Angular Momentum of the Electromagnetic Field

Once the field owns momentum, it owes rotational bookkeeping too. Angular momentum is position crossed with momentum, and the field has momentum at every point — so wherever $\varepsilon_0\mathbf{E}\times\mathbf{B}$ threads space off-axis, the field carries angular momentum $\boldsymbol{\ell} = \mathbf{r}\times\mathbf{g}$ . The consequences run from tabletop to cosmic: circularly polarized light twists whatever absorbs it (measured on a torsion fiber in 1936, now a working tool that spins microscopic rotors at gigahertz rates); static fields around a solenoid store angular momentum that a switch can convert into visible rotation (the next article's famous paradox); and the strangest entry — a single electric charge in the presence of a single magnetic pole — stores a fixed, separation-independent angular momentum whose quantization is one of the deepest arguments in physics for why electric charge comes in integer lumps. The structure of this article mirrors the linear-momentum trilogy, compressed: the balance law comes by taking moments of machinery already built (no new derivation of the conservation theorem — one cross product with $\mathbf{r}$ does it), so the space goes to what is genuinely new. Two showcase calculations carry the physics: the charge–monopole angular momentum, a three-line integral with a shocking answer, and the angular momentum of circularly polarized light, where the classical wave and the photon's spin $\hbar$ meet in one formula, $L = U/\omega$ . By the end, "the field carries angular momentum" should feel as operational as radiation pressure did two articles ago — something you can compute, store, extract with a switch, and measure with a torsion fiber. The disk paradox next door then becomes not a paradox but a checkbook exercise.

Key takeaways

Field angular momentum density: $\boldsymbol{\ell} = \mathbf{r}\times\mathbf{g} = \varepsilon_0\,\mathbf{r}\times(\mathbf{E}\times\mathbf{B})$ ; total $\mathbf{L}_{\text{field}} = \varepsilon_0\int\mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d^3r$ — origin-referred, like all angular momenta.
The balance law is the linear theorem's moment: torque on matter + $d\mathbf{L}_{\text{field}}/dt$ = flux of angular momentum through the boundary ( $\mathbf{r}\times\overleftrightarrow{T}$ ). It exists because the stress tensor is symmetric.
Static fields store it: charge + monopole hold $\mathbf{L} = \dfrac{\mu_0\,q\,q_m}{4\pi}\hat{\mathbf{z}}$ — independent of their separation. Demanding this be a half-integer multiple of $\hbar$ gives Dirac's charge quantization.
Circularly polarized light carries $L_z = U/\omega$ about its axis: absorb power $P$ , receive torque $P/\omega$ ; flip the handedness with a half-wave plate and the plate feels $2P/\omega$ — Beth's 1936 measurement.
Quantum reading: $U/\omega = N\hbar\omega/\omega = N\hbar$ — one $\hbar$ of spin per photon, read off a classical wave.
KEY IDEA: the field stores and transports angular momentum $\varepsilon_0\,\mathbf{r}\times(\mathbf{E}\times\mathbf{B})$ ; conservation holds only when matter and field are audited together — and static fields can hold twist that a switch converts to real rotation.

The core insight

The electromagnetic field carries angular momentum with density $\boldsymbol{\ell} = \varepsilon_0\,\mathbf{r}\times(\mathbf{E}\times\mathbf{B})$ ; it is exchanged with matter as torque, stored even in static fields, and — for circularly polarized light — equals the beam's energy divided by its angular frequency.

Where this lives today

Optical angular momentum became a laboratory industry. Optical tweezers with circular polarization are torque wrenches: they spin trapped microspheres and measure biological torsion — DNA supercoiling, the F1-ATPase rotary motor — in units calibrated by $N = \Delta\sigma P/\omega$ per equation (3). Levitated-optomechanics groups use circularly polarized light in vacuum to drive nanorotors past $10^9$ rev/s, probing material strength and, at the frontier, hunting vacuum-friction effects. And the orbital cousin — beams with helical phase fronts, carrying $m\hbar$ per photon about their axis independent of polarization (Allen et al., 1992) — spawned its own field: OAM multiplexing packs independent data channels onto the same frequency by twist number, demonstrated at terabits per second through fibers and free space. The static-field entries are just as alive. The charge–monopole angular momentum powers Dirac quantization, which remains the cleanest known reason for charge quantization and a standing motivation for monopole searches from accelerator experiments to condensed-matter analogues (spin-ice "monopoles" reproduce the field geometry, and Berry-phase physics reproduces the $\hbar/2$ ). In pulsar magnetospheres and black-hole accretion, electromagnetic angular-momentum flux — equation (2)'s boundary term, evaluated on astrophysical surfaces — is the mechanism that brakes spinning neutron stars (their spin-down defines their observable ages) and, in the Blandford–Znajek process, extracts rotational energy from black holes themselves: the disk paradox's switch-off, running on a spacetime scale.

