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electrical_engineering_2:the_magnetostatic_field [2024/04/16 01:34] – [Bearbeiten - Panel] mexleadminelectrical_engineering_2:the_magnetostatic_field [2025/04/29 02:45] (aktuell) – mexleadmin
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   * $\vec{B}$-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 ====
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 </callout> </callout>
  
-Please have a look at the German contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.5.2/xcontent2.html|KIT-Brückenkurs >> 5.2.3 Lorentz-Kraft]]. Make sure that ''Gesamt'' is selected in the selection bar at the top. The last part "Magnetic field within matter" can be skipped.+Please have a look at the German contents (text, videos, exercises) on the page of the [[https://obkp.mint-kolleg.kit.edu/#OBKP_EDYNAMIK_LADUNGSBEWEGUNG|KIT-Brückenkurs >> Lorentz-Kraft]]. The last part "Magnetic field within matter" can be skipped.
  
 ===== 3.4 Matter in the Magnetic Field ===== ===== 3.4 Matter in the Magnetic Field =====
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 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 ====
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 \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 890: Zeile 897:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 3.2.3 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Task 3.2.3 Magnetic Potential Difference"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 <WRAP> <WRAP>
Zeile 900: Zeile 907:
 Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05>). Let $I_1 = 2~\rm A$ and $I_2 = 4.5~\rm A$ be valid. Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05>). Let $I_1 = 2~\rm A$ and $I_2 = 4.5~\rm A$ be valid.
  
-In each case, the magnetic voltage $V_{\rm m}$ along the drawn path is sought.+In each case, the magnetic potential difference $V_{\rm m}$ along the drawn path is sought.
  
  
 #@HiddenBegin_HTML~323100,Path~@# #@HiddenBegin_HTML~323100,Path~@#
  
-  * The magnetic voltage is given as the **sum of the current through the area within a closed path**.+  * The magnetic potential difference is given as the **sum of the current through the area within a closed path**.
   * The direction of the current and the path have to be considered with the righthand rule.   * The direction of the current and the path have to be considered with the righthand rule.
  
Zeile 911: Zeile 918:
  
 #@HiddenBegin_HTML~323102,Result a)~@# #@HiddenBegin_HTML~323102,Result a)~@#
-a) $\theta_\rm a = - I_1 = - 2~\rm A$ \\+a) $V_{\rm m,a} = - I_1 = - 2~\rm A$ \\
 #@HiddenEnd_HTML~323102,Result~@# #@HiddenEnd_HTML~323102,Result~@#
  
 #@HiddenBegin_HTML~323103,Result b)~@# #@HiddenBegin_HTML~323103,Result b)~@#
-b) $\theta_\rm b = - I_2 = - 4.5~\rm A$ \\+b) $V_{\rm m,b} = - I_2 = - 4.5~\rm A$ \\
 #@HiddenEnd_HTML~323103,Result~@# #@HiddenEnd_HTML~323103,Result~@#
  
 #@HiddenBegin_HTML~323104,Result c)~@# #@HiddenBegin_HTML~323104,Result c)~@#
-c) $\theta_\rm c = 0 $ \\+c) $V_{\rm m,c} = 0 $ \\
 #@HiddenEnd_HTML~323104,Result~@# #@HiddenEnd_HTML~323104,Result~@#
  
 #@HiddenBegin_HTML~323105,Result d)~@# #@HiddenBegin_HTML~323105,Result d)~@#
-d) $\theta_\rm d = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\+d) $V_{\rm m,d} = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\
 #@HiddenEnd_HTML~323105,Result~@# #@HiddenEnd_HTML~323105,Result~@#
  
 #@HiddenBegin_HTML~323106,Result e)~@# #@HiddenBegin_HTML~323106,Result e)~@#
-e) $\theta_\rm e = + I_1 = + 2~\rm A$ \\+e) $V_{\rm m,e} = + I_1 = + 2~\rm A$ \\
 #@HiddenEnd_HTML~323106,Result~@# #@HiddenEnd_HTML~323106,Result~@#
  
 #@HiddenBegin_HTML~323107,Result f)~@# #@HiddenBegin_HTML~323107,Result f)~@#
-f) $\theta_\rm f = 2 \cdot (- I_1) = - 4~\rm A$ \\+f) $V_{\rm m,f} = 2 \cdot (- I_1) = - 4~\rm A$ \\
 #@HiddenEnd_HTML~323107,Result~@# #@HiddenEnd_HTML~323107,Result~@#
  
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>
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 <panel type="info" title="Task 3.3.2 Electron in Plate Capacitor with magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 3.3.2 Electron in Plate Capacitor with magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-An electron shall move with the velocity $\vec{v}$ in a plate capacitor parallel to the plates, which have a potential difference $U$ and a distance $d$.  +An electron enters a plate capacitor on a trajectory parallel to the plates.  
-In the vacuum in between the plates acts additionally a magnetic field $\vec{B}$. +It shall move with the velocity $\vec{v}$ in the plate capacitor parallel to the plates.  
 +The plates have a potential difference $U$ and a distance $d$.  
 +In the vacuum in between the plates, there is also a magnetic field $\vec{B}$ present. 
  
 <WRAP> <WRAP>
Zeile 962: Zeile 1025:
  
 Calculate the velocity depending on the other parameters $\vec{v} = f(U, |\vec{B}|, d) $! Calculate the velocity depending on the other parameters $\vec{v} = f(U, |\vec{B}|, d) $!
 +
 +<button size="xs" type="link" collapse="Loesung_3_1_0_Tipps">{{icon>eye}} Path</button><collapse id="Loesung_3_1_0_Tipps" collapsed="true">
 +  * Think about the two forces on the electron from the fields - gravity is ignored. \\ Write their definitions down.
 +  * With which relationship between these two forces does the electron moves through the plate capacitor __parallel__ to the plates? \\ So the trajectory neither get bent up nor down.
 +  * What is the relationship between the $E$-field in the plate capacitor and the electric voltage $U$?
 +</collapse>
  
 <button size="xs" type="link" collapse="Loesung_3_1_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_3_1_2_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_3_1_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_3_1_2_Lösungsweg" collapsed="true">
Zeile 1011: Zeile 1080:
 <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|