Vector Potential and Bound Currents

Derivation

Vector Potential and Bound Currents

D-288 Home PU-305 Threads fields · matter Depends on Ampère's Circuital Law from Biot–Savart

Statement

Because \(\nabla\!\cdot\!\mathbf{B}=0\) everywhere, the field admits a vector potential \(\mathbf{B}=\nabla\times\mathbf{A}\). Applying this to the field of magnetized matter, whose magnetization is \(\mathbf{M}\) (magnetic dipole moment per unit volume), the potential reduces exactly to that of a bound volume current \(\mathbf{J}_b=\nabla\times\mathbf{M}\) together with a bound surface current \(\mathbf{K}_b=\mathbf{M}\times\hat{\mathbf n}\). Absorbing the bound part of \(\nabla\times\mathbf{B}=\mu_0(\mathbf{J}_f+\mathbf{J}_b)\) defines the auxiliary field \(\mathbf{H}=\tfrac{1}{\mu_0}\mathbf{B}-\mathbf{M}\), obeying \(\nabla\times\mathbf{H}=\mathbf{J}_f\).

Why it matters

Inside real magnetic materials the true microscopic current is a hopeless tangle of atomic orbital and spin currents. This derivation shows that, at the macroscopic scale, that tangle is captured completely by a single smooth field \(\mathbf{M}\), and its magnetic effect is identical to two ordinary currents \(\mathbf{J}_b\) and \(\mathbf{K}_b\). Magnetostatics of matter therefore needs no new laws — only bookkeeping.

The reward is \(\mathbf{H}\): a field whose curl sees only the currents we control in the lab (free currents in wires). It plays the role for magnetism that \(\mathbf{D}\) plays for dielectrics, and it is what makes boundary-value problems in magnetic media tractable.

Assumptions

The field is magnetostaticif \(\partial_t\mathbf{E}\neq0\), Ampère's law gains the displacement-current term and \(\nabla\times\mathbf{H}=\mathbf{J}_f\) acquires \(\partial_t\mathbf{D}\).

\(\nabla\!\cdot\!\mathbf{B}=0\) holds with no exceptionthis is what guarantees a single-valued \(\mathbf{A}\) exists; a magnetic monopole density would forbid \(\mathbf{B}=\nabla\times\mathbf{A}\) globally.

\(\mathbf{M}\) is a well-defined macroscopic averageif we probe below the averaging scale (individual atoms), \(\mathbf{M}\) is not smooth and \(\mathbf{J}_b=\nabla\times\mathbf{M}\) is meaningless.

Each volume element is a pure point dipoleretaining quadrupole and higher moments of the atomic current adds correction terms beyond \(\mathbf{J}_b\); these are utterly negligible macroscopically but nonzero in principle.

Derivation

\[ \nabla\!\cdot\!\mathbf{B}=0 \quad\Longrightarrow\quad \mathbf{B}=\nabla\times\mathbf{A} \]

The divergence of any curl vanishes identically, so a divergence-free field is expressible as a curl (Helmholtz / Poincaré, on a simply connected domain). A

\[ \mathbf{A}\to\mathbf{A}+\nabla\chi \quad\Rightarrow\quad \mathbf{B}\ \text{unchanged},\qquad \text{choose }\nabla\!\cdot\!\mathbf{A}=0 \]

The curl of a gradient is zero, so \(\mathbf{A}\) is fixed only up to a gauge \(\chi\); the Coulomb gauge \(\nabla\!\cdot\!\mathbf{A}=0\) is always attainable and simplifies what follows. B

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

Prior result: the vector potential of a point magnetic dipole \(\mathbf{m}\) at \(\mathbf{r}'\). Superpose over the dipole distribution \(\mathbf{m}=\mathbf{M}\,d^3r'\). A

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

Linear superposition of the potentials of all volume elements (magnetostatics is linear in sources). A

\[ \nabla'\!\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)=\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^{3}} \quad\Rightarrow\quad \mathbf{A}=\frac{\mu_0}{4\pi}\int_V \mathbf{M}\times\nabla'\!\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)d^3r' \]

