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184_notes:ind_graphs [2022/11/26 14:33] – valen176 | 184_notes:ind_graphs [2022/12/07 14:37] – valen176 | ||
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- | In these notes, we will examine a few examples of changing magnetic fluxes and associated induced voltages. First let's consider when $\Phi_B$ rises and falls linearly with the same magnitude of slope: | + | ===== Induction Graphs ===== |
+ | |||
+ | In these notes, we will examine a few examples of changing magnetic fluxes and associated induced voltages. | ||
+ | |||
+ | $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | ||
+ | |||
+ | This is saying that the induced current is the **negative slope** of the magnetic flux. In other words, if the magnetic flux is increasing, then $V_{ind}$ will be negative, if the magnetic flux is decreasing, then $V_{ind}$ will be positive, and if the magnetic flux is constant, then $V_{ind} = 0$ | ||
+ | |||
+ | First let's consider when $\Phi_B$ rises and falls linearly with the same magnitude of slope: | ||
[{{184_notes: | [{{184_notes: | ||
- | From t = 0 to t = 5, $\Phi_B$ has a positive, | + | |
+ | From $t = 0$ to $t = 5$, $\Phi_B(t)$ has a constant | ||
+ | |||
+ | Specifically, | ||
$$ | $$ | ||
\Phi_B(t)= | \Phi_B(t)= | ||
\begin{cases} | \begin{cases} | ||
- | 2t & \text{if } x \in \mathbb{Q}\\ | + | 2t & \text{if } 0<t<5\\ |
- | -2t & \text{if } x \in \mathbb{R}\setminus\mathbb{Q} | + | -2t & \text{if } 5< |
\end{cases} | \end{cases} | ||
$$ | $$ | ||
+ | Which means $\frac{d \Phi_B}{dt}$ is: | ||
+ | $$ | ||
+ | \frac{d \Phi_B}{dt}= | ||
+ | \begin{cases} | ||
+ | 2 & \text{if } 0< | ||
+ | -2 & \text{if } 5< | ||
+ | \end{cases} | ||
+ | $$ | ||
+ | Which finally means that $V_{ind}$ is: | ||
+ | $$ | ||
+ | V_{ind}= | ||
+ | \begin{cases} | ||
+ | -2 & \text{if } 0< | ||
+ | 2 & \text{if } 5< | ||
+ | \end{cases} | ||
+ | $$ | ||
+ | |||
+ | |||
+ | Next, let's consider an example with a few different slopes: | ||
[{{184_notes: | [{{184_notes: | ||
- | More Words | + | 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. Specifically we have that: |
+ | |||
+ | |||
+ | $$ | ||
+ | \Phi_B(t)= | ||
+ | \begin{cases} | ||
+ | 2t & \text{if } 0< | ||
+ | 5t -15 & \text{if } 5< | ||
+ | -10t + 135 & \text{if } 10< | ||
+ | \end{cases} | ||
+ | $$ | ||
+ | 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: | ||
+ | $$ | ||
+ | V_{ind}= | ||
+ | \begin{cases} | ||
+ | -2 & \text{if } 0< | ||
+ | -5 & \text{if } 5< | ||
+ | 10 & \text{if } 10< | ||
+ | \end{cases} | ||
+ | $$ | ||
+ | |||
+ | Finally, let's look at an example with a non-linear $\Phi_B(t)$: | ||
[{{184_notes: | [{{184_notes: | ||
- | Some final words | + | $\Phi_B(t)$ 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: | ||
+ | |||
+ | $$V_{ind} = -2(t-2)$$ |