Chapter 10 — Magnetostatics

Quote

“At every point in space at which there is a finite magnetic force there is … a magnetic field.”

— William Thomson (1851)

Motivation

We wish to calculate the forces and torques (energy in Chapter 12) on a current distribution ${\mathbf{j}}^{\ast }(\mathbf{r})$ caused by the magnetic field of another current distribution $\mathbf{j}(\mathbf{r})$:

$$\mathbf{F}=\int {\mathrm{d}}^{3}r{\mathbf{j}}^{\ast }(\mathbf{r})\times \mathbf{B}(\mathbf{r}),\mathbf{N}=\int {\mathrm{d}}^{3}r\mathbf{r}\times [{\mathbf{j}}^{\ast }(\mathbf{r})\times \mathbf{B}(\mathbf{r})]$$

Under magnetostatic conditions:

$$\nabla \cdot \mathbf{B}=0,\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$$

Taking the divergence of the curl equation gives $\nabla \cdot \mathbf{j}=0$ — so magnetostatics requires steady currents (Chapter 9).

Magnetostatics is electrostatics on steroids. Technically, the cross product (instead of scalar) complicates field-line geometry, matching conditions, and superposition. Physically, electric currents from moving charges plus magnetization from spin make magnetizable matter far more varied than polarizable matter — ferromagnets alone have no electric analogue of comparable practical importance.

Field Lines

Field lines of $\mathbf{B}$ are integral curves of the ODE $\mathrm{d}\mathbf{r}/\mathrm{d}s\propto \mathbf{B}(\mathbf{r})$ — formally identical to electrostatics. In practice, things are less straightforward: $\mathbf{B}$-lines never start or end in space, but are not forced to close either.

The magnetic flux through a surface $S$ is:

$${\Phi }_B={\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{B}$$

For a closed surface, ${\Phi }_B={\int }_V{\mathrm{d}}^{3}r\nabla \cdot \mathbf{B}=0$. Shrinking $V$ arbitrarily small, this forbids field lines from beginning or ending at isolated points — the three possibilities are:

1. Close on themselves (textbook case, rare in general).
2. Begin and end at infinity (e.g., infinite straight wire).
3. Fill a surface or volume ergodically (neither close nor go to infinity — more on this below).

By analogy with electrostatics, $\nabla \cdot \mathbf{B}=0$ encodes the absence of magnetic monopoles and disallows being able to rotate between $\mathbf{E}$ and $\mathbf{B}$ freely.

---

Thomson’s Theorem

Thomson's Theorem

$|\mathbf{B}(\mathbf{r})|$ can have a local minimum, but never a local maximum, in a current-free volume.

Proof

Suppose $\mathbf{B}\cdot \mathbf{B}$ has a local maximum at $P$, with $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$ in a small ball $V$ around $P$. Then:

$${\int }_S\mathrm{d}\mathbf{S}\cdot \nabla (\mathbf{B}\cdot \mathbf{B})<0$$

By the divergence theorem:

$${\int }_S\mathrm{d}\mathbf{S}\cdot \nabla (\mathbf{B}\cdot \mathbf{B})=2{\int }_V{\mathrm{d}}^{3}r({\nabla }_i{B}_j{)}^2+2{\int }_V{\mathrm{d}}^{3}r{B}_j{\nabla }^2{B}_j$$

The second integral vanishes: $\nabla \times \mathbf{B}=0$ gives

$${\nabla }_i{B}_j={\nabla }_j{B}_i⟹B_j{\nabla }^2{B}_j=B_j{\nabla }_i{\nabla }_j{B}_i=B_j{\nabla }_j(\nabla \cdot \mathbf{B})=0$$

Hence ${\int }_S\mathrm{d}\mathbf{S}\cdot \nabla (\mathbf{B}\cdot \mathbf{B})\ge 0$, contradicting the assumption.

Why the Asymmetry — Harmonic vs. Subharmonic

Earnshaw is about $\phi$, which is harmonic (${\nabla }^2\phi =0$) in charge-free regions — so $\phi$ has neither local maxima nor local minima. Thomson is about $|\mathbf{B}{|}^2$, which is subharmonic (${\nabla }^2|\mathbf{B}{|}^2\ge 0$) in current-free regions — so it has no local maxima, but local minima are allowed. The asymmetry comes directly from the $({\nabla }_i{B}_j{)}^2\ge 0$ term in the proof: it forces ${\nabla }^2|\mathbf{B}{|}^2\ge 0$, which rules out maxima but is perfectly consistent with minima (where the field vanishes or bottoms out). These are genuinely different statements about different quantities — not two faces of the same “harmonic” machinery.

Physical Consequence — Magnetic Traps

A paramagnet has energy $U\propto -|\mathbf{B}{|}^2$ (it seeks maxima of $|\mathbf{B}|$); a diamagnet has $U\propto +|\mathbf{B}{|}^2$ (it seeks minima). Thomson’s theorem therefore forbids purely magnetostatic traps for paramagnets — there are no free-space maxima to trap them at — but permits traps for diamagnets, since minima are allowed. This is why diamagnetic levitation of water, graphite, and living frogs works, and why “magnetic bottles” for low-field-seeking cold atoms and plasmas are possible. Stable paramagnetic trapping requires time-dependent fields or active feedback.

Electric Analogue — No Trap for Polarizable Atoms

The same argument applies to $|\mathbf{E}{|}^2$ in a charge-free region: ${\nabla }^2(E_i{E}_i)=2({\nabla }_j{E}_i{)}^2+2E_i{\nabla }^2{E}_i=2({\nabla }_j{E}_i{)}^2\ge 0$, since each Cartesian component of $\mathbf{E}$ satisfies Laplace’s equation. So $|\mathbf{E}{|}^2$ is subharmonic, and the maximum principle forbids local maxima in the interior. A neutral atom with polarizability $\alpha >0$ has energy $U=-\frac{1}{2}\alpha |\mathbf{E}{|}^2$ — stable equilibrium would require a maximum of $|\mathbf{E}{|}^2$, which cannot exist. This is the electrostatic twin of Thomson’s theorem: the atom with $\alpha >0$ plays the role of the paramagnet (attracted to strong-field regions), and no trap is possible for the same reason.

---

Biot–Savart Law

With $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$, the Helmholtz theorem gives:

$$\mathbf{B}(\mathbf{r})=\nabla \times \frac{{\mu }_0}{4\pi }\int {\mathrm{d}}^3{r}^{\prime }\frac{\mathbf{j}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

This integral converges for sufficiently well-behaved localized currents, and limits carefully taken for infinite distributions (planes, lines) exactly as in electrostatics. Moving the curl inside and swapping to primed coordinates yields the empirically discovered Biot–Savart law:

$$\mathbf{B}(\mathbf{r})=\frac{{\mu }_0}{4\pi }\int {\mathrm{d}}^3{r}^{\prime }\frac{\mathbf{j}({\mathbf{r}}^{\prime })\times (\mathbf{r}-{\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}$$

Specializing to surface currents $\mathbf{K}$ or filamentary currents $I$ via $\mathbf{j}{\mathrm{d}}^{3}r\to \mathbf{K}\mathrm{d}S\to I\mathrm{d}\mathbf{\ell }$:

$$\mathbf{B}(\mathbf{r})=\frac{{\mu }_0}{4\pi }{\int }_S\mathrm{d}S\frac{\mathbf{K}({\mathbf{r}}_S)\times (\mathbf{r}-{\mathbf{r}}_S)}{|\mathbf{r}-{\mathbf{r}}_S{|}^3}=\frac{{\mu }_{0}I}{4\pi }\oint \frac{\mathrm{d}\mathbf{\ell }\times (\mathbf{r}-\mathbf{\ell })}{|\mathbf{r}-\mathbf{\ell }{|}^3}$$

Magnetic Fields Care About Circulation

Irrotational Currents Produce No Field

A localized curl-free current distribution ($\nabla \times \mathbf{j}=0$) produces $\mathbf{B}=0$ everywhere.

Proof

Rewrite Biot–Savart as $\mathbf{B}(\mathbf{r})=\frac{{\mu }_0}{4\pi }\int {\mathrm{d}}^3{r}^{\prime }\mathbf{j}({\mathbf{r}}^{\prime })\times {\nabla }^{\prime }\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$. Integrate by parts:

$$\mathbf{B}(\mathbf{r})=\frac{{\mu }_0}{4\pi }\int {\mathrm{d}}^3{r}^{\prime }\frac{{\nabla }^{\prime }\times \mathbf{j}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}-\frac{{\mu }_0}{4\pi }{\int }_S\mathrm{d}{\mathbf{S}}^{\prime }\times \frac{\mathbf{j}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

For sufficiently localized $\mathbf{j}$, the surface term vanishes and the first term vanishes by $\nabla \times \mathbf{j}=0$.

The deep point: Biot–Savart looks like it cares about the current itself, but the rewritten form shows it really cares about the curl of the current — the part that swirls. An irrotational current (e.g., purely radial flow from a point source) has no circulation and produces no field.

Why Localization Matters

An infinite straight wire has $\nabla \times \mathbf{j}=0$ everywhere, yet clearly produces $\mathbf{B}\ne 0$. The resolution: the wire is not localized, so the surface term above does not vanish. Physically, a localized curl-free distribution decomposes into tiny radial “explosion” sources, each producing zero field by spherical symmetry. An infinite wire cannot be decomposed this way — it borrows its magnetic field from the return current “at infinity.” Close the circuit finitely and you introduce $\nabla \times \mathbf{j}\ne 0$ at the bends, which is where the field truly comes from.

---

Worked Examples

Current Ring on the Symmetry Axis

Derivation

A circular loop of radius $R$ in the $z=0$ plane with current $I\hat{\mathbf{\varphi }}$. For a point $z\hat{\mathbf{z}}$ on the axis, $\mathbf{r}-{\mathbf{r}}^{\prime }=-R{\hat{\mathbf{\rho }}}^{\prime }+z\hat{\mathbf{z}}$ and $|\mathbf{r}-{\mathbf{r}}^{\prime }|=\sqrt{R^2+z^2}$. With $\mathrm{d}{\mathbf{\ell }}^{\prime }=R\mathrm{d}\varphi {\hat{\mathbf{\varphi }}}^{\prime }$:

$$\mathrm{d}{\mathbf{\ell }}^{\prime }\times (\mathbf{r}-{\mathbf{r}}^{\prime })=R\mathrm{d}\varphi {\hat{\mathbf{\varphi }}}^{\prime }\times (-R{\hat{\mathbf{\rho }}}^{\prime }+z\hat{\mathbf{z}})=R^2\mathrm{d}\varphi \hat{\mathbf{z}}+Rz\mathrm{d}\varphi {\hat{\mathbf{\rho }}}^{\prime }$$

The ${\hat{\mathbf{\rho }}}^{\prime }=\cos \varphi \hat{\mathbf{x}}+\sin \varphi \hat{\mathbf{y}}$ component integrates to zero over $\varphi \in [0,2\pi ]$; only the $\hat{\mathbf{z}}$ part survives:

$$\mathbf{B}(z)=\frac{{\mu }_{0}I{R}^2}{2(R^2+z^2{)}^{3/2}}\hat{\mathbf{z}}$$

No analytic closed form exists off-axis — solving Laplace with matching conditions below will give a series expansion.

