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The Hall effect is a transverse electric response to a magnetic force on moving charge carriers. Take conventional current density $\mathbf{J}=J_x\hat{\mathbf{x}}$ and magnetic field $\mathbf{B}=B_z\hat{\mathbf{z}}$ in a conducting slab. For one carrier type of number density $n$ and signed electric charge $q$, the drift velocity is ${\mathbf{v}}_d=\mathbf{J}/(nq)$. Carriers initially deflect sideways until a Hall electric field balances the magnetic Lorentz force:The Hall coefficient is thereforeFor electrons it is $-1/(ne)$, whereas a simple hole-dominated conductor gives $+1/(pe)$. The sign reveals the effective carrier sign; the magnitude estimates their density in the one-carrier approximation.

For width $w$ along $y$ and thickness $t$ along $z$, $I=J_{x}wt$. Define the signed Hall voltage by $V_H=E_{y}w$, namely the electric potential at the lower-$y$ edge minus that at the higher-$y$ edge. ThenThis explicit contact convention fixes the voltage sign. A Hall bar with longitudinal and transverse contacts measures both electrical resistivity and the Hall response. Reverse the field and take the part of the transverse voltage odd in $B_z$ to suppress longitudinal-voltage pickup from misaligned contacts; reversing current provides another useful check.

A Hall measurement is useful for carrier identification, semiconductor doping measurements and magnetic-field sensing. Combining a one-carrier $R_H$ with electrical conductivity gives a mobility magnitude $\mu =\sigma |R_H|$. However, mixed electron and hole conduction is not described by a single density: in weak fields,The response weights mobilities differently from ordinary conductivity. Complicated energy bands, anisotropic Fermi surfaces and multiple bands can likewise change the simple interpretation in metals; a positive Hall coefficient does not mean that the material contains positively charged fundamental electrons.

A Stern-Gerlach experiment sends a collimated beam through an inhomogeneous magnetic field and then measures its transverse position after a drift region. The neutron is electrically neutral but has a nonzero magnetic moment proportional to its spin, $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A predominantly $z$-directed field gives the interaction Hamiltonian operator and transverse forceSince a neutron has spin one-half, its spin component along the local field has only two eigenvalues, $S_z=\pm \hslash /2$. Hence there are two opposite force branches, corresponding to ${\mu }_z=\pm {\gamma }_n\hslash /2$, and a beam containing both spin components gives two displaced detector spots. This is neutron Stern-Gerlach splitting, not an electric-charge Lorentz force. A uniform magnetic field can cause Larmor precession but does not provide the required separating magnetic field gradient.

The neutron's gyromagnetic ratio is negative, so its magnetic moment points opposite to its spin; this reverses which spin label goes to which spot without changing the two-way splitting. Equal populations occur for an initially unpolarized beam. A beam prepared entirely in one field-axis spin eigenstate gives only its corresponding branch, explaining the qualification “in general”.

For a fixed circuit, Faraday's law of induction readsSurface orientation and contour direction obey the right-hand rule. The negative sign is Lenz's law: a passive induced current acts against the flux change, consistently with energy conservation. The induced electric field generally has nonzero curl and cannot be represented by one single-valued electrostatic potential.

For a material circuit moving with local velocity $\mathbf{u}$, the measured electromotive force includes the magnetic Lorentz force on its carriers:This distinguishes transformer induction in a stationary coil from motional induction in a changing circuit. A straight sliding rod of length $l$ moving perpendicular to itself and to uniform $B$ has $|\mathcal{E}|=Blv$. With circuit resistance $R$, the electrical power $(Blv{)}^2/R$ equals the external mechanical power needed to oppose magnetic drag. An ideal transformer gives $V_2/V_1=N_2/N_1$ in magnitude from common flux linkage; real leakage, resistance and core losses limit that simple result. pickup coils measure flux change rather than an absolute static magnetic field, so integration and a reference state are needed to recover a field history.

Self-inductance relates current to its own magnetic flux linkage: $\Lambda =LI$, so the induced electromotive force is $-L\dot{I}$ for fixed geometry. A long solenoid has $B={\mu }_{0}NI/l$ and area $\pi a^2$. Its $N$ turns giveThe homopolar generator uses a disc rotating through this axial field; integrate the motional Lorentz force from shaft to rim:For a homopolar disc dynamo, choose the wiring and rotation sense giving positive magnetic feedback. Kirchhoff's voltage law isfor prescribed constant $\omega$. ThusA nonzero seed grows when $\omega >{\omega }_c$, decays when $\omega <{\omega }_c$, and is marginal at equality. This explains the printed $\omega \ge {\omega }_c$ as the threshold including its boundary; equality does not itself give exponential growth. An exactly zero initial current remains zero in the ideal homogeneous equation, and reversing the feedback sense prevents the instability.

