Lorentz Force Law

Twenty-seven kilometres under the Franco-Swiss border, protons carrying the energy of a speeding train are held on a circular track by nothing but magnetic fields — while electric fields, pulsing in perfect rhythm, nudge them faster on every lap. Why the division of labour? Why doesn't the LHC just use stronger magnets to speed the protons up? Because it can't. No magnetic field, however monstrous, has ever added a single joule to a particle's kinetic energy. That prohibition is written into one compact equation. The Lorentz force, $\vec F = q(\vec E + \vec v\times\vec B)$ , is the contract between electromagnetic fields and matter. Maxwell's equations tell fields how charges create them; the Lorentz force closes the loop and tells charges what the fields do back. Its two terms could not be more different in character. The electric term pushes along the field whether the charge moves or not, and it does work — it is the accelerator pedal. The magnetic term switches on only when the charge moves, and the cross product locks it perpendicular to the velocity forever — a steering wheel welded at right angles, able to bend the path into circles and helices but never to change the speed. This article opens the magnetostatics chapter by unpacking that asymmetry to full depth: the no-work theorem and the paradox it creates (a magnetic crane visibly lifts iron — who paid for the lift?), the law's continuum form $\rho\vec E + \vec J\times\vec B$ that will power everything from wire forces to plasmas, and the relativistic packaging in which the electric–magnetic split dissolves into a single field tensor.

Key takeaways

A charge $q$ moving at velocity $\vec v$ through fields $\vec E$ and $\vec B$ feels the Lorentz force $\vec F = q(\vec E + \vec v\times\vec B)$ — the operational definition of what the fields mean.
The electric part $q\vec E$ acts regardless of motion and does work; the magnetic part $q\vec v\times\vec B$ is always perpendicular to $\vec v$ and does exactly zero work.
When a magnetic force seems to do work — a crane lifting scrap, a motor turning — the energy is always smuggled in from elsewhere (the battery, the source of the current); the magnetic force only redirects it. This article audits the books line by line.
For extended matter the law becomes a force density $\vec f = \rho\vec E + \vec J\times\vec B$ ; integrating it over a wire gives $\vec F = I\!\int\mathrm{d}\vec\ell\times\vec B$ , the workhorse of the next articles.
In a uniform magnetic field a charge circles with radius $r = mv/(qB)$ — quoted here from a one-line force balance, and developed into the full theory of orbits, helices and accelerators in Cyclotron Motion.
The law is relativistically exact as $\mathrm{d}\vec p/\mathrm{d}t = q(\vec E + \vec v\times\vec B)$ with $\vec p = \gamma m \vec v$ ; covariantly it is $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ , and the electric/magnetic split becomes frame-dependent.
KEY IDEA: electric fields accelerate, magnetic fields steer — the cross product makes the magnetic force forever perpendicular to the motion, so it bends paths and redirects energy without ever supplying any.

The core insight

The fields act on charge through $\vec F = q(\vec E + \vec v\times\vec B)$ : the electric part pushes along $\vec E$ and does work, while the magnetic part deflects sideways and does none — it curves the trajectory without ever touching the speed.

Where this lives today

Every working accelerator is the Lorentz force with a budget. In the LHC, 1232 superconducting dipoles at $8.09\ \mathrm{T}$ enforce $r = p/(qB)$ on $6.8\ \mathrm{TeV}$ protons — plug in the numbers and the bending radius comes out near $2.8\ \mathrm{km}$ , so the dipoles trace $2\pi r \approx 17.6\ \mathrm{km}$ , about two-thirds of the $26.7\ \mathrm{km}$ ring; the rest is straight sections for RF, the experiments, injection and beam dump — while the actual accelerating is done by RF cavities whose electric fields deliver the energy, exactly as equation (5) demands. Fusion machines are the confinement side of the same law: a tokamak's plasma is held by the volume force $\vec J\times\vec B$ of equation (4), the continuum Lorentz force balancing the plasma's pressure gradient — magnetohydrodynamics is this article scaled up to a hundred million kelvin. The sky runs the experiment for free. Solar-wind electrons and protons corkscrew along Earth's magnetic field lines into the polar atmosphere — the aurora is Lorentz-force steering made visible — and the Van Allen belts are populations of charges trapped for years on bouncing orbits, steered every second of their lives by a force that has never done a joule of work on them. Even your DC motor obeys the audit of Derivation B: the "back-EMF" every electrical engineer subtracts from the supply voltage is the drag term $quB$ wearing its circuit-theory clothes, and motor efficiency accounting is the rising-bar example with commutators attached.

