Magnetized Matter, Bound Currents and H

Derivation

Magnetized Matter, Bound Currents and H

D-056 Home PU-102 Threads fields · matter Depends on Magnetic Dipole: Field, Torque and Energy, Ampère's Circuital Law from Biot–Savart

Statement

A body carrying magnetization \(\vec{M}(\vec{r})\) (magnetic dipole moment per unit volume) produces exactly the same vector potential, and hence the same \(\vec{B}\), as an ordinary bound volume current \(\vec{J}_b=\nabla\times\vec{M}\) together with a bound surface current \(\vec{K}_b=\vec{M}\times\hat{n}\). Consequently Ampère's law reads \(\nabla\times\vec{B}=\mu_0(\vec{J}_f+\vec{J}_b)\), and the auxiliary field \(\vec{H}\equiv\dfrac{\vec{B}}{\mu_0}-\vec{M}\) obeys \(\nabla\times\vec{H}=\vec{J}_f\), i.e. \(\oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}}\), with only the free current on the right.

Why it matters

Inside real matter the microscopic magnetization currents are hopelessly detailed, yet they enter Maxwell's equations only through the smooth field \(\vec{M}\). Recognising that their macroscopic effect is exactly two effective currents lets us solve magnetostatics in media with the same tools used in vacuum, and it isolates the current we actually control — the free current in wires — into a single field \(\vec{H}\).

This is the magnetic counterpart of bound charge and the displacement field \(\vec{D}\). It underpins every calculation with permeable cores, permanent magnets, and boundary conditions at magnetic interfaces, and it makes the constitutive relation \(\vec{B}=\mu_0(\vec{H}+\vec{M})\) the organising identity of magnetism in matter.

Assumptions

Magnetostatics.Currents are steady, so \(\partial\vec{E}/\partial t=0\) and Ampère's law carries no displacement-current term; drop this and \(\nabla\times\vec{H}=\vec{J}_f\) acquires a \(\partial\vec{D}/\partial t\) correction.

Continuum / dipole picture.Each volume element carries a net moment \(d\vec{m}=\vec{M}\,dV'\) and higher multipoles are negligible over the scale on which \(\vec{M}\) varies; drop this and \(\vec{J}_b=\nabla\times\vec{M}\) is only the leading term.

Piecewise-smooth \(\vec{M}\) with a definite boundary.\(\nabla\times\vec{M}\) must exist in the interior and the body must have a well-defined surface with outward normal \(\hat{n}\); if \(\vec{M}\) jumps inside the volume, that discontinuity is itself a surface with its own \(\vec{K}_b=(\vec{M}_1-\vec{M}_2)\times\hat{n}\), otherwise a current is silently lost.

\(\vec{H}\) is not automatically curl-free.\(\nabla\times\vec{H}=\vec{J}_f\) holds always, but \(\oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}}\) is only useful when symmetry fixes \(\vec{H}\); in general \(\nabla\cdot\vec{H}=-\nabla\cdot\vec{M}\neq0\), so \(\vec{H}\) also carries a "magnetic-charge" part Ampère's law alone cannot supply.

Derivation

\[ \vec{A}(\vec{r})=\frac{\mu_0}{4\pi}\,\frac{\vec{m}\times(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^{3}} \]

Vector potential of a single point dipole \(\vec{m}\) at \(\vec{r}'\) (prior result: magnetic dipole moment and torque). A

\[ \vec{A}(\vec{r})=\frac{\mu_0}{4\pi}\int_V \frac{\vec{M}(\vec{r}')\times(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^{3}}\,d^3r' \]

Superpose dipoles: each element carries \(d\vec{m}=\vec{M}\,d^3r'\) and potentials add linearly. A

\[ \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}=\nabla'\frac{1}{|\vec{r}-\vec{r}'|},\qquad \vec{A}=\frac{\mu_0}{4\pi}\int_V \vec{M}\times\left(\nabla'\frac{1}{|\vec{r}-\vec{r}'|}\right)d^3r' \]