In the torque-controlled homopolar dynamo equilibrium, a radial current element exerts a braking torque $d{G}_B=IBrdr$, so $G_B=IB{a}^2/2=C{I}^2$. For steady rotation and nonzero steady current, $G=C{I}^2$ and $\omega ={\omega }_c$. Hence the field intensity magnitude isThe two current signs give opposite magnetic polarities. Finally,The steady magnetic and rotational energies do not change. This is a steady-state balance, not an assertion that every dynamical trajectory must approach that state.

For a steady thin-wire electric current, the Biot-Savart law iswhere the source element follows positive current. For a circular loop, source points are $(R\cos \varphi ,R\sin \varphi ,0)$ and the observer is $(0,0,z)$. Their cross product with $d{\mathbf{\ell }}^{\prime }$ has axial component $R^{2}d\varphi$; transverse components integrate to zero. Thus the on-axis magnetic field of a circular current loop, multiplied by the number of coincident turns, isTake positive current counterclockwise as seen from $+z$. magnetic field lines pass upward through the loop and return around its exterior, closing rather than terminating. They fan outward above its centre and converge below it. The original numerical Biot-Savart law integration below illustrates these directions; the wire singularities are masked.

The axisymmetric magnetic field near a regular axis follows from Gauss's law for magnetism. In cylindrical coordinates, regularity gives $B_{\rho }(0,z)=0$, while symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. Integrate ${\partial }_{\rho }(\rho B_{\rho })=-\rho B_0^{\prime }(z)$ from the axis:A formal $C(z)/\rho$ term is excluded by regularity. For $z>0$ the axial field decreases upward, so the radial component is outward, consistent with the sketch.

On the small receiving coil, $d\mathbf{\ell }=ad\varphi \hat{\mathbf{\varphi }}$. The Lorentz force $I_{2}d\mathbf{\ell }\times \mathbf{B}$ has axial component $-I_{2}a{B}_{\rho }(a,d)d\varphi$. Contributions from the axial field cancel radially around the loop. ThereforeWith both currents giving magnetic dipole moments along $+z$ and $d>0$, this is attraction. Reversing one current reverses the force. The small coil has $m_2=N_2{I}_2\pi a^2$, so the result is exactly the leading magnetic dipole force $F_z=m_2{B}_0^{\prime }(d)$. Equivalently $U=-{\mathbf{m}}_2\cdot {\mathbf{B}}_0$. The approximation requires $a\ll R$ and negligible variation over the receiving coil; corrections begin at relative order $a^2/(R^2+d^2)$.

Use peak-amplitude phasors with time factor $e^{-i\omega t}$ and $\mu ={\mu }_0$. Faraday's law of induction and the Ampere-Maxwell law giveIn the good-conductor limit $\sigma \gg \omega ϵ$, take the root with positive attenuation into the slab:The real physical field is $\mathrm{Re}[{\tilde{E}}_y{e}^{-i\omega t}]$. A thick slab excludes a returning wave from its rear face. The skin depth is the electric-field amplitude decay length; the intensity decay length is $\delta /2$.

Faraday's law of induction fixes ${\tilde{H}}_z=k{\tilde{E}}_y/({\mu }_0\omega )$. The surface impedance of a good conductor and the complete transmitted magnetic field are consequentlyThe sign of the imaginary part belongs to the stated time convention. Average the Poynting vector using complex conjugation and the factor one half appropriate to peak phasors:As an independent energy check, $-{\partial }_x⟨N_x⟩=\sigma |E_0{|}^2{e}^{-2x/\delta }/2$, the mean Joule heating per unit volume.

