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electrical_engineering_and_electronics_1:block17 [2025/11/22 20:04] – angelegt mexleadminelectrical_engineering_and_electronics_1:block17 [2025/12/02 17:39] (aktuell) – [Applications of the Lorentz Force: Two parallel Conductors] mexleadmin
Zeile 1: Zeile 1:
-====== Block xx - xxx ======+====== Block 17 — Magnetic Flux Density and Forces ======
  
 ===== Learning objectives ===== ===== Learning objectives =====
Zeile 31: Zeile 31:
 ===== Core content ===== ===== Core content =====
  
-...+We know from [[block11]] that a static charge $Q_1$ generate a static electric field $D$. \\ 
 +Before in [[block09]], we developed that a static electric field $E= {{1}\over{\varepsilon}} D$ effects a force $F_C$ on a static charge $Q_2$  
 + 
 +From the last chapter ([[block16]]) we got, that moving charges ${{d}\over{dt}}Q_1 = I_1$ generate a static magnetic field $H$. \\  
 +So, how does an acting magnetic field effects a force on a moving charge ${{d}\over{dt}}Q_2 = I_2$? 
 + 
 + 
 +==== Definition of the Magnetic Flux Density ==== 
 + 
 +To derive the forces, we do a step back to the images of field lines. \\ 
 +In <imgref BildNr01> 1) the field lines of a single current-carrying wire is shown. \\ 
 +<imgref BildNr01> 2) depicts the homogenous field of a coil. 
 + 
 +When a current-carrying wire is within the homogenous field, we get the superimposed picture of both fields ( <imgref BildNr01> 3) ). \\ 
 +This leads to an enrichment of magnetic field on the left and an depletion on the right. \\ 
 +With the knowledge, that the field lines usually do not like to stay next to each other, one can conclude that there will be a force to the right. 
 + 
 +<WRAP> 
 +<imgcaption BildNr01 | Force in magnetic field> 
 +</imgcaption> 
 +{{drawio>magnForce01.svg}} \\ 
 +</WRAP> 
 + 
 +When no current is flowing through the conductor the force is equal to zero. \\  
 +The following is detectable: 
 +  - $|\vec{F}| \sim I$ : The stronger the current, the stronger the force $F$. 
 +  - $|\vec{F}| \sim l$ : As longer the conductor length $l$, as stronger the force $F$ gets. 
 +  - $|\vec{F}| \sim {H}$ : As more current through the coil, as stronger the $H$-field. And a stronger the $H$-field leads to stronger force $F$. 
 + 
 +To summarize: 
 +\begin{align*} 
 +F \sim H \cdot I \cdot l 
 +\end{align*} 
 + 
 +The proportionality factor is $\mu_0$, the **magnetic field constant**, **permeability** or {{wp>vacuum permeability}}: $\mu_0 = 4\pi \cdot 10^{-7} {\rm {Vs}\over{Am}}$. 
 + 
 +\begin{align*} 
 +F = \mu_0 \cdot H \cdot I \cdot l 
 +\end{align*} 
 + 
 +When adding an iron core into the coil the force $F$ gets stronger. Therefore, we include a material-dependent constant $\mu_r$, the so-called relative permeability 
 + 
 +\begin{align*} 
 +F = \mu_0 \mu_r \cdot H \cdot I \cdot l \\ 
 +\end{align*} 
 + 
 +The new field quantity is $B$ the **magnetic flux density**: 
 + 
 +\begin{align*} 
 +\boxed{ \vec{B} = \mu \cdot \vec{H}  } \quad | \text{with }  \mu =  \mu_0  \mu_r 
 +\end{align*} 
 + 
 +Investigating the vectorial behaviour leads to the cross-product, and to the so-called **Lorentz force**  
 + 
 +\begin{align*} 
 +\boxed{ \vec{F}_L = I \cdot \vec{l} \times \vec{B} } 
 +\end{align*} 
 + 
 +With \vec{l} pointing in the direction of the positive current $I$. 
 +The absolute value can be calculated by  
