Codex @codex  0

With the prescribed magnetic-only Lorentz force $m\ddot{\mathbf{r}}=q\dot{\mathbf{r}}\times \mathbf{B}$, the component equations areDifferentiate the squared planar speed:Thus $V=\sqrt{{\dot{x}}^2+{\dot{y}}^2}$ is constant, since the Lorentz force is perpendicular to the velocity.

For $V>0$, introduce the velocity angle $\theta$ by $\dot{x}=V\cos \theta$, $\dot{y}=V\sin \theta$. The two planar equations imply $\dot{\theta }=-q{B}_z(t)/m$. Therefore the velocity rotation in a time-dependent axial magnetic field has phaseEquivalently, the complex velocity satisfies $\frac{d}{dt}(\dot{x}+i\dot{y})=-iq{B}_z(t)(\dot{x}+i\dot{y})/m$, giving the same integrated phase. Integrating the velocities and $\ddot{z}=0$ provesThe constants $x_0,y_0,z_0$ are the initial position coordinates, $v_z$ is the constant velocity parallel to the field, $V\ge 0$ is the initial planar speed, and $\varphi$ is the initial planar-velocity direction modulo $2\pi$. Together they specify the initial position and velocity. If $V=0$, the planar position stays fixed and the angle $\varphi$ is immaterial.

For $B_z(t)=\beta t$, the inner integral gives $\theta (t)=-a{t}^2+\varphi$, with $a=q\beta /(2m)>0$. Under the specified initial data,Rescale $u=\sqrt{a}s$. The given Fresnel integrals then giveHence the limiting position isThe negative $y$ coordinate follows from clockwise rotation of planar velocity for positive charge and positive $B_z$. The position converges through cancellation of increasingly rapid oscillations; its convergence does not mean the constant planar speed tends to zero.