---

Infinitely Long Solenoid

Setup

An azimuthal surface current $\mathbf{K}=K\hat{\mathbf{\varphi }}$ on a cylinder of arbitrary but uniform cross section, extending to $z=\pm \infty$.

TODO: Add TikZ picture showing $\mathbf{R}$ sweeping around the cross section at the $z$ closest to $P$.

Derivation

Factor the Biot–Savart surface integral into a $z$-integral (along the solenoid) and a perimeter integral around the cross section closest to $P$:

$$\mathbf{B}(P)=\frac{{\mu }_{0}K}{4\pi }\oint \mathrm{d}{\mathbf{\ell }}^{\prime }{\int }_{-\infty }^{\infty }\mathrm{d}z\frac{{\hat{\mathbf{\varphi }}}^{\prime }\times \mathbf{R}}{R^3}$$

Write $\mathbf{R}={\mathbf{R}}_{\perp }-z\hat{\mathbf{z}}$ (with ${\mathbf{R}}_{\perp }$ in the cross-sectional plane) and $R^2=R_{\perp }^2+z^2$. The $z$-odd piece integrates to zero; the $z$-even piece gives $\int \mathrm{d}z/(R_{\perp }^2+z^2{)}^{3/2}=2/R_{\perp }^2$:

$$\mathbf{B}(P)=\frac{{\mu }_{0}K}{2\pi }\oint \frac{{\hat{\mathbf{\varphi }}}^{\prime }\times {\mathbf{R}}_{\perp }}{R_{\perp }^2}\mathrm{d}{\ell }^{\prime }$$

Now parametrize the perimeter by the angle $\theta$ that ${\mathbf{R}}_{\perp }$ subtends at $P$. Geometric reasoning: as ${\mathbf{R}}_{\perp }$ sweeps around the cross section, ${\hat{\mathbf{\varphi }}}^{\prime }\times {\mathbf{R}}_{\perp }/R_{\perp }^2\mathrm{d}{\ell }^{\prime }$ reduces to $\pm \hat{\mathbf{z}}\mathrm{d}\theta$ — the projected angular swept element.

• If $P$ is inside the cross section, ${\mathbf{R}}_{\perp }$ sweeps out a total angle $2\pi$ as ${\mathbf{\ell }}^{\prime }$ traces the boundary.
• If $P$ is outside, the net sweep is $0$ (the contribution from the near face cancels the far face).

$$\mathbf{B}(P)=\{\begin{array}{ll}{\mu }_{0}K\hat{\mathbf{z}}& P\text{ inside}\\ 0& P\text{ outside}\end{array}$$

The only assumptions are: uniform cross section, $\mathbf{K}$ constant in magnitude, $\mathbf{K}$ parallel to the azimuthal tangent of the cross section. (A $K_z$ component would produce additional axial current, effectively a separate current tube with its own geometry.)

---

Finite-Length Wire Segment

Derivation via Affine Parametrization

Choose the origin at the closest point on the wire to the observation point $P$, so $\mathrm{d}\mathbf{\ell }$ is along $\hat{\mathbf{I}}$ and $\mathrm{d}\mathbf{\ell }\times \mathbf{\ell }=0$. Let $\mathbf{a}$ be the unit vector along the wire, $\mathbf{b}$ and $\mathbf{c}$ the endpoints. Since $\mathrm{d}\mathbf{\ell }\times \mathbf{r}$ and $\mathbf{c}\times \mathbf{a}$ are parallel:

$$\frac{\mathrm{d}\mathbf{\ell }\times (\mathbf{r}-\mathbf{\ell })}{|\mathbf{r}-\mathbf{\ell }{|}^3}=\frac{\mathrm{d}\ell r}{(r^2+{\ell }^2{)}^{3/2}}\frac{\mathbf{c}\times \mathbf{a}}{|\mathbf{c}\times \mathbf{a}|}$$

with $r=|\mathbf{c}\times \mathbf{a}|/a$. Integrating $\ell$ from $\mathbf{a}\cdot \mathbf{b}/a$ to $\mathbf{a}\cdot \mathbf{c}/a$:

$$\mathbf{B}=\frac{{\mu }_{0}I}{4\pi }\frac{\mathbf{c}\times \mathbf{a}}{|\mathbf{c}\times \mathbf{a}{|}^2}\left[\frac{\mathbf{a}\cdot \mathbf{c}}{c}-\frac{\mathbf{a}\cdot \mathbf{b}}{b}\right]$$

Infinite-Wire Limit

As $a,b,c\to \infty$ with $r$ fixed: $\mathbf{c}\times \mathbf{a}/|\mathbf{c}\times \mathbf{a}{|}^2\to \hat{\mathbf{\varphi }}/(ra)$ and $\mathbf{a}\cdot \mathbf{c}/c-\mathbf{a}\cdot \mathbf{b}/b\to 2a$, recovering the standard result:

$$\mathbf{B}=\frac{{\mu }_{0}I}{2\pi r}\hat{\mathbf{\varphi }}$$

---

Ampère’s Law

Integrating $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ over any open surface $S$ bounded by a closed curve $C$:

$${\oint }_C\mathrm{d}\mathbf{\ell }\cdot \mathbf{B}={\mu }_0{\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{j}={\mu }_0{I}_{\text{enc}}$$

What "Enclosed" Means

For a given loop $C$, many surfaces $S$ can be bounded by it. Ampère’s law requires that the answer not depend on which $S$ you choose. This is automatic provided $\nabla \cdot \mathbf{j}=0$: differences between two surfaces form a closed region, through which net current vanishes by conservation. Violate $\nabla \cdot \mathbf{j}=0$ (e.g., a current that begins and ends abruptly in the bulk) and Ampère’s law becomes ambiguous (Problem 10.13 makes this dramatic).

Sign convention (right-hand rule): thumb along $I$, fingers curl in the direction of $\mathbf{B}$.

Symmetry + Ampère: Formal Arguments

$\mathbf{B}$ Is Axial — Reflection Picks Up an Extra Sign

Under reflection, polar vectors flip the perpendicular component; $\mathbf{B}$ (axial) flips the parallel component instead — equivalently, transform as a polar vector and multiply by $\det M=-1$. Forgetting this collapses every Ampère/symmetry argument: e.g. it would let you “prove” $B_z=0$ outside an infinite wire by reflection through a plane that obviously preserves the wire. When in doubt, derive the transformation rule from $\mathbf{B}=\nabla \times \mathbf{A}$ ($\mathbf{A}$ polar) or from Biot–Savart and a concrete current.

Infinite Line of Current

Translation invariance in $z$ and rotation invariance about $\hat{\mathbf{z}}$: $\mathbf{B}(\rho ,\varphi ,z)=\mathbf{B}(\rho )$.

$\nabla \cdot \mathbf{B}=0$ combined with $\mathbf{B}(\rho )$ forces $B_{\rho }=0$ (else field lines would start/end on the axis).

Reversing $I$ is equivalent to reflecting in the $xy$-plane. Both operations send $\mathbf{B}\to -\mathbf{B}$ by linearity. But $\mathbf{B}$ is an axial vector, so under $z\to -z$: ${\mathbf{B}}^{\prime }=-B_x\hat{\mathbf{x}}-B_y\hat{\mathbf{y}}+B_z\hat{\mathbf{z}}$. Forcing ${\mathbf{B}}^{\prime }=-\mathbf{B}$ gives $B_z=0$. Hence $\mathbf{B}=B(\rho )\hat{\mathbf{\varphi }}$, and Ampère delivers $\mathbf{B}={\mu }_{0}I/(2\pi \rho )\hat{\mathbf{\varphi }}$.

Infinite Current Sheet

$\mathbf{K}=K\hat{\mathbf{y}}$ at $z=0$. Translation invariance in $x,y$: $\mathbf{B}(\mathbf{r})=\mathbf{B}(z)$.

Two operations leave the current invariant and send $z\to -z$:

1. Reflection in the $z=0$ plane: ${\mathbf{B}}^{\prime }=-B_x\hat{\mathbf{x}}-B_y\hat{\mathbf{y}}+B_z\hat{\mathbf{z}}$ (axial).
2. $\pi$-rotation around $\hat{\mathbf{y}}$: ${\mathbf{B}}^{\prime \prime }=-B_x\hat{\mathbf{x}}+B_y\hat{\mathbf{y}}-B_z\hat{\mathbf{z}}$.

Requiring $\mathbf{B}(-z)={\mathbf{B}}^{\prime }(-z)={\mathbf{B}}^{\prime \prime }(-z)$ forces $B_y=B_z=0$, and $B_x(-z)=-B_x(z)$. An Amperian rectangle then gives:

$$\mathbf{B}=\frac{1}{2}{\mu }_{0}K\mathrm{sgn}(z)\hat{\mathbf{x}}$$

More generally, for a current sheet with normal $\hat{\mathbf{n}}$ pointing from sheet to observation point:

$$\mathbf{B}=\frac{{\mu }_0}{2}\mathbf{K}\times \hat{\mathbf{n}}$$

Axial Vectors — Reflection Intuition

Under reflection through a mirror plane:

• Polar vectors (position, velocity, force) transform like arrows — the component perpendicular to the mirror flips.
• Axial vectors (angular momentum, $\mathbf{B}$, torque) pick up an extra $\det M=-1$ — the component parallel to the mirror flips.

Picture this: take $\oint I\mathrm{d}\mathbf{\ell }\times {\hat{\mathbf{r}}}^{\prime }/|{\mathbf{r}}^{\prime }{|}^2$ and reflect through a vertical mirror. For the original wire (current rightward, ${\mathbf{r}}^{\prime }$ up), the integrand points out of the page. For the mirror image (current leftward, ${\mathbf{r}}^{\prime }$ up), the integrand points into the page. The Biot–Savart integrand flips sign under reflection — component parallel to the mirror (out-of-page) reverses. TODO: TikZ figure.

Physically, axial vectors represent oriented circulations (or planes) disguised as arrows via the right-hand rule; polar vectors are real arrows. Reflecting an oriented circulation flips its orientation, hence the sign.

Transformations in Curvilinear Coordinates

When doing a reflection in cylindrical or spherical coordinates, two things happen:

1. Physical object transforms. Polar vectors as arrows; axial vectors pick up $\det M=-1$ in addition.
2. Coordinate basis at the new point is different. $\hat{\mathbf{\rho }},\hat{\mathbf{\varphi }},\hat{\mathbf{r}},\hat{\mathbf{\theta }}$ change from point to point, while $\hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}}$ do not.

Confusion arises because Cartesian avoids the second step entirely. In cylindrical/spherical, you must first apply the (active) physical transformation to the vector (using the global Cartesian basis), then re-express the result in the (passive) local curvilinear basis at the new point.

