Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S7 / 1

Only $A_z$ depends on position. Taking the curl,The field reverses across $x=0$. Its divergence is zero, as required for a magnetic field; the vanishing field on the midplane is important for the crossing and separatrix trajectories below.

With zero scalar potential, the nonrelativistic charged-particle Lagrangian in SI units is kinetic energy plus $q\mathbf{A}\cdot \dot{\mathbf{r}}$. ThusThe velocity-dependent magnetic term changes canonical momentum but does not contribute to the mechanical energy.

Define $a=q{B}_0/(2L)>0$. Differentiating the Lagrangian with respect to each generalized velocity givesThe coordinates $y$ and $z$ are cyclic, so $p_y$ and $p_z$ are conserved. The momentum $p_x$ is generally not conserved: its equation is ${\dot{p}}_x=-2ax\dot{z}$. In particular, conserved canonical $p_z$ is not the same as conserved mechanical $m\dot{z}$, which changes as the particle moves across the magnetic-field gradient.

The Legendre transform cancels the term linear in velocity and givesIt equals the mechanical kinetic energy. The Hamiltonian has no explicit time dependence, so $H$ is conserved. This is also a direct consequence of the magnetic force doing no work: $q\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})=0$.

Since $p_y=0$ initially and remains constant, the motion stays in a plane of constant $y$. Setting $P=\sqrt{2m{H}_0}$ and $p=p_{z0}$ givesThe allowed interval obeys $|p+a{x}^2|\le P$. The sign of $\dot{x}$ is set by the initial direction and changes at a turning point. The specified $x_0$ fixes the starting position on the energy orbit, but the data do not specify which sign of the initial transverse velocity is chosen.

Dividing the two velocities givesHere $\dot{z}=a{x}^2/m\ge 0$: the orbit always advances toward increasing $z$, except for instantaneous zeros at each midplane crossing $x=0$. The transverse motion oscillates between $-X$ and $X$. At these outer turning points $\dot{x}=0$ and the trajectory is tangent to the $z$ direction; at the midplane $\dot{x}\ne 0$ while $\dot{z}=0$, so the tangent is along $x$. The sign of the initial $\dot{x}$ only chooses the phase or reflected trace.

With $p=-P$, the allowed range is $|x|\le X_2$, where $X_2=\sqrt{2P/a}$. But an orbit starting with $x\ne 0$ never crosses the midplane in finite time, becauseOne positive-side separatrix, with $u=\omega (t-t_c)$ and $\omega =\sqrt{2Pa}/m$, isThe negative-side orbit is its reflection in $x=0$. For the positive branch, $x$ rises from zero asymptotically, reaches $X_2$ at $u=0$, then returns asymptotically to zero. Thus the infimum is zero but it is not attained in finite time on this nontrivial branch. Since $\dot{z}=(P/m)(2{\mathrm{sech}}^{2}u-1)$, the two axial-velocity zeros occur at $x=\sqrt{P/a}$. Motion is toward decreasing $z$ near either asymptotic midplane tail and toward increasing $z$ near the outer excursion. These directions are marked in part i.

The exactly midplane initial state $x=0$, $\dot{x}=0$ is a separate straight trajectory with $\dot{z}=-P/m$; it must not be described as a finite-time crossing of the separatrix.