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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 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.

With no electric field, the Lorentz force is $F=q\dot{r}\times B\hat{z}$. The equation from Newton's second law and its first integral areThe cyclotron frequency is $|\omega |$; the sign of $\omega$ specifies the sense of rotation. Since $\hat{x}\times \hat{z}=-\hat{y}$, the initial data giveThe component equations are $\dot{x}=\omega y$, $\dot{y}+\omega x=u+a\omega$ and $\dot{z}=v$. For $\omega \ne 0$, differentiating the first givesSolving this linear differential equation and then using $y=\dot{x}/\omega$ givesThe perpendicular projection is a circle of radius $|u/\omega |$ centred at $(a+u/\omega ,0)$, the guiding centre. The velocity parallel to the magnetic field stays constant, producing helical motion in a uniform magnetic field.

When $a\omega +u=0$, this simplifies toThus the trajectory is a helix about the $z$ axis, with radius $|a|$ and axial advance $2\pi v/|\omega |$ per complete revolution in increasing time. Its projection moves clockwise when viewed from positive $z$ for $\omega >0$. If $v=0$ it is a circle; if $a=0$ it is a straight line along the axis.

For $\omega =0$, the Lorentz force vanishes and the solution is instead $r(t)=a\hat{x}+ut\hat{y}+vt\hat{z}$, also the limit of the general formula. The special condition then forces $u=0$.

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.

In Cartesian coordinates the Lagrangian isIntroduce a Lagrange multiplier $\lambda$ for $x^2+y^2+z^2=c^{-2}$. The Euler-Lagrange equations becomeorDifferentiating the constraint twice givesTaking the scalar product of the vector equation with $\mathbf{x}$ therefore yieldsSubstitution gives

For $c=0$, the equations have the first integralsWriting $X=x+B$ and $Y=y-A$ gives $\dot{X}=-Y$ and $\dot{Y}=X$. Hence every trajectory isThese are helices of unit angular frequency about a line parallel to the $z$ axis, including circles when $C=0$ and a straight line when $R=0$.

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.