---

Matching Conditions

Applying the small-pillbox + small-loop trick to a surface current $\mathbf{K}({\mathbf{r}}_S)$, using the result $\mathbf{B}=\frac{1}{2}{\mu }_0\mathbf{K}\times \hat{\mathbf{n}}$ for the self-field of the small patch:

$${\mathbf{B}}_1={\mathbf{B}}_S+\frac{1}{2}{\mu }_0\mathbf{K}\times {\hat{\mathbf{n}}}_2,{\mathbf{B}}_2={\mathbf{B}}_S-\frac{1}{2}{\mu }_0\mathbf{K}\times {\hat{\mathbf{n}}}_2$$

Subtracting and taking dot/cross products with ${\hat{\mathbf{n}}}_2$:

$${\hat{\mathbf{n}}}_2\cdot [{\mathbf{B}}_1-{\mathbf{B}}_2]=0,{\hat{\mathbf{n}}}_2\times [{\mathbf{B}}_1-{\mathbf{B}}_2]={\mu }_0\mathbf{K}$$

Normal component of $\mathbf{B}$ is continuous; tangential component can jump. Combining the tangential condition with ${\mathbf{B}}_{\text{in}}=0$ inside a perfect conductor gives ${\mu }_0\mathbf{K}=\hat{\mathbf{n}}\times {\mathbf{B}}_{\text{out}}$.

Force on a current sheet. Adding the two equations gives ${\mathbf{B}}_S=\frac{1}{2}({\mathbf{B}}_1+{\mathbf{B}}_2)$ — the local field on the sheet is the average of the two sides (the sheet cannot exert force on itself). So:

$$\mathbf{f}({\mathbf{r}}_S)=\frac{1}{2}\mathbf{K}({\mathbf{r}}_S)\times ({\mathbf{B}}_1+{\mathbf{B}}_2)$$

Toroidal Solenoid

$N$ turns carrying current $I$ wound on a torus. By symmetry (rotational around the axis), any field must be azimuthal: $\mathbf{B}(\rho ,z)=B(\rho ,z)\hat{\mathbf{\varphi }}$. An Amperian circle at radius $\rho$ inside the torus encloses $NI$; outside, zero. Hence:

$$\mathbf{B}=\{\begin{array}{ll}{\mu }_{0}NI/(2\pi \rho )\hat{\mathbf{\varphi }}& \text{inside}\\ 0& \text{outside}\end{array}$$

---

The Magnetic Scalar Potential

In a current-free region $V$, $\nabla \times \mathbf{B}=0$ permits:

$$\mathbf{B}(\mathbf{r})=-\nabla \psi (\mathbf{r}),{\nabla }^2\psi =0$$

All of Chapter 7’s Laplace-equation machinery applies. The difference from electrostatics: the matching condition $\hat{\mathbf{n}}\cdot ({\mathbf{B}}_1-{\mathbf{B}}_2)=0$ gives continuity of the normal derivative $\partial \psi /\partial n$, not of $\psi$ itself.

Off-Axis Field of a Current Ring

Using the On-Axis Field to Build the General Solution

Azimuthal symmetry gives expansions regular in each sub-volume:

$$\psi (r,\theta )=\{\begin{array}{ll}{\sum }_{\ell =1}^{\infty }{A}_{\ell }(r/R{)}^{\ell }{P}_{\ell }(\cos \theta )& r<R\\ {\sum }_{\ell =1}^{\infty }{B}_{\ell }(R/r{)}^{\ell +1}{P}_{\ell }(\cos \theta )& r>R\end{array}$$

Sums start at $\ell =1$ because the normal-derivative matching forces the $\ell =0$ coefficient to zero. The tangential matching would pull in associated Legendre functions — instead, we “go off the axis” (cf. Problem 7.20): compute $\psi$ on the axis and match via the Legendre generating function.

On axis: integrating $B_z(z)={\mu }_{0}I{R}^2/[2(R^2+z^2{)}^{3/2}]$:

$$\psi (z)=-\frac{{\mu }_{0}I}{2}\frac{z}{\sqrt{R^2+z^2}}$$

Using $(1-2xt+t^2{)}^{-1/2}=\sum t^{\ell }{P}_{\ell }(x)$ to expand in $z/R$ (for $z<R$) and differentiate:

$$\psi (r,\theta )=-\frac{{\mu }_{0}I}{2}\sum _{\ell =1}^{\infty }[P_{\ell -1}(0)-P_{\ell +1}(0)]\{\begin{array}{ll}(r/R{)}^{\ell }{P}_{\ell }(\cos \theta )& r<R\\ -(R/r{)}^{\ell +1}{P}_{\ell }(\cos \theta )& r>R\end{array}$$

Far-Field = Dipole

Keeping only $\ell =1$: $\psi (r,\theta )\approx ({\mu }_0/4\pi )\pi R^{2}I\cos \theta /r^2$. Identifying $m=I\pi R^2$ as the magnetic dipole moment, this is structurally identical to the electric dipole potential (Chapter 4) under $\mathbf{p}/{ϵ}_0\to {\mu }_0\mathbf{m}$. The $\mathbf{B}$ field lines far from any localized loop are identical to the $\mathbf{E}$ field lines of an electric dipole.

---

Historical: Helmholtz Coils

In 1853 Gaugain noted that a coil of radius $R$ in the $z=0$ plane produces an approximately uniform field at $z=R/2$ along the axis (where $B_z^{\prime \prime }(R/2)=0$ — an inflection point). Helmholtz pointed out that placing a second identical coil in the $z=R$ plane dramatically improves uniformity: by reflection symmetry about $z=R/2$, the field $B_z(u)$ with $u:=z-R/2$ satisfies $B_z(-u)=B_z(u)$, so all odd derivatives vanish at $u=0$. Combined with $B_z^{\prime \prime }(R/2)=0$ from the single-coil inflection point, the first nonvanishing derivative is the fourth — extremely flat near the midpoint.

Helmholtz's Lament

Helmholtz had lectured on the principle in 1849 but never published; Gaugain’s 1853 paper preempted him. His friend du Bois-Raymond consoled: “Your priority in the matter of the Gaugain [coil] is irretrievably lost… This is a shame, but it is a new warning not to hide one’s talents.” Helmholtz chose not to publish anyway — yet Wiedemann’s 1861 textbook attributed the design to him, and “Helmholtz coil” stuck.

---

Application: MRI Active Shielding

Physical Setup

MRI exploits nuclear magnetic resonance on proton spins. Image formation requires controlled gradient fields (from coils) that must not penetrate the superconducting magnet generating the main alignment field. Active shielding: given a current ${\mathbf{K}}_1(\varphi ,z)$ on an inner cylinder $(R_1)$, choose ${\mathbf{K}}_2$ on an outer cylinder $(R_2)$ so $\mathbf{B}=0$ for $\rho >R_2$.

Solution Sketch

Scalar potential with modified Bessel functions: Solutions of ${\nabla }^2\psi =0$ in cylindrical coordinates with Fourier integrals in $z$ and modified Bessel functions $I_m,K_m$ in $\rho$. Building in continuity of $B_{\rho }=-\partial \psi /\partial \rho$ at $\rho =R$ and boundedness:

$$\psi (\rho ,\varphi ,z)=\sum _m{e}^{im\varphi }\int \mathrm{d}\kappa A_m(\kappa )e^{i\kappa z}\times \{\begin{array}{ll}{K}_m^{\prime }(|\kappa |R)I_m(|\kappa |\rho )& \rho <R\\ I_m^{\prime }(|\kappa |R)K_m(|\kappa |\rho )& \rho >R\end{array}$$

The $\varphi$-component of the tangential matching condition, combined with the Bessel Wronskian $x[I_m^{\prime }(x)K_m(x)-I_m(x)K_m^{\prime }(x)]=1$, fixes $A_m(\kappa )$ in terms of the Fourier components $K_{\varphi }^m(\kappa )$.

For the two-cylinder problem, superpose two potentials. Outside the outer cylinder the potential vanishes iff:

$$K_{2\varphi }^m(\kappa )=-\frac{R_1{I}_m^{\prime }(|\kappa |R_1)}{R_2{I}_m^{\prime }(|\kappa |R_2)}{K}_{1\varphi }^m(\kappa )$$

Each Fourier mode shields independently — the physicist has full freedom to engineer ${\mathbf{K}}_1$ to produce the desired interior gradient, and ${\mathbf{K}}_2$ is then determined mode-by-mode. Used in virtually all commercial MRI scanners.

One Component Suffices

Steady surface currents satisfy ${\nabla }_{\text{surf}}\cdot \mathbf{K}=0$, i.e. $\frac{1}{R}{\partial }_{\varphi }{K}_{\varphi }+{\partial }_z{K}_z=0$. So specifying $K_{\varphi }$ determines $K_z$ (up to integration constants) — only one scalar component is independent.

---

Multi-Valuedness of $\psi$

$\psi$ is generally not single-valued. Integrating around a closed loop that encircles a line current:

$$\psi (A)-\psi (B)=\pm {\mu }_{0}I,A,B\text{ infinitesimally close}$$

This occurs whenever $V$ is not simply connected — there exists at least one loop that cannot continuously be deformed to a point. Analogy: $\log z$ picks up $2\pi i$ as you wind around $z=0$. More precisely, $\mathbf{B}$ is closed ($\nabla \times \mathbf{B}=0$) but not exact (not globally $-\nabla \psi$) whenever the first de Rham cohomology of the region is nontrivial — this is also at the heart of the Aharonov–Bohm effect.

Why Electrostatics Is Different

$\nabla \times \mathbf{E}=0$ holds everywhere including at sources (charges), so the domain of definition of $\phi$ is always simply connected, and $\phi$ is always single-valued. In contrast, $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ excludes current-carrying regions from $\psi$‘s domain, which can make $V$ multiply connected.

Barriers (Cut Surfaces)

Just as complex analysis uses branch cuts, we introduce a barrier (cut surface) that transforms $V$ into a simply connected domain. Solve ${\nabla }^2\psi =0$ in the cut domain; the matching condition across the barrier is:

$${\psi }_{+}-{\psi }_{-}=\pm {\mu }_{0}I,{\partial }_n{\psi }_{+}-{\partial }_n{\psi }_{-}=0$$

(Normal derivative continuous because $\mathbf{B}=-\nabla \psi$ is perfectly smooth there — no physical source at the barrier.) The barrier location is arbitrary; the physical $\mathbf{B}$ does not depend on where you put it.

Solid-Angle Representation

Derivation

Start from $\mathbf{B}(\mathbf{r})=({\mu }_{0}I/4\pi ){\oint }_C\mathrm{d}\mathbf{\ell }\times (\mathbf{r}-\mathbf{\ell })/|\mathbf{r}-\mathbf{\ell }{|}^3$ and apply the integral identity ${\oint }_C\mathrm{d}\mathbf{\ell }\times \mathbf{f}={\int }_S\mathrm{d}{S}_k\nabla f_k-{\int }_S\mathrm{d}\mathbf{S}\nabla \cdot \mathbf{f}$ to any surface $S$ bounded by $C$. The divergence term is a delta function (zero for $\mathbf{r}\notin S$), and the remaining integral is a gradient of the solid angle:

$$\mathbf{B}(\mathbf{r})=\frac{{\mu }_{0}I}{4\pi }\nabla {\int }_S\mathrm{d}{\mathbf{S}}^{\prime }\cdot \frac{\mathbf{r}-{\mathbf{r}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}=\frac{{\mu }_{0}I}{4\pi }\nabla {\Omega }_S(\mathbf{r})$$

Hence

$$\psi (\mathbf{r})=-\frac{{\mu }_{0}I}{4\pi }{\Omega }_S(\mathbf{r})$$

Solid Angle Jumps and the Matching Condition

The solid angle ${\Omega }_S$ jumps by $\pm 4\pi$ as the observation point crosses $S$ (from seeing $\sim +2\pi$ of the front to $\sim -2\pi$ of the back). Hence $\Delta \psi =\mp {\mu }_{0}I$, precisely the barrier matching condition.

