Unterschiede

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--- electrical_engineering_and_electronics_1:block15 [2025/11/02 22:07] – angelegt mexleadmin
+++ electrical_engineering_and_electronics_1:block15 [2025/11/23 11:47] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== Block 15 — Magnets and their Effects ======
+====== Block 15 — Magnets, their Effects and Fieldline Images ======
 ===== Learning objectives =====
@@ Zeile 60: / Zeile 60: @@
   - So magnets experience a force in the vicinity of other magnets.
   - A compass is a small rotating "sample" magnet and is also called a magnetic needle. This sample magnet can thus represent the effect of a magnet. This is also similar to the sample charge of the electric field.
-  - The naming of the magnetic poles was done by the part of the compass which points to the geographic north pole. The reason for this is that the magnetic south pole is found at the geographic north pole.
+  - The naming of the magnetic poles was done by the part of the compass which points to the geographic north pole. The reason for this is that the magnetic south pole is found at the geographic north pole, see <imgref BildNr02> leftside.
   - Magnetic poles are not isolatable. Even the smallest fraction of a magnet shows either no magnetism or both north and south poles.
+Interestingly, even non-magnetized, ferromagnetic materials experience a force effect in the magnetic field. A nail - which is not a magnet itself - is attracted by a permanent magnet. This even happens independently of the magnetic pole. This also explains the visualization about iron filings (= small ferromagnetic parts), see <imgref BildNr02> rightside. Also here there is a force effect and a torque, which aligns the iron filings. The visible field seems to form field lines here.
 <WRAP>
@@ Zeile 68: / Zeile 70: @@
 </WRAP>
-Interestingly, even non-magnetized, ferromagnetic materials experience a force effect in the magnetic field. A nail - which is not a magnet itself - is attracted by a permanent magnet. This even happens independently of the magnetic pole. This also explains the visualization about iron filings (= small ferromagnetic parts), see <imgref BildNr02>. Also here there is a force effect and a torque, which aligns the iron filings. The visible field seems to form field lines here.
 <callout icon="fa fa-exclamation" color="red" title="Notice:">
@@ Zeile 139: / Zeile 140: @@
 ~~PAGEBREAK~~~~CLEARFIX~~
 ==== Superposition of the magnetostatic Field ====
+In the electric field, the field line density was a measure of the strength of the field. This is also used for the magnetic field. Looking at the simulations in Falstad (e.g. <imgref BildNr105>) with this understanding, one notices an inconsistency: contrary to the relationship just given, the field line density in the Falstad simulation __**not**__ indicates the strength of the field. A realistic simulation is shown in <imgref BildNr106> for comparison, which makes the difference clear: the field is stronger near the conductor. Thus the field line density must also be stronger there.
+<WRAP>
+<imgcaption BildNr106 | correct Picture of Magnetic Field Lines around a Conductor>
+</imgcaption> \\
+{{url>https://www.geogebra.org/material/iframe/id/fy0yrGYK/width/500/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 700,350 noborder}}
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+<callout icon="fa fa-exclamation" color="red" title="Attention:">
+  * The density of the field lines is a measure of the field strength.
+  * The simulation in Falstad cannot represent this in this way. \\ Here the field strength is coded by the color intensity (dark green = low field strength, light green to white = high field strength).
+</callout>
 Before the magnetic field strength will be considered in more detail, the simulation and superposition of the magnetic field will be discussed in more detail here.
@@ Zeile 144: / Zeile 159: @@
 Magnetostatic fields can be superposed, just like electrostatic fields. This allows the fields of several current-carrying lines to be combined into a single one.
 This trick is used in the following chapter to examine the magnetic field in more detail.
+Below, the magnetic field of a single current-carrying conductor is shown. This was already derived in the previous chapter by symmetry considerations. The representation in the simulation can be simplified a bit here to see the conditions more clearly: Currently, the field lines are displayed in 3D, which is done by selecting ''Display: Field Lines'' and ''No Slicing''. If you change the selection to ''Show Z Slice'' instead of ''No Slicing'', you can switch to a 2D display. In this display, small compass needles can also show the magnetic field. To do this, select ''Display: Field Vectors'' instead of ''Display: Field Lines''. In addition, a "magnetic sample", i.e. a moving compass, can be found at the mouse pointer in the 2D display.
