 Solution

Use nonrelativistic dynamics, take $B>0$ along $+y$, and neglect radiation. The Lorentz force $q\mathbf{v}\times \mathbf{B}$ initially points along $+z$. It does no work, so the speed stays $v$ and the trajectory inside the field is circular, withStarting from the entry point, the orbit can be parametrized as $x=r\sin \theta$, $z=r(1-\cos \theta )$, where $\theta =\Omega t$. Its largest possible $x$ before turning back is $r$. Therefore it enters the strict region $x>b$ exactly whenFor that case, the first boundary crossing satisfies $\sin {\theta }_b=b/r$. The velocity components are $v_x=v\cos {\theta }_b$ and $v_z=v\sin {\theta }_b$, so the outgoing angle above the $+x$ direction isThe zero-field limit is straight motion. At $r=b$ the particle only touches the far boundary tangentially and does not enter $x>b$; for $r<b$ it turns around before reaching that boundary and eventually returns through the entrance side. The equality case must therefore not be included in a crossing condition.