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The magnetostatic Ampère's law, in vacuum, isThe path direction and surface normal are related by the right-hand rule. Taking the divergence of the differential law gives $\nabla \cdot \mathbf{J}=0$, since the divergence of a curl vanishes. However, charge conservation requires the continuity equation $\nabla \cdot \mathbf{J}=-\partial \rho /\partial t$, which need not vanish during charging.

Using Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the combination $\mathbf{J}+{ϵ}_0\partial \mathbf{E}/\partial t$ has zero divergence. Replacing $\mathbf{J}$ by this conserved total current density gives the Ampère-Maxwell equation:Its integral form is $\oint \mathbf{B}\cdot d\mathbf{\ell }={\mu }_0[I_{\mathrm{enc}}+{ϵ}_{0}d{\Phi }_E/dt]$. The extra term is the displacement current. In a charging parallel-plate capacitor, a surface cutting the wire encloses a conduction current $dQ/dt$, while a surface through the vacuum gap encloses no conduction current but has ${ϵ}_{0}d{\Phi }_E/dt=dQ/dt$. This modification makes the magnetic circulation independent of which spanning surface is chosen. The vacuum displacement current represents a changing electric field, not literal transport of charge across the gap.

The Lorentz force isFor $\mathbf{E}=0$, the magnetic part is perpendicular to the velocity, so its instantaneous power $\mathbf{F}\cdot \mathbf{v}$ vanishes. The speed and kinetic energy therefore remain constant. Decompose the initial velocity into $v_{\parallel }$ along the uniform magnetic field and $v_{\perp }$ perpendicular to it. The parallel component is unchanged, while the perpendicular component rotates uniformly.

For a nonrelativistic particle with $qB\ne 0$, the resulting helical motion hasHere $r_L$ is the Larmor radius, ${\omega }_c$ the cyclotron frequency in angular units, and $h$ the axial advance per turn. Opposite signs of electric charge give opposite senses of rotation. Pure perpendicular motion gives a circle; pure parallel motion gives a straight line. If $q=0$ or $B=0$, there is no magnetic deflection. At relativistic speeds, replace $m$ in the radius and period by $\gamma m$; $\gamma$ stays constant because the magnetic force does no work.