A rod of mass 40 g slides down two vertical smooth rails 0.40 m apart in a field of 0.90 T at right…

Step 1. The extra mass. $CB^2l^2 = 0.35\ \text{F} \times (0.90\ \text{T})^2 \times (0.40\ \text{m})^2 = 0.0454\ \text{kg}$.

Step 2. The acceleration. $mg - CB^2l^2a = ma$, so $a = \dfrac{mg}{m + CB^2l^2} = \dfrac{0.040\ \text{kg} \times 9.8\ \text{m/s}^2}{0.040\ \text{kg} + 0.0454\ \text{kg}} = 4.59\ \text{m/s}^2$.

Step 3. The current. $I = CBla = 0.35\ \text{F} \times 0.90\ \text{T} \times 0.40\ \text{m} \times 4.59\ \text{m/s}^2 = 0.579\ \text{A}$. It is steady.

Step 4. After $0.50$ s. $v = at = 4.59\ \text{m/s}^2 \times 0.50\ \text{s} = 2.30\ \text{m/s}$. The charge is $q = CBlv = 0.35 \times 0.90 \times 0.40 \times 2.30 = 0.289\ \text{C}$. The energy is $\dfrac{q^2}{2C} = \dfrac{(0.289\ \text{C})^2}{2 \times 0.35\ \text{F}} = 0.120\ \text{J}$.