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electrical_engineering_and_electronics_1:block17 [2025/11/23 16:26] – mexleadminelectrical_engineering_and_electronics_1:block17 [2025/12/06 13:45] (aktuell) – mexleadmin
Zeile 14: Zeile 14:
  
 For checking your understanding please do the following exercises: For checking your understanding please do the following exercises:
-  * ...+  * Exercise E2 Toroidal Coil 
 +  * Task 3.2.1 Magnetic Field Strength around a horizontal straight Conductor 
 +  * Task 3.3.2 Electron in Plate Capacitor with magnetic Field
  
 ===== 90-minute plan ===== ===== 90-minute plan =====
Zeile 41: Zeile 43:
  
 To derive the forces, we do a step back to the images of field lines. \\ To derive the forces, we do a step back to the images of field lines. \\
-In <imgref BildNr01> a) the field lines of a single current-carrying wire is shown. \\ +In <imgref BildNr01> 1) the field lines of a single current-carrying wire is shown. \\ 
-<imgref BildNr01> b) depicts the homogenous field of a coil.+<imgref BildNr01> 2) depicts the homogenous field of a coil.
  
-When a current-carrying wire is within the homogenous field, we get the superimposed picture of both fields. \\+When a current-carrying wire is within the homogenous field, we get the superimposed picture of both fields ( <imgref BildNr01> 3) ). \\
 This leads to an enrichment of magnetic field on the left and an depletion on the right. \\ This leads to an enrichment of magnetic field on the left and an depletion on the right. \\
 With the knowledge, that the field lines usually do not like to stay next to each other, one can conclude that there will be a force to the right. With the knowledge, that the field lines usually do not like to stay next to each other, one can conclude that there will be a force to the right.
Zeile 132: Zeile 134:
 </WRAP> </WRAP>
  
 +\\ \\
 <WRAP group> <WRAP group>
-<WRAP column third>+<WRAP>
  
 === Diamagnetic Materials === === Diamagnetic Materials ===
Zeile 163: Zeile 166:
 </WRAP> </WRAP>
  
-</WRAP><WRAP column third>+</WRAP><WRAP >
  
 === Paramagnetic Materials === === Paramagnetic Materials ===
Zeile 191: Zeile 194:
 </WRAP> </WRAP>
  
-</WRAP><WRAP column third>+</WRAP><WRAP>
  
 === Ferromagnetic Materials === === Ferromagnetic Materials ===
  
   * Ferromagnetic materials strengthen the magnetic field strongly, compared to the vacuum.   * Ferromagnetic materials strengthen the magnetic field strongly, compared to the vacuum.
-  * The strengthening can create a field multiple times stronger than in a vacuum.+  * The strengthening can create a field multiple times stronger than in a vacuum (see <imgref BildNr53>). 
   * For ferromagnetic materials applies $\mu_{\rm r}  \gg 1$    * For ferromagnetic materials applies $\mu_{\rm r}  \gg 1$ 
   * Ferromagnetic materials are characterized by the magnetization curve (see <imgref BildNr24>)   * Ferromagnetic materials are characterized by the magnetization curve (see <imgref BildNr24>)
Zeile 204: Zeile 207:
     * In order to eliminate the stored field the counteracting **coercive field strength** $H_{\rm C}$ (also called coercivity) has to be applied.     * In order to eliminate the stored field the counteracting **coercive field strength** $H_{\rm C}$ (also called coercivity) has to be applied.
     * The **saturation flux density** $B_{\rm sat}$ is the maximum possible magnetic flux density (at the maximum possible field strength $H_{\rm sat}$)     * The **saturation flux density** $B_{\rm sat}$ is the maximum possible magnetic flux density (at the maximum possible field strength $H_{\rm sat}$)
 +
 +<WRAP>
 +<imgcaption BildNr53 | Magnetic field in paramagnetic materials></imgcaption>
 +{{drawio>Ferromagnets.svg}}
 +</WRAP>
 +
  
 <WRAP> <WRAP>
Zeile 212: Zeile 221:
 </WRAP></WRAP> </WRAP></WRAP>
  
-==== Applications of the Lorentz Force ==== +==== Applications of the Lorentz Force: Two parallel Conductors ===
- +
-We want to apply the Lorentz force for two common situations. +
- +
-<WRAP group><WRAP half column> +
- +
-=== Two parallel Conductors ===+
  
 The Lorentz force can be applied to two parallel conductors. \\  The Lorentz force can be applied to two parallel conductors. \\ 
Zeile 245: Zeile 248:
 \end{align*} \end{align*}
  
