advanced32 min studyLesson 250 in Physics

Hysteresis and magnetic work cycles

Interpret a material's magnetic history and compute cycle work from a B–H loop with the correct density units and sign convention.

Before you begin

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The question for today

What you’ll understand

Distinguish remanence, coercivity, and differential response on a hysteresis loop.
Calculate magnetic work density and cycle loss from an oriented B–H path.
Separate reversible stored energy from hysteresis and frequency-dependent eddy-current loss.

A magnetic material can return to the same applied field with a different magnetization because its internal domain configuration remembers the route taken. A B–H loop records this history in a useful macroscopic form. The area represents work per volume per cycle under specified magnetic and thermodynamic assumptions. It is not a permeability, a total energy without a volume factor, or proof that every measured loss mechanism is independent of drive frequency.

One field value does not select a state

In a ferromagnet, domain-wall motion and moment rotation encounter energetic barriers and pinning. Decreasing the drive after magnetization can therefore follow a different branch from increasing it. Remanent B is the flux density left when H returns to zero on the selected loop. B coercivity is the reverse H needed to bring B to zero. The reverse field needed to make M zero can differ, so measurements must specify which coercive quantity is reported.

Saturation means the magnetization approaches its available aligned moment; B can still increase through its vacuum-permeability H contribution. A secant ratio B divided by H and a differential slope along a branch answer different questions, and neither uniquely describes a whole hysteresis loop. Minor loops depend on the previous reversal points. A measured graph should therefore identify its drive amplitude, temperature, direction, and frequency rather than being treated as a universal function B of H.

B=\mu_0(H+M),\qquad \mu_{\mathrm{sec}}=\frac{B}{H},\qquad \mu_{\mathrm{diff}}=\frac{dB}{dH}\ \text{along a specified branch}

B is in T and H and M are in A/m, so both permeability measures have units H/m. This scalar notation assumes collinear fields. The secant ratio is undefined at H=0 and a branch slope need not specify history or reversibility.↔ Scroll wide equations horizontally.

A work differential becomes an oriented loop area

For a fixed magnetic body with no mechanical work, the magnetic work supplied per unit volume is H times the increment of B in the collinear quasistatic description. Integrating it along a reversible single-valued branch yields a stored-energy change. In a linear reversible material this reduces to one half BH. For a nonlinear reversible relation the integral must actually be evaluated; one half of the endpoint product generally gives the wrong energy.

Over a periodic hysteresis cycle returning the material to the same macroscopic state, the net work becomes dissipated energy when no other net storage or work channel remains. Its positive value is the oriented integral of H dB for the physical traversal. On a graph with H horizontal and B vertical, a counterclockwise traversal gives positive supplied work. Taking an unsigned geometric area is acceptable for loss magnitude only after the coordinate and traversal conventions have been identified.

dw_{\mathrm{mag}}=H\,dB,\qquad w_{\mathrm{loss}}=\oint H\,dB=\mu_0\oint H\,dM,\qquad P=fVw_{\mathrm{loss}}

Work density w is in J/m³ because (A/m)·T=J/m³. Cycle loss becomes total energy after multiplication by sample volume V in m³, and power after multiplying by frequency f in Hz. The equality removes the zero closed-cycle integral of μ₀H dH; the loss interpretation assumes a closed material cycle and no net mechanical work.↔ Scroll wide equations horizontally.

What a measured loop does and does not isolate

The integral of B dH has the opposite sign because the closed-cycle change of BH is zero. Mixing these two work conventions without changing a sign can turn a lossy material into an apparent energy source. Similarly, plotting magnetization rather than B changes the vertical units and requires the vacuum-permeability factor. Numerically estimating a loop requires ordered points around the cycle, not a scatter plot whose points have lost their time order.

A very slow loop can isolate approximately rate-independent hysteresis, but a finite-frequency measured loop may include eddy-current and other dynamic losses. These depend on conductivity, sample dimensions, and drive conditions, so multiplying one measured area by an unrelated frequency is not a generally valid extrapolation. When using frequency times loop area, assume the area is appropriate to that frequency. Soft magnetic cores seek small losses under their operating cycle, whereas useful permanent magnets need resistance to reversal as well as remanent magnetization.

Work it through

From loop area to core heating

A hypothetical collinear core follows a counterclockwise rectangular B–H loop spanning H=−100 to +100 A/m and B=−0.80 to +0.80 T. Its volume is $2.0\times10^{-5}$ m³. Find loss density, energy per cycle, and power at 50 Hz, assuming this same loop applies at that frequency.

The horizontal width is 200 A/m and the vertical height is 1.60 T. Multiplying gives loop area $w_{\mathrm{loss}}=320\,\mathrm{J/m^3}$ . On horizontal branches dB is zero; the two vertical branches each supply positive work for the stated counterclockwise traversal.
Multiply by the actual magnetic volume: $W_{\mathrm{cycle}}=320(2.0\times10^{-5})=0.0064$ J. The loop area itself was an energy density, so omitting this volume would give the wrong physical quantity and units.
At 50 cycles per second, $P=50(0.0064)=0.32$ W. This assumes a periodic material cycle and a frequency-appropriate loop, with the net work converted to heat rather than mechanical motion or accumulating stored energy.

The result

The cycle loss density is 320 J/m³, the core loses 6.4 mJ per cycle, and its average heating power is 0.32 W under the specified 50 Hz loop assumption.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Integrate a reversible nonlinear branch

Use the lab's y=a x² only as a scaled reversible constitutive segment H=B² for B≥0. Reset and set both controls to Coefficient a=1 and Position x=1; predict area 1/3. For a fresh comparison, Reset and set both controls to a=1 and x=2; predict area 8/3, eight times larger. These are independent evaluations of a single-valued branch, not a continuous magnetic drive protocol.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNonlinear branch work is an integral, not generally half the endpoint product. This memoryless quadratic lab has no hysteresis state, closed-loop area readout, coercivity, or magnetic loss model.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The product of the axis units is T A/m=J/m³. Multiplying by volume gives energy per cycle; multiplying the density by the frequency appropriate to that loop gives power density. The area is not a permeability or a frequency-independent power.

Your turn

Try it yourself

A reversible collinear material follows $H=aB+bB^3$ from B=0 to 0.50 T, with $a=200\,\mathrm{A/(m\,T)}$ and $b=400\,\mathrm{A/(m\,T^3)}$ . Compute supplied work density and net work if the same branch is retraced to zero.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate H with respect to B term by term. Reversible return on the same branch reverses the integration limits.

Compare with the solution

The outward work density is $aB^2/2+bB^4/4=200(0.50)^2/2+400(0.50)^4/4=25+6.25=31.25\,\mathrm{J/m^3}$ . The endpoint H is 150 A/m, so one half BH would instead give 37.5 J/m³ and is not correct for this nonlinear law. Retracing the identical branch gives −31.25 J/m³, leaving zero net cycle work. Nonlinearity alone therefore does not imply hysteresis loss; different outgoing and returning branches are needed for a nonzero enclosed area.

Keep the idea

Take it with you

Hysteresis requires a specified history and cannot be summarized by one constant permeability.
The physical H dB loop integral gives cycle loss density, with units J/m³.
Convert area to energy using volume and to power using the frequency appropriate to the measured loop.

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