Ohm's law gives ${\tilde{J}}_y=\sigma {\tilde{E}}_y$ and ${\tilde{B}}_z={\mu }_0{\tilde{H}}_z$. Hence the averaged Lorentz force per unit volume isIntegrating through the illuminated half-space gives the positive inward radiation pressureFinally let $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$ be the vacuum wave impedance. Tangential-field boundary conditions at normal incidence give the electric-amplitude reflection coefficient $r=(Z-Z_0)/(Z+Z_0)$. Since $|Z|\ll Z_0$,Here $N_0=(1-R)I_{\mathrm{inc}}$ by surface energy balance. Combining the leading formulas gives $p\simeq 2I_{\mathrm{inc}}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident vacuum wave, $\omega {ϵ}_0/\sigma \ll 1$.

In microscopic SI form, Maxwell's equations couple charge density $\rho$ and current density $\mathbf{J}$ to electric and magnetic fields:Gauss's electric law expresses electric flux from charge and reduces to the inverse-square Coulomb field for a point charge. Gauss's magnetic law gives zero net magnetic flux through every closed surface; magnetic field lines do not begin or end on magnetic monopoles in this theory.

Faraday's induction law says changing magnetic flux produces a circulating electric field. For a fixed loop, its circulation is minus the rate of change of enclosed magnetic flux, explaining the sign of induction and Lenz's law. For example, changing a primary coil's electric current changes the flux through a stationary secondary coil and induces a voltage. A moving wire also has the magnetic part of the Lorentz force on its charges; the fixed-loop field equation and the full motional-emf calculation must be distinguished.

The Ampère-Maxwell law relates magnetic circulation to conduction current and displacement current. The latter is essential around a charging capacitor: a surface cutting the lead and another passing through the gap must predict the same circulation. Taking the divergence givesusing Gauss's electric law. Charge conservation therefore fixes the consistency of the time-dependent equations. In matter, the equivalent macroscopic equations use $\mathbf{D}$ and $\mathbf{H}$, free sources and constitutive relations. The Lorentz force density $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ describes exchange of momentum with charges.

In source-free vacuum, take the curl of Faraday's law and use $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. Both fields satisfyFor a plane wave, $\omega =ck$, $\mathbf{k}\cdot \mathbf{E}=\mathbf{k}\cdot \mathbf{B}=0$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. The fields are transverse, mutually perpendicular and in phase. Their energy density and energy flux areThis is Poynting's theorem: field energy is transported and can do work on matter. Electromagnetic radiation is a propagating, energy-carrying solution of the coupled electric and magnetic field equations.

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.

Field law and centre value. The Biot-Savart law for a steady thin wire isHere $d{\mathbf{\ell }}^{\prime }$ follows the conventional electric current and the displacement points from source to observation point. For a circular loop of radius $a$, every element at the centre has distance $a$ and contributes in the same axial direction. Choose $+z$ by the right-hand rule. Then

On-axis integration. At axial distance $z$, every wire element has distance $\sqrt{a^2+z^2}$. Azimuthal symmetry cancels transverse components. Parameterizing ${\mathbf{r}}^{\prime }=(a\cos \phi ,a\sin \phi ,0)$ gives axial cross-product element $a^{2}d\phi$, so the on-axis magnetic field of a circular current loop isIt points the same way on the two halves of the axis; its strength is even in $z$.

Directions near the axial point. At a point on the positive axis the field is axial, but nearby lines spread outward because $B_0$ decreases as $z$ increases. On the negative-axis side they converge toward the axis while still progressing in the positive axial direction. There is no azimuthal field. The near-axis sketch shows this local spreading; the complete magnetic field lines return around the outside of the loop and form closed curves.

Radial component from zero divergence. Smooth axial symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$ and $B_{\rho }=O(\rho )$. By Gauss's law for magnetism,Integrate radially from zero, excluding a singular $1/\rho$ term, to obtain the near-axis radial magnetic field, $B_{\rho }=-\rho B_0^{\prime }(z)/2+O({\rho }^3)$. Differentiating the axial field therefore givesFor $z>0$ it is outward; for $z<0$ it is inward. This supplies the direction reasoning used in the sketch.

Force on the second loop. Put the second loop at $z=d$ and radius $\rho =a$. For $d\gg a$, the radial field due to the first loop is $B_{\rho }\simeq 3{\mu }_{0}I{a}^3/(4d^4)$. The element $d\mathbf{\ell }=ad\phi \hat{\mathbf{\phi }}$ experiences Lorentz forceThe radial contributions cancel as vectors around the loop. The axial contributions add, giving the force between distant coaxial current loops,The sign follows $\hat{\mathbf{\phi }}\times \hat{\mathbf{\rho }}=-\hat{\mathbf{z}}$. Both the small radial coordinate relative to $d$ and the dipole approximation are justified by $a/d\ll 1$.

