Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- electrical_engineering_2:magnetic_circuits [2023/05/09 13:59] – mexleadmin
+++ electrical_engineering_2:magnetic_circuits [2024/07/11 18:54] (aktuell) – [Effects in the electric Circuits] mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== 5. Magnetic Circuits ======
+====== 5 Magnetic Circuits ======
 <callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout>
@@ Zeile 9: / Zeile 9: @@
 In this chapter, we will investigate, how far we come with such an analogy and where it can be practically applied.
-===== 5.1 Linear magnetic Circuits =====
+===== 5.1 Linear Magnetic Circuits =====
 For the upcoming calculations, the following assumptions are made
@@ Zeile 168: / Zeile 168: @@
 Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions:
-  - $l=35.8~\rm cm$, $d=1.9~\rm cm$
+  - $l=35.8~\rm cm$, $d=1.90~\rm cm$
-  - $l=22.5~\rm cm$, $d=1.5~\rm cm$
+  - $l=11.1~\rm cm$, $d=1.50~\rm cm$
-<button size="xs" type="link" collapse="Solution_5_1_2_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_2_1_Result" collapsed="true">
+#@HiddenBegin_HTML~5_1_2s,Solution~@#
-  - $1.5\cdot 10^5 ~\rm {{1}\over{H}}$
+The magnetic resistance is given by:
-  - $3.0\cdot 10^5 ~\rm {{1}\over{H}}$
+\begin{align*}
+\ R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}
+\end{align*}
-</collapse>
+With
+  * the area $ A = \left({{d}\over{2}}\right)^2 \cdot \pi $
+  * the vacuum magnetic permeability $\mu_{0}=4\pi\cdot 10^{-7} ~\rm H/m$, and
+  * the relative permeability $\mu_{\rm r}=1$.
+#@HiddenEnd_HTML~5_1_2s,Solution ~@#
+#@HiddenBegin_HTML~5_1_2r,Result~@#
+  - $1.00\cdot 10^9 ~\rm {{1}\over{H}}$
+  - $0.50\cdot 10^9 ~\rm {{1}\over{H}}$
+#@HiddenEnd_HTML~5_1_2r,Result~@#
 </WRAP></WRAP></panel>
@@ Zeile 296: / Zeile 308: @@
 ===== 5.3 Mutual Induction and Coupling =====
-Situation: Two coils $1$ and $2$ near each other. \\ Questions:
+Imagine charging your phone wirelessly by simply placing it on a charging pad.
+This seamless experience is made possible by the fascinating phenomenon of mutual induction and coupling between two coils.
+This situation is depicted in <imgref ImgNr09>:
+When an alternating current flows through one coil (Coil $1$), it creates a time-varying magnetic field that induces a voltage in the nearby coil (Coil $2$), even though they are not physically connected.
+This mutual influence is governed by the principle of electromagnetic induction.
+<WRAP center 35%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP>
+The key factor determining the strength of mutual induction is the mutual inductance ($M$) between the coils.
+It quantifies the magnetic flux linkage and depends on factors like the number of turns, current, and relative orientation of the coils.
+While geometric properties play a role, the fundamental principle can be described using electric properties alone, making mutual induction a versatile concept with numerous applications, including:
+  * Wireless power transfer
+  * Transformers
+  * Inductive coupling in communication systems
+  * Inductive sensors
+As we explore this chapter, we'll delve into the mathematical models, equations, and practical considerations of mutual induction and coupling, unlocking a world of innovative technologies that shape our modern lives.
+We explicitly try to answer the following questions:
   * Which effect do the coils have on each other?
   * Can we describe the effects with mainly electric properties (i.e. no geometric properties)
-<WRAP center 35%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP>
 ==== Effect of Coils on each other ====
@@ Zeile 317: / Zeile 348: @@
 For the single coil, we got the relationship between the linked flux $\Psi$ and the current $i$ as: $\Psi = L \cdot i$. \\
-Now the coils also are interacting with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$:
+Now the coils also interact with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$:
 \begin{align*}
 \Psi_1 &= &\Psi_{11}        &+ \Psi_{12} \\
@@ Zeile 332: / Zeile 363: @@
 The formula can also be described as:
-\begin{align*}
+{{drawio>VectorialformulaOfInduction.svg}}
-\left( \begin{array}{c} \Psi_1 \\          \Psi_2 \end{array}           \right) =
-\left( \begin{array}{c} L_{11} & M_{12} \\  M_{21} & L_{22} \end{array} \right)
-\cdot
-\left( \begin{array}{c} i_1 \\              i_2 \end{array}             \right)
-\end{align*}
-The view of the magnetic flux is sometimes good when effects like an acting Lorentz force in of interest.
