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electrical_engineering_2:the_magnetostatic_field [2023/05/02 13:03]
mexleadmin 		electrical_engineering_2:the_magnetostatic_field [2024/04/23 02:26]
mexleadmin
Zeile 1: Zeile 1:
-====== 3. The magnetostatic Field ======+====== 3 The magnetostatic Field ======
  
 <callout> <callout>
Zeile 644: Zeile 644:
  
 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 728:
 <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 815:
 </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>
  
Zeile 757: Zeile 904:
  
 In each case, the magnetic voltage $V_{\rm m}$ along the drawn path is sought. In each case, the magnetic voltage $V_{\rm m}$ along the drawn path is sought.
 +
 +
 +#@HiddenBegin_HTML~323100,Path~@#
 +
 +  * The magnetic voltage 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) $\theta_\rm a = - I_1 = - 2~\rm A$ \\
 +#@HiddenEnd_HTML~323102,Result~@#
 +
 +#@HiddenBegin_HTML~323103,Result b)~@#
 +b) $\theta_\rm b = - I_2 = - 4.5~\rm A$ \\
 +#@HiddenEnd_HTML~323103,Result~@#
 +
 +#@HiddenBegin_HTML~323104,Result c)~@#
 +c) $\theta_\rm c = 0 $ \\
 +#@HiddenEnd_HTML~323104,Result~@#
 +
 +#@HiddenBegin_HTML~323105,Result d)~@#
 +d) $\theta_\rm d = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\
 +#@HiddenEnd_HTML~323105,Result~@#
 +
 +#@HiddenBegin_HTML~323106,Result e)~@#
 +e) $\theta_\rm e = + I_1 = + 2~\rm A$ \\
 +#@HiddenEnd_HTML~323106,Result~@#
 +
 +#@HiddenBegin_HTML~323107,Result f)~@#
 +f) $\theta_\rm f = 2 \cdot (- I_1) = - 4~\rm A$ \\
 +#@HiddenEnd_HTML~323107,Result~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 767: Zeile 946:
 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 835: Zeile 1068:
 <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|