-</WRAP><WRAP half column> 
-=== Moving single Charge  === 
  
-The true Lorentz force is not the force on the whole conductor but the single force onto an (elementary) charge. \\ 
-To find this force the previous force onto a conductor can be used as a start. However, the formula will be investigated infinitesimally for small parts ${\rm d} \vec{l}$ of the conductor: 
  
 +===== Common pitfalls =====
 +  * ...
 +
 +===== Exercises =====
 +
 +{{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}}
 +{{page>electrical_engineering_and_electronics:task_ti7loik6aurfewkb_with_calculation&nofooter}}
 +{{page>electrical_engineering_and_electronics:task_w3m7fo4hjahkzogw_with_calculation&nofooter}}
 +{{page>electrical_engineering_and_electronics:task_elndbo3xwi2klxuu_with_calculation&nofooter}}
 +
 +
 +
 +
 +<panel type="info" title="Task 3.2.1 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +The current $I_0 = 100~\rm A$ flows in a long straight conductor with a round cross-section.
 +The conductor shall have constant electric properties everywhere. 
 +The radius of the conductor is $r_{\rm L}= 4~\rm mm$.
 +
 +1. What is the magnetic field strength $H_1$ at a point $P_1$, which is __outside__ the conductor at a distance of $r_1 = 10~\rm cm$ from the conductor axis?
 +
 +#@HiddenBegin_HTML~100,Path~@#
 +
 +  * The $H$-field is given as: the current $I$ through an area divided by the "specific" length $l$ of the closed path around the area. This shall give you the formula (when not in already known)
 +  * The relevant current is the given $I_0$.
 +
 +#@HiddenEnd_HTML~100,Path~@#
 +
 +#@HiddenBegin_HTML~101,Solution~@#
 +
 +The $H$-field is given as:
 \begin{align*} \begin{align*}
-\vec{{\rm d}F}_{\rm L} = I \cdot {\rm d}\vec{l} \times \vec{B}+H(r) &= {{I_0}\over{2\pi \cdot r}} \\ 
 +  &= {{100~\rm A}\over{2\pi \cdot 0.1 ~\rm m}} \\
 \end{align*} \end{align*}
  
-The current is now substituted by $I = {\rm d}Q/{\rm d}t$, where ${\rm d}Q$ is the small charge packet in the length $\vec{{\rm d}l}$ of the conductor.+#@HiddenEnd_HTML~101,Solution ~@#
  
 +#@HiddenBegin_HTML~102,Result~@#
 \begin{align*} \begin{align*}
-\vec{{\rm d}F}_{\rm L} = {{{\rm d}Q}\over{{\rm d}t}} \cdot {\rm d}\vec{l} \times \vec{B}+            H(10~\rm cm) &= 159.15... ~\rm{{A}\over{m}} \\  
 +\rightarrow H(10~\rm cm) &= 159 ~\rm{{A}\over{m}} 
 \end{align*} \end{align*}
  
-Mathematically not quite correct, but in a physical way true the following rearrangement can be done:+#@HiddenEnd_HTML~102,Result~@# 
 + 
 +2. What is the magnetic field strength $H_2$ at a point $P_2$, which is __inside__ the conductor at a distance of $r_2 = 3~\rm mm$ from the conductor axis? 
 + 
 +#@HiddenBegin_HTML~200,Path~@# 
 + 
 +  * Again, the $H$-field is given as: the current $I$ through an area divided by the "specific" length $l$ of the closed path around the area.  
 +  * Here, the relevant current is **not** the given one. There is only a fraction of the current flowing through the part of the conductor inside the $r_2$ 
 + 
 +#@HiddenEnd_HTML~200,Path~@# 
 + 
 +#@HiddenBegin_HTML~201,Solution~@#
  
 +The $H$-field is given as:
 \begin{align*} \begin{align*}
-\vec{{\rm d}F}_{\rm L} &= {{{\rm d}Q \cdot   {\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ +H(r) &= {{I}\over{2\pi \cdot r}} 
-                       &= {\rm d}Q   \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ +
-                       &= {\rm d}Q   \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\+
 \end{align*} \end{align*}
  
-Here, the part ${{{\rm d}\vec{l}}\over{{\rm d}t}}$ represents the speed $\vec{v}$ of the small charge packet ${\rm d}Q$.+But now $I$ is not $I_0$ anymore, but only a fraction, so $\Delta I$.  
 +$I_0$ is evenly distributed over the cross-section $A$ of the conductor.  
 +The cross-sectional area is given as $A= r^2 \cdot \pi$
  