Rewrite the geometric kernel as a primed gradient; the gradient acts on source coordinates \(\mathbf{r}'\), preparing an integration by parts. B

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

Product rule \(\nabla\times(f\mathbf{M})=f\,\nabla\times\mathbf{M}-\mathbf{M}\times\nabla f\). Solve for the integrand of Step 5. C

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

Substitute Step 6 into Step 5, splitting into a "curl of source" term and a total-curl term. B

\[ \int_V \nabla'\times\mathbf{F}\,d^3r' = -\oint_S \mathbf{F}\times d\mathbf{a}' \quad\Rightarrow\quad -\int_V \nabla'\times\!\left(\tfrac{\mathbf{M}}{|\mathbf{r}-\mathbf{r}'|}\right)d^3r'=\oint_S \frac{\mathbf{M}\times\hat{\mathbf n}}{|\mathbf{r}-\mathbf{r}'|}\,da' \]

The curl theorem converts the second volume integral to a surface integral over the boundary of the magnetized body. C

\[ \mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V \frac{\nabla'\times\mathbf{M}}{|\mathbf{r}-\mathbf{r}'|}\,d^3r' \;+\;\frac{\mu_0}{4\pi}\oint_S \frac{\mathbf{M}\times\hat{\mathbf n}}{|\mathbf{r}-\mathbf{r}'|}\,da' \]

Combine Steps 7–8. Each term has the form of the potential of an ordinary current, letting us read off the effective sources. A

\[ \boxed{\;\mathbf{J}_b=\nabla\times\mathbf{M}\;},\qquad \boxed{\;\mathbf{K}_b=\mathbf{M}\times\hat{\mathbf n}\;} \]

Compare Step 9 term-by-term with \(\mathbf{A}=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}}{|\mathbf{r}-\mathbf{r}'|}d^3r'+\frac{\mu_0}{4\pi}\oint\frac{\mathbf{K}}{|\mathbf{r}-\mathbf{r}'|}da'\). Identification is unique. A

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

Ampère's law (prior result) with the total current split into free plus bound; substitute Step 10. A

\[ \nabla\times\!\left(\frac{1}{\mu_0}\mathbf{B}-\mathbf{M}\right)=\mathbf{J}_f \quad\Longrightarrow\quad \mathbf{H}\equiv\frac{1}{\mu_0}\mathbf{B}-\mathbf{M} \]

Move \(\mu_0\nabla\times\mathbf{M}\) to the left and divide by \(\mu_0\); the bracket is a new field whose curl contains only free current. A

Result

\[ \mathbf{J}_b=\nabla\times\mathbf{M},\qquad \mathbf{K}_b=\mathbf{M}\times\hat{\mathbf n},\qquad \mathbf{H}=\frac{1}{\mu_0}\mathbf{B}-\mathbf{M},\qquad \nabla\times\mathbf{H}=\mathbf{J}_f \]

Reading. A magnetized body produces exactly the field of a current \(\nabla\times\mathbf{M}\) threading its interior plus a sheet current \(\mathbf{M}\times\hat{\mathbf n}\) wrapping its surface — no more, no less. The field \(\mathbf{H}\) is engineered so that only the currents you drive through wires appear as its source; the material's own response is hidden inside \(\mathbf{H}\)'s definition through \(\mathbf{M}\).

Units check. \([\mathbf{M}]=\mathrm{A\,m^{-1}}\); then \([\nabla\times\mathbf{M}]=\mathrm{A\,m^{-2}}=[\mathbf{J}]\) ✓, and \([\mathbf{M}\times\hat{\mathbf n}]=\mathrm{A\,m^{-1}}=[\mathbf{K}]\) (surface current per unit length) ✓. \([\mathbf{B}/\mu_0]=\mathrm{T}/(\mathrm{T\,m\,A^{-1}})=\mathrm{A\,m^{-1}}=[\mathbf{H}]\) ✓, so \(\mathbf{H}\) and \(\mathbf{M}\) are commensurate.

