1. Mar 27, 2010

### Denver Dang

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:
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

$$\frac{F}{L}=\frac{{{\mu }_{0}}II'}{2\pi r}$$

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:

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

and

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

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:

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

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

$${{F}_{tot}}=\frac{\sqrt{10}{{\mu }_{0}}{{I}^{2}}L}{4\pi a}$$

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. Mar 27, 2010

### gabbagabbahey

You said it yourself, the individual forces are vectors. What are their directions?

3. Mar 28, 2010

### Denver Dang

Well, something like this:

And sorry for the not so pretty paint picture :)

Last edited by a moderator: May 4, 2017
4. Mar 28, 2010

### gabbagabbahey

Okay, so add the components of the individual forces to get the compoents of the total force, and then find its magnitude.

5. Mar 28, 2010

### Denver Dang

But how ? That's kinda my problem.
I have totally forgot how to do this :S

6. Mar 28, 2010

### gabbagabbahey

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. Mar 28, 2010

### Denver Dang

So, the x-component of F2-1 is:

$$${{F}_{x}}=-\frac{{{\mu }_{0}}{{I}^{2}}}{\pi a}$$$

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

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

\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}

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

$$$F=\sqrt{{{\left( \sum{{{F}_{x}}} \right)}^{2}}+{{\left( \sum{{{F}_{y}}} \right)}^{2}}}$$$

Argh... So simple !

Thank you :)