Origins · 1 minHendrik Lorentz and how this was discovered

Hendrik Lorentz

1853–1928 · Leiden

Lorentz pictured matter the way we now know it to be — swarms of tiny charged particles — decades before anyone had seen an electron. Within his "electron theory" of the 1890s he wrote the force law in its definitive form, welding the electric and magnetic pushes into one expression. The payoff came fast: when Pieter Zeeman found spectral lines splitting in a magnetic field, Lorentz used the force law to predict the splitting's size and polarization — and to estimate the charge-to-mass ratio of the unseen oscillating particles a year before Thomson's electron. The two shared the 1902 Nobel Prize. Chasing how his force law behaves for observers in relative motion, Lorentz then built the transformation equations that Einstein would reinterpret as the structure of space and time itself — and that make the covariant form of this article's law possible.

History

1820

— Hans Christian Ørsted's compass needle twitches beside a current-carrying wire; André-Marie Ampère quantifies the forces between currents within months.

1879

— Edwin Hall discovers that a magnetic field pushes the moving charges inside a conductor sideways — the Hall effect, the force law caught in the act.

1892

— Hendrik Lorentz publishes the force law in its modern form,

\vec F = q(\vec E + \vec v\times\vec B)

, as the heart of his electron theory.

1897

— J.J. Thomson balances the electric term against the magnetic one in crossed fields, measures

e/m

, and discovers the electron.

1905

— Einstein's special relativity reveals the deeper truth: electric and magnetic forces are one frame-dependent phenomenon, packaged today as

F^{\mu\nu}

1932

— Ernest Lawrence and M. Stanley Livingston run the first cyclotron — accelerators have compounded on the law's two terms ever since.

The physical picture

Drop a charge at rest into the fields and only one term wakes up: $q\vec E$ tugs it along the field line, and as it gains speed the magnetic term switches on in proportion. The cross product $\vec v\times\vec B$ then enforces a strange discipline — point your right hand's fingers along $\vec v$ , curl toward $\vec B$ , and the thumb gives the force on a positive charge: always sideways, never fore or aft. A force that is perpetually perpendicular to your motion is like a steering wheel jammed at full lock: you can drive in circles all day, but the speedometer never moves. That is the whole design philosophy of particle machines in two sentences. When an engineer wants a beam faster, she builds an electric field along its path (an RF cavity). When she wants it bent, focused, or trapped, she builds a magnetic field across it (a dipole or quadrupole magnet). The Lorentz force never lets the two jobs blur.

Figure

Two fields, two jobs: the pedal and the steering wheel

One more picture to keep. Why doesn't the circling charge spiral inward, like a satellite decaying in the atmosphere? Because decay requires losing energy, and the magnetic force cannot take any away — the same theorem that stops it giving energy. With speed frozen and force magnitude fixed at $qvB$ , the orbit is a perfect circle, closed and eternal (until the particle radiates — a story the radiation chapter owns). One line of Newton fixes its size: the magnetic force supplies the centripetal demand, $qvB = mv^2/r$ , so $r = mv/(qB)$ . That single formula already runs mass spectrometers and beam pipes; the full dynamics — frequencies, helices, drifts, and the cyclotron's timing miracle — is the business of the next articles, Magnetic Force on Moving Charges and Cyclotron Motion. But there is a puzzle hiding in plain sight, and this chapter refuses to tiptoe past it. A junkyard electromagnet hoists half a tonne of scrap iron three metres into the air. Lifting is work — force through a distance, joules by the hundred thousand. Yet we have just argued, from pure geometry, that magnetic forces never do work. Both statements are true. Resolving them — finding who actually pays for the lift — is Derivation B below, and it is one of the most instructive audits in electromagnetism.

The law and what it forbids

The picture needs an equation to stand on, so ask the operational question: release a test charge into the fields and watch its acceleration — what law reproduces what you see? Every experiment since Ørsted's twitching compass gives the same answer, and it compresses a century of deflected beams and Hall voltages into a single line: one term for the push that is always there, and one for the push that motion switches on.

