ET2_Uebung_BEI/ET2_L_B20_A1.tex

\section {Ringspule}
Durch das Zentrum einer Ringspule mit rechteckigem Querschnitt (Länge
$l=12\,\milli\metre$, $r_i=10\,\milli\metre$, $r_a=20\,\milli\metre$;
Windungszahl $N= 1800$) wird ein Leiter geführt, in dem ein Wechselstrom fließt.
($I_{eff}=40\,\ampere$; $f=50\,\hertz$).\\
Berechnen Sie den Effektivwert $U$ der Spannung $u$.\\
$\mu=\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
\begin{align*}
	\begin{tikzpicture}[scale=2.5, z=0.2pt]
	\begin{scope}[>=latex, xshift=0, yshift=0]
		\foreach \z in {0,0.2,...,1}
			\fill [black!25!](0,0,\z cm) circle (0.7cm);% Zylinder außen
		\draw [black!75!](0,0,1 cm) circle (0.7cm);%Hintere äußere Ringkante
%\foreach \z in {0,0.2,...,0.5}
%\fill [black!50!](0,0,\z cm) circle (0.5cm);% Zylinder innen
		\fill [black!25!](0,0) circle (0.7cm);% Stirnseitenfüllung
		\fill [black!35](0,0) circle (0.5cm); %Füllung innen
	\begin{scope}%Clip Füllung
		\clip (0,0,1cm) circle (0.5cm);%Clip mit hinterer inneren Ringkante
		\fill [black!0!](0,0) circle (0.5cm);%Innere Ringkante
	\end{scope}
	\begin{scope}%Clip hintere innere Kante
		\clip (0,0) circle (0.5cm);%Clip mit hinterer inneren Ringkante
		\draw [black!75!](0,0,1cm) circle (0.5cm);%Hintere innere Ringkante
	\end{scope}
			\draw [black!75!](0,0) circle (0.5cm);% Innere Ringkante
			\draw [black!75!](0,0) circle (0.7cm);% Äußere Ringkante
			\foreach \a in {-45,-30,-15,15,30,45,60,75,90,105,120,135} %Rechte Drähte
				\draw [red!60!black, ultra thick] (\a:0.5)--(\a:0.7)--+(45:.3);
			\foreach \a in {150,165,...,300} %Linke Drähte
				\draw [red!60!black, ultra thick] (\a:0.7)--(\a:0.5)--+(45:.3);
			\draw [red!60!black, ultra thick] (0.5,0,0)--(1.5,0,0)circle(0.01);%Anschußdrähte
			\draw [red!60!black, ultra thick] (0.7,0,1cm)--(1.5,0,1cm)circle(0.01);%Anschußdrähte
			\draw [->,blue, thick] (1.6,0,1cm) -- (1.6,0,0) node at (1.6,0,.5cm)[right] {$u$};
			\draw [blue, ultra thick] (0,0,3.5cm) -- (0,0,7cm);
			\draw [->,red, thick] (0,0,7.5cm) -- (0,0,8cm)node [right] {$i$};
			\draw [blue, ultra thick] (0,0,1.75cm) -- (0,0,-5cm)node [above left]{Leiter};
		\end{scope}
	\end{tikzpicture}
\begin{tikzpicture}[scale=1.4]%Querschnitt
	\begin{scope}[>=latex,xshift=0,yshift=0]
		\fill [black!25!] (0,0)rectangle(1,1);
	\end{scope}
	\begin{scope}[>=latex,xshift=0,yshift=3cm]
		\fill [black!25!] (0,0)rectangle(1,1);
	\end{scope}	
	\begin{scope}[>=latex,xshift=0,yshift=0cm]
		\draw [very thick] (0,0)rectangle(1,4);
	\end{scope}
	\begin{scope}[>=latex,xshift=-1cm,yshift=1.9cm]
