Calculating Magnetic Forces Between Infinite Wires

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Homework Help Overview

The discussion revolves around calculating the magnetic forces between three infinite long wires carrying currents. The original poster expresses difficulty in determining the total force acting on one wire due to the others, specifically in handling the vector nature of the forces involved.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the vector addition of forces, questioning how to properly account for the directions of the individual forces. There is an emphasis on breaking down the forces into components and using trigonometry to find the resultant force.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting the addition of force components. The original poster acknowledges their confusion and seeks further clarification on the process.

Contextual Notes

The original poster mentions a specific formula for magnetic force per unit length and refers to a diagram for the configuration of the wires, indicating that visual representation is part of the problem context.

Denver Dang
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Adding two magnetic forces

Homework Statement


Well, it's kinda stupid, but I've truly 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/"
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:
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You said it yourself, the individual forces are vectors. What are their directions?
 
Well, something like this:

http://www.gratisupload.dk/download/41871/"

And sorry for the not so pretty paint picture :)
 
Last edited by a moderator:
Okay, so add the components of the individual forces to get the compoents of the total force, and then find its magnitude.
 
But how ? That's kinda my problem.
I have totally forgot how to do this :S
 
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.
 
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]\[\begin{align}<br /> & {{F}_{x}}=\frac{{{\mu }_{0}}{{I}^{2}}\sqrt{2}}{4\pi a}\cos \left( \pi /4 \right) \\ <br /> & {{F}_{y}}=\frac{{{\mu }_{0}}{{I}^{2}}\sqrt{2}}{4\pi a}\sin \left( \pi /4 \right) \\ <br /> \end{align}\][/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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