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electrical_engineering_2:the_magnetostatic_field [2024/04/28 17:43] – mexleadminelectrical_engineering_2:the_magnetostatic_field [2025/04/29 02:45] (aktuell) – mexleadmin
Zeile 520: Zeile 520:
 </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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 </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 907: 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 918: 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 971: Zeile 971:
 \begin{align*} \begin{align*}
 I &= {{B \cdot l}\over{\mu \cdot N}} \\ 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}} +  &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{4\pi\cdot 10^{-7} {\rm{Vs}\over{Am}}  \cdot 10'000}} 
 \end{align*} \end{align*}
  
Zeile 994: Zeile 994:
 \begin{align*} \begin{align*}
 I &= {{B \cdot l}\over{\mu \cdot N}} \\ 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}} +  &= {{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*} \end{align*}
  
Zeile 1013: Zeile 1013:
 <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 1023: 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">