advanced31 min studyLesson 249 in Physics

Magnetization and bound-current description

Convert a prescribed magnetization into equivalent currents and keep internal H, applied H, and physical B distinct.

Before you begin

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The question for today

What you’ll understand

Compute volume and surface bound currents from magnetization.
Use the SI relation among B, H, and M without interchanging their roles.
Include sample shape through a demagnetizing field when determining internal response.

Magnetization describes the magnetic dipole moment per unit volume after averaging over microscopic structure. A prescribed magnetization can be represented by equivalent volume and surface currents, allowing ordinary magnetic field methods to be used. Introducing H moves those bound currents out of the free-current Ampère equation. This bookkeeping does not make H identical to the applied field, nor does it imply that a magnetized body with no free current has zero physical B.

Cancellation inside leaves currents where magnetization changes

Imagine adjacent small current loops representing an averaged magnetization. Currents on shared interior edges cancel when the loop moments are uniform, whereas uncancelled currents remain at the sample surface. A spatial variation can also leave a net volume current, expressed by the curl of magnetization. The surface current is magnetization crossed with the body's outward normal. Its units are amperes per metre, while the volume curl has amperes per square metre.

Uniform magnetization therefore does not mean that every equivalent current is zero. Its volume curl vanishes but its surface current generally survives. On a cylinder magnetized along its axis, the cylindrical side carries azimuthal bound current and the end faces carry none, since their normals are parallel to the moment. This solenoidal current pattern produces an interior field. The representation includes spin magnetism macroscopically without claiming that each microscopic moment is literally a tiny classical wire loop.

\mathbf J_b=\nabla\times\mathbf M,\qquad \mathbf K_b=\mathbf M\times\hat{\mathbf n}_{\mathrm{out}},\qquad \mathbf m_{\mathrm{total}}=\int_V\mathbf M\,dV

M is in A/m, Jb in A/m², Kb in A/m, and total moment in A m². The normal is outward from the magnetized body. These are the macroscopic current representation of a prescribed magnetization and do not determine M without a material law.↔ Scroll wide equations horizontally.

H separates sources but still feels geometry

In SI, H is defined as B divided by vacuum permeability minus magnetization. For magnetostatics its curl is the free current density. Zero free current makes H curl free, not necessarily zero: boundaries and the magnetization's spatial variation determine its divergence and boundary values. B remains the field entering the magnetic Lorentz force on a test charge. Both H and M use amperes per metre, whereas B uses teslas.

A linear isotropic susceptibility relates M to the internal H, not automatically to the field that would exist if the sample were removed. Sample shape produces a demagnetizing contribution to H. For a uniformly magnetized ellipsoid aligned with a principal axis, one scalar demagnetizing factor can describe that contribution. A sphere has factor one third. This result is a special geometry theorem; a general shaped body need not have a uniform internal field or a single usable factor.

\mathbf B=\mu_0(\mathbf H+\mathbf M),\qquad \nabla\times\mathbf H=\mathbf J_f,\qquad H_{\mathrm{int}}=H_0-N_dM,\quad M=\chi_mH_{\mathrm{int}}

The first relation defines SI H and the second assumes magnetostatics. The scalar demagnetization equations additionally assume an ellipsoid, uniform aligned magnetization, and a linear isotropic susceptibility χm. Nd is dimensionless, H0 is the applied field without the sample, and Hint is the internal field.↔ Scroll wide equations horizontally.

A constitutive law is additional physics

Prescribing M and calculating its field is a source problem. Predicting M from an applied field is a response problem requiring a material law and boundary solution. In a linear isotropic medium, permeability equals vacuum permeability times one plus susceptibility, with susceptibility defined against internal H. In anisotropic media a tensor can be necessary. In ferromagnets a history-dependent response can replace any single-valued permeability relation, so dividing B by H does not produce a universal material constant.