 +So the current $\Delta I$ is given as: current divided by the full area and then times the fractional area:
 \begin{align*} \begin{align*}
-\vec{{\rm d}F}_{\rm L} &= {\rm d}Q \cdot \vec{v} \times \vec{B} +\Delta I &= I_0 \cdot {{r_2^2 \cdot \pi}\over{r_{\rm L}^2 \cdot \pi}} \\ 
 +         &= I_0 \cdot {{r_2^2          }\over{r_{\rm L}^2          }} 
 \end{align*} \end{align*}
  
-The **Lorenz Force** on a finite charge packet is the integration:+Therefore, the $H$-field is: 
 +\begin{align*} 
 +H(r) &= {{\Delta I}\over{2\pi \cdot r_2}}  
 +     &&= {{I_0 \cdot {{ r_2^2}\over{r_{\rm L}^2}} }\over{2\pi \cdot r_2}} \\ 
 +     &= {{I_0 \cdot {{ r_2}\over{r_{\rm L}^2}} }\over{2\pi}}  
 +     &&= {{1}\over{2\pi}} I_0 \cdot {{ r_2}\over{r_{\rm L}^2}}   
 +\end{align*} 
 + 
 + 
 +#@HiddenEnd_HTML~201,Solution ~@#
  
 +#@HiddenBegin_HTML~202,Result~@#
 \begin{align*} \begin{align*}
-\boxed{\vec{F}_{\rm L} = Q \cdot \vec{v} \times \vec{B}}+            H(3~\rm mm) &= 2984.1... ~\rm{{A}\over{m}} \\  
 +\rightarrow H(3~\rm mm) &= 3.0 ~\rm{{kA}\over{m}} 
 \end{align*} \end{align*}
  
 +#@HiddenEnd_HTML~202,Result~@#
  
 +</WRAP></WRAP></panel>
  
-<callout icon="fa fa-exclamation" color="red" title="Notice:">+~~PAGEBREAK~~~~CLEARFIX~~
  
-  * A charge $Q$ moving with a velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force of $\vec{F_{\rm L}}$. 
-  * The direction of the force is given by the right-hand rule. 
  
-</callout>+<panel type="info" title="Task 3.3.1 magnetic Flux Density"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-</WRAP></WRAP>+A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface. 
  
 +1. For comparison, the same flux density shall be created inside a toroidal coil with $10'000$ windings and a toroidal diameter for the average field line of $d = 1~\rm m$. \\ How much current $I$ is necessary for one of the windings of the toroidal coil?
  
-Please have a look at the German contents (text, videos, exercises) on the page of the [[https://obkp.mint-kolleg.kit.edu/#OBKP_EDYNAMIK_LADUNGSBEWEGUNG|KIT-Brückenkurs >> Lorentz-Kraft]]. The last part "Magnetic field within matter" can be skipped.+#@HiddenBegin_HTML~331100,Path~@#
  
 +  * The $B$-field can be calculated by the $H$-field.
 +  * The $H$-field is given as: the current $I$ through an area divided by the "specific" length $l$ of the closed path around the area. This shall give you the formula (when not already known)
 +  * The current is number of windings times $I$.
  