The barrier is any $S$ bounded by $C$. Different choices of $S$ give different ${\Omega }_S$, but all give the same $\mathbf{B}=-\nabla \psi$ — the choice is gauge-like.

Ring Field on Axis via Solid Angle

Center a sphere on the observation point on the axis so that part of the sphere (a spherical cap) is bounded by the ring. The solid angle of the cap:

$$\Omega (z)=-{\int }_0^{2\pi }\mathrm{d}\varphi {\int }_{{\theta }_0}^{\pi }\mathrm{d}\theta \sin \theta =2\pi \left[\frac{z}{\sqrt{R^2+z^2}}-1\right]$$

Then:

$$\mathbf{B}(z)=-\nabla \psi =\frac{{\mu }_{0}I}{4\pi }\frac{\mathrm{d}\Omega }{\mathrm{d}z}\hat{\mathbf{z}}=\frac{{\mu }_{0}I{R}^2}{2(R^2+z^2{)}^{3/2}}\hat{\mathbf{z}}$$

---

The Vector Potential

$\mathbf{B}=-\nabla \psi$ fails where $\mathbf{j}\ne 0$. Since $\nabla \cdot \mathbf{B}=0$ holds everywhere, we instead introduce:

$$\mathbf{B}(\mathbf{r})=\nabla \times \mathbf{A}(\mathbf{r})$$

Immediate consequence — Stokes’ theorem for the flux:

$${\Phi }_B={\int }_S\mathrm{d}\mathbf{S}\cdot \mathbf{B}={\int }_S\mathrm{d}\mathbf{S}\cdot (\nabla \times \mathbf{A})={\oint }_C\mathrm{d}\mathbf{\ell }\cdot \mathbf{A}$$

essential in the quantum-mechanical treatment of charged particles in magnetic fields (Aharonov–Bohm).

Gauge Freedom

$\mathbf{A}\to \mathbf{A}+\nabla \chi$ leaves $\mathbf{B}$ invariant (since $\nabla \times \nabla \chi =0$). We impose one scalar constraint on $\mathbf{A}$ to fix this ambiguity — but only constraints that are achievable (solvable for $\chi$). Examples:

• Coulomb gauge: $\nabla \cdot \mathbf{A}=0$ gives ${\nabla }^2\chi =-\nabla \cdot {\mathbf{A}}_{\text{old}}$ (Poisson, solvable).
• Lorenz gauge: $\nabla \cdot \mathbf{A}+{\mu }_0{ϵ}_0{\partial }_t\phi =0$ gives a wave equation for $\chi$ (solvable).
• Not allowed example: simultaneously $\nabla \cdot \mathbf{A}=0$ and $A_z=0$ everywhere — two conditions on one function generically has no solution.

Coulomb Gauge and Vector Poisson

In Coulomb gauge, $\nabla \times (\nabla \times \mathbf{A})={\mu }_0\mathbf{j}$ becomes:

$${\nabla }^2\mathbf{A}(\mathbf{r})=-{\mu }_0\mathbf{j}(\mathbf{r})$$

Component-wise this is electrostatic Poisson, giving:

$$\mathbf{A}(\mathbf{r})=\frac{{\mu }_0}{4\pi }\int {\mathrm{d}}^3{r}^{\prime }\frac{\mathbf{j}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

with analogous surface and line versions. This is the Helmholtz-theorem expression with Coulomb gauge implicitly chosen.

Cartesian Caveat

Each Cartesian component of $\mathbf{A}$ is related to the corresponding component of $\mathbf{j}$ exactly as $\phi$ is related to $\sigma$. This does not hold for curvilinear components because curvilinear unit vectors vary with position. Exceptions:

$$\mathbf{j}=j(\rho ,z)\hat{\mathbf{\varphi }}\Rightarrow \mathbf{A}=A(\rho ,z)\hat{\mathbf{\varphi }}$$

and

$$\mathbf{j}=j(r,\theta )\hat{\mathbf{\phi }}\Rightarrow \mathbf{A}=A(r,\theta )\hat{\mathbf{\phi }}$$

(For example, a current ring.)

Matching Conditions on $\mathbf{A}$

Since each Cartesian component of $\mathbf{A}$ satisfies a Poisson-like equation with source $\mathbf{K}$ on surfaces, $\mathbf{A}$ itself is continuous across current sheets:

$${\mathbf{A}}_1({\mathbf{r}}_S)={\mathbf{A}}_2({\mathbf{r}}_S)$$

Independent check: $\mathbf{B}$ has only finite tangential jumps (not delta functions), so $\nabla \times \mathbf{A}$ contains no deltas, so $\mathbf{A}$ must be continuous.

Normal derivative jump (Problem 10.19): analogous to the electrostatic $\partial \phi /\partial n$ jump $=-\sigma /{ϵ}_0$ is:

$$\left(\frac{\partial {\mathbf{A}}_1}{\partial n_1}-\frac{\partial {\mathbf{A}}_2}{\partial n_1}\right){|}_S=-{\mu }_0\mathbf{K}({\mathbf{r}}_S)$$

(Derivable by integrating ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{j}$ across the sheet.)

Coulomb Gauge = Transverse Part of Helmholtz Decomposition

Helmholtz says $\mathbf{A}=-\nabla \Phi +\nabla \times \mathbf{W}$, with $\Phi$ determined by $\nabla \cdot \mathbf{A}$ and $\mathbf{W}$ by $\nabla \times \mathbf{A}=\mathbf{B}$. The longitudinal part contributes nothing to $\mathbf{B}$ (pure gauge). Coulomb gauge ($\nabla \cdot \mathbf{A}=0$) simply kills the longitudinal piece, leaving $\mathbf{A}$ purely transverse — the minimal $\mathbf{A}$ producing the required $\mathbf{B}$.

Direct Solution of $\nabla \times (\nabla \times \mathbf{A})={\mu }_0\mathbf{j}$

When symmetry reduces $\mathbf{A}$ to a single component, direct PDE solution can be simpler than the integral.

Uniform Cylindrical Wire

$\mathbf{j}=j\hat{\mathbf{z}}$ in $\rho \le a$, $\mathbf{j}=0$ outside. Coulomb gauge gives $\mathbf{A}=A_z(\rho )\hat{\mathbf{z}}$, satisfying:

$$\frac{1}{\rho }\frac{\mathrm{d}}{\mathrm{d}\rho }\left(\rho \frac{\mathrm{d}{A}_z}{\mathrm{d}\rho }\right)=-{\mu }_{0}j{\chi }_{[0,a]}$$

Inside: $A_z=-\frac{1}{4}{\mu }_{0}j{\rho }^2+D_0$ (the $\ln \rho$ piece excluded by regularity). Outside: $A_z=-\frac{1}{4}{\mu }_{0}j{a}^2+C\ln (\rho /a)+D_0$. Continuity of $A_z$ and tangential $\mathbf{B}$ matching at $\rho =a$ fix $C=-\frac{1}{2}{\mu }_{0}j{a}^2$. With $I=j\pi a^2$:

$$A_z(\rho )=\{\begin{array}{ll}-\frac{{\mu }_{0}I}{4\pi }\frac{{\rho }^2}{a^2}& \rho \le a\\ -\frac{{\mu }_{0}I}{4\pi }\left[1+2\ln (\rho /a)\right]& \rho \ge a\end{array},\mathbf{B}=B_{\varphi }(\rho )\hat{\mathbf{\varphi }}=\{\begin{array}{ll}\frac{{\mu }_{0}I\rho }{2\pi a^2}\hat{\mathbf{\varphi }}& \rho \le a\\ \frac{{\mu }_{0}I}{2\pi \rho }\hat{\mathbf{\varphi }}& \rho \ge a\end{array}$$

Current Ring via Vector Potential

With $\mathbf{K}=I\delta (\rho -R)\hat{\mathbf{\varphi }}$ at $z=0$, take $\mathbf{A}=A(\rho ,z)\hat{\mathbf{\varphi }}$. The double-curl equation becomes:

$$\frac{{\partial }^{2}A}{\partial z^2}+\frac{{\partial }^{2}A}{\partial {\rho }^2}+\frac{1}{\rho }\frac{\partial A}{\partial \rho }-\frac{A}{{\rho }^2}=0$$

Separating $A=R(\rho )Z(z)$ with $Z(z)=e^{\mp k|z|}$ gives Bessel’s equation of order one for $R(\rho )$: regularity selects $J_1(k\rho )$. The tangential-$\mathbf{B}$ jump at $z=0$ together with Bessel completeness gives:

$$A(\rho ,z)=\frac{{\mu }_{0}IR}{2}{\int }_0^{\infty }\mathrm{d}k{J}_1(k\rho )J_1(kR)e^{-k|z|}\hat{\mathbf{\varphi }}$$

---

Field-Line Topology

Lines That Never Close

Consider a uniform ring current plus an infinite coaxial line current. For a single ring, field lines are tangent to nested tori. Adding the line current puts a helical twist on each torus: a given line wraps around the torus helically with some rotation number. For generic current ratios this rotation number is irrational — the line winds forever without closing, filling the torus surface densely (ergodically).

Key Point

$\nabla \cdot \mathbf{B}=0$ forbids field lines from starting or ending, but does not force them to close. “Never starting or ending” is consistent with ergodically filling a surface forever. The simple closed-loop textbook pictures are special cases where extra symmetry constrains the topology.

TODO: Figure 10.19 adapted from McDonald (1954) — line current threading a ring current.

Magnetic Reconnection

Magnetic field line topology is normally fixed, but when sources move quasistatically, field lines can be brought together to touch at null points where $\mathbf{B}=0$. There lines can disconnect and reconnect, changing the overall topology — reconnection.

Concrete example: two horseshoe magnets far apart have separate field-line topologies (lines run N-to-S on each). As they approach, a null point forms at a critical separation, lines touch, and past that distance connectivity switches — lines now run N-of-one to S-of-the-other. The topology change is abrupt.

Physical relevance:

• Solar flares and coronal mass ejections: convection tangles the sun’s field over years, building up helicity and magnetic stress. Reconnection releases it violently.
• Earth’s magnetosphere: the interplanetary field (solar wind) meets the geomagnetic dipole. Where antiparallel, null points form; reconnection transiently links the IMF to Earth’s field lines before they disconnect downstream. Drives geomagnetic storms.
• Prominences, stellar dynamos, tokamak disruptions — all dominated by reconnection physics.

