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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

A Lorentz transformation preserves the Minkowski metric. For a Lorentz boost, let $S^{\prime }$ move at constant velocity $+v$ along the $x$-axis of $S$, with parallel spatial axes and coincident origins at $t=t^{\prime }=0$. With $\beta =v/c$ and the Lorentz factor $\gamma$, the coordinates of the same spacetime event obey

The inverse Lorentz transformation replaces $v$ by $-v$. In particular, $x^{\prime }=0$ traces the moving origin $x=vt$, fixing the sign convention. The time formula is $t^{\prime }=\gamma (t-vx/c^2)$; dropping its position term would lose the relativity of simultaneity. ^[ref].

For the moving-clock example in time dilation, the second tick has $(t,x)=(5\mu \mathrm{s},3c\mu \mathrm{s})$. At $v=0.6c$, substitution gives $t^{\prime }=1.25(5-0.6\times 3)\mu \mathrm{s}=4\mu \mathrm{s}$ and $x^{\prime }=1.25(3-0.6\times 5)c\mu \mathrm{s}=0$. The clock stays at its own origin, and its coordinate time there equals its proper time.

For a periodic thin mucus gap $h(x)$ translating as a backward wave relative to a force-free foot, put $I_j={\int }_0^L{h}^{-j}dx$. In the wave frame, the plane and foot have horizontal velocities $V-U$ and $V$. A Couette-Poiseuille flow with periodic pressure has flux $q=(V-U/2)I_2/I_3$. Zero net plane force gives the displayed crawling speed. The leading viscous dissipation per wavelength is $\Phi =2\mu UV{I}_1$.

The general ideal magnetic-energy balance has outward-surface contribution ${\mu }_0^{-1}\oint [(u\cdot B)B-B^{2}u]\cdot dS$ and volume contribution $-\int u\cdot (J\times B)dV$. The latter vanishes for a force-free magnetic field. For an axisymmetric boundary rotating tangentially, $u\cdot B=\Omega f(\psi )$ and the directed flux of a thin tube is $2\pi d\psi$. Pair its exit and entry endpoints to obtain the displayed energy injection, counting each boundary endpoint once. Equal endpoint angular velocities inject no magnetic energy.

A magnetic field is a vector field $\mathbf{B}(\mathbf{r},t)$ whose local value determines the magnetic part of the Lorentz force on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units,where $v$ is the speed, $B=|\mathbf{B}|$, and $\alpha$ is the angle between the velocity and the magnetic field. The cross product makes a nonzero magnetic force perpendicular to both the velocity and the magnetic field. The right-hand rule gives its direction for positive electric charge; negative electric charge reverses it. ^[ref].

The SI unit of magnetic field is the tesla:For a stipulated example in SI units, take electric charge $q=2\mathrm{C}$, velocity $\mathbf{v}=(3,0,0)\mathrm{m}{\mathrm{s}}^{-1}$, and magnetic field $\mathbf{B}=(0,0,4)\mathrm{T}$. The cross product gives magnetic force ${\mathbf{F}}_B=(0,-24,0)\mathrm{N}$. With the same speed directed along $+z$, the magnetic force is zero: motion parallel to a magnetic field produces no magnetic acceleration.

Magnetic field lines visualize the direction of a magnetic field by being tangent to it at each point. They are not generally the path of a body with electric charge: in a uniform magnetic field along $+z$, the magnetic field lines are straight while transverse velocity produces circular motion, and adding parallel velocity produces helical motion in a uniform magnetic field. The worked Lorentz force example below illustrates this distinction.

A magnetic field need not be static. Magnetostatics is the restricted case with no dependence on time; an electromagnetic wave has changing electric field and magnetic field even in vacuum. A changing magnetic field is coupled to the electric field through Faraday's law of induction. ^[ref]. The useful separation is between local bending, described by the Lorentz force, and the amount of magnetic field crossing a surface, described by magnetic flux.

The magnetic part of the Lorentz force is perpendicular to the particle velocity. Its power is therefore zero, so with no other forces the kinetic energy remains constant. This remains true relativistically because $dE/dt=\mathbf{F}\cdot \mathbf{v}$; a magnetic force can change momentum direction without changing total energy.

Magnetic relaxation converts magnetic energy into material motion through the Lorentz force density, and dissipates the resulting kinetic energy through viscosity. In a perfect conductor, magnetic flux freezing preserves field-line connectivity throughout every smooth stage. A regular limiting state has zero velocity and is a magnetostatic equilibrium, but its magnetic field can contain current sheets. Conservation of one global magnetic helicity is weaker than preservation of the complete frozen field-line structure.

In a thin magnetic skin layer next to a rigid no-slip boundary condition, vertical pressure balance gives $p=P-C^2{e}^{-2y/\delta }/(4{\mu }_0)$. With negligible inertia in the layer, the tangential Stokes flow balance is $\rho \nu u_{yy}=p_x$. Matching zero leading shear outside the layer gives $u=U(1-e^{-2y/\delta })$ and the displayed effective tangential slip. The physical wall still has zero velocity; the slip is a boundary condition for the outer flow extrapolated through the unresolved layer. It points toward larger squared boundary-field amplitude.

A polarization specifies the direction or trajectory of the oscillating velocity and magnetic field in a magnetohydrodynamic wave. The linear Alfvén wave has velocity perpendicular to the plane of the background magnetic field and wavevector; the magnetosonic waves have velocity in that plane. A real fixed polarization vector gives a linear oscillation. Two orthogonal components of equal amplitude in quadrature give a circular trajectory. For propagation parallel to a uniform field, combining the two degenerate transverse linear polarizations produces circular polarization; at finite amplitude its constant transverse magnetic-field magnitude removes the longitudinal magnetic pressure gradient.