darkenlighten

Damn good question actualy!

 

I know electrons take every possible path from A to B, the Photon may do this also, although it makes me wonder about Lasers and coherence.

I'm going to have to disagree.

I don't believe electrons take every possible path (I would like to know where you heard that from or why you know this), though they have the ability to take certain paths and based on its wave function you can find the probability of where it will be, but it does have a definite location.

The wave function (or its probability wave) is not the actual wave-like property of light or matter. The wave function is merely a mathematical model of what you are describing, in the case of an electron, its usually position, but can also be spin or something of the like. Depending on how you construct the hamiltonian (energy of the system) of Schroedinger's equation is what determines the physical entity you are describing.

So as far as I know: No light is not a probability wave, because the probability wave is not something physical, but a description of something physical.

Edit: Also, anything can have a probability wave if you want it to.

lol how did this one get get pulled out of the stack, just realized this is from 2007, gah I dislike posting on old stuff.

I like how he completely ignored my post: postcount 111

ambros, you are still not getting the correct results. The force on one wire is not 1 x 10^-7, it is 2 x 10^-7.

Taken from http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law

"The best-known and simplest example of Ampère's force law, which underlies the definition of the ampere, the SI unit of current, is as follows: For two thin, straight, infinitely long, stationary, parallel wires, the force per unit length one wire exerts upon the other in the vacuum of free space is:

[math] F_m = 2k_A\frac{I_1 I_2}{r} [/math]"

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.

Stop repeating someone else's opinion. What do you imagine how words "is wrong" can have impact on me, and what it can possibly contribute to the discussion? So lame, pathetic... go away!

Hmm I'm sure I was first to say you were wrong, but what do I know its not like I've spent the last 7 months at The Ohio State University taking an Electrodynamics course, nah no not me, I'm not the Engineering Physics student. Oh wait...

I challenge you this ambros:

Obtain the exact fields of both wires. Then obtain the force felt on each wire from the external magnetic fields.

I believe this will be best to understand what is going on is to be exact, no approximations.

I will do the same and provide reference and compare.

Big manz on the internetz.

But really, you are not integrating right and have the wrong B field. That is your mistake. I did it correctly. There is not just 1 charge in a wire, you need integrate properly in order to solve for a wire.

And I never argued that you couldn't use the Biot-Savart Law, but for an infinite wire, it is heck alot easier to solve using Ampere's Law.

Oh my!!! Open your eyes man....

Read the last post again. The equation from the BIPM did not just mysteriously come into existence. I showed you how it came to be!!!! Using the Lorentz Force Law and Biot-Savart Law and combining them you get what you are so fond of. And I have solved that.

Edit: Your original post was more funny

You are dreaming. Go ahead, show us that post where we can see how you put REAL VARIABLES of: I1= 1A, I2= 1A and r=1m, in THIS EQUATION:

 

[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]

 

What post, what number? What say you?

postcount 47 for the correct derivation of the B field (just look at the scanned text)

postcount 59 Shows the straight forward method

postcount 122 showing you how the equation from BIPM was derived.

Q1: Did you ever used this equation to calculated the attraction of two parallel wires 1m apart with both having steady current of 1A:

 

[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]

 

YES/NO?

Yes I did.

Q2: Are you hallucinating, laying and fulling yourself, or you actually can SHOW YOUR WORK so we can see your LOGIC and MATHEMATICS?

 

Until then, you're only amusing me and fooling yourself.

I wish I was.

I showed you how this equation was derived. You ignored it. I showed you the steps of integration. You ignored it. I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. You ignored that.

I'm feeling a strong trend going on here.

There is no such thing as zero distance, just like there is no edge to the universe. I explained this in previous thread. -- It is 'line integral', why is that under question? The trick is that if you do "point integral" it still works.

The only thing I know that is close to a "point integral" is an impulse. And this is clearly not what we have. This is the ENTIRE argument ambros. It has nothing to do with which equation is right or wrong. It's the fact you are calculating it WRONG, end of story.

How bout this look: [math] I \int d\mathbf{l} = I x [/math] where x is the length of your straight wire. We will take out the integral since we are working with a straight path. Try it now, you will see the problem for the infinite wire.

CAN YOU RESPOND TO THIS:

 

[math]

I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}

[/math]

 

What speed is that of 1C/s?

[math] I\int d\mathbf{l} \Rightarrow 1 A\cdot meter [/math]

dl is not dimensionless, because it is the length of the path.

You also have to remember that just because you have the correct dimensions, it does not necessarily mean you have the correct answer. It is just a step towards making sure you do. The next step is checking the math.

You do not understand what BETWEEN TWO wires mean.

 

If the wanted to say "the force acting on ONE wire", they would have said so.

 

If I hit you, and you hit me back with equal force from opposite direction, then BETWEEN TWO of us we exchanged two hits, and if someone told you the actual force felt "BETWEEN US" only "count" when I hit you, then you would object, wouldn't you?

This is just silly. You do realize that one force is negative and the other is positive, if you add them together, you get zero??? This is not what they mean by the force between the wires, they say that only because both wires are involved in making a force on ONE wire.

So you do not understand what [math] q\mathbf{v} = I \int d\mathbf{l} [/math] means

I'll go over it:

First of all it's not going to be just q of one charge, but the charge density. So for a wire it is [math] \lambda [/math] therefore [math] \lambda\mathbf{v} = \lambda \frac{dx}{dt}[/math], now when you take it further you get [math] \frac{d\lambda}{dt} x [/math], this means the change in charge over time multiplied by the distance; hence [math] I \int d\mathbf{l} [/math] is the change in charge with respect to time multiplied by the length of the path.

And you are not getting the right answer. It should be 2 x 10^-7 N/m on ONE wire.

I know what you are saying, and that is not what I meant. I'm not meaning to argue, because of course you are correct. I merely meant the magnitude could be changing and still be following a circular path with radius r, for example: if the angular frequency was changing, but the distance from the center. It's really a mute point and no longer helps the poster.

absolutely wrong.

Not true. In uniform circular motion the magnitude of the velocity vector is, by definition, constant. In non-uniform circular motion the magnitude of the velocity vector is not constant.

I meant to say the vector v is not constant, while of course the magnitude is. And that is what I said but nonetheless we are both correct in what we have stated.

But remember it just really depends how v is changing, because here v is NOT constant. The magnitude is constant, but the direction is always changing, otherwise we would not have an acceleration. Now you could get a v where the the magnitude was changing, but the direction was the same (circular motion) and [math] a = \frac{v^2}{r}[/math] would still hold, but the v (magnitude) could be a function of something else. Either way the conditions will not be exactly the same.

What is it your were talking about, again? Equations give the same result for any length, and how about conspicuous absence of LENGTH variable in Ampere's law? Am I dreaming right now, or was this all a part of _your hallucination? Distances change, angles change and fields superimpose over length, right?

 

Apparently, you are mistaken - the length of the wire is not even a part of the Ampere's force law equation, and so it obviously does not contribute, nor it even makes any difference in affecting the amount of the total force, per unit length.

Yea once again you are taking things out of context. As swansont said, the one we are giving is for an infinite wire, it gives you the force per unit length.

Also because it makes it a function of time, since [math] \theta = \omega t [/math] this is a more convenient notation when working with a physical system.

This is my main objection , how do we prove that v is perpendicular to r

I proved it here:

[math]\dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

and then you can take a dot product of v and r to see they are perpendicular.

ambros stop being so hasty. You are reading it wrong.

I was not saying it was my answer divided by a length, but the force per unit length so:

[math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}}[/math]

So [math] F_1 [/math] per unit length [math] \Rightarrow F_1 \ \ (N/m) = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} \ \ (N/m) [/math]

Thanks ,but you have to explain each step because i am a bit confused:

 

How do you get :

 

[math] {\mathbf{v}} = v (-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

yea we will go step by step, just keep asking questions.

So this is just vector v, which consists of its magnitude and direction, since v is moving in a circular path the sin and cos show which direction it is pointing:

For example if it is at the angle [math] \theta = 90 [/math] (at the top) the vector is [math] \mathbf{v} = v (-\hat{\mathbf{x}}) [/math] pointing only left in the x direction.

Why the middle equation has units in Newtons and the next one in N/m?

Because you left out the rest of it:

[math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}}[/math] per unit length, which it does.

Since we are looking at a infinite wire you can't just say the force, since [math] L = \int d\mathbf{l} = \infty [/math]. So it needs to be force per unit length.