 + 
 +\begin{align*} 
 +\boxed{|\vec{F_L}| = I \cdot |l| \cdot |B| \cdot \sin(\angle \vec{l},\vec{B} )} 
 +\end{align*} 
 + 
 +For the orientation of the vectors, another right-hand rule can be applied. 
 + 
 +<callout icon="fa fa-exclamation" color="red" title="Notice:"> 
 +Right-hand rule for the Lorentz Force: 
 +  * The causing current $I$ is on the thumb. Since the current is not a vector, the direction is given by the direction of the conductor $\vec{l}$ 
 +  * The mediating external magnetic field $\vec{B}$ is on the index finger 
 +  * The resulting force $\vec{F}$ on the conductor is on the middle finger 
 +This is shown in <imgref BildNr06>. 
 + 
 +A way to remember the orientation is the mnemonic **FBI** (from middle finger to thumb): 
 +  * $\vec{F}$orce on middle finger 
 +  * $\vec{B}$-Field on index finger 
 +  * 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> 
 +<imgcaption BildNr06 | Force onto a single Conductor in a B-Field> 
 +</imgcaption> 
 +{{drawio>SingleConductorInBField.svg}} \\ 
 +</WRAP> 
 +</callout> 
 + 
 +==== Materials ==== 
 + 
 +The material can be divided into different types by looking at its relative permeability. 
 +<imgref BildNr00> shows the relative permeability in the **magnetization curve** (also called $B$-$H$-curve). 
 +In this diagram, the different effect ($B$-field on $y$-axis) based on the causing external $H$-field (on $x$-axis) for different materials is shown. 
 +The three most important material types shall be discussed shortly. 
 + 
 +<WRAP> 
 +<imgcaption BildNr00 | Magnetization Curve of different materials></imgcaption> 
 +{{drawio>MagnetizationCurveDifferent.svg}} 
 +</WRAP> 
 + 
 +\\ \\ 
 +<WRAP group> 
 +<WRAP> 
 + 
 +=== Diamagnetic Materials === 
 + 
 +  * Diamagnetic materials weaken the magnetic field, compared to the vacuum. 
 +  * The weakening is very low (see <tabref tab01>). 
 +  * For diamagnetic materials applies $0<\mu_{\rm r}<1$  
 +  * The principle behind the effect is based on quantum mechanics (see <imgref BildNr22>):  
 +    * Without the external field no counteracting field is generated by the matter. 
 +    * With an external magnetic field an antiparallel-orientated magnet is induced. 
 +    * The reaction weakens the external field. This is similar to the weakening of the electric field by the dipoles of materials. 
 +  * Due to the repulsion of the outer magnetic field the material tends to move out of a magnetic field. \\ For very strong magnetic fields small objects can be levitated (see clip). 
 + 
 +<WRAP> 
 +<tabcaption tab01| Diamagnetic Materials> 
 +^ Material  ^ Symbol      ^ $\mu_{\rm r}$  ^ 
 +| Antimon   | $\rm Sb$    | $0.999 946$    | 
 +| Copper    | $\rm Cu$    | $0.999 990$    | 
 +| Mercury   | $\rm Hg$    | $0.999 975$    | 
 +| Silver    | $\rm Ag$    | $0.999 981$    | 
 +| Water     | $\rm H_2O$  | $0.999 946$    | 
 +| Bismut    | $\rm Bi$    | $0.999 830$    | 
 +</tabcaption> 
 +</WRAP> 
 + 
 +<WRAP> 
 +<imgcaption BildNr22 | Magnetic field in diamagnetic materials></imgcaption> 
 +{{drawio>Diamagnets.svg}} 
 +</WRAP> 
 + 
 +</WRAP><WRAP > 
 + 
 +=== Paramagnetic Materials === 
 + 
 +  * Paramagnetic materials strengthen the magnetic field, compared to the vacuum. 
 +  * The strengthening is very low (see <tabref tab02>). 
 +  * For paramagnetic materials applies $\mu_{\rm r}>1$  
