Maxwell's equations: meaning, derivation and applicability

[swansont]

It is silly to be waving hands around if you can not solve this, so will you?

 

How many frakking times do you need to be shown the solution? It is monstrously disingenuous to ask for the solution after it has been provided.

[ambros]

How many frakking times do you need to be shown the solution? It is monstrously disingenuous to ask for the solution after it has been provided.

 

Never did you or anyone, except me did dimensional analysis of 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]

 

 

...so, I'm asking you again to show your work and make it clear what are the units of this equation, or point exactly to what post are you referring to. Can you do that please?

[Cap'n Refsmmat]

http://www.scienceforums.net/forum/showpost.php?p=555883&postcount=119

http://www.scienceforums.net/forum/showpost.php?p=556340&postcount=148 (explains the first link)

http://www.scienceforums.net/forum/showpost.php?p=555865&postcount=115 (does dimensional analysis of the resulting equation)

 

There you go.

[ambros]

I already told you that your "solution" is not how equations are used, you made a mess of mathematics and physics there and you are explaining by referencing nonexistent and presumed information...

 

...you only needed to plug in the numbers GIVEN and CALCULATE the result. That equation is EXACT, it MUST NOT BE MODIFIED IN ANY WAY, it is what DEFINES 99% of all the electronics and electrics. Only for infinite wires?!?! Sheesh!

 

 

Do you know what "radians" are?

Yes, thanks for asking. Do you know you did not explain that step you wanted me to tell you about?

 

[math]

\oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}

[/math]

 

...and explain how you got this: [math]=> \frac{\cos \theta_1 - \cos \theta_2}{R}[/math]

 

 

----------------------------------------------------------------------------
                                           \ theta   L  |
                                            \            |
                                         r    \          |  R
                                                \        |
                                                  \      |
--------------------------------------------------------A------------------

 

That's the triangle. You can see that the trig identities in post 72 work, assuming you know what radians are.

 

(L is s in the textbook's example in 72.)

 

As L goes to infinity in either direction, theta goes to either zero or pi.

 

Q: Is that "theta" - really theta or theta_1, or theta_2? You got from theta to having theta_1 and theta_2, explain that please.

 

 

Q: Are you saying it is Ampere's force law or BIPM magnetic force equation, or both, that do not apply to real real world situations?

 

 

Q: What would then be the actual result if wires were indeed 1 meter long, less or more?

 

 

Q: Previously you said in post #80:

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

[math]\frac{F}{L} = IB \, \sin \alpha[/math]

 

and consider that the magnetic field must be perpendicular to the wire, so sin(alpha) is 1.

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

 

...so, why your magnetic fields must not be perpendicular anymore? Where is that "alpha" angle now, is that not supposed to be "theta" and why don't you just cancel it like you did then? That was supposed to be the same situation, right?

 

 

Q: What units do you say this equation has and can you do dimensional analysis of it, please?

 

[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]

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Consecutive posts merged

postcount=115[/url] (does dimensional analysis of the resulting equation)

 

No' date=' that is not the same equation.

 

[math']\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math]

 

 

The whole point of this discussion is for you to realize that equation is WRONG according to 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]

 

 

...you will never realize they are different if you do not perform dimensional analysis and by refusing to do so you are only making it more interesting for me. So, what units that equation has, what say you?

Edited April 7, 2010 by ambros
Consecutive posts merged.

[Cap'n Refsmmat]

Q: Is that "theta" - really theta or theta_1, or theta_2? You got from theta to having theta_1 and theta_2, explain that please.

http://www.scienceforums.net/forum/showpost.php?p=555940&postcount=131

 

Definite integration. I integrate along [math]\theta[/math] from [math]\theta_1[/math] to [math]\theta_2[/math], just like when I do:

 

[math]\int_0^4 1 \, dx = x |_0^4 = (4) - (0) = 4[/math]

 

Fundamental theorem of calculus. If you do not understand this, you clearly do not understand vector calculus at all.

