= Solution
Take positive current around $D\to E\to F\to G\to D$, so it runs upward through the bar. With $\mathbf B=B_0\mathbf e_z$, the oriented <magnetic flux> is $B_0Ls(t)$ and the <motional electromotive force> is $-B_0L\dot s$. <Ohm's law> therefore gives
$$
\boxed{I(t)=\frac{\mathcal E_0-B_0L\dot s(t)}R.}
$$
\b[If $I>0$, the bar current is $E\to F$; if $I<0$, it is $F\to E$.] The induced part opposes the battery for a bar moving to the right when $B_0>0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-3-moving-bar.png]
{title=Sliding-bar circuit with positive current upward and magnetic field out of the page}
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