Two horizontal rails 0.41 m apart are joined to a battery of emf 9.0 V…

Step 1. The start. The rod is at rest, so there is no back emf. $I = \dfrac{\mathcal{E}}{R} = \dfrac{9.0\ \text{V}}{1.5\ \Omega} = 6.0\ \text{A}$. The force is $F = BIl = 0.62\ \text{T} \times 6.0\ \text{A} \times 0.41\ \text{m} = 1.53\ \text{N}$.

Step 2. The greatest speed. The current is zero when $Blv = \mathcal{E}$. We want $v$, so $v = \dfrac{\mathcal{E}}{Bl} = \dfrac{9.0\ \text{V}}{0.62\ \text{T} \times 0.41\ \text{m}} = 35.4\ \text{m/s}$.

Step 3. At $20$ m/s. The back emf is $Blv = 0.62\ \text{T} \times 0.41\ \text{m} \times 20\ \text{m/s} = 5.08\ \text{V}$. So $I = \dfrac{9.0\ \text{V} - 5.08\ \text{V}}{1.5\ \Omega} = 2.61\ \text{A}$.