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ee2:task_ddjurcpk494go2q1_with_calculation [2024/07/15 19:09]
mexleadmin angelegt 		ee2:task_ddjurcpk494go2q1_with_calculation [2024/07/15 21:37] (aktuell)
mexleadmin
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 {{tag>electric_field magnetic_field exam_ee2_SS2024}}{{include_n>1040}} {{tag>electric_field magnetic_field exam_ee2_SS2024}}{{include_n>1040}}
    
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Capacitor \\+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Fields of an coax Cable\\
 <fs medium>(written test, approx. 12 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@# <fs medium>(written test, approx. 12 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
  
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 {{drawio>ee2:ddjurcpk494go2q1_question1.svg}} {{drawio>ee2:ddjurcpk494go2q1_question1.svg}}
  
-1. What is the magnitude of the magnetic field strength $H$ at $\rm (-0.1 ~mm |  0)$ and $\rm (0.55 ~mm |  0)$?+1. What is the magnitude of the magnetic field strength $H$ at $\rm (0.1 ~mm |  0)$ and $\rm (0.55 ~mm |  0)$?
  
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 +
 +The magnitude of the magnetic field strength $H$ can be calculated by: $H = {{I}\over{2 \pi \cdot r}} $ \\
 +So, we get for $H_{\rm i}$ at $\rm (0.1 ~mm |  0)$, and $H_{\rm o}$ at $\rm (0.55 ~mm |  0)$:
 +
 \begin{align*} \begin{align*}
-C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\ +H_{\rm i} &= {{I}\over{2 \pi \cdot r_{\rm i}}} \\ 
-  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }}+          &= {{+3.3 A}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m}}} \\ 
 +H_{\rm o} &= {{I}\over{2 \pi \cdot r_{\rm o}}} \\ 
 +          &= {{+3.3 A}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}}} \\
 \end{align*} \end{align*}
  
 +Hint: For the direction, one has to consider the right-hand rule. 
 +By this, we see that the $H$-field on the right side points downwards. \\
 +Therefore, the sign of the $H$-field is negative. \\
 +But here, only the magnitude was questioned!
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-$C = 1.1 ~\rm pF$+  * for $(0.1 ~/rm mm | 0)$ : $H_{\rm i} = 5.25... ~\rm A/m$ 
 +  * for $(0.55 ~/rm mm| 0)$ : $H_{\rm o} = 0.955... ~\rm A/m$
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-2. Plot the graph of the magnitude of $H(x)$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!+2. Plot the graph of the magnitude of $H(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!
  
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-\begin{align*} +  * In general, the $H$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors. 
-C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\ +  * For $H(x)$ with $x$ within the inner conductor, one has to consider how much current is conducted within a circle with the radius $x$. \\ This is proportional to the area within this radius. Therefore, Ther formula $H = {{I}\over{2 \pi \cdot r}} $ gets $H(x) = {{I}\over{2 \pi \cdot x}} \cdot {{\pi x^2}\over{\pi (0.1~\rm mm)^2}}$. This leads to a formula proportional to $x$. 
-  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }} +  * For $x$ within the outer conductor one also gets a linear proportionality with a similar approach.
-\end{align*} +
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-$C = 1.1 ~\rm pF$+{{drawio>ee2:ddjurcpk494go2q1_answer1.svg}}
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 +The magnitude of the electric displacement field $D$ can be calculated by: $\int D {\rm d}A = Q$. \\
 +Here, for any position radial to the center, the surrounding area is the surface of a cylindrical shape (here for simplicity without the round endings). \\
 +This leads to:
 \begin{align*} \begin{align*}
-C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\ +D(x) &= {{Q}\over{A}} \\ 
-  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }}+     &= {{Q}\over{l \cdot 2\pi \cdot x}} \\
 \end{align*} \end{align*}
 +
 +
 +So, we get for $D_{\rm i}$ at $\rm (0.1 ~mm |  0)$, and $D_{\rm o}$ at $\rm (0.55 ~mm |  0)$:
 +
 +\begin{align*}
 +D_{\rm i} &= {{Q                 }\over{2 \pi \cdot r_{\rm i} \cdot l}} \\
 +          &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m} \cdot 0.5 ~\rm m}} \\
 +D_{\rm o} &= {{Q                 }\over{2 \pi \cdot r_{\rm o} \cdot l}} \\
 +          &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}\cdot 0.5 ~\rm m}} \\
 +\end{align*}
 +
 +Hint: For the direction, one has to consider the sign of the enclosed charge. 
 +By this, we see that the $D$-field is positive. \\
 +But here, again only the magnitude was questioned!
  
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-$C = 1.1 ~\rm pF$+  * for $(0.1 ~\rm mm | 0)$ : $D_{\rm i} = 31.8... ~\rm uC/m^2$ 
 +  * for $(0.55 ~\rm mm| 0)$ : $D_{\rm o} = 5.78... ~\rm uC/m^2$
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-4. Plot the graph of the magnitude of $D(x)$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!+4. Plot the graph of the magnitude of $D(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm |  0)$ to $\rm (0.6 ~mm |  0)$ in one diagram. Use proper dimensions and labels for the diagram!
  
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-\begin{align*} +  * In general, the $D$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors. 
-C &= \varepsilon_0 \varepsilon_r {{A}\over{d}} \\ +  * Since the charges are on the surface of the conductor, there is no $D$-field within the conductor.
-  &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1  \cdot {{25 \cdot 10^{-6} {~\rm m} }\over{200 \cdot 10^{-6} {~\rm m} }} +
-\end{align*} +
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-$C = 1.1 ~\rm pF$+{{drawio>ee2:ddjurcpk494go2q1_answer2.svg}}
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