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ee2:task_elndbo3xwi2klxuu_with_calculation [2024/07/03 11:42] – angelegt mexleadminee2:task_elndbo3xwi2klxuu_with_calculation [2024/07/03 15:21] (aktuell) – angelegt mexleadmin
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 Considering the right-hand rule (and the cross product), the vertical field $B_{\rm v}$ generates a horizontal force $F_{\rm h}$ and vice versa. Considering the right-hand rule (and the cross product), the vertical field $B_{\rm v}$ generates a horizontal force $F_{\rm h}$ and vice versa.
  
-The horizontal component is given by+The **__horizontal component__** is given by 
 + 
 +{{drawio>ee2:ELndBo3XWI2kLXuu_answer1.svg}} 
 \begin{align*} \begin{align*}
-F_{\rm h} = I \cdot (l \cdot B_{\rm v})+F_{\rm h} &= I \cdot (l \cdot B_{\rm v}) \\ 
 +          &= 1′200 {~\rm A} \cdot 300 \cdot 10^3{~\rm m} \cdot 40 \cdot 10^{-6}{~\rm {{Vs}\over{m²}}} \\ 
 +          &= 14'400 ~\rm {{VAs}\over{m}} 
 +           = 14'400 ~\rm {{Ws}\over{m}} 
 +           = 14'400 ~\rm N 
 +          
 \end{align*} \end{align*}
  
 +For the **__vertical component__** the angle &\alpha& has to be considered. \\ For the maximum $F_{\rm v}$ the angle &\alpha& has to be $90°$, therefore the $\sin$ has to be used.
  
 +{{drawio>ee2:ELndBo3XWI2kLXuu_answer2.svg}}
  
-#@HiddenEnd_HTML~ELndBo3XWI2kLXuu_11,Path~@#+\begin{align*} 
 +F_{\rm v} &= I \cdot l \cdot B_{\rm h} \cdot \sin\alpha \\ 
 +          &= 1′200 {~\rm A} \cdot 300 \cdot 10^3{~\rm m} \cdot 40 \cdot 10^{-6}{~\rm {{Vs}\over{m²}}} \cdot \sin 20° \\ 
 +          &= 2'462.545... ~\rm N 
 +\end{align*} 
 + 
 +For the **__overall force__** $F$ the Pythagorean theorem has to be used:
  
-#@HiddenBegin_HTML~ELndBo3XWI2kLXuu_12,Result~@# 
 \begin{align*} \begin{align*}
-\theta_{(1)} &= -4 {~\rm A} \\ +F &= \sqrt{F_{\rm v}^2 +F_{\rm h}^2} \\ 
-\theta_{(2)} &= 0 {~\rm A} \\ +  &= \sqrt{({14'400 ~\rm N})^2 +({2'462.545... ~\rm N})^2} \\ 
-\theta_{(3)} &= -5 {~\rm A} \\+  &= 14'609.04... ~\rm N
 \end{align*} \end{align*}
  
 +#@HiddenEnd_HTML~ELndBo3XWI2kLXuu_11,Path~@#
 +
 +#@HiddenBegin_HTML~ELndBo3XWI2kLXuu_12,Result~@#
 +$F = 14'609 ~\rm N$
 #@HiddenEnd_HTML~ELndBo3XWI2kLXuu_12,Result~@# #@HiddenEnd_HTML~ELndBo3XWI2kLXuu_12,Result~@#
  
-b) The picture below shows the top view again. In which of the directions shown does the horizontal component of the resulting force act?  +b) The picture below shows the top view again. In which of the directions shown does the horizontal component $F_{\rm h}$ of the resulting force act?  
 (Independent)  (Independent) 
  
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 #@HiddenBegin_HTML~ELndBo3XWI2kLXuu_21,Path~@# #@HiddenBegin_HTML~ELndBo3XWI2kLXuu_21,Path~@#
-For the resulting current the direction of the path has to be considered with the right-hand rule: +  * The horizontal component $\vec{F}_{\rm h}$ of the force is based on the vertical component $\vec{B}_{\rm v}$ of the magnetic field. \\ 
-  * $I_{(1)} = +I_2 - I_1 - I_3 \quad \rightarrow \quad \theta_{(1)} = 2 {~\rm A} - 5 {~\rm A} - 1 {~\rm A} $  +  * The vertical component $\vec{B}_{\rm v}$ of the magnetic field is not shown in the image but is pointing into the ground. \\ 
-  * $I_{(2)} = +I_3 + I_4 - I_1 \quad \rightarrow \quad \theta_{(2)} = 1 {~\rm A} + 4 {~\rm A} - 5 {~\rm A} $   +  * It has to be perpendicular to $\vec{B}_{\rm v}$ and to $\vec{l}$. The right-hand rule has to be applied.
-  * $I_{(3)} = +I_3 - I_4 - I_2 \quad \rightarrow \quad \theta_{(3)} = 1 {~\rm A} - 4 {~\rm A} - 2 {~\rm A} $ +
 #@HiddenEnd_HTML~ELndBo3XWI2kLXuu_21,Path~@# #@HiddenEnd_HTML~ELndBo3XWI2kLXuu_21,Path~@#
  
 #@HiddenBegin_HTML~ELndBo3XWI2kLXuu_22,Result~@# #@HiddenBegin_HTML~ELndBo3XWI2kLXuu_22,Result~@#
-\begin{align*} +Only option $7.$ is perpendicular to $\vec{B}_{\rm v}$ and to $\vec{l}$ and points in the right direction by the right-hand rule.
-\theta_{(1)} &= -4 {~\rm A} \\ +
-\theta_{(2)} &= 0 {~\rm A} \\ +
-\theta_{(3)} &= -5 {~\rm A} \\ +
-\end{align*} +
 #@HiddenEnd_HTML~ELndBo3XWI2kLXuu_22,Result~@# #@HiddenEnd_HTML~ELndBo3XWI2kLXuu_22,Result~@#
  
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#