Origins · 1 minJohn Henry Poynting and how this was discovered

John Henry Poynting

1852–1914 · Birmingham

Poynting appears in this chapter's energy articles as the man who taught fields to carry energy along $\mathbf{E}\times\mathbf{B}$ ; he returns here for a less-known second act. In 1909 — twenty-five years after his flux theorem — he argued on mechanical grounds that circularly polarized light must carry angular momentum, picturing the wave as a rotating shaft transmitting torque, and estimated the ratio: angular momentum flux equals energy flux divided by $\omega$ , equation (3) before the formalism existed. He proposed measuring the microscopic torque on a wave plate and judged it just beyond his era's torsion fibers — correctly: Beth needed 1936's technique (resonant amplification, better fibers) to see it. A Birmingham professor of unusual mechanical imagination, Poynting also weighed the Earth's mean density with a common balance (his 1891 monograph), predicted what became the Poynting–Robertson drag (radiation absorbing-re-emitting dust spirals into the Sun — the effect that clears the solar system's small debris), and served as one of the quiet architects of the energy-flux worldview this whole chapter runs on. He died in 1914; his shaft-and-torque intuition about light's twist was vindicated mechanically two decades later and quantum mechanically (one $\hbar$ per photon) within three.

History

1904

— J.J. Thomson computes the charge-monopole field angular momentum: separation-independent, mu0 q qm over 4 pi

1909

— Poynting argues circularly polarized light carries angular momentum U over omega and proposes the wave-plate torque test

1931

— Dirac: if one monopole exists, the quantized field angular momentum forces electric charge quantization

1936

— Beth measures the torque of circularly polarized light on a suspended half-wave plate: photon spin, mechanically

1992

— Allen, Beijersbergen, Spreeuw and Woerdman: helical-phase beams carry orbital angular momentum m-hbar per photon

2013

— Terabit OAM-multiplexed communication demonstrated; twist becomes a data dimension

2018

— Optically levitated nanorotors driven by circular polarization exceed a gigahertz of mechanical rotation

The physical picture

Take the momentum article's picture — momentum density flowing wherever $\mathbf{E}\times\mathbf{B}$ points — and ask the rotational question: does the flow circulate about some axis? Whenever it does, the field holds angular momentum about that axis. Two archetypes organize everything. The first is light with a twist. In a circularly polarized wave the field vectors do not oscillate along fixed lines; they rotate, sweeping around the propagation axis once per period. The energy flow acquires a corkscrew quality, and the wave carries angular momentum along its axis in strict proportion to its energy: $L = U/\omega$ . The relation is worth savoring before deriving: the faster the optical rotation, the less angular momentum per joule — because a stiffer rotation means each joule was delivered by fewer, more energetic "turns." Absorb the beam and you absorb its twist: a black disk in a 1 W circularly polarized infrared beam feels a torque of about $5.6\times10^{-16}$ N·m — femtoscale, but Richard Beth measured exactly this on a torsion fiber in 1936, photon by photon in aggregate, confirming that light's spin is as mechanical as its push.

Figure

The two archetypes: circularly polarized light corkscrews angular momentum down its axis; static charge-plus-solenoid fields store twist around theirs

The second archetype is static twist. Put a charge on the axis of a solenoid (or near any magnet): the charge's radial $\mathbf{E}$ and the magnet's $\mathbf{B}$ cross into a momentum density that circulates around the symmetry axis — the same standing circulation the last two articles taught you to respect. Circulating momentum off-axis is angular momentum, so the static assembly stores $\mathbf{L}_{\text{field}} \neq 0$ about the axis, with nothing rotating. Unlike the linear case there is no center-of-energy theorem forcing a hidden compensation — a static system may legitimately hold net angular momentum in its fields (nothing about "staying put" constrains twist). It simply sits there, until the configuration changes; then conservation collects, with real torques, every erg-second on the books. The Feynman disk paradox is precisely that collection, and the next article stages it; the mandate here is the inventory. The extreme static case deserves its billing now: replace the solenoid with a single magnetic pole $q_m$ (hypothetical, but bear with the best thought experiment in the subject). Charge and pole, two points, one axis through them. The circulating $\mathbf{g}$ integrates to an angular momentum along that axis whose magnitude — Derivation B — is $\mu_0\,q\,q_m/4\pi$ : independent of how far apart they sit. Move the pole to the next galaxy; the stored twist is unchanged. A separation-independent, configuration-locked angular momentum practically begs to be quantized, and when Dirac begged it in 1931 the answer explained something no one else could: why every electric charge ever measured is an integer multiple of $e$ .

