 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 $B_{0}Ls(t)$ and the motional electromotive force is $-B_{0}L\dot{s}$. Ohm's law therefore givesIf $I>0$, 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 $B_0>0$.