===== Induction Graphs =====

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's Law** which says:

$$V_{ind} = -\frac{d\Phi_b}{dt}$$

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 an example where $\Phi_B$ rises and falls linearly with the same magnitude of slope:


[{{184_notes:examples:ind_graph1.png?800|  }}]

From $t = 0$ to $t = 5$, $\Phi_B(t)$ has a constant positive slope, so  $V_{ind}$ will be constant and **negative**. Conversely, from $t = 5$ to $t = 10$, $\Phi_B(t)$ has a constant negative slope, so $V_{ind}$ will be constant and **positive**.

Specifically, in this case $\Phi_B(t)$ is defined as:
$$
\Phi_B(t)=
    \begin{cases}
        2t & \text{if } 0<t<5\\
        -2t & \text{if } 5<t<10
    \end{cases}
$$
Which means $\frac{d \Phi_B}{dt}$ is:
$$
\frac{d \Phi_B}{dt}=
    \begin{cases}
        2 & \text{if } 0<t<5\\
        -2 & \text{if } 5<t<10
    \end{cases}
$$
Now we can multiply by $-1$ because of the negative sign in Faraday's law to find $V_{ind}$:
$$
V_{ind}=
    \begin{cases}
        -2 & \text{if } 0<t<5\\
         2 & \text{if } 5<t<10
    \end{cases}
$$


Next, let's consider an example with a few different slopes:
[{{184_notes:examples:ind_graph2.png?800|  }}]

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:


$$
\Phi_B(t)=
    \begin{cases}
        2t & \text{if } 0<t<5\\
        5t -15 & \text{if } 5<t<10\\
        -10t + 135 & \text{if } 10<t<15
    \end{cases}
$$
Which means $\frac{d \Phi_B}{dt}$ is:
$$
\frac{d \Phi_B}{dt}=
    \begin{cases}
        2 & \text{if } 0<t<5\\
        5 & \text{if } 5<t<10\\
        -10 & \text{if } 10<t<15
    \end{cases}
$$
Which finally means that $V_{ind}$ is:
$$
V_{ind}=
    \begin{cases}
        -2 & \text{if } 0<t<5\\
        -5 & \text{if } 5<t<10\\
        10 & \text{if } 10<t<15
    \end{cases}
$$

Finally, let's look at an example with a non-linear $\Phi_B(t)$:

[{{184_notes:examples:ind_graph3.png?800|  }}]

$\Phi_B(t)$ looks like a quadratic centered about t = 2. We can see that while $\Phi_B(t)$ is decreasing ($0<t<2$), $V_{ind}$ is positive, and while $\Phi_B(t)$ is increasing ($2<t<8$), $V_{ind}$ is negative. 

Specifically, in this case we have:

$$\Phi_B(t) = (t-2)^2$$

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:

$$V_{ind} = -2(t-2)$$