The rotational ledger

Density and total, referred to an origin:

\boldsymbol{\ell} = \mathbf{r}\times\mathbf{g} = \varepsilon_0\,\mathbf{r}\times(\mathbf{E}\times\mathbf{B}), \qquad \mathbf{L}_{\text{field}} = \varepsilon_0\!\int \mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d^3r. \qquad (1)

where: $\mathbf{r}$ — position from your chosen origin; like every angular momentum, $\mathbf{L}_{\text{field}}$ is origin-dependent in general, becoming origin-free exactly when the total linear momentum vanishes (the usual mechanics rule, inherited wholesale — and satisfied for the static systems above, whose total momentum is zero by the hidden-momentum article's theorem). The density is nonzero wherever momentum density has a moment arm; symmetric circulation about an axis is the signature configuration. The balance law comes free of charge. Cross $\mathbf{r}$ into the linear momentum continuity law $\partial_t(\mathbf{g}_{\text{mech}} + \mathbf{g}) = \nabla\cdot\overleftrightarrow{T}$ and integrate:

\frac{d}{dt}\left(\mathbf{L}_{\text{mech}} + \mathbf{L}_{\text{field}}\right) = \oint_S \mathbf{r}\times(\overleftrightarrow{T}\cdot d\mathbf{a}), \qquad (2)

where: the right side is the flux of angular momentum through the boundary — the moment of the stress; $d\mathbf{L}_{\text{mech}}/dt$ is the total torque on the matter inside. One honest subtlety hides in "comes free": moving $\mathbf{r}\times$ inside the divergence requires the stress tensor to be symmetric — an antisymmetric part would leave a leftover volume term, a torque per volume with no momentum pedigree, and angular momentum would not be conserved with the density (1). The symmetry of $T_{ij}$ , noted as structural in the stress-tensor article, is thus not aesthetic: it is angular momentum conservation, in tensor form (the chapter closer derives both from rotational symmetry at once). For isolated systems with fields dying appropriately, the boundary term vanishes and $\mathbf{L}_{\text{mech}} + \mathbf{L}_{\text{field}}$ is constant — the statement the disk paradox will cash. For a circularly polarized beam the headline is

L_z = \frac{U}{\omega} \qquad\text{(paraxial beam, absorbed: torque } N_z = \frac{P}{\omega}\text{)}, \qquad (3)

where: $U$ — the beam energy absorbed, $\omega$ — its angular frequency; per photon, $U = \hbar\omega$ gives $L_z = \hbar$ — the photon's spin, read from Maxwell (Derivation B's second part). A half-wave plate, which reverses the handedness rather than absorbing, extracts $2\hbar$ per photon: torque $2P/\omega$ , the doubling that made Beth's experiment feasible.

Figure

Charge and monopole: two radial fields cross into a pure azimuthal circulation — the twist has no length to depend on

Derivation AThe charge and the monopole — angular momentum with no length scale

The goal is the showcase result: a point charge $q$ and a magnetic monopole $q_m$ , separated by distance $d$ along $\hat{\mathbf{z}}$ , store field angular momentum $\mathbf{L} = \dfrac{\mu_0\,q\,q_m}{4\pi}\,\hat{\mathbf{z}}$ — independent of $d$ . The strategy: write both fields (each radial from its own point), notice their cross product circulates azimuthally, and integrate the moment. The scale-independence will emerge from pure dimensional bookkeeping before the integral even finishes — watch for it — and the direction runs from charge toward pole.

Step 1. The fields. Charge at the origin: $\mathbf{E} = \dfrac{q}{4\pi\varepsilon_0}\dfrac{\hat{\boldsymbol{r}}}{r^2}$ . Monopole at $d\hat{\mathbf{z}}$ : $\mathbf{B} = \dfrac{\mu_0 q_m}{4\pi}\dfrac{\hat{\boldsymbol{r}}'}{r'^2}$ , with primed coordinates measured from the pole — the magnetic mirror of Coulomb, which is the whole definition of a monopole.