Limiting cases

\(\mathbf{M}=0\) (vacuum): \(\mathbf{J}_b=0\), \(\mathbf{K}_b=0\), and \(\mathbf{H}=\mathbf{B}/\mu_0\) — the auxiliary field collapses to \(\mathbf{B}\) scaled by \(\mu_0\).
Uniform \(\mathbf{M}\) inside a body: \(\nabla\times\mathbf{M}=0\) so \(\mathbf{J}_b=0\); all the current lives on the surface as \(\mathbf{K}_b=\mathbf{M}\times\hat{\mathbf n}\) — a uniformly magnetized bar behaves like a solenoid.
Linear medium \(\mathbf{M}=\chi_m\mathbf{H}\): \(\mathbf{B}=\mu_0(1+\chi_m)\mathbf{H}=\mu\mathbf{H}\), recovering the constitutive relation as a special case.
No free current, \(\mathbf{J}_f=0\): \(\nabla\times\mathbf{H}=0\), so \(\mathbf{H}=-\nabla\phi_M\) admits a magnetic scalar potential — the magnetostatic analogue of electrostatics.

Breaks when

Time-dependent fields. With \(\partial_t\mathbf{D}\neq0\), \(\nabla\times\mathbf{H}=\mathbf{J}_f\) is incomplete; the displacement current \(\partial_t\mathbf{D}\) must be added, and \(\mathbf{A}\) then also couples to the scalar potential through the gauge condition.
Below the averaging scale. At atomic resolution \(\mathbf{M}\) is not a smooth field; \(\nabla\times\mathbf{M}\) has no meaning and the real microscopic currents must be used instead.
Nonlinear / hysteretic media. When \(\mathbf{M}\) is a history-dependent, multivalued function of \(\mathbf{H}\) (ferromagnets), \(\mathbf{M}(\mathbf{H})\) is not a state function; \(\mathbf{H}\) is still well-defined but \(\mathbf{B}(\mathbf{H})\) is not single-valued.
Magnetic monopoles present. If \(\nabla\!\cdot\!\mathbf{B}=\mu_0\rho_m\neq0\), then \(\mathbf{B}\neq\nabla\times\mathbf{A}\) globally and the entire construction fails at the outset.

Failure modes

Treating \(\mathbf{H}\) as "the field caused by free currents." \(\mathbf{H}\) is not determined by \(\mathbf{J}_f\) alone — only its curl is. Its divergence \(\nabla\!\cdot\!\mathbf{H}=-\nabla\!\cdot\!\mathbf{M}\) is sourced by magnetization, so \(\mathbf{H}\) generally has a longitudinal part too.
Forgetting the surface current \(\mathbf{K}_b\). Computing only \(\mathbf{J}_b=\nabla\times\mathbf{M}\) for a uniformly magnetized bar gives zero everywhere and predicts no field — the entire effect is in \(\mathbf{K}_b\).
Sign / order errors in \(\mathbf{M}\times\hat{\mathbf n}\). Writing \(\hat{\mathbf n}\times\mathbf{M}\) flips the surface current direction; \(\hat{\mathbf n}\) must be the outward normal.
Applying Ampère's law to \(\mathbf{H}\) with a symmetry it doesn't have. \(\oint\mathbf{H}\!\cdot d\boldsymbol\ell=I_{f,\text{enc}}\) only pins down \(\mathbf{H}\) when the geometry forces \(\mathbf{H}\) to be uniform along the loop — not true near edges of finite magnets.
Confusing \(\chi_m\) sign conventions. Diamagnets have \(\chi_m<0\); using \(\mu=\mu_0(1+\chi_m)\) with the wrong sign gives \(\mu<\mu_0\) reversed.

Discussion

The deepest content of this derivation is a statement of equivalence, not of new physics: the magnetic field of any distribution of oriented atomic dipoles is indistinguishable from that of the two macroscopic bound currents. This is why a permanent magnet and a suitably wound solenoid produce identical external fields. The volume current \(\nabla\times\mathbf{M}\) arises where neighbouring dipoles fail to cancel — a spatial gradient or curl of \(\mathbf{M}\) leaves a net circulation — while at a surface the abrupt drop of \(\mathbf{M}\) to zero leaves the outer loops of the atomic currents uncancelled, giving the sheet \(\mathbf{M}\times\hat{\mathbf n}\).

