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electrical_engineering_2:the_magnetostatic_field [2023/09/19 23:51] – mexleadminelectrical_engineering_2:the_magnetostatic_field [2025/04/29 02:45] (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 516: 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 =====
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 725: Zeile 732:
 <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%> <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 $I = 100~\rm A$ flows in a long straight conductor with a round cross-section. The radius of the conductor is $r_{\rm L}= 4~\rm mm$.+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~@#
  
-  * 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? 
-  * 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? 
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 740: Zeile 819:
 </WRAP> </WRAP>
  
-Three long straight conductors are arranged in a vacuum so that they 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 $I = 2 ~\rm A$.+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> 
 +<imgcaption BildNr02 | Conductor Arrangement> 
 +</imgcaption> 
 +{{drawio>Solution1.svg}} \\ 
 +</WRAP> 
 +#@HiddenEnd_HTML~322101,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~322102,Result~@# 
 +\begin{align*} 
 +            H &= 0 ~\rm{{A}\over{m}}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~322102,Result~@# 
 + 
 +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~@# 
 + 
 +#@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~@# 
  
-  - What is the magnetic field strength $H({\rm P})$ at the center of the equilateral triangle? 
-  - Now, the current in one of the conductors is reversed. To which value does the magnetic field strength $H({\rm P})$ change? 
 </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 756: 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~@# 
 + 
 +  * 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. 
 + 
 +#@HiddenEnd_HTML~323100,Path~@# 
 + 
 +#@HiddenBegin_HTML~323102,Result a)~@# 
 +a) $V_{\rm m,a} = - I_1 = - 2~\rm A$ \\ 
 +#@HiddenEnd_HTML~323102,Result~@# 
 + 
 +#@HiddenBegin_HTML~323103,Result b)~@# 
 +b) $V_{\rm m,b} = - I_2 = - 4.5~\rm A$ \\ 
 +#@HiddenEnd_HTML~323103,Result~@# 
 + 
 +#@HiddenBegin_HTML~323104,Result c)~@# 
 +c) $V_{\rm m,c} = 0 $ \\ 
 +#@HiddenEnd_HTML~323104,Result~@# 
 + 
 +#@HiddenBegin_HTML~323105,Result d)~@# 
 +d) $V_{\rm m,d} = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\ 
 +#@HiddenEnd_HTML~323105,Result~@# 
 + 
 +#@HiddenBegin_HTML~323106,Result e)~@# 
 +e) $V_{\rm m,e} = + I_1 = + 2~\rm A$ \\ 
 +#@HiddenEnd_HTML~323106,Result~@# 
 + 
 +#@HiddenBegin_HTML~323107,Result f)~@# 
 +f) $V_{\rm m,f} = 2 \cdot (- I_1) = - 4~\rm A$ \\ 
 +#@HiddenEnd_HTML~323107,Result~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 767: 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 776: 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 786: 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 835: 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|