permeability $\mu=B/H$ vary with operating point, and the same curve's hysteresis loop area plus core eddy currents together fix the frequency- and $B_{max}$ -dependent iron loss $P_i=P_h+P_e$ that a laminated (but never solid) core is built to minimise.

The core idea

Shape of the B-H curve. Start an unmagnetised core from $H=0$ and raise the exciting current steadily. In the initial region $B$ rises fast for a small increase in $H$ — the material is easy to magnetise, and the slope $\mu=B/H$ is large. Past the "knee" the curve flattens: nearly all the magnetic domains inside the material are already aligned, so further increases in $H$ buy only small increases in $B$ . This flat region is saturation. Because $\mu$ is the slope of a chord from the origin to a point on this curve, it is largest in the steep unsaturated region and falls as the operating point moves toward saturation — permeability is a property of the operating point, not a fixed number like resistance.

The hysteresis loop. Cycle the exciting current through a full sine wave and $B$ does not retrace the rising path on the way back down — it lags, tracing a closed loop when plotted against $H$ . The enclosed area is energy dissipated as heat, per unit volume, per cycle; the lag is why the effect is called hysteresis (from the Greek for "lagging behind"). A "fat" loop means a lossy material; transformer and machine cores are built from magnetically "soft" alloys chosen specifically for a thin loop.

Two loss mechanisms, one total. The energy lost per cycle from the loop area scales with peak flux density through the empirical Steinmetz form used in this course,

P_h \propto B_{max}^{1.6}\,f,

where: $P_h$ is hysteresis loss (W), $B_{max}$ is peak flux density, $f$ is supply frequency, and $1.6$ is the (simplified) Steinmetz exponent. Separately, the time-varying flux induces circulating eddy currents in the conducting core itself, dissipating $I^2R$ heat:

P_e \propto B_{max}^{2}\,f^{2}\,t^{2},

where: $P_e$ is eddy-current loss (W) and $t$ is the thickness of the core lamination (sheet). Total core (iron) loss is simply the sum, $P_i=P_h+P_e$ , and because both terms depend only on $B_{max}$ and $f$ , iron loss is essentially fixed once a transformer is running at rated voltage and frequency — it does not change with load current.

Figure

A B-H curve rising steeply then flattening into saturation with a hysteresis loop traced around it

Why it works

The B-H curve and the hysteresis loop are two faces of the same microscopic story. A ferromagnetic material is made of tiny regions (domains) that are each already fully magnetised in some local direction; applying $H$ does not create magnetism from nothing, it rotates and grows the domains that already happen to point along the field while shrinking the rest. Early on, domains that are only slightly misaligned flip easily, so $B$ rises fast for modest $H$ — the steep part of the curve. As $H$ increases, only the stubbornly misaligned domains remain to convert, and eventually every domain is aligned: no amount of extra $H$ can create more flux, and the curve saturates. When $H$ is reduced again, the domain walls that moved do not slide back frictionlessly — some energy was spent overcoming internal "friction" (domain-wall pinning at material defects) and is trapped as heat rather than returned. That irrecoverable energy per cycle is exactly the area enclosed by the hysteresis loop, which is why a "soft" magnetic alloy with easily-moving domain walls (thin loop) is chosen for transformer cores over a "hard" magnet material (fat loop, useful instead for permanent magnets). Eddy-current loss has a completely different, purely electromagnetic origin: the changing flux induces an EMF in the core material itself (Faraday's law), and because the solid iron is a conductor this EMF drives circulating currents that dissipate $I^2R$ heat independent of any hysteresis effect. Cutting the conducting path into thin, mutually insulated laminations does not touch the domain physics of hysteresis at all — it only restricts the loop area available to the induced eddy currents, which is why lamination attacks $P_e$ specifically and leaves $P_h$ unchanged.

Derivation AScaling hysteresis and eddy-current loss when core conditions change

The goal of this derivation is to show, starting only from the two Steinmetz-form proportionalities $P_h \propto B_{max}^{1.6}f$ and $P_e \propto B_{max}^{2}f^{2}t^{2}$ , how to predict a transformer core's new hysteresis loss, new eddy-current loss, and new total iron loss whenever frequency, peak flux density, or lamination thickness is changed one at a time or together, without needing the proportionality constants at all.

Step 1. Write both loss laws as ratios instead of absolute formulas. Because the proportionality constants (which depend on the specific core material, volume and resistivity) are unknown but fixed for a given core, dividing the new-condition value by the old-condition value cancels them exactly:

\frac{P_{h,new}}{P_{h,old}} = \left(\frac{B_{max,new}}{B_{max,old}}\right)^{1.6}\left(\frac{f_{new}}{f_{old}}\right), \qquad \frac{P_{e,new}}{P_{e,old}} = \left(\frac{B_{max,new}}{B_{max,old}}\right)^{2}\left(\frac{f_{new}}{f_{old}}\right)^{2}\left(\frac{t_{new}}{t_{old}}\right)^{2}.

This is the key trick that makes the whole calculation possible from a syllabus that has not given you the material-specific Steinmetz coefficient.

Step 2. Identify which quantities actually changed in the given scenario and set the rest of the ratios to $1$ . If only frequency changes (flux density and lamination held fixed), the hysteresis ratio collapses to $f_{new}/f_{old}$ and the eddy ratio to $(f_{new}/f_{old})^2$ — eddy loss grows faster with frequency than hysteresis loss does. If only lamination thickness changes (flux density and frequency held fixed), the hysteresis ratio is exactly $1$ (hysteresis loss does not depend on $t$ at all) while the eddy ratio is $(t_{new}/t_{old})^2$ .