Rewrite the geometric kernel as a gradient with respect to the source coordinate \(\vec{r}'\); the primed gradient makes the sign positive. B

\[ \vec{M}\times\left(\nabla'\frac{1}{|\vec{r}-\vec{r}'|}\right)=\frac{\nabla'\times\vec{M}}{|\vec{r}-\vec{r}'|}-\nabla'\times\left(\frac{\vec{M}}{|\vec{r}-\vec{r}'|}\right) \]

Product rule \(\nabla'\times(f\vec{M})=f\,\nabla'\times\vec{M}+(\nabla'f)\times\vec{M}\) with \(f=1/|\vec{r}-\vec{r}'|\), rearranged using \(\vec{M}\times\nabla'f=-(\nabla'f)\times\vec{M}\). B

\[ \vec{A}=\frac{\mu_0}{4\pi}\int_V \frac{\nabla'\times\vec{M}}{|\vec{r}-\vec{r}'|}\,d^3r'-\frac{\mu_0}{4\pi}\int_V \nabla'\times\left(\frac{\vec{M}}{|\vec{r}-\vec{r}'|}\right)d^3r' \]

Insert step 4 into step 2 and split the linear integral. A

\[ -\int_V \nabla'\times\left(\frac{\vec{M}}{|\vec{r}-\vec{r}'|}\right)d^3r'=\oint_S \frac{\vec{M}\times\hat{n}'}{|\vec{r}-\vec{r}'|}\,da' \]

Curl theorem \(\int_V(\nabla\times\vec{F})\,dV=-\oint_S \vec{F}\times\hat{n}\,da\) (corollary of the divergence theorem); here \(\vec{F}=\vec{M}/|\vec{r}-\vec{r}'|\) and \(-\vec{F}\times\hat{n}'=\vec{M}\times\hat{n}'/|\vec{r}-\vec{r}'|\). C

\[ \vec{A}=\frac{\mu_0}{4\pi}\int_V \frac{\vec{J}_b}{|\vec{r}-\vec{r}'|}\,d^3r'+\frac{\mu_0}{4\pi}\oint_S \frac{\vec{K}_b}{|\vec{r}-\vec{r}'|}\,da'\;\Longrightarrow\;\vec{J}_b=\nabla\times\vec{M},\ \ \vec{K}_b=\vec{M}\times\hat{n} \]

Match term-by-term against the standard potential of a volume-plus-surface current; the source-to-\(\vec{A}\) map is unique, so the identifications follow. B

\[ \nabla\times\vec{B}=\mu_0(\vec{J}_f+\vec{J}_b)=\mu_0\vec{J}_f+\mu_0\,(\nabla\times\vec{M}) \]

Ampère's law (prior result: Ampère from Biot–Savart) with total current split into free and bound; substitute \(\vec{J}_b\) from step 7. A

\[ \nabla\times\left(\frac{\vec{B}}{\mu_0}-\vec{M}\right)=\vec{J}_f\;\Longrightarrow\;\vec{H}\equiv\frac{\vec{B}}{\mu_0}-\vec{M} \]

Divide by \(\mu_0\), move \(\nabla\times\vec{M}\) to the left, and combine the curls by linearity; define the bracket as \(\vec{H}\). A

\[ \oint_{\mathcal C}\vec{H}\cdot d\vec{l}=I_{f,\text{enc}} \]

Integrate \(\nabla\times\vec{H}=\vec{J}_f\) over any surface bounded by \(\mathcal C\) and apply Stokes' theorem. B

Result

\[ \vec{J}_b=\nabla\times\vec{M},\quad \vec{K}_b=\vec{M}\times\hat{n}\;\Big|\;\vec{H}=\frac{\vec{B}}{\mu_0}-\vec{M},\quad \nabla\times\vec{H}=\vec{J}_f,\quad \oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}} \]

Reading. Magnetization that curls in the interior (\(\nabla\times\vec{M}\neq0\)) is indistinguishable from a real volume current; magnetization that terminates at a surface (\(\vec{M}\times\hat{n}\neq0\)) acts as a surface sheet of current, exactly like the wound surface of a solenoid. The field \(\vec{H}\) is engineered to subtract the bound part, so its curl "sees" only the current we put in by hand.