-===== Common pitfalls ===== +#@HiddenEnd_HTML~331100,Path~@#
-  * ...+
  
-===== Exercises =====+#@HiddenBegin_HTML~331101,Solution~@#
  
-{{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}} +The $B$-field is given as: 
-{{page>electrical_engineering_and_electronics:task_ti7loik6aurfewkb_with_calculation&nofooter}} +\begin{align*} 
-{{page>electrical_engineering_and_electronics:task_w3m7fo4hjahkzogw_with_calculation&nofooter}} +B &= \mu \cdot H \\ 
-{{page>electrical_engineering_and_electronics:task_elndbo3xwi2klxuu_with_calculation&nofooter}}+  &= \mu \cdot {{I \cdot N}\over{l}} \\ 
 +\end{align*} 
 + 
 +This can be rearranged to the current $I$: 
 +\begin{align*} 
 +I &= {{B \cdot l}\over{\mu \cdot N}} \\ 
 +  &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{4\pi\cdot 10^{-7} {\rm{Vs}\over{Am}}  \cdot 10'000}}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331101,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~331102,Result~@# 
 +\begin{align*} 
 +            I &= 95.49... ~\rm A \\  
 +\rightarrow I &= 95.5 ~\rm A  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331102,Result~@# 
 + 
 +2. What would be the current $I_{\rm Fe}$, when a iron core with $\varepsilon_{\rm Fe,r} = 10'000$? 
 + 
 + 
 +#@HiddenBegin_HTML~331201,Solution~@# 
 + 
 +Now $\mu$ has to be given as $\mu_r \cdot \mu_0$: 
 + 
 +This can be rearranged to the current $I$: 
 +\begin{align*} 
 +I &= {{B \cdot l}\over{\mu \cdot N}} \\ 
 +  &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{10'000 \cdot 4\pi\cdot 10^{-7} {\rm{Vs}\over{Am}}  \cdot 10'000}}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331201,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~331202,Result~@# 
 +\begin{align*} 
 +            I &= 0.009549... ~\rm A \\  
 +\rightarrow I &= 9.55 ~\rm mA  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331202,Result~@# 
 + 
 +</WRAP></WRAP></panel> 
 + 
 +<wrap #task3_3_2 /> 
 + 
 +<panel type="info" title="Task 3.3.2 Electron in Plate Capacitor with magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +An electron enters a plate capacitor on a trajectory parallel to the plates.  
 +It shall move with the velocity $\vec{v}$ in the plate capacitor parallel to the plates.  
 +The plates have a potential difference $U$ and a distance $d$.  
 +In the vacuum in between the plates, there is also a magnetic field $\vec{B}$ present.  
 + 
 +<WRAP> 
 +<imgcaption BildNr71 | Electron in B- and E-Field> 
 +</imgcaption> 
 +{{drawio>ElectronInBandEfield.svg}} \\ 
 +</WRAP> 
 + 
 +Calculate the velocity depending on the other parameters $\vec{v} = f(U, |\vec{B}|, d) $! 
 + 
 +<button size="xs" type="link" collapse="Loesung_3_1_0_Tipps">{{icon>eye}} Path</button><collapse id="Loesung_3_1_0_Tipps" collapsed="true"> 
 +  * Think about the two forces on the electron from the fields - gravity is ignored. \\ Write their definitions down. 
 +  * With which relationship between these two forces does the electron moves through the plate capacitor __parallel__ to the plates? \\ So the trajectory neither get bent up nor down. 
 +  * What is the relationship between the $E$-field in the plate capacitor and the electric voltage $U$? 
 +</collapse> 
 + 
 +<button size="xs" type="link" collapse="Loesung_3_1_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_3_1_2_Lösungsweg" collapsed="true"> 
 + 
 +Within the electric field, the Coulomb force acts on the electron: 
 + 
 +\begin{align*} 
 +\vec{F}_C = q_e \cdot \vec{E} 
 +\end{align*} 
 + 
 +Within the magnetic field, also the Lorentz force acts on the electron: 
 + 
 +\begin{align*} 
 +\vec{F}_L = q_e \cdot \vec{v} \times \vec{B} 
 +\end{align*} 
 + 
 +The absolute value of both forces must be equal to compensate each other:  
 + 
 +\begin{align*} 
 +|\vec{F}_C|          &= |\vec{F}_L|\\ 
 +|q_e \cdot \vec{E}|  &= |q_e \cdot \vec{v} \times \vec{B}| \\ 
 +q_e \cdot |\vec{E}|  &= q_e \cdot |\vec{v} \times \vec{B}| \\ 
 +          |\vec{E}|  &=           |\vec{v} \times \vec{B}| \\ 
 +\end{align*} 
 + 
 +Since $\vec{v}$ is perpendicular to $\vec{B}$ the right side is equal to $|\vec{v}| \cdot |\vec{B}| = v \cdot B$. \\ 
 +Additionally, for the plate capacitor $|\vec{E}|= U/d$. \\ 
 + 
 +Therefore, it leads to the following: 
 + 
 +\begin{align*} 
 +          {{U}\over{d}}  &=           v \cdot B \\ 
 +          v              &=          {{U}\over{B \cdot d}}  
 +\end{align*} 
 + 
 +</collapse> 
 + 
 +<button size="xs" type="link" collapse="Loesung_3_1_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_3_1_2_Endergebnis" collapsed="true"> 
 +\begin{align*} 
 +v = {{U}\over{B\cdot d}} 
 +\end{align*} 
 + 
 +</collapse> 
 + 
 +</WRAP></WRAP></panel> 
 + 
 + **<fs large>Task 1</fs>** 
 + 
 +<WRAP group> <WRAP half column> 
 + 
 +<quizlib id="quiz" rightanswers="[['a0'],['a2'], ['a1'], ['a2'], ['a1'], ['a2']]" submit="Check Answers"> 
 +<question title="1. Which hand can be used to infer magnetic field direction from currents?" type="radio"> 
 +The right hand| 
 +The left hand 
 +</question> 
 +<question title="2. In the derivation from 1. how are the fingers to be assigned?" type="radio">  
 +Thumb for current direction, remaining fingers for magnetic field direction |  
 +Thumb for magnetic field direction, remaining fingers for current direction |  
 +both possibilities are correct  
 +</question> 
 +<question title="3.  Two conductors carrying current are parallel and close to each other. The current in both is flowing in the same direction. What force effect can be seen?" type="radio">  
 +none |  
 +The conductors attract |  
 +The conductors repel 
 +</question> 
 +<question title="4. Two conductors carrying current are at right angles to each other. Current flows through both of them. What force effect can be seen?" type="radio">  
 +none |  
 +The conductors attract |  
 +The conductors repel 
 +</question> 
 +<question title="5. What is the magnetic field inside the earth or a permanent magnet?" type="radio">  
 +from the magnetic north pole to the south pole |  
 +from the magnetic south pole to the north pole |  
 +the inside is free of field 
 +</question> 
 +<question title="6. At which location of a current-carrying coil are the field lines densest?" type="radio">  
 +at the magnetic north pole |  
 +at the magnetic south pole |  
 +inside the coil |  
 +at both poles  
 +</question> 
 + 
 +</quizlib> 
 + 
 +</WRAP> <WRAP half column> 
 + 
 +++++Tip for 1| 
 +For the current, you use which hand? 
 +++++ 
 + 
 +++++Tip for 2| 
 +  * Imagine a coil with a winding pictorially, or draw it on. 
 +  * Now think of a generated field through this to it. What direction must the current flow, that causes the field? Does this fit the rule of thumb? 
 +  * Then try it the other way round: If a current is given, where do the field lines go in and where out? What poles are created there? 
 +++++ 
 + 
 +++++Tip for 3| See 3rd video. 
 +  * Picture the two wires, or draw them on. 
 +  * In which direction would the outer field run in each case? 
 +  * The field is a linear vector field. So the total field can be created from several individual fields by adding them together. Does adding the field in between make it larger, or smaller? 
 +++++ 
 + 
 +++++Tip for 4| 
 +  * First imagine the parallel wires again. What happens when the current flows in the same direction and what happens when the current flows in opposite directions? Are the resulting forces equal in magnitude? 
 +  * The reversal of the direction of the current can now also be produced by turning the wire instead of changing the current - just so that the wires are perpendicular to each other in the meantime when turning. 
 +  * With parallel wires and different current directions, the amount-wise same force arises. So, this is also true for every angle in between (in detail given by integration of the force over single wire pieces). 
 +  * But then there must be a point at which the force becomes 0. 
 +++++ 
 + 
 +++++Tip for 5| 
 +  * The magnetic field lines must be closed. 
 +  * Compare the field curve between the coil and permanent magnet. 
 +++++ 
 +++++Tip for 6| 
 +  * In video 1 you can see the course outside and inside the coil. 
 +++++
  