 ~~PAGEBREAK~~~~CLEARFIX~~
@@ Zeile 151: / Zeile 169: @@
 </WRAP>
-On the right side, the magnetic field of a single current-carrying conductor is shown. This was already derived in the previous chapter by symmetry considerations. The representation in the simulation can be simplified a bit here to see the conditions more clearly: Currently, the field lines are displayed in 3D, which is done by selecting ''Display: Field Lines'' and ''No Slicing''. If you change the selection to ''Show Z Slice'' instead of ''No Slicing'', you can switch to a 2D display. In this display, small compass needles can also show the magnetic field. To do this, select ''Display: Field Vectors'' instead of ''Display: Field Lines''. In addition, a "magnetic sample", i.e. a moving compass, can be found at the mouse pointer in the 2D display.
+If there is another current-carrying conductor near the first conductor, the fields overlap. In the simulation below, the current of both conductors is directed in the same direction. The field between the conductors overlaps just enough to weaken. This can also be deduced by previous knowledge if just the middle point between both conductors is considered: There, for the left conductor the right-hand rule results in a vector directed towards the observer. For the right conductor, it results in a vector that is directed away from the observer. These just cancel each other out. Further outward field lines go around both conductors. The North and south poles here are not fixed localized toward the outside.
 ~~PAGEBREAK~~~~CLEARFIX~~
@@ Zeile 159: / Zeile 177: @@
 </WRAP>
-If there is another current-carrying conductor near the first conductor, the fields overlap. In the simulation below, the current of both conductors is directed in the same direction. The field between the conductors overlaps just enough to weaken. This can also be deduced by previous knowledge if just the middle point between both conductors is considered: There, for the left conductor the right-hand rule results in a vector directed towards the observer. For the right conductor, it results in a vector that is directed away from the observer. These just cancel each other out. Further outward field lines go around both conductors. The North and south poles here are not fixed localized toward the outside.
+If, on the other hand, the current in the second conductor is directed in the opposite direction to the current in the first conductor, the picture changes: Here there is a reinforcing superposition between the two conductors.
+Using the nomenclature from the previous chapter, it is also possible to assign north and south poles locally. Towards the outside, one pole appears to be located in front of the two conductors and another one behind.
+in both simulations, the distances between the conductors can also be changed using the ''Line Separation'' slider. What do you notice in each case when the two lines are brought close together?
 ~~PAGEBREAK~~~~CLEARFIX~~
 <WRAP>
@@ Zeile 166: / Zeile 187: @@
 <WRAP>
-If, on the other hand, the current in the second conductor is directed in the opposite direction to the current in the first conductor, the picture changes: Here there is a reinforcing superposition between the two conductors.
-Using the nomenclature from the previous chapter, it is also possible to assign north and south poles locally. Towards the outside, one pole appears to be located in front of the two conductors and another one behind.
-in both simulations, the distances between the conductors can also be changed using the ''Line Separation'' slider. What do you notice in each case when the two lines are brought close together?
+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). \\
+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**.
-~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Magnetic Field Strength part 1: toroidal Coil ====
-<WRAP>
-<imgcaption BildNr04 | Magnetic field in a toroidal coil></imgcaption> \\
-{{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}}
-<WRAP>
-So far the magnetic field was defined quite pragmatically by the effect on a compass.
-For a deeper analysis of the magnetic field, the field is now to be considered again - as with the electric field - from __two__ directions.
-The magnetic field will also be considered a "causer field" (a field caused by magnets) and an "acting field" (a field acting on a magnet).
-This chapter will first discuss the acting magnetic field. For this, it is convenient to consider the effects inside a toroidal coil (= donut-like setup).
-This can be seen in <imgref BildNr04>. For reasons of symmetry, it is also clear here that the field lines form concentric circles.
-In an experiment, a magnetic needle inside the toroidal coil is now to be aligned perpendicular to the field lines.
+===== Common pitfalls =====
-Then, the magnetic field will generate a torque $M$ which tries to align the magnetic needle in the field direction.
+  * ...
+===== 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 BildNr21 | Toroidal Coil>
+<imgcaption BildNr01 | Conductor Arrangement>
 </imgcaption>
-{{drawio>toroidalcoil.svg}} \\
+{{drawio>Task1LadderArrangement.svg}} \\
 </WRAP>
-It now follows:
+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 $I = 2 ~\rm A$.
-  - $M = {\rm const.} \neq f(\varphi)$: For the same distance from the axis of symmetry, the torque $M$ is independent of the angle $\phi$.
-  - $M \sim I$: The stronger the current flowing through a winding, the stronger the effect, i.e. the stronger the torque.