\vec F = q\left(\vec E + \vec v\times\vec B\right) \qquad (1)

where: $q$ — the particle's charge, sign included; $\vec E$ — electric field at the particle's location ( $\mathrm{V/m}$ ); $\vec B$ — magnetic flux density ( $\mathrm{T}$ ); $\vec v$ — the particle's velocity. This is an experimental law, not a theorem — and it is the operational definition of $\vec E$ and $\vec B$ : put a test charge somewhere, watch its acceleration at rest and then at two independent velocities, and you can read off both fields. Whatever else "the electromagnetic field" may mean, equation (1) is how it is measured. How hard does the magnetic term actually push? A cross product's magnitude carries the sine of the angle between its factors, so the answer depends on how the velocity is aimed relative to the field — the same speed can buy anything from zero force to the full maximum. Isolating the magnetic piece of (1) and taking magnitudes:

|\vec F_B| = qvB\sin\theta \qquad (2)

where: $\theta$ — the angle between $\vec v$ and $\vec B$ . Motion along the field ( $\theta = 0$ ) feels nothing; motion across it ( $\theta = 90^\circ$ ) feels the full $qvB$ . The geometry of this term — directions, signs, and what crossed $\vec E$ and $\vec B$ fields can do in concert — gets its own article next. One piece remains before the law can run an accelerator: what dynamics does the force feed? A force changes momentum — that is Newton's second law in the form that survives at every speed. Feed the Lorentz force into it and, remarkably, the resulting equation is relativistically exact as written — no correction terms appear even as $v$ approaches $c$ :

\frac{\mathrm{d}\vec p}{\mathrm{d}t} = q\left(\vec E + \vec v\times\vec B\right), \qquad \vec p = \gamma m \vec v \qquad (3)

where: $\gamma = 1/\sqrt{1 - v^2/c^2}$ — the Lorentz factor; at everyday speeds $\gamma \to 1$ and $\vec p \to m\vec v$ . The law predates special relativity by a decade, yet needed no correction — a strong hint, cashed at the end of this article, that magnetism is relativity in disguise. Matter in bulk is not one charge but a smeared density of them, so the law must also come in a continuum edition. Replace the single $q$ by the charge $\rho\,\mathrm{d}\tau$ in each volume element and the single $q\vec v$ by the current density $\vec J\,\mathrm{d}\tau$ , and the force per unit volume reads

\vec f = \rho\,\vec E + \vec J\times\vec B, \qquad \vec F = \int_V\left(\rho\,\vec E + \vec J\times\vec B\right)\mathrm{d}\tau \qquad (4)

where: $\rho$ — charge density ( $\mathrm{C/m^3}$ ); $\vec J$ — current density ( $\mathrm{A/m^2}$ ); the integral runs over the body. This is the form that grips wires, plasmas and motors: specialized to a thin wire it becomes $\vec F = I\!\int\mathrm{d}\vec\ell\times\vec B$ — the carrier-by-carrier bookkeeping that turns $\vec J\times\vec B$ into $I\,\mathrm{d}\vec\ell\times\vec B$ is owned by Electric Current and Current Density, and the applications (parallel wires, torques, motors) by Magnetic Force on Current-Carrying Wires. Here it is the law itself that matters: equation (4) is still just (1), summed.

Derivation AThe magnetic force does no work

The goal is the single most consequential theorem about equation (1): the magnetic term never changes a particle's kinetic energy — at any speed, in any field, along any trajectory. The strategy is embarrassingly short: compute the power delivered by the force and watch the magnetic piece die by pure vector geometry. The thing to appreciate is how little physics the proof uses — no trajectory is solved, no field specified — which is exactly why the theorem admits no exceptions.

Step 1. Write the power delivered to the particle:

P = \vec F\cdot\vec v.

Why: force dotted into velocity is the rate of change of kinetic energy — the work–energy theorem in rate form, valid relativistically when $W$ is read as the kinetic energy attached to $\vec p = \gamma m\vec v$ .

Step 2. Insert the Lorentz force and distribute the dot product:

P = q\,\vec E\cdot\vec v + q\,(\vec v\times\vec B)\cdot\vec v.