Units check. \([\vec{M}]=\text{A·m}^2/\text{m}^3=\text{A/m}\). Then \([\nabla\times\vec{M}]=(\text{A/m})/\text{m}=\text{A/m}^2=[\vec{J}]\ \checkmark\), and \([\vec{M}\times\hat{n}]=\text{A/m}=[\vec{K}]\ \checkmark\) (surface current density). Also \([\vec{B}/\mu_0]=\text{T}\cdot\text{A}/(\text{T·m})=\text{A/m}=[\vec{M}]=[\vec{H}]\ \checkmark\); and \([\oint\vec{H}\cdot d\vec{l}]=(\text{A/m})\cdot\text{m}=\text{A}\ \checkmark\).

Limiting cases

Uniform \(\vec{M}\): \(\nabla\times\vec{M}=0\), so \(\vec{J}_b=0\) and the entire effect lives in the surface sheet \(\vec{K}_b=\vec{M}\times\hat{n}\) — a uniformly magnetized cylinder is a bound solenoid.
Linear medium \(\vec{M}=\chi_m\vec{H}\): then \(\vec{B}=\mu_0(1+\chi_m)\vec{H}=\mu\vec{H}\), and Ampère's law in \(\vec{H}\) closes without ever computing bound currents.
No free current (permanent magnet): \(\vec{J}_f=0\) gives \(\nabla\times\vec{H}=0\), so \(\vec{H}=-\nabla\Phi_M\) is fixed entirely by "magnetic charge" \(\rho_M=-\nabla\cdot\vec{M}\).
Vacuum (\(\vec{M}=0\)): \(\vec{H}=\vec{B}/\mu_0\) and the two Ampère laws coincide.

Breaks when

Time-varying fields. With \(\partial\vec{D}/\partial t\) present, \(\nabla\times\vec{H}=\vec{J}_f+\partial\vec{D}/\partial t\); the bound picture also gains a polarization current, so the static statement fails at high frequency.
Sharp discontinuities in \(\vec{M}\). Where \(\vec{M}\) jumps (a domain wall or buried interface) \(\nabla\times\vec{M}\) contains a delta function; treating it as a smooth volume current loses the interfacial \(\vec{K}_b=(\vec{M}_1-\vec{M}_2)\times\hat{n}\).
Atomic / mesoscopic scale. If the sampling volume is not large compared with the lattice, \(\vec{M}\) is not a smooth field and the dipole superposition (step 2) is meaningless; the true microscopic currents must be used.
Hysteretic / nonlinear media. If \(\vec{M}\) is not single-valued in \(\vec{H}\) (ferromagnets), the constitutive relation cannot be inverted and knowing \(\vec{H}\) alone does not fix \(\vec{B}\).

Failure modes

"\(\vec{H}\) is caused only by free current." False — \(\oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}}\) constrains the circulation of \(\vec{H}\), but \(\vec{H}\) also has a divergence part \(-\nabla\cdot\vec{M}\). A bar magnet with \(I_f=0\) still has nonzero \(\vec{H}\).
Sign/order error in \(\vec{K}_b=\vec{M}\times\hat{n}\). Writing \(\hat{n}\times\vec{M}\) reverses the surface current and flips the field; the outward normal and the order matter.
Forgetting the surface current. Computing only \(\vec{J}_b=\nabla\times\vec{M}\) for a uniformly magnetized object gives zero everywhere and the false conclusion \(\vec{B}=0\); the whole field comes from \(\vec{K}_b\).
Assuming \(\vec{B}=\mu_0\vec{H}\) inside matter. This drops \(\vec{M}\); the correct relation is \(\vec{B}=\mu_0(\vec{H}+\vec{M})\).
Mixing free and bound current in Ampère loops. Enclosing bound current when computing \(\vec{H}\) (or free current when computing \(\vec{B}\) in matter) double-counts or omits sources.

