Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 No, those equations are different and you will only know that if you actually use it. I repeat, "your" formula is not the same as THE DEFINITION GIVEN by the BIPM SI Units brochure, 8th Edition, PAGE 105. You also said my integration of it was wrong, so you need to show what you believe is the "correct" integration by applying it to the two parallel wires and unit of ampere setup, and see if it will indeed give the same result as "your" formula. -- Prove your assertion, argument your opinion, demonstrate your theory and show your work. It's a few lines of math, can you that please? So Wikipedia and the three sources it cites are wrong, yes? To say that the equation I used tells us the force that we're looking for? Just curious. I'll work on doing the integral; I have other work to do at the moment.
ambros Posted April 5, 2010 Author Posted April 5, 2010 (edited) So Wikipedia and the three sources it cites are wrong, yes? To say that the equation I used tells us the force that we're looking for? Just curious. I'll work on doing the integral; I have other work to do at the moment. Ok, but there is nothing to integrate really, especially in this example as all those are unit vectors and magnitudes of value one so everything on the right becomes "one", and then all what's left is magnetic constant. Wikipedia is actually very clear as in the very first paragraph it states that this force is defined by Lorentz force and Biot-Savart law. It is really a gentle way of saying that Ampere's force law itself is not good enough. Then, further down the page they give this equation we are now talking about as is defined by BIPM, but if you look at it you will see it is not really Ampere's law, but fusion of proper Biot-Savart law and Lorentz force: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} \ \ \&\ \ \ F = I*L \times B\ => \ F_{12} = \frac {\mu_0} {4 \pi} \frac {I_1*L_1 \mathbf{ \times} \ (I_2*L_2 \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] Merged post follows: Consecutive posts mergedAlso 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] Yes, and do you realize now we are actually solving the same problem from the very beginning with SINGLE ELECTRON. Yes, we are solving for INFINITESIMAL segments "dl", and that means we are solving for point charges again, only these magnetic fields in wires theoretically do NOT EXIST in front or behind their perpendicular plane, not in any of our equations anyway. In two parallel wires, these "magnetic slices" do not interact with any other slices but only the ones of the other wire whose slices are in that same plane. There are no varying angles and no superposition here. Edited April 5, 2010 by ambros Consecutive posts merged.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 Ok, but there is nothing to integrate really, especially in this example as all those are unit vectors and magnitudes of value one so everything on the right becomes "one", and then all what's left is magnetic constant. This is where you've screwed up. Here's how the double integral is working: ----1------2--------3----4------5--- -------------------A---------------- For each point on the upper wire, calculate the force caused by the current there on point A, and sum them. You will notice that r varies. Next, move A along the entire lower wire, repeating the sum each time. This is how the integral works. That's the only way you're going to find the complete influence of one wire on the other -- remember, magnetic fields extend an infinite distance, so the force on point A depends on the field from every point on the upper wire, and the force on the entire lower wire depends on the forces at every point on the lower wire. That is why your integral disagrees with every other source I have pointed you to. It's also why I don't yet know how to perform the integration correctly -- that's one nasty integral. But every source I point to shows that my formula ([imath]\frac{\mu_0 I}{2\pi R}[/imath]) works, so I think the integral will work out. Now, again, can you cite sources to support your claim? Ones that directly state that your formula is the correct formula for the force experienced by a wire?
