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electrical_engineering_and_electronics_1:block18 [2025/12/02 18:45] – mexleadminelectrical_engineering_and_electronics_1:block18 [2025/12/06 13:47] (aktuell) – mexleadmin
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 For checking your understanding please do the following exercises: For checking your understanding please do the following exercises:
-  * ...+  * Exercise E3 Coil in a magnetic Field 
 +  * Exercise 4.1.2 Magnetic Field Strength around a horizontal straight Conductor 
 +  * Exercise 4.1.4 Effects of induction I 
  
 ===== 90-minute plan ===== ===== 90-minute plan =====
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 ===== Exercises ===== ===== Exercises =====
 +
 +{{page>electrical_engineering_and_electronics:task_rdz03rspbwusy7wk_with_calculation&nofooter}}
 +{{page>electrical_engineering_and_electronics:task_ludzwiuhjxitz85b_with_calculation&nofooter}}
 +
 +
 +<panel type="info" title="Exercise 4.1.4 Effects of induction I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +A change of magnetic flux is passing a coil with a single winding. The following pictures <imgref ImgNrEx01> show different flux-time-diagrams as examples.
 +
 +  * Create for each $\Phi(t)$-diagram the corresponding $u_{\rm ind}(t)$-diagram!
 +  * Write down each maximum value of $u_{\rm ind}(t)$
 +
 +<WRAP> <imgcaption ImgNrEx01| Flux-Time-Diagrams> </imgcaption> <WRAP> {{drawio>FluxTimeDia1.svg}} \\ </WRAP></WRAP>
 +
 +<button size="xs" type="link" collapse="Solution_4_1_4_1_Solution">{{icon>eye}} Solution for (a)</button><collapse id="Solution_4_1_4_1_Solution" collapsed="true">
 +
 +For partwise linear $u_{\rm ind}$ one can derive: 
 +\begin{align*} 
 +u_{\rm ind} &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\ 
 +            &= -{{\Delta \Phi}\over{\Delta t}} 
 +\end{align*}
 +
 +For diagram (a):
 +
 +  * $t= 0.0 ... 0.6 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$
 +  * $t= 0.6 ... 1.5 ~\rm s$: $u_{\rm ind} = -{{-3.75\cdot 10^{-3} ~\rm Vs}\over{0.9 ~\rm s}}= +4.17 ~\rm mV$
 +  * $t= 1.5 ... 2.1 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$
 +
 +</collapse>
 +
 +<button size="xs" type="link" collapse="Solution_4_1_4_1_Finalresult">
 +{{icon>eye}} Final result for (a)</button><collapse id="Solution_4_1_4_1_Finalresult" collapsed="true"> 
 +<WRAP> <imgcaption ImgNrEx01| Flux-Time-Diagrams Solution> </imgcaption> <WRAP> {{drawio>FluxTimeDia1Solution.svg}} \\ 
 +</WRAP></WRAP> \\ 
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 4.1.5 Effects of induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +A changing of magnetic flux is passing a coil with a single winding and induces the voltage $u_{\rm ind}(t)$. 
 +The following pictures <imgref ImgNrEx02> show different voltage-time diagrams as examples.
 +
 +  * Create for each $u_{\rm ind}(t)$-diagram the corresponding $\Phi(t)$-diagram!
 +  * Write down each maximum value of $\Phi(t)$
 +
 +Note the given start value $\Phi_0$ for each flux.
 +
 +<WRAP> <imgcaption ImgNrEx02| Voltage-Time-Diagrams> </imgcaption> <WRAP> {{drawio>FluxTimeDia2.svg}} \\ </WRAP></WRAP>
 +
 +#@HiddenBegin_HTML~415_1S,Solution for (a)~@#
 +
 +For partwise linear $u_{\rm ind}$ one can derive: 
 +\begin{align*} 
 +u_{\rm ind}        &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\ 
 +\rightarrow  \Phi  &= -\int_0^t{ u_{\rm ind} \;{\rm d}t} \\
 +\Phi               &= \Phi_0 -\sum_k {u_{{\rm ind},~k} \; \Delta t} \\
 +\end{align*}
 +
 +For diagram (a):
 +
 +  * $t= 0.00 ... 0.04 ~\rm s\quad$: $\quad \Phi = \Phi_0 - {0 \cdot \; \Delta t} \quad\quad\quad\quad\quad\quad\quad= 0 ~\rm Wb$
 +  * $t= 0.04 ... 0.10 ~\rm s\quad$: $\quad \Phi =   0 {~\rm Wb} - {{30 ~\rm mV} \cdot \; (t - 0.04 ~\rm s)} = \quad {1.2 ~\rm mWb} - t \cdot 30 ~\rm mV$
 +  * $t= 0.10 ... 0.14 ~\rm s\quad$: $\quad \Phi =   {1.2 ~\rm mWb} - {0.10 ~\rm s} \cdot 30 ~\rm mV \quad = - {1.8 ~\rm mWb}$
 +
 +#@HiddenEnd_HTML~415_1S,Solution ~@#
 +
 +#@HiddenBegin_HTML~415_1R,Result for (a)~@#
 +{{drawio>FluxTimeDia2Res.svg}} 
 +#@HiddenEnd_HTML~415_1R,Result~@#
 +
 +
 +</WRAP></WRAP></panel>
 +
 +
  
 <panel type="info" title="Exercise 4.1.1 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.1 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
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 </collapse> </WRAP></WRAP></panel> </collapse> </WRAP></WRAP></panel>
  
