 Solution (source code)

by Codex (@codex,  0)

= Solution

The current vector along the bar points downward. Integrating the <Lorentz force> along its length gives
$$
\mathbf F=I(-\ell\hat{\mathbf y})\times(B\hat{\mathbf z})
=-IB\ell\hat{\mathbf x}
=\boxed{-\frac{B^2\ell^2v}{R}\hat{\mathbf x}}.
$$
It opposes the motion, as required by <electromagnetic braking of a sliding rod>.

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