Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

Link zu dieser Vergleichsansicht

Beide Seiten der vorigen Revision Vorhergehende Überarbeitung
Nächste Überarbeitung 		Vorhergehende Überarbeitung
Nächste Überarbeitung Beide Seiten der Revision
electrical_engineering_2:the_magnetostatic_field [2024/04/16 01:34]
mexleadmin [Bearbeiten - Panel] 		electrical_engineering_2:the_magnetostatic_field [2024/04/23 02:24]
mexleadmin
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 743: Zeile 746:
 \begin{align*} \begin{align*}
 H(r) &= {{I_0}\over{2\pi \cdot r}} \\ H(r) &= {{I_0}\over{2\pi \cdot r}} \\
-  &= {{100~\rm A}\over{2\pi \cdot 0.01 ~\rm m}} \\+  &= {{100~\rm A}\over{2\pi \cdot 0.1 ~\rm m}} \\
 \end{align*} \end{align*}
  
Zeile 750: Zeile 753:
 #@HiddenBegin_HTML~102,Result~@# #@HiddenBegin_HTML~102,Result~@#
 \begin{align*} \begin{align*}
-            H(10~\rm cm) &= 1591.5... ~\rm{{A}\over{m}} \\  +            H(10~\rm cm) &= 159.15... ~\rm{{A}\over{m}} \\  
-\rightarrow H(10~\rm cm) &= 1.6 ~\rm{{kA}\over{m}} +\rightarrow H(10~\rm cm) &= 159 ~\rm{{A}\over{m}} 
 \end{align*} \end{align*}
  
Zeile 943: 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>