Discussion

The deep content of the derivation is that \(\vec{B}\) does not care whether a current is "free" or "bound": both enter \(\nabla\times\vec{B}=\mu_0\vec{J}_{\text{total}}\) on equal footing, because \(\vec{B}\) is the physically fundamental field — it deflects charges and threads flux. The split into free and bound is a bookkeeping choice about which currents we track, and \(\vec{H}\) is the field that results from that choice. This is why \(\vec{H}\) is called auxiliary: convenient, symmetry-adapted, and tied to laboratory currents, but not more fundamental than \(\vec{B}\).

The parallel with electrostatics is exact and worth holding in mind. Bound charge \(\rho_b=-\nabla\cdot\vec{P}\) and \(\sigma_b=\vec{P}\cdot\hat{n}\) mirror \(\vec{J}_b=\nabla\times\vec{M}\) and \(\vec{K}_b=\vec{M}\times\hat{n}\); the relation \(\vec{D}=\varepsilon_0\vec{E}+\vec{P}\) mirrors \(\vec{H}=\vec{B}/\mu_0-\vec{M}\). But the mirror is twisted: polarization couples through a divergence, magnetization through a curl. Hence \(\vec{D}\) is sourced by free charge (a divergence law) while \(\vec{H}\) is sourced by free current (a curl law), and \(\vec{H}\) — unlike \(\vec{D}\) — generally has both a curl and a divergence.

This asymmetry is the origin of the two complementary methods for magnets. When free currents are present and symmetric, one uses the Ampèrian route \(\oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}}\). When they are absent, \(\nabla\times\vec{H}=0\) permits a magnetic scalar potential \(\vec{H}=-\nabla\Phi_M\) with \(\nabla^2\Phi_M=\nabla\cdot\vec{M}\), so the demagnetizing field of a permanent magnet is solved by the same Poisson machinery as electrostatics, with \(-\nabla\cdot\vec{M}\) playing the role of charge density. The uniformly magnetized sphere, with uniform interior \(\vec{H}=-\vec{M}/3\) and a dipolar exterior, is the canonical instance of this correspondence.

Common misconceptions. Bound currents are not fictitious — they are real circulating currents (Ampèrian loops of orbiting and spinning electrons); "bound" only means they cannot be led off down a wire. And \(\vec{H}\) is not "the field in matter" versus \(\vec{B}\) "in vacuum": both fields exist everywhere, and it is \(\vec{B}\), not \(\vec{H}\), whose flux is conserved (\(\nabla\cdot\vec{B}=0\)) and that enters the Lorentz force.

Worked examples

Example 1 — Uniformly magnetized cylinder as a bound solenoid.

\[ \vec{M}=M\,\hat{z}\quad(\text{uniform, along the axis}) \]

A long circular cylinder magnetized uniformly parallel to its axis; outward normal on the curved wall is \(\hat{s}\). A

\[ \vec{J}_b=\nabla\times\vec{M}=0 \]

\(\vec{M}\) is constant, so its curl vanishes; there is no volume bound current. A

\[ \vec{K}_b=\vec{M}\times\hat{n}=M\,\hat{z}\times\hat{s}=M\,\hat{\phi} \]

On the curved surface \(\hat{n}=\hat{s}\) and \(\hat{z}\times\hat{s}=\hat{\phi}\): an azimuthal sheet of magnitude \(M\). B

\[ B_{\text{in}}=\mu_0 K_b=\mu_0 M \]