TODO: Add figures/examples of magnetic reconnection.

Chaotic Field Lines

Hamiltonian Analogy

Consider $\mathbf{B}(\mathbf{r})=B_0\hat{\mathbf{z}}+\hat{\mathbf{z}}\times \nabla f(\mathbf{r})$ — divergence-free by construction. Field-line ODEs:

$$\frac{\mathrm{d}x}{\mathrm{d}z}=-\frac{1}{B_0}\frac{\partial f}{\partial y},\frac{\mathrm{d}y}{\mathrm{d}z}=+\frac{1}{B_0}\frac{\partial f}{\partial x}$$

Identify $x=q$, $y=p$, $z=t$, $H=-f/B_0$ — these are exactly Hamilton’s equations.

The punchline: any divergence-free field with a guide field $B_0\hat{\mathbf{z}}$ describes how field lines wander in the $xy$-plane as one follows them along $z$. Since lines never terminate, this defines a flow on ${\mathbb{R}}^2$, and $\nabla \cdot \mathbf{B}=0$ makes it area-preserving. Area-preserving planar flows are Hamiltonian dynamics with one degree of freedom — a theorem, not an analogy.

Generic Hamiltonian systems with $\ge 2$ degrees of freedom are chaotic (KAM theory). Here one effective degree of freedom still allows integrability, but adding a third spatial variation (broken symmetry) gives two degrees of freedom and generically chaotic field lines — lines that never close, don’t lie on tori, fill volumes ergodically. Matters in plasma confinement: chaotic field lines let plasma wander across the device and escape.

Dimensionality and KAM

For $n=1$ d.o.f., phase space is 2D, trajectories are 1D curves, and conserved-quantity level sets act as barriers: no chaos topologically. For $n\ge 2$, tori have too low dimension to separate the $\ge 4$D phase space, and generic perturbations break them into chaotic seas.

TODO: Blog post on chaos in higher dimensions and KAM theory.

Helicity

$$h:=\int {\mathrm{d}}^{3}r\mathbf{A}\cdot \mathbf{B}$$

is a quantitative measure of the topological complexity of a $\mathbf{B}$-field configuration — used extensively in solar physics and plasma dynamo theory.

Linking of Flux Tubes

Define a flux tube as a bundle of field lines carrying common flux $\Phi$ through every cross section. Two tubes: $\mathbf{A}={\mathbf{A}}_1+{\mathbf{A}}_2$. Inside each tube, $\mathbf{B}\parallel \mathrm{d}\mathbf{S}\parallel \mathrm{d}\mathbf{\ell }$, so ${\mathrm{d}}^{3}r=\mathrm{d}\mathbf{S}\cdot \mathrm{d}\mathbf{\ell }$. The crucial algebraic rearrangement:

$$(\mathrm{d}\mathbf{S}\cdot \mathrm{d}\mathbf{\ell })(\mathbf{A}\cdot \mathbf{B})=(\mathrm{d}\mathbf{\ell }\cdot \mathbf{A})(\mathrm{d}\mathbf{S}\cdot \mathbf{B})$$

separates $h$ into line integrals of $\mathbf{A}$ times cross-sectional flux integrals:

$$h={\Phi }_1{\oint }_{C_1}\mathrm{d}\mathbf{\ell }\cdot \mathbf{A}+{\Phi }_2{\oint }_{C_2}\mathrm{d}\mathbf{\ell }\cdot \mathbf{A}$$

${\oint }_{C_1}\mathrm{d}\mathbf{\ell }\cdot \mathbf{A}$ is the flux linked by tube 1. If the tubes are unlinked, no flux links: $h=0$. If linked once, full ${\Phi }_2$ threads $C_1$ (and vice versa): $h=2{\Phi }_1{\Phi }_2$.

Gauge Invariance

Under $\mathbf{A}\to \mathbf{A}+\nabla \chi$: $h\to h+\int \mathrm{d}\mathbf{S}\cdot \mathbf{B}\chi$. So $h$ is gauge-invariant iff $\hat{\mathbf{n}}\cdot \mathbf{B}=0$ on the boundary — i.e., the system is topologically self-contained (no flux leaks out). Otherwise linking is undefined and helicity loses physical meaning.

Physical Significance

Helicity counts linking and twisting of field lines — a topological invariant unchanged by smooth deformations (you can’t unlink rings without cutting).

• Ideal plasma: in a highly conducting plasma, $\mathbf{B}$ is “frozen into” the fluid; helicity is approximately conserved even as the field deforms wildly.
• Sun: convective motions twist the magnetic field, building helicity over the cycle. Since helicity is conserved, stress accumulates until reconnection releases it as flares and coronal mass ejections.
• Tokamaks: plasma relaxes toward minimum-energy states at fixed helicity (Taylor states) — precisely the helical twisted field configurations needed for stable confinement. Helicity conservation explains why these configurations persist.
• Fluid analogy: kinetic helicity $\int \mathbf{v}\cdot (\nabla \times \mathbf{v}){\mathrm{d}}^{3}r$ measures linking of vortex tubes and is conserved in inviscid flow — identical mathematical structure tied to the frozen-in property.

Intuition: Why $\mathbf{A}\cdot \mathbf{B}$ Measures Topology

$\mathbf{B}$ points along the local field-line direction. $\mathbf{A}$ encodes flux threading loops around the point (via $\oint \mathrm{d}\mathbf{\ell }\cdot \mathbf{A}=\Phi$). Their dot product asks: as this field line passes through this point, how much other flux is wrapping around it?

If $\mathbf{A}\parallel \mathbf{B}$, the line threads circulating flux — linking. If $\mathbf{A}\perp \mathbf{B}$, the line moves sideways past the circulation — no linking. The volume integral sums this across all space.

Making it precise: the Gauss linking number

$$n_{12}=\frac{1}{4\pi }{\oint }_{C_1}{\oint }_{C_2}\frac{\mathrm{d}{\mathbf{\ell }}_1\times \mathrm{d}{\mathbf{\ell }}_2\cdot ({\mathbf{r}}_1-{\mathbf{r}}_2)}{|{\mathbf{r}}_1-{\mathbf{r}}_2{|}^3}$$

counts crossings. Helicity is the flux-weighted sum of pairwise linking numbers for all field lines in the volume.

TODO: Blog post on what solar flares and plasma physics have to do with topology.

---

Problems

10.3 — Finite-Length Solenoid I

(a) For a semi-infinite tightly wound circular solenoid, prove that the magnetic flux out the open end equals half the flux through a cross section deep inside. (b) For a solenoid of length $L=5R$, sketch the field lines near the open ends. Do any field lines pass through the walls?

Solution Sketch

(a) Superposition of two semi-infinite solenoids. Place two identical semi-infinite solenoids end-to-end with the same winding direction. The combined field is that of an infinite solenoid. By symmetry, the field at the common plane is the longitudinal infinite-solenoid field, which has flux ${\Phi }_{\infty }$ — this is the sum of contributions from each half. Each must contribute ${\Phi }_{\infty }/2$ through its open end.

(b) Yes, some field lines pass through the walls. All field lines form closed loops. From (a), only half the interior flux exits through the open end; the other half must escape through the walls. The matching conditions ($\hat{\mathbf{n}}\cdot ({\mathbf{B}}_{\text{in}}-{\mathbf{B}}_{\text{out}})=0$, $\hat{\mathbf{n}}\times ({\mathbf{B}}_{\text{in}}-{\mathbf{B}}_{\text{out}})={\mu }_0\mathbf{K}$) force the field lines to execute a sharp “hairpin” turn where they cross, since the interior and exterior fields near the wall are nearly antiparallel to the wall direction.

---

10.5 — A Step Off the Symmetry Axis

Given $B_z(0,z)={\mu }_{0}I{R}^2/[2(R^2+z^2{)}^{3/2}]$ for a current ring, use Maxwell’s equations to find $B_{\rho }(\rho ,z)$ and the first correction to $B_z(\rho ,z)$ for small $\rho$.

Solution Sketch

$\nabla \cdot \mathbf{B}=0$ in cylindrical: $\frac{1}{\rho }{\partial }_{\rho }(\rho B_{\rho })+{\partial }_z{B}_z=0$. For $\rho \ll R$: $B_{\rho }\approx f(z)\rho$, giving $2f(z)=-B_z^{\prime }(0,z)$, hence:

$$B_{\rho }(\rho ,z)\approx \frac{3{\mu }_{0}I{R}^2}{4}\frac{z\rho }{(R^2+z^2{)}^{5/2}}$$

Then $\nabla \times \mathbf{B}=0$ off the current ring gives ${\partial }_z{B}_{\rho }-{\partial }_{\rho }{B}_z=0$, integrating for the ${\rho }^2$ correction:

$$B_z(\rho ,z)\approx B_z(0,z)+\frac{3{\mu }_{0}I{R}^2}{8}{\rho }^2\frac{R^2-4z^2}{(R^2+z^2{)}^{7/2}}$$

Iteration note: Alternating $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$ generates successive higher-$\rho$ corrections — there’s mild arbitrariness in the order of application, but all give the same full expansion.

---

10.6 — Two Approaches to the Field of a Current Sheet

Use (a) Biot–Savart directly and (b) superposition of infinite straight wires to find $\mathbf{B}(\mathbf{r})$ for a current sheet at $x=0$ with $\mathbf{K}=K\hat{\mathbf{z}}$.

Solution Sketch

(a) Biot–Savart:

$$\mathbf{B}(\mathbf{r})=\frac{{\mu }_{0}K}{4\pi }\iint \mathrm{d}{y}^{\prime }\mathrm{d}{z}^{\prime }\frac{\hat{\mathbf{z}}\times (x\hat{\mathbf{x}}+(y-y^{\prime })\hat{\mathbf{y}})}{[x^2+(y-y^{\prime }{)}^2+z^{\prime 2}{]}^{3/2}}$$

The $\hat{\mathbf{x}}$-contribution is odd in $y-y^{\prime }$ and vanishes. The $z^{\prime }$-integral gives $2/[x^2+(y-y^{\prime }{)}^2]$, and the $y^{\prime }$-integral then gives $\pi \mathrm{sgn}(x)$:

$$\mathbf{B}=\frac{1}{2}{\mu }_{0}K\mathrm{sgn}(x)\hat{\mathbf{y}}$$

(b) Superposition of infinite wires: Each wire at $y^{\prime }$ gives $B={\mu }_0(K\mathrm{d}{y}^{\prime })/(2\pi R)$ azimuthally. $\hat{\mathbf{x}}$-contributions from $\pm s$ cancel; with $R^2=x^2+(y-y^{\prime }{)}^2$ and substitution $y-y^{\prime }=x\tan \theta$:

$$|\mathbf{B}|=\frac{{\mu }_{0}K}{2\pi }{\int }_{-\pi /2}^{\pi /2}\mathrm{d}\theta =\frac{1}{2}{\mu }_{0}K$$

---

10.7 — The Geometry of Biot and Savart

Biot and Savart’s original derivation used a wire bent at an angle. Find $\mathbf{B}$ in the plane of the wire at distance $d$ from the bend along the axis of symmetry.

