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--- electrical_engineering_and_electronics_1:block19 [2025/12/02 17:12] – mexleadmin
+++ electrical_engineering_and_electronics_1:block19 [2025/12/02 19:42] (aktuell) – mexleadmin
@@ Zeile 42: / Zeile 42: @@
 In this chapter, we will investigate how far we have come with such an analogy and where it can be practically applied.
-==== Flux ====
-The magnetic flux is a measurement of the amount of magnetic field lines through a given surface area, as seen in <imgref ImgNr03>. The magnetic flux is the amount of magnetic field lines cutting through a surface area defined by the surface vector $\vec{A}$. If the angle between the $\vec{A} = A\cdot \vec{n}$ and magnetic field vector $\vec{B}$ is parallel or antiparallel, as shown in the diagram, the absolute value of the magnetic flux is the highest possible value given the values of the area and the magnetic field.
-<WRAP> <imgcaption ImgNr03 | The magnetic flux is the amount of magnetic field lines cutting through a surface area A.> </imgcaption> {{drawio>MagneticFieldAndFlux.svg}} </WRAP>
-This definition leads to a magnetic flux similar to the electric flux studied earlier:
-\begin{align*}
-\Phi_{\rm m} = \iint_A \vec{B} \cdot {\rm d} \vec{A}
-\end{align*}
-<imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume.
-Since the magnetic field is a source-free vortex field, the flux over a closed area is always zero:
-\begin{align*}
-\Phi_{\rm m} = {\rlap{\Large \rlap{\int} \int} \; \Huge\circ}_{\!\!\! A} \vec{B} \cdot {\rm d} \vec{A} = 0
-\end{align*}
-By this, it can be shown that any open surface bounded by the circuit in question can be used to evaluate $\Phi_{\rm m}$.
-For example, $\Phi_{\rm m}$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure.
-<WRAP> <imgcaption ImgNr05 | A circuit bounding an arbitrary open surface S. > </imgcaption> {{drawio>FluxThroughSurfaces.svg}} </WRAP>
-The SI unit for magnetic flux is the $\rm Weber$ (Wb), \begin{align*} [\Phi_{\rm m}] = [B] \cdot [A] = 1 ~\rm  T \cdot m^2 = 1 ~ Wb \end{align*}
-If the $B$ field is homogenous and $B$ is perpendicular to the surface $S$, then the fomula simplifies to:
-\begin{align*}
-\Phi_{\rm m} = B \cdot A
-\end{align*}
-Based on this definition, the magnetic field unit is occasionally expressed as Weber per square meter ($\rm Wb/m^2$) instead of teslas.
-In many practical applications, the circuit of interest consists of a number $N$ of tightly wound turns (similar to <imgref ImgNr06>).
-Each turn experiences the same magnetic flux $\Phi_{\rm m}$.
-Therefore, the net magnetic flux through the circuits is $N$ times the flux through one turn, and Faraday’s law is written as
-==== Linked Flux ====
-When looking at the magnetic field in a coil multiple windings capture the passing flux, see <imgref ImgNr14> (a).
-It can also be interpreted in such a way that the flux is going through the closed surface of the circuit multiple times (in picture (b)).
-<WRAP> <imgcaption ImgNr14 | Example for a linked Flux> </imgcaption> {{drawio>LinkedFlux.svg}} </WRAP>
-The **linked flux** $\Psi$ is defined as the resulting flux given by the sum of the partial fluxes of the closed circuit.
-\begin{align*}
-\boxed{ \Psi = \sum_{i=1}^n \Phi_{i} }
-\end{align*}
-For the case of a coil with $N$ windings, this leads to:
-\begin{align*}
-\boxed{ \Psi = N \cdot \Phi }
-\end{align*}
-<callout icon="fa fa-exclamation" color="red" title="Notice:">
-When calculating the force, use the flux within the material ($\Phi$). \\
-When investigating effects in the coil, use the linked flux ($\Psi$). \\
-</callout>
 ==== Basics for Linear Magnetic Circuits ====
@@ Zeile 298: / Zeile 230: @@
 {{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}}
+<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 $l_R = 300 ~\rm mm$.
+The $N=400$ windings are evenly distributed along the circumference.
+The diameter on the cross-section of the plastic ring is $d = 10 ~\rm mm$. In the windings, a current of $I=500 ~\rm mA$ is flowing.
+Calculate
+  - the magnetic field strength $H$ in the middle of the ring cross-section.
+  - the magnetic flux density $B$ in the middle of the ring cross-section.
+  - the magnetic resistance $R_{\rm m}$ of the plastic ring.
+  - the magnetic flux $\Phi$.