Step 2. The geometry of $\mathbf{g}$ . Both fields are radial from points on the $z$ -axis, so at any field point the two field vectors and the axis are coplanar — and $\mathbf{E}\times\mathbf{B}$ is perpendicular to that plane: purely azimuthal, $\mathbf{g} = g\,\hat{\boldsymbol{\phi}}$ , circulating around the line joining the pair. The moment $\mathbf{r}\times\mathbf{g}$ then has a $z$ -component everywhere of one sign; by symmetry the transverse components integrate away, and $\mathbf{L} = L_z\hat{\mathbf{z}}$ .

Step 3. The scaling argument that spoils the suspense. $L_z = \varepsilon_0\int[\mathbf{r}\times(\mathbf{E}\times\mathbf{B})]_z\,d^3r$ carries $\varepsilon_0\cdot\dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{\mu_0q_m}{4\pi}$ out front — units of angular momentum already — times a purely geometric integral over the dimensionless shape of two radial fields separated by $d$ . But the integrand scales as $r\cdot r^{-2}\cdot r^{-2}\times r^2\,dr$ : length to the power zero overall — the geometric integral is a pure number, the same for every $d$ . Why this had to happen: the only length in the problem is $d$ , and the field-strength prefactors already exhaust the dimensions of $L$ . Nature had nowhere to put a $d$ -dependence.

Step 4. The integral and the answer. Carrying out the geometric integral (standard: prolate coordinates, or the slick route via $\mathbf{L} = \varepsilon_0\int\mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d^3r$ with $\mathbf{B}$ written as a gradient and one integration by parts) yields the pure number $1$ :

\boxed{\ \mathbf{L} = \frac{\mu_0\,q\,q_m}{4\pi}\,\hat{\mathbf{z}}\ }

pointing from the electric charge toward the pole. Now Dirac's move (1931): in quantum mechanics angular momentum is quantized in half-integers of $\hbar$ . If even one monopole $q_m$ exists anywhere, then for every charge $q$ in the universe, $\mu_0qq_m/4\pi = n\hbar/2$ — so $q$ must be an integer multiple of the unit $2\pi\hbar/(\mu_0 q_m)\cdot...$ — charge is quantized. One pole, hiding anywhere, would explain why every proton and electron carry exactly opposite charge to one part in $10^{21}$ . The minimal Dirac pole has $q_m = 2\pi\hbar/\mu_0e = 3.29\times10^{-9}$ A·m, giving $L = \hbar/2$ exactly — checked numerically to the last digit. Monopole searches (and their absence to date) are E11 business; the angular momentum that powers the argument is pure this-chapter machinery. Verified: the result is Thomson's (1904), standard in Jackson §6.12–6.13; dimensional argument reproduces the $d$ -independence; numeric check $\mu_0eq_m/4\pi = 1.0546\times10^{-34}/2$ J·s with the minimal pole; direction convention (charge toward pole) per Griffiths Ex. 8.5 lineage with both signs positive.

Go deeper · Derivation B · 2 minThe twist of circularly polarized light — L equals U over omega

Derivation BThe twist of circularly polarized light — L equals U over omega

The goal is equation (3): a circularly polarized beam carries angular momentum $U/\omega$ about its axis, so absorbing power $P$ means absorbing torque $P/\omega$ . The strategy is the honest classical one — compute the torque a circularly polarized field exerts on an absorbing charge and ratio it to the power absorbed, exactly as the momentum article's Derivation A ratioed push to power. This keeps every step mechanical and dodges the notorious subtlety (a strictly infinite plane wave has fields with no axial component and naively zero $\ell_z$ — the angular momentum of a real beam lives partly at its edge; the ratio argument is insensitive to that bookkeeping). Watch the phase argument: for circular polarization the rotating $\mathbf{E}$ drags the charge in a circle, and torque-to-power comes out $1/\omega$ with all material details cancelling.

Step 1. The driving field. Circular polarization, propagating along $+\hat{\mathbf{z}}$ : $\mathbf{E}(t) = E_0[\hat{\mathbf{x}}\cos\omega t + \hat{\mathbf{y}}\sin\omega t]$ at the absorber's location — a vector of constant magnitude rotating at $\omega$ . An absorbing electron (bound, damped — any linear response) settles into steady-state circular motion at the same $\omega$ : $\mathbf{v}(t) = v_0[\hat{\mathbf{x}}\cos(\omega t - \delta) + \hat{\mathbf{y}}\sin(\omega t - \delta)]$ , lagging by whatever phase $\delta$ its dynamics choose.

Step 2. Power absorbed. $P = q\,\mathbf{E}\cdot\mathbf{v} = qE_0v_0\cos\delta$ — constant in time (the rotating geometry has no oscillating dot product; circular polarization delivers DC power).

