Dimensional analysis

[Mr Skeptic]

You can not really comment on my result if you can not show how do you obtain your result. So what is your result, can you show your work here with equations and mathematics, or simply give a code for Mathematica?

 

Showing you the work has been done plenty and doesn't seem to help much. Mostly because you are trying to do the wrong work.

 

http://www.bipm.org/en/si/si_brochure/chapter2/2-1/ampere.html

See the infinitely long wire? The problem is that you aren't using one.

 

Let me see your solution for parallel and perpendicular wire, eh? I wonder if you get zero force with perpendicular wires or perpendicular loop around straight wire... actually, with infinity, you can hardly get anything else but infinity, which means - ERROR, DOES NOT COMPUTE.

 

Nope. Infinity is perfectly easy to deal with. Read the definition carefully, it uses infinitely long wires.

 

That's good, because I have been repeatedly saying it... though I can do that in many different ways, as I also demonstrated previously. -- So, you are complaining that I use "1m"? But, how else do you expect me to tell you the force PER METER, if I do not integrate OVER ONE METER?

 

So then go ahead, do dimensional analysis: is your answer in force per meter or not? If not, you are solving the wrong problem. Correctly solving the wrong problem, but that doesn't make it the right problem.

 

The integral itself is what give us this another dimension and with it - a new UNIT. It is absolutely necessary to ingrate over 1m, because that is how ampere unit is defined - force BETWEEN TWO wires of 2*1-^-7 Newtons per meter of length.

 

Infinite wires, my friend. The wires are infinitely long.

 

I solved the problem in five different ways now, you have the math of it all right in front of you. I do not want to argue as I think this can not be said and demonstrated any more clear than this, but if not, then tell me - what do I need to do, or show you, or demonstrate to convince you?

 

Your answer is a correct answer to the wrong problem. You have to solve the right problem correctly, then you will get the right answer to the right problem.

 

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If you have trouble working with infinity, we can help with that (probably best to do in a new thread)

[Cap'n Refsmmat]

Ok, let's see that now, your turn.

 

Can you do it the way you solved the equation before, just using the formula? I mean, you didn't integrate it before.

[darkenlighten]

What I would like to see ambros is the exact fields of those two wires. The field of a loop of circumference of 1m and a straight wire of length of 1m.

[mooeypoo]

You 're hallucinating my young friend. What you said is a lie, amusing and interesting lie, disgusting too. -- Can you say what step exactly you imagined might be incorrect in my calculation? Why can you not print out that solution in Mathematica code so everyone can see it online?

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Ok, let's see that now, your turn.

I am not going to get into the science, you have enough help in that department. I will say, however, that it is a matter of common decency as well as SFN rules that you drop the attitude and start cooperating with people in a civil manner.

 

You can agree or disagree with people all you want, but if you don't stop being obnoxious about it, you won't remain here.

 

Please go over our rules and our etiquette guidelines, as they are not a recommendation. They are law in this forum, which you chose to participate in.

 

~moo

[Mr Skeptic]

To clarify what I said earlier, the thing about the infinitely long wires is this: if you calculate the force between two wires 1 meter long, then the force between two wires 2 meters long, then 3 m, and so on, each time dividing by the length of the wires, each time you will get a different result. This is a result of how the magnetic field is generated: the current 10 m away contributes to the field at 1 m away. Eventually (I assume based on the different answers), this difference comes out to double.

[mooeypoo]

You came here, you agreed to our rules, I think it's time you read them and abide by them, or you will be the one to go away.

 

~moo

[swansont]

What I would like to see ambros is the exact fields of those two wires. The field of a loop of circumference of 1m and a straight wire of length of 1m.

 

Excellent observation. If the length does not matter, as ambros incorrectly claims, then they should give the same result.

[ambros]

You came here, you agreed to our rules, I think it's time you read them and abide by them, or you will be the one to go away.

 

~moo

 

Ok, but before I leave, can you please just copy/paste that solution for parallel (perpendicular) wires as calculated with BIPM magnetic force equation everyone is talking about? -- If that supposed solution can not be expressed in Mathematica code, then I do not accept it as solution, is that reasonable?

[Cap'n Refsmmat]

It has been shown to you numerous times. Why do you want to see it again?

 

I even showed you the Mathematica code.

[ambros]

Excellent observation. If the length does not matter, as ambros incorrectly claims, then they should give the same result.

