= Solution
For injected <velocity> $\mathbf v=v\widehat{\mathbf x}$, the <Lorentz force> is
$$
\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)=q\left(-\frac{IR}{d}+v\frac{\mu_0I}{w}\right)\widehat{\mathbf z}.
$$
For nonzero <electric charge> and nonzero operating current, the <velocity selector> condition is
$$
\boxed{v=\frac EB=\frac{Rw}{\mu_0d}.}
$$
The requested nonrelativistic <kinetic energy> is consequently
$$
\boxed{T=\frac m2\left(\frac{Rw}{\mu_0d}\right)^2.}
$$
Both <electric current> and <electric charge> cancel. Reversing the sign of $q$ reverses both component <forces> and preserves their cancellation. The dimensions are consistent because $R/\mu_0$ has units of <speed>, while $w/d$ is dimensionless. The assumed <nonrelativistic particle> regime requires $Rw/(\mu_0d)\ll c$; the problem supplies no values with which to evaluate a numerical <energy>.
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