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--- 184_notes:ind_graphs [2022/11/26 15:18] – valen176
+++ 184_notes:ind_graphs [2022/12/07 14:39] – valen176
@@ Line 1: / Line 1: @@
 ===== Induction Graphs =====
-In these notes, we will examine a few examples of changing magnetic fluxes and associated induced voltages.
+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 $\Phi_B$ rises and falls linearly with the same magnitude of slope:
@@ Line 26: / 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 39: / Line 43: @@
 [{{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.
+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 $\Phi_B(t)$:
@@ Line 45: / Line 77: @@
 [{{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 ($2<t<8$), $V_{ind} is negative.
+$\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: