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electrical_engineering_2:magnetic_circuits [2024/05/10 20:36] – mexleadminelectrical_engineering_2:magnetic_circuits [2025/05/27 07:56] (aktuell) – mexleadmin
Zeile 7: Zeile 7:
 <WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP> <WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP>
  
-In this chapter, we will investigate, how far we come with such an analogy and where it can be practically applied.+In this chapter, we will investigate how far we have come with such an analogy and where it can be practically applied.
  
 ===== 5.1 Linear Magnetic Circuits ===== ===== 5.1 Linear Magnetic Circuits =====
Zeile 17: Zeile 17:
   - The fields inside of airgaps are homogeneous. This is true for small air gaps.   - The fields inside of airgaps are homogeneous. This is true for small air gaps.
  
-One can calculate a lot of simple magnetic circuits when these assumptions and focusing on the average field line are applied.+One can calculate a lot of simple magnetic circuits when these assumptions are applied and focusing on the average field line are applied.
  
 <WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP> <WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP>
Zeile 140: Zeile 140:
 |Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) | |Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) |
  
-<panel type="info" title="Task 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$.  A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$. 
Zeile 164: Zeile 164:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions: Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions:
Zeile 192: Zeile 192:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Calculate the magnetic resistances of an airgap with the following dimensions: Calculate the magnetic resistances of an airgap with the following dimensions:
Zeile 208: Zeile 208:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Calculate the magnetic voltage necessary to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions: Calculate the magnetic voltage necessary to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions:
Zeile 224: Zeile 224:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Calculate the magnetic flux created on a magnetic resistance of $R_m = 2.5 \cdot 10^6 ~\rm {{1}\over{H}}$ with the following magnetic voltages: Calculate the magnetic flux created on a magnetic resistance of $R_m = 2.5 \cdot 10^6 ~\rm {{1}\over{H}}$ with the following magnetic voltages:
Zeile 242: Zeile 242:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A core shall consist of two parts as seen in <imgref ImgExNr08>. +A core shall consist of two parts, as seen in <imgref ImgExNr08>. 
 In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$. In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$.
  
Zeile 250: Zeile 250:
 The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$. The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$.
  
-The air gaps on the coupling joint between both parts have the length $\delta=0.23 ~\rm mm$ each. +The air gaps on the coupling joint between both parts have the length $\delta = 0.23 ~\rm mm$ each. 
 The permeability of the ferrite is $\mu_r = 3000$.  The permeability of the ferrite is $\mu_r = 3000$. 
 The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$ The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$
Zeile 270: Zeile 270:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 The magnetic circuit in <imgref ImgExNr09> passes a magnetic flux density of $0.4 ~\rm T$ given by an excitation current of $0.50 ~\rm A$ in $400$ windings.  The magnetic circuit in <imgref ImgExNr09> passes a magnetic flux density of $0.4 ~\rm T$ given by an excitation current of $0.50 ~\rm A$ in $400$ windings. 
Zeile 289: Zeile 289:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 The choke coil shown in <imgref ImgExNr10> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$.  The choke coil shown in <imgref ImgExNr10> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$. 
Zeile 458: Zeile 458:
  
  
-<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%>+<panel type="info" title="Exercise 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 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 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 $N_1 = 400$, $N_2=300$.
- 
-Calculate  
-  * the self inductions $L_{11}$, $L_{22}$,  
-  * the mutual inductions $M_{12}$, and $M_{21}$, 
-  * the coupling factors $k_{12}$ and $k_{21}$. 
  
 <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP> <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP>
  
-=== Step 1: Draw the problem as a network ===+1.  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> <WRAP> <imgcaption ExImgNr11 | Equivalent Network> </imgcaption> {{drawio>CoreWithTwoCoils2network.svg}} </WRAP>
 +#@HiddenEnd_HTML~5311,Result~@#
  
 +2. Calculate all magnetic resistances. Additionally, calculate the magnetic resistances $R_{\rm m1}$  and $R_{\rm m2}$ seen from the magnetic voltage source $1$ and $2$.
  