The auxiliary field \(\mathbf{H}\) is a convenience, not a fundamental field: \(\mathbf{B}\) is what deflects charges and threads Faraday loops. \(\mathbf{H}\) earns its keep because its curl is blind to bound current, so Ampère loops enclosing only free current determine it directly. But \(\mathbf{H}\) is not curl-free-plus-nothing: because \(\nabla\!\cdot\!\mathbf{H}=-\nabla\!\cdot\!\mathbf{M}\), a bar magnet has "magnetic charge" surfaces that source \(\mathbf{H}\) inside, where \(\mathbf{H}\) and \(\mathbf{B}\) actually point in opposite directions.

The parallel with dielectrics is exact in structure: \(\mathbf{P}\leftrightarrow\mathbf{M}\), bound charge \(-\nabla\!\cdot\!\mathbf{P}\leftrightarrow\) bound current \(\nabla\times\mathbf{M}\), and \(\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}\leftrightarrow\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M}\). The sign asymmetry (\(+\mathbf{P}\) but \(-\mathbf{M}\)) reflects that polarization adds to \(\mathbf{D}\)'s source while magnetization is subtracted out of \(\mathbf{H}\) — a consequence of \(\mathbf{M}\) sitting on the same side of Ampère's law as \(\mathbf{J}\).

At a more careful level, the reduction to a pure point dipole in Step 3 discards the internal structure of each atomic current loop. Retaining the full multipole expansion of the bound current shows that \(\mathbf{J}_b=\nabla\times\mathbf{M}\) is only the leading term; the next correction involves the magnetic quadrupole density and is of order \((a/L)^2\) smaller, where \(a\) is the atomic scale and \(L\) the macroscopic scale. For any laboratory sample this is \(\sim10^{-16}\), which is why the bound-current picture is treated as exact. It also clarifies that \(\mathbf{M}\) itself is defined as the dipole moment per volume — higher moments are simply not part of \(\mathbf{M}\).

Common misconceptions. Bound currents are real currents (real moving charge), not a mathematical fiction — they dissipate no energy only because they are the persistent atomic currents, but they source \(\mathbf{B}\) exactly as free currents do. And \(\mathbf{H}\), despite its name "magnetic field" in older texts, is the auxiliary quantity; the physically primary field is \(\mathbf{B}\).

Worked examples

Example 1 — Uniformly magnetized iron rod (surface current only). A long cylindrical rod is uniformly magnetized along its axis, \(\mathbf{M}=M\,\hat{\mathbf z}\), with \(M=8.0\times10^{5}\ \mathrm{A\,m^{-1}}\) (typical for saturated iron). Find the bound currents and the interior field.

\[ \mathbf{J}_b=\nabla\times\mathbf{M}=\nabla\times(M\,\hat{\mathbf z})=0 \]

\(M\) is constant, so its curl vanishes — no volume bound current. A

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

On the curved surface the outward normal is \(\hat{\mathbf s}\); the cross product gives an azimuthal sheet current, exactly like a solenoid winding. A

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

A surface current \(K\) per unit length gives \(B=\mu_0 K\) inside a long solenoid (Ampère). B

\[ B_{\text{in}}=(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}})(8.0\times10^{5}\ \mathrm{A\,m^{-1}}) \]

Insert numbers only now. A

\[ \mathbf{J}_b=0,\quad K_b=8.0\times10^{5}\ \mathrm{A\,m^{-1}},\quad B_{\text{in}}\approx1.0\ \mathrm{T} \]

Reading. The magnetization alone, with no external winding, produces a tesla-scale field inside the rod. Since \(\mathbf{J}_f=0\), \(\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M}=0\) inside — a useful check.

Units check. \(\mathrm{T\,m\,A^{-1}}\times\mathrm{A\,m^{-1}}=\mathrm{T}\) ✓.

