Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- electrical_engineering_and_electronics_1:block17 [2025/12/02 02:00] – [Materials] mexleadmin
+++ electrical_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:
-  * ...
+  * 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 =====
@@ Zeile 134: / Zeile 136: @@
 \\ \\
 <WRAP group>
-<WRAP column third>
+<WRAP>
 === Diamagnetic Materials ===
@@ Zeile 164: / Zeile 166: @@
 </WRAP>
-</WRAP><WRAP column third>
+</WRAP><WRAP >
 === Paramagnetic Materials ===
@@ Zeile 192: / Zeile 194: @@
 </WRAP>
-</WRAP><WRAP column third>
+</WRAP><WRAP>
 === Ferromagnetic Materials ===
@@ Zeile 208: / Zeile 210: @@
 <WRAP>
 <imgcaption BildNr53 | Magnetic field in paramagnetic materials></imgcaption>
-{{drawio>Paramagnets.svg}}
+{{drawio>Ferromagnets.svg}}
 </WRAP>
@@ Zeile 219: / Zeile 221: @@
 </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. \\
@@ Zeile 252: / Zeile 248: @@
 \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:
-\begin{align*}
-\vec{{\rm d}F}_{\rm L} = I \cdot {\rm d}\vec{l} \times \vec{B}
-\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.
-\begin{align*}
-\vec{{\rm d}F}_{\rm L} = {{{\rm d}Q}\over{{\rm d}t}} \cdot {\rm d}\vec{l} \times \vec{B}
-\end{align*}
-Mathematically not quite correct, but in a physical way true the following rearrangement can be done:
-\begin{align*}
-\vec{{\rm d}F}_{\rm L} &= {{{\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} \\
-                       &= {\rm d}Q   \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\
-\end{align*}
-Here, the part ${{{\rm d}\vec{l}}\over{{\rm d}t}}$ represents the speed $\vec{v}$ of the small charge packet ${\rm d}Q$.
-\begin{align*}
-\vec{{\rm d}F}_{\rm L} &= {\rm d}Q \cdot \vec{v} \times \vec{B}
-\end{align*}
-The **Lorenz Force** on a finite charge packet is the integration:
-\begin{align*}
-\boxed{\vec{F}_{\rm L} = Q \cdot \vec{v} \times \vec{B}}
-\end{align*}
-<callout icon="fa fa-exclamation" color="red" title="Notice:">
-  * 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>
-</WRAP></WRAP>
@@ Zeile 390: / Zeile 340: @@
 #@HiddenEnd_HTML~202,Result~@#
-</WRAP></WRAP></panel>
-<panel type="info" title="Task 3.2.2 Superposition"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP>
-<imgcaption BildNr01 | Conductor Arrangement>
-</imgcaption>
-{{drawio>Task1LadderArrangement.svg}} \\
-</WRAP>
-Three long straight conductors are arranged in a vacuum to lie at the vertices of an equilateral triangle (see <imgref BildNr01>). The radius of the circumcircle is $r = 2 ~\rm cm$; the current is given by $I = 2 ~\rm A$.
-. What is the magnetic field strength $H({\rm P})$ at the center of the equilateral triangle?
-#@HiddenBegin_HTML~322100,Path~@#
-  * The formula for a single wire can calculate the field of a single conductor.
-  * For the resulting field, the single wire fields have to be superimposed.
-  * Since it is symmetric the resulting field has to be neutral.
-#@HiddenEnd_HTML~322100,Path~@#
-#@HiddenBegin_HTML~322101,Solution~@#
-In general, the $H$-field of the single conductor is given as:
-\begin{align*}
-H &= {{I}\over{2\pi \cdot r}} \\
-  &= {{2~\rm A}\over{2\pi \cdot 0.02 ~\rm m}} \\
-\end{align*}
-  * However, even without calculation, the constant distance between point $\rm P$ and the three conductors dictates, that the $H$-field has a similar magnitude.
-  * By the symmetry of the conductor, the angles of the $H$-field vectors are defined and evenly distributed on the revolution:
-<WRAP>
-<imgcaption BildNr02 | Conductor Arrangement>
-</imgcaption>
-{{drawio>Solution1.svg}} \\
-</WRAP>
-#@HiddenEnd_HTML~322101,Solution ~@#
-#@HiddenBegin_HTML~322102,Result~@#
-\begin{align*}
-            H &= 0 ~\rm{{A}\over{m}}
-\end{align*}
-#@HiddenEnd_HTML~322102,Result~@#
-. Now, the current in one of the conductors is reversed. To which value does the magnetic field strength $H({\rm P})$ change?