ambros Posted April 5, 2010 Author Posted April 5, 2010 (edited) This is where you've screwed up. For each point on the upper wire, calculate the force caused by the current there on point A, and sum them. You will notice that r varies. Next, move A along the entire lower wire, repeating the sum each time. This is how the integral works. That's the only way you're going to find the complete influence of one wire on the other -- remember, magnetic fields extend an infinite distance, so the force on point A depends on the field from every point on the upper wire, and the force on the entire lower wire depends on the forces at every point on the lower wire. I know your opinion, but there is no need to assume - just calculate it. That is why your integral disagrees with every other source I have pointed you to. It's also why I don't yet know how to perform the integration correctly -- that's one nasty integral. But every source I point to shows that my formula ([imath]\frac{\mu_0 I}{2\pi R}[/imath]) works, so I think the integral will work out. Now, again, can you cite sources to support your claim? Ones that directly state that your formula is the correct formula for the force experienced by a wire? I cited them several times and will do it again. -- But, it is you who instead of expressing opinion how these integrals might work out, should use the math and ACTUALLY CALCULATE IT. We have perfect example to apply this and it will definitively prove or disprove validity of either equation, so please, just solve those integrals with given variables as defined by the two parallel wires and ampere unit scenario. 1.) I say this is correct formula to calculate magnetic field of a straight current carrying wire in relation to distance, and I quote my source: http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law - "The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is: [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} [/math] I also say this "other" equation: [imath]\frac{\mu_0 I}{2\pi R}[/imath] will not produce same results as above equation, and which one is generally correct can only be determined by the reality, and/or our definition of unit system, that is by solving the example of two parallel wires as defined by the ampere unit. Since you do not believe me, you then absolutely must solve that equation given below yourself, especially if you mean to keep saying my integration of it was wrong. --- CONFIRMATION EXPERIMENT --- 2.) In the case of two parallel wires, I say this is correct formula to calculate magnetic force exerted on ONE wire due to the B field of the other wire, and I quote my source (in integral form): http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [math]F_{12} = \frac {\mu_0} {4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*d \mathbf{l_1}\ \mathbf{ \times} \ (I_2*d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {|r|^2}[/math] I also say that to calculate the force BETWEEN the two parallel wires, as defined by the unit ampere, we need to take into account BOTH forces, that means: F(m) = F(12)+F(21) = 2 * 10^-7 N/m. -- Both of these claims can be refuted or confirmed with one thing, and all you need to do is to solve that one equation for that particular case scenario given by the BIPM and ampere unit, and then see what result will you actually get. Can you do that please? ...alternatively, you may use "double differential" form as given by the BIPM, or this one with slightly different notation: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law [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] Edited April 5, 2010 by ambros
swansont Posted April 5, 2010 Posted April 5, 2010 (edited) I checked everything. Now, you have to check it yourself. -- There is no need to be wondering about units, both equations have the same units but there is only one valid relation to correctly describe the magnitude of magnetic field B, of straight current carrying wire, in relation to distance: [math]B® = \frac{\mu_0 I}{2\pi r}[/math] -OR- [math]B® = \frac{\mu_0 I}{4\pi r^2}[/math] ...and to find out which one is correct you have to use THE DEFINITION given by the -BIPM SI Units brochure, 8th Edition, PAGE 105- solve it for the given values that define the unit of ampere, and see for yourself which one can produce that exact result. OK, since you are either unwilling or incapable of doing this, I will. We know that the force per unit length is given in Newtons per meter, so F/L = I x B. B must have units of Newtons/(meter*Ampere) The permeability has units of Newtons/Ampere^2 http://en.wikipedia.org/wiki/Vacuum_permeability [math]B® = \frac{\mu_0 I}{4\pi r}[/math] This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter = Newtons/(meter*Ampere) [math]B® = \frac{\mu_0 I}{4\pi r^2}[/math] This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter^2 = Newtons/(meter^2*Ampere) Oops. Dimensional analysis fail. Edited April 5, 2010 by swansont typo
ambros Posted April 5, 2010 Author Posted April 5, 2010 [math]B® = \frac{\mu_0 I}{4\pi r}[/math] This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter = Newtons/(meter*Ampere) [math]B® = \frac{\mu_0 I}{4\pi r^2}[/math] This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter^2 = Newtons/(meter^2*Ampere) Oops. Dimensional analysis fail. Interesting' date=' I would have never realized that, thank you. [math']\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I*d\mathbf{l}*sin(alpha)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}*sin(90)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}}{4\pi r^2} \ N/A*m [/math] ...which makes me wonder where did that term go from "your" equation? OK, since you are either unwilling or incapable of doing this, I will. I never noticed you were insisting on anything, especially since I have trouble getting answers to my own questions. -- Anyway, can you now solve the force between two parallel wires and see if the dimensional analysis still hold for your equation? I wonder how will you enter those angles and wire segments length without breaking up your units, can you do that?
