Materials Science · Magnetism and coupled material responses

Soft and hard magnetic behaviour

Useful background: Atoms, bonding, structure and defects; Energy, stability and environmental context; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish magnetic hardness from mechanical hardness.
Compare remanence, B-zero coercive field and loop loss under a stated field cycle.
Use a descending branch to test a retained-flux requirement rather than ranking materials by coercivity alone.
Explain why internal field, specimen geometry and operating frequency limit transfer of a loop comparison.

Easy and difficult reversal serve different uses

Soft magnetic materials reverse magnetisation relatively readily, useful where changing fields should incur limited hysteretic loss. Hard magnetic materials resist reversal and can retain useful magnetisation. “Hard” here does not mean resistance to indentation.

“Magnetic moments and collective order” separates local magnetic order from the net response of a specimen. “Domains, hysteresis and magnetic losses” introduces the path followed as a field changes. A material can have ordered domains yet a small net moment because their contributions cancel; a large net moment also need not reverse easily.

In a multidomain solid, a field can favour growth of some domains by motion of their boundaries. Resistance to that motion helps make the return path differ from the outward path, and irreversible changes dissipate energy. Soft behaviour favours relatively easy repeated reversal; hard behaviour favours retention against a reverse field. These are functional descriptions, not a claim that one class has greater indentation hardness or is preferable for every use.

Compare the same loop quantities on the same axes

Let H mean the magnetic field strength inside the specimen, in amperes per metre (A/m), B the magnetic flux density in tesla (T), and M the magnetisation, or magnetic moment per volume, also in A/m. In SI, B = μ₀(H + M), where μ₀ is the vacuum permeability. H and B are different quantities with different units. This lesson uses collinear signed components along the test direction.

The figure contains original invented piecewise-straight B–H loops, S and P. Assume matched 300 K specimens in closed ring tests, with leakage and variation across each ring’s cross-section negligible. The field is cycled repeatedly between +1000 and −1000 A/m until each shown path repeats. The model represents a slow-cycle, rate-independent comparison. Begin at (+1000 A/m, +1.00 T), follow the upper path while H decreases, then the lower path as H increases. Neither endpoint is asserted to be magnetic saturation, and the letters are not named commercial grades.

On the descending path, the retained B at H = 0 is Bᵣ: 0.60 T for S and 0.80 T for P in this cycle. The magnitude of reverse H where B first reaches zero is labelled H_cB here: 100 A/m for S and 600 A/m for P. Thus P has the harder response by this B-zero comparison, while S reverses its flux density with a smaller reverse field. These are cycle-specific values, not universal numerical boundaries between soft and hard materials.

A B-zero point is not generally an M-zero point. At B = 0, the SI relation gives M = −H, which can be nonzero. Coercivity defined by zero M, often called intrinsic coercivity, must be distinguished from H_cB. Reports must state which response reaches zero and whether the field excursion established a saturation loop or a smaller loop.

Invented matched slow-cycle B–H comparison at 300 K, between ±1000 A/m. Bᵣ and H_cB refer to the descending branch from the positive endpoint; neither endpoint is claimed to be saturation.
Candidate	Bᵣ at H = 0 (T)	H_cB (A/m)	Enclosed loop area (J/m³ per cycle)
S	0.60	100	260
P	0.80	600	1880

Original invented piecewise-straight B versus internal H loops at 300 kelvin, after repeated slow cycling between plus and minus 1000 amperes per metre in ideal closed rings. H is horizontal from minus 1000 to plus 1000 amperes per metre; B is vertical in tesla. S is a solid blue curve with circular intercept markers. Its descending vertices in H, B order are (1000, 1), (200, 0.8), (0, 0.6), (minus 100, 0), (minus 200, minus 0.8), and (minus 1000, minus 1). P is an orange long-dashed curve with square intercept markers, descending through (1000, 1), (0, 0.8), (minus 600, 0), and (minus 1000, minus 1). Ascending branches are the sign-reversed descending branches. Both start at the positive endpoint, traverse the upper branch toward decreasing H and return along the lower branch. S has retained B of 0.60 tesla and B-zero reverse-field magnitude 100 amperes per metre. P has 0.80 tesla and 600 amperes per metre. Enclosed areas are 260 and 1880 joules per cubic metre per cycle respectively. Endpoints are not asserted to be saturation; these are teaching models, not commercial grade specifications.

An area measures an energy loss under that cycle

For a uniform collinear specimen returning to the same magnetic state after a cycle, the energy dissipated per magnetic-material volume is the closed-loop integral w = ∮ H dB. The symbol means adding H times each small change in B around the complete path; energy supplied during some portions can be partly returned during others. The remaining positive amount is the enclosed B–H area. A simple product BᵣH_cB is not the area of an arbitrary loop.

The units give a useful check: (A/m) × T = J/m³. The constructed straight segments give w_S = 260 J/m³ and w_P = 1880 J/m³. At magnetic-material volume V and cycle frequency f, the corresponding power is P_loss = fVw. Frequency counts complete cycles per second, so (1/s) × m³ × J/m³ gives watts. This comparison favours S when repeated cycling with this flux excursion and small dissipation is the objective.