 

Funny, whenever I try to integrate over infinity - Mathematica tells me that I get infinity. So, please, is Mathematica wrong or is your solution wrong, because Mathematica says - IT DOES NOT COMPUTE. Maybe I do not know how to properly do it, especially since I never saw anything even close to what you be pretending to have as a "solution. -- So, can you write down that solution of yours in Mathematica code, please? What result do you get for the loop wire perpendicularly placed around the straight wire?

[Mr Skeptic]

Here's something that will compute: make the wires longer (but still finite), and then divide by the length.

[ambros]

It has been shown to you numerous times. Why do you want to see it again?

 

I even showed you the Mathematica code.

 

That code does not compute. I do not think you can use Mathematica properly if you can not make it work online, IT IS THE SAME ENGINE. -- Yes, I would like if that mess you made can actually get any result other than ERROR. Now, give us that solution for loop and wire you said you have ready, will you?

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Here's something that will compute: make the wires longer (but still finite), and then divide by the length.

 

Please demonstrate what are you talking about. Write down that integral/equation exactly, so I can try it in Mathematica, or you write it down in Mathematica code yourself, please.

[Mr Skeptic]

Please demonstrate what are you talking about. Write down that integral/equation exactly, so I can try it in Mathematica, or you write it down in Mathematica code yourself, please.

 

From:

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

Define wires of various lengths, parallel and separated by one meter:

 

[math]\frac {\mu_0} {4 \pi} I_1 I_2 \int_{<0,0,0>}^{<1,0,0>} \int_{<0,1,0>}^{<1,1,0>} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

 

[math]\frac{1}{2} \frac {\mu_0} {4 \pi} I_1 I_2 \int_{<0,0,0>}^{<2,0,0>} \int_{<0,1,0>}^{<2,1,0>} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

 

[math]\frac{1}{3} \frac {\mu_0} {4 \pi} I_1 I_2 \int_{<0,0,0>}^{<3,0,0>} \int_{<0,1,0>}^{<3,1,0>} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

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---TWO PARALLEL WIRES: i1=i2= 1A, 1m distance ---

integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0}

 

[math]\int_0^1( \int_0^1((1*10^-7 N) {1,0,0} \times ({1,0,0} \times {0,-1,0}))/1^2 dx) dx = {0, \ 1*10^-7 N, \ 0}[/math]

 

And just so you know, this is not the equation for two parallel wires -- it is a bunch of gibberish. You're taking the integral of a constant, a funny looking constant that bears no relation to electromagnetism nor to the above equations. If you don't know mathamatica's language well enough, you can do the cross product by hand and insert the proper trigonometric functions yourself.

Edited April 9, 2010 by Mr Skeptic
Consecutive posts merged.

[Cap'n Refsmmat]

That code does not compute. I do not think you can use Mathematica properly if you can not make it work online, IT IS THE SAME ENGINE. -- Yes, I would like if that mess you made can actually get any result other than ERROR. Now, give us that solution for loop and wire you said you have ready, will you?

 

http://www.wolfram.com/products/player/

 

Download that.

 

Next, download this:

 

http://www.wolfram.com/solutions/interactivedeployment/publish/download.jsp?id=4029150501&filename=ambros+and+ampere.nbp

 

Wolfram Alpha is not just a free Mathematica online -- they sell Mathematica for hundreds of bucks. It's a simplified version modified for web use.

 

With the Player I just linked to, you can see the Mathematica notebook I used. There's several integral expressions in there. I'm not sure how the Mathematica player works; if you can, re-evaluate it line-by-line to be sure the cells have the correct answers. (When I uploaded, the first integral's result was different because I had used it on a different situation. Re-evaluate it and you'll get the correct result.)

 

The second integral is the same thing, but written out.

 

The third integral is for the situation I asked you to work out.

 

Enjoy!

[ambros]

And just so you know' date=' this is not the equation for two parallel wires -- it is a bunch of gibberish.

[/quote']

 

How do you explain it gives correct results?

 

It is actually complete program that you can apply to any situation by simply changing vector orientation and magnitude according to given amperes and geometry. You can enter any lengths and any geometry possible. Any number can be substituted with any variable, including dl length aka "integral limits", try it.