-=== Step 2: Calculate the magnetic resistances === +#@HiddenBegin_HTML~5312,Path~@#
-\\+
  
 <WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP> <WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP>
Zeile 492: Zeile 489:
 \end{align*} \end{align*}
  
-With the given geometry this leads to +With the given geometry, this leads to 
 \begin{align*}  \begin{align*} 
 R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\  R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\ 
Zeile 503: Zeile 500:
 \end{align*} \end{align*}
  
-== Step 3: Calculate the self-induction == +#@HiddenEnd_HTML~5312,Path~@# 
-\\+ 
 +3. 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. 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*}  \begin{align*} 
Zeile 510: Zeile 510:
 L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\  L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\ 
 \end{align*} \end{align*}
 +#@HiddenEnd_HTML~5313,Path~@#
 +
 +4. Calculate the coupling factors $k_{12}$ and $k_{21}$.
  
-== Step 4: Calculate the coupling factors == +#@HiddenBegin_HTML~5314,Path~@#
-\\+
 <WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP> <WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>
  
Zeile 526: Zeile 528:
  
 A similar approach leads to $k_{12}$ with $k_{12}= 1/4$. A similar approach leads to $k_{12}$ with $k_{12}= 1/4$.
 +#@HiddenEnd_HTML~5314,Path~@#
  
 +5. Calculate the mutual inductions $M_{12}$, and $M_{21}$,
 +
 +#@HiddenBegin_HTML~5315,Path~@#
 \begin{align*}  \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_{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\\ \\  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*} \end{align*}
 +#@HiddenEnd_HTML~5315,Path~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
 +
 +#@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.
 +
 +1. 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 ~@#
 +
 +2. 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@#
 +
 +
 +
 ==== Effects in the electric Circuits ==== ==== Effects in the electric Circuits ====
  
Zeile 548: Zeile 588:
  
   * the direction of the windings, and   * the direction of the windings, and
-  * the orientation/counting of the current in the circuit.+  * The orientation/counting of the current in the circuit.
  
 <WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP> <WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP>
Zeile 558: Zeile 598:
 <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}} <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}}
  
-In this case, the **mutual induction is positiv $(M>0)$**.+In this case, the **mutual induction is positive $(M>0)$**.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
Zeile 566: Zeile 606:
 \end{align*} \end{align*}
  
-=== negative Polarity ===+=== Negative Polarity ===
  
-The polarity is negative when only one current either flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13>).+The polarity is negative when only one current flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13>).
  
 <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP> <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP>
  
-In this case, the **mutual induction is negativ $(M<0)$***.+In this case, the **mutual induction is negative $(M<0)$***.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
Zeile 580: Zeile 620:
 \end{align*} \end{align*}
  
-<panel type="info" title="Task 5.3.2 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 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 $l=280 ~\rm mm$.  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 $I_2=300 ~\rm mA$.+At 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 image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ 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?
Zeile 596: Zeile 636:
 **Step 1 - Draw an equivalent magnetic circuit** **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>).+Since there are no branches, all of the core can be lumped into a single magnetic resistance (see <imgref ImgEx14circ>).
 <WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP> <WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP>
  
Zeile 602: Zeile 642:
  
 Hopkinson's Law can be used here as a starting point. \\ 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 $R_m$. \\+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 $\theta_x$. \\ It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\
  
Zeile 634: Zeile 674:
  
 #@HiddenBegin_HTML~5_3_2r,Result~@# #@HiddenBegin_HTML~5_3_2r,Result~@#
-  - $0.10 ~\rm  mVs$+  - $0.10 ~\rm mVs$
   - $0.40 ~\rm mVs$   - $0.40 ~\rm mVs$
 #@HiddenEnd_HTML~5_3_2r,Result~@# #@HiddenEnd_HTML~5_3_2r,Result~@#
Zeile 643: Zeile 683:
  
 The magnetic field of a coil stores magnetic energy.  The magnetic field of a coil stores magnetic energy. 
-The energy transfer from the electric circuit to the magnetic field is also the cause of the "current dampening" effect of the inductor. +The energy transfer from the electric circuit to the magnetic field is also the cause of the "current-damping" effect of the inductor. 
 The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by: The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by:
  