-  - $M \sim N$: The greater the number $N$ of windings, the stronger the torque $M$.
-  - $M \sim {1 \over l}$ : The smaller the average coil circumference $l$ the greater the torque. The average coil circumference $l$ is equal to the **mean magnetic path length** (=average field line length).
-To summarize:
+. 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*}
-M \sim {{I \cdot N}\over{l}}
+H &= {{I}\over{2\pi \cdot r}} \\
+  &= {{2~\rm A}\over{2\pi \cdot 0.02 ~\rm m}} \\
 \end{align*}
-The **magnetic field strength** $H$ inside the toroidal coil is given as:
+  * 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*}
-\boxed{H ={{I \cdot N}\over{l}}} \quad \quad | \quad \text{applies to toroidal coil only}
+            H &= 0 ~\rm{{A}\over{m}}
 \end{align*}
-For the unit of the magnetic field strength $H$ we get $[H] = {{[I]}\over{[l]}}= \rm 1~{{A}\over{m}}$
+#@HiddenEnd_HTML~322102,Result~@#
-==== Magnetic Field Strength part 2: straight conductor ====
+. Now, the current in one of the conductors is reversed. To which value does the magnetic field strength $H({\rm P})$ change?
-The previous derivation from the toroidal coil is now used to derive the field strength around a long, straight conductor. For a single conductor the part $N \cdot I$ of the formula can be reduced to $ N \cdot I = 1 \cdot I = I$ since there is only one conductor. For the toroidal coil, the magnetic field strength was given by this current(s) divided by the (average) field line length. Because of the (same rotational) symmetry, this is also true for the single conductor. Also here the field line length has to be taken into account.
+#@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.
-The length of a field line around the conductor is given by the distance $r$ of the field line from the conductor: $l = l(r) = 2 \cdot \pi \cdot r$. \\ For the magnetic field strength of the single conductor we then get:
+#@HiddenEnd_HTML~322200,Path~@#
+#@HiddenBegin_HTML~322201,Solution~@#
+The $H$-field of the single reversed conductor $I_3$ is given as:
 \begin{align*}
-\boxed{H ={I\over{l}} = {{I}\over{2 \cdot \pi \cdot r}}} \quad \quad | \quad \text{applies only to the long, straight conductor}
+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 BildNr105 | magnetic Field Lines around a Conductor>
+<imgcaption BildNr02 | Conductor Arrangement>
-</imgcaption> \\
+</imgcaption>
-{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=InverseRotational&d=streamlines&sl=z&st=19&ld=12&rx=0&ry=0&rz=0&zm=1.8 700,350 noborder}}
+{{drawio>Solution2.svg}} \\
 </WRAP>
-In the electric field, the field line density was a measure of the strength of the field. This is also used for the magnetic field. Looking at the simulations in Falstad (e.g. <imgref BildNr105>) with this understanding, one notices an inconsistency: contrary to the relationship just given, the field line density in the Falstad simulation __**not**__ indicates the strength of the field. A realistic simulation is shown in <imgref BildNr106> for comparison, which makes the difference clear: the field is stronger near the conductor. Thus the field line density must also be stronger there.
+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 $H(I_3)$.
-<WRAP>
-<imgcaption BildNr106 | correct Picture of Magnetic Field Lines around a Conductor>
-</imgcaption> \\
-{{url>https://www.geogebra.org/material/iframe/id/fy0yrGYK/width/500/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 700,350 noborder}}
-</WRAP>
-~~PAGEBREAK~~ ~~CLEARFIX~~
+#@HiddenEnd_HTML~322201,Solution ~@#
-<callout icon="fa fa-exclamation" color="red" title="Attention:">
-  * The density of the field lines is a measure of the field strength.
-  * The simulation in Falstad cannot represent this in this way. Here the field strength is coded by the color intensity (dark green = low field strength, light green to white = high field strength).
-</callout>
+#@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>
-===== Common pitfalls =====
+{{page>electrical_engineering_and_electronics:task_76ksbc114ylxftfl_with_calculation&nofooter}}
-  * ...
-===== Exercises =====
-==== Worked examples ====
-...
 ===== Embedded resources =====
@@ Zeile 263: / Zeile 381: @@
 Superposition of magnetic fields
 {{youtube>xB8J-NaNYc4}}
+</WRAP>
+<WRAP column half>
+{{youtube>VkSQX5VpYpQ}}
 </WRAP>