Dies ist eine alte Version des Dokuments!
Exercise E1 effect of induction
(written test, approx. 5 % of a 120-minute written test, SS2021)
A simple 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.
Based on Faraday's Law of Induction the induced voltage is given by: \begin{align*} U_{\rm ind} = - {{ {\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:
- For the intervals $\rm I$, $\rm III$, and $\rm V$ , the flux $\Phi(t)$ is constant. Therefore, $\Delta \Phi(t)=0$ and $U_{\rm ind}(t)=0{~\rm V}$
- 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: $U_{\rm ind}(t) = + {{3.0 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 1.5 {~\rm mV}$
- 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: $U_{\rm ind}(t) = + {{1.5 \cdot 10^{-4} {~\rm Vs}} \over {0.2 ~\rm s} } = 0.75 {~\rm mV}$