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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 [2017/11/28 16:51] – tallpaul | ||
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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 | + | 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 of the alternating current |
- | $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$ | + | The period of the graph is given by the time between peaks. One can see that the graph can be split into vertical slices as shown below. |
- | Usually, we would pull the $B$ out of the integral, but we cannot do that in this case! That is because $B$ varies for different points in the rectangle' | + | {{ 184_notes:14_ac_graph_slices.png? |
- | $$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$$ | + | 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{ |
- | In the first equality, we just changed | + | The frequency is just the reciprocal of the period: |
- | {{ 184_notes: | + | $$f = \frac{1}{\tau} = \frac{1}{0.04\text{ s}} = 25\text{ Hz}$$ |
- | At this point, our integral is set up enough that we can crunch through | + | 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 begins at $I=0$, |
- | \begin{align*} | + | $$I=I_0\sin(2\pi \cdot f \cdot t) = (0.3\text{ |
- | \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 | + | |
- | &= \frac{\mu_0 I}{2 \pi} \left[\log x \right]_{d}^{d+w} \left[ y \right]_{0}^{y} \\ | + | |
- | &= \frac{\mu_0 I}{2 \pi} \left(\log(d+w) - \log(d)\right) \left( h-0 \right) \\ | + | |
- | &= \frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right) | + | |
- | \end{align*} | + | |
- | + | ||
- | At this point, we are equipped to find the induced voltage. Notice that everything in the magnetic flux expression is constant in time, //except// for to current, $I$. Here is the induced voltage: | + | |
- | + | ||
- | $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$ | + | |
- | + | ||
- | Notice that the induced voltage is negative. This means that the induced current produces a magnetic fields whose corresponding flux is negative flux. Since the area-vector was defined as into the page ($-\hat{z}$), | + | |
- | + | ||
- | {{ 184_notes: | + |