Step 3. Kill the magnetic term. $\vec v\times\vec B$ is perpendicular to $\vec v$ by the very definition of the cross product, and the dot product of perpendicular vectors vanishes identically — for any $\vec v$ , any $\vec B$ , at any speed. Why this cannot be rescued: the statement is pure geometry, prior to any dynamics. No cleverness of field design can make a vector overlap with a direction it is constructed to be perpendicular to.

Step 4. Read the result:

\frac{\mathrm{d}W}{\mathrm{d}t} = q\,\vec E\cdot\vec v \qquad (5)

Why it matters: every joule a charged particle ever gains or loses electromagnetically is delivered by the electric field. In a betatron or transformer, where "the magnetic field" seems to accelerate charges, the true agent is the electric field induced by the changing $\vec B$ . *Note: equation (5) is precisely the time component of the covariant equation $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ derived below — the relativistic bookkeeping reproduces the same statement exactly.*

The paradox: B does no work — so what lifts the bar?

Now collide the theorem with the junkyard. A horizontal conducting bar carries current $I$ along its length and sits in a magnetic field; the force $I\vec\ell\times\vec B$ points up, and the bar rises. Force up, displacement up: the magnetic force appears to do work $F\Delta h$ on the bar, in flat contradiction of Derivation A. Magnetic cranes, loudspeaker cones and every electric motor stage the same crime scene daily. The resolution is not a technicality — it is the correct mental model of what magnetic forces are: brokers, not payers. Watch a single charge carrier inside the rising bar.

Figure

Inside the rising bar — one carrier's force budget

Go deeper · Derivation B · 2 minWho pays for the lift — the rising-bar energy audit

Derivation BWho pays for the lift — the rising-bar energy audit

The goal is to resolve the work paradox quantitatively: show that the magnetic force on the carriers in a rising current-carrying bar does exactly zero net work, and identify the true source of the lifting energy. The strategy is to follow one carrier and keep honest books — the trap this derivation exists to expose is using the carrier's drift velocity in the force law when its true velocity also contains the bar's rise. Watch the two components of one cross product pay each other off.

Step 1. Set the stage. Current $I$ flows along the bar (positive carriers of charge $q$ drift at $w\,\hat{\vec x}$ ); the field points into the page, $\vec B = -B\,\hat{\vec z}$ ; the bar rises at $u\,\hat{\vec y}$ . A carrier locked in the bar therefore truly moves at

\vec v = w\,\hat{\vec x} + u\,\hat{\vec y}.

Why this is the crux: the carrier is not just drifting — it is also being carried upward. Forgetting the $u\hat{\vec y}$ piece is what makes the paradox look paradoxical.

Step 2. Compute the full magnetic force on the carrier, using $\hat{\vec x}\times\hat{\vec z} = -\hat{\vec y}$ and $\hat{\vec y}\times\hat{\vec z} = +\hat{\vec x}$ :

q\,\vec v\times\vec B = q\left(w\,\hat{\vec x} + u\,\hat{\vec y}\right)\times(-B\,\hat{\vec z}) = qwB\,\hat{\vec y} - quB\,\hat{\vec x}.

Two clean pieces: $qwB$ pointing vertically up (the lift — born of the drift) and $quB$ pointing against the drift (the drag — born of the rise). The lift component, summed over all carriers in the bar, is the familiar $BI\ell$ that raises it.

Step 3. Audit the power, carrier by carrier. The lift component $qwB$ acts along the rise $u$ : rate of work $+qwB\,u$ . The drag component $quB$ acts against the drift $w$ : rate of work $-quB\,w$ . Add them:

P_{\rm mag} = qwB\,u - quB\,w = 0.

Why it had to vanish: the total force $q\vec v\times\vec B$ is perpendicular to the total velocity $\vec v$ — Derivation A applied honestly. The "work done by the magnetic force" on the bar's rise is real, but it is exactly financed by magnetic work extracted from the carriers' drift motion.

Step 4. Find the payer. The drag $quB$ per carrier would kill the current in microseconds unless something drives the carriers forward against it — that something is the battery (or supply) maintaining $I$ . Per unit charge, the drag force integrated along the bar is $uB\ell$ — precisely the motional EMF the moving bar generates. The battery therefore does extra electrical work at rate

P_{\rm batt} = (\text{EMF})\,I = B\ell u\,I = (BI\ell)\,u = F_{\rm lift}\,u = P_{\rm mech},

exactly the mechanical power delivered to the rising bar. Energy flows battery → (against $quB$ ) → carriers → (via $qwB$ ) → bar. The magnetic field takes a commission of zero. Worked example 2 runs this audit with numbers — a 20 g bar held by 7.85 A in 0.25 T, lifted at 0.50 m/s: mechanical power 0.098 W, back-EMF 12.5 mV, battery surcharge 12.5 mV × 7.85 A = 0.098 W, matching to the last digit.

