{{tag>lorentz_force exam_ee2_SS2021}}{{include_n>1220}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Lorentz Force (hard!) \\
<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#

A $300 ~\rm km$ long, straight high-voltage direct current transmission line shall be analyzed. A current of $I = 1′200 ~\rm A$ flows through it. \\
A homogeneous geomagnetic field is assumed. The magnetic field strength has a vertical component of $B_{\rm v} = 40 ~\rm \mu T$ and a horizontal component of $B_{\rm h} = 20 ~\rm \mu T$. \\
The angle between the transmission line and the horizontal component of the field strength is $\alpha = 20°$. \\
The picture on the right shows the line (black), the field strength components, and the angle in front and top view for illustration purposes. 

{{drawio>ee2:ELndBo3XWI2kLXuu_question1.svg}}
 
a) Calculate the force that results from the current flow on the entire conductor. \\
First, calculate the vertical and horizontal components and combine them accordingly. 

#@HiddenBegin_HTML~ELndBo3XWI2kLXuu_11,Path~@#

The force on the transmission line can be calculated via the Lorentz force $\vec{F}_\rm L$:
\begin{align*}
\vec{F} = I \cdot (\vec{l} \times \vec{B})
\end{align*}

Here, we have two components for the current - and therefore for the force - to evaluate. \\
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

{{drawio>ee2:ELndBo3XWI2kLXuu_answer1.svg}}

\begin{align*}
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*}

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}}

\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:

\begin{align*}
F &= \sqrt{F_{\rm v}^2 +F_{\rm h}^2} \\
  &= \sqrt{({14'400 ~\rm N})^2 +({2'462.545... ~\rm N})^2} \\
  &= 14'609.04... ~\rm N
\end{align*}

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#@HiddenBegin_HTML~ELndBo3XWI2kLXuu_12,Result~@#
$F = 14'609 ~\rm N$
#@HiddenEnd_HTML~ELndBo3XWI2kLXuu_12,Result~@#

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) 

{{drawio>ee2:ELndBo3XWI2kLXuu_question2.svg}}

#@HiddenBegin_HTML~ELndBo3XWI2kLXuu_21,Path~@#
  * The horizontal component $\vec{F}_{\rm h}$ of the force is based on the vertical component $\vec{B}_{\rm v}$ of the magnetic field. \\
  * The vertical component $\vec{B}_{\rm v}$ of the magnetic field is not shown in the image but is pointing into the ground. \\
  * It has to be perpendicular to $\vec{B}_{\rm v}$ and to $\vec{l}$. The right-hand rule has to be applied.
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#@HiddenBegin_HTML~ELndBo3XWI2kLXuu_22,Result~@#
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.
#@HiddenEnd_HTML~ELndBo3XWI2kLXuu_22,Result~@#

#@TaskEnd_HTML@#