A slow loop is not a complete operating-loss specification at every frequency. Changing flux can induce circulating electrical currents in a conductive core; their resistance dissipates energy. Electrically insulated thin laminations can interrupt large current paths, and higher resistivity can reduce such currents under suitable conditions. These measures do not remove all magnetic losses. Faster cycling can also change the loop itself. A measured dynamic loop already includes the dissipation represented by that measured path: do not add the same contribution again.

Geometry and operation also matter

Demagnetising fields depend on specimen shape. Frequency introduces additional loss processes, and temperature changes operating margins. Select a magnetic material using its intended field cycle, geometry and environment rather than coercivity alone.

The closed ring model reduces the demagnetising-field problem by keeping the flux path within the magnetic material. An open magnet can experience a field opposing its magnetisation even when the externally imposed field is zero. Removing the applied field therefore does not automatically reproduce the ring’s H = 0 point.

For an ideal uniformly magnetised ellipsoid along a principal axis, a scalar model gives H_int = H_ext − NM. Here H_int is internal field, H_ext is the uniform field that would be imposed at that location without the specimen, and N is the dimensionless demagnetising factor for that axis in this SI convention, between 0 and 1. The −NM contribution opposes positive M. This uniform model is not a general local-field formula for a rectangular part with corners.

M is part of the material’s response to internal field and history. Supplying M allows a pointwise field calculation; it does not establish that M will stay unchanged if the shape or applied field changes. Predicting a new operating state requires a compatible response curve and field solution. A permanent-magnet part also needs adequate flux at its actual operating state, not simply a large tabulated coercivity.

Original signed field account at one supplied uniformly magnetised ellipsoid state along a principal axis. The dimensionless SI demagnetising factor N is 0.25; the supplied magnetisation is plus 120 kiloamperes per metre. Three horizontal bars share a field scale from minus 40 to plus 20 kiloamperes per metre. External field is plus 10, the demagnetising contribution minus NM is minus 30, and internal field is their sum minus 20 kiloamperes per metre. Positive bars extend right from zero; negative bars extend left. Distinct row names and hatching identify the contributions. This is a calculation at the specified state, not a prediction that magnetisation stays fixed after a shape change.

Match the comparison to the intended task

For repeated reversal, compare loop loss at the required B excursion, frequency, waveform and temperature, with a specimen or component geometry relevant to the use. For retention, first state the required response and the adverse internal-field range, then examine the appropriate descending branch. A larger remanence or coercive field alone does not test that entire requirement.

Retain the material state, specimen dimensions and direction, magnetic preparation, maximum field, response convention, temperature, waveform and rate with the result. Direction can change a measured loop, and two methods can report different zero-response definitions. “Specimens, dimensions and test conditions” explains why the tested object and actual conditions belong in the evidence; “Comparable results and independent reproduction” develops how to investigate apparently conflicting values.

The following cases compare central values in deliberately complete invented models. They supply neither measurement uncertainties nor a real component qualification. When a predicted response lies close to a requirement, the relevant uncertainty, variation and operating margins must be assessed rather than hidden by rounding.

Worked examples

Work through the case

A transformer material and a permanent-magnet material require different coercivities. Is one universally superior?

A transformer repeatedly reverses its response.
A permanent magnet must resist unwanted reversal.
The useful property depends on function.

No. Their objectives differ.

Use the required response, not a single ranking

After the positive endpoint, an application is stipulated to require B ≥ 0.20 T at internal H = −400 A/m. For P the descending straight segment runs from (0 A/m, 0.80 T) to (−600 A/m, 0 T). For S the relevant segment runs from (−200 A/m, −0.80 T) to (−1000 A/m, −1.00 T). Which candidate meets this model requirement? Would P still meet it at −500 A/m?

Follow the descending branch from the prepared positive endpoint. Do not use the ascending branch, the initial magnetisation curve or the external field in place of the stipulated internal field.
For P at −400 A/m, the fraction of the segment traversed is 400/600 = 2/3. Thus B_P = 0.80 + (2/3)(0 − 0.80) = 0.2667 T, above 0.20 T.
For S, the fraction from −200 to −400 A/m is 200/800 = 1/4. Thus B_S = −0.80 + (1/4)(−1.00 + 0.80) = −0.85 T. S has reversed and fails the positive retained-flux requirement.
At −500 A/m, P gives 0.80(1 − 500/600) = 0.1333 T, which fails. P still has the larger H_cB, but that ranking does not guarantee the required B at every reverse field.

P alone meets the stipulated −400 A/m requirement in this model. At −500 A/m it fails. A real part requires its internal-field solution, response history, conditions and margins.

Turn cycle loss into a bounded power comparison

Use the table’s areas for magnetic-material volume 20 cm³ at 10 complete cycles per second. Assume these same rate-independent loops apply, with no additional loss contribution in this illustration. What power does each dissipate?