swansont Posted April 5, 2010 Posted April 5, 2010 Interesting, I would have never realized that, thank you. [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I*d\mathbf{l}*sin(alpha)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}*sin(90)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}}{4\pi r^2} \ N/A*m [/math] ...which makes me wonder where did that term go from "your" equation? Alpha is a variable, and depends on dl, so you cannot substitute a constant in for it. dl is the variable of integration, and is evaluated when doing the integral, so it should not appear in a formula after the integral has been performed. You will notice that since there is a length term in the numerator, that the units will cancel one of the length terms in the denominator. It turns out that the integral of dl is proportional to r (since the angle and location of dl form a triangle), so one of the r's cancels. I never noticed you were insisting on anything, especially since I have trouble getting answers to my own questions. -- Anyway, can you now solve the force between two parallel wires and see if the dimensional analysis still hold for your equation? I wonder how will you enter those angles and wire segments length without breaking up your units, can you do that? I asked twice, and you said you "checked everything." Dimensional analysis of the force equation is just the reverse of what I did before. [math]B = \frac{\mu_0 I}{4\pi r}[/math] has units of N/mA F = ILB , and we solve for F/L = IB which gives units of N/m There is no need for any angles here, since the angles were evaluated in the limits of the integral. The final equation has no angle in it, since the net field is in the azimuthal direction — all of the contributions in the z direction cancel (symmetry tells you this has to happen)
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 (edited) [math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I*d\mathbf{l}*sin(alpha)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}*sin(90)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}}{4\pi r^2} \ N/A*m [/math] This is, roughly, like saying this: [math]\int a \, dx = a \, dx[/math] which is patently false. What physical quantity does [math]d\mathbf{l}[/math] represent in your final equation? And I'm still waiting on those sources. I asked for sources that explicitly state and derive "your" equation, not ones that just state the Biot-Savart law. I'd like so see how they derive it so I can work this out correctly. edit: hmm. I'm attempting to evaluate the integral. Does anyone know if [math]\int \mathbf{a} \times \mathbf{x} \, dx = \mathbf{a} \times \int \mathbf{x} \, dx[/math]? (That integral probably doesn't work, I'm just curious if constant cross product terms can be pulled out like constant scalar multiples can be.) Edited April 5, 2010 by Cap'n Refsmmat
darkenlighten Posted April 5, 2010 Posted April 5, 2010 Yes, and do you realize now we are actually solving the same problem from the very beginning with SINGLE ELECTRON. Yes, we are solving for INFINITESIMAL segments "dl", and that means we are solving for point charges again, only these magnetic fields in wires theoretically do NOT EXIST in front or behind their perpendicular plane, not in any of our equations anyway. In two parallel wires, these "magnetic slices" do not interact with any other slices but only the ones of the other wire whose slices are in that same plane. There are no varying angles and no superposition here. 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.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 [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] [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] (Reordering multiple integrals is not a problem AFAIK.) Now, let's do this chunk by itself: [math]\oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] Let's consider only its magnitude: [math]\oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}[/math] Using the same steps as in post #72 in this thread, we can turn this integral into: [math]\frac{\cos \theta_1 - \cos \theta_2}{R}[/math] where R is the distance between the wires. And since [math]\theta_1 = 0[/math] and [math]\theta_2 = \pi[/math] for an infinite wire, as in post #72 again, we get: [math]\frac{2}{R}[/math] Now, plug in to our original double integral... [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \frac{2}{R}[/math] [math]d\mathbf{s_2}[/math] is always perpendicular to the force it experiences, so: [math]\frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \frac{2}{R} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} \frac{2}{R} \, ds_2[/math] Integrating: [math]\frac {\mu_0} {4 \pi} I_1 I_2 \frac{2s}{R}[/math] (s is simply a length, so let's call it L for consistency) [math]F = \frac{\mu_0 I_1 I_2 L}{2\pi R}[/math] so [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Done!
