The electric current through a wire varies with time as $I=I_0+\beta t$, where $I_0=20 \mathrm{~A}$ and $\beta=3 \mathrm{~A} / \mathrm{s}$. The amount of electric charge crossed through a section of the wire in $20 \mathrm{~s}$ is :

<p>To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I = I_0 + \beta t$, where $I_0 = 20$ A (initial current) and $\beta = 3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.</p> <p>The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds. The formula for $Q$ is given by:</p> <p>$Q = \int_{0}^{T} I(t) \, dt$</p> <p>Where $I(t) = I_0 + \beta t$ and $T = 20$ s. Substituting the given values:</p> <p>$Q = \int_{0}^{20} (20 + 3t) \, dt$</p> <p>This integral can be solved in two parts:</p> <p>1. The integral of the constant term $20$:</p> <p>$\int 20 \, dt = 20t$</p> <p>2. The integral of the linear term $3t$:</p> <p>$\int 3t \, dt = \frac{3}{2}t^2$</p> <p>Thus, combining these and evaluating from $0$ to $20$ seconds:</p> <p>$Q = \left[20t + \frac{3}{2}t^2\right]_{0}^{20}$</p> <p>$Q = \left[20(20) + \frac{3}{2}(20)^2\right] - \left[20(0) + \frac{3}{2}(0)^2\right]$</p> <p>$Q = 400 + 600 = 1000$ C</p> <p>Therefore, the amount of electric charge that crosses through a section of the wire in $20$ seconds is $1000$ C.</p>