 +</WRAP> </WRAP>
  
 ===== Embedded resources ===== ===== Embedded resources =====
-A living insect ("diamagnet") floats in a very strong magnetic field+Please have a look at the German contents (text, videos, exercises) on the page of the [[https://obkp.mint-kolleg.kit.edu/#OBKP_EDYNAMIK_LADUNGSBEWEGUNG|KIT-Brückenkurs >> Lorentz-Kraft]]. The last part "Magnetic field within matter" can be skipped. 
 +\\ \\ \\ 
 +<WRAP column half> 
 +The rotating flux (density) in the stator of a motor, source/CC-BY 3.0 see {{https://commons.wikimedia.org/wiki/File:2HP_1500RPM_induction_motor_no_slip.gif|wikipedia}} 
 +{{electrical_engineering_and_electronics_1:2hp_1500rpm_induction_motor_no_slip.gif}} 
 +</WRAP> 
 + 
 +<WRAP column half> 
 +A living insect ("diamagnet") floats in a very strong magnetic field \\
 {{youtube>KlJsVqc0ywM?start=45}} {{youtube>KlJsVqc0ywM?start=45}}
 +</WRAP>
  
-Explanation of diamagnetism and paramagnetism +\\ \\ 
-<WRAP> +<WRAP column half> 
-<WRAP column half>{{ youtube>u36QpPvEh2c }}         </WRAP> +Explanation of diamagnetism and paramagnetism \\ 
-<WRAP column half>{{ youtube>pniES3kKHvY?300x500 }} </WRAP>+{{ youtube>u36QpPvEh2c }}         
 </WRAP> </WRAP>
 +<WRAP column half>
 +In Oxygen magnetic? \\
 +{{ youtube>pniES3kKHvY?300x500 }} 
 +</WRAP>
 +
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~