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electrical_engineering_and_electronics_1:block15 [2025/11/02 22:49] – mexleadminelectrical_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 ===== ===== Learning objectives =====
Zeile 60: Zeile 60:
   - So magnets experience a force in the vicinity of other magnets.   - 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.   - 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.   - 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> <WRAP>
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 </WRAP> </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:"> <callout icon="fa fa-exclamation" color="red" title="Notice:">
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-~~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 $r_{\rm L}= 4~\rm mm$. 
 + 
 +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*} 
 +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~@# 
 + 
 +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*} 
 +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> <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 $I = 2 ~\rm A$. 
 + 
 +1. 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> <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~@#
  
-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).+2. 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~@#
  
-===== 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 ===== ===== Embedded resources =====
Zeile 222: Zeile 381:
 Superposition of magnetic fields Superposition of magnetic fields
 {{youtube>xB8J-NaNYc4}} {{youtube>xB8J-NaNYc4}}
 +</WRAP>
 +
 +<WRAP column half>
 +{{youtube>VkSQX5VpYpQ}}
 </WRAP> </WRAP>