The current in a conductor is expressed as $I=3 t^2+4 t^3$, where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t=1 \mathrm{~s}$ to $t=2 \mathrm{~s}$ is __________ C.

<p>To find the amount of electric charge that flows through a section of the conductor, we have to integrate the current over the given time interval. The current $I(t)$ as a function of time $t$ is given by:</p> $I=3t^2+4t^3$ <p>The electric charge $Q$ that flows through the conductor from time $t = 1$ s to $t = 2$ s is calculated by integrating the current $I(t)$ with respect to time over this interval:</p> $Q = \int_{t_1}^{t_2} I(t) \, dt$ <p>Substituting the given limits ($t_1=1$ and $t_2=2$) and the expression for $I(t)$, we get:</p> $Q = \int_{1}^{2} (3t^2+4t^3) \, dt$ <p>Now we'll integrate the function with respect to $t$:</p> $Q = \left[ \frac{3}{3}t^3 + \frac{4}{4}t^4 \right]_{1}^{2}$ <p>Simplifying the integrated function:</p> $Q = \left[ t^3 + t^4 \right]_{1}^{2}$ <p>Substitute the upper and lower limits of the integration:</p> $Q = \left[ (2)^3 + (2)^4 \right] - \left[ (1)^3 + (1)^4 \right]$ $Q = \left[ 8 + 16 \right] - \left[ 1 + 1 \right]$ $Q = 24 - 2$ $Q = 22 \text{ C}$ <p>Therefore, the amount of electric charge that flows through the section of the conductor from $t=1$ s to $t=2$ s is 22 Coulombs.</p>