Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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?

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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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:

    & {{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) \\

    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook