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184_notes:examples:week14_ac_graph [2017/11/28 16:43] – [Changing Current Induces Voltage in Rectangular Loop] tallpaul | 184_notes:examples:week14_ac_graph [2018/08/09 19:19] (current) – curdemma | ||
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===== Analyzing an Alternating Current Graph ===== | ===== Analyzing an Alternating Current Graph ===== | ||
Suppose you are given the following graph of current over time. You can see that the first peak is at the point where $t=0.01\text{ s}$, and $I=0.3\text{ A}$. The graph is shown below. Find the amplitude, period, and frequency of the current, and give an equation that describes the alternating current. | Suppose you are given the following graph of current over time. You can see that the first peak is at the point where $t=0.01\text{ s}$, and $I=0.3\text{ A}$. The graph is shown below. Find the amplitude, period, and frequency of the current, and give an equation that describes the alternating current. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
===Facts=== | ===Facts=== | ||
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* We represent the graph as given in the example statement. | * We represent the graph as given in the example statement. | ||
====Solution==== | ====Solution==== | ||
- | In order to find the induced voltage, we will need the magnetic flux. This requires defining an area-vector and determining the magnetic field. We can use the right hand rule to determine the the magnetic field from the wire is into the page ($-\hat{z}$) near the rectangle. For convenience, | + | Since the graph has not been shifted vertically, we can say that the peak we see is also the amplitude of the graph. So the amplitude |
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- | $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$ | + | |
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- | Usually, we would pull the $B$ out of the integral, but we cannot do that in this case! That is because | + | |
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- | $$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$$ | + | |
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- | In the first equality, we just changed the integral from over the " | + | |
- | {{ 184_notes: | + | The period |
- | At this point, our integral is set up enough that we can crunch through the analysis. We can pull out constants to the front and calculate to the end: | + | [{{ 184_notes:14_ac_graph_slices.png? |
- | \begin{align*} | + | It should be easy to see based on the visual above that the time between peaks is just four times the time value at the first peak. So then the period is $0.04\text{ |
- | \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y &= \frac{\mu_0 I}{2 \pi} \int_{d}^{d+w} \frac{\text{d}x}{x} \int_{0}^{y} \text{d}y \\ | + | |
- | &= \frac{\mu_0 I}{2 \pi} \left[\log x \right]_{d}^{d+w} \left[ y \right]_{0}^{y} \\ | + | |
- | &= \frac{\mu_0 I}{2 \pi} \left(\log(d+w) - \log(d)\right) \left( h-0 \right) \\ | + | |
- | &= \frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right) | + | |
- | \end{align*} | + | |
- | At this point, we are equipped to find the induced voltage. Notice that everything in the magnetic flux expression is constant in time, //except// for to current, $I$. Here is the induced voltage: | + | The frequency is just the reciprocal of the period: |
- | $$V_{ind} | + | $$f = \frac{1}{\tau} = \frac{1}{0.04\text{ |
- | Notice that the induced voltage is negative. This means that the induced current produces | + | We are also now equipped to write a equation for the alternating current. We will use a sine function rather than a cosine function, since the current |
- | {{ 184_notes: | + | $$I=I_0\sin(2\pi \cdot f \cdot t) = (0.3\text{ A}) \cdot \sin(2\pi \cdot 25\text{ Hz} \cdot t)$$ |