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--- electrical_engineering_and_electronics_1:block15 [2025/11/22 13:39] – mexleadmin
+++ electrical_engineering_and_electronics_1:block15 [2025/11/23 11:47] (aktuell) – mexleadmin
@@ Zeile 188: / Zeile 188: @@
-~~PAGEBREAK~~ ~~CLEARFIX~~
+In the following chapters the magnetic field will also be divided into a "causer field" (a field caused by magnets/currents) and an "acting field" (a field acting on a magnet/current). \\
-=== Complex Geometry: toroidal Coil ===
+Since the $H$-field focuses on the causing current ($M  \sim I$) and does not give us a force depending on the value of second magnet/current, it is the **causer field**.
+===== Common pitfalls =====
+  * ...
+===== Exercises =====
+{{page>electrical_engineering_and_electronics:task_8a117vmnbbmsbfz3_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 $r_{\rm L}= 4~\rm mm$.
+. 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*}
+H(r) &= {{I_0}\over{2\pi \cdot r}} \\
+  &= {{100~\rm A}\over{2\pi \cdot 0.1 ~\rm m}} \\
+\end{align*}
+#@HiddenEnd_HTML~101,Solution ~@#
+#@HiddenBegin_HTML~102,Result~@#
+\begin{align*}
+            H(10~\rm cm) &= 159.15... ~\rm{{A}\over{m}} \\
+\rightarrow H(10~\rm cm) &= 159 ~\rm{{A}\over{m}}
+\end{align*}
+#@HiddenEnd_HTML~102,Result~@#
+. 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*}
+H(r) &= {{I}\over{2\pi \cdot r}}
+\end{align*}
+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*}
+\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*}
+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*}
+            H(3~\rm mm) &= 2984.1... ~\rm{{A}\over{m}} \\
+\rightarrow H(3~\rm mm) &= 3.0 ~\rm{{kA}\over{m}}
+\end{align*}
+#@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 BildNr04 | Magnetic field in a toroidal coil></imgcaption> \\
+<imgcaption BildNr01 | Conductor Arrangement>
-{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=ToroidalSolenoidField&d=streamlines&sl=none&st=1&ld=8&a1=77&a2=26&a3=100&rx=0&ry=0&rz=0&zm=1.8 700,450 noborder}}
+</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 ~@#
-A toroidal coil has a donut-like setup. This can be seen in <imgref BildNr04>. \\
+#@HiddenBegin_HTML~322102,Result~@#
-For reasons of symmetry, it shall get clear that the field lines form concentric circles. \\
+\begin{align*}
-The magnetic field in a toroidal coil is often considered as homogenious.
+            H &= 0 ~\rm{{A}\over{m}}
+\end{align*}
+#@HiddenEnd_HTML~322102,Result~@#
-In the following chapters the magnetic field will also be divided into a "causer field" (a field caused by magnets/currents) and an "acting field" (a field acting on a magnet/current). \\
+. Now, the current in one of the conductors is reversed. To which value does the magnetic field strength $H({\rm P})$ change?
-Since the $H$-field focuses on the cause ($M  \sim I$) and does not give us a force, it is the causer field.
+#@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~@#
-===== Common pitfalls =====
+#@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>
+{{page>electrical_engineering_and_electronics:task_76ksbc114ylxftfl_with_calculation&nofooter}}
-===== Exercises =====
-==== Worked examples ====
-...
 ===== Embedded resources =====
@@ Zeile 224: / Zeile 381: @@
 Superposition of magnetic fields
 {{youtube>xB8J-NaNYc4}}
+</WRAP>
+<WRAP column half>
+{{youtube>VkSQX5VpYpQ}}
 </WRAP>