184_notes:ind_graphs

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184_notes:ind_graphs [2022/12/05 16:11] valen176184_notes:ind_graphs [2022/12/07 14:43] (current) valen176
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 $$V_{ind} = -\frac{d\Phi_b}{dt}$$ $$V_{ind} = -\frac{d\Phi_b}{dt}$$
  
-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:
  
  
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     \end{cases}     \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}= V_{ind}=
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 [{{184_notes:examples:ind_graph2.png?800|  }}] [{{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:
  
  
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     \begin{cases}     \begin{cases}
         2t & \text{if } 0<t<5\\         2t & \text{if } 0<t<5\\
-        5t -15 & \text{if } 5<t<10+        5t -15 & \text{if } 5<t<10\\
         -10t + 135 & \text{if } 10<t<15         -10t + 135 & \text{if } 10<t<15
     \end{cases}     \end{cases}
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     \begin{cases}     \begin{cases}
         2 & \text{if } 0<t<5\\         2 & \text{if } 0<t<5\\
-        5 & \text{if } 5<t<10+        5 & \text{if } 5<t<10\\
         -10 & \text{if } 10<t<15         -10 & \text{if } 10<t<15
     \end{cases}     \end{cases}
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 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<t<5\\         -2 & \text{if } 0<t<5\\
-        -5 & \text{if } 5<t<10+        -5 & \text{if } 5<t<10\\
         10 & \text{if } 10<t<15         10 & \text{if } 10<t<15
     \end{cases}     \end{cases}
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