= Solution
The bar's motion increases positive $z$-directed <magnetic flux> at rate $B\ell v$. By <Lenz's law>, induced current circulates clockwise as viewed from positive $z$, producing a field into the page. Equivalently $\mathbf v\times\mathbf B=-vB\hat{\mathbf y}$ drives positive charge down the moving bar. The <motional electromotive force> has magnitude $B\ell v$, so
$$
\boxed{I=\frac{B\ell v}{R}.}
$$
The current flows downward in the bar, left along the lower rail, upward through the resistor, and right along the upper rail. The ideal circuit approximation neglects self-inductance and the bar's resistance.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-2-sliding-rod.png]
{title=Clockwise induced current and magnetic braking for a rod moving right through an outward magnetic field}
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