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--- 184_notes:ind_graphs [2022/11/26 14:51] – valen176
+++ 184_notes:ind_graphs [2022/12/07 14:43] (current) – valen176
@@ Line 1: / Line 1: @@
-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. 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 $t = 5$, $\Phi_B$ has a constant positive slope, so  $V_{ind}$ will be constant and **negative**. Conversely, from $t = 5$ to $t = 10$, $\Phi_B$ has a constant negative slope, so $V_{ind}$ will be constant and **positive**.
-Specifically, in this case $\Phi_b(t)$ is defined as:
+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:
 $$
 \Phi_B(t)=
@@ Line 21: / Line 30: @@
     \end{cases}
 $$
-Which finally means that $V_{ind}$ is:
+Now we can multiply by $-1$ because of the negative sign in Faraday's law to find $V_{ind}$:
 $$
 V_{ind}=
@@ Line 31: / Line 40: @@
+Next, let's consider an example with a few different slopes:
 [{{184_notes:examples:ind_graph2.png?800|  }}]
-More Words
+We can see that from $t=0$ to $t = 10$, $\Phi_B(t)$ has a positive slope, so $V_{ind}$ is negative on that time interval. However, $\Phi_B(t)$ is steeper from $t=5$ to $t=10$, so $V_{ind}$ is **more negative** on that time interval than from $t = 0$ to $t = 5$. From $t = 10$ to $t = 15$, $\Phi_B(t)$ has a constant and negative slope, so $V_{ind}$ is constant and positive on that time interval. Specifically we have that:
+$$
+\Phi_B(t)=
+    \begin{cases}
+t & \text{if } 0<t<5\\
+t -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}
+& \text{if } 0<t<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\\
+& \text{if } 10<t<15
+    \end{cases}
+$$
+Finally, let's look at an example with a non-linear $\Phi_B(t)$:
 [{{184_notes:examples:ind_graph3.png?800|  }}]
-Some final words
+$\Phi_B(t)$ looks like a quadratic centered about t = 2. We can see that while $\Phi_B(t)$ is decreasing ($0<t<2$), $V_{ind}$ is positive, and while $\Phi_B(t)$ is increasing ($2<t<8$), $V_{ind}$ is negative.
+Specifically, in this case we have:
+$$\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)$$