+The view of the magnetic flux is advantageous when effects like an acting Lorentz force is of interest.
-More often the coils are coupling two electric circuits linked in a transformer or a wireless charger.
+However, more often the coils couple electric circuits, like in a transformer or a wireless charger.
 Here, the effect on the circuits is of interest. This can be calculated with the induced electric voltages $u_{\rm ind,1}$ and $u_{\rm ind,2}$ in each circuit.
 They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$:
@@ Zeile 353: / Zeile 379: @@
 The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$?
-==== Magnetic Circuit with 2 Sources ====
+==== Magnetic Circuit with two Sources ====
 To get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08>). \\
@@ Zeile 360: / Zeile 386: @@
 <WRAP> <imgcaption ImgNr08 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils.svg}} </WRAP>
-The <imgref ImgNr08> shows the fluxes on each part. The black dots nearby the windings mark the direction of the windings: \\
+The <imgref ImgNr08> shows the fluxes on each part. The black dots near the windings mark the direction of the windings, and therefore the sign of the generated flux. \\
-All the fluxes created from currents that flow into the marked pins are summed up positively in the core. \\
+All the fluxes caused by currents flowing into the __marked pins__ are summed up __positively__ in the core. \\
-When there is a current flowing into a non-marked pin, its flux has to be subtracted from the others.
+When there is a current flowing into a __non-marked pin__, its flux has to be __subtracted__ from the others.
 To get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\
@@ Zeile 393: / Zeile 419: @@
 $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$):
-\begin{align*} k_{21} = {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}
+\begin{align*} k_{21} = \pm {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}
-When $k_{21}=100~\%$, there is no flux in the middle leg but only in the second coil. \\
+The sign of $k_{21}$ depends on the direction of $\Phi_{21}$ relative to $\Phi_{22}$! If the directions are the same, the positive sign applies, if the directions are opposite, the minus sign applies.
+When $k_{21}=+100~\%$, there is no flux in the middle leg but only in the second coil and in the same direction as the flux that originates from the second coil. \\
+When $k_{21}=-100~\%$, there is no flux in the middle leg but only in the second coil and in the opposite direction as the flux that originates from the second coil. \\
 For  $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling.
@@ Zeile 409: / Zeile 438: @@
        &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\
 \end{align*}
+Note, that also $M_{21}$ and $M_{12}$ can be either positive or negative, depending on the sign of the coupling coefficients.
 The formula is finally:
@@ Zeile 418: / Zeile 449: @@
 \left( \begin{array}{c} i_1                                                                        \\ i_2                                                                        \end{array} \right)
 \end{align*}
+For most of the applications the induction matrix has to be symmetric((This can be derived from energy considerations, when only electric circuits are coupled without additional flow of mechanical energy. This is, for example  not the case for motors with a mechanical load.)). \\ Therefore, the following applies:
+  * In General: the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$
+  * For symmetric induction matrix: The mutual inductances are equal: $M_{12} = M_{21} = M$
+  * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}
 <panel type="info" title="Task 5.3.1 Example for magnetic Circuit with two Sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 The magnetical configuration in <imgref ExImgNr01> shall be given. \\
 The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length $l=12 ~\rm cm$ and the number of windings $N_1 = 400$, $N_2=300$.
-The coupling factors are $k_{12}=0.6$ and $k_{21}=0.8$.
-Calculate $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$.
 <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP>
-=== Step 1: Draw the problem as a network ===
+.  Simplify the configuration into three magnetic resistors and 2 voltage sources. Draw the problem as an equivalent circuit
+#@HiddenBegin_HTML~5311,Result~@#
+<WRAP> <imgcaption ExImgNr11 | Equivalent Network> </imgcaption> {{drawio>CoreWithTwoCoils2network.svg}} </WRAP>
+#@HiddenEnd_HTML~5311,Result~@#
+. Calculate all magnetic resistances. Additionally, calculate the magnetic resistances $R_{\rm m1}$  and $R_{\rm m2}$ seen from the magnetic voltage source $1$ and $2$.
+#@HiddenBegin_HTML~5312,Path~@#
-=== Step 2: Calculate the magnetic resistances ===
+<WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP>
-The magnetic resistance is summed up by looking at the circuit from the source $1$:
+The magnetic resistance is summed up by looking at the circuit from the source $1$ (see <imgref ExImgNr13>):
 \begin{align*}
 R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\
@@ Zeile 440: / Zeile 484: @@
 where the parts are given as
 \begin{align*}
-R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} \\
+R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} &&= 398 \cdot 10^{3} ~\rm {{1}\over{H}} \\
-R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} \\
+R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} &&= 133 \cdot 10^{3} ~\rm {{1}\over{H}} \\
-R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} \\
+R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} &&= 265 \cdot 10^{3} ~\rm {{1}\over{H}} \\
 \end{align*}
 With the given geometry this leads to
 \begin{align*}
-R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot (3 + {{1\cdot 2}\over{1 + 2}}) \\
+R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\
-           &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{3}} \\
+           &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{3}} &&= 486 \cdot 10^{3} ~\rm {{1}\over{H}}\\
-           &= 133 \cdot 10^{3}                          \cdot {{11}\over{3}} ~\rm {{1}\over{H}}\\
 \end{align*}
 Similarly, the magnetic resistance $R_{m2}$ is
 \begin{align*}
-R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} \\
+R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} &&= 365 \cdot 10^{3} ~\rm {{1}\over{H}}\\
-           &= 133 \cdot 10^{3}                          \cdot {{11}\over{4}} ~\rm {{1}\over{H}}\\
 \end{align*}
-== Step 3: Calculate the magnetic inductances ==
+#@HiddenEnd_HTML~5312,Path~@#
+. Calculate the self-inductions $L_{11}$ and $L_{22}$
+#@HiddenBegin_HTML~5313,Path~@#
+For the self-induction the effect on the electrical circuit is relevant. That is why the number of windings has to be considered.
 \begin{align*}
 L_{11} &= {{N_1^2}\over{R_{\rm m1}}}                    &= 329 ~\rm mH\\ \\
 L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\
-M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &= 197 ~\rm mH\\ \\
-M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &= 197 ~\rm mH\\
 \end{align*}
+#@HiddenEnd_HTML~5313,Path~@#
+. Calculate the coupling factors $k_{12}$ and $k_{21}$.
+#@HiddenBegin_HTML~5314,Path~@#
+<WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>
+The coupling factor $k_{21}$ is defined as "how much of the flux created by one coil ($\Phi_{11}$) crosses the other coil ($\Phi_{21}$) ":
+\begin{align*}
+k_{21} &= {{\Phi_{21}}\over{\Phi_{11}}}
+\end{align*}
+For this, we look at the circuit considering only one coil ("magnetic voltage source"), see <imgref ExImgNr12>. Based on the image, we see that the flux $\Phi_{11}$ divides into $\Phi_{S1}$ and $\Phi_{21}$ based on the magnetic resistance $\Phi_{\rm m,SS}$ and $\Phi_{\rm m,22}$.
+In step 2, we have calculated that $R_{\rm m,22}$ is twice $R_{\rm m,SS}$.  So from the incoming flux, only $1/3$ reaches $R_{\rm m,22}$ and - by this - "crosses the other coil".
+Therefore, the coupling factor $k_{21}$ is: $k_{21}= 1/3$.
+A similar approach leads to $k_{12}$ with $k_{12}= 1/4$.