Example 2 — Non-uniform magnetization (volume current, and total-current cancellation). Inside a cylinder of radius \(R=0.10\ \mathrm{m}\) the magnetization is azimuthal and grows with radius: \(\mathbf{M}=k\,s\,\hat{\boldsymbol\phi}\), with \(k=5.0\times10^{2}\ \mathrm{A\,m^{-2}}\). Find \(\mathbf{J}_b\), \(\mathbf{K}_b\), and verify the net bound current is zero.

\[ \mathbf{J}_b=\nabla\times\mathbf{M}=\frac{1}{s}\frac{\partial}{\partial s}\big(s\,M_\phi\big)\,\hat{\mathbf z}=\frac{1}{s}\frac{\partial}{\partial s}\big(s\cdot k s\big)\,\hat{\mathbf z} \]

For a purely azimuthal field depending on \(s\), the curl in cylindrical coordinates reduces to this single \(z\)-component. B

\[ \mathbf{J}_b=\frac{1}{s}\,(2ks)\,\hat{\mathbf z}=2k\,\hat{\mathbf z} \]

Differentiate \(k s^2\); the volume bound current is uniform along the axis. A

\[ \mathbf{K}_b=\mathbf{M}\times\hat{\mathbf n}\big|_{s=R}=kR\,\hat{\boldsymbol\phi}\times\hat{\mathbf s}=-kR\,\hat{\mathbf z} \]

At the outer surface \(\hat{\mathbf n}=\hat{\mathbf s}\), and \(\hat{\boldsymbol\phi}\times\hat{\mathbf s}=-\hat{\mathbf z}\). A

\[ I_{\text{vol}}=J_b\,\pi R^2=2k\pi R^2,\qquad I_{\text{surf}}=K_b\,(2\pi R)=-kR\,(2\pi R)=-2k\pi R^2 \]

Total axial current from each contribution; they are equal and opposite. B

\[ J_b=2(5.0\times10^{2})=1.0\times10^{3}\ \mathrm{A\,m^{-2}},\qquad K_b=-(5.0\times10^{2})(0.10)=-50\ \mathrm{A\,m^{-1}} \]

Insert numbers. A

\[ \mathbf{J}_b=1.0\times10^{3}\,\hat{\mathbf z}\ \mathrm{A\,m^{-2}},\quad \mathbf{K}_b=-50\,\hat{\mathbf z}\ \mathrm{A\,m^{-1}},\quad I_{\text{net}}=0 \]

Reading. Bound currents always carry zero net current through any cross-section — the outward-flowing volume current is exactly returned by the surface sheet. This is guaranteed because \(\oint\mathbf{M}\!\cdot d\boldsymbol\ell\) around a loop enclosing the whole body equals the enclosed bound current, and \(\mathbf{M}=0\) outside.

Units check. \([2k]=\mathrm{A\,m^{-2}}\) ✓; \([kR]=\mathrm{A\,m^{-2}}\cdot\mathrm{m}=\mathrm{A\,m^{-1}}\) ✓.

