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Adding to magnetic forces

  1. Mar 27, 2010 #1
    Adding two magnetic forces

    1. The problem statement, all variables and given/known data
    Well, it's kinda stupid, but I've truely forgot how to do this.

    I've got 3 infinite long wires that is located as in this image:
    http://www.gratisupload.dk/download/41857/" [Broken]
    where the wires is, ofc, inifinite.

    Well, I have to calculate the force acting on wire 1 from wire 2 and 3.
    And the current running through the wires are as stated in the picture.

    2. Relevant equations

    [tex]\frac{F}{L}=\frac{{{\mu }_{0}}II'}{2\pi r}[/tex]

    3. The attempt at a solution

    So I calculate the force from wire 2 to wire 1, and from wire 3 to wire 1, which gives:

    [tex]\frac{{{F}_{2-1}}}{L}=\frac{{{\mu }_{0}}I\left( -2I \right)}{2\pi a}=\frac{-{{\mu }_{0}}{{I}^{2}}}{\pi a}[/tex]

    and

    [tex]\frac{{{F}_{3-1}}}{L}=\frac{{{\mu }_{0}}II}{2\pi \left( \sqrt{{{a}^{2}}+{{a}^{2}}} \right)}=\frac{{{\mu }_{0}}{{I}^{2}}\sqrt{2}}{4\pi a}[/tex]

    My problem is, that I know that the total force is just not by adding the two expressions, but I need to do it vectorstyle - I think.

    And the only way I can think of is:

    [tex]\frac{{{F}_{tot}}}{L}=\sqrt{{{\left( \frac{{{F}_{2-1}}}{L} \right)}^{2}}+{{\left( \frac{{{F}_{3-1}}}{L} \right)}^{2}}}[/tex]

    But that doesn't give the right result, which should be:

    [tex]{{F}_{tot}}=\frac{\sqrt{10}{{\mu }_{0}}{{I}^{2}}L}{4\pi a}[/tex]

    So what am I doing wrong? I've been looking through my book, but I haven't found anything that could solve this for me. And if I'm correct it's pretty simple, but even so, I just can't remember it or figure it out...

    So can anyone give me a hint?


    Regards
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 27, 2010 #2

    gabbagabbahey

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    You said it yourself, the individual forces are vectors. What are their directions?
     
  4. Mar 28, 2010 #3
    Well, something like this:

    http://www.gratisupload.dk/download/41871/" [Broken]

    And sorry for the not so pretty paint picture :)
     
    Last edited by a moderator: May 4, 2017
  5. Mar 28, 2010 #4

    gabbagabbahey

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    Okay, so add the components of the individual forces to get the compoents of the total force, and then find its magnitude.
     
  6. Mar 28, 2010 #5
    But how ? That's kinda my problem.
    I have totally forgot how to do this :S
     
  7. Mar 28, 2010 #6

    gabbagabbahey

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    Well, what are the x and y-components of each individual force? You know their magnitudes and directions, so use a little trig and find them.
     
  8. Mar 28, 2010 #7
    So, the x-component of F2-1 is:

    [tex]\[{{F}_{x}}=-\frac{{{\mu }_{0}}{{I}^{2}}}{\pi a}\][/tex]

    and the y-component is 0, since it's only in the x-direction.

    The x- and y-components of F3-1 is:

    [tex]\[\begin{align}
    & {{F}_{x}}=\frac{{{\mu }_{0}}{{I}^{2}}\sqrt{2}}{4\pi a}\cos \left( \pi /4 \right) \\
    & {{F}_{y}}=\frac{{{\mu }_{0}}{{I}^{2}}\sqrt{2}}{4\pi a}\sin \left( \pi /4 \right) \\
    \end{align}\]
    [/tex]

    Then add the x-components, and finding the magnitude by saying:

    [tex]\[F=\sqrt{{{\left( \sum{{{F}_{x}}} \right)}^{2}}+{{\left( \sum{{{F}_{y}}} \right)}^{2}}}\][/tex]


    Argh... So simple !

    Thank you :)
     
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