The bound sheet is exactly a solenoid with surface current per length \(K_b=M\); the long-solenoid interior field is \(\mu_0\) times current-per-length. Equivalently \(H_{\text{in}}=B/\mu_0-M=0\) (no free current). B

\[ M=8.0\times10^{5}\ \tfrac{\text{A}}{\text{m}}\;\Rightarrow\;B_{\text{in}}=(4\pi\times10^{-7})(8.0\times10^{5}) \]

Insert numbers for a moderately magnetized ferromagnet. A

\[ \vec{K}_b=8.0\times10^{5}\ \tfrac{\text{A}}{\text{m}}\ (\text{azimuthal}),\qquad B_{\text{in}}\approx 1.0\ \text{T},\qquad H_{\text{in}}=0 \]

Reading. A permanent magnet with no wires produces a tesla-scale interior field purely from its surface bound current, and \(\vec{H}\) vanishes inside the long cylinder because there is neither free current nor (locally) magnetic surface charge.

Example 2 — Solenoid with a linear magnetic core: use \(\vec{H}\) to bypass bound currents.

\[ \oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}}\;\Rightarrow\;H=nI_f \]

Long solenoid, \(n\) turns per metre carrying free current \(I_f\); the Ampèrian loop encloses \(n\) turns per unit length. Only free current appears, so \(H\) is independent of the core. B

\[ \vec{M}=\chi_m\vec{H},\qquad \vec{B}=\mu_0(\vec{H}+\vec{M})=\mu_0(1+\chi_m)\vec{H} \]

Linear soft-magnetic core of susceptibility \(\chi_m\); substitute the constitutive relation. A

\[ n=1000\ \text{m}^{-1},\ I_f=2.0\ \text{A}\;\Rightarrow\;H=2.0\times10^{3}\ \tfrac{\text{A}}{\text{m}} \]

Numbers for the free-current field. A

\[ \chi_m=200\;\Rightarrow\;M=200\times2.0\times10^{3}=4.0\times10^{5}\ \tfrac{\text{A}}{\text{m}},\quad B=\mu_0(201)(2.0\times10^{3}) \]

Compute magnetization and total field. A

\[ H=2.0\times10^{3}\ \tfrac{\text{A}}{\text{m}},\quad M=4.0\times10^{5}\ \tfrac{\text{A}}{\text{m}},\quad B\approx 0.51\ \text{T} \]

Reading. The same 2 A produces \(H=2000\ \text{A/m}\) whether or not the core is present; the core amplifies \(B\) by \(\mu_r=201\) entirely through its bound currents (\(K_b=M=4\times10^{5}\ \text{A/m}\) at the core surface), which we never had to compute directly.

