= Solution
Let $U>0$ be the drop in <electric potential> along the positive <electric charge>'s path. By <conservation of energy>, $qU=mv^2/2$, so the nonrelativistic exit <speed> is $v=\sqrt{2qU/m}$. Apply a uniform <magnetic field> perpendicular to the incident <velocity> and to the desired orbital plane. The magnetic <Lorentz force> is perpendicular to the <velocity>, does no <work>, and supplies the <centripetal force>:
$$
qvB=\frac{mv^2}{R},\qquad
\boxed{R=\frac{mv}{qB}=\frac1B\sqrt{\frac{2U}{q/m}}.}
$$
For example, $\mathbf v=v\widehat{\mathbf x}$ and $\mathbf B=B\widehat{\mathbf z}$ give initial deflection along $-y$ for positive $q$. Reversing the <magnetic field> reverses the bending. A uniform <electric field> alone would give constant <acceleration> and a parabolic path, not uniform <circular motion>. The displayed result assumes $qU\ll mc^2$; the exact relativistic <momentum> relation would give $R=\sqrt{q^2U^2/c^2+2mqU}/(qB)$.
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