 

VARIABLE:  c1             c2               r      I1        I2       r_hat
integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0}

 

 

 

Go ahead, take your text-book or whatever experimental measurements and any example you want, then I will actually use that one and the same code to solve any problem regrading magnetic force you can think of. ANY PROBLEM. Now, if that equation is gibberish, then it certainly would be impossible for it to solve all the problems, imaginary and the real world ones alike. How is that for a challenge?

 

 

If you don't know mathamatica's language well enough, you can do the cross product by hand and insert the proper trigonometric functions yourself.

 

That equation does exactly that. You can change all the parameters, any number you see you can use as variable, though unit vector should stay 1.

 

                                                |\                          
                                               |                           
   y.                                    y.    |I2= 1A*1m                 
                                               |                           
    .        I2= 1A*1m                    .    |                           
          -------------->                      |___                        
    .  z /\                               .  z /\                          
        /                                     /                            
    .  /r= 1m                             .  /r= 1m                        
      /                                     /
    .--------------> x                    .--------------> x               
        I= 1A*1m                              I= 1A*1m

 {1,0,0}x{1,0,0}x{0,-1,0}              {1,0,0}x{0,1,0}x{0,0,-1}

 

PARALLEL (LEFT):

integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0}

 

PERPENDICULAR (RIGHT):

integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {0,1,0} x {0,0,-1}

 

 

Say, for two parallel wires, you wanna input: C1=3.4m, C2= 2.5m, I1= 34A, I2= 12A, r = 0.3m

-------------------------------------------------------------------------------------------

 

Write this:

integrate[0,3.4] integrate[0,2.5] 10^-7Newtons/0.3^2 {34,0,0} x {12,0,0} x {0,-1,0}

-------------------------------------------------------------------------------------------

 

 

 

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

Define wires of various lengths, parallel and separated by one meter:

 

[math]\frac {\mu_0} {4 \pi} I_1 I_2 \int_{<0,0,0>}^{<1,0,0>} \int_{<0,1,0>}^{<1,1,0>} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

 

[math]\frac{1}{2} \frac {\mu_0} {4 \pi} I_1 I_2 \int_{<0,0,0>}^{<2,0,0>} \int_{<0,1,0>}^{<2,1,0>} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

 

[math]\frac{1}{3} \frac {\mu_0} {4 \pi} I_1 I_2 \int_{<0,0,0>}^{<3,0,0>} \int_{<0,1,0>}^{<3,1,0>} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math]

 

 

Have you tested that? That does not compute here. You seem to believe to know this better than me, so please just write one of those in Mathmatica code. Can you do that please?

Edited April 9, 2010 by ambros

[mooeypoo]

Have you tested that? That does not compute here. You seem to believe to know this better than me, so please just write one of those in Mathmatica code. Can you do that please?

It does not compute where? In your notebook? Because it does compute if you know how to compute it.

 

If these mathematical expressions are above your level, it's fine, ambros, it happens, I would just recommend not making hasty generalized judgments on something you apparently need to brush up on. Or learn more about. Either way, it's quite clear that your idea of a mathematical expression is lacking, as people pointed out, and you don't quite seem to understand the expressions people give you.

 

That's not a shortcoming of the (many) people who speak with you and their (many different) attempts to explain this to you.

 

~moo

[Mr Skeptic]

How do you explain it gives correct results?

 

Giving correct results is no guarantee that either the problem is what it purports to be, nor that the calculations done were correct. In your case, you also give the wrong result which definitely means you did something wrong.

 

Have you tested that? That does not compute here. You seem to believe to know this better than me, so please just write one of those in Mathmatica code. Can you do that please?

 

No, I know I know better than you. Also, I have run out of patience. It's a simple fact that without understanding multi-dimensional calculus, you can't really properly use multi-dimensional calculus. I do encourage you to continue trying to figure things out, but I will allocate my time to someone who actually wants to learn rather than inflate their ego pretending they know more than they do. Spoon-feeding answers to people who don't want to learn is not why I came to this forum.

 

Why don't you take a break from arguing, and maybe wait for a reply to the quarry you sent to the standards folks, or perhaps you can review what we have said and compare it to other sources.

[ambros]

 

Enjoy!

 

You want me to try to make a sense of your mess that does not compute? If you can not make it work, that only suggests you never actually used it and so your advices and opinions are just plain unnecessary.