		\fill [blue!50!] (0,0)rectangle(3,.2);
	\end{scope}
	\begin{scope}[>=latex,xshift=0cm,yshift=4cm]%Hilfslinien
		\draw [very thin] (-1,0)--(0,0)--(0,.5) (1,0)--(1,.5)(-.5,-1)--(0,-1);
		\draw [very thin, dashed] (-1.5,-2)--(2.5,-2);		
		\draw [->,very thin] (-.9,-2)--(-.9,0)node at(-.9,-1)[right]{$r_a$};
		\draw [->,very thin] (-.4,-2)--(-.4,-1)node at(-.4,-1.5)[right]{$r_i$};
		\draw [<->,very thin] (0,0.4)--(1,0.4)node at(.5,0.4)[above]{$l$};
		\draw node at(.5,-.5){$\mu=\mu_0$};
		\draw [->,red,thin] (1.25,-1.75)--(1.5,-1.75) node [right]{$i$};		
	\end{scope}			
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
u&=N\cdot \frac{d\phi(t)}{dt}
\end{align}
Berechnung:
Strom im Leiter bewirkt Fluss durch die Spule:
\begin{align*}
\phi&=\widehat{\phi}_0\cdot \sin(\omega t)\\
\widehat{\phi}_0&=\int_{r_i}^{r_a}l\cdot \widehat{B}(r)\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \int_{r_i}^{r_a}\frac{1}{r}\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \ln\frac{r_a}{r_i}\\
\text{mit }\widehat{i}&=\sqrt{2}\cdot I\\
\Rightarrow \widehat{\phi}_0&=\frac{1{,}26\cdot \power{10}{-6}\, \frac{\volt\second}{\ampere\metre}\cdot 12\cdot \power{10}{-3}\,\metre\cdot \sqrt{2}\cdot 40\,\ampere}{2\pi}\cdot \ln(2)=9{,}44\cdot \power{10}{-8}\,\volt\second\\
\phi_{eff}&=\frac{\widehat{\phi}_0}{\sqrt{2}}=6{,}672\cdot \power{10}{-8}\,\volt\second\\
\intertext{Veränderlicher Fluss induziert Spannung:}
u&=N\cdot \frac{d\phi(t)}{dt}=N\cdot \widehat{\phi}_0\cdot \omega\cdot \cos(\omega t)=\widehat{u}\cdot \cos(\omega t)\\
\widehat{u}&=N\cdot \phi_0\cdot \omega=1800\cdot 9{,}44\cdot \power{10}{-8}\,\volt\second\cdot 2\pi\cdot 50\,\power{\second}{-1}=53{,}36\,\milli\volt\\
U&=\frac{\widehat{u}}{\sqrt{2}}=\frac{53{,}36\,\milli\volt}{\sqrt{2}}=\uuline{37{,}73\,\milli\volt}
\intertext{Anmerkung:}
u_i&=2\cdot a\cdot r\cdot N\cdot B\cdot \omega\cdot \sin(\omega t)\\
&=2\cdot a\cdot r\cdot N\cdot \frac{\Phi}{A}\cdot \omega\cdot \sin(\omega t)\\
&\text{mit $2\cdot a\cdot r=A$}\\
&=N\cdot \Phi \cdot \omega\cdot \sin(\omega t)\\
&=\widehat{u}_i\cdot \sin(\omega t)\\
\intertext{oder}
U&=\mu_0\cdot l\cdot \ln\frac{r_a}{r_i}\cdot N\cdot f\cdot I\\
&=1{,}26\cdot \power{10}{-6}\,\frac{\volt\cancel{\second}}{\cancel{\ampere}\cancel{\metre}}\cdot 12\cdot \power{10}{-3}\,\cancel{\metre}\cdot \ln\frac{20\,\cancel{\milli\metre}}{10\,\cancel{\milli\metre}}\cdot 1800\cdot 50\,\cancel{\frac{1}{\second}}\cdot 40\,\cancel{\ampere}=\uuline{37{,}73\,\milli\volt}
\end{align*}
\clearpage
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