Problems

Show explicitly that \(\nabla\!\cdot\!\mathbf{J}_b=0\) for any \(\mathbf{M}\), and explain why this must hold for a magnetostatic bound current.
Solution
\(\nabla\!\cdot\!\mathbf{J}_b=\nabla\!\cdot\!(\nabla\times\mathbf{M})=0\) identically, since the divergence of a curl vanishes. Physically, magnetostatics requires \(\nabla\!\cdot\!\mathbf{J}=0\) (steady currents form closed loops with no charge accumulation), so any valid bound current must be divergenceless — and \(\nabla\times\mathbf{M}\) automatically is. This is a consistency check that the dipole model produces only steady currents.
A sphere of radius \(R\) is uniformly magnetized, \(\mathbf{M}=M\hat{\mathbf z}\). Find the bound surface current \(\mathbf{K}_b(\theta)\) and its maximum value for \(M=4.0\times10^{5}\ \mathrm{A\,m^{-1}}\).
Solution
Outward normal \(\hat{\mathbf n}=\hat{\mathbf r}\). \(\mathbf{K}_b=\mathbf{M}\times\hat{\mathbf r}=M\,\hat{\mathbf z}\times\hat{\mathbf r}\). Using \(\hat{\mathbf z}=\cos\theta\,\hat{\mathbf r}-\sin\theta\,\hat{\boldsymbol\theta}\), we get \(\hat{\mathbf z}\times\hat{\mathbf r}=-\sin\theta\,(\hat{\boldsymbol\theta}\times\hat{\mathbf r})=\sin\theta\,\hat{\boldsymbol\phi}\). So \(\mathbf{K}_b=M\sin\theta\,\hat{\boldsymbol\phi}\), maximal at the equator \(\theta=\pi/2\): \(K_{b,\max}=M=4.0\times10^{5}\ \mathrm{A\,m^{-1}}\). (This \(\sin\theta\) sheet is exactly a spinning uniformly charged sphere, giving a uniform interior field \(\mathbf{B}=\tfrac{2}{3}\mu_0\mathbf{M}\).)
A linear magnetic medium (\(\mathbf{M}=\chi_m\mathbf{H}\), \(\chi_m=1500\)) fills a toroid wound with \(n=1000\) turns/m carrying free current \(I_f=2.0\ \mathrm{A}\). Find \(H\), \(B\), and \(M\) inside.
Solution
Ampère's law for \(\mathbf{H}\): \(\oint\mathbf{H}\!\cdot d\boldsymbol\ell=I_{f,\text{enc}}\Rightarrow H=nI_f=1000\times2.0=2.0\times10^{3}\ \mathrm{A\,m^{-1}}\). Then \(B=\mu_0(1+\chi_m)H=(4\pi\times10^{-7})(1501)(2.0\times10^{3})\approx3.77\ \mathrm{T}\). And \(M=\chi_m H=1500\times2.0\times10^{3}=3.0\times10^{6}\ \mathrm{A\,m^{-1}}\). The medium amplifies \(B\) by the factor \((1+\chi_m)\approx1501\) over the vacuum value \(\mu_0 H\approx2.5\ \mathrm{mT}\).
Inside a cylinder of radius \(R=0.05\ \mathrm{m}\) the magnetization points along the axis but varies with radius: \(\mathbf{M}=M_0(1-s^2/R^2)\,\hat{\mathbf z}\), \(M_0=6.0\times10^{5}\ \mathrm{A\,m^{-1}}\). Find \(\mathbf{J}_b(s)\) and its value at \(s=R/2\).
Solution
For \(\mathbf{M}=M_z(s)\hat{\mathbf z}\), \(\nabla\times\mathbf{M}=-\dfrac{\partial M_z}{\partial s}\,\hat{\boldsymbol\phi}\). Here \(\dfrac{\partial M_z}{\partial s}=M_0(-2s/R^2)\), so \(\mathbf{J}_b=+\dfrac{2M_0 s}{R^2}\,\hat{\boldsymbol\phi}\). At \(s=R/2\): \(J_b=\dfrac{2M_0(R/2)}{R^2}=\dfrac{M_0}{R}=\dfrac{6.0\times10^{5}}{0.05}=1.2\times10^{7}\ \mathrm{A\,m^{-2}}\), directed azimuthally. (Surface current \(\mathbf{K}_b=\mathbf{M}\times\hat{\mathbf s}\big|_R=0\) since \(M_z(R)=0\) — the magnetization already vanishes at the boundary.)
Starting from \(\mathbf{H}=\mathbf{B}/\mu_0-\mathbf{M}\), derive the boundary conditions on the normal component of \(\mathbf{B}\) and the tangential component of \(\mathbf{H}\) at an interface carrying free surface current \(\mathbf{K}_f\).
Solution
From \(\nabla\!\cdot\!\mathbf{B}=0\), a Gaussian pillbox gives \(B_\perp^{\text{above}}-B_\perp^{\text{below}}=0\): the normal component of \(\mathbf{B}\) is continuous. From \(\nabla\times\mathbf{H}=\mathbf{J}_f\), an Amperian rectangle straddling the surface gives \(\mathbf{H}_\parallel^{\text{above}}-\mathbf{H}_\parallel^{\text{below}}=\mathbf{K}_f\times\hat{\mathbf n}\): the tangential component of \(\mathbf{H}\) is discontinuous by the free surface current (and continuous where \(\mathbf{K}_f=0\)). Note the bound surface current \(\mathbf{K}_b\) does not appear — that is precisely the advantage of \(\mathbf{H}\).