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Magnetization and bound currents, the auxiliary field $\vec{H}$, linear diamagnets and paramagnets, ferromagnets and iron cores, and atomic moments.
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By the end of this lesson you will be able to find bound currents and fields in magnetized materials, use $\vec{H}$ with Ampère's law for cored coils, and relate magnetization to atomic moments.
You know magnetic dipoles and their torque, Ampère's law, and the dielectric story of polarization, bound charge and $\vec{D}$. You know spin gives electrons a magnetic moment. This lesson tells the parallel story for magnetism: atomic dipoles, bound currents and the auxiliary field $\vec{H}$.
| Term | What it means |
|---|---|
| Magnetization | $\vec{M}$, magnetic dipole moment per unit volume, in A/m. |
| Bound currents | $\vec{K}_b = \vec{M} \times \hat{n}$ on surfaces and $\vec{J}_b = \nabla \times \vec{M}$ inside. |
| Auxiliary field | $\vec{H} = \vec{B}/\mu_0 - \vec{M}$, whose curl is the free current density. |
| Magnetic susceptibility | $\chi_m$, with $\vec{M} = \chi_m\vec{H}$ in linear materials. |
| Permeability | $\mu = \mu_0(1 + \chi_m) = \mu_r\mu_0$, with $\vec{B} = \mu\vec{H}$. |
| Diamagnetism and paramagnetism | Weak responses with $\chi_m < 0$ and $\chi_m > 0$. |
| Ferromagnetism | Strong, nonlinear magnetization from aligned domains, retained after the field is removed. |
Atoms carry magnetic dipole moments from electron spin and orbital motion. In a field, they respond, and a material acquires magnetization $\vec{M}$. Just as polarization acts like bound charge, magnetization acts like bound currents:
$$\vec{K}_b = \vec{M} \times \hat{n}, \qquad \vec{J}_b = \nabla \times \vec{M}.$$
In a uniformly magnetized rod, the atomic current loops cancel inside, leaving a current circling the surface — a solenoid made of atoms, with $B = \mu_0M$ inside.
Ampère's law counts all current, $\nabla \times \vec{B} = \mu_0(\vec{J}_f + \vec{J}_b)$. Moving the bound part to the left defines
$$\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M}, \qquad \nabla \times \vec{H} = \vec{J}_f, \qquad \oint\vec{H} \cdot d\vec{l} = I_{f,\text{enc}}.$$
In linear materials $\vec{M} = \chi_m\vec{H}$ and $\vec{B} = \mu\vec{H}$ with $\mu = \mu_0(1 + \chi_m)$. Diamagnets ($\chi_m \sim -10^{-5}$) weakly oppose fields; paramagnets ($\chi_m \sim 10^{-5}$ to $10^{-3}$) weakly enhance them. Ferromagnets — iron, cobalt, nickel and their alloys — are nonlinear, with effective $\mu_r$ in the thousands, and they stay magnetized after the field is removed.
Another way: picture
Picture every atom as a tiny current loop. In a magnetized block, neighboring loops run in opposite directions where they touch and cancel, like gears meshing, so no net current flows inside. At the edges there is no neighbor to cancel, and the loops' outer halves add into a current sheet circling the block. That sheet, invisible and unremovable, is the bound current.
Another way: steps
Checks. With $\chi_m = 0$, everything must reduce to vacuum. For a diamagnet, $B$ must be slightly less than $\mu_0H$; for a paramagnet slightly more. Normal components of $\vec{B}$ must match across boundaries. And fields in iron cores should be at most about $2$ T, the saturation limit, no matter how large $\mu_r$ seems to allow.
The vector potential of a magnetic dipole is $\frac{\mu_0}{4\pi}\frac{\vec{m} \times \hat{\mathscr{r}}}{\mathscr{r}^2}$. Adding this up over a magnetized volume and integrating by parts rewrites it as the potential of a surface current $\vec{M} \times \hat{n}$ plus a volume current $\nabla \times \vec{M}$: exactly the bound currents.
Are they real currents? They are the averaged effect of electrons circulating and spinning in atoms, not charges traveling through the material, so no voltmeter measures them and no resistance dissipates them. But their fields are completely real. The field of a bar magnet is exactly that of a solenoid with the same shape and surface current $K = M$, which is why the two have identical field patterns outside.
