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## Main Question or Discussion Point

Where can I find a proof for the Ampere Circuital law? Wherever I look, I just find a proof for an infinitely long current carrying conductor.

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Where can I find a proof for the Ampere Circuital law? Wherever I look, I just find a proof for an infinitely long current carrying conductor.

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Dale

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There is no proof for it, it is an empirically measured law.

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So is it an axiom? I always thought they declared a law only after some mathematical proof. I mean, for all we know, the experiment might be erroneous.There is no proof for it, it is an empirically measured law.

Edit:Were all Maxwell's equations experimentally determined? Gauss's law seems pretty intuitive. I can't understand the others.

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Dale

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Ampère's circuital law is the magnetic version of Gauss's law, if Gauss's law is intuitive, why isn't Ampère's?So is it an axiom? I always thought they declared a law only after some mathematical proof. I mean, for all we know, the experiment might be erroneous.

Edit:Were all Maxwell's equations experimentally determined? Gauss's law seems pretty intuitive. I can't understand the others.

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WannabeNewton

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I suppose you are right. I always imagined that the basic axioms of the universe would be much simpler than this.

Gauss's law is simply based on the fact that if any curve enters/exits a closed Gaussian surface, it must exit/enter it as well, as long as it does not have an end/origin bounded by the closed surface. Ampere's circuital law is completely different.Ampère's circuital law is the magnetic version of Gauss's law, if Gauss's law is intuitive, why isn't Ampère's?

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AlephZero

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Sure. And if you figure out how to do anI mean, for all we know, the experiment might be erroneous.

Sorry, but there is no law of Physics which says "the universe is so simple that everybody can understand it". Keep trying - you will probably get there eventually.Gauss's law seems pretty intuitive. I can't understand the others.

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BruceW

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We are really talking about two mathematical theorems here (as WannabeNewton mentioned). The divergence theorem and the Kelvin-Stokes theorem. Now you say they are completely different, but actually, they are both special cases of the same theorem: the generalised Stokes' theorem http://en.wikipedia.org/wiki/Stokes'_theoremGauss's law is simply based on the fact that if any curve enters/exits a closed Gaussian surface, it must exit/enter it as well, as long as it does not have an end/origin bounded by the closed surface. Ampere's circuital law is completely different.

That may not be much help to you now if you have not learned much differential geometry. (I haven't really either). But you can sleep soundly, safe in the knowledge that there is an elegant generalisation :)

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These are of course the integral forms.

To get to the differential forms you do need the vector calculus identities in addition to a continuum approximation.

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BruceW

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[tex]\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j},[/tex]

for the case of stationary, i.e., time-independent fields. Then the local form reads

[tex]\vec{\nabla} \times \vec{B}=\vec{j},[/tex]

and you get the circuital law by integrating over a surface, using Stokes integral theorem,

[tex]\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{B}=\int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{j}=i_{F}.[/tex]

Here, the convention of the orientation is the standard one, i.e., given the orientation of [itex]F[/itex], i.e., the direction of the area-element vectors [itex]\mathrm{d}^2 \vec{F}[/itex], the direction of the tangent vectors [itex]\mathrm{d} \vec{x}[/itex] of its boundary curve [itex]\partial F[/itex] is according to the right-hand rule. The direction of the area-element vector determines the sign of the current through the area, [itex]i_F[/itex].

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Law of continuity: ∂ρ/∂t + ∇⋅j=0

∇xB=∇x(∇xA)=∇(∇⋅A)-∇

=-∇(μ/4π*∫(∇'⋅(j'/(r-r')) -∇'⋅j'/(r-r'))dv') + μj=-∇(μ/4π*∫(∇'⋅(j'/(r-r')) +∂ρ'/∂t/(r-r')dv') - μj=-∇(μ/4π(∫j'/(r-r')⋅dS' +∫∂ρ'/∂t/(r-r')dv')) + μj=

(Now we assume that the body is closed, so there is no current in the surface. Then, the integral over the surface is 0(j'⋅dS'=0))

=-∇(μ/4π*∂/∂t(∫ρ'/(r-r')dv'))+ μj=με∂/∂t(-∇(1/4πε*∫ρ'/(r-r')dv')) + μj=με∂/∂t(-∇Φ) + μj=μj + με∂E/∂t→∇xB=μj + με∂E/∂t

As you can see, it can be proved. I can't believe that some teachers see it as an experimental poof instead of trying to prove it.

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