Convert the magnetic-material volume: 1 cm³ = 10⁻⁶ m³, so V = 20 × 10⁻⁶ = 2.0 × 10⁻⁵ m³. Do not use the surrounding air or winding volume.
S loses Vw_S = (2.0 × 10⁻⁵)(260) = 0.0052 J per cycle; P loses Vw_P = (2.0 × 10⁻⁵)(1880) = 0.0376 J per cycle.
Multiply by 10 s⁻¹: P_loss,S = 0.052 W and P_loss,P = 0.376 W. The loss ratio is 1880/260 ≈ 7.23 for these loops and equal volume/frequency.
Doubling frequency would double these powers only if the same loop areas still applied. Real frequency-dependent loss evidence is needed before extending the comparison to a different operating rate.

S dissipates 0.052 W and P 0.376 W under the stipulated model. This supports the lower-loss candidate for this cycling task, not a universal performance ranking.

A positive external field can coexist with a negative internal field

An ideal uniformly magnetised ellipsoid has N = 0.25 along the stated axis. At one specified state, M = +120 kA/m and H_ext = +10 kA/m. Find H_int. Does it follow that the specimen’s magnetisation has already become negative?

Keep the signed components along the same axis. The demagnetising contribution is −NM = −0.25 × 120 = −30 kA/m.
The internal field is H_int = 10 − 30 = −20 kA/m. The second figure shows this signed sum on a common scale.
No negative magnetisation follows: the stated state has M = +120 kA/m. A positive magnetisation can persist against a reverse internal field because the response depends on history.
This is a field account at a supplied state. To predict a different shape or applied field, solve for its compatible M rather than treating the supplied value as an unchanging material constant.

H_int = −20 kA/m while the supplied M remains positive. External field alone does not identify the internal state.

Try the reasoning

Does magnetic hardness establish scratch resistance?
Show the answer and reasoning
No.
The terms describe different properties.
Why include component shape?
Show the answer and reasoning
It affects internal magnetic fields.
The applied field alone may not describe the local state.
At P’s descending B-zero point, H_int = −600 A/m. Use B = μ₀(H_int + M) to find M there. Has zero B established zero magnetisation?
Show the answer and reasoning
M = +600 A/m. Zero B has not established zero M.
Since μ₀ is nonzero, 0 = H_int + M gives M = −H_int. The internal-field and magnetisation terms cancel in B. A reported zero-M coercivity therefore cannot silently replace H_cB.
A second use cycles the same S loop at 40 Hz with 5.0 cm³ of magnetic material. Assume the same area 260 J/m³ applies. Calculate power. May this result be carried straight to 400 Hz by multiplying by ten?
Show the answer and reasoning
The stipulated 40 Hz power is 0.052 W. A tenfold prediction at 400 Hz needs evidence that the loop area and other relevant losses remain represented by the model.
V = 5.0 × 10⁻⁶ m³; fVw = 40 × (5.0 × 10⁻⁶) × 260 = 0.052 W. Its smaller volume offsets its higher frequency relative to the worked S case. Frequency-dependent loop changes and eddy-current effects can invalidate a simple rate extrapolation.
At a supplied state of a different ideal ellipsoid, N = 0.10, M = +80 kA/m and H_ext = +6 kA/m along its principal axis. Find H_int. A designer then changes N to 0.05 and asserts M must still be 80 kA/m. Which part of that claim needs new evidence?
Show the answer and reasoning
At the supplied state H_int = 6 − 0.10 × 80 = −2 kA/m. The assertion that M stays at 80 kA/m after the shape change needs the compatible response and field solution.
If M were supplied as 80 kA/m at the new state, its pointwise field account would give 6 − 0.05 × 80 = +2 kA/m. That conditional arithmetic does not predict that the altered specimen actually takes the same M; its internal field and magnetic history affect its response.
One report gives a zero-M coercive field measured along a film’s plane after a large field excursion. Another gives a zero-B coercive field for a ring after a smaller excursion at another temperature. Both use A/m. Can their numbers alone rank magnetic hardness for a proposed part? State the comparison needed.
Show the answer and reasoning
No. Match the zero-response definition, magnetic preparation/excursion, direction, internal-field treatment, temperature and intended operating task before making that comparison.
A common field unit does not make the endpoints identical. Shape and direction affect the measured response, and a smaller loop can differ from a saturation loop. Compare the relevant branch or operating-cycle loss under justified conditions; neither report by itself qualifies the proposed part.

What to carry forward

Select magnetic reversal behaviour for the job.
Keep magnetic and mechanical hardness separate.
Label B-zero and M-zero coercive fields separately and retain the field cycle and test conditions.
Use the branch value needed by the requirement; a larger coercivity alone does not supply that value.
B–H loop area is energy per material volume per cycle; transferring it to another rate requires a justified loop/loss model.
An open specimen can have an internal field opposing its magnetisation; shape and magnetic response must be compatible.

References

Teaching resources and research supporting this lesson’s explanations and examples.