A rod of mass 45 g slides down two vertical rails 0.60 m apart, in a field of 0.40 T at right…

Step 1. The factor $B^2l^2$. $B^2l^2 = (0.40\ \text{T})^2 \times (0.60\ \text{m})^2 = 0.0576\ \text{T}^2\text{m}^2$.

Step 2. The terminal speed. $v_t = \dfrac{mgR}{B^2l^2} = \dfrac{0.045\ \text{kg} \times 9.8\ \text{m/s}^2 \times 0.30\ \Omega}{0.0576} = 2.30\ \text{m/s}$.

Step 3. The time constant. $\tau = \dfrac{mR}{B^2l^2} = \dfrac{0.045\ \text{kg} \times 0.30\ \Omega}{0.0576} = 0.234\ \text{s}$.

Step 4. Time to $90\ \%$. Set $1 - e^{-t/\tau} = \dfrac{9}{10}$. Then $e^{-t/\tau} = \dfrac{1}{10}$. We want $t$, so $t = \tau\ln 10 = 0.234\ \text{s} \times 2.303 = 0.540\ \text{s}$.