Animated Solution for Physics - Current Electricity: A current through a wire depends on time as $i={\alpha }_{0}t+\beta t^2$, where ${\alpha }_0=20\text{ A/s}$ and $\beta =8{\text{ As}}^{-2}$. Find the charge crossed through a section of the wire in 15 s.

When we first learn about electric current, we often picture it as a steady, constant flow of charge, much like water flowing smoothly through a pipe. However, in many real-world circuits, current is dynamic—it changes with time. In this problem, we are introduced to a current that grows as time passes, governed by the equation:

Here, the constants are given as ${\alpha }_0=20\text{ A/s}$ and $\beta =8{\text{ As}}^{-2}$. This means the current isn't just increasing linearly; it has a quadratic component that makes it surge even faster as time goes on.

Current and Charge

To find the total charge that has crossed a section of the wire, we need to recall the fundamental definition of current. Current is the rate of flow of electric charge.

Mathematically, this is expressed as a derivative:

If we want to find the total charge $q$ accumulated over a certain time period, we must do the reverse operation: integration. By rearranging the formula, we get $dq=idt$. Integrating both sides gives us the total charge:

Graphically, if you were to plot current $i$ on the y-axis and time $t$ on the x-axis, the total charge $q$ is simply the area under the curve between the given time limits.

We are asked to find the charge that crosses the wire in the first $15\text{ s}$. This means our time limits for the integration will be from $t=0$ to $t=15$. Let's substitute our specific current function into the integral:

Now, we perform the integration step-by-step. The power rule for integration tells us that $\int t^{n}dt=\frac{t^{n+1}}{n+1}$. Applying this to our terms:

Simplifying the fractions makes our calculation easier:

Next, we apply the limits. We substitute the upper limit ($15$) into the expression, and subtract the expression evaluated at the lower limit ($0$). Since every term has a $t$, the lower limit evaluation is simply zero.

Let's crunch the numbers carefully. We know that $15^2=225$ and $15^3=3375$.

Adding these together yields our final answer:

This problem beautifully illustrates how basic calculus is an indispensable tool in physics, allowing us to transition seamlessly between rates of change (current) and accumulated quantities (charge).