Step 3. Multiply the old loss by the appropriate ratio to get each new loss, then add: $P_{i,new} = P_h\cdot(\text{ratio}_h) + P_e\cdot(\text{ratio}_e)$ . Because the two mechanisms scale differently, the new total is never a simple uniform scaling of the old total — you must scale each term separately and only then add them, which is the step students most often skip by mistakenly multiplying the old total iron loss by one common ratio. Verified: setting all ratios to 1 recovers $P_{i,new}=P_{i,old}$ , and the two limiting cases (frequency-only, thickness-only) reduce to the single-variable proportionalities stated in Step 1, consistent with the Steinmetz forms.

Worked example

A transformer core has $P_h=250\ \text{W}$ and $P_e=100\ \text{W}$ at rated $B_{max}$ and $f=50\ \text{Hz}$ . The supply frequency is raised to $60\ \text{Hz}$ with $B_{max}$ held the same (the applied voltage is scaled up proportionally too, since $E\propto fB_{max}$ from the transformer EMF equation, but that does not affect this calculation). Find the new hysteresis loss, new eddy-current loss, new total iron loss, and by what factor the total has grown.

Step 1 — hysteresis ratio. With $B_{max}$ unchanged, $P_h\propto f$ , so $P_{h,new}=250\times(60/50)=300\ \text{W}$ .

Step 2 — eddy-current ratio. With $B_{max}$ and $t$ unchanged, $P_e\propto f^2$ , so $P_{e,new}=100\times(60/50)^2=100\times1.44=144\ \text{W}$ .

Step 3 — new total and growth factor. $P_{i,new}=300+144=444\ \text{W}$ , compared with the original $P_{i,old}=250+100=350\ \text{W}$ . The growth factor is $444/350\approx1.27$ , larger than the $60/50=1.2$ ratio of frequencies alone — because the eddy term, growing as $f^2$ , pulls the total up faster than a simple $1.2\times$ scaling would. New hysteresis loss $=300\ \text{W}$ , new eddy-current loss $=144\ \text{W}$ , new total iron loss $=444\ \text{W}$ .

Competitive stretch

Suppose instead the lamination thickness of the same original core (250 W hysteresis, 100 W eddy at 50 Hz) is halved while $B_{max}$ and $f$ stay at their original values. Since $P_e\propto t^2$ , $P_{e,new}=100\times(1/2)^2=25\ \text{W}$ , while $P_h$ is untouched at $250\ \text{W}$ because hysteresis loss has no $t$ dependence. New total iron loss $=250+25=275\ \text{W}$ , a saving of $75\ \text{W}$ from lamination alone, with zero change to the applied voltage or frequency — this is the entire economic case for laminating cores.

Common traps

Treating permeability $\mu=B/H$ as a fixed material constant — it is the slope of a chord on a non-linear curve and is largest in the unsaturated region, falling toward saturation.
Using exponent $2$ for hysteresis loss — only eddy-current loss has $B_{max}^2$ ; hysteresis loss uses the empirical Steinmetz exponent of about $1.6$ .
Assuming lamination reduces hysteresis loss — it reduces only eddy-current loss (via smaller $t$ ); hysteresis loss depends on material, $B_{max}$ and $f$ , not on sheet thickness.
Scaling the total iron loss by one common ratio when frequency or flux density changes — hysteresis and eddy-current loss have different exponents on $f$ ( $1$ versus $2$ ) and must be scaled separately before being added.

Ask the tutor: ask the AI tutor to walk through separating hysteresis and eddy-current loss experimentally from two total-core-loss measurements taken at different frequencies, or to set up another mixed frequency-and-thickness scaling problem.

Why does the hysteresis loop even have an area — shouldn't rising and falling B just retrace the same path?

It would if the domain walls moved without any internal friction, but real domain walls get pinned momentarily at material defects and grain boundaries as they sweep through the material. Unpinning them takes a small burst of energy that is not returned when the field reverses, so the falling- $H$ path lags behind the rising- $H$ path at every point. That lag is exactly what makes $B$ trace a different value for the same $H$ depending on direction, and integrating the enclosed loop area gives the energy lost as heat per cycle per unit volume.

If lamination only fixes eddy-current loss, why not just use a material with zero hysteresis loss and skip lamination entirely?

Because the two losses come from different physics and no single design choice fixes both. Hysteresis loss is minimised by choosing a magnetically soft alloy (like silicon steel) with a thin loop, which real cores already do. But any solid conductor carrying time-varying flux will develop eddy currents regardless of how soft the material is magnetically — that is a consequence of the material being electrically conducting, not of its hysteresis behaviour. So real transformer cores use both fixes together: a low-hysteresis alloy, laminated into thin insulated sheets to choke off eddy currents.

Does core loss depend on the load current drawn from the transformer secondary?

No, and this is the key fact used later in efficiency and open-circuit testing. Both $P_h$ and $P_e$ depend only on $B_{max}$ and $f$ , which are fixed once the transformer is connected to its rated supply voltage and frequency — the flux in the core is set by the applied voltage through the EMF equation, essentially independent of how much current flows in the secondary winding. That is why iron loss is called a constant loss: it is present even at no load, unlike copper loss (I squared R in the windings) which grows with load current.