+#@HiddenEnd_HTML~5314,Path~@#
+. Calculate the mutual inductions $M_{12}$, and $M_{21}$,
+#@HiddenBegin_HTML~5315,Path~@#
+\begin{align*}
+M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &&= {{1}\over{3}}\cdot{{400 \cdot 300}\over{ 486 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\
+M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &&= {{1}\over{4}}\cdot{{400 \cdot 300}\over{ 365 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\
+\end{align*}
+#@HiddenEnd_HTML~5315,Path~@#
 </WRAP></WRAP></panel>
-For symmetrical magnetic structures and $\mu_{\rm r} = \rm const.$ the following applies:
+#@TaskTitle_HTML@# 5.3.2 Wireles Charging #@TaskText_HTML@#
+For Electric vehicles sometimes wireless charging systems are employed. These use the principle of mutual inductance to transfer power from a charging pad on the ground to the vehicle's battery pack. \\
+This system consists of two coils: a transmitter coil embedded in the charging pad and a receiver coil mounted on the underside of the vehicle.
+  * The transmitter coil has a self-inductance of $L_{\rm T} = 200 ~\rm \mu H$.
+  * The receiver coil has a self-inductance of $L_{\rm R} = 150 ~\rm \mu H$.
+  * The mutual inductance between the coils at this distance is measured to be  $M = 20 ~\rm \mu H$ - when the vehicle is properly aligned over the charging pad.
+. Calculate the coupling coefficient $k$ between the transmitter and receiver coils when the vehicle is properly aligned over the charging pad.
+#@HiddenBegin_HTML~53211,Path ~@#
+The given self-inductances are $L_{\rm T} = L_{11}$, $L_{\rm R} = L_{22}$. \\
+By this, the following formula can be applied:
+\begin{align*}
+M = k \cdot \sqrt{L_{\rm T} \cdot L_{\rm R}}
+\end{align*}
+Therefore, $k$ is given as:
+\begin{align*}
+k = {{M}\over{ \sqrt{ L_{\rm T} \cdot L_{\rm R} } }}
+\end{align*}
+#@HiddenEnd_HTML~53211,Path ~@#
+. If the vehicle is misaligned by 10 cm from the center of the charging pad, the mutual inductance drops to $M = 12 ~\rm \mu H$. Calculate the new coupling coefficient in this misaligned position.
+#@TaskEnd_HTML@#
-  * the mutual inductances are equal: $M_{12} = M_{21} = M$
-  * the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$
-  * The resulting *total coupling* $k$ is given as \begin{align*} k = \sqrt{k_{12}\cdot k_{21}} \end{align*}
 ==== Effects in the electric Circuits ====
@@ Zeile 499: / Zeile 598: @@
 <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}}
-In this case, the **mutual induction added positively**.
+In this case, the **mutual induction is positiv $(M>0)$**.
 The formula of the shown circuitry is then:
@@ Zeile 513: / Zeile 612: @@
 <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP>
-In this case, the **mutual induction added negatively**.
+In this case, the **mutual induction is negativ $(M<0)$***.
 The formula of the shown circuitry is then:
 \begin{align*}
-u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\
+u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\
-u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\
+u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\
 \end{align*}
-<panel type="info" title="Task 5.3.1 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Task 5.3.3 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$.
 On the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$.
-  - The coils shall pass the currents with positive polarity (see the top image in <imgref ImgEx14>).    What is the magnetic flux in the coil?
+  - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ in the coil?
-  - The coils shall pass the currents with negative polarity (see the bottom image in <imgref ImgEx14>). What is the magnetic flux in the coil?
+  - The coils shall pass the currents with negative polarity (see the image **B** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm B}$ in the coil?
 <WRAP> <imgcaption ImgEx14 | toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg.svg}} </WRAP>
-<button size="xs" type="link" collapse="Solution_5_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_3_1_1_Result" collapsed="true">
+#@HiddenBegin_HTML~5_3_2s,Solution~@#
-  - $0.40 ~\rm  mVs$
+The resulting flux can be derived from a superposition of the individual fluxes $\Phi_1(I_1)$ and $\Phi_2(I_2)$, or alternatively by summing the magnetic voltages in the loop ($\sum_x \theta_x = 0$).
-  - $0.10 ~\rm mVs$
-</collapse>
+**Step 1 - Draw an equivalent magnetic circuit**
+Since there are no branches all of the core can be lumped to a single magnetic resistance (see <imgref ImgEx14circ>).
