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electrical_engineering_2:the_magnetostatic_field [2023/05/02 12:52] – [Tabelle] mexleadminelectrical_engineering_2:the_magnetostatic_field [2025/04/29 02:45] (aktuell) – mexleadmin
Zeile 1: Zeile 1:
-====== 3. The magnetostatic Field ======+====== 3 The magnetostatic Field ======
  
 <callout> <callout>
Zeile 207: Zeile 207:
  
 <WRAP> <WRAP>
-<imgcaption BildNr106 | correct Picture of magnetic Field Lines around a Conductor>+<imgcaption BildNr106 | correct Picture of Magnetic Field Lines around a Conductor>
 </imgcaption> \\ </imgcaption> \\
 {{url>https://www.geogebra.org/material/iframe/id/fy0yrGYK/width/500/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 700,350 noborder}} {{url>https://www.geogebra.org/material/iframe/id/fy0yrGYK/width/500/height/500/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 700,350 noborder}}
Zeile 236: Zeile 236:
  
 <callout icon="fa fa-exclamation" color="red" title="Notice:"> <callout icon="fa fa-exclamation" color="red" title="Notice:">
-  * For the sign of the magnetic voltage, one has to consider the orientation of the current and way on the enclosing path. The <imgref BildNr05> shows the positive orientation: The positive orientation is given, when the currents show out of the drawing plane and the path shows a counterclockwise orientation. +  * For the sign of the magnetic voltage, one has to consider the orientation of the current and way on the enclosing path. The <imgref BildNr05> shows the positive orientation: The positive orientation is given when the currents show out of the drawing plane and the path shows a counterclockwise orientation. 
   * This is again given as the right-hand rule (see <imgref BildNr76>): For the positive orientation the current shows along the thumb of the right hand, while the path is counted along the direction of the fingers of the right hand.   * This is again given as the right-hand rule (see <imgref BildNr76>): For the positive orientation the current shows along the thumb of the right hand, while the path is counted along the direction of the fingers of the right hand.
  
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 688: Zeile 695:
     * {{wp>relay}}s, {{wp>contactor}}s, {{wp>transformer}}s, compact antennas     * {{wp>relay}}s, {{wp>contactor}}s, {{wp>transformer}}s, compact antennas
  
-=== magnetically hard ferromagnetic Materials ===+=== Magnetically hard ferromagnetic Materials ===
  
   * low permeability    * low permeability 
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 813: Zeile 1058:
 Additionally, for the plate capacitor $|\vec{E}|= U/d$. \\ Additionally, for the plate capacitor $|\vec{E}|= U/d$. \\
  
-Therefore, it leads to:+Therefore, it leads to the following:
  
 \begin{align*} \begin{align*}
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">The right hand|The left hand</question>. +<question title="1. Which hand can be used to infer magnetic field direction from currents?" type="radio"> 
-    <question title="2. In the derivation from 1. how are the fingers to be assigned?" type="radio"> Thumb for current direction, remaining fingers for magnetic field direction | Thumb for magnetic field direction, remaining fingers for current direction| both possibilities are correct </question> +The right hand| 
-    <question title="3. Two conductors carrying current are parallel and close to each other. The current in both is flowing in the same direction. What force effect is seen?" type=" radio"> none | The conductors attract | The conductors repel</question>. +The left hand 
-    <question title="4. Two conductors carrying current are at right angles to each other. Current flows through both of them. What force effect can be seen?" type=" radio"> none | The conductors attract | The conductors repel</question>. +</question> 
-    <question title="5. What is the magnetic field inside the earth or a permanent magnet?" type="radio"> from the magnetic north pole to the south pole | from the magnetic south pole to the north pole | the inside is free of field</question>. +<question title="2. In the derivation from 1. how are the fingers to be assigned?" type="radio">  
-    <question title="6. At which location of a current-carrying coil are the field lines densest?" type="radio"> at the magnetic north pole | at the magnetic south pole | inside the coil | at both poles </question>+Thumb for current direction, remaining fingers for magnetic field direction |  
 +Thumb for magnetic field direction, remaining fingers for current direction |  
 +both possibilities are correct  
 +</question> 
 +<question title="3.  Two conductors carrying current are parallel and close to each other. The current in both is flowing in the same direction. What force effect can be seen?" type="radio">  
 +none |  
 +The conductors attract |  
 +The conductors repel 
 +</question> 
 +<question title="4. Two conductors carrying current are at right angles to each other. Current flows through both of them. What force effect can be seen?" type="radio">  
 +none |  
 +The conductors attract |  
 +The conductors repel 
 +</question> 
 +<question title="5. What is the magnetic field inside the earth or a permanent magnet?" type="radio">  
 +from the magnetic north pole to the south pole |  
 +from the magnetic south pole to the north pole |  
 +the inside is free of field 
 +</question> 
 +<question title="6. At which location of a current-carrying coil are the field lines densest?" type="radio">  
 +at the magnetic north pole |  
 +at the magnetic south pole |  
 +inside the coil |  
 +at both poles  
 +</question>
  
 </quizlib> </quizlib>
  
 </WRAP> <WRAP half column> </WRAP> <WRAP half column>
-++++Tip to 1|+ 
 +++++Tip for 1|
 For the current, you use which hand? For the current, you use which hand?
 ++++ ++++
Zeile 852: Zeile 1122:
 ++++Tip for 2| ++++Tip for 2|
   * Imagine a coil with a winding pictorially, or draw it on.   * Imagine a coil with a winding pictorially, or draw it on.
-  * Now think of a generated field through this to it. What direction must the current flow, that cause the field? Does this fit the rule of thumb?+  * Now think of a generated field through this to it. What direction must the current flow, that causes the field? Does this fit the rule of thumb?
   * Then try it the other way round: If a current is given, where do the field lines go in and where out? What poles are created there?   * Then try it the other way round: If a current is given, where do the field lines go in and where out? What poles are created there?
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Zeile 862: Zeile 1132:
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-++++Tip to 4|+++++Tip for 4|
   * First imagine the parallel wires again. What happens when the current flows in the same direction and what happens when the current flows in opposite directions? Are the resulting forces equal in magnitude?   * First imagine the parallel wires again. What happens when the current flows in the same direction and what happens when the current flows in opposite directions? Are the resulting forces equal in magnitude?
   * The reversal of the direction of the current can now also be produced by turning the wire instead of changing the current - just so that the wires are perpendicular to each other in the meantime when turning.   * The reversal of the direction of the current can now also be produced by turning the wire instead of changing the current - just so that the wires are perpendicular to each other in the meantime when turning.