The $\mathrm{R}-\mathrm{L}$ circuit with $\mathrm{R}=10 ~ \mathrm{k} \Omega$ and $\mathrm{L}=1 \mathrm{mH}$ is excited by a step current $\mathrm{I}_{\mathrm{u}} \mathrm{u}(\mathrm{t})$. At $\mathrm{t}=0^{-}$, there is a current $\mathrm{I}_L=\mathrm{I}_0 / 5$ flowing through the inductor. The minimum time taken for the current through the inductor to reach $99 \%$ of its final value is___________$\mu \mathrm{s}$ (rounded off to two decimal places).
