= Solution
For $B_0\ne0$, <Newton's second law> gives
$$
m\ddot s=\frac{B_0L\mathcal E_0}{R}-\frac{B_0^2L^2}{R}\dot s.
$$
Define the <exponential relaxation time> $\tau=mR/(B_0^2L^2)$ and <terminal velocity> $v_\infty=\mathcal E_0/(B_0L)$. Solving the first-order equation for $v=\dot s$ and then integrating gives
$$
\boxed{\dot s(t)=v_\infty+(v_0-v_\infty)e^{-t/\tau},\qquad
s(t)=s_0+v_\infty t+\tau(v_0-v_\infty)(1-e^{-t/\tau}).}
$$
Both prescribed initial values hold. The current approaches zero as the bar approaches terminal velocity. If $B_0=0$, there is no magnetic force and instead $\dot s=v_0$, $s=s_0+v_0t$; the constant current is $\mathcal E_0/R$.
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