+<button size="xs" type="link" collapse="Solution_5_1_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_1_1_Result" collapsed="true">
+  - $H = 667 ~\rm {{A}\over{m}}$
+  - $B = 0.84 ~\rm mT$
+  - $R_m = 3 \cdot 10^9 ~\rm {{1}\over{H}}$
+  - $\Phi = 66 ~\rm nVs$
+</collapse>
+</WRAP></WRAP></panel>
+<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:
+  - $l=35.8~\rm cm$, $d=1.90~\rm cm$
+  - $l=11.1~\rm cm$, $d=1.50~\rm cm$
+#@HiddenBegin_HTML~5_1_2s,Solution~@#
+The magnetic resistance is given by:
+\begin{align*}
+\ R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}
+\end{align*}
+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>
+<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:
+  - $\delta=0.5~\rm mm$, $A=10.2~\rm cm^2$
+  - $\delta=3.0~\rm mm$, $A=11.9~\rm cm^2$
+<button size="xs" type="link" collapse="Solution_5_1_3_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_3_1_Result" collapsed="true">
+  - $3.9\cdot 10^5 ~\rm {{1}\over{H}}$
+  - $2.0\cdot 10^6 ~\rm {{1}\over{H}}$
+</collapse>
+</WRAP></WRAP></panel>
+<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:
+  - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$
+  - $\delta=5.0~\rm mm$, $A=7.1~\rm cm^2$
+<button size="xs" type="link" collapse="Solution_5_1_4_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_4_1_Result" collapsed="true">
+  - $\theta = 1.5\cdot 10^3 ~\rm A$, or $1000$ windings with $1.5~\rm A$
+  - $\theta = 2.8\cdot 10^3 ~\rm A$, or $1000$ windings with $2.8~\rm A$
+</collapse>
+</WRAP></WRAP></panel>
+<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:
+  - $\theta = 35   ~\rm A$
+  - $\theta = 950  ~\rm A$
+  - $\theta = 2750 ~\rm A$
+<button size="xs" type="link" collapse="Solution_5_1_5_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_5_1_Result" collapsed="true">
+  - $\Phi =14  ~\rm µVs$
+  - $\Phi =0.38~\rm mVs$
+  - $\Phi =1.1 ~\rm mVs$
+</collapse>
+</WRAP></WRAP></panel>
+<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>.
+In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$.
+The cross sections are $A_1=530 ~\rm mm^2$ and $A_2=460 ~\rm mm^2$.
+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 permeability of the ferrite is $\mu_r = 3000$.
+The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$
+<WRAP> <imgcaption ImgExNr08 | Two-parted ferrite Core> </imgcaption> {{drawio>TwoPartedCoil.svg}} </WRAP>
+  - Draw the lumped circuit of the magnetic system
+  - Calculate all magnetic resistances $R_{\rm m,i}$
+  - Calculate the flux in the circuit
+<button size="xs" type="link" collapse="Solution_5_1_6_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_6_1_Result" collapsed="true">
+  - -
+  - magnetic resistances: $R_{\rm m,1} = 100 \cdot 10^3 ~\rm {{1}\over{H}}$, $R_{\rm m,1} = 75 \cdot 10^3 ~\rm {{1}\over{H}}$, $R_{{\rm m},\delta} = 400 \cdot 10^3 ~\rm {{1}\over{H}}$
+  - magnetic flux: $\Phi = 0.80 ~\rm mVs$
+</collapse>
+</WRAP></WRAP></panel>
+<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.
+At position $\rm A-B$, an air gap will be inserted. After this, the same flux density will be reached with $3.70 ~\rm A$
+<WRAP> <imgcaption ImgExNr09 | Example of a magnetic circuit> </imgcaption> {{drawio>ExMagncirc01.svg}} </WRAP>
+  - Calculate the length of the airgap $\delta$ with the simplification $\mu_{\rm r} \gg 1$
+  - Calculate the length of the airgap $\delta$ exactly with $\mu_{\rm r} = 1000$
+<button size="xs" type="link" collapse="Solution_5_1_7_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_7_1_Result" collapsed="true">
+  - $\delta = 4.02(12) ~\rm mm$
+  - $\delta = 4.02(52) ~\rm mm$
+</collapse>
+</WRAP></WRAP></panel>
+<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 number of windings shall be $N$ and the current through a single winding $I$.
+<WRAP> <imgcaption ImgExNr10 | Example for a Choke Coil> </imgcaption> {{drawio>ChokeCoilEx1.svg}} </WRAP>
+  - Draw the lumped circuit of the magnetic system
+  - Calculate all magnetic resistances $R_{{\rm m},i}$
+  - Calculate the partial fluxes in all the legs of the circuit
+</WRAP></WRAP></panel>
+<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 $l=280 ~\rm mm$.
+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 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>
+#@HiddenBegin_HTML~5_3_2s,Solution~@#
+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 ($\sum_x \theta_x = 0$).
+**Step 1 - Draw an equivalent magnetic circuit**
+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>
+**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>
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