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ee2:task_elndbo3xwi2klxuu_with_calculation [2024/07/03 13:05]
mexleadmin angelegt 		ee2:task_elndbo3xwi2klxuu_with_calculation [2024/07/03 15:21] (aktuell)
mexleadmin angelegt
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-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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-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} $ +
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-\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*} +
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