-{{page>electrical_engineering_and_electronics:task_rdz03rspbwusy7wk_with_calculation&nofooter}} + 
-{{page>electrical_engineering_and_electronics:task_ludzwiuhjxitz85b_with_calculation&nofooter}}+ 
 +<panel type="info" title="Exercise 4.1.6 Coil in magnetic Field I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +A single winding is located in a homogenous magnetic field ($B = 0.5 ~\rm T$) between the pole pieces.  
 +The winding has a length of $150 ~\rm mm$ and a distance between the conductors of $50 ~\rm mm$ (see <imgref ImgNrEx03>). 
 + 
 +  * Determine the function $u_{\rm ind}(t)$, when the coil is rotating with $3000 ~\rm min^{-1}$. 
 +  * Given a current of $20 ~\rm A$ through the winding: What is the torque $M(\varphi)$ depending on the angle between the surface vector of the winding and the magnetic field? 
 + 
 +<WRAP> <imgcaption ImgNrEx03| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces.svg}} \\ </WRAP></WRAP> 
 + 
 +<button size="xs" type="link" collapse="Solution_4_1_6_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_6_1_Solution" collapsed="true">  
 +\begin{align*}  
 +u_{\rm ind} &= -   {{{\rm d}\Phi}\over{{\rm d}t}} \\  
 +            &= -         {{\rm d}\over{{\rm d}t}} B \cdot A \\ 
 +            &= - B \cdot {{\rm d}\over{{\rm d}t}} A\\ 
 +            &= - B \cdot {{\rm d}\over{{\rm d}t}} \left(l \cdot d \cdot \cos(\omega t) \right)\\ 
 +            &= + B \cdot l \cdot d \cdot \omega \cdot \sin(\omega t)\\  
 +\end{align*} 
 + 
 +</collapse> </WRAP></WRAP></panel> 
 + 
 +<panel type="info" title="Exercise 4.1.7 Coil in magnetic Field II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +A rectangular coil is given by the sizes $a=10 ~\rm cm$, $b=4 ~\rm cm$, and the number of windings $N=200$.  
 +This coil moves with a constant speed of $v=2 ~\rm m/s$ perpendicular to a homogeneous magnetic field ($B=1.3 ~\rm T$ on a length of $l=5 ~\rm cm$).  
 +The area of the coil is tilted with regard to the field in $\alpha=60°$ and enters the field from the left side (see <imgref ImgNrEx04>). 
 + 
 +  * Determine the function $u_{\rm ind}(t)$ on the coil along the given path. Sketch of the $u_{\rm ind}(t)$ diagram. 
 +  * What is the maximum induced voltage $u_{\rm ind,Max}$? 
 + 
 +<WRAP> <imgcaption ImgNrEx04| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces2.svg}} \\ </WRAP></WRAP> 
 + 
 +<button size="xs" type="link" collapse="Solution_4_1_7_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_7_1_Solution" collapsed="true">  
 + 
 +Let assume, that $l$ is in the $x$-axis, $\vec{B}$ in the $y$-axis and $a$.  
 +\\ \\ 
 + 
 +**Step 1**: Calculate the effective area, perpendicular to the $\vec{B}$-field (independent from whether the area is in the $\vec{B}$-field or not). 
 + 
 +For this $b$ has to be projected onto the plane perpendicular to the $\vec{B}$-field: $b_{\rm eff}= b \cdot \cos \alpha$ 
 +\begin{align*}  
 +A_{\rm eff} &= a \cdot b \cdot \cos \alpha 
 +\end{align*} 
 + 
 +**Step 2**: Focus on entering and exiting the $\vec{B}$-field. \\ 
 +Induction only occurs for ${{\rm d}\over{{\rm d}t}}(A\cdot B)\neq 0$, so here: when the area $A_{\rm eff}$ enters and leave the constant $\vec{B}$-field.  
 + 
 +When entering the $\vec{B}$-field the area $A$ with $0<A<A_{eff}$ is in the field.  
 +The area moves with $v$. Therefore, after $\Delta t = b_{\rm eff} \cdot v$ the full $\vec{B}$-field is provided onto the area $A_{\rm eff}$:  
 +\begin{align*}  
 +u_{\rm ind} &= -        {{{\rm d}\Psi}\over{{\rm d}t}} \\  
 +            &= -N \cdot       {{\rm d}\over{{\rm d}t}} B \cdot A \\  
 +            &= -N \cdot           {{1}\over{\Delta t}} B \cdot A_{\rm eff} \\  
 +            &= -N \cdot {{1}\over{b \cdot \cos \alpha \cdot v}} B \cdot a \cdot b \cdot \cos \alpha \\  
 +            &= -N \cdot B \cdot {{a}\over{v}}\\  
 +\end{align*} 
 + 
 +The following diagram shows ... 
 +  * ... how one can derive the effective width $b_{\rm eff}$, which is projected onto the plane perpendicular to the $\vec{B}$-field: $b_{\rm eff}= b \cdot \cos \alpha$ 
 +  * ... what happens on the effective area $A_{\rm eff}$: there is a change of the field lines in the area only for entering and leaving the $\vec{B}$-field.  
 +  * ... how the $u_{\rm ind}(t)$ looks as a graph: the part of $A_{\rm eff}$, where the $\vec{B}$-field passes through increase linearly due to the constant speed $v$ 
 +Be aware, that the task did not provide a clue for the direction of windings and therefore it provides no clue for the polarization of the induced voltage. \\  
 +So, the course of the voltage when entering or exiting is not uniquely given. 
 + 
 +<WRAP> <imgcaption ImgNrEx04s| Solution> </imgcaption> <WRAP>{{drawio>WindingPolePieces2solution.svg}}  \\ </WRAP></WRAP> 
 + 
 + 
 +</collapse> </WRAP></WRAP></panel>
  
 ===== Embedded resources ===== ===== Embedded resources =====