ambros Posted April 5, 2010 Author Posted April 5, 2010 ...it's that I know I am right. Good' date=' can you show you work please: r=1m; L1=L2=1m; I1=I2=1A; µ0= 4π*10^-7 -------------------------------------------- |----- C2 -----| | | ========|-----ds2----->|=============infinite wire===>> I2 | | |r ^ | | r_hat | | ========|-----ds1----->|=============infinite wire===>> I1 | | |<---- L ----->| |----- C1 -----| [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] * The double line integration sums the force upon each element of circuit 2 due to each element of circuit 1 * ds1 and ds2 (dl1 and dl2 alternatively) are infinitesimal vector elements of the paths C1 and C2, measured in metres * The vector [math]\hat{\mathbf{r}}_{12}[/math] is a unit vector along the line connecting the element pair [from s1 to s2] * r is the distance separating these elements http://en.wikipedia.org/wiki/Ampere%27s_force_law If there is any objection to the diagram or definition of terms then feel free to correct it. -- We should at least get our terms straight, and please, less of prose and words, bring on some math and numbers.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 * The vector [math]\hat{\mathbf{r}}_{12}[/math] is a unit vector along the line connecting the element pair [from s1 to s2] s1 and s2 are not necessarily directly across from each other. This is a double integral, not a single integral of two wires at a time, so s1 and s2 can have different values. Hence [math]\hat{\mathbf{r}}_{12}[/math] is not constant. Imagine integrating [math]\int_0^4 \int_{-2}^7 xy\, dx \, dy[/math]. [math]x \neq y[/math], clearly.
swansont Posted April 5, 2010 Posted April 5, 2010 Good, can you show you work please: r=1m; L1=L2=1m; I1=I2=1A; µ0= 4π*10^-7 -------------------------------------------- Cap'n did this in post 110. The equation gives 2 x 10^-7 N/m when you put these numbers in.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 Cap'n did this in post 110. The equation gives 2 x 10^-7 N/m when you put these numbers in. For additional verification: [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Using the values ambros gives... [math]\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1} = 2 \times 10^{-7}[/math] Checking units, next. [imath]\frac{F}{L}[/imath] should be in units of Newtons/meter. [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R} = \frac{\mbox{N} \mbox{ A}^2}{\mbox{A}^2 \mbox{ m}} = \frac{\mbox{N}}{\mbox{m}}[/math] Physics works!
ambros Posted April 5, 2010 Author Posted April 5, 2010 Alpha is a variable' date=' and depends on dl, so you cannot substitute a constant in for it. dl is the variable of integration, and is evaluated when doing the integral, so it should not appear in a formula after the integral has been performed. You will notice that since there is a length term in the numerator, that the units will cancel one of the length terms in the denominator. It turns out that the integral of dl is proportional to r (since the angle and location of dl form a triangle), so one of the r's cancels. [/quote'] I do not follow, can you show your math please? I do not understand why would you be convincing me something is wrong if all you need to do is actually show me - plug in the numbers, or whatever you are talking about, and point out what part does not compute, ok? I asked twice, and you said you "checked everything." Ahhm, by "everything" I meant everything except that. I'm not a computer, nor Chuck Norris... I can make a slip or two by writing this BOOK that this thread has become, so just instead of asking, please show the numbers right away, that is the most convincing, and highly appreciated.
darkenlighten Posted April 5, 2010 Posted April 5, 2010 Because it has already been done, refer to post #49 with the derivation (the math) that shows how to properly setup and integrate the Biot-Savart Law for a wire of length L and steady current I.
ambros Posted April 5, 2010 Author Posted April 5, 2010 (edited) Cap'n did this in post 110. The equation gives 2 x 10^-7 N/m when you put these numbers in. Yes, I saw it, I'll get to that, but I can say now it very unclear. There is no left term on most of those equations, so we can not analyze units as we do not know whether it is being solved for field or force, or something else. I would like darkenlighten and you show your work too. So, again, can you, yourself, solve that problem and make it more consistent in notation and more clear what is being evaluated so we can do dimensional analysis, can you solve it please? Merged post follows: Consecutive posts mergedBecause it has already been done, refer to post #49 with the derivation (the math) that shows how to properly setup and integrate the Biot-Savart Law for a wire of length L and steady current I. You did not solve this equation given by the BIPM: http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [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] Can you show your work? Merged post follows: Consecutive posts mergedFor additional verification: [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Using the values ambros gives... [math]\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1} = 2 \times 10^{-7}[/math] Checking units, next. [imath]\frac{F}{L}[/imath] should be in units of Newtons/meter. [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R} = \frac{\mbox{N} \mbox{ A}^2}{\mbox{A}^2 \mbox{ m}} = \frac{\mbox{N}}{\mbox{m}}[/math] Physics works! Is that force F(12)? The force with which one of these two wires will move towards the other, or in other words - the force exerted on only ONE wire due to magnetic field of the other wire? Where are the left terms in your derivation/integration? You start with F12, then you take out some part and is not known what is all that supposed to represent - field or force, or what units it has. Can you show where are those angles theta1=0 and theta2=3.14, where do you see those, do you have any diagram to explain where did you get all those values and what do they represent? It's very unclear and all over the place, I'll get back to that post and be more specific. Edited April 5, 2010 by ambros Consecutive posts merged.