Solution Sketch

Trick: Place the origin at the bend. Then $\mathrm{d}\mathbf{\ell }=\mathrm{d}\ell \hat{\mathbf{I}}$ with $\mathbf{\ell }$ along the same direction, so $\mathrm{d}\mathbf{\ell }\times \mathbf{\ell }=0$ in the Biot–Savart integrand — only the $\mathrm{d}\mathbf{\ell }\times \mathbf{r}$ piece contributes. The integral simplifies dramatically.

Alternative (angular parametrization): With $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$, $\mathrm{d}s\sin \theta =R\mathrm{d}\theta$ and $\sin \theta /\sin \alpha =d/R$. Both legs contribute equally into the page:

$$B=\frac{{\mu }_{0}I}{2\pi d\sin \alpha }{\int }_0^{\alpha }\mathrm{d}\theta \sin \theta =\frac{{\mu }_{0}I}{2\pi d}\frac{1-\cos \alpha }{\sin \alpha }=\frac{{\mu }_{0}I}{2\pi d}\tan (\alpha /2)$$

TODO: Add a figure or more explicit geometric discussion.

---

10.8 — The Magnetic Field of Planar Circuits

(a) For a wire bent into a planar loop with observation point $P$ in the plane at the origin, show:

$$B(P)=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{2\pi }\frac{\mathrm{d}\varphi }{r(\varphi )}$$

(b) Apply to an ellipse with axes $2a,2b$ (reduces to elliptic integral). (c) Planar coil with $\mathbf{K}=K\hat{\mathbf{\varphi }}$.

Solution Sketch

Key trick: In the plane of the loop with $P$ at origin, $\mathrm{d}\mathbf{\ell }\times {\mathbf{r}}^{\prime }=r^2\mathrm{d}\varphi \hat{\mathbf{z}}$. Writing ${\mathbf{r}}^{\prime }=r(\varphi )(\cos \varphi ,\sin \varphi ,0)$ and $\mathrm{d}\mathbf{\ell }=\mathrm{d}r(\cos \varphi ,\sin \varphi ,0)+r\mathrm{d}\varphi (-\sin \varphi ,\cos \varphi ,0)$: the cross product gives $r^2\mathrm{d}\varphi \hat{\mathbf{z}}$ (the $\mathrm{d}r$ term is parallel to ${\mathbf{r}}^{\prime }$, no contribution). Then:

$$B=\frac{{\mu }_{0}I}{4\pi }\int \frac{r^2\mathrm{d}\varphi }{r^3}=\frac{{\mu }_{0}I}{4\pi }\int \frac{\mathrm{d}\varphi }{r(\varphi )}$$

(b) Ellipse: $r(\varphi )=ab/\sqrt{b^2{\cos }^2\varphi +a^2{\sin }^2\varphi }$ → elliptic integral of the second kind $E(e)$. Circle ($a=b$): $B={\mu }_{0}I/(2a)$. Infinite wire ($a\to \infty$, $b$ fixed): recovers ${\mu }_{0}I/(2\pi b)\cdot 2$ for the two parallel runs.

(c) $B={\mu }_{0}K(1-\cos \alpha )/2$ on axis, where $\alpha$ is the half-angle subtended.

---

10.9 — Invert the Biot–Savart Law

For $\mathbf{K}(y,z)=K(z)\hat{\mathbf{y}}$ on the $x=x_0$ plane, show Biot–Savart reduces to a 1D convolution and invert to recover $K(z)$ from $\mathbf{B}(x,z)$ for $x\ne x_0$.

Solution Sketch

The $y$-integration collapses to leave a 1D convolution in $z$:

$$B_j(x,z)=\int \mathrm{d}{z}^{\prime }{G}_j(x-x_0,z-z^{\prime })K(z^{\prime })$$

Fourier-transforming diagonalizes the convolution; the kernel ${\tilde{G}}_j(x-x_0,k_z)$ is algebraic in modified Bessel / exponential form. Inverting gives $\tilde{K}(k_z)={\tilde{B}}_j(x,k_z)/{\tilde{G}}_j$ — essentially a Wiener-like deconvolution.

Physical significance: reconstruct an unknown surface current from magnetic measurements at a standoff distance — used in radiation therapy dosimetry and magnetoencephalography (MEG).

---

10.10 — Symmetry and Ampère’s Law

Current $I$ flows down the $z$-axis from $+\infty$ and spreads radially in the $z=0$ plane. Use reflection and rotation symmetries of $\mathbf{B}$ (axial vector) to show $\mathbf{B}=B_{\varphi }(\rho ,z)\hat{\mathbf{\varphi }}$, then apply Ampère.

Solution Sketch

(a) Reflection through $yz$-plane: axial $\mathbf{B}$ transforms with $\det M=-1$ extra, so ${\mathbf{B}}^{\prime }=B_x\hat{\mathbf{x}}-B_y\hat{\mathbf{y}}-B_z\hat{\mathbf{z}}$. The reflection sends $\varphi \to \pi -\varphi$, so $\sin \varphi$ is invariant and $\cos \varphi \to -\cos \varphi$. Combining: $B_{\rho }^{\prime }=-B_{\rho }$, $B_{\varphi }^{\prime }=B_{\varphi }$, $B_z^{\prime }=-B_z$.

(b) A $\pi$-rotation about $\hat{\mathbf{z}}$: $\tilde{\mathbf{B}}=-B_x\hat{\mathbf{x}}-B_y\hat{\mathbf{y}}+B_z\hat{\mathbf{z}}$, with $\varphi \to \varphi +\pi$, so ${\tilde{B}}_{\rho }=B_{\rho }$, ${\tilde{B}}_{\varphi }=B_{\varphi }$, ${\tilde{B}}_z=B_z$. For consistency: $B_{\rho }=0$ and $B_z=0$. Rotational symmetry gives $\mathbf{B}=B_{\varphi }(\rho ,z)\hat{\mathbf{\varphi }}$.

(c) Ampère on a circular loop at fixed $z$, radius $\rho$: encloses $I$ for $z>0$, zero for $z<0$:

$$B_{\varphi }(\rho ,z)=\{\begin{array}{ll}{\mu }_{0}I/(2\pi \rho )& z>0\\ 0& z<0\end{array}$$

(d) Matching at $z=0$: $\hat{\mathbf{z}}\times [{\mathbf{B}}_{+}-{\mathbf{B}}_{-}]={\mu }_0\mathbf{K}$ gives $\mathbf{K}=I/(2\pi \rho )\hat{\mathbf{\rho }}$ — matches the uniform radial spreading.

---

10.12 — Finite-Length Solenoid II

(a) Superpose ring fields to find $B_z$ at the midpoint of a finite solenoid ($R$, $L$, $n$ turns/unit length, current $I$). (b) For $L\gg R$, estimate $B_{\text{out}}$ near the wall (far from ends) via Ampère.

Solution Sketch

(a) Each ring contributes $\mathrm{d}I=nI\mathrm{d}z$; integrate $B_z={\mu }_0\mathrm{d}I{R}^2/[2(R^2+z^2{)}^{3/2}]$ from $-L/2$ to $L/2$:

$$B_z(0)=\frac{{\mu }_{0}nIL}{\sqrt{L^2+4R^2}}$$

(b) For $L\gg R$: $B_{\text{in}}\approx {\mu }_{0}nI$. An Amperian rectangle straddling the wall (two long sides parallel to the axis, short sides normal): only the long sides contribute. $B_{\text{out}}\ell +B_{\text{in}}\ell ={\mu }_{0}nI\ell$ would give zero exterior field, but the finite solenoid has a correction:

$$B_{\text{out}}\approx {\mu }_{0}nI\left[1-\frac{L}{\sqrt{L^2+4R^2}}\right]\approx -\frac{2{\mu }_{0}nI{R}^2}{L^2}$$

The minus sign (opposite to $B_{\text{in}}$) makes physical sense: field lines form closed loops, so near the wall the exterior field is anti-parallel to the interior.

---

10.13 — How Biot–Savart Differs From Ampère

A current $I_0$ flows up the $z$-axis from $z_1$ to $z_2$ (isolated segment — not a closed circuit). (a) Biot–Savart gives $\mathbf{B}(\rho ,0)=({\mu }_0{I}_0/4\pi \rho )[\cos {\theta }_2-\cos {\theta }_1]\hat{\mathbf{\varphi }}$. (b) Ampère in the $z=0$ plane naively gives zero because $\nabla \cdot \mathbf{j}\ne 0$. Add radial current densities ${\mathbf{j}}_{1,2}$ to close the circuit at infinity and reconcile.

Solution Sketch

(a) Biot–Savart: Parametrize $\mathrm{d}\mathbf{\ell }=\mathrm{d}{z}^{\prime }\hat{\mathbf{z}}$, $\mathbf{r}-\mathbf{\ell }=\rho \hat{\mathbf{\rho }}-z^{\prime }\hat{\mathbf{z}}$:

$$\mathbf{B}=\frac{{\mu }_0{I}_0}{4\pi }\hat{\mathbf{\varphi }}{\int }_{z_1}^{z_2}\frac{\rho \mathrm{d}{z}^{\prime }}{({\rho }^2+z^{\prime 2}{)}^{3/2}}=\frac{{\mu }_0{I}_0}{4\pi \rho }[\cos {\theta }_2-\cos {\theta }_1]\hat{\mathbf{\varphi }}$$

where $\cos {\theta }_i=z_i/\sqrt{{\rho }^2+z_i^2}$.

(b) Closure of the current: ${\mathbf{j}}_1=-(I_0/4\pi )(\mathbf{r}-{\mathbf{r}}_1)/|\mathbf{r}-{\mathbf{r}}_1{|}^3$ is a radial sink at ${\mathbf{r}}_1$ (current flowing inward from infinity); ${\mathbf{j}}_2$ is a radial source at ${\mathbf{r}}_2$. Check: $\nabla \cdot {\mathbf{j}}_1=-I_0{\delta }^3(\mathbf{r}-{\mathbf{r}}_1)$, which exactly cancels the $+I_0{\delta }^3(\mathbf{r}-{\mathbf{r}}_1)$ from $\nabla \cdot (\text{segment})$ at the bottom endpoint. Similarly for ${\mathbf{j}}_2$ at the top. Together the three currents form a topologically closed circuit (through infinity).

(c) Ampère with all three sources: An Amperian circle at $z=0$ encloses $I_0$ (all of it, since the segment pierces $z=0$ when $z_1<0<z_2$) plus contributions from ${\mathbf{j}}_1,{\mathbf{j}}_2$. The current through the disk from ${\mathbf{j}}_i$: use $\int \mathrm{d}\mathbf{S}\cdot {\mathbf{j}}_i=\mp (I_0/2)[z_i/\sqrt{{\rho }^2+z_i^2}-\mathrm{sgn}(z_i)]$.

For $z_1<0<z_2$: the $\mathrm{sgn}$ terms give $(-I_0/2)+(-I_0/2)=-I_0$, killing the $I_0$ that Ampère naively enclosed. The remaining cosines match Biot–Savart exactly. For $z_1,z_2$ same sign, the $\mathrm{sgn}$ terms cancel.

