darkenlighten
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Everything posted by darkenlighten

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.

Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
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. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
The issue is that you are not solving the integral properly. If you are not sure and would like to know how to solve it properly look at http://www.scienceforums.net/forum/showpost.php?p=554213&postcount=47 This will make it clear as to why you cannot say dl is 1 meter, but it is the integration variable. It is simpler to solve for the magnetic field first (especially so you can see the correct way to solve the integral) and then substitute it in the force equation. And to be clear, I originally solved the magnetic field for an infinite wire using Ampere's Law out of Maxwell's equations, but used the BiotSavart Law to show you where that double integral came from. 
Okay so this is what I was able to do: [math] {\mathbf{r}} = r \hat{r} = r ( cos\theta\hat{\mathbf{x}} + sin\theta\hat{\mathbf{y}}) [/math] [math] {\mathbf{v}} = v (sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math] [math] \mathbf{a} = \frac{d\mathbf{v}}{dt} = v (\dot{\theta} cos\theta\hat{\mathbf{x}}  \dot{\theta}sin\theta\hat{\mathbf{y}}) [/math] 1) [math] \mathbf{a} =  v\dot{\theta}\hat{\mathbf{r}} [/math] 2) [math] \dot{\theta} = \frac{d( \omega t)}{dt} = \omega [/math] [math] \dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math] 3) [math] \mathbf{v} = v = \sqrt{ (r \dot{\theta})^2((sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}))^2} = r \dot{\theta} [/math] Then using 1), 2) and 3 you get [math] a = \frac{v^2}{r} [/math] Is this what you were wanting?

It would still be geometry/trig. What is your concern with it?

Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
I did, the equation above is a derivation from the Force equation: [math] \mathbf{F}_{12} = I_2 \int d \mathbf{l} \times \mathbf{B}_1 ;[/math] where [math] d\mathbf{l} = d\mathbf{s}_2[/math] and [math]\mathbf{B}_1 = \int\frac{\mu_0}{4\pi} \frac{I_1 d\mathbf{l} \times \mathbf{\hat r}}{\mathbf{r}^2};[/math] where [math] d\mathbf{l} = d\mathbf{s}_1[/math] and [math] \mathbf{r} = r_{12}\hat{\mathbf{r}}_{12} [/math] [math] \Rightarrow \mathbf{F}_{12} = I_2 \int d \mathbf{s}_2 \times \int\frac{\mu_0}{4\pi} \frac{I_1 d\mathbf{s}_1 \times \mathbf{\hat r}_{12}}{\mathbf{r}_{12}^2}[/math] cleaning it up and we get: [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \int \int \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}[/math] As you can see I solved for B and then used that in the [math] \mathbf{F} = I \int d \mathbf{l} \times \mathbf{B} [/math] in my example in post # 59. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
Because it has already been done, refer to post #49 with the derivation (the math) that shows how to properly setup and integrate the BiotSavart Law for a wire of length L and steady current I. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
And so did I back on page 3, post #59 lol 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
No this is not true. If you look at the derivation again. [math] q \mathbf{v} = q \frac{d\mathbf{x}}{dt} = \frac{dq}{dt}\mathbf{x} = I \int{d\mathbf{l}} [/math] This means the Current times the LENGTH of the wire, which is [math] x = \int_0^L d\mathbf{l} [/math] where L is the length of the wire. dl is not the length of the wire, you've stated that yourself, it is an infinitesimal piece of the whole length. You know vector calculus, so you should have no trouble understanding how a line integral works. And since dl IS a vector, it will have a magnitude and direction, which means it can change with respect to the distance away. Therefore the contribution from the whole length is not confined to the perpendicular plane. I think the picture is confusing you. As swansont said, due to symmetry the field only points in the azimuthal direction, but is caused by the rest of the wire. It's unfortunate you've misunderstood me for being arrogant, you might have stopped to think about what I've been posting and realized your mistake. There is a fine line between arrogance and confidence. I have been studying this subject for almost 3 quarters now, it's not that I'm arrogant, it's that I know I am right. Just because you don't believe me does not make me arrogant, because you think that I think I'm right. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