The moral generalizes: whenever magnetism appears to do work — cranes, motors, two magnets snapping together — trace the energy and you will find an electric field, a battery, or (for permanent magnets) the quantum machinery that maintains atomic currents doing the actual paying, with the magnetic force acting as a perfect, frictionless transmission. The FAQ returns to the permanent-magnet case, which is subtler and honestly beyond pure classical magnetostatics.

Relativity built in

The law's two terms look like separate phenomena, but hold equation (3) up to the light of special relativity and the seam disappears. Package the fields into the antisymmetric field tensor $F^{\mu\nu}$ — whose components are just $\vec E$ and $\vec B$ filed in a 4×4 grid — and the entire Lorentz force law, power equation included, collapses into one covariant line. This is graduate-register bookkeeping arriving early, but the E4 chapter earns it: it explains why the law needed no relativistic correction.

\frac{\mathrm{d}p^{\mu}}{\mathrm{d}\tau} = q\,F^{\mu\nu}u_{\nu} \qquad (6)

where: $p^{\mu} = (W/c,\ \vec p)$ — the four-momentum; $\tau$ — proper time; $u_{\nu}$ — the four-velocity; $F^{\mu\nu}$ — the field tensor, with $F^{i0} = E^i/c$ and $F^{ij} = -\epsilon^{ijk}B_k$ . Read it physically: one tensor, one equation, no "electric force" and "magnetic force" as separate citizens — the split depends on who is watching, because a Lorentz boost mixes $\vec E$ and $\vec B$ into each other. A force that is purely magnetic in the lab is partly electric to the moving particle.

Go deeper · Derivation C · 1 minUnpacking the covariant law — three-vector force and the power equation from one tensor line

Derivation CUnpacking the covariant law — three-vector force and the power equation from one tensor line

The goal is to verify that equation (6) is the Lorentz force law: its space components must reproduce $\mathrm{d}\vec p/\mathrm{d}t = q(\vec E + \vec v\times\vec B)$ , and its time component must reproduce the no-work bookkeeping $\mathrm{d}W/\mathrm{d}t = q\vec E\cdot\vec v$ of Derivation A. The strategy is direct component expansion using the tensor dictionary $F^{i0} = E^i/c$ , $F^{ij} = -\epsilon^{ijk}B_k$ and the four-velocity $u^\nu = \gamma(c, \vec v)$ . Watch the factor of $\gamma$ — it enters through proper time and leaves again, which is the whole magic.

Step 1. Expand a space component ( $\mu = i$ ), splitting the sum over $\nu$ into time and space parts:

\frac{\mathrm{d}p^{i}}{\mathrm{d}\tau} = q\left(F^{i0}u_0 + F^{ij}u_j\right).

Step 2. Substitute the dictionary. With $u_0 = \gamma c$ and $u_j = -\gamma v_j$ (lowered index flips the sign of the space part in the $({+},{-},{-},{-})$ convention):

\frac{\mathrm{d}p^{i}}{\mathrm{d}\tau} = q\left(\frac{E^i}{c}\,\gamma c + \epsilon^{ijk}B_k\,\gamma v_j\right) = q\gamma\left(E^i + (\vec v\times\vec B)^i\right).

Why the cross product appears: $\epsilon^{ijk}v_jB_k$ is precisely the $i$ -th component of $\vec v\times\vec B$ — the antisymmetry of $F^{\mu\nu}$ is where the cross product has been hiding all along.

Step 3. Convert proper time to lab time. Since $\mathrm{d}\tau = \mathrm{d}t/\gamma$ , the left side is $\gamma\,\mathrm{d}p^i/\mathrm{d}t$ ; the $\gamma$ 's cancel:

\frac{\mathrm{d}\vec p}{\mathrm{d}t} = q\left(\vec E + \vec v\times\vec B\right).