+<WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP>
+**Step 2 - Get the absolute values of the individual fluxes**
+Hopkinson's Law can be used here as a starting point. \\
+It connects the magnetic flux $\Phi$ and the magnetic voltage $\theta$ on the single magnetic resistor $R_\rm m$. \\
+It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\
+\begin{align*}
+\theta_x             &= R_{\rm m}                                  \cdot \Phi_x \\
+N_x \cdot I_x        &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} \cdot \Phi_x \\
+\rightarrow \Phi_x   &= \mu_0 \mu_{\rm r} {{A}\over{l}} \cdot N_x \cdot I_x
+                      = {{1}\over{R_{\rm m} }}          \cdot N_x \cdot I_x \\
+\end{align*}
+With the given values we get: $R_{\rm m} = 495 {\rm {kA}\over{Vs}}$
+**Step 3 - Get the signs/directions of the fluxes**
+The <imgref5_3_2_Solution> shows how to get the correct direction for every single flux by use of the right-hand rule. \\
+The fluxes have to be added regarding these directions and the given direction of the flux in question.
+<WRAP> <imgcaption 5_3_2_Solution| toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg_solution.svg}} </WRAP>
+Therefore, the formulas are
+\begin{align*}
+\Phi_{\rm A}   &= \Phi_{1} - \Phi_{2} \\
+               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  -  N_2 \cdot I_2 \right) \\
+               & = 0.25 ~{\rm mVs} - 0.15 ~{\rm mVs} \\
+\Phi_{\rm B}   &= \Phi_{1} + \Phi_{2} \\
+               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  +  N_2 \cdot I_2 \right) \\
+               & = 0.25 ~{\rm mVs} + 0.15 ~{\rm mVs}
+\end{align*}
+#@HiddenEnd_HTML~5_3_2s,Solution~@#
+#@HiddenBegin_HTML~5_3_2r,Result~@#
+  - $0.10 ~\rm mVs$
+  - $0.40 ~\rm mVs$
+#@HiddenEnd_HTML~5_3_2r,Result~@#
 </WRAP></WRAP></panel>
@@ Zeile 558: / Zeile 698: @@
 \boxed{W_m = {{1}\over{2}}L\cdot I^2 }
 \end{align*}
 ==== magnetic Energy of a magnetic Circuit ====
-With this formula also the stored energy in a magnetic circuit can be calculated.
+With this formula also the stored energy in a magnetic circuit can be calculated. For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: \begin{align*} \boxed{W_{\rm m} = {{1}\over{2}} \Psi \cdot I = {{1}\over{2}} {{\Psi^2}\over{L}}= {{1}\over{2}}{{\Phi^2 }\over{N^2 \cdot L}} = {{1}\over{2}} \Phi^2 \cdot R_{\rm m} = {{1}\over{2}}{{\theta^2 }\over{R_{\rm m}}}} \end{align*}
-For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit:
-\begin{align*}
-\boxed{W_{\rm m} = {{1}\over{2}}  \Psi\cdot I
-                 = {{1}\over{2}}{{\Psi^2 }\over{L}}}
-\end{align*}
 ==== magnetic Energy of a toroid Coil ====
@@ Zeile 693: / Zeile 832: @@
 <panel type="info" title="Task 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-The <imgref ImgTask01> and <imgref ImgTask01> show a shaded pole motor of a commercial oven.
+The <imgref ImgTask01> and <imgref ImgTask02> show a shaded pole motor of a commercial oven.
   * Find out how this motor works - explicitly: why is there a preferred direction of the motor?
@@ Zeile 717: / Zeile 856: @@
 The big difference there is, that there the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$.
 This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https://en.wikipedia.org/wiki/Magnetic_circuit#Resistance–reluctance_model|Magnetic_circuit}}, which was also presented in this chapter.
+==== Moving a Plate into an Air Gap ====
+<WRAP> <imgcaption ImgNr81 | a magnetic circuit with a moving plate> {{electrical_engineering_2:plate_in_airgap_50_.gif}} {{electrical_engineering_2:stiftt_in_luftspalt_2.3.gif}} </imgcaption> </WRAP>
 ==== Switch Reluctance Motor ====