Also to make it clear: [math] q \mathbf{v} = q \frac{d\mathbf{x}}{dt} = \frac{dq}{dt}\mathbf{x} = I \int{d\mathbf{l}} [/math] 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
I'm going to state this again...( I said this in a previous post) Hopefully this will understand why you cannot look at just 1 element of dl, but the sum of all dl's (hence an integral) At point "P" away from the wire there is a magnitude of the magnetic field that is caused by each segment of wire, not just by the segment below it, otherwise it would make it act no different than one moving charge, which is obviously not true. For example: take multiple moving charges equally spaced apart traveling at the same speed (a steady current) and confine them in 1 dimension (a straight line). They all have their own magnetic field, but if you look at a point in space, the total magnetic field at that point will be due to the superposition of all of the moving charges magnetic fields. Since their magnetic fields depend on distance away, this correlates to the length of a wire. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
Okay, though I've repeated myself. The first 2 equations, the E field I have there can be ignored, since we have already corrected that saying that the E field is zero, E = 0. The B field there is correct with the reference of the scanned pages of Griffith's text showing the full derivation using the BiotSavart Law. I used Maxwell's equations to derive everything there, I don't need a reference using Maxwell's because my math is not wrong, so the reference I have is to show I have the correct answer. As far as changing B field scenario, I do not have a reference since that was a scenario I made up, but since you do not believe the previous answer, you will not believe the last 2 I gave here. Which you apparently had a problem with a changing Magnetic field creating an Electric field...which is silly. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
No its not clear to you, because this is incorrect. The length contributes and yes there is such a variable [math] \int d\mathbf{l} [/math]. You claim to understand vector calculus but you have been wrong about this the whole time. At point "P" away from the wire there is a magnitude of the magnetic field that is caused by each segment of wire, not just by the segment below it, otherwise it would make it act no different than one moving charge, which is obviously not true. For example: take multiple moving charges equally spaced apart traveling at the same speed (a steady current) and confine them in 1 dimension (a straight line). They all have their own magnetic field, but if you look at a point in space, the total magnetic field at that point will be due to the superposition of all of the moving charges magnetic fields. Since their magnetic fields depend on distance away, this correlates to the length of a wire. Technically it is more than one field, due to the example above (about multiple moving charges). That is not what it means when it says the forces between the wires. Is one force not left and the other right, hence one negative and one positive? If you sum them, you will get zero. No you are misunderstanding what it means "the force between the wires". What it means is the force that one wire feels due to the external magnetic field of the other wire. Once again I have already provided reference, but you refuse to accept it, why? 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
Two Questions ambros: 1) Do you think that a wire of length 1m with steady constant current I will have a different magnetic field than a wire of length 2m? 2) When you are solving for, what you call, the total magnetic force, why are you adding up the individual forces from 2 different objects and why is one of them not negative? Answering these straight forward will put us back on the correct path. If you choose to ignore them and not answer directly, this discussion is going to continue to be circular. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