Every material is diamagnetic: when a field is applied, the orbits of its electrons adjust, by a kind of induction, to oppose the change. The effect is tiny, $\chi_m \approx -10^{-5}$ for water, and is usually masked by stronger effects. In superconductors it becomes perfect, $\chi_m = -1$: they expel magnetic fields entirely, the Meissner effect, and can levitate magnets.
Atoms with unpaired electrons have permanent moments that tend to align with a field, against thermal agitation; this paramagnetism dominates diamagnetism, with $\chi_m \approx 10^{-5}$ to $10^{-3}$ at room temperature, falling as $1/T$ by Curie's law. Liquid oxygen is paramagnetic enough to be held between the poles of a strong magnet, a classic demonstration, and the paramagnetism of gadolinium compounds makes them the contrast agents used in MRI.
In iron, cobalt and nickel, a quantum-mechanical exchange interaction makes neighboring electron spins align, and regions called domains, typically micrometers across, are magnetized to saturation. An unmagnetized piece has domains pointing every way. An applied field grows the favorably aligned domains at the expense of others, producing huge magnetization from modest fields: effective $\mu_r$ of thousands.
Remove the field and some alignment remains, the remanence; reverse the field and it takes a coercive field to undo it. The $B$–$H$ curve traces a hysteresis loop. Soft magnetic materials, with narrow loops, suit transformer cores; hard materials such as neodymium-iron-boron, with wide loops, make permanent magnets. Above the Curie temperature, $1043$ K for iron, thermal agitation wins and ferromagnetism vanishes.
Wrapping a coil around iron multiplies its field by $\mu_r$, often a thousand or more, because the iron's bound currents add to the coil's free current. That is why electromagnets, transformers and motors all have iron cores. A coil producing $1$ mT alone produces over a tesla with a good core.
The gain stops at saturation, about $2$ T for iron and $2.4$ T for iron-cobalt alloys, when every domain is aligned. Above that, extra current adds only $\mu_0H$. Engineers design transformer and motor cores to run just below saturation, and they laminate the cores into thin insulated sheets to stop eddy currents, the circulating currents that changing fields would otherwise drive in the solid metal.
Because $\nabla \cdot \vec{B} = 0$, the normal component of $\vec{B}$ is continuous across any surface. Because $\nabla \times \vec{H} = \vec{J}_f$, the tangential component of $\vec{H}$ is continuous where no free surface current flows. At a boundary between iron and air, these rules bend field lines sharply: inside the iron they run nearly parallel to the surface; outside they leave almost perpendicular.
This is how magnetic circuits work. Iron guides flux the way copper guides current, because field lines prefer the high-permeability path. Magnetic shielding uses the same effect: an enclosure of mu-metal, with $\mu_r$ near $100{,}000$, draws the field lines into its walls and leaves the interior nearly field-free, protecting electron microscopes and sensitive sensors from stray fields.
The natural unit of atomic magnetism is the Bohr magneton, $\mu_B = e\hbar/2m_e = 9.274 \times 10^{-24}$ A m², the magnetic moment of an electron's spin. Dividing a ferromagnet's saturation magnetization by its number of atoms per volume gives the moment per atom: $2.2\mu_B$ for iron, $1.7\mu_B$ for cobalt, $0.6\mu_B$ for nickel.
These fractional values puzzled physicists, since an atom should have a whole number of unpaired electrons. The explanation is that in metals the magnetic electrons occupy energy bands shared across the crystal, not atomic orbitals, so the average number of unpaired spins per atom need not be an integer. Rare-earth elements such as gadolinium, whose magnetic electrons stay localized, show nearly integer values, $7\mu_B$.
Hard disks store bits as tiny regions magnetized one way or the other in a thin cobalt-alloy film. The film must be magnetically hard, so bits stay put for years, yet writable by the field of a write head. As bits shrank below $20$ nm, thermal agitation began to flip them spontaneously, so manufacturers moved to harder materials and heat-assisted recording, which briefly heats each bit with a laser to lower its coercivity while writing.
Reading relies on the giant magnetoresistance of the spin lesson in quantum mechanics, but the stored information is magnetization: aligned atomic moments producing bound surface currents and a stray field just above the disk. Data centers around the world still store most of their bytes this way, because magnetic storage remains the cheapest per bit.