The ideal infinitely long axially magnetized cylinder illustrates the distinction sharply. With no applied field, its interior H vanishes away from ends, while B equals vacuum permeability times M. For a finite cylinder, end effects create a demagnetizing field and an exterior return field. Using the long-cylinder approximation near an end, or asserting that its exterior field vanishes everywhere for a finite body, ignores the geometry that closes the physical magnetic flux.

Work it through

A uniformly magnetized long cylinder

A cylinder of radius 0.020 m and length 0.30 m has prescribed uniform magnetization $1000\hat{\mathbf z}$ A/m, no free current, and no applied field. Find its bound currents and total moment, and estimate the interior B far from its ends using the infinite-cylinder approximation.

Uniform M has zero curl in the bulk, so Jb=0. On the cylindrical side, $\hat{\mathbf z}\times\hat{\mathbf r}=\hat{\boldsymbol\phi}$ , giving $\mathbf K_b=1000\hat{\boldsymbol\phi}$ A/m. On the flat end faces the cross product is zero.
The prescribed total moment is magnetization times volume: $m=1000\pi(0.020)^2(0.30)=0.377\,\mathrm{A\,m^2}$ , directed along +z. This calculation does not need the long-cylinder field approximation.
Approximating the side current as an infinite solenoid gives $\mathbf B\approx\mu_0\mathbf M=1.26\hat{\mathbf z}$ mT in the central interior. In this limit $\mathbf H=\mathbf B/\mu_0-\mathbf M=0$ . A finite sample has corrections from its ends, even though no free current flows.

The result

Only the side carries bound sheet current, of magnitude 1000 A/m. The total moment is 0.377 A m² and the central long-cylinder estimate is +1.26 mT; H≈0 here does not imply B≈0.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

Curl as an equivalent volume current

Interpret the plotted vector as a scaled planar magnetization. Reset and set both controls to Source strength q=0 and Rotation Ω=0.5; predict readouts Fy=0.5 and magnitude 0.5 at (1,0). For a fresh comparison, Reset and set both controls to q=0 and Ω=−1; predict Fy=−1 and magnitude 1 there. The formula Ω(−y,x) gives analytic curl +1 then −2, which is not a readout. Use readouts for magnitude comparisons because arrow lengths are compressed nonlinearly.

Local field directionSource strength

Fx at (1, 0): 0.833
Fy at (1, 0): 0.5
Field magnitude there: 0.972

Source strength q1Rotation Ω0.5

t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. 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 forCurl measures the volume bound-current contribution of an assumed magnetization field. This planar illustration has no bounded magnetic body, surface current, demagnetizing-field calculation, or physical B readout.

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Check your understanding

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Read the checkpoint explanation

Uniform magnetization eliminates the bulk curl but leaves the side surface current. The infinite-solenoid equivalent produces B=μ₀M inside. Subtracting M in the definition of H then gives zero, consistently with the absence of free current.

Your turn

Try it yourself

A linear isotropic sphere has susceptibility χm=2.0 and is placed in an applied field H0=300 A/m along z. Using demagnetizing factor Nd=1/3, find its internal H, magnetization, and B.

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

Give me a hint

Solve $H_{\mathrm{int}}=H_0-\chi_mH_{\mathrm{int}}/3$ before computing magnetization. Using H0 directly in the constitutive law would ignore the sphere's field.

Compare with the solution

Collecting internal-field terms gives $H_{\mathrm{int}}=300/(1+2/3)=180$ A/m. Hence $M=2(180)=360$ A/m, and $B=\mu_0(180+360)=\mu_0(540)=0.679$ mT along z. The internal H is smaller than the applied H, while B is enhanced relative to the empty-space value μ₀H0. These conclusions use the uniform sphere solution and a fixed linear susceptibility; they need not survive saturation or a different sample shape.

Keep the idea

Take it with you

Uniform magnetization can have zero volume current and nonzero surface current.
H separates free-current sources, while B remains the magnetic Lorentz-force field.
A susceptibility law uses internal H, which depends on sample geometry and demagnetization.

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