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--- 184_notes:ind_graphs [2022/11/26 15:00] – 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:
@@ Line 22: / 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 35: / Line 43: @@
 [{{184_notes:examples:ind_graph2.png?800|  }}]
-Here, 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  $t = 0$  t0  $t = 5$ .
+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:
+ $V_{ind} = -2(t-2)$