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Nächste Überarbeitung 		Vorhergehende Überarbeitung
electrical_engineering_2:the_magnetostatic_field [2024/04/16 01:34]
mexleadmin [Bearbeiten - Panel] 		electrical_engineering_2:the_magnetostatic_field [2024/04/28 17:43] (aktuell)
mexleadmin
Zeile 464: Zeile 464:
   * $\vec{B}$-Field on index finger   * $\vec{B}$-Field on index finger
   * Current $I$ on thumb (direction with length $\vec{l}$)   * Current $I$ on thumb (direction with length $\vec{l}$)
 + \\ \\ 
 +<collapse id="openAni1" collapsed="true"><well> {{url>https://www.geogebra.org/material/iframe/id/apafjxqh/width/730/height/400/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false 450,250 noborder}} </well></collapse> 
 +<collapse id="openAni2" collapsed="false"> <button type="warning" collapse="openAni1">To view the animation: click here!</button> </collapse> 
 + \\
 <WRAP> <WRAP>
 <imgcaption BildNr06 | Force onto a single Conductor in a B-Field> <imgcaption BildNr06 | Force onto a single Conductor in a B-Field>
Zeile 470: Zeile 473:
 {{drawio>SingleConductorInBField.svg}} \\ {{drawio>SingleConductorInBField.svg}} \\
 </WRAP> </WRAP>
- 
 </callout> </callout>
 +
 +
  
 ==== Lorentz Law and Lorentz Force ==== ==== Lorentz Law and Lorentz Force ====
Zeile 644: Zeile 648:
  
 Explanation of diamagnetism and paramagnetism Explanation of diamagnetism and paramagnetism
-{{youtube>u36QpPvEh2c}}+<WRAP> 
 +<WRAP column half>{{ youtube>u36QpPvEh2c }}         </WRAP> 
 +<WRAP column half>{{ youtube>pniES3kKHvY?300x500 }} </WRAP> 
 +</WRAP>
  
 ==== Ferromagnetic Materials ==== ==== Ferromagnetic Materials ====
Zeile 743: Zeile 750:
 \begin{align*} \begin{align*}
 H(r) &= {{I_0}\over{2\pi \cdot r}} \\ H(r) &= {{I_0}\over{2\pi \cdot r}} \\
-  &= {{100~\rm A}\over{2\pi \cdot 0.01 ~\rm m}} \\+  &= {{100~\rm A}\over{2\pi \cdot 0.1 ~\rm m}} \\
 \end{align*} \end{align*}
  
Zeile 750: Zeile 757:
 #@HiddenBegin_HTML~102,Result~@# #@HiddenBegin_HTML~102,Result~@#
 \begin{align*} \begin{align*}
-            H(10~\rm cm) &= 1591.5... ~\rm{{A}\over{m}} \\  +            H(10~\rm cm) &= 159.15... ~\rm{{A}\over{m}} \\  
-\rightarrow H(10~\rm cm) &= 1.6 ~\rm{{kA}\over{m}} +\rightarrow H(10~\rm cm) &= 159 ~\rm{{A}\over{m}} 
 \end{align*} \end{align*}
  
Zeile 943: Zeile 950:
 A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface.  A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface. 
  
-  - For comparison, the same flux density shall be created on the inside of a toroidal coil with $10'000$ windings and a toroidal diameter for the average field line of $d = 1~\rm m$. \\ How much current $I$ is necessary for one of the windings of the toroidal coil? +1. For comparison, the same flux density shall be created inside a toroidal coil with $10'000$ windings and a toroidal diameter for the average field line of $d = 1~\rm m$. \\ How much current $I$ is necessary for one of the windings of the toroidal coil? 
-  - What would be the current $I_{\rm Fe}$, when a iron core with $\varepsilon_{\rm Fe,r} = 10'000$?+ 
 +#@HiddenBegin_HTML~331100,Path~@# 
 + 
 +  * The $B$-field can be calculated by the $H$-field. 
 +  * 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 already known) 
 +  * The current is number of windings times $I$. 
 + 
 +#@HiddenEnd_HTML~331100,Path~@# 
 + 
 +#@HiddenBegin_HTML~331101,Solution~@# 
 + 
 +The $B$-field is given as: 
 +\begin{align*} 
 +B &= \mu \cdot H \\ 
 +  &= \mu \cdot {{I \cdot N}\over{l}} \\ 
 +\end{align*} 
 + 
 +This can be rearranged to the current $I$: 
 +\begin{align*} 
 +I &= {{B \cdot l}\over{\mu \cdot N}} \\ 
 +  &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{4\pi\cdot 10^-7 {\rm{Vs}\over{Am}}  \cdot 10'000}}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331101,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~331102,Result~@# 
 +\begin{align*} 
 +            I &= 95.49... ~\rm A \\  
 +\rightarrow I &= 95.5 ~\rm A  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331102,Result~@# 
 + 
 +2. What would be the current $I_{\rm Fe}$, when a iron core with $\varepsilon_{\rm Fe,r} = 10'000$? 
 + 
 + 
 +#@HiddenBegin_HTML~331201,Solution~@# 
 + 
 +Now $\mu$ has to be given as $\mu_r \cdot \mu_0$: 
 + 
 +This can be rearranged to the current $I$: 
 +\begin{align*} 
 +I &= {{B \cdot l}\over{\mu \cdot N}} \\ 
 +  &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{10'000 \cdot 4\pi\cdot 10^-7 {\rm{Vs}\over{Am}}  \cdot 10'000}}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331201,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~331202,Result~@# 
 +\begin{align*} 
 +            I &= 0.009549... ~\rm A \\  
 +\rightarrow I &= 9.55 ~\rm mA  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~331202,Result~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 1011: Zeile 1072:
 <WRAP group> <WRAP half column> <WRAP group> <WRAP half column>
  
-<quizlib id="quiz" rightanswers="[['a0'],['a2'], ['a2'], ['a0'], ['a1'], ['a2']]" submit="Check Answers">+<quizlib id="quiz" rightanswers="[['a0'],['a2'], ['a1'], ['a2'], ['a1'], ['a2']]" submit="Check Answers">
 <question title="1. Which hand can be used to infer magnetic field direction from currents?" type="radio"> <question title="1. Which hand can be used to infer magnetic field direction from currents?" type="radio">
 The right hand| The right hand|