{{tag>induction flux induced_voltage exam_ee2_SS2021}}{{include_n>1200}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ effect of induction\\
<fs medium>(written test, approx. 5 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#

A single conductor loop is penetrated by a changing magnetic flux. \\
The following figure shows the variation of the flux $\Phi(t)$ over time.  \\ \\
Calculate the variation of the induced voltage $u_{\rm ind}(t)$ over time and draw it in a separate diagram. 

{{drawio>ee2:LUdzWiUhjxITZ85B_diagram1.svg}}
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Based on Faraday's Law of Induction the induced voltage is given by:
\begin{align*}
u_{\rm ind} =& - {{ {\rm d} }\over{ {\rm d}t}} \Psi(t) \bigg\rvert_{n=1}\\
            =& - {{ {\rm d} }\over{ {\rm d}t}} \Phi(t) \\
\end{align*}

For a linear function, the derivative can be substituted by Deltas ($\rm d \rightarrow \Delta$): \\
\begin{align*}
u_{\rm ind} = - {{ \Delta \Phi(t)}\over{ \Delta t}} = - { { \Phi(t_{\rm n+1} ) - \Phi(t_{\rm n} ) } \over { t_{\rm n+1} - t_{\rm n} } } \\
\end{align*}

For a piece-wise linear function, the induced voltage can be calculated for each interval. \\
Here, there are 5 different intervals - in the following called $\rm I$ to $\rm V$ from left to right:
{{drawio>ee2:LUdzWiUhjxITZ85B_path1.svg}}

  * For the intervals $\rm I$, $\rm III$, and $\rm V$ , the flux $\Phi(t)$ is constant. Therefore, $\Delta \Phi(t)=0$ and $u_{\rm ind}(t)=0{~\rm V}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path2.svg}}
  * For the interval  $\rm II$: 
    * The change in the flux is: $ \Delta \Phi(t) = 1.5 \cdot 10^{-4} {~\rm Vs} - 4.5 \cdot 10^{-4} {~\rm Vs}= - 3.0 \cdot 10^{-4} {~\rm Vs}$
    * The time span is: $0.2 ~\rm s$
    * Conclusively, the induced voltage is: $u_{\rm ind}(t) = + {{3.0 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 1.5 {~\rm mV}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path3.svg}}
  * For the interval  $\rm IV$: 
    * The change in the flux is: $ \Delta \Phi(t) = 0 \cdot 10^{-4} {~\rm Vs} - 1.5 \cdot 10^{-4} {~\rm Vs}= - 1.5 \cdot 10^{-4} {~\rm Vs}$
    * The time span is: $0.2 ~\rm s$
    * Conclusively, the induced voltage is: $u_{\rm ind}(t) = + {{1.5 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 0.75 {~\rm mV}$ \\ {{drawio>ee2:LUdzWiUhjxITZ85B_path4.svg}}

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{{drawio>ee2:LUdzWiUhjxITZ85B_solution1.svg}}
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