184_notes:ind_graphs

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184_notes:ind_graphs [2022/12/05 16:08] valen176184_notes:ind_graphs [2022/12/07 14:37] 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 $\Phi_B$ rises and falls linearly with the same magnitude of slope:
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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: 
 + 
 + 
 +$$ 
 +\Phi_B(t)= 
 +    \begin{cases} 
 +        2t & \text{if } 0<t<5\\ 
 +        5t -15 & \text{if } 5<t<10\\ 
 +        -10t + 135 & \text{if } 10<t<15 
 +    \end{cases} 
 +$$ 
 +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 $\Phi_B(t)$: Finally, let's look at an example with a non-linear $\Phi_B(t)$:
  • 184_notes/ind_graphs.txt
  • Last modified: 2022/12/07 14:43
  • by valen176