 

Q: Are you saying it is Ampere's force law or BIPM magnetic force equation, or both, that do not apply to real real world situations?

I could apply both to real world situations by adjusting the paths I integrate along accordingly.

 

Q: What would then be the actual result if wires were indeed 1 meter long, less or more?

I'd have to work out the geometry and integrate this. It would be a rather complicated integration.

 

 

...so, why your magnetic fields must not be perpendicular anymore? Where is that "alpha" angle now, is that not supposed to be "theta" and why don't you just cancel it like you did then? That was supposed to be the same situation, right?

Theta is not the angle of the magnetic field. It is the angle from a point on one wire to a point on another wire.

 

Q: What units do you say this equation has and can you do dimensional analysis of it, please?

 

[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]

 

Check my previous post for links.

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Consecutive posts merged

The whole point of this discussion is for you to realize that equation is WRONG according to this equation:

 

Then show me exactly where I made a mistake, rather than demonstrating your lack of understanding of basic definite integration, and of the concept of "radians" (fundamental to any vector calculus course).

[ambros]

Then show me exactly where I made a mistake' date=' rather than demonstrating your lack of understanding of basic definite integration, and of the concept of "radians" (fundamental to any vector calculus course).[/quote']

 

Your mistake is to derive the EXACT equation when you only needed to plug in the NUMBERS and CALCULATE. Your mistake is that you never solved anything nor you performed dimensional analysis, like this:

 

 

[math]

F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

 

[math]=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

 

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

 

 

[math]=> \frac {4 \pi*10^-7_{N/A^-2}*1_A*1_m \times (1_A*1_m)}{4 \pi*1_{m^2}} \ =OR= \ 10^-7_{N/A^-2}*1_{C/s} \times (1_{C/s})[/math]

 

 

[math]=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N[/math]

 

 

Can you now tell me EXACTLY what do you believe is mistake here?

 

 

 

Q: Are you saying it is Ampere's force law or BIPM magnetic force equation, or both, that do not apply to real real world situations?

I could apply both to real world situations by adjusting the paths I integrate along accordingly.

 

[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]

 

Did you just say that this equation given by the BIPM that DEFINES all the electronics and electrics needs some corrections and can not be applied to the real world as it is?

[swansont]

Did you just say that this equation given by the BIPM that DEFINES all the electronics and electrics needs some corrections and can not be applied to the real world as it is?

 

Do you really think that there is a standards lab that has a device with infinite-length wires? "Realizing" a standard is not the same thing as the definition; the definition for the second, for example, assumes zero magnetic field and zero temperature. Which is impossible to achieve, and all Cs devices that try and realize the second put a magnetic field on the system, on purpose (because that way it's controllable) and are at some nonzero temperature.

 

From the definition of ampere one can immediately occupy that this serves only as "directive" on the realisation of unit of current. The same helpfulness has also the definition of metre. It is also characterized as an "absolute" unit, since the dimensions of electric units could have been determined only via the clearly theoretical estimates, while they couldn't have been determined in a practical level.

http://www.eim.gr/html/english/metrology/measures/si/ampere.html

[Mr Skeptic]

Sadly, it appears that the discussion on this interesting topic is going nowhere. Thread locked.

 

For those who wish to continue to pursue this topic, here is some reading material:

Maxwell's equations - everyone loves wiki

http://www.sfu.ca/physics/associate/emeriti/cochran/MAXoutline.html -- has a 400 page free textbook on Maxwell's equations, problems, and the solutions to the problems

http://www.engr.uconn.edu/~lanbo/DeriveMaxwell.pdf --derivation of Maxwell Equations and the form of the boundary value problem

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/wirfor.html#c1 -- the magnetic field of an infinitely long wire.

 

Sadly calculus is required for directly using Maxwell's equations, which is rather hard even on those who know how to do it. The various special cases are much easier to use, and need not involve calculus.