Force direction. The second loop is pulled back toward the first: the force is attractive for the same current sense. Reversing one current reverses the force. The much larger axial field chiefly produces radial element forces; retaining the smaller radial field is essential for the net axial attraction.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.

Induction and its sign. The Maxwell-Faraday equation is $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. For a fixed loop, Stokes theorem givesThis is Faraday's law of induction. The orientation of the surface normal and positive circulation are linked by the right-hand rule. The minus sign is Lenz's law: the induced current's linked flux opposes the imposed flux change, rather than reinforcing it.

The inductive current. Let the loop normal make angle $\theta$ with the applied field, with positive current defined by that normal. The external magnetic flux is ${\Phi }_{\mathrm{ext}}=A{B}_0\sin \omega t\cos \theta$. With negligible electrical resistance, the circuit equation is $L\dot{I}=-{\dot{\Phi }}_{\mathrm{ext}}$, so total linked flux $LI+{\Phi }_{\mathrm{ext}}$ is constant. For zero initial trapped flux,The printed result needs this initial-flux assumption; in general an additional constant current is allowed. The units are amperes because a weber divided by a henry is an ampere.

Current and forces at one instant. Take $B>0$ along $+x$ and $0<\theta <\pi /2$, so $I<0$ relative to the chosen normal. The loop magnetic dipole moment $\mathbf{\mu }=IA\mathbf{n}$ therefore has a component opposite the applied field. Each wire element experiences Lorentz force $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Opposite sides have equal and opposite forces, giving zero net force in the uniform field but a nonzero couple.

The horizontal top and bottom wire segments also experience equal and opposite vertical forces. They cancel without adding torque about the vertical rotation axis; the displayed force pair supplies that torque.

Average torque. Since $\mathbf{\tau }=\mathbf{\mu }\times \mathbf{B}$, its generalized component for increasing $\theta$ is ${\tau }_{\theta }=-IAB\sin \theta$. Substitution givesand, using $⟨{\sin }^2\omega t⟩=1/2$,The orientation torque on a flux-conserving loop drives its normal away from the field for the stated initial angle. Its magnitude has units joules, or newton metres.

Free motion with large inertia. Write $J$ for the mechanical moment of inertia. If its orientation changes little during each field cycle, the slow equation is $J\ddot{\theta }=⟨{\tau }_{\theta }⟩$, with effective potentialThe stable orientation is $\theta =\pi /2$, where the loop plane contains the applied field. Released from rest between $0$ and $\pi /2$, the loop accelerates toward that orientation, overshoots and librates around it. In the leading averaged, undamped model its turning angles are ${\theta }_i$ and $\pi -{\theta }_i$. It does not settle without a dissipative mechanism. For small librations the squared angular frequency is $A^2{B}_0^2/(2LJ)$.

Negligible inductance and finite resistance. Now the current is proportional to the induced electromotive force, not to the flux. Including slow rotation,The rotational damping of a resistive loop in an oscillating field follows fromAt fixed orientation, the first term alternates at $2\omega$ and has zero mean. The loop can therefore rock slightly, but there is no leading mean aligning torque like that in the inductive case. Slow rotation giveswhich opposes motion and produces Joule heating. Initial-phase transients can initiate motion, and the instantaneous torque is not zero. This description retains the large-inertia, slow-motion approximation; it does not assert an exact stationary solution at every starting angle.

For steady currents in vacuum, Ampère's law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enclosed}}$. A rectangular loop with one long side inside an ideal long solenoid and the other outside gives $B\ell ={\mu }_0(n\ell )I$, because the exterior field is negligible and the perpendicular sides contribute zero. Thus $B={\mu }_{0}nI$, approximately uniform away from the ends.

Write the electron electric charge as $q_e=-e$, with $e>0$. The Lorentz force is $\mathbf{F}=-e\mathbf{v}\times \mathbf{B}$; it does no work. For perpendicular motion, $m{v}^2/r_L=evB$ givesHere ${\omega }_c$ is the positive angular speed; the rotation sense is determined by the negative electron electric charge.