Problems

A sphere of radius \(R\) is uniformly magnetized, \(\vec{M}=M_0\hat{z}\). Find the bound volume and surface current densities.
Solution
\(\vec{J}_b=\nabla\times\vec{M}=0\) since \(\vec{M}\) is uniform. On the surface \(\hat{n}=\hat{r}\), so \(\vec{K}_b=\vec{M}\times\hat{n}=M_0(\hat{z}\times\hat{r})\). With \(\hat{z}=\cos\theta\,\hat{r}-\sin\theta\,\hat{\theta}\) one gets \(\hat{z}\times\hat{r}=\sin\theta\,\hat{\phi}\). Hence \(\vec{K}_b=M_0\sin\theta\,\hat{\phi}\) — azimuthal, largest at the equator, zero at the poles. This is exactly the surface current that yields the uniform interior \(\vec{B}=\tfrac{2}{3}\mu_0\vec{M}\).
A long cylinder of radius \(a\) carries the azimuthal magnetization \(\vec{M}=k\,s^{2}\,\hat{\phi}\) for \(0\le s\le a\). Find the bound volume current density, and confirm the total bound current through a cross-section is zero.
Solution
For \(\vec{M}=M_\phi(s)\,\hat{\phi}\) the axial curl is \(J_{b,z}=\frac{1}{s}\frac{d}{ds}(s\,M_\phi)=\frac{1}{s}\frac{d}{ds}(k s^{3})=\frac{1}{s}(3k s^{2})=3ks\), so \(\vec{J}_b=3ks\,\hat{z}\). Surface current \(\vec{K}_b=\vec{M}\times\hat{n}=k a^{2}\,\hat{\phi}\times\hat{s}=-k a^{2}\,\hat{z}\). Volume total \(\int J_{b,z}\,dA=\int_0^{a}(3ks)(2\pi s)\,ds=6\pi k\,\frac{a^{3}}{3}=2\pi k a^{3}\). Surface total \(K_b(2\pi a)=-k a^{2}(2\pi a)=-2\pi k a^{3}\). They cancel, so the net bound current is zero — as required since \(\nabla\cdot\vec{J}_b=\nabla\cdot(\nabla\times\vec{M})=0\).
An iron rod of radius \(r=0.50\ \text{cm}\) is uniformly magnetized along its axis with \(M=1.5\times10^{6}\ \text{A/m}\). Treating it as a long cylinder, find the bound surface current per unit length, the interior \(B\), and the total circulating bound current over a \(2.0\ \text{cm}\) length.
Solution
\(K_b=M=1.5\times10^{6}\ \text{A/m}\) (azimuthal). Interior \(B=\mu_0 M=(4\pi\times10^{-7})(1.5\times10^{6})=1.88\ \text{T}\). Total circulating bound current over \(L=2.0\ \text{cm}\): \(I=K_b L=(1.5\times10^{6})(0.020)=3.0\times10^{4}\ \text{A}\). The radius affects the field only near the ends, not the long-cylinder interior value.
A toroid of \(N=500\) turns and mean radius \(R=8.0\ \text{cm}\) carries \(I_f=3.0\ \text{A}\) around a linear core with \(\chi_m=1500\). Find \(H\), \(M\) and \(B\) at the mean radius, and the bound surface current density on the core.
Solution
\(\oint\vec{H}\cdot d\vec{l}=I_{f,\text{enc}}\Rightarrow H(2\pi R)=NI_f\), so \(H=\frac{NI_f}{2\pi R}=\frac{500\times3.0}{2\pi(0.080)}=\frac{1500}{0.5027}=2.98\times10^{3}\ \text{A/m}\). Then \(M=\chi_m H=1500\times2.98\times10^{3}=4.48\times10^{6}\ \text{A/m}\), and \(B=\mu_0(1+\chi_m)H=(4\pi\times10^{-7})(1501)(2.98\times10^{3})=5.6\ \text{T}\). Bound surface current density \(K_b=M=4.48\times10^{6}\ \text{A/m}\), running around the core surface parallel to the windings. (Such a large \(B\) would saturate a real core; this treats \(\chi_m\) as constant for illustration.)
Show that for a permanent magnet carrying no free current \(\vec{H}\) can be written \(\vec{H}=-\nabla\Phi_M\), and state the equation obeyed by \(\Phi_M\) inside the material.
Solution
With \(\vec{J}_f=0\), \(\nabla\times\vec{H}=\vec{J}_f=0\), so \(\vec{H}\) is curl-free and admits a scalar potential \(\vec{H}=-\nabla\Phi_M\). Taking the divergence, \(\nabla\cdot\vec{H}=-\nabla^2\Phi_M\). But \(\vec{H}=\vec{B}/\mu_0-\vec{M}\) with \(\nabla\cdot\vec{B}=0\), so \(\nabla\cdot\vec{H}=-\nabla\cdot\vec{M}\). Therefore \(\nabla^2\Phi_M=\nabla\cdot\vec{M}=-\rho_M\), a Poisson equation with effective magnetic-charge density \(\rho_M=-\nabla\cdot\vec{M}\) (and surface density \(\sigma_M=\vec{M}\cdot\hat{n}\)). This is the magnetic-scalar-potential method, formally identical to electrostatics.