(d) Why ${\mathbf{j}}_{1,2}$ don’t spoil Biot–Savart: They’re spherically symmetric (curl-free) and sufficiently localized around their sources that the surface term in the $\mathbf{B}=({\mu }_0/4\pi )\int (\nabla \times \mathbf{j})/|\mathbf{r}-{\mathbf{r}}^{\prime }|-\text{surf}$ identity vanishes. Alternatively: for any observation point, each radial current element has a diametrically opposite counterpart, and their Biot–Savart contributions cancel.

---

10.15 — A Spinning Spherical Shell of Charge

Charge $Q$ uniform on a sphere of radius $R$, spinning at $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Find $\mathbf{B}$ everywhere using $\mathbf{B}=-\nabla \psi$.

Solution Sketch

$\mathbf{v}({\mathbf{r}}_S)=\mathbf{\omega }\times {\mathbf{r}}_S=\omega R\sin \theta \hat{\mathbf{\varphi }}$, so $\mathbf{K}=\sigma \omega R\sin \theta \hat{\mathbf{\varphi }}$.

Expand $\psi =\sum A_{\ell }{r}^{\ell }{P}_{\ell }$ inside, $\sum B_{\ell }{r}^{-\ell -1}{P}_{\ell }$ outside. Normal-derivative continuity at $r=R$: $B_{\ell }=-A_{\ell }\ell R^{2\ell +1}/(\ell +1)$. Tangential-$\mathbf{B}$ jump: $\hat{\mathbf{n}}\times \Delta \mathbf{B}={\mu }_0\mathbf{K}$ gives $A_{\ell }\propto {\delta }_{\ell 1}$:

$$A_1=-\frac{2}{3}{\mu }_0\omega R\sigma ,B_1=\frac{1}{3}{\mu }_0\omega R^4\sigma$$

So ${\psi }_{<}(r,\theta )=-\frac{2}{3}{\mu }_0\omega R\sigma r\cos \theta$ and ${\psi }_{>}(r,\theta )=\frac{1}{3}{\mu }_0\omega R^4\sigma \cos \theta /r^2$, giving:

$${\mathbf{B}}_{<}=\frac{2}{3}{\mu }_{0}R\sigma \mathbf{\omega },{\mathbf{B}}_{>}=\frac{{\mu }_0}{4\pi }\frac{3(\mathbf{m}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}}{r^3}$$

with magnetic dipole moment $\mathbf{m}=\frac{4}{3}\pi R^4\sigma \mathbf{\omega }$. Inside: uniform. Outside: pure dipole. (Same structure as the uniformly polarized sphere in electrostatics with $1/{ϵ}_0\leftrightarrow {\mu }_0$.)

---

10.16 — Distant Field of a Helical Coil

Infinite current filament wound as a helix of radius $R$ and pitch $\ell$ (one wind per length $\ell$ along $z$). Find $\mathbf{B}(\rho ,\varphi ,z)$ dependence for $\rho \gg \ell$. Does $\ell \to 0$ make sense?

Solution Sketch

Key symmetry — screw invariance: The helix is invariant under $\Delta z=a$ combined with $\Delta \varphi =2\pi a/\ell$. Infinitesimally: ${\partial }_z\psi =(2\pi /\ell ){\partial }_{\varphi }\psi$ — more cleanly, $\psi$ depends on $z-(\ell /2\pi )\varphi$ rather than $z$ and $\varphi$ separately.

Substituting $\psi (\rho ,\varphi ,z)={\sum }_m{e}^{im\varphi }{e}^{i{k}_{m}z}{R}_m(\rho )$ with $k_m=2\pi m/\ell$ (because there is true $\ell$ periodicity in $z$ direction) into Laplace reduces to modified Bessel’s equation: $R_m(\rho )\propto K_m(k_m\rho )$ for exterior decay.

Large-$\rho$ behavior: $K_m(x)\sim e^{-x}/\sqrt{x}$, so:

$$\mathbf{B}(\rho ,\varphi ,z)\sim \frac{e^{-2\pi \rho /\ell }}{\sqrt{\rho /\ell }}\times \text{oscillation}$$

$\ell \to 0$ limit: The decay rate $e^{-2\pi \rho /\ell }\to 0$ — no field outside. Consistent: a tightly wound helix approaches an infinite solenoid, which has zero external field.

---

10.17 — Distant Field of Helmholtz Coils

Two coaxial circular loops radius $R$, separation $R$, same-direction current $I$. (a) Show each asymptotic component $\sim f(\theta )/r^3+g(\theta )/r^7+\dots$ at $r\to \infty$ (dipole + hexadecapole, the $1/r^5$ octopole vanishes by mirror symmetry). (b) Show a second Helmholtz pair can cancel the dipole and leave only hexadecapole.

Solution Sketch

(a) Expand ${\psi }_{>}={\sum }_{\ell }{a}_{\ell }{P}_{\ell }(\cos \theta )/r^{\ell +1}$. The Helmholtz coil has mirror symmetry about $z=R/2$ (the midplane), which for the magnetic scalar potential centered at the midplane kills all even-$\ell$ terms. Additionally, the Helmholtz spacing was designed so that $B_z^{\prime \prime }$ vanishes on axis at the midpoint — this kills the $\ell =3$ term (octopole) in the axial expansion. So the leading terms are $\ell =1$ (dipole, $\sim 1/r^3$) and $\ell =5$ (hexadecapole, $\sim 1/r^7$).

(b) Cancelling the dipole: Place a second Helmholtz pair of opposite-signed current at a different radius/separation, tuned so its dipole moment equals minus that of the first. The ratio of dipole to hexadecapole scales differently with coil size, so a two-pair combination can be arranged to null the dipole while leaving a net hexadecapole.

---

10.18 — Solid Angles for Magnetic Fields

Use the solid-angle form $\psi =-({\mu }_{0}I/4\pi ){\Omega }_S$ to find $\mathbf{B}$ for an infinite straight current $I$. Specify the cut surface.

Solution Sketch

Place origin at the observation point so the current is parallel to $\hat{\mathbf{z}}$. Choose the cut $S$ as a semi-infinite half-plane extending from the wire out to infinity (closing via a “giant square” at infinity).

Solid angle at origin subtended by this half-plane: varies azimuthally. For an observation point at azimuth $\alpha$ (measured from the cut):

$$\Omega ={\int }_0^{2\alpha }\mathrm{d}\varphi {\int }_0^{\pi }\sin \theta \mathrm{d}\theta =2\cdot 2\alpha =-4\alpha$$

(sign from cut orientation). Hence $\psi =({\mu }_{0}I/\pi )\alpha$, and:

$$\mathbf{B}=-\nabla \psi =-\frac{{\mu }_{0}I}{\pi }\frac{1}{\rho }\hat{\mathbf{\alpha }}\cdot \frac{1}{r}\to \frac{{\mu }_{0}I}{2\pi \rho }\hat{\mathbf{\varphi }}$$

recovering the standard result. (The overall factor comes from careful bookkeeping of the solid-angle differential.)

Subtlety: the cut choice breaks the rotational symmetry of the original current — we have to “pick a branch” just as with $\log z$ — but $\mathbf{B}$ itself is cut-independent.

---

10.19 — A Matching Condition for $\mathbf{A}$

Show that $\hat{\mathbf{n}}\cdot \nabla \mathbf{A}$ jumps across a surface current.

Solution Sketch

In Coulomb gauge, each Cartesian component $A_k$ satisfies ${\nabla }^2{A}_k=-{\mu }_0{j}_k$ — identical to electrostatic Poisson with $j_k\leftrightarrow \rho /{ϵ}_0$. The electrostatic jump condition $\hat{\mathbf{n}}\cdot (\nabla {\phi }_1-\nabla {\phi }_2)=-\sigma /{ϵ}_0$ translates to:

$$\hat{\mathbf{n}}\cdot (\nabla A_{1,k}-\nabla A_{2,k})=-{\mu }_0{K}_k$$

i.e., $({\partial }_n{\mathbf{A}}_1-{\partial }_n{\mathbf{A}}_2){|}_S=-{\mu }_0\mathbf{K}$.

Direct derivation via pillbox: Integrate ${\nabla }^2{A}_k=-{\mu }_0{j}_k$ across the sheet. ${\nabla }^2{A}_k={\partial }_n^2{A}_k+{\nabla }_{\text{surf}}^2{A}_k$; the surface-tangent part contributes nothing in the $ϵ\to 0$ limit because $A_k$ is continuous. So ${\int }_{-ϵ}^{+ϵ}{\partial }_n^2{A}_k={\partial }_n{A}_k{|}_{+}-{\partial }_n{A}_k{|}_{-}=-{\mu }_0\int j_k\mathrm{d}n=-{\mu }_0{K}_k$.

Note: This matching condition is gauge-dependent — in non-Coulomb gauges the relationship between $\mathbf{A}$ and $\mathbf{j}$ changes. The physical (gauge-invariant) matching is always on $\mathbf{B}$.

---

10.22 — Magnetic Field of Charge in Uniform Motion

(a) Show that for a charge distribution $\rho (\mathbf{r})$ moving rigidly with velocity $\mathbf{v}$: $\mathbf{B}=(\mathbf{v}/c^2)\times \mathbf{E}$. (b) Apply to an infinite line and an infinite sheet.

Solution Sketch

(a) The current is $\mathbf{j}(\mathbf{r})=\rho (\mathbf{r})\mathbf{v}$ with $\mathbf{v}$ uniform (crucial for promoting $\mathbf{v}$ outside the integral):

$$\mathbf{B}(\mathbf{r})=\frac{{\mu }_0}{4\pi }\int {\mathrm{d}}^3{r}^{\prime }\frac{\rho ({\mathbf{r}}^{\prime })\mathbf{v}\times (\mathbf{r}-{\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}=\frac{{\mu }_0}{4\pi }\mathbf{v}\times \int {\mathrm{d}}^3{r}^{\prime }\rho ({\mathbf{r}}^{\prime })\frac{\mathbf{r}-{\mathbf{r}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}={\mu }_0{ϵ}_0\mathbf{v}\times \mathbf{E}=\frac{\mathbf{v}}{c^2}\times \mathbf{E}$$

(b) Infinite line: $\mathbf{E}=(\lambda /2\pi {ϵ}_0)\hat{\mathbf{\rho }}/\rho$, $\mathbf{v}=v\hat{\mathbf{z}}$ gives $\mathbf{B}={\mu }_{0}I/(2\pi \rho )\hat{\mathbf{\varphi }}$ with $I=\lambda v$. Sheet: $\mathbf{E}=(\sigma /2{ϵ}_0)\mathrm{sgn}(z)\hat{\mathbf{z}}$, $\mathbf{v}=v\hat{\mathbf{y}}$ gives $\mathbf{B}=({\mu }_{0}K/2)\mathrm{sgn}(z)\hat{\mathbf{x}}$ with $K=\sigma v$.