The parallel velocity is unaffected, while the perpendicular component rotates. Hence the path is a helix with $v_{\parallel }=v\cos \theta$ and $r_L=mv\sin \theta /(eB)$. For on-axis entry into a solenoidal tube, the transverse circle passes through that axis: its guiding centre is displaced by $r_L$, and its maximum distance from the tube axis is $2r_L$, not $r_L$.

Using the nonrelativistic approximation appropriate to $1\mathrm{keV}$,The cyclotron period is $2\pi /{\omega }_c=0.893\mathrm{ns}$, so the flight lasts $4.783$ revolutions. The electron therefore crosses the axis again four times before leaving. At $3^{\circ }$, $2r_L=0.279\mathrm{mm}<a$. Wall clearance requires $2mv\sin \theta /(eB)\le a$, givingThe transit is longer than a full orbit, so the maximum transverse excursion really is sampled before exit.

For the solenoidal time-of-flight energy window, the prescribed time fixes $v_{\parallel }=l/t_0=v_0=\sqrt{2E_0/m}$, while $0\le v_{\perp }\le eBa/(2m)$. ThereforeThese are total kinetic energies. The upper value is the ideal grazing limit; if touching the wall counts as rejection, it is a supremum approached from below.

The Lorentz force determines the trajectory:Thus a magnetic field changes momentum direction but does no work; an electric field can change the particle's kinetic energy. In a uniform electric field with negligible magnetic field and nonrelativistic speed, $\mathbf{a}=q\mathbf{E}/m$ is constant. A transverse initial velocity then gives a parabola.

In a uniform magnetic field, the parallel velocity is unchanged and the perpendicular motion is circular. Nonrelativistically,the combined trajectory is a helix, with circulation reversed when the charge sign reverses. Relativistically $r_L=p_{\perp }/(|q|B)$ and the gyrofrequency is $|q|B/(\gamma m)$.

For perpendicular uniform fields in the magnetic-dominated case, the drift velocity is ${\mathbf{v}}_d=\mathbf{E}\times \mathbf{B}/B^2$, which obeys $\mathbf{E}+{\mathbf{v}}_d\times \mathbf{B}=0$. A particle at this velocity is undeflected: the crossed-field velocity selector selects speed $E/B$ in the appropriate direction, independent of mass and charge. Radiation reaction and collisions are neglected in these ideal trajectories.

A field assigns a physical quantity to each point. Gravitational field is force per unit test mass; electric field is force per unit test charge. In a conservative, time-independent field,Write $U=m\Phi$ for gravitational potential $\Phi$, or $U=q\varphi$ for electric potential $\varphi$, giving $\mathbf{g}=-\nabla \Phi$ and $\mathbf{E}=-\nabla \varphi$. A point mass has $U=-GMm/r$ with zero at infinity, while two point charges have $U=Qq/(4\pi {ϵ}_{0}r)$. Only potential-energy differences matter; a constant can be added freely. The mechanical energy $K+U$ is conserved if no nonconservative forces do work. A magnetic Lorentz force $q\mathbf{v}\times \mathbf{B}$ does no work on a point charge and generally cannot be represented by a position-only scalar potential energy.

In SI units, with microscopic charge density ${\rho }_e$ and current density $\mathbf{J}$, Maxwell's equations can be writtenThe first relates electric flux to charge. The second states that the net magnetic flux through a closed surface is zero. Faraday's law relates electric circulation to changing magnetic flux. The Ampère-Maxwell equation includes both conduction current and displacement current; the latter is essential to time-dependent fields and local charge conservation.

For fixed contours and spanning surfaces, the integral curl equations are $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ and $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}+{\mu }_0{\epsilon }_{0}d{\Phi }_E/dt$. For a moving material contour the first must include the motional force term; its total emf is $\oint (\mathbf{E}+{\mathbf{v}}_{\mathrm{loop}}\times \mathbf{B})\cdot d\mathbf{l}$. This prevents confusing a changing circuit area with a partial time derivative of a static magnetic field.

Taking the divergence of the Ampère-Maxwell equation and using electric Gauss's law gives ${\partial }_t{\rho }_e+\nabla \cdot \mathbf{J}=0$. In a source-free vacuum, combining the two curl equations gives ${\nabla }^2\mathbf{E}-{\mu }_0{\epsilon }_0{\partial }_t^2\mathbf{E}=0$, and similarly for $\mathbf{B}$, predicting electromagnetic waves of speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$. The Lorentz force law then specifies how these fields act on a charge. Macroscopic matter descriptions may instead use $\mathbf{D}$ and $\mathbf{H}$ with free sources and constitutive relations; those should not be silently mixed with the microscopic-source equations above.

