Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
ee2:task_elndbo3xwi2klxuu_with_calculation [2024/07/03 11:42] mexleadmin angelegt |
ee2:task_elndbo3xwi2klxuu_with_calculation [2024/07/03 15:21] (aktuell) mexleadmin angelegt |
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Zeile 24: | Zeile 24: | ||
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 | + | The **__horizontal component__** |
+ | |||
+ | {{drawio> | ||
\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 & | ||
+ | {{drawio> | ||
- | # | + | \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' | ||
+ | \end{align*} | ||
+ | |||
+ | For the **__overall force__** $F$ the Pythagorean theorem has to be used: | ||
- | # | ||
\begin{align*} | \begin{align*} | ||
- | \theta_{(1)} &= -4 {~\rm A} \\ | + | F & |
- | \theta_{(2)} &= 0 {~\rm A} \\ | + | & |
- | \theta_{(3)} | + | & |
\end{align*} | \end{align*} | ||
+ | # | ||
+ | |||
+ | # | ||
+ | $F = 14'609 ~\rm N$ | ||
# | # | ||
- | 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 |
(Independent) | (Independent) | ||
Zeile 48: | Zeile 67: | ||
# | # | ||
- | For the resulting current the direction of the path has to be considered with the right-hand rule: | + | |
- | | + | * The vertical component |
- | * $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} $ | + | |
# | # | ||
# | # | ||
- | \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)} & | + | |
- | \end{align*} | + | |
# | # | ||
# | # |