Why this is remarkable: the naive 1892 law, with $\vec p = \gamma m\vec v$ , is term-for-term the exact relativistic equation — no velocity-dependent corrections, no expansion in $v/c$ . Lorentz wrote better than he could have known.

Step 4. Expand the time component ( $\mu = 0$ ): $\mathrm{d}p^{0}/\mathrm{d}\tau = qF^{0j}u_j = q(-E^j/c)(-\gamma v_j) = (q\gamma/c)\,\vec E\cdot\vec v$ . With $p^0 = W/c$ and $\mathrm{d}\tau = \mathrm{d}t/\gamma$ :

\frac{\mathrm{d}W}{\mathrm{d}t} = q\,\vec E\cdot\vec v.

Why this closes the circle: the no-work theorem of Derivation A is not an accident of three-vector geometry — it is the time component of the covariant law. $\vec B$ never appears in the energy row of the tensor equation because the antisymmetric $F^{\mu\nu}$ has no slot for it there. *Note: the metric-convention sign flip in $u_j$ is the standard $({+},{-},{-},{-})$ bookkeeping, and the final three-vector law is independent of that convention, as it must be.*

Limiting cases

Case	Result	Meaning
$\vec B = 0$ or $\vec v = 0$	$\vec F = q\vec E$	pure electric push — accelerates, does work
$\vec E = 0$ , $\vec v \perp \vec B$	circle: $r = mv/qB$	steering only — dynamics developed in Cyclotron Motion
$\vec v \parallel \vec B$	$\vec F_B = 0$	charges slide along field lines untouched
$v \ll c$	$m\,\mathrm{d}\vec v/\mathrm{d}t = q(\vec E + \vec v\times\vec B)$	Newtonian limit, $\gamma \to 1$
$v \to c$	same law, $\vec p = \gamma m\vec v$	exact — no correction terms, ever
continuous matter	$\vec f = \rho\vec E + \vec J\times\vec B$	force per volume — wires, plasmas, motors
covariant package	$\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$	E/B split becomes frame-dependent

Common trapThe magnetic field sped the particle up

It never has and never will. Because $\vec v\times\vec B \perp \vec v$ , the magnetic force delivers zero power — equation (5) — so it produces only centripetal acceleration: direction changes, speed does not. Whenever a magnetic device seems to energize matter (betatron, transformer, crane, motor), audit the books: the working agent is an electric field, a battery fighting the magnetic back-force (Derivation B), or the internal machinery of magnetized matter. Two smaller traps ride along: the order of the cross product matters ( $\vec v\times\vec B$ , not $\vec B\times\vec v$ — reversing it flips the force), and $q$ carries its sign, so an electron curves the opposite way to a proton in the same field. And one subtle one: in the force law, use the particle's total velocity — a carrier in a moving wire drifts and rides the wire, and dropping either piece breaks the energy audit.

Worked examples

1 · An electron beam bent into an arc (confidence builder)

An electron is accelerated from rest through a potential difference of $2.0\ \mathrm{kV}$ , then enters a uniform field $B = 1.0\ \mathrm{mT}$ perpendicular to its velocity. Find its speed and the radius of its circular path, and audit who did the work. ( $m_e = 9.11\times10^{-31}\ \mathrm{kg}$ , $e = 1.60\times10^{-19}\ \mathrm{C}$ .)

Step 1. The electric stage does the work: $eV = \tfrac12 m v^2$ , so $v = \sqrt{2eV/m} = \sqrt{2(1.60\times10^{-19})(2000)/(9.11\times10^{-31})} = 2.65\times10^{7}\ \mathrm{m/s}$ — about $0.09c$ , safely non-relativistic ( $\gamma \approx 1.004$ ).

Step 2. The magnetic stage only bends. Force balance $evB = mv^2/r$ gives $r = \dfrac{mv}{eB} = \dfrac{(9.11\times10^{-31})(2.65\times10^{7})}{(1.60\times10^{-19})(1.0\times10^{-3})} = \boxed{0.15\ \mathrm{m}}$ .

Step 3. The audit: the electron's kinetic energy at entry is $2.0\ \mathrm{keV}$ — all of it banked in the accelerating gap, at rate $q\vec E\cdot\vec v$ per equation (5). After a thousand laps in the magnetic field it is still exactly $2.0\ \mathrm{keV}$ : the field has bent the path through $6280$ radians and contributed zero joules ✓.