The Lorentz force determines the motion of a charged particle:At nonrelativistic speeds, $\mathbf{p}=m\mathbf{v}$. The electric field does work at rate $q\mathbf{E}\cdot \mathbf{v}$, while the magnetic part of the Lorentz force is perpendicular to velocity and does no work. A uniform electric field gives constant acceleration $q\mathbf{E}/m$; with a transverse initial velocity, the trajectory is a parabola.

In a uniform magnetic field, the parallel velocity is constant, while the perpendicular component undergoes circular motion. The angular cyclotron frequency, orbit radius and helical pitch areThe sign of $q$ determines the sense of rotation. For perpendicular uniform electric fields and magnetic fields, the guiding centre has the electric drift velocityIndeed, $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$, so the remaining motion is cyclotron rotation about that drifting centre, with possible motion along the magnetic field. The drift is independent of electric charge sign and mass. At relativistic energies, retain $\mathbf{p}=\Gamma m\mathbf{v}$ and the corresponding relativistic orbit radius rather than assuming constant nonrelativistic acceleration.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

For injected velocity $\mathbf{v}=v\hat{\mathbf{x}}$, the Lorentz force isFor nonzero electric charge and nonzero operating current, the velocity selector condition isThe requested nonrelativistic kinetic energy is consequentlyBoth electric current and electric charge cancel. Reversing the sign of $q$ reverses both component forces and preserves their cancellation. The dimensions are consistent because $R/{\mu }_0$ has units of speed, while $w/d$ is dimensionless. The assumed nonrelativistic particle regime requires $Rw/({\mu }_{0}d)\ll c$; the problem supplies no values with which to evaluate a numerical energy.

For the ultrarelativistic magnetic bending radius, the Lorentz force supplies the transverse momentum change, so $r=p/(eB)$. At this energy the proton is ultrarelativistic, giving $pc\simeq E$ andThe electronvolt-to-joule conversion cancels the elementary-charge factor numerically. This uses external CODATA light speed and charge definitions.^[ref] If the stated beam energy means total energy, $pc=\sqrt{E^2-m_p^2{c}^4}$; if it means kinetic energy, $pc=\sqrt{T(T+2m_p{c}^2)}$. Either interpretation gives the quoted radius at the precision of the supplied numbers.

Take the two-dimensional electron gas in the $xy$ plane, the Hall bar along $x$, and $\mathbf{B}=B\hat{\mathbf{z}}$ with $B>0$ perpendicular to it. An electron has charge $-e$. The confining electron potential energy $U(y)$ is approximately flat inside a wide strip and rises at either boundary; smooth quenched disorder adds hills and valleys. For a slowly varying edge, each spin-resolved Landau level approximately followsHere $s=\pm 1$, and $Y$ is the guiding centre. For $\mathbf{A}=(-By,0,0)$, $Y={\ell }_B^2{k}_x$ and the edge group velocity isThus electrons move along $+x$ at the upper edge and $-x$ at the lower edge. The Lorentz force gives counterclockwise bulk cyclotron motion in this convention. At a hard boundary its interrupted circles become directed skipping orbits; a smooth boundary gives the corresponding guiding-centre drift. Smooth bulk quenched disorder generally gives closed drift contours around localized hills or valleys, with extended trajectories appearing near a broadened level's critical energy. Realistic electrostatic screening can produce alternating compressible quantum Hall edge strips and incompressible quantum Hall edge strips, so the unscreened bending curves are an effective sketch rather than an exact self-consistent edge profile.

In the original diagram, contacts are ordered around the electron propagation direction: upper-left $1\to 2\to 3$, then lower-right $4\to 5\to 6$. Let $I_1=I$, $I_4=-I$ and the other leads be ideal equilibrating voltage probes. Define $R_H=(V_2-V_6)/I$ and $R_{xx}=(V_2-V_3)/I$. These choices fix the voltage sign; reversing the field reverses chirality and the Hall polarity. For spin-resolved plateau index $N$,At fixed sheet carrier density and increasing field, $N$ decreases, giving successively higher Hall plateaus; longitudinal peaks occur between them. Unresolved spin branches can suppress odd plateaus. The sketch uses resistance magnitudes, and every integer step is not promised in an arbitrary sample.