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A photograph records a path but supplies no arrow of time. The Lorentz force depends on $q\mathbf{v}\times \mathbf{B}$: reversing both electric charge and direction of travel leaves the spatial bending compatible with the same path. Thus a particular arc can represent a Positron travelling one way or an electron travelling the other. Knowing the magnetic field's polarity is necessary but insufficient. Gravity does not justify assuming that every cosmic ray track travels downward.

The lead plate adds a physical ordering. For a single particle crossing a passive absorber without an accelerating interaction, the outgoing momentum is smaller, so the outgoing gyroradius is smaller in the same magnetic field. The less curved segment comes first. Once this direction is assigned, the cross product in the Lorentz force determines the sign of electric charge. For the convention below, $\mathbf{B}=B\hat{\mathbf{z}}$ points out of the page; at upward velocity $\mathbf{v}=v\hat{\mathbf{y}}$, $\mathbf{v}\times \mathbf{B}=vB\hat{\mathbf{x}}$. Bending to the right therefore means positive electric charge.

Magnetic forces do no work: the change of kinetic energy occurs through interaction with the material, not through the bending magnetic field. The ideal construction suppresses multiple scattering, kinetic energy fluctuations, and loss of kinetic energy in the gas. In an actual analysis one compares arcs away from the plate and allows a change of tangent from multiple scattering; one cannot interpret every kink as a change of electric charge. Short visible arcs, uncertain projection and small bending also limit a determination of magnetic rigidity.

Electromagnetism describes how electric charge and electric current interact through the electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$. The Maxwell equations describe the fields and their sources; the Lorentz force describes the response of a particle of electric charge $q$ and velocity $\mathbf{v}$:The electric field can change the particle's energy. The magnetic part of the Lorentz force is perpendicular to velocity, so magnetic forces do no work; they can bend a trajectory. These are local interactions: a particle responds to the electromagnetic field at its position. ^[ref].

The remarkable unification is that the same Maxwell equations connect Coulomb's law, Faraday's law of induction, and freely propagating electromagnetic waves. A changing electric field contributes to magnetic circulation through displacement current; together with Faraday's law, this permits waves in vacuum without conducting matter. The Poynting theorem then explains how the fields store and transport energy. This connects the force on a charge to the energy carried by light.

In his 1865 paper, James Clerk Maxwell derived transverse disturbances and compared the propagation speed inferred from electrical measurements with independently measured light speeds. Part VI, especially §§95–97, argues that their agreement supports identifying light as an electromagnetic wave. This was a comparison with finite experimental accuracy, rather than a numerical identity between the historical measurements. ^[ref]; ^[ref].

Lenz's law expresses the minus sign in Faraday's law of induction: the induced electromotive force drives a passive resistive loop's electric current in the direction whose own magnetic field opposes the imposed change in magnetic flux. It opposes the change of flux, not invariably the external magnetic field. Choose an oriented surface and its positive boundary direction first, find the sign of $d{\Phi }_{\mathrm{ext}}/dt$, and then use the right-hand rule to translate the signed electric current into a physical direction. ^[ref].

In the fixed-loop example in Faraday's law of induction, out-of-page magnetic flux increases at $0.0060Wb/s$. With counterclockwise positive, $I=-0.0030\mathrm{A}$ is clockwise, producing an in-page magnetic field through the loop. Reverse the ramp to $B(t)=0.40\mathrm{T}-(0.30T/s)t$ on the same one-second interval: now $\mathcal{E}=+0.0060\mathrm{V}$ and $I=+0.0030\mathrm{A}$. The counterclockwise electric current produces an out-of-page magnetic field, supporting the decreasing external magnetic flux. A steady out-of-page magnetic field with fixed loop geometry gives no induced electromotive force in either direction.

For the increasing-area example in motional electromotive force, clockwise electric current runs down the right-hand rod. Its magnetic force on a current-carrying wire points left, opposing the imposed motion to the right. Writing $I_{\mathrm{mag}}=B\ell v/R$ for the current magnitude, the power required to maintain constant speed isThus the sign agrees with conservation of energy: the external drive supplies the Joule heating. The magnetic forces do no work on individual charges; mechanical work by the drive supplies the electrical energy. This also explains the usefulness of electromagnetic braking of a sliding rod and the terminal speed of a falling conducting loop. ^[ref].