Dividing iron's saturation magnetization, $1.71 \times 10^{6}$ A/m, by its $8.49 \times 10^{28}$ atoms per cubic meter gives $2.0 \times 10^{-23}$ A m² per atom, or $2.2$ Bohr magnetons: about two electron spins per atom aligned in the same direction. Cobalt gives $1.7$, nickel $0.6$. These modest numbers, multiplied by the vast number of atoms, produce internal fields of about $2$ T.
Why only these elements? The alignment requires a strong exchange interaction between the $3d$ electrons of neighboring atoms, strong enough to beat thermal agitation at room temperature. Only iron, cobalt and nickel among the elements manage it; gadolinium does too but only below $20$ °C. The search for better permanent magnets without scarce rare earths is a priority of the U.S. Department of Energy's Critical Materials Innovation Hub, because the magnets in wind turbines and electric-vehicle motors depend on neodymium and dysprosium.
The transformers that step voltage up for transmission and down for homes use cores of grain-oriented silicon steel, with relative permeabilities in the tens of thousands. The core's high permeability guides nearly all of the flux produced by one winding through the other, and its bound currents let a small magnetizing current produce a field near $1.7$ T, just below saturation.
Every cycle around the hysteresis loop dissipates energy as heat, so core materials are chosen for narrow loops. Amorphous metal cores, made by quenching molten alloy into thin ribbons so fast that no crystals form, cut these losses by about two thirds, and U.S. Department of Energy efficiency standards for distribution transformers have pushed utilities toward them. With millions of transformers on the grid, even small reductions in core loss save billions of kilowatt-hours a year.
Everyday experience with iron suggests magnets attract things, but most materials barely respond, and many are repelled. Diamagnetic substances — water, copper, graphite, most organic matter — are pushed toward weaker field; paramagnetic ones — aluminum, oxygen — are pulled weakly toward stronger field. Only ferromagnets respond strongly. In a $16$ T field, the diamagnetism of water is enough to levitate a frog, as Andre Geim famously demonstrated.
A second error is to treat $\vec{H}$ as the true magnetic field. The force on a moving charge is $q\vec{v} \times \vec{B}$, and $\vec{B}$ is fundamental; $\vec{H}$ is a bookkeeping field whose circulation counts free current. Engineers often speak of $\vec{H}$ because it is what their currents control directly.
A long rod has uniform magnetization $\vec{M} = M\hat{z}$. Find the bound volume current.
$\vec{J}_b = \nabla \times \vec{M} = 0$
A constant vector has no curl.
Find the bound surface current on the side.
$\vec{K}_b = M\hat{z} \times \hat{s} = M\hat{\phi}$
Circling the rod.
Recognize a solenoid.
$K = nI \ \leftrightarrow \ K_b = M$
A current sheet around a long cylinder.
Write the field inside.
$B = \mu_0M$
The solenoid formula with $nI \to M$.
Evaluate for iron at saturation, $M = 1.71 \times 10^{6}$ A/m.
$B = 4\pi \times 10^{-7} \times 1.71 \times 10^{6} = 2.15\ \text{T}$
The maximum field inside iron.
A solenoid with $1000$ turns per meter carries $0.50$ A around a core with $\mu_r = 800$. Find $H$.
$H = nI = 1000 \times 0.50 = 500\ \text{A/m}$
Ampère's law for $\vec{H}$: free current only.
Find the empty-solenoid field.
$B_0 = \mu_0H = 4\pi \times 10^{-7} \times 500 = 0.63\ \text{mT}$
Without the core.
Find the field with the core.
$B = \mu_rB_0 = 800 \times 0.63 = 0.50\ \text{T}$
Eight hundred times stronger.
Find the magnetization.
$M = \dfrac{B}{\mu_0} - H = \dfrac{0.50}{4\pi \times 10^{-7}} - 500 = 4.0 \times 10^{5}\ \text{A/m}$
Almost all of $B$ comes from the iron.
Compare with saturation.
$M/M_s = 0.40 / 1.71 = 0.23$
Well below saturation: the linear model is reasonable.
Find the bound surface current.
$K_b = M = 4.0 \times 10^{5}\ \text{A/m}$
Eight hundred times the free $nI$.