2 · The levitating bar — the paradox with numbers (exam level)

A rigid bar of mass $20\ \mathrm{g}$ and length $\ell = 10\ \mathrm{cm}$ can slide vertically on frictionless rails in a horizontal field $B = 0.25\ \mathrm{T}$ perpendicular to the bar. (a) What current levitates it? (b) With that current flowing, the bar is allowed to rise at a steady $u = 0.50\ \mathrm{m/s}$ . Find the mechanical power delivered, the motional EMF, and show the battery pays for everything.

Step 1. Levitation current. The lift force is $F = BI\ell$ (the vertical component $qwB$ summed over carriers — Derivation B, Step 2). Setting $BI\ell = mg$ : $I = \dfrac{mg}{B\ell} = \dfrac{(0.020)(9.81)}{(0.25)(0.10)} = \boxed{7.85\ \mathrm{A}}$ .

Step 2. Mechanical power during the rise: $P_{\rm mech} = mgu = (0.020)(9.81)(0.50) = \boxed{0.098\ \mathrm{W}}$ — this is the rate at which something must be doing work on the bar-plus-gravity system.

Step 3. The motional EMF. As the bar rises, each unit of charge in it feels the drag component $uB$ along the wire (Derivation B, Step 3), integrating to $\mathcal{E} = B\ell u = (0.25)(0.10)(0.50) = \boxed{12.5\ \mathrm{mV}}$ opposing the current.

Step 4. The battery's surcharge: to keep $I = 7.85\ \mathrm{A}$ flowing against this EMF, the supply must furnish extra power $\mathcal{E}I = (0.0125)(7.85) = \boxed{0.098\ \mathrm{W}}$ — equal to $P_{\rm mech}$ to the last digit. The magnetic force transmitted 98 milliwatts from the battery to the lifted bar and kept none: net magnetic work zero, paradox closed.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §5.1 · Jackson, Classical Electrodynamics, 3rd ed., §5.1 and §11.5 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 5–6 (the relativistic origin of the magnetic force) · Zangwill, Modern Electrodynamics, §2.1 and §22.2. All equations are standard results. Key quantities used throughout: the Lorentz force $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ and force density $\rho\mathbf{E}+\mathbf{J}\times\mathbf{B}$ ; the power identity $\mathrm{d}W/\mathrm{d}t=q\mathbf{E}\cdot\mathbf{v}$ ; the cyclotron results $\omega_c=qB/m$ , $r=mv/(qB)$ ; the covariant form $\mathrm{d}p^\mu/\mathrm{d}\tau=qF^{\mu\nu}u_\nu$ with tensor components $F^{i0}=E^i/c$ , $F^{ij}=-\epsilon^{ijk}B_k$ ; and the proton-to-electron mass ratio $m_p/m_e\approx1836$ . Worked numbers: electron gun — $v = 2.65\times10^{7}$ m/s, $r = 0.151$ m; levitating bar — $I = 7.85$ A, $P_{\rm mech} = 0.098$ W, $\mathcal{E} = 12.5$ mV, $\mathcal{E}I = 0.098$ W; LHC bending radius $p/(qB) = 2.7$ km at 6.8 TeV, 8.3 T.

Builds on: Electric Current and Current Density · Vector cross product and Newton's second law

Leads to: Magnetic Force on Moving Charges · Cyclotron Motion and Magnetic Confinement · Magnetic Force on Current-Carrying Wires · Biot-Savart Law

If magnetic fields never do work, how do two permanent magnets snap together?

Honestly: classical magnetostatics alone cannot close this case, which is why the fridge magnet is a better paradox than the crane. At the classical level a permanent magnet is modeled as bound atomic current loops, and when two magnets attract, the work done on the approaching magnet is financed by changes in those internal currents — the same broker-not-payer structure as the rising bar, with the "battery" role played by whatever maintains the atomic currents. But atomic magnetic moments are quantum objects (spin has no classical drift velocity to drag), and the full accounting needs quantum mechanics. Griffiths flags exactly this in §5.1.3: the theorem "magnetic forces do no work" is airtight for the classical Lorentz force on charges; its application to intrinsic-moment matter is a subtler, partly quantum story.

Is the Lorentz force law derived from Maxwell's equations, or is it separate?

