Hysteresis and electromagnets

If you magnetise a piece of iron and then switch the magnetising field back to zero, the iron does not fully "forget" — it stays partly magnetised. Cycle the field forward, backward and forward again, and the iron's magnetisation traces out a closed loop rather than retracing the same line: the field $B$ inside always lags behind the applied field $H$ . This lag is called hysteresis, and it is not a curiosity — it is the single idea that tells an engineer which material to use for a permanent magnet (a compass needle, a loudspeaker magnet) and which to use for an electromagnet core (a relay, a transformer, a crane magnet) that must switch cleanly on and off. This article builds the B–H loop qualitatively, defines retentivity and coercivity, and closes with the practical formula for the magnetising field of a solenoid-wound electromagnet.

The core insight

A ferromagnetic material's $B$ lags behind the cycled applied field $H$ , tracing a closed hysteresis loop whose retained field (retentivity) and reversal field (coercivity) decide whether the material suits a permanent magnet or an electromagnet core.

The core idea

Magnetising an unmagnetised ferromagnet. Start with a completely unmagnetised sample and increase the applied field $H$ from zero. The internal field $B$ rises steeply at first, then bends over and flattens near saturation — beyond this point nearly all the microscopic magnetic domains are aligned with $H$ , so pushing $H$ higher adds almost nothing more to $B$ .

Reducing $H$ back to zero — retentivity. Now reduce $H$ back to zero. $B$ does not fall back to zero along the same curve; it drops along a different, higher path and settles at a nonzero value when $H = 0$ . This leftover field is the retentivity (or remanence) of the material — and it is the entire physical reason a permanent magnet stays magnetised once the magnetising current or field is removed.

Reversing $H$ — coercivity. To force $B$ all the way down to zero, $H$ must be pushed in the reverse direction to a magnitude called the coercivity. Continue cycling $H$ symmetrically forward and back, and the plot of $B$ against $H$ closes into a loop — every later cycle retraces this same loop, not the original zero-to-saturation curve. That is hysteresis: the material's state depends on where it has been, not only on the present value of $H$ .

Reading the loop's shape. The syllabus asks only for a qualitative reading of the loop's shape, not a calculation of its enclosed area as energy loss.

A fat loop (high retentivity and high coercivity) — materials such as steel or alnico — resists demagnetisation strongly. That resistance is exactly what a permanent magnet needs.
A thin, tall loop (high retentivity but low coercivity) — soft iron is the classic example — reaches a high $B$ easily and also loses it easily when $H$ is removed or reversed. That combination, with little energy dissipated per cycle, is exactly what an electromagnet core or a galvanometer/transformer core needs, since these must be magnetised and demagnetised repeatedly.

Electromagnets versus permanent magnets. An electromagnet is a temporary magnet: current flowing through a coil wound on a soft-iron core produces the field, so the field can be switched on or off, and its strength varied, simply by changing the current — something a permanent magnet cannot do. Inside the core, the relation between the field the coil alone would produce and the amplified field the iron core actually delivers is

B = \mu_r\,\mu_0\,H

where: $H$ is the magnetising field set up by the coil's current, $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space, and $\mu_r$ is the material's relative permeability — a large number (hundreds to thousands) for soft iron, which is why a coil wound on an iron core produces a far stronger field than the same coil with an air core. An electromagnet's strength therefore depends on the current, the number of turns per unit length of the coil, and the core's permeability. A permanent magnet's strength, by contrast, depends only on the material's intrinsic retentivity and coercivity and on how thoroughly it was magnetised in the first place — there is no current left flowing to adjust.

Figure

A B against H hysteresis loop with saturation at the top, retentivity marked where the loop crosses the vertical B axis and coercivity marked where it crosses the horizontal H axis

Derivation AThe magnetising field of a solenoid-wound electromagnet

The goal is to obtain a usable, exam-ready expression for the magnetising field $H$ that a coil of wire produces inside its core, since $H$ is the quantity that then decides how strongly the core magnetises through $B=\mu_r\mu_0H$ , and to show how it depends only on the coil's geometry and current, never on the core material itself.

Step 1. Model the electromagnet as a long, tightly wound solenoid — a coil of $N$ turns spread uniformly over a length $l$ , carrying a steady current $I$ . For such a solenoid, the magnetising field along its axis is essentially uniform and depends on the coil alone, not on whatever is placed inside it.