 

My equations can solve any example in your textbook, so there is no way I will waste time looking into something you yourself do not know what it is. You could just as well offered me to drink cactus, no thanks! - Make it work and only then I will look at it, sorry.

[Cap'n Refsmmat]

It does work. Try it before you jump to conclusions.

 

Remember, Wolfram Alpha is not full Mathematica. It won't handle multiple lines of code.

[ambros]

It does not compute where? In your notebook? Because it does compute if you know how to compute it.

 

If these mathematical expressions are above your level, it's fine, ambros, it happens, I would just recommend not making hasty generalized judgments on something you apparently need to brush up on. Or learn more about. Either way, it's quite clear that your idea of a mathematical expression is lacking, as people pointed out, and you don't quite seem to understand the expressions people give you.

 

That's not a shortcoming of the (many) people who speak with you and their (many different) attempts to explain this to you.

 

~moo

 

I can demonstrate it any way you want. The question is what is your level and have you got ant idea what you talking about.

 

 

TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m

*DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.3

================================================

 

[math]B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}[/math]

 

 

[math]=> \frac{4\pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times 1}{4\pi*1_{m^2}}[/math]

 

[math]=> \frac{10^-7_{N/A^-2}*1_A*1_m }{1_{m^2}} \ =OR= \ \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m}[/math]

 

[math]=> \frac{10^-7_{N/A}}{1_m} \ = \ 10^-7 \ N/A*m[/math]

 

 

[math]F = Q*v \times B[/math]

 

[math]=> 1_C*1_{m/s} \times 10^-7_{N/A*m}[/math]

 

[math]=>1_{C/s}*1_m \times 10^-7_{N/A*m} \ =OR= \ 1_A*1_m \times 10^-7_{N/A*m} \ = \ 10^-7 \ N[/math]

 

 

I hope this is not above your level, so please do you see a mistake and can you point it out exactly what term in what step do you believe is not correct?

[Cap'n Refsmmat]

Okay, so you calculated the magnetic field generated by one coulomb of charge moving at 1m/s. But there can be more than one coulomb in a wire at a given moment; it's just that one coulomb has to pass a given point in one second to make an ampere.

 

In other words, you found the magnetic field from one packet of charge. But there are many identical packets in a wire, at different points in the wire.

[mooeypoo]

Think of it like water coming out of a faucet. Calculating the mass of a single drop traveling with the stream doesn't solve for the entire stream of water. So this isn't "wrong answer", it's just the wrong question. You're talking about the ENTIRE stream -- the current, not just one single 'packet' of charge moving and making its own little current -- or not just a single drop of water in the stream.

 

~moo

[ambros]

Okay, so you calculated the magnetic field generated by one coulomb of charge moving at 1m/s. But there can be more than one coulomb in a wire at a given moment; it's just that one coulomb has to pass a given point in one second to make an ampere.

 

In other words, you found the magnetic field from one packet of charge. But there are many identical packets in a wire, at different points in the wire.

 

But how do you explain 2X coulomb can produce the same force as 2x 1meter of wire? Because the former is "per point", and the later is "per meter", that's it.

 

You are actually close when talking about dividing distance with distance, but I explained all that long ago, and really, the only possible way to do it correctly is that little piece of code I gave above. You can seriously use it in real life and go on and solve all the problems that have to do with the magnetic force, try it.

[Cap'n Refsmmat]

But how do you explain 2X coulomb can produce the same force as 2x 1meter of wire? Because the former is "per point", and the later is "per meter", that's it.

Yes, and that's why your above calculation is wrong for a wire. You say it's for two parallel wires, but it's in fact for two point-like packets of charge. Hence you don't get the same answer.

[ambros]

Think of it like water coming out of a faucet. Calculating the mass of a single drop traveling with the stream doesn't solve for the entire stream of water. So this isn't "wrong answer", it's just the wrong question. You're talking about the ENTIRE stream -- the current, not just one single 'packet' of charge moving and making its own little current -- or not just a single drop of water in the stream.

 

~moo

 

My little Mathematica code for Lorentz force can solve any example in your text book that has to do with any magnetic forces, it can solve it CORRECTLY, so there is no way you can call such equation "wrong".

 

 

To call it wrong you should first provide what you believe is correct, or at least you should be able to find some example for which my equation will give the wrong answer, but If it always gives correct answer, then it is not wrong, right?