Domains, hysteresis and magnetic losses

Field history changes the state

Magnetic domains can grow or rotate as applied field changes. Obstacles to domain-wall motion and other processes create path dependence. A hysteresis loop records that the response on increasing field differs from the return path.

Consider a specimen at the same temperature and the same zero applied field on two visits. It can retain a positive response after a positive excursion and a negative response after a negative one. Field value alone has not specified the state: the path matters. We will first see how a domain can grow, then follow an explicitly supplied loop. Every numerical record below is an invented teaching case, not a measured specification of a named magnetic material.

Changing domain volume changes the net moment

A ferromagnetic domain is a region with a substantially common magnetisation direction. A specimen can contain differently oriented domains whose contributions partly cancel, even though each remains locally ordered. Magnetic moments and collective order explains that distinction. A domain wall is a transition region where the magnetisation direction changes; it is not an empty crack separating two rigid blocks.

An applied field can favour one orientation. As a wall moves, a region previously belonging to one domain adopts the neighbouring orientation. The favoured domain grows at the expense of the other. The atoms do not migrate across the specimen with the wall. Magnetisation can also rotate within a domain; nucleation of new domains and disappearance of old walls can participate in reversal.

Use a separate two-orientation account. Each domain has the same local magnetisation magnitude M_s and points exactly along +x or −x. Let f be the fraction of specimen volume pointing +x. The volume average is M_x = f M_s − (1 − f)M_s = (2f − 1)M_s. This is a signed sum, not a count of magnetic atoms appearing or disappearing. If M_s = 800 kA/m, changing f from 1/2 to 3/4 changes M_x from zero to +400 kA/m while M_s remains unchanged.

The first diagram draws equal-volume cells only to make this account visible. Common depth and equal face areas make the drawn area fractions equal the assumed volume fractions. Real walls have finite structure, and real domains need not have these two directions or this geometry. The account alone supplies neither a coercive field nor a loop: a rule for motion and reversal is still missing.

A return path need not undo the earlier motion

Domain structures reflect competing energies. Alignment within a domain, preferred crystal directions, fields produced by the specimen and elastic strain can favour different arrangements; making more walls also has an energy cost. This is why a large specimen is not necessarily one uniformly magnetised domain at zero applied field. Shape, texture and stress are part of the conditions.

An inclusion, a boundary or a local change in anisotropy or stress can make some wall positions energetically preferable. A wall may bend reversibly near a preferred position, then move irreversibly when the drive allows it to leave that region. On reducing the field it need not follow the same sequence in reverse. This provides one physical route to pinning and memory. It does not mean every hysteresis loop comes only from pinned walls: nucleation and rotation barriers can matter too.

Conditioning specifies how a measurement approaches its repeated cycle. Repeated excursions between stated turning fields may eventually give the same closed path; an initial magnetisation trace or an incompletely settled first cycle need not coincide with it. A smaller reversal can trace a minor loop. Do not draw its missing return path merely by cutting a piece out of a larger loop.

Temperature, field-change rate and waiting time can change the opportunity to cross barriers. A measured loop change is therefore not automatically a count of defects. Compare geometry, direction, temperature, prior treatment, turning fields, waveform and timing before attributing a change to pinning.

Loop quantities require definitions

Magnetisation M describes magnetic moment per volume, H is magnetic field strength, and B is magnetic flux density or induction. Remanence describes retained M or B at a specified zero-field condition; coercivity describes a reverse H associated with a specified zero M or B. Those two zero-response conditions generally differ, so identify whether the loop is B–H or M–H. Loop area relates to dissipated energy with the appropriate variables and units.

In SI, M and H are measured in A/m; B is measured in teslas (T). One kA/m is 1000 A/m. Here M is magnetic moment per volume, not magnetic flux. For the collinear scalar components used below, B = μ₀(H + M). Use μ₀ ≈ 4π × 10⁻⁷ H/m (henries per metre) for the numerical examples; this is an adequate approximation here, not a claim of an exact present-day SI value.

H in that relation is the field inside the material. In an open specimen, the specimen’s own field can make it differ from the field inferred from the external source alone. Zero applied field therefore need not mean zero internal H. The scaled example stipulates a closed, ungapped, thin-ring approximation with negligible leakage and sufficiently uniform field along the magnetic path; its H is the internal field. These assumptions remove end effects from the teaching model, not from every measurement.

For a stated positive excursion followed back to H = 0, call the retained induction B_r and retained magnetisation M_r. On further reversal, the H where B becomes zero defines a B-based coercive crossing. The H where M becomes zero defines a different crossing in general. We quote positive coercive magnitudes when saying how large the reverse field is, and signed H coordinates when locating points. The qualitative M–H sketch has no numeric scale; its area cannot be read as a loss in joules.

Follow one supplied cycle in order

The table supplies a closed, repeatable B–H path at fixed temperature and one reference waveform. Join successive points by straight segments. Start at P+, decrease H through R+ to P−, then increase H through R− to P+. Values between table points are defined by that linear interpolation. The turning points are the specified limits, not independently established saturation. This is not a prediction from the two-domain account.

At R+ the field has returned to zero but B = +0.600 T. At R− the same H = 0 instead has B = −0.600 T. These are different states with the same field. On the descending segment R+ → P−, B falls by 1.60 T over a field change of −100 kA/m, so B = 0.600 T + (0.0160 T per kA/m)H. Its zero occurs at H = −37.5 kA/m. The ascending B-zero crossing is +37.5 kA/m.

Do not jump from a point on one branch to the other merely because H is equal. To go from R+ to R− along this declared cycle, continue to P− and then return to zero. That direction is drawn with arrows; the descending path is solid and the ascending path dashed. An initial state at H = B = 0 is not on either of these zero-field branch points.

Convert a closed path into a conditional loss

A field source can supply energy while the magnetic state changes, and some energy can be returned on another part of the path. For an ideal winding around the uniform fixed ring, let n be turn count, A the magnetic cross-sectional area in m² and ℓ the mean path length in metres. The field contribution to winding voltage is v = nA dB/dt in volts and its current satisfies i = Hℓ/n in amperes. These use the passive sign convention, with vi positive for input to the magnetic port. Thus vi/(Aℓ) = H dB/dt. This isolates the magnetic port; winding resistance and leakage are excluded from this relation.

Integrating over one complete repeated cycle gives magnetic input per volume w_cyc = ∮ H dB. The closed integral means follow the path once in its physical direction. In the stipulated stationary, periodic, fixed-geometry cycle, with no net change of stored energy or mechanical output, the net input is dissipated. With H in A/m and B in T, the result is J/m³ per cycle. A strip under only one branch is not the whole-cycle loss.

For a straight segment, its contribution is the mean of its two H values multiplied by its change in B. The four segments give −20.0, +80.0, −20.0 and +80.0 kJ/m³, so the total is +120 kJ/m³. Negative segment input means energy is returned through the magnetic port over that segment; it is not negative heat production. The same total is the geometric area: the vertical line from R− to R+ splits the polygon into two triangles, each with vertical base 1.20 T and perpendicular horizontal height 100 kA/m. Each contributes 60.0 kJ/m³.

Using B = μ₀(H + M) also gives ∮ H dB = μ₀∮ H dM, because the closed integral of H dH is zero. An M–H area therefore needs the μ₀ factor and consistent A/m units. Neither M times H at one point nor the height of the retained schematic gives the cycle loss. The familiar expression BH/2 belongs to an appropriate reversible linear energy-storage case, not this hysteretic path.

For a uniform magnetic volume V, the loss per complete cycle is Vw_cyc. If cycles repeat at frequency f_cyc and the same loop is actually maintained, mean magnetic power is P = f_cyc Vw_cyc. With V = 2.00 cm³ = 2.00 × 10⁻⁶ m³ and f_cyc = 10.0 Hz, the example gives 0.240 J per cycle and 2.40 W. This does not predict a temperature rise without a thermal model.

A slow-loop area is not automatically the loss at a new frequency. Induced electrical currents and dynamic reversal can alter a measured loop and add loss; the field distribution may cease to be uniform. A dynamic loop measured with a complete, justified power boundary can already include such effects. Do not then add the same loss again. Winding heating is another account. The source and measurement conditions determine which quantity is being reported.

Separate the measured change from its explanation

Suppose the fictional 10.0 Hz record has w_cyc = 120 kJ/m³. Repeating an unchanged loop at 20.0 Hz would give 4.80 W in the same 2.00 cm³. A separate complete record at 20.0 Hz instead supplies w_cyc = 150 kJ/m³, so its magnetic power is 6.00 W. The extra factor comes from the changed cycle area. It cannot be calculated by frequency scaling alone, and the two totals do not separate wall loss from electrical or other dynamic contributions.