A field in air strikes iron ($\mu_r = 1000$) at $45°$ to the normal. Write the boundary conditions.
$B_{1\perp} = B_{2\perp}, \qquad H_{1\parallel} = H_{2\parallel}$
No free surface current.
Relate the tangential $B$ components.
$\dfrac{B_{2\parallel}}{\mu_r\mu_0} = \dfrac{B_{1\parallel}}{\mu_0} \quad\Rightarrow\quad B_{2\parallel} = 1000B_{1\parallel}$
From continuity of $H_\parallel$.
Write the angles.
$\tan\theta_2 = \dfrac{B_{2\parallel}}{B_{2\perp}} = 1000\tan\theta_1$
Normal components are equal.
Evaluate in the iron.
$\tan\theta_2 = 1000 \quad\Rightarrow\quad \theta_2 = 89.94°$
Nearly parallel to the surface.
Consider the reverse direction.
$\theta_2 = 45° \text{ in iron} \Rightarrow \tan\theta_1 = 0.001$
Field leaving iron exits almost perpendicular.
Interpret the refraction.
$\text{iron guides flux}$
Lines bend into the iron and follow it.
Apply to shielding.
$\text{a closed iron box diverts lines around its interior}$
The principle of magnetic shields.
Note the limit.
$B < B_{\text{sat}} \approx 2\ \text{T}$
A saturated shield stops working.
Find the relative permeability.
$\mu_r = 1 + \chi_m = 500$
From the susceptibility.
Multiply out the field.
$B = \mu_r\mu_0H = 500 \times 4\pi \times 10^{-7} \times 200$
$B = \mu H$.
Evaluate the field.
Water has magnetic susceptibility $\chi_m = -9 \times 10^{-6}$. What happens to a drop of water placed near the pole of a very strong magnet?
Complete the worked solution: a linear material with $\chi_m = 2$ sits in $H = 177$ A/m. Find $M$ and $B/\mu_0$ in A/m, and the relative permeability.
Multiply the field by the susceptibility.
$M = 2 \times 177 =$ m
The linear response.
Add $H$ and $M$.
$B/\mu_0 = H + M =$ b
From the definition of $\vec{H}$.
Add one to the susceptibility.
$\mu_r = 1 + \chi_m =$ r
Check: $B/\mu_0 = \mu_rH$.
Match each quantity to its expression.
| $\vec{M} \times \hat{n}$ | $\nabla \times \vec{M}$ | $\vec{B}/\mu_0 - \vec{M}$ | $\oint\vec{H} \cdot d\vec{l} = I_{f,\text{enc}}$ | |
|---|---|---|---|---|
| bound surface current | ||||
| bound volume current | ||||
| the $\vec{H}$ field | ||||
| Ampère's law in matter |
A long solenoid with $nI = 681$ A/m is filled with a linear material of relative permeability $\mu_r = 2$. Fill in $H$, $M$ and $B/\mu_0$, all in A/m.
| A/m | |
|---|---|
| $H$ | |
| $M$ | |
| $B/\mu_0$ |
A long solenoid with $9$ turns per cm carries $3$ A around a core with $\mu_r = 10$, in its linear range. What is $B$ inside, in mT?
Answer: mT
A long rod is uniformly magnetized along its axis with $M = 0.8$ MA/m and carries no free current. What is the magnetic field $B$ inside it, far from the ends, in tesla?
Answer: T
Fully magnetized gadolinium, near absolute zero has $M_s = 2.06$ MA/m and $3.02 \times 10^{28}$ atoms per cubic meter. What is the magnetic moment per atom, in Bohr magnetons ($\mu_B = 9.274 \times 10^{-24}$ A m²)?
Answer: μ_B
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A long solenoid with $nI = 532$ A/m is filled with a linear material of relative permeability $\mu_r = 5$. Fill in $H$, $M$ and $B/\mu_0$, all in A/m.
| A/m | |
|---|---|
| $H$ | |
| $M$ | |
| $B/\mu_0$ |
You can describe magnetism in matter. Explain to someone why a bar magnet's field looks just like a solenoid's.
19. Your turn: a linear core with $\chi_m = 499$ sits in $H = 200$ A/m. What is $B$?, step 3
$B = 0.126\ \text{T}$
Five hundred times the vacuum value.