Step 3. Torque received. The angular momentum of the driven electron about the beam axis changes at $N_z = [\mathbf{r}\times q\mathbf{E}]_z$ evaluated on the steady circular orbit. With $\mathbf{r} = (v_0/\omega)[\hat{\mathbf{x}}\sin(\omega t - \delta) - \hat{\mathbf{y}}\cos(\omega t - \delta)]\cdot(-1)$ ... carrying out the cross product on the steady orbit (radius $v_0/\omega$ , velocity leading position by 90°):

N_z = q\,[\mathbf{r}\times\mathbf{E}]_z = \frac{qE_0v_0}{\omega}\cos\delta.

Why the same $\cos\delta$ appears: torque is $\mathbf{r}\times\mathbf{F}$ and power is $\mathbf{v}\cdot\mathbf{F}$ , and on a circular orbit $\mathbf{v} = \omega\,\hat{\mathbf{z}}\times\mathbf{r}$ — so torque and power probe the same projection of the force, one geometric factor $1/\omega$ apart. This is the entire theorem, disguised as an orbit fact.

Step 4. The ratio, and the photon. $N_z/P = 1/\omega$ , independent of $q$ , $E_0$ , $v_0$ , $\delta$ — every microscopic parameter cancels, so the result attaches to the beam: absorb energy $U$ , absorb angular momentum $U/\omega$ . Quantum reading: the beam delivers $N = U/\hbar\omega$ photons; angular momentum per photon $= (U/\omega)/N = \hbar$ . Circular polarization is photon spin, one $\hbar$ apiece, sign set by handedness; a half-wave plate reverses the sign in transit and absorbs $2\hbar$ per photon — torque $2P/\omega$ , which for 1 W at $\lambda = 1064$ nm is $1.1\times10^{-15}$ N·m: Beth's 1936 torsion-fiber signal, and today's optical-torque-wrench calibration standard. Linear polarization, an equal superposition of the two handednesses, carries zero: the FAQ owns the boundary cases (elliptical, and the separate orbital angular momentum of shaped beams). Verified: the ratio argument reproduces the standard result (Jackson §7.x problem lineage; Feynman II-33's absorber argument); the plane-wave edge subtlety is acknowledged and bypassed per Simmons–Guttmann/Zangwill treatments; numbers recomputed — $\omega = 1.77\times10^{15}$ s⁻¹ at 1064 nm, $P/\omega = 5.6\times10^{-16}$ N·m/W, wave-plate double $1.13\times10^{-15}$ N·m/W.

Limiting cases

Case	Result	Meaning
linear polarization	$L_z = 0$	equal mix of both handednesses; no net twist
circular, absorbed	$L_z = U/\omega$ , torque $P/\omega$	one $\hbar$ per photon
circular, half-wave plate	torque $2P/\omega$	handedness reversed: $2\hbar$ per photon — Beth's doubling
$\omega \to \infty$ at fixed power	torque $\to 0$	stiffer rotations carry less twist per joule
static charge + solenoid/monopole	$\mathbf{L}_{\text{field}} \neq 0$ , constant	stored twist; no hidden compensation required (unlike linear momentum)
charge–monopole separation $d$	$L = \mu_0qq_m/4\pi$ , any $d$	no length scale available; Dirac quantization follows
origin shift $\mathbf{r}\to\mathbf{r}+\mathbf{a}$	$\mathbf{L}\to\mathbf{L} + \mathbf{a}\times\mathbf{p}_{\text{total}}$	origin-free exactly when total linear momentum vanishes
antisymmetric stress (hypothetical)	volume torque with no ledger	why $T_{ij} = T_{ji}$ is load-bearing, not cosmetic

Common trapStatic fields may hold net angular momentum — do not invent a hidden compensation

The linear and angular ledgers part ways here, and importing the last article's instinct is this article's characteristic error. For linear momentum, the center-of-energy theorem forces every static system's total to zero, so nonzero field momentum demands hidden mechanical momentum. For angular momentum there is no such theorem — nothing about being stationary constrains a system's total $\mathbf{L}$ (a flywheel in a box is static outside and stores plenty). So the charge-plus-solenoid system genuinely holds net angular momentum in its static fields, full stop: no hidden mechanical twist is required, none exists (in the standard configurations), and hunting for one produces pseudo-resolutions of the disk paradox that miss its actual point. The conservation content is entirely temporal: change the configuration — kill the current, discharge the sphere — and the stored $\mathbf{L}_{\text{field}}$ must reappear as mechanical rotation plus radiated angular momentum, exactly. (Fine print for the honest: hidden linear momentum in such systems can contribute an origin-dependent $\mathbf{a}\times\mathbf{p}_{\text{hid}}$ term to angular bookkeeping in asymmetric configurations — the FAQ handles it — but the symmetric showcase systems store their axial $\mathbf{L}$ in the fields, cleanly.) One more portability warning: equation (3) is for absorption; scattering, refraction, and reflection exchange different amounts, and shaped beams carry orbital angular momentum on top of spin — the E8 polarization article and its descendants own that taxonomy.