Zeile 669: Zeile 709:
 ==== magnetic Energy of a toroid Coil ==== ==== magnetic Energy of a toroid Coil ====
  
-The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. +The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. \\
 By this, the following formulas can be used:  By this, the following formulas can be used: 
-\begin{align*}  +  - For the magnetic voltage: $\theta = H \cdot l = N \cdot I $ \\  
-\theta = H \cdot l = N \cdot I \\  +  - For the magnetic flux: $\Phi   = B \cdot A $
-\Phi   = B \cdot A  +
-\end{align*}+
  
 With the above-mentioned formulas of the magnetic circuit, we get:  With the above-mentioned formulas of the magnetic circuit, we get: 
Zeile 695: Zeile 733:
 ==== generalized magnetic Energy ==== ==== generalized magnetic Energy ====
  
-The general term to find the magnetic energy (e.g. for inhomogeneous magnetic fields) is given by +The general term to find the magnetic energy (e.g., for inhomogeneous magnetic fields) is given by 
 \begin{align*}  \begin{align*} 
 W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\  W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\ 
Zeile 715: Zeile 753:
 \end{align*} \end{align*}
  
-Multiplying with $i$ and with $dt$ we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step.+Multiplying with $i$ and with $dt$, we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step.
  
 \begin{align*}  \begin{align*} 
Zeile 725: Zeile 763:
 \begin{align*}  \begin{align*} 
 dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}}          \cdot i \cdot {\rm d}t \\  dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}}          \cdot i \cdot {\rm d}t \\ 
- W_{\rm d} &=   \int                                                  {\rm d}W_{\rm m} \\ + W_{\rm m} &=   \int                                                  {\rm d}W_{\rm m} \\ 
            &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\             &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\ 
            &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\             &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\ 
Zeile 747: Zeile 785:
  
 In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown.  In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown. 
-Given by the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. +Given the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. 
 In the circuit in <imgref ImgNr14>, the inductor will experience increasing and decreasing current.  In the circuit in <imgref ImgNr14>, the inductor will experience increasing and decreasing current. 
 Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$. Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$.
Zeile 754: Zeile 792:
  
 The situation for integrating the area in the graph is also shown:  The situation for integrating the area in the graph is also shown: 
-For each step ${\rm d}B$ the corresponding value of the field strength $H$ has to be integrated. +For each step ${\rm d}B$, the corresponding value of the field strength $H$ has to be integrated. 
 For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is
  
Zeile 777: Zeile 815:
 As an example, the situation of the field strength $H(t_1)=H_1$ is shown.  As an example, the situation of the field strength $H(t_1)=H_1$ is shown. 
 This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow).  This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow). 
-The magnetization corresponds to an energy intake to the magnetic field and the demagnetization to an energy outtake.+The magnetization corresponds to an energy intake into the magnetic field, and the demagnetization to an energy outtake.
  
-Moving along the $H$-$B$-curve, one can see, that the energy intake and outtake are the same when coming back to a start point. +Moving along the $H$-$B$-curve, one can see that the energy intake and outtake are the same when coming back to a start point. 
 This means that the magnetization and demagnetization take place lossless in this example.  This means that the magnetization and demagnetization take place lossless in this example. 
-This is a good approximation for magnetically soft materials, however, does not work for magnetically hard materials like a permanent magnet. +This is a good approximation for magnetically soft materials; however, it does not work for magnetically hard materials like a permanent magnet. 
 Here, hysteresis also has to be considered. Here, hysteresis also has to be considered.
  
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 <WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP> <WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP>
  
-===== Tasks =====+===== Exercise =====
  
-<panel type="info" title="Task 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 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?   * Find out how this motor works - explicitly: why is there a preferred direction of the motor?
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The book [[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|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercise for beginning in the topic of magnetic circuits.+The book [[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|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercises for beginning in the topic of magnetic circuits.
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
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 An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}.  An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}. 
-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 $Q$.  +The big difference there is that 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.+This model can solve more questions; however, it 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 ==== ==== Moving a Plate into an Air Gap ====