= 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, with
$$
r=\frac{mv}{qB},\qquad\Omega=\frac{qB}{m}.
$$
Starting 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 when
$$
\boxed{r>b,\qquad B<\frac{mv}{qb}}.
$$
For 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 is
$$
\boxed{\theta_b=\sin^{-1}\!\left(\frac{qBb}{mv}\right),\qquad0\le\theta_b<\frac\pi2}.
$$
The 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.
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