The sign of induced electromotive force alone does not determine an arbitrary circuit's instantaneous electric current. Retaining constant self-inductance $L$ gives $LdI/dt+RI=-d{\Phi }_{\mathrm{ext}}/dt$, so a pre-existing electric current can persist while the induced drive opposes its change. The distinction matters in the lossless limit of magnetic loop braking. The resistive toy examples use $L=0$ as an approximation, not a general consequence of Lenz's law.

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

A particle accelerator increases the kinetic energy of charged particles through the work of an electric field. A single electric potential difference $V$ gives an energy gain of magnitude $|q|V$ to a particle of electric charge $q$ crossing in the accelerating direction. Reusing a smaller electric potential difference many times can give a much larger final kinetic energy without requiring the whole apparatus to withstand the equivalent single voltage. A magnetic field can return the particles to the accelerating region: magnetic forces do no work, but they make repeated energy gains possible. The cyclotron turns this separation between bending and acceleration into a particularly elegant machine.

Taking the scalar product of the Lorentz force equation with the velocity givesThe triple product vanishes because the cross product is perpendicular to its first factor. Thus the kinetic energy is constant, even if the magnetic field varies with position: magnetic forces do no work.

In cylindrical coordinates, the radial, angular and vertical equations areFor a nonstationary circular orbit with constant $r$ and $z$, the first equation gives $-m{\dot{\theta }}^2=qB(r)\dot{\theta }$. Hence its signed angular velocity is $\boxed{\dot{\theta }=-qB(r)/m}$. The sign gives the sense of circulation. The zero root describes a stationary transverse position, not a moving circular orbit. A constant vertical velocity can be superposed to produce a helix.

Define the mechanical angular momentum about the axis by $L=m{r}^2\dot{\theta }$. Multiplying the angular equation by $r$ givesFor the inverse-radius magnetic field this becomes $\dot{L}=-q{B}_0\dot{r}$. Integrating from an initial radius $r_0$ provesLet $c=q{B}_0/m$ and $\ell =(L_0+q{B}_0{r}_0)/m$, so that $L/m=\ell -cr$. Since $\dot{z}$ is constant and the total kinetic energy is constant, the transverse speed squaredis also constant. Substituting the angular momentum relation gives the required radial curve:Only the region $W(r)\ge 0$ is accessible. For $\ell \ne 0$, $W(r)\to -\infty$ as $r\to 0$, and its horizontal asymptote is $v_{\perp }^2-c^2$. Its derivative is $W^{\prime }(r)=2\ell (\ell -cr)/r^3$. Thus, when $\ell c>0$, it has a single maximum $W=v_{\perp }^2$ at $r=\ell /c$. The positive roots of $W$ are the turning points obtained from $r=\ell /(c\pm v_{\perp })$.

For nonzero $c$ and $\ell c>0$, write $r_c=|\ell /c|$. When $0<v_{\perp }<|c|$, two turning points occur atand the radius is confined between them. At $v_{\perp }=|c|$, there is only the finite lower turning point $r_c/2$; $W$ tends to zero from above and an outgoing particle reaches arbitrarily large radius. For $v_{\perp }>|c|$, only the lower turning point remains and the outgoing radial velocity approaches $\sqrt{v_{\perp }^2-c^2}$. The sketch shows these three cases and omits inaccessible negative values of ${\dot{r}}^2$ from the physical interpretation.

For completeness, if $\ell c<0$, $W$ rises monotonically from $-\infty$ to its asymptote; admissible motion then requires $v_{\perp }>|c|$ and is radially unbound after any inward branch turns. If $\ell =0$, $W=v_{\perp }^2-c^2$ is constant; equality gives a circular orbit rather than escape. Above threshold the outgoing branch escapes, whereas an incoming branch reaches the singular axis in finite time and cannot be continued through it without specifying a nonsingular core field. These cases explain why a borderline escape assertion needs an initial-direction qualification.

The transverse escape velocity threshold for radial escape in an inverse-radius axial magnetic field isFor a radial launch at $r_0$, $\ell =c{r}_0$, and equality permits marginal escape; for $\ell \ne 0$, larger transverse speed guarantees radial escape in forward time after any initial inward leg turns. The singular incoming $\ell =0$ branch is the exception described above. More general launches at equality can instead give the circular exception $\ell =0$. A vertical velocity adds to the total speed but cannot pay the transverse escape-energy requirement; if escape merely means unbounded $z$, any nonzero $\dot{z}$ already achieves that. If $q{B}_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.