Limitations: Requires uniform rigid motion. For non-uniform $\mathbf{v}(\mathbf{r})$ the promotion outside the integral fails; for accelerated motion, retardation (radiation fields) enters. More deeply, $\mathbf{B}=\mathbf{v}/c^2\times \mathbf{E}$ is the low-velocity limit of the relativistic Lorentz transformation of fields — developed fully in later chapters.

---

10.23 — An Aharonov–Bohm Geometry

(a) Find $\mathbf{A}$ inside and outside an infinite solenoid producing $\mathbf{B}=B\hat{\mathbf{z}}$ for $\rho <R$. Use Coulomb gauge. (b) Show that ${\mathbf{A}}^{\prime }=\mathbf{A}+\nabla \chi$ with $\chi =-\varphi {\Phi }_B/2\pi$ (where ${\Phi }_B=\pi R^{2}B$) gives ${\mathbf{A}}_{\text{out}}^{\prime }=0$. (c) Show the new ${\mathbf{B}}^{\prime }$ corresponds to a different physical system: ${\mathbf{B}}^{\prime }=B-\hat{\mathbf{z}}{\Phi }_B\delta (\rho )/(2\pi \rho )\cdot ?$

Solution Sketch

(a) Cylindrical symmetry + Coulomb gauge: $\mathbf{A}=A_{\varphi }(\rho )\hat{\mathbf{\varphi }}$ with $\rho B_z={\partial }_{\rho }(\rho A_{\varphi })$:

$$\mathbf{A}=\hat{\mathbf{\varphi }}\times \{\begin{array}{ll}B\rho /2& \rho \le R\\ B{R}^2/(2\rho )& \rho \ge R\end{array}$$

(b) $\nabla \chi =-({\Phi }_B/2\pi )\hat{\mathbf{\varphi }}/\rho =-(B{R}^2/2\rho )\hat{\mathbf{\varphi }}$. Adding to ${\mathbf{A}}_{\text{out}}$ gives zero. Adding to ${\mathbf{A}}_{\text{in}}$ gives ${\mathbf{A}}_{\text{in}}^{\prime }=\hat{\mathbf{\varphi }}(B\rho /2-B{R}^2/2\rho )$.

(c) Flux inconsistency: ${\oint }_{\rho >R}\mathrm{d}\mathbf{\ell }\cdot {\mathbf{A}}^{\prime }=0$ (from (b)) but physically ${\Phi }_B\ne 0$. So ${\mathbf{B}}^{\prime }=\nabla \times {\mathbf{A}}^{\prime }$ must acquire a delta-function concentrated flux at $\rho =0$:

$${\mathbf{B}}^{\prime }=B\hat{\mathbf{z}}{\chi }_{[\rho \le R]}-\hat{\mathbf{z}}\frac{{\Phi }_B}{2\pi \rho }\delta (\rho )$$

where the delta-flux tube at the axis exactly compensates. Computing $\nabla \times \nabla \chi$ directly (naively zero by identity) via a regularization: integrating $\nabla \times \nabla \chi$ over any small disk at $\rho =0$ gives $-{\Phi }_B$ by Stokes — a delta function.

Physical lesson: The gauge transformation $\chi =-\varphi {\Phi }_B/2\pi$ is not single-valued (jumps by $-{\Phi }_B$ each azimuthal circuit), so $\nabla \chi$ has a delta-function curl at the axis. The “gauge-transformed” problem has a genuinely different $\mathbf{B}$-field with a singular flux line — not an admissible gauge transformation. The Aharonov–Bohm effect therefore cannot be gauged away by any legitimate transformation.

---

10.24 — Lamb’s Formula

A quantum particle with charge $q$, mass $m$, momentum $\mathbf{p}$ in field $\mathbf{B}=\nabla \times \mathbf{A}$ has velocity $\mathbf{v}=\mathbf{p}/m-(q/m)\mathbf{A}$. A charge distribution $\rho (\mathbf{r})$ generates a diamagnetic current $\mathbf{j}=-(q/m)\rho \mathbf{A}$. (a) Show $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$ works for uniform $\mathbf{B}$. (b) Find the induced ${\mathbf{A}}_{\text{ind}}$ for atomic $\rho (r)$ in uniform $\mathbf{B}$. (c) Derive Lamb’s formula ${\mathbf{B}}_{\text{ind}}(0)=e\phi (0)/(3m{c}^2)\mathbf{B}$.

Solution Sketch

(a) $\nabla \times (\frac{1}{2}\mathbf{B}\times \mathbf{r})=\frac{1}{2}[\mathbf{B}(\nabla \cdot \mathbf{r})-(\mathbf{B}\cdot \nabla )\mathbf{r}]=\frac{1}{2}[3\mathbf{B}-\mathbf{B}]=\mathbf{B}$. ✓ Also $\nabla \cdot \mathbf{A}=0$ (Coulomb gauge).

(b) With $\mathbf{j}=-(e/2m)\rho ({\mathbf{r}}^{\prime })(\mathbf{B}\times {\mathbf{r}}^{\prime })$ (electron, charge $-e$), the Coulomb-gauge vector potential:

$${\mathbf{A}}_{\text{ind}}(\mathbf{r})=-\frac{{\mu }_{0}e}{8\pi m}\mathbf{B}\times \int {\mathrm{d}}^3{r}^{\prime }\rho (r^{\prime })\frac{{\mathbf{r}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Key step: Orient $\mathbf{r}=r\hat{\mathbf{z}}$ so the integrand’s ${\mathbf{r}}^{\prime }$ decomposes using ${\mathbf{r}}^{\prime }=r^{\prime }\cos {\theta }^{\prime }\hat{\mathbf{z}}+\dots$ — only the $\ell =1$ Legendre piece survives under $\rho (r^{\prime })$‘s spherical symmetry. Using the $\ell =1$ expansion of $1/|\mathbf{r}-{\mathbf{r}}^{\prime }|$ and angular integrals:

$${\mathbf{A}}_{\text{ind}}(\mathbf{r})=\frac{{\mu }_0}{4\pi }\frac{e\mathbf{B}\times \mathbf{r}}{6m}\left[\frac{1}{r^3}{\int }_0^r{r}^{\prime 2}\rho (r^{\prime })r^{\prime }\mathrm{d}{r}^{\prime }+{\int }_r^{\infty }\rho (r^{\prime })\mathrm{d}{r}^{\prime }\right]$$

(c) Expand for small $r$: the first bracket term vanishes as $r\to 0$; the second becomes ${\int }_0^{\infty }\rho (r^{\prime })\mathrm{d}{r}^{\prime }$, which relates to $\phi (0)=\frac{1}{4\pi {ϵ}_0}\int \rho (r^{\prime })/r^{\prime }{\mathrm{d}}^3{r}^{\prime }=\frac{1}{{ϵ}_0}{\int }_0^{\infty }\rho (r^{\prime })r^{\prime }\mathrm{d}{r}^{\prime }$. Using $\nabla \times (\mathbf{B}\times \mathbf{r})=2\mathbf{B}$:

$${\mathbf{B}}_{\text{ind}}(0)=\frac{e\phi (0)}{3m{c}^2}\mathbf{B}$$

Historical significance: Lamb derived this in 1941 at Rabi’s request to determine if diamagnetic corrections mattered for extracting nuclear magnetic moments from molecular-beam data. They did.

---

10.25 — Toroidal and Poloidal Magnetic Fields

Any divergence-free $\mathbf{V}(\mathbf{r})$ decomposes uniquely as $\mathbf{V}=\mathbf{L}\psi +\nabla \times \mathbf{L}\gamma$, with $\mathbf{L}=-i\mathbf{r}\times \nabla$. The first term is toroidal $\mathbf{T}$, the second poloidal $\mathbf{P}$. (a) Verify $\nabla \cdot \mathbf{V}=0$. (b) Poloidal current → toroidal $\mathbf{B}$ and vice versa. (c) $\mathbf{B}$ is toroidal for a toroidal solenoid. (d) ${\nabla }^2\mathbf{B}=0$ in current-free $V$. (e) In Coulomb gauge, $\mathbf{A}$ is purely toroidal when $\psi ,\gamma$ are chosen so ${\nabla }^2\mathbf{B}=0$.

Solution Sketch

(a) $\nabla \cdot (\mathbf{L}\psi )=0$ and $\nabla \cdot (\nabla \times \cdot )=0$ trivially.

(b) Poloidal $\mathbf{B}$ → toroidal $\mathbf{j}$: Let $\mathbf{B}=\nabla \times \mathbf{L}\gamma$. Then ${\mu }_0\mathbf{j}=\nabla \times \mathbf{B}=\nabla \times \nabla \times \mathbf{L}\gamma =\nabla (\nabla \cdot \mathbf{L}\gamma )-{\nabla }^2\mathbf{L}\gamma =-\mathbf{L}{\nabla }^2\gamma$, which is purely toroidal. The key identity ${\nabla }^2\mathbf{L}=\mathbf{L}{\nabla }^2$ (index manipulation using $[{\nabla }_i,L_j]$).

(c) Toroidal solenoid: $\mathbf{j}$ is poloidal (goes around the torus the “short way”), so $\mathbf{B}$ is toroidal. Compatible with $\psi =(i/r)\ln \tan (\theta /2)$ or similar.

(d) $\nabla \times \nabla \times \mathbf{B}=\nabla (\nabla \cdot \mathbf{B})-{\nabla }^2\mathbf{B}=-{\nabla }^2\mathbf{B}$; in current-free regions this is zero.

(e) $\mathbf{B}=\nabla \times \mathbf{A}$. If $\mathbf{A}=\mathbf{L}\psi +\nabla \times \mathbf{L}\gamma$, then $\nabla \times \mathbf{A}=\nabla \times \mathbf{L}\psi -\mathbf{L}{\nabla }^2\gamma$. The choice ${\nabla }^2\gamma =0$ makes the second piece vanish, and Coulomb-gauge $\mathbf{A}$ reduces to $\mathbf{L}\psi$ (purely toroidal).

Intuition — Toroidal vs. Poloidal

Picture a donut (torus). Two inequivalent ways to draw a circle:

• Poloidal: the small circle through the hole (cross-section of the donut).
• Toroidal: the big circle around the hole (following the ring).

Correspondingly:

• Toroidal field circulates around the donut — purely tangential to spheres (no radial component), since $\mathbf{L}\psi =-i\mathbf{r}\times \nabla \psi \perp \hat{\mathbf{r}}$. Example: field inside a toroidal solenoid.
• Poloidal field threads through spheres — has a radial component. Example: magnetic dipole.

Result (b) — poloidal ↔ toroidal swap under curl — is physically intuitive. Current going around the donut the short way (poloidal) produces a field going the long way (toroidal), by Ampère’s law.

Terminology origin: Earth and Sun. Earth’s main dipole is poloidal. The solar dynamo cycles energy between poloidal (from meridional flow) and toroidal (wound up by differential rotation) components, which is why the decomposition is central to dynamo theory.

The decomposition is really about spheres, not tori. At each radius, the tangential part of $\mathbf{V}$ on the sphere decomposes (by Helmholtz-on-the-sphere) into a “curl” piece (toroidal) and a “gradient” piece (poloidal). The names are inherited from the axisymmetric (torus-adjacent) case where the decomposition was historically central in plasma physics.