Separate — and that separateness is the architecture of the whole theory. Maxwell's equations determine the fields produced by given charges and currents; the Lorentz force law determines the forces given fields exert on charges. Neither implies the other: together they form the complete closed loop of classical electrodynamics (charges → fields → forces → motion of charges → …). In more advanced formulations both emerge from a single action principle, and the force law can be packaged with field momentum conservation via the Maxwell stress tensor — but at this chapter's level, treat equation (1) as an independent experimental pillar, on the same footing as Maxwell's equations themselves.

How can one equation define two fields? Measuring F at one velocity gives one datum, but E and B are six unknowns.

Use the velocity dependence. Park the test charge ( $\vec v = 0$ ): the measured force is $q\vec E$ — three components, $\vec E$ done. Now throw it in two different directions and subtract the electric part: each throw measures $q\vec v\times\vec B$ , and two independent throws pin all three components of $\vec B$ (one throw alone leaves the component of $\vec B$ parallel to $\vec v$ invisible, since it drops out of the cross product). That is the operational content of "the Lorentz force defines the fields": a systematic protocol of tosses that reads out all six numbers at a point.

Does the no-work theorem survive relativity, where forces and energies change between frames?

Perfectly — in fact relativity explains it. The covariant law $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ carries the energy equation as its time component, and expanding it (Derivation C, Step 4) gives $\mathrm{d}W/\mathrm{d}t = q\vec E\cdot\vec v$ with no magnetic contribution, at any speed. What does change between frames is the split: a force that is purely magnetic (zero work) in the lab may be partly electric (nonzero work) for another observer — and the energy transferred is different in different frames, exactly as kinetic energy itself is. Each observer's books balance internally; relativity guarantees they are consistent translations of one another.

In the rising-bar audit, the battery does the work — but the battery pushes charges along the wire, perpendicular to the lift. How does its energy end up vertical?

Through the magnetic force acting as a rigid, frictionless linkage. The battery pushes carriers along the wire against the drag $quB$ ; the same carriers, because they drift, feel the vertical $qwB$ which pushes the lattice of the bar upward through collisions and the constraint forces binding carriers to the wire. Think of a bicycle: your legs push down on pedals, yet the bike moves forward — the chain redirects the work without doing net work itself. The magnetic force is the chain. The bookkeeping in Derivation B shows the redirection is exact: power absorbed from the drift channel equals power delivered to the lift channel, always.

Why is the force qv×B and not qB×v? How do I stop getting the sign wrong?

The order is a convention locked into the definition of $\vec B$ 's direction (via the right-hand rule for cross products); flip the order and you flip the force, which is a physics error, not a style choice. Reliable ritual: fingers point along $\vec v$ , curl them toward $\vec B$ through the smaller angle, thumb gives $\vec v\times\vec B$ ; then multiply by the sign of $q$ — for electrons the force is opposite your thumb. And keep one memorized anchor to calibrate every new problem against: a proton moving along $+\hat{\vec x}$ in $\vec B = B\hat{\vec z}$ feels a force along $-\hat{\vec y}$ , since $\hat{\vec x}\times\hat{\vec z} = -\hat{\vec y}$ . Work one cross product carefully, then trust the anchor forever.

The article says the law is exact even near light speed. Doesn't something have to break eventually?

Two honest boundaries. First, at extreme accelerations a charge radiates electromagnetic energy, and the recoil of that radiation (radiation reaction) adds a correction to the equation of motion that the plain Lorentz law omits — classical electrodynamics patches this with the Abraham–Lorentz force, a story owned by the advanced-electrodynamics tier. Second, at the quantum level forces give way to interaction vertices in QED; the Lorentz force re-emerges as the classical limit. But within classical, point-charge electrodynamics — LHC energies emphatically included — equation (3) with $\vec p = \gamma m\vec v$ is exact, and accelerator engineers use it without correction terms every day.

Do E and B fields of the same source push on the charge twice — is there double counting in the two terms?

No — the two terms are read off simultaneously from whatever the total local fields are, whoever made them, and each does a different job on the velocity vector. There is, however, one charge whose fields never enter: the particle's own. A point charge does not exert a net Lorentz force on itself in this framework (self-interaction is the radiation-reaction subtlety of the previous FAQ). Practical rule: assemble $\vec E$ and $\vec B$ at the particle's location from all other sources, then apply equation (1) once.