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184_notes:ind_graphs [2022/11/26 15:14] – valen176 | 184_notes:ind_graphs [2022/12/05 16:11] – valen176 | ||
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===== Induction Graphs ===== | ===== Induction Graphs ===== | ||
- | In these notes, we will examine a few examples of changing magnetic fluxes and associated induced voltages. | + | In these notes, we will examine a few examples of changing magnetic fluxes and associated induced voltages. Recall from the previous notes that these are related by **Faraday' |
+ | |||
+ | $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | ||
+ | |||
+ | This is saying that the induced current is the **negative slope** of the magnetic flux. | ||
First let's consider when $\Phi_B$ rises and falls linearly with the same magnitude of slope: | First let's consider when $\Phi_B$ rises and falls linearly with the same magnitude of slope: | ||
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We can see that from $t=0$ to $t = 10$, $\Phi_B(t)$ has a positive slope, so $V_{ind}$ is negative on that time interval. However, $\Phi_B(t)$ is steeper from $t=5$ to $t=10$, so $V_{ind}$ is **more negative** on that time interval than from $t = 0$ to $t = 5$. From $t = 10$ to $t = 15$, $\Phi_B(t)$ has a constant and negative slope, so $V_{ind}$ is constant and positive on that time interval. | We can see that from $t=0$ to $t = 10$, $\Phi_B(t)$ has a positive slope, so $V_{ind}$ is negative on that time interval. However, $\Phi_B(t)$ is steeper from $t=5$ to $t=10$, so $V_{ind}$ is **more negative** on that time interval than from $t = 0$ to $t = 5$. From $t = 10$ to $t = 15$, $\Phi_B(t)$ has a constant and negative slope, so $V_{ind}$ is constant and positive on that time interval. | ||
+ | |||
+ | |||
+ | $$ | ||
+ | \Phi_B(t)= | ||
+ | \begin{cases} | ||
+ | 2t & \text{if } 0< | ||
+ | 5t -15 & \text{if } 5< | ||
+ | -10t + 135 & \text{if } 10< | ||
+ | \end{cases} | ||
+ | $$ | ||
+ | Which means $\frac{d \Phi_B}{dt}$ is: | ||
+ | $$ | ||
+ | \frac{d \Phi_B}{dt}= | ||
+ | \begin{cases} | ||
+ | 2 & \text{if } 0< | ||
+ | 5 & \text{if } 5< | ||
+ | -10 & \text{if } 10< | ||
+ | \end{cases} | ||
+ | $$ | ||
+ | Which finally means that $V_{ind}$ is: | ||
+ | $$ | ||
+ | \frac{d \Phi_B}{dt}= | ||
+ | \begin{cases} | ||
+ | -2 & \text{if } 0< | ||
+ | -5 & \text{if } 5< | ||
+ | 10 & \text{if } 10< | ||
+ | \end{cases} | ||
+ | $$ | ||
Finally, let's look at an example with a non-linear $\Phi_B(t)$: | Finally, let's look at an example with a non-linear $\Phi_B(t)$: | ||
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[{{184_notes: | [{{184_notes: | ||
- | $\Phi_B(t)$ looks like a quadratic centered about t = 2. We can see that while $\Phi_B(t)$ is decreasing ($0< | + | $\Phi_B(t)$ looks like a quadratic centered about t = 2. We can see that while $\Phi_B(t)$ is decreasing ($0< |
+ | |||
+ | Specifically, | ||
+ | |||
+ | $$\Phi_B(t) = (t-2)^2$$ | ||
+ | |||
+ | Taking a first derivative with respect to time yields: | ||
+ | |||
+ | $$\frac{d \Phi_B}{dt} = 2(t-2)$$ | ||
+ | |||
+ | Multiplying by $-1$ to find $V_{ind}$ gives: | ||
+ | |||
+ | $$V_{ind} = -2(t-2)$$ |