= Solution
The <electron> follows a semicircle under the transverse <Lorentz force>. Parallel, oppositely directed incoming and outgoing paths are separated by a diameter, so the orbit radius is $r=1.00\,\mathrm{cm}$. Acceleration from rest through the <electric potential difference> $V$ gives <kinetic energy> $m_ev^2/2=eV$, where $e>0$ is the elementary <electric charge>. Equating the <Lorentz force> magnitude to the required <centripetal force> gives
$$
e v B=\frac{m_ev^2}{r},\qquad
\boxed{B=\frac1r\sqrt{\frac{2m_eV}{e}}\simeq3.372\times10^{-2}\,\mathrm T.}
$$
The <magnetic field> must be perpendicular to the plane of the turn, with its sign chosen for the <electron>'s negative <electric charge>. In the illustration the <electron> enters rightwards at the upper endpoint, so a <magnetic field> into the page bends it downwards.
The nonrelativistic estimate has $v\simeq0.198c$. If the small <special relativity> correction is retained, use $pc=\sqrt{K(K+2m_ec^2)}$ with $K=eV$ and $B=p/(er)$. This gives $B\simeq3.389\times10^{-2}\,\mathrm T$, about $0.49\%$ larger. The leading answer is adequate at the precision implicit in the classical treatment.
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