= Solution
The <electric field> does work $qEd$ on the initially stationary particle. Thus the entry <speed> into the magnetic region is
$$
v_0=\sqrt{\frac{2qEd}{m}}.
$$
In the source geometry the <magnetic field> is perpendicular to the page, and the particle enters perpendicular to the straight interface. Its return to that interface occurs after a semicircle. The separation between exit and return points is its orbit diameter, so the required separation $d$ gives $r=d/2$. Consequently
$$
\boxed{|\mathbf B|=\frac{2mv_0}{qd}=\sqrt{\frac{8mE}{qd}}}.
$$
For the drawn field into the page and a positive charge entering upwards, the trajectory bends left. Its <speed> is constant while in the <magnetic field>, so the time there is the semicircle length divided by $v_0$:
$$
\boxed{t_B=\frac{\pi r}{v_0}=\frac{\pi d}{2v_0}=\pi\sqrt{\frac{md}{8qE}}=\frac{\pi m}{q|\mathbf B|}}.
$$
The last form is half the <cyclotron period> and independently checks the geometric calculation.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit-magnetic-particle-trajectories.png]
{title=Finite magnetic-strip transmission and a semicircular return to the electric region}
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