 +  * The principle behind the effect is again based on quantum mechanics (see <imgref BildNr23>):  
 +    * Without the external field no counteracting field is generated by the matter. 
 +    * With an external magnetic field internal "tiny magnets" based on the electrons in their orbitals are orientated similarly. 
 +    * This reaction strengthens the external field. 
 + 
 +<WRAP> 
 +<tabcaption tab02| Paramagnetic Materials> 
 +^ Material   ^ Symbol ^ $\mu_{\rm r}$        ^ 
 +| Aluminum   | $\rm Al$     | $1.000 022$        | 
 +| Air        |              | $1.000 000 4$      | 
 +| Oxygen     | $\rm O_2$    | $1.000 001 3$      | 
 +| Platinum   | $\rm Pt$     | $1.000 36$         | 
 +| Tin        | $\rm Sn$     | $1.000 003 8$      | 
 +</tabcaption> 
 +</WRAP> 
 + 
 +<WRAP> 
 +<imgcaption BildNr23 | Magnetic field in paramagnetic materials></imgcaption> 
 +{{drawio>Paramagnets.svg}} 
 +</WRAP> 
 + 
 +</WRAP><WRAP> 
 + 
 +=== Ferromagnetic Materials === 
 + 
 +  * Ferromagnetic materials strengthen the magnetic field strongly, compared to the vacuum. 
 +  * The strengthening can create a field multiple times stronger than in a vacuum (see <imgref BildNr53>).  
 +  * For ferromagnetic materials applies $\mu_{\rm r}  \gg 1$  
 +  * Ferromagnetic materials are characterized by the magnetization curve (see <imgref BildNr24>) 
 +    * Non-magnetized ferromagnets are located in the origin. 
 +    * With an external field $H$ the initial magnetization curve (in German: //Neukurve//, dashed in <imgref BildNr24>) is passed. 
 +    * Even without an external field ($H=0$) and the internal field is stable. \\ The stored field without external field is called **remanence** $B(H=0) = B_{\rm R}$ (or remanent magnetization). 
 +    * In order to eliminate the stored field the counteracting **coercive field strength** $H_{\rm C}$ (also called coercivity) has to be applied. 
 +    * The **saturation flux density** $B_{\rm sat}$ is the maximum possible magnetic flux density (at the maximum possible field strength $H_{\rm sat}$) 
 + 
 +<WRAP> 
 +<imgcaption BildNr53 | Magnetic field in paramagnetic materials></imgcaption> 
 +{{drawio>Ferromagnets.svg}} 
 +</WRAP> 
 + 
 + 
 +<WRAP> 
 +<imgcaption BildNr24 | Magnetization Curve></imgcaption> 
 +{{drawio>MagnetizationCurvePrinciple.svg}} 
 +</WRAP> 
 + 
 +</WRAP></WRAP> 
 + 
 +==== Applications of the Lorentz Force: Two parallel Conductors === 
 + 
 +The Lorentz force can be applied to two parallel conductors. \\  
 +The experiment consists of a part $l$ of two very long((ideally: infinite long; in reality much longer, than the distance between them)) and parallel conductors with the currents $I_1, I_2$ and the distance $r$  (see <imgref BildNr06>).  
 + 
 +<WRAP> 
 +<imgcaption BildNr06 | Forces between two Conductors> 
 +</imgcaption> 
 +{{drawio>TwoConductorsAlt.svg}} \\ 
 +</WRAP> 
 + 
 +Here, we get for the $B$ field caused by $I_2$:  
 +\begin{align*} 
 +B_2 &= \mu \cdot H_2 \\ 
 +    &= \mu \cdot {{ I_2 }\over{2\pi \cdot r}} 
 +\end{align*} 
 + 
 +We insert this into the formula of the Lorenz force 
 +\begin{align*} 
 +\vec{F}_L = I \cdot \vec{l} \times \vec{B} 
 +\end{align*} 
 + 
 +This leads to the so-called **Ampere's Force Law**, applied on long and parallel conductors: 
 +\begin{align*} 
 +\boxed{ |\vec{F}_{12}| = {{\mu}\over{2 \pi}} \cdot {{I_1 \cdot I_2 }\over{r}} \cdot l }  
 +\end{align*} 
 + 
  