-#@HiddenBegin_HTML~322200,Path~@#
-  * Now, the formula for a single wire has to be used to calculate the field of a single conductor.
-  * For the resulting field, the single wire fields again have to be superimposed.
-  * The symmetry and the result of question 1 give a strong hint about how much stronger the resulting field has to be compared to the field of the reversed one.
-#@HiddenEnd_HTML~322200,Path~@#
-#@HiddenBegin_HTML~322201,Solution~@#
-The $H$-field of the single reversed conductor $I_3$ is given as:
-\begin{align*}
-H(I_3) &= {{I}\over{2\pi \cdot r}} \\
-  &= {{2~\rm A}\over{2\pi \cdot 0.02 ~\rm m}} \\
-\end{align*}
-Once again, one can try to sketch the situation of the field vectors:
-<WRAP>
-<imgcaption BildNr02 | Conductor Arrangement>
-</imgcaption>
-{{drawio>Solution2.svg}} \\
-</WRAP>
-Therefore, it is visible, that the resulting field is twice the value of $H(I_3)$: \\
-The vectors of $H(I_1)$ plus $H(I_2)$ had in the task 1 just the length of $H(I_3)$.
-#@HiddenEnd_HTML~322201,Solution ~@#
-#@HiddenBegin_HTML~322202,Result~@#
-\begin{align*}
-            H &= 31.830... ~\rm{{A}\over{m}} \\
-\rightarrow H &= 31.8 ~\rm{{A}\over{m}} \\
-\end{align*}
-#@HiddenEnd_HTML~322202,Result~@#
-</WRAP></WRAP></panel>
-<panel type="info" title="Task 3.2.3 Magnetic Potential Difference"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP>
-<imgcaption BildNr05 | different trajectories around current-carrying conductors>
-</imgcaption>
-{{drawio>Task3MagneticFieldCurrentFlowingConductor.svg}} \\
-</WRAP>
-Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05>). Let $I_1 = 2~\rm A$ and $I_2 = 4.5~\rm A$ be valid.
-In each case, the magnetic potential difference $V_{\rm m}$ along the drawn path is sought.
-#@HiddenBegin_HTML~323100,Path~@#
-  * The magnetic potential difference is given as the **sum of the current through the area within a closed path**.
-  * The direction of the current and the path have to be considered with the righthand rule.
-#@HiddenEnd_HTML~323100,Path~@#
-#@HiddenBegin_HTML~323102,Result a)~@#
-a) $V_{\rm m,a} = - I_1 = - 2~\rm A$ \\
-#@HiddenEnd_HTML~323102,Result~@#
-#@HiddenBegin_HTML~323103,Result b)~@#
-b) $V_{\rm m,b} = - I_2 = - 4.5~\rm A$ \\
-#@HiddenEnd_HTML~323103,Result~@#
-#@HiddenBegin_HTML~323104,Result c)~@#
-c) $V_{\rm m,c} = 0 $ \\
-#@HiddenEnd_HTML~323104,Result~@#
-#@HiddenBegin_HTML~323105,Result d)~@#
-d) $V_{\rm m,d} = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\
-#@HiddenEnd_HTML~323105,Result~@#
-#@HiddenBegin_HTML~323106,Result e)~@#
-e) $V_{\rm m,e} = + I_1 = + 2~\rm A$ \\
-#@HiddenEnd_HTML~323106,Result~@#
-#@HiddenBegin_HTML~323107,Result f)~@#
-f) $V_{\rm m,f} = 2 \cdot (- I_1) = - 4~\rm A$ \\
-#@HiddenEnd_HTML~323107,Result~@#
 </WRAP></WRAP></panel>
@@ Zeile 734: / Zeile 551: @@
 ===== Embedded resources =====
 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
+A living insect ("diamagnet") floats in a very strong magnetic field \\
 {{youtube>KlJsVqc0ywM?start=45}}
+</WRAP>
 \\ \\
-Explanation of diamagnetism and paramagnetism
+<WRAP column half>
-<WRAP>
+Explanation of diamagnetism and paramagnetism \\
-<WRAP column half>{{ youtube>u36QpPvEh2c }}         </WRAP>
+{{ youtube>u36QpPvEh2c }}
-<WRAP column half>{{ youtube>pniES3kKHvY?300x500 }} </WRAP>
 </WRAP>
+<WRAP column half>
+In Oxygen magnetic? \\
+{{ youtube>pniES3kKHvY?300x500 }}
+</WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~