= Lorentz force
{c}
{title2=$\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)$}
{wiki}
= Lorentz force law
{c}
{synonym}
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>, is
$$
\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B),\qquad
m\frac{d\mathbf v}{dt}=q(\mathbf E+\mathbf v\times\mathbf B).
$$
Both 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>. https://openstax.org/books/university-physics-volume-2/pages/5-4-electric-field[OpenStax, electric force and field]{ref}; https://openstax.org/books/university-physics-volume-2/pages/11-2-magnetic-fields-and-lines[OpenStax, magnetic force on moving charge]{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> is
$$
\mathbf r(t)=(\sin t,\cos t-1,0),\qquad
\mathbf v(t)=(\cos t,-\sin t,0),\qquad
\mathbf a(t)=(-\sin t,-\cos t,0).
$$
At $t=\pi/2$ 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.
\Image[/lorentz-force-orbits.png]
{title=<Lorentz force>: <circular motion> and a <helix> in the same uniform <magnetic field>}
{description=Stipulated rescaled $m=q=B=1$ and zero <electric field>. Left: positive <electric charge> follows the clockwise <circle> from initial <velocity> $(1,0,0)$; the <velocity> is tangent and the magnetic <force> points inward. Dots denote a <magnetic field> out of the <plane>. Right: adding parallel <velocity> $v_z=0.5$ gives <helical motion in a uniform magnetic field>, with <position> $z=0.5t$ and the same transverse <radius> one. Straight arrows show <magnetic field lines>, distinct from the curved path.}
Adding initial <velocity> $v_z=0.5$ leaves the transverse <circular motion> unchanged and gives
$$
\mathbf r(t)=(\sin t,\cos t-1,0.5t),\qquad
\mathbf v(t)=(\cos t,-\sin t,0.5).
$$
This 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. https://openstax.org/books/university-physics-volume-2/pages/11-3-motion-of-a-charged-particle-in-a-magnetic-field[OpenStax, circular and helical motion]{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>:
$$
r=\frac{mv_\perp}{|q|B},\qquad
\omega_c=\frac{|q|B}{m},\qquad
T=\frac{2\pi}{\omega_c}=\frac{2\pi m}{|q|B}.
$$
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 $mv_\perp^2/r=|q|v_\perp B$; 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 gives
$$
\frac{dK}{dt}=\mathbf F\cdot\mathbf v
=q\mathbf E\cdot\mathbf v
+q\mathbf v\cdot(\mathbf v\times\mathbf B)
=q\mathbf E\cdot\mathbf v,
\qquad K=\tfrac12m|\mathbf v|^2.
$$
The <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> is
$$
\frac{dp^\mu}{d\tau}=qF^{\mu\nu}u_\nu.
$$
Its spatial and temporal components are
$$
\frac{d\mathbf p}{dt}=q(\mathbf E+\mathbf v\times\mathbf B),
\qquad
\frac{d\mathcal E}{dt}=q\mathbf E\cdot\mathbf v.
$$
Back to article page