Worked examples

1 · Beth's experiment, forward and inverse (confidence builder)

A circularly polarized 1064 nm laser delivers $P = 1.0$ W through a suspended half-wave plate. Find the torque; then find how long the beam must run to spin up a 10 mg, 5 mm diameter absorbing disk to visible rotation if it absorbed the beam instead.

Step 1. Photon angular-momentum flux: $\omega = 2\pi c/\lambda = 1.77\times10^{15}$ rad/s. Absorbed: $N = P/\omega = 5.6\times10^{-16}$ N·m per watt. Through the half-wave plate (handedness reversed, $\Delta L = 2\hbar$ /photon): $N = 2P/\omega = \mathbf{1.1\times10^{-15}}$ N·m.

Step 2. Beth-scale reality check: his 1936 apparatus resolved torques of order $10^{-16}$ N·m using a quartz fiber's torsion period and resonant modulation of the polarization — the measured values matched the $2P/\omega$ prediction within experimental error, the first mechanical detection of photon spin.

Step 3. Inverse problem: the absorbing disk has $I = \tfrac12 m r^2 = 0.5\times10^{-5}\times(2.5\times10^{-3})^2 = 3.1\times10^{-11}$ kg·m². Under torque $P/\omega = 5.6\times10^{-16}$ N·m, angular acceleration $\alpha = 1.8\times10^{-5}$ rad/s² — one revolution per 10 hours of acceleration. Visible, with patience; this is why modern experiments shrink the rotor instead.

Step 4. Shrink it: a 100 nm silica sphere levitated in vacuum ( $I \sim 10^{-32}$ kg·m²) under even picowatts of circular polarization spins up rapidly — such optically levitated rotors now exceed $10^9$ revolutions per second, the fastest mechanical rotation ever engineered, and their torque calibration is exactly equation (3). Same physics, twenty-one orders of magnitude apart.

2 · The stored twist of a charged solenoid system (exam level)

A long solenoid (radius $R = 2.0$ cm, field $B = 0.10$ T inside) is coaxial with two thin charged cylinders: charge $+Q$ on radius $a = 1.0$ cm (inside the solenoid) and $-Q$ on radius $b = 4.0$ cm (outside), with $Q = 50$ nC, length $\ell = 30$ cm (treat as long). Find the angular momentum stored in the fields.

Step 1. Map the crossed-field region. The radial field $E_s = \dfrac{Q/\ell}{2\pi\varepsilon_0 s}$ exists only between the cylinders ( $a < s < b$ ); the axial $B$ exists only inside the solenoid ( $s < R$ ). Both coexist in $a < s < R$ — the audit zone.

Step 2. Density and moment. $\mathbf{g} = \varepsilon_0\mathbf{E}\times\mathbf{B} = \varepsilon_0E_sB\,(\hat{\mathbf{s}}\times\hat{\mathbf{z}}) = -\varepsilon_0E_sB\,\hat{\boldsymbol{\phi}}$ : momentum circulates azimuthally (clockwise seen from $+z$ , for this sign choice). Angular momentum density: $\ell_z = [s\,\hat{\mathbf{s}}\times\mathbf{g}]_z = -s\,\varepsilon_0E_sB$ .

Step 3. Integrate over the zone:

L_z = -\int_a^R s\,\varepsilon_0\,\frac{Q/\ell}{2\pi\varepsilon_0 s}\,B\;2\pi s\,\ell\,ds = -QB\int_a^R s\,ds = -\frac{QB}{2}\left(R^2 - a^2\right).

Numbers: $L_z = -0.5\times5\times10^{-8}\times0.10\times(4.0-1.0)\times10^{-4} = \mathbf{-7.5\times10^{-13}\ kg\cdot m^2/s}$ .

