Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
184_notes:ind_graphs [2022/12/05 16:11] – valen176 | 184_notes:ind_graphs [2022/12/07 14:43] (current) – valen176 | ||
---|---|---|---|
Line 5: | Line 5: | ||
$$V_{ind} = -\frac{d\Phi_b}{dt}$$ | $$V_{ind} = -\frac{d\Phi_b}{dt}$$ | ||
- | 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 an example where $\Phi_B$ rises and falls linearly with the same magnitude of slope: |
Line 30: | Line 30: | ||
\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}= | ||
Line 43: | Line 43: | ||
[{{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} |