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The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are

Magnetic rigidity measures the resistance of a charged particle's momentum to bending by a magnetic field. The relativistic cyclotron orbit gives $p_{\perp }=|q|Br$, where $r$ is the transverse gyroradius. For motion perpendicular to a uniform magnetic field, $p_{\perp }=p$ is the whole momentum. A measured magnetic rigidity fixes momentum per magnitude of electric charge; it does not by itself fix mass, kinetic energy, or the sign of the electric charge. This distinction makes the Anderson positron track experiment an especially revealing application of the Lorentz force.

For $A^{\mu }=(0,0,Bx,0)$, the only independent nonzero component of the electromagnetic field tensor isThe temporal and $z$ components of the Lorentz force show that $u^0$ and $u^3$ are constant. Taking $u^3=0$, the transverse equations arewhere dots now denote proper time derivatives. Witha solution isDirect differentiation verifies both equations. Hence the trajectory is a relativistic cyclotron orbit with proper angular frequency $omega=qB/m$; the sign fixes its orientation.

Let $\mathbf{b}=\mathbf{B}/B$ for a nonzero constant magnetic field. Since $\gamma$ is constant, the relativistic Lorentz force reduces toThe projection along $\mathbf{b}$ is constant. In the perpendicular plane define the signed cyclotron frequency $\Omega =eB/(m{\gamma }_0)$; thenThe explicit rotating velocity isThis is the relativistic cyclotron orbit frequency in laboratory time, reduced from the nonrelativistic frequency by $1/{\gamma }_0$. With nonzero parallel velocity, the trajectory is a helix. If ${\mathbf{v}}_{\perp 0}=0$, that perpendicular vector stays zero and has no nontrivial fundamental period; the displayed $T$ is the period of a nonzero transverse orbit.

At larger kinetic energy, special relativity changes the relation between momentum and velocity. With $m$ the invariant mass, the Lorentz factor is $\gamma =1+K/(m{c}^2)$, and a relativistic cyclotron orbit in a fixed uniform magnetic field satisfiesThe cyclotron frequency decreases as the kinetic energy increases. If the drive stays at the classical value ${\omega }_0=|q|B/m$, the half-turn time is $\pi \gamma /{\omega }_0$. During it the drive advances through $\pi \gamma$ rather than $\pi$ radians. Its excess phase per half-turn is thereforeEven a small fractional frequency mismatch accumulates across many turns. A particle eventually reaches a weak or decelerating gap phase, so the classical fixed-field, fixed-frequency design has a practical kinetic energy limit. The limit depends on the kick size and acceptable accumulated phase error, rather than one universal cutoff energy. Changing drive frequency or shaping the magnetic field can restore synchronism; the argument does not forbid relativistic circular accelerators.

For the illustrative proton at $1.00\mathrm{MeV}$, $\gamma -1\simeq 0.00106579$: the frequency is lower by about $0.1065\%$, and the next half-turn adds about $0.19184^{\circ }$ of drive phase excess. To see cumulative sensitivity, prescribe an ideal ramp $K_j=jϵ$ with $ϵ=10\mathrm{keV}$ and sum the errors of the first $N$ coasts after kicks:This gives ${\Phi }_{100}\simeq 9.688^{\circ }$ after the coast following kick $100$; the arrival at kick $100$ instead has ${\Phi }_{99}\simeq 9.496^{\circ }$. The prescribed ramp diagnoses detuning but does not feed the resulting phase error back into the kicks. A self-consistent thin-gap model uses the cyclotron gap energy gain with $\cos {\varphi }_n$, then advances the phase with the new kinetic energy during the following coast.

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.