That doesn't mean I can't consider this whole system, but that is hardly the point. There is only 1 Force acting on each object and that force is 2 x 10^7 N/m derived as I have shown. Um yea it does. It solved the infinite wire with steady current I using the BiotSavart Law...And once again do you understand why you look at dl and r? Think of it this way, say you start with a piece of wire that is 1L long (that somehow has a steady current), it will have a certain magnetic field. Then look at another piece of wire that is 2L, it will have another magnetic field different than that of the wire of length 1L. If you agree with that, then you should realize that dl x r is not 1 and should refer to the correct derivation 14 posts before this. While these sources are correct, you do not solve them correctly, which is the half the point here. And from the source http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law it shows what I am saying read it: [math]F_m = 2 k_A \frac {I_1 I_2 } {r} [/math] Not quite. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
On the contrary: You would still be wrong if that meant the net force, but the net force here is zero, and that is not what it means when saying the force between the wires. And you are off by a half on each and since you are working with a unit of 1m it being r^2 or r will not make a difference for this example. This is the correct answer: [math] \mathbf{B}_2 = \frac{\mu_0 I_2}{2\pi s} \hat{\phi} ;[/math] [math] \mathbf{F}_1 = I_1 \int{d\mathbf{l} \times \mathbf{B}_2} [/math] [math]\Rightarrow F_1 = \frac{\mu_0 I_1 I_2}{2\pi s} \mathbf{\hat{r}} [/math] per unit length [math] \mu_0 = 4\pi\times 10^{7};[/math] [math] I_1 = I_2 = 1 A;[/math] [math] s = 1m [/math] [math]\Rightarrow F_1 = 2\times 10^{7}N[/math] per unit length One force is negative one force is positive, look at your own diagram. So [math] F_{net} = F_1 + F_2 = 2\times 10^{7}  2\times 10^{7} = 0[/math] Please answer me again when I ask: Why do you not trust my source? Also, where you got the picture from even states I am correct: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
So as stated before, we already determined that [math] \mathbf{E} = 0 [/math] and the B field I gave there is correct. This is not correct, reason being is that you are dealing with a wire and all of the pieces contribute towards the total magnetic field, therefore there has to be an integral to sum up all of the pieces, which also means that [math] d\mathbf{l} \times \mathbf{r} [/math] is not 1 but what is shown in my previous post with the reference from Griffith's text. So why do you refuse to believe a highly reputable source? I've already gave enough information for why these are correct. I am using correct notation, as cylindrical coordinates are best, if you are not familiar with them, read about them and it will make things easier. Also I think the underlying issue is that you do not believe that a changing B field creates an electric field??? Which I don't understand considering that's electric motors work. A changing magnetic field induces an emf (electromotive force) which is a vector potential (a voltage) that corresponds to an electric field. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
I'm sorry for being rash. This is just kind of frustrating, so I will probably stop here. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
No. I already stated that was an example for a changing B field. Stop being a fool and read back through. I am done. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
The reference in my previous post. 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
Call it what you will, doesn't bother me any. Lets start with this. First that is not what I gave for this specific example. If you would read back through, which I feel you fail to do. This was the example I gave for a changing magnetic field, where I was NOT constant. And we already determined for this E = 0 for a wire, due to its net charge, but this is not true if the B field is changing, which was my example. Yes it can, but not the E field that is due to an electrostatic charge, but due to changing B field, hence a changing current or moving magnet. You are wrong. So lets take a look at the derivation and as to why you are wrong: Taken from "Introduction to Electrodynamics" by David J. Griffiths Page 216217 Looking at Faraday's equation, you can see that an Electric Field can exist if there is a changing B Field. Therefore, in MY example for a changing current [math] I = I_0 cos(\omega t )[/math], there will be an electric field around the wire due to its changing magnetic field. Hate to break it to you but [math] \hat{r} [/math] is a polar coordinate, hence a cylindrical coordinate, so we are in the same format, you just don't see it. Once again due to the changing B in MY example. I have gave another example where an E field exists due to a changing B Field. And once again in YOUR example with constant I, E = 0 . 
Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
I am saying what you have, [math] \mathbf{B}®= \frac{\mu_0 I}{4\pi r^2} [/math], is wrong. You cannot ignore the [math] d\mathbf{l} \times \mathbf{r} [/math]. So I will repeat, that is not the simplified solution. Please refer back to my earlier example showing what the actual solution is. I don't feel like scanning my book again, but trust me it is a legitimate source. Heck it's even what wikipedia has sourced. I really don't feel like it lol So the new example I gave had current as [math] I = I_o cos(\omega t) [/math], so therefore this is not a constant current, but it is a steady current. This is what type of current comes out of the wall with a frequency of 60hz, aka alternating current. Okay so it might not be easy to tell, but that is not correct. Flux is actually denoted by capital phi: [math] \Phi [/math] and the lower case phi: [math] \hat{\phi} [/math] is the azimuthal direction in cylindrical coordinates. So in my case it is a unit vector describing a direction. And as I stated before, [math] d\mathbf{l} [/math] is not just a direction, but a vector, just dropping it out and assigning its direction is not the whole story. which is why the equation you gave is wrong for our earlier example with constant current I. Once again it has to do with the [math] d\mathbf{l} [/math] Not at all...how do you figure this, I don't know how many times I need to stress this. Do you know what the derivative of [math] cos(\omega t) [/math] is? It is definitely not zero (spoiler it is [math]  \omega sin(\omega t) [/math]). And yes there will be an E field in this scenario since the B field is changing!!! That is the whole point. If there is changing B, there will be an Electric Field and since the Magnetic Field depends on current, the Electric Field will also. If the current is constant means no electric field (no change), if the current is changing, there will bean electric field. Put it this way ambros, I don't feel that you will really understand what is going on here until you further your studies in vectors, spherical and cylindrical coordinates and integrals. Because I have more than enough shown you when and why Maxwell's equations are extremely useful and that they are also not always useful, and sometimes you need to use Coulomb's or BiotSavart Laws. But you are lacking if you don't know when to use what and why. I'm sorry I cannot help you further if my explanations do not do it justice. Thank you though. I'm not sure how much more effort I am going to put into this. We will see how much time I have, Spring Quarter is starting tomorrow and I need some sleep 
Sounds like you should have went to Ohio State and gone for Engineering Physics with a concentration in Electrical and Computer Engineering, which is what I am doing. I wanted the same approach. I was initially thinking Electrical Engineering, but really liked physics and wanted to understand why things work and not just how things worked. Its been really good, but as stated before, if you are good at what you do, you should get far. I would still see what other colleges offer you, and what it would take to get into one of your liking.

Maxwell's equations: meaning, derivation and applicability
darkenlighten replied to ambros's topic in Classical Physics
To start, the reason I (and if you work with these equations) do not use the constant k, is because it does not always appears because some things cancel out the 4 or the pi. This has to do with the geometry of the electric and magnetic fields. And yes what you get is necessary for vector form because it describes the whole picture. The reason it looks different (and the field is different), is because it is due to the geometry of the object. All of the examples I have given have been different scenarios with different geometry. Here a couple of scenarios: A magnet in motion near a stationary object, causing a change in the B field Or a timevarying current causing a changing B field. The reason you don't "see" any constants is because you can derive the changing B field or you might already know it. Which would be done from Ampere's Law. Take a look: Say you have wire with current [math] I = I_0 cos(\omega t)[/math] what E field is it producing? So [math]\triangledown \cdot \mathbf{E} = \frac{\partial \mathbf{B}}{\partial t}[/math] [math] \Rightarrow \oint \mathbf{E} \cdot d\mathbf{l} =  \int \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{a}[/math] and using Ampere's Law you get [math] \mathbf{B} = \frac{\mu_0 I_0 cos(\omega t)}{2 \pi s} \hat{\phi} [/math] [math] \Rightarrow \mathbf{E} =  \frac{\partial}{\partial t}[\frac{\mu_0 I_0 cos(\omega t)}{4\pi}] \hat{s}[/math] [math] \Rightarrow \mathbf{E} = \frac{\mu_0 I_0 \omega sin(\omega t)}{4\pi} \hat{s} [/math] which makes sense for an infinite wire that E would not depend on distance away The equation you have is wrong, I can show you the derivation from "Introduction to Electrodynamics" if you would like. And as far as the second equation goes. I am unfortunately not experienced with that, from what I've seen it is more of a reason why these equations work the way they do rather than using it to actually solve for the B field. So I do not currently have an example for the 2nd equation that can solve directly the B field, but don't worry we have Ampere's Law mainly for that. Once again please, please look and understand why we are using loops when solving for B, it shows how the Magnetic Field is behaving. As for why I had a t in there for the electric field for the most recent example was due to that fact the electric field is caused by charges (or a changing B), so if you have a current I that is equal to [math] I = \lambda v[/math], where [math]\lambda[/math] is a line charge. So that was my attempt to get it into terms of charge, since I is Coulomb's per second, the charge in your example for 1 ampere would have been 1 Coulomb. So my equation was like a snapshot or an observation in a certain time interval, since the charge was changing at a rate proportional to current. What are you asking about when say "why is this important"? 
Ah crap we are doomed...