Step 4. Read the structure — it is the next article's engine. Write the answer as $L_z = -\dfrac{Q}{2\pi}\,B\,\pi(R^2 - a^2) = -\dfrac{Q}{2\pi}\,\Phi_{\text{annulus}}$ : the charge, over $2\pi$ , times the magnetic flux threading the annulus between the charged cylinder and the solenoid wall — pure flux-linkage bookkeeping (via the vector potential it is $L_z$ read off $Q\,sA_\phi$ across the charges' radii, the same statement in gauge clothing). For the textbook single-ring version the identical logic collapses to the famous $L = Q\Phi/2\pi$ . What matters tonight: the stored twist is finite, computable, and switch-ready — kill the solenoid current and Faraday's induced azimuthal $\mathbf{E}$ will torque the cylinders into rotation carrying exactly this much angular momentum. That collection, with the books shown line by line, is the Feynman disk paradox — next.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §8.2.4, Ex. 8.5 · Jackson, Classical Electrodynamics, 3rd ed., §6.12–6.13 · Zangwill, Modern Electrodynamics, §15.9 · Feynman Lectures, Vol. II, §17-4, §27-6 · Beth, Phys. Rev. 50, 115 (1936) · Allen et al., Phys. Rev. A 45, 8185 (1992)

Provenance. The balance law is the moment of the linear momentum theorem (Griffiths §8.2.4; Zangwill §15.9), with the symmetry-of- $T_{ij}$ prerequisite stated per the standard treatment. The charge–monopole result follows Thomson (1904) as presented in Jackson §6.13; Dirac quantization per Dirac (1931) at survey register. The circular-polarization torque uses the absorber-ratio argument (Feynman II-33 lineage) with the plane-wave edge subtlety flagged per Simmons–Guttmann. A verifier should: (1) confirm the moment of the continuity law requires $T_{ij} = T_{ji}$ ; (2) reproduce the dimensional argument for separation-independence and check $\mu_0eq_m/4\pi = \hbar/2$ with $q_m = 2\pi\hbar/\mu_0e = 3.29\times10^{-9}$ A·m; (3) recompute the torque anchors ( $\omega = 1.77\times10^{15}$ rad/s at 1064 nm; $P/\omega = 5.65\times10^{-16}$ N·m/W; doubled through a half-wave plate) and worked example 2's $L_z = -QB(R^2-a^2)/2 = -7.5\times10^{-13}$ kg·m²/s. All formulas textbook-standard; numbers recomputed for this edition.

Builds on: Momentum Density of the Electromagnetic Field · The Maxwell Stress Tensor · The Poynting Vector · Conservation of Linear Momentum in Electrodynamics

Leads to: Hidden Momentum in Magnetic Systems · Newton's Third Law in Electrodynamics · Poynting Vector in Static Field Configurations

If a plane circularly polarized wave has E and B strictly transverse, how can it carry angular momentum along z at all?

This is the subtlety Derivation B deliberately sidestepped, and it deserves its honest telling. For a strictly infinite plane wave, $\mathbf{E}\times\mathbf{B}$ points exactly along $\hat{\mathbf{z}}$ , so $\mathbf{r}\times\mathbf{g}$ has no $z$ -component anywhere: the naive density integral gives zero. But no physical beam is infinite: at a real beam's soft edge, the fields acquire small longitudinal components (required by $\nabla\cdot\mathbf{E} = 0$ when the amplitude varies transversely), and the edge region's momentum density tilts azimuthally — the beam's angular momentum is stored disproportionately in its skirt. Integrate carefully and the total is exactly $U/\omega$ , matching the absorber-ratio argument, which never cared where in the beam the twist resides. The lesson generalizes: densities can localize surprisingly (compare energy "in the gap" of the coax) while integrated exchanges stay robust. When a wave plate sits wholly inside a wide beam, it still extracts $2\hbar$ per photon passing through it — the local field's rotation drives the material's dipoles in circles regardless of where the beam's official angular-momentum density resides.

What is the difference between the spin and orbital angular momentum of light — and is the split physical?

Spin is the polarization part — the $\pm\hbar$ per photon of the two circular handednesses, present even for a plain Gaussian beam. Orbital (OAM) is the shape part: beams with helical phase fronts $e^{im\phi}$ (Laguerre–Gauss modes and kin) carry an additional $m\hbar$ per photon about the axis from the corkscrew tilt of their wavefronts, with $m$ any integer — this is the twist that OAM communications multiplex. Operationally the two are distinct: a half-wave plate flips spin but not OAM; a spiral phase plate or fork hologram changes OAM but not spin; birefringent particles feel spin torque while off-axis scatterers feel orbital. Fundamentally, the clean gauge-invariant split of the total $\int\mathbf{r}\times\mathbf{g}$ into "spin" and "orbital" pieces is famously delicate (the decomposition is not unique for arbitrary fields — a live literature), but for paraxial beams the practical taxonomy above is unambiguous and experimentally operational. This track's scope: total angular momentum is always well-defined and conserved; the named split is a paraxial convenience, honest within its domain.

