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Adding to magnetic forces

  • #1
148
1
Adding two magnetic forces

Homework Statement


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.

Homework Equations



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

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:

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6
You said it yourself, the individual forces are vectors. What are their directions?
 
  • #3
148
1
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:
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Okay, so add the components of the individual forces to get the compoents of the total force, and then find its magnitude.
 
  • #5
148
1
But how ? That's kinda my problem.
I have totally forgot how to do this :S
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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.
 
  • #7
148
1
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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