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electrical_engineering_and_electronics_1:block15 [2025/11/22 13:21] – mexleadminelectrical_engineering_and_electronics_1:block15 [2025/11/23 11:47] (aktuell) – mexleadmin
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-====== Block 15 — Magnets and their Effects ======+====== Block 15 — Magnets, their Effects and Fieldline Images ======
  
 ===== Learning objectives ===== ===== Learning objectives =====
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 $r_{\rm L}= 4~\rm mm$. 
 + 
 +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 223: Zeile 381:
 Superposition of magnetic fields Superposition of magnetic fields
 {{youtube>xB8J-NaNYc4}} {{youtube>xB8J-NaNYc4}}
 +</WRAP>
 +
 +<WRAP column half>
 +{{youtube>VkSQX5VpYpQ}}
 </WRAP> </WRAP>