Differences
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| Both sides previous revision Previous revision Next revision | Previous revisionLast revisionBoth sides next revision | ||
| 184_notes:ind_graphs [2022/12/05 16:11] – valen176 | 184_notes:ind_graphs [2022/12/07 14:39] – valen176 | ||
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| $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | ||
| - | This is saying that the induced current is the **negative slope** of the magnetic flux. | + | 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: | First let's consider when $\Phi_B$ rises and falls linearly with the same magnitude of slope: | ||
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| \end{cases} | \end{cases} | ||
| $$ | $$ | ||
| - | Which finally means that $V_{ind}$ | + | Now we can multiply by $-1$ because of the negative sign in Faraday' |
| $$ | $$ | ||
| V_{ind}= | V_{ind}= | ||
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| [{{184_notes: | [{{184_notes: | ||
| - | 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. |
| Line 50: | Line 50: | ||
| \begin{cases} | \begin{cases} | ||
| 2t & \text{if } 0< | 2t & \text{if } 0< | ||
| - | 5t -15 & \text{if } 5< | + | 5t -15 & \text{if } 5< |
| -10t + 135 & \text{if } 10< | -10t + 135 & \text{if } 10< | ||
| \end{cases} | \end{cases} | ||
| Line 59: | Line 59: | ||
| \begin{cases} | \begin{cases} | ||
| 2 & \text{if } 0< | 2 & \text{if } 0< | ||
| - | 5 & \text{if } 5< | + | 5 & \text{if } 5< |
| -10 & \text{if } 10< | -10 & \text{if } 10< | ||
| \end{cases} | \end{cases} | ||
| Line 65: | Line 65: | ||
| Which finally means that $V_{ind}$ is: | Which finally means that $V_{ind}$ is: | ||
| $$ | $$ | ||
| - | \frac{d \Phi_B}{dt}= | + | V_{ind}= |
| \begin{cases} | \begin{cases} | ||
| -2 & \text{if } 0< | -2 & \text{if } 0< | ||
| - | -5 & \text{if } 5< | + | -5 & \text{if } 5< |
| 10 & \text{if } 10< | 10 & \text{if } 10< | ||
| \end{cases} | \end{cases} | ||