At matched temperature, waveform, turning fields, direction and geometry, a treatment that reduces the complete area from 120 to 90.0 kJ/m³ reduces this model’s per-cycle loss by 25.0%. That comparison still does not identify which barrier or defect changed. A smaller excursion with lower area is a different loading history, not by itself proof of an improved material. Soft and hard magnetic behaviour develops the distinct application choices.

Useful discriminating evidence includes repeated full paths after stated conditioning, field/flux calibration, checks on internal versus applied field, matched frequency and temperature, and complementary domain or structural observations. A scalar net loop can constrain reversal and loss while several microscopic arrangements produce similar traces. Domain imaging also samples a specified region and may not reveal the complete three-dimensional interior. Neither a remanent intercept nor a loop area establishes chemical composition, permanent-magnet suitability or safe operating limits.

Worked examples

Try the reasoning

1. Does one initial magnetisation point describe a full loop?

   Show the answer and reasoning

   No.

   History-dependent reversal has not been measured.

2. Can defects affect hysteresis?

   Show the answer and reasoning

   Yes.

   They can impede domain-wall processes.

3. The scaled specimen is at R+ with H = 0 and B = +0.600 T. Another record shows H = 0 and B = −0.600 T. Does the second point contradict the first, and which route on the supplied cycle connects them?

   Show the answer and reasoning

   No. The same field occurs on different branches; the declared route is R+ → P− → R−.

   The intermediate negative excursion changes the state. H alone does not select B, and a direct unmeasured jump between the two zero-field states is not supplied.

4. In the separate two-orientation account, f = 5/8 and M_s = 800 kA/m. Find M_x. Does this fraction tell you the reverse field required to erase it?

   Show the answer and reasoning

   M_x = +200 kA/m; no erasing field follows from the fraction alone.

   (2 × 5/8 − 1)800 = 200. A motion/barrier law and geometry are missing, even though the signed volume average is known.

5. At the descending B-zero crossing H = −37.5 kA/m, a report labels the specimen “zero M”. Check that statement from the stipulated B–H–M relation.

   Show the answer and reasoning

   It is incorrect here: M = +37.5 kA/m.

   At B = 0, H + M = 0. A nonzero reverse H is balanced by nonzero positive M. Use the zero-response convention actually measured.

6. A separate 20.0 Hz complete record supplies 150 kJ/m³ per cycle in the same 2.00 cm³. Find power and compare it with simply doubling the original 10.0 Hz power. Does the difference isolate one microscopic loss mechanism?

   Show the answer and reasoning

   6.00 W, versus 4.80 W from the unchanged-loop assumption; it does not isolate one mechanism.

   20 × 2.00 × 10⁻⁶ × 150000 = 6.00. The actual area increased by 25%; rate, field distribution and several loss processes may contribute. No unique defect count follows.

7. A measurement stops at R+ after starting from P+. Its magnetic port integral is −20.0 kJ/m³. Has it established a negative full-cycle dissipation?

   Show the answer and reasoning

   No. It is an open segment that returns energy through the port, not a closed cycle.

   The state and stored energy have changed. The full loop has positive net input. Inferring heat from a partial magnetic work segment discards that storage account.

8. Another supplied closed path is the entire original B–H polygon scaled by one half in B and unchanged in H. Find its area and power at 10.0 Hz in 2.00 cm³. Does its lower power establish a better treatment for the original ±1 T task?

   Show the answer and reasoning

   60.0 kJ/m³ per cycle and 1.20 W; the changed excursion does not establish improvement for the original task.

   All dB values halve, so ∮H dB halves. The new path reaches only ±0.5 T. Comparing unequal operating cycles cannot by itself rank loss for the same required response.

What to carry forward

• Read the path, not just endpoints.
• Keep magnetic variables and definitions explicit.
• Wall motion, rotation and nucleation are different reversal processes; net cancellation does not erase local order.
• Specify internal/applied field, magnetic response, conditioning, geometry, temperature and time dependence.
• A B-zero crossing and an M-zero crossing generally differ. A schematic M–H area has no numeric loss without units and the correct factor.
• Use the signed complete-cycle integral ∮H dB and the actual maintained loop before multiplying by volume and frequency.
• A loop constrains a response under conditions; it does not uniquely identify a defect mechanism or an operating limit.