 ===== Common pitfalls ===== ===== Common pitfalls =====
Zeile 38: Zeile 253:
 ===== Exercises ===== ===== Exercises =====
  
-{{page>electrical_engineering_and_electronics:task_ddjurcpk494go2q1_with_calculation&nofooter}}+{{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}}
 {{page>electrical_engineering_and_electronics:task_ti7loik6aurfewkb_with_calculation&nofooter}} {{page>electrical_engineering_and_electronics:task_ti7loik6aurfewkb_with_calculation&nofooter}}
 {{page>electrical_engineering_and_electronics:task_w3m7fo4hjahkzogw_with_calculation&nofooter}} {{page>electrical_engineering_and_electronics:task_w3m7fo4hjahkzogw_with_calculation&nofooter}}
-{{page>electrical_engineering_and_electronics:task_w3m7fo4hjahkzogw_with_calculation&nofooter}}+{{page>electrical_engineering_and_electronics:task_elndbo3xwi2klxuu_with_calculation&nofooter}}
  
 +
 +
 +
 +<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 $I_0 = 100~\rm A$ 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~@#
 +
 +</WRAP></WRAP></panel>
 +
 +~~PAGEBREAK~~~~CLEARFIX~~
 +
 +
 +<panel type="info" title="Task 3.3.1 magnetic Flux Density"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface. 
 +
 +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?
 +
 +#@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 #task3_3_2 />
 +
 +<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 enters a plate capacitor on a trajectory parallel to the plates. 
 +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>
 +<imgcaption BildNr71 | Electron in B- and E-Field>
 +</imgcaption>
 +{{drawio>ElectronInBandEfield.svg}} \\
 +</WRAP>
 +
 +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">
 +
 +Within the electric field, the Coulomb force acts on the electron:
 +
 +\begin{align*}
 +\vec{F}_C = q_e \cdot \vec{E}
 +\end{align*}
 +
 +Within the magnetic field, also the Lorentz force acts on the electron:
 +
 +\begin{align*}
 +\vec{F}_L = q_e \cdot \vec{v} \times \vec{B}
 +\end{align*}
 +
 +The absolute value of both forces must be equal to compensate each other: 
 +
 +\begin{align*}
 +|\vec{F}_C|          &= |\vec{F}_L|\\
 +|q_e \cdot \vec{E}|  &= |q_e \cdot \vec{v} \times \vec{B}| \\
 +q_e \cdot |\vec{E}|  &= q_e \cdot |\vec{v} \times \vec{B}| \\
 +          |\vec{E}|  &=           |\vec{v} \times \vec{B}| \\
 +\end{align*}
 +
 +Since $\vec{v}$ is perpendicular to $\vec{B}$ the right side is equal to $|\vec{v}| \cdot |\vec{B}| = v \cdot B$. \\
 +Additionally, for the plate capacitor $|\vec{E}|= U/d$. \\
 +
 +Therefore, it leads to the following:
 +
 +\begin{align*}
 +          {{U}\over{d}}  &=           v \cdot B \\
 +          v              &=          {{U}\over{B \cdot d}} 
 +\end{align*}
 +
 +</collapse>
 +
 +<button size="xs" type="link" collapse="Loesung_3_1_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_3_1_2_Endergebnis" collapsed="true">
 +\begin{align*}
 +v = {{U}\over{B\cdot d}}
 +\end{align*}
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 + **<fs large>Task 1</fs>**
 +
 +<WRAP group> <WRAP half column>
 +
 +<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="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>
 +<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>
 +
 +</WRAP> <WRAP half column>
 +
 +++++Tip for 1|
 +For the current, you use which hand?
 +++++
 +
 +++++Tip for 2|
 +  * 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 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?
 +++++
 +
 +++++Tip for 3| See 3rd video.
 +  * Picture the two wires, or draw them on.
 +  * In which direction would the outer field run in each case?
 +  * The field is a linear vector field. So the total field can be created from several individual fields by adding them together. Does adding the field in between make it larger, or smaller?
 +++++
 +
 +++++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?
 +  * 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.
 +  * With parallel wires and different current directions, the amount-wise same force arises. So, this is also true for every angle in between (in detail given by integration of the force over single wire pieces).
 +  * But then there must be a point at which the force becomes 0.
 +++++
 +
 +++++Tip for 5|
 +  * The magnetic field lines must be closed.
 +  * Compare the field curve between the coil and permanent magnet.
 +++++
 +++++Tip for 6|
 +  * In video 1 you can see the course outside and inside the coil.
 +++++
 +
 +</WRAP> </WRAP>
  
 ===== Embedded resources ===== ===== Embedded resources =====
 +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.
 +\\ \\ \\
 <WRAP column half> <WRAP column half>
-Explanation (video): ...+The rotating flux (density) in the stator of a motor, source/CC-BY 3.0 see {{https://commons.wikimedia.org/wiki/File:2HP_1500RPM_induction_motor_no_slip.gif|wikipedia}} 
 +{{electrical_engineering_and_electronics_1:2hp_1500rpm_induction_motor_no_slip.gif}}
 </WRAP> </WRAP>
 +
 +<WRAP column half>
 +A living insect ("diamagnet") floats in a very strong magnetic field \\
 +{{youtube>KlJsVqc0ywM?start=45}}
 +</WRAP>
 +
 +\\ \\
 +<WRAP column half>
 +Explanation of diamagnetism and paramagnetism \\
 +{{ youtube>u36QpPvEh2c }}         
 +</WRAP>
 +<WRAP column half>
 +In Oxygen magnetic? \\
 +{{ youtube>pniES3kKHvY?300x500 }} 
 +</WRAP>
 +
  
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