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 (edited) Perhaps this is a bit clearer. [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] [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] (Reordering multiple integrals is not a problem AFAIK.) Now, let's do this chunk by itself: [math]\mathbf{u} = \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] (so [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \mathbf{u}[/math]) Let's consider only its magnitude: [math]u = \oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}[/math] Using the same steps as in post #72 in this thread, we can turn this integral into: [math]u = \frac{\cos \theta_1 - \cos \theta_2}{R}[/math] where R is the distance between the wires. And since [math]\theta_1 = 0[/math] and [math]\theta_2 = \pi[/math] for an infinite wire, as in post #72 again, we get: [math]u = \frac{2}{R}[/math] Now, plug in to our original double integral... (I know we only found the magnitude of u, but we're only interested in the magnitude of the force anyway.) [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \mathbf{u} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \frac{2}{R}[/math] [math]d\mathbf{s_2}[/math] is always perpendicular to the force it experiences, so we can turn this into an easy magnitude: [math]F_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} \frac{2}{R} \, ds_2[/math] Integrating: [math]F_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \frac{2s}{R}[/math] (s is simply a length, so let's call it L for consistency) [math]F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R}[/math] so [math]\frac{F_{12}}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Done! Merged post follows: Consecutive posts mergedCan you show where are those angles theta1=0 and theta2=3.14, where do you see those, do you have any diagram to explain where did you get all those values and what do they represent? --1--------------------------------------------------------2-- theta \ ----------------------------A--------------------------------- Imagine here that theta is the angle between the wire, A, and point 1. If each wire is infinitely long, theta will approach 0. Now look at the angle between the wire, A, and point 2. As the wire becomes infinitely long, it approaches [imath]\pi[/imath]. Edited April 5, 2010 by Cap'n Refsmmat clarify 1
swansont Posted April 5, 2010 Posted April 5, 2010 Post #47 has all the setup for the solution, in the scan, figure 5.18. The problem has been solved a number of times already in this thread.
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 Post #47 has all the setup for the solution, in the scan, figure 5.18. The problem has been solved a number of times already in this thread. Indeed. Note that the scan uses sine, while I use cosine -- but I choose a different angle to be theta, namely the opposite angle in the triangle. Since [imath]\sin (\frac{\pi}{2} - \theta) = \cos \theta[/imath], it works.
darkenlighten Posted April 5, 2010 Posted April 5, 2010 You did not solve this equation given by the BIPM: http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf [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] Can you show your work? 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.
ambros Posted April 5, 2010 Author Posted April 5, 2010 What physical quantity does [math]d\mathbf{l}[/math] represent in your final equation? And I'm still waiting on those sources. I asked for sources that explicitly state and derive "your" equation' date=' not ones that just state the Biot-Savart law. I'd like so see how they derive it so I can work this out correctly. [/quote'] I do not have any special equations and we are both now actually using "my" equations as darkenlighten just demonstrated. It is only that we obviously have different ways of SOLVING them. -- My best guess is that you're objecting as to where did the integral sign go from my equation, right? "dl" in my final equation represents a vector of 1 meter in length, which is why I do not have integral sign, I do not integrate - I can solve this in one step, with only one derivative. -- Anyway, as all that will be automatically sorted out when we resolve the ultimate equation, so I'd rather concentrate on this: [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]
darkenlighten Posted April 5, 2010 Posted April 5, 2010 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 Biot-Savart Law to show you where that double integral came from.
ambros Posted April 5, 2010 Author Posted April 5, 2010 I did' date=' the equation above is a derivation from the Force equation: [/quote'] Yes, but with the different equation for B than this one. The whole point of this new equation is for you to demonstrate whether that 1st solution was indeed correct, or not. [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] Will you solve this equation now, which means - to plug in the numbers given by the two parallel wires and ampere unit scenario and obtain the value for F(12)? darkenlighten, post #59: [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] Why the middle equation has units in Newtons and the next one in N/m?
Recommended Posts