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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.

The Lorentz force in the uniform magnetic field giveswhere the last step is the vector triple product identity. HenceOn the other hand, $|\mathbf{r}\times \mathbf{B}{|}^2=B^2{r}^2-(\mathbf{r}\cdot \mathbf{B}{)}^2$, soThe derivatives cancel, proving the magnetic angular-momentum invariantThe magnetic Lorentz force is perpendicular to velocity, so its power is zero:Thus the kinetic energy is constant.

For the alternative kinetic energy expression, work on an interval where $r>0$ so that the radial unit vector $\mathbf{u}=\mathbf{r}/r$ is defined. Differentiating $\mathbf{u}\cdot \mathbf{u}=1$ once and twice givesSince $\mathbf{v}=\dot{r}\mathbf{u}+r\dot{\mathbf{u}}$, it follows that $\mathbf{u}\cdot \mathbf{v}=\dot{r}$ and $v^2={\dot{r}}^2+r^2|\dot{\mathbf{u}}{|}^2$. Therefore the unit-direction kinetic-energy identity isThis identity holds for any twice differentiable trajectory away from the origin, not only for motion in a magnetic field.

The solenoidal constraint on the magnetic field reduces to ${\partial }_x{B}_x+{\partial }_z{B}_z=0$. Locally, or globally in a simply connected cross-section, it permits a Cartesian magnetic flux function satisfyingThe transverse magnetic field lines are contours of $\psi$, because $\mathbf{B}\cdot \nabla \psi =0$. The independent component $B_y$ supplies the twist of the flux tube; it is not restricted by the solenoidal constraint.

For the Lorentz force, take the curl explicitly:Its cross product with $\mathbf{B}$ has transverse components $-({\nabla }^2\psi )\nabla \psi -B_y\nabla B_y$ and $y$ component ${\psi }_x{B}_{y,z}-{\psi }_z{B}_{y,x}$. ConsequentlyThe term $-B_y\nabla B_y/{\mu }_0$ is the transverse gradient of the magnetic pressure associated with the axial field, while the other terms include magnetic tension.

In magnetostatics, $-\nabla p-\rho \nabla \Phi +{\mathbf{f}}_L=0$. The plane-parallel Newtonian gravitational potential and all thermodynamic fields are independent of $y$, so the $y$ component of the Lorentz force must vanish:Away from a null of $\nabla \psi$, this says that $B_y$ is constant along each connected contour of the Cartesian magnetic flux function. Thus, on a regular flux region,This statement is local; disconnected contours with the same numerical flux label need not share one global function without the usual flux-tube connectivity assumption.

Since $\nabla B_y=b^{\prime }(\psi )\nabla \psi$, the transverse magnetostatic equilibrium equation becomesThis is a Cartesian magnetostatic flux-function equilibrium. Projecting it along a transverse magnetic field line gives hydrostatic balance along that line; projecting across the line balances the gas pressure and weight against magnetic pressure and magnetic tension.

The MHD induction equation can be written in material form asIts $y$ component immediately givesFor the Cartesian magnetic flux function, the $x,z$ components of the MHD induction equation, or the $y$ component of the magnetic vector potential equation, giveThe additive function of time in $\psi$ has no effect on $\mathbf{B}$. Choose this gauge so that $C=0$. Then the flux label is materially conserved:This is magnetic flux freezing in the two-dimensional geometry.

To prove the absence of axial motion, construct an invariant solution with $u_y=0$. Along each Lagrangian trajectory, $\psi$ is fixed, and the assumed divergence givesThus $B_y$ remains a function of $\psi$ and time alone. Its gradient stays parallel to $\nabla \psi$, so the $y$ component of the Lorentz force remains zero. The axial momentum equation isbecause $(B_x{\partial }_x+B_z{\partial }_z)\psi =0$. With $u_y=0$ initially, it remains zero. Hence neither axial force nor axial motion is generated. This flux-surface preservation during magnetic-tube expansion argument assumes the smooth ideal evolution for which the initial-value problem is unique; the particular in-plane rising motion need not be calculated.

For the magnetic part of the Lorentz force, $\mathbf{F}=q\mathbf{v}\times \mathbf{B}$. Since velocity has physical dimension $L{T}^{-1}$, the magnetic field has dimension

The Lorentz force on the current-carrying bar isThere is no friction force along the rails. The battery part accelerates the bar rightward for positive ${\mathcal{E}}_0{B}_0$, whereas the induced-current part is the opposing force $-B_0^2{L}^2\dot{s}/R$.