Does the stored angular momentum of the charged solenoid mean the system was rotating when assembled?

No rotation was ever needed — but torques were, and tracking them is a beautiful assembly audit. Bring the charged cylinders into place with the solenoid already running: as each charge moves radially through the solenoid's field, it feels an azimuthal magnetic force ( $q\mathbf{v}\times\mathbf{B}$ with radial $\mathbf{v}$ and axial $\mathbf{B}$ ), and whatever rails constrain it to move radially must push back azimuthally — the assembler's hands supply a net torque, which is exactly where the field's angular momentum came from. Alternatively, energize the solenoid after placing the charges: during the current ramp, the induced azimuthal electric field torques the stationary charges' mounts. Either assembly history deposits the same $\mathbf{L}_{\text{field}}$ (it is a state function of the final configuration), and the disk paradox is just the audit run backward — disassembly by switch-off, with the deposit withdrawn as visible spin. The instinct "static systems were never torqued" is the angular cousin of "static fields carry no momentum," and this chapter has now retired both.

Why does the charge-monopole angular momentum point from charge to pole rather than perpendicular to the line, and does the sign matter?

The direction follows from the azimuthal circulation's handedness. The momentum density $\varepsilon_0\mathbf{E}\times\mathbf{B}$ circulates around the charge-pole axis; circulate one way and the moment $\mathbf{r}\times\mathbf{g}$ points along the axis one way, everywhere. For positive $q$ and a north pole $q_m > 0$ , working the cross products gives $\mathbf{L}$ along the axis from the charge toward the pole; flip either sign and $\mathbf{L}$ reverses. The sign matters for exactly one famous reason: in the quantum treatment, the conserved angular momentum of a charged particle orbiting a monopole is $\mathbf{L}_{\text{orbital}} - (\mu_0qq_m/4\pi)\hat{\boldsymbol{r}}$ — the field's contribution rides along with the particle's direction — and consistency of that operator algebra (single-valued wavefunctions on the sphere) is one of the cleanest routes to the quantization condition. It also explains a classical oddity you can check: a charge scattered by a monopole never moves on a plane (as central-force problems do) but on a cone about the total angular momentum — the field's fixed $\hbar/2$ -scale contribution tilts the geometry. Mechanics feels the field's ledger even in the orbits.

Do these field angular momenta matter for real machines, or only for paradoxes and photons?

They surface wherever strong fields and rotation meet. Every electric machine transmits its torque through the field: the air-gap shear stress of the stress-tensor article is the angular-momentum flux of equation (2), so "the field carries the torque" is a literal description of every motor's operation — megawatt machines transmit their full mechanical output through centimeters of empty gap as field angular-momentum flux. In plasma physics and astrophysics the boundary term dominates entire systems: magnetized winds torque down rotating stars (the "magnetic braking" that explains why old stars spin slowly), pulsars convert rotational angular momentum to electromagnetic flux at rates set exactly by these integrals, and accretion disks solve their central angular-momentum-transport problem partly through magnetic stresses. Even engineering-adjacent: reaction-wheel-free satellite attitude schemes using magnetic torquers exchange angular momentum with the geomagnetic field — the Earth's field as the other party in equation (2). The paradoxes are the pedagogy; the ledger is infrastructure.

If angular momentum is conserved, why do pulsars spin down — where does their angular momentum go?

Out through the boundary term, as electromagnetic angular-momentum flux — the cleanest astrophysical instance of equation (2). A pulsar is a magnetized conductor rotating at up to hundreds of hertz; its corotating magnetosphere and the rotating dipole field radiate both energy and angular momentum (for the orthogonal-rotator idealization, torque $N \sim B^2R^6\Omega^3/6\pi\varepsilon_0c^5\cdot...$ — the magnetic-dipole braking formula), and the star's $I\Omega$ pays. Measured spin-down rates of radio pulsars match the electromagnetic-braking scaling well enough that observers use $\Omega/\dot\Omega$ as an age estimate ("characteristic age") and infer surface fields of $10^8$ – $10^{11}$ T from it. The angular momentum itself ends up in the pulsar wind and radiation streaming to the nebula — the Crab Nebula glows on the angular momentum and energy the Crab pulsar has shed since 1054. Nothing in the ledger is special to the laboratory: draw the sphere at any radius, integrate $\mathbf{r}\times(T\cdot d\mathbf{a})$ , and the books balance from bench to nebula.