With no electric field, the Lorentz force isFor nonrelativistic motion, the work-energy theorem givesConsequently $\boxed{K(t)=K(0)}$. The reason is geometric: magnetic forces do no work because they are perpendicular to the instantaneous velocity. They can change its direction but not its magnitude. The same energy conclusion holds relativistically, using $dE/dt=\mathbf{F}\cdot \mathbf{v}$ and subtracting the constant rest energy to obtain kinetic energy.

In the Lorentz law, $q$ is signed electric charge, $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic fields at the bead, and the velocity in $q\mathbf{V}\times \mathbf{B}$ is its full laboratory velocity. Here $\mathbf{V}=v\hat{\mathbf{r}}+w\hat{\mathbf{\theta }}$. Since $\hat{r}\times \hat{z}=-\hat{\theta }$ and $\hat{\theta }\times \hat{z}=\hat{r}$,The tangential component integrated around an instantaneous circle of nearly fixed radius givesThis is the stated tangential circulation work expression in the $w\gg v$ limit. It is tangential circulation work. The tangential circulation work versus particle work distinction matters because the bead actually follows an expanding spiral. The radial displacement may be small, but its force is proportional to $w$, so its power is of the same order as the tangential power. In factThe cancellation expresses the fact that magnetic forces do no work. Ignoring radial work while retaining the first-order tangential effect would incorrectly conclude that a magnetic field supplies net particle energy.

The electric field giving the same tangential force would be $\boxed{{\mathbf{E}}_{\mathrm{eq}}=-vB\hat{\theta }}$. This is a force-equivalent field for the moving guide, not evidence that a real circulating electric field has appeared in the laboratory merely because the loop expands in a static magnetic field. An electromotive force is the tangential driving-force circulation per unit charge, with units of energy per charge. For a material circuit moving with local velocity ${\mathbf{v}}_{\mathrm{loop}}$,Orient the ring counterclockwise with positive normal $+\hat{z}$. Its flux is $\Phi =B\pi R^2$, so $\dot{\Phi }=2\pi BRv$. The tangential circulation therefore givesIt is the motional term that matters here. For a fixed loop, the circulation of the actual electric field instead follows the partial time derivative of the magnetic field. The distinction between moving-wire magnetic forces and induction in a stationary loop is essential to the flux rule.^[ref]

For the magnetic power balance on an expanding circular guide, the guide must supply an additional inward reaction $-qBw\hat{r}$ opposing the radial magnetic force. With radial guide speed $v$, this contribution to its power on the bead is $P_{\mathrm{guide}}^{(B)}=-qBvw$, equal to the tangential magnetic power. The radial magnetic power is positive by the same amount, so the total magnetic power still vanishes; the mechanical exchange is associated with the moving mechanical constraint.

If “the ring's work” is meant as its full work on a finite-mass bead, the source statement needs a further qualification. For a frictionless bead of mass $m_b$ with constant imposed radial speed, its radial acceleration is $-w^2/R$, so the complete radial guide reaction and power areThe first term is present even when $B=0$ and describes the familiar change of tangential kinetic energy on an expanding guide. The requested $-qBvw$ is exactly the additional magnetic-field-dependent term, not that full reaction power. No bead mass or extra agency fixing $w$ is supplied, so it would be unjustified to assign a numerical total mechanical power or silently discard its centripetal contribution.

For an axial magnetic field $\mathbf{B}=(B_0/r){\mathbf{e}}_z$, let $L=m{r}^2\dot{\theta }$. The Lorentz force gives $\dot{L}=-q{B}_0\dot{r}$, so $\ell =(L+q{B}_{0}r)/m$ is constant. Since magnetic forces do no work and $\dot{z}$ is constant, the transverse speed $v_{\perp }$ is constant and ${\dot{r}}^2=v_{\perp }^2-(\ell /r-c{)}^2$. If $c\ell >0$ and $0<v_{\perp }<|c|$, two finite radial turning points confine the motion. If $v_{\perp }>|c|$, an outward branch escapes to infinite radius. At equality a marginal escaping branch exists when $c\ell >0$, but $\ell =0$ gives a circular orbit instead. The transverse escape velocity threshold is $|q{B}_0|/m$; motion parallel to the magnetic field cannot provide transverse escape energy.