Step 2. Define the turns density $n$ , the number of turns per unit length of the coil: $n = N/l$ . This single number captures how tightly the wire is wound — a coil with more loops packed into the same length produces a proportionally stronger field for the same current.

Step 3. The magnetising field inside the solenoid is then

H = nI = \frac{NI}{l}

where: $N$ is the total number of turns, $l$ is the coil's length in metres, $I$ is the current in amperes, and $H$ comes out in amperes per metre (A/m). This $H$ is set entirely by the coil — swapping an air core for a soft-iron core changes the resulting $B$ enormously (through $\mu_r$ ) but leaves $H$ itself unchanged, since $H$ measures what the current supplies, not what the core does with it.

Step 4. Once $H$ is known, the core's material decides the payoff: a soft-iron core (large $\mu_r$ , low coercivity) converts a modest $H$ into a large $B$ that also collapses again as soon as the current is switched off — precisely the behaviour an electromagnet needs.

Worked example

An electromagnet's coil has $N = 500$ turns wound over a length $l = 0.4\ \text{m}$ and carries a current $I = 3\ \text{A}$ around a soft-iron core. Find the magnetising field $H$ inside the core.

Step 1. Compute the turns per unit length: $n = N/l = 500/0.4 = 1250\ \text{turns/m}$ .

Step 2. Apply $H = nI$ : $H = 1250 \times 3 = 3750\ \text{A/m}$ .

Step 3. Because the core is soft iron — high permeability, low coercivity — the actual $B$ produced inside is far larger than $\mu_0 H$ alone would give with an air core, and it will also fall back to near zero once the current is switched off, since soft iron's loop is thin. This amplify-then-release behaviour is exactly why real electromagnets use ferromagnetic cores rather than empty coils. The magnetising field inside the core is $H = 3750\ \text{A/m}$ .

Competitive stretch

Suppose the current is doubled to $I = 6\ \text{A}$ with the same coil. Since $H = nI$ and $n$ is unchanged, $H$ simply doubles: $H = 1250 \times 6 = 7500\ \text{A/m}$ . But $B$ does not necessarily double — if the core is already near saturation (as described in "The core idea"), further increases in $H$ produce only a small further rise in $B$ . This is why doubling the current in a heavily used electromagnet gives diminishing returns once saturation is approached.

Common traps

Assuming $B$ always returns to exactly zero once $H$ is removed — some magnetisation (the retentivity) survives, and that survival is the whole basis of a permanent magnet.
Swapping the definitions of retentivity (the leftover $B$ at $H=0$ ) and coercivity (the reversed $H$ needed to bring $B$ down to zero) — they sit on different axes of the loop.
Assuming a good permanent-magnet material automatically makes a good electromagnet core — the requirements are opposite: high coercivity for one, low coercivity for the other.
Forgetting that $H = nI$ depends only on the coil (turns, length, current), while it is the core's permeability $\mu_r$ that decides how large the resulting $B$ becomes.

Ask the tutor: Want to trace the B-H loop point by point as the field cycles, or practise plugging different N, l and I values into H = nI?

Why does an iron nail stay magnetised after I pull the magnet away, but a steel nail stays magnetised much longer?

Both nails acquire some retentivity while the magnet is near them, but soft iron has low coercivity, so its magnetisation collapses almost as soon as the external field is gone. Steel has much higher coercivity, so its domains resist being disturbed once aligned, and the magnetisation persists for a long time — that is precisely why steel, not soft iron, is chosen for permanent magnets like those in compasses and loudspeakers.

Does the area inside the B-H loop matter for this syllabus?

The full picture is that the enclosed area represents the energy dissipated as heat per magnetisation cycle, which is why a fat loop (large area) is undesirable in a transformer core that gets cycled fifty or sixty times a second. However, this syllabus asks only for the qualitative shape comparison — fat loop implies large retentivity, coercivity and energy loss, thin loop implies the opposite — and does not require you to calculate the area as an energy quantity.

If H depends only on the coil, why does changing the core material matter at all?

Because $H=nI$ only measures what the coil's current supplies, but the field actually felt inside the material is $B=\mu_r\mu_0H$ . A soft-iron core has a relative permeability $\mu_r$ in the hundreds or thousands, so it turns a modest $H$ into a large $B$ ; an air core, with $\mu_r\approx1$ , gives almost nothing extra. The coil sets $H$ ; the core decides how much of a magnet that $H$ actually makes.

Hysteresis and electromagnets

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads