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Can someone please explain Ampere's law to me?

  1. Oct 2, 2014 #1
    The closed line integral of the B field equals m*I where m is permeability and I is the current enclosed.

    Does this mean amperes law only holds for infinitesimal current elements (or a moving charge?) Because how else can a current be "enclosed"?
     
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  3. Oct 3, 2014 #2

    jtbell

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    You have a closed loop that you're integrating ##\vec B \cdot d\vec l## around. Imagine a surface whose boundary (edge) is that loop. This surface can be any shape, so long as it doesn't have any "holes" in it, and its boundary is the same loop. The current is the total current that passes through ("pierces") the surface.

    Example: if the loop is a circle, then the surface might be (a) a flat disk with the circle being its edge, (b) a hemisphere with the circle being the "equator" along which a sphere was divided in order to produce the hemisphere, (c) a butterfly-net or wind-sock with the circle being its "mouth", etc.
     
  4. Oct 3, 2014 #3
    So what im getting from this is I apply stokes theorem to the curl of the magnetic field to get the line integral of B.
    But im having trouble here because the surface is open so no current is actually enclosed.
    wouldnt the current piercing the surface be infinitesimal?
     
  5. Oct 3, 2014 #4

    vanhees71

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    First of all one must emphasize that Ampere's Law only holds for static situations, i.e., both the magnetic field and the current density under consideration must be time independent. Then the Ampere-Maxwell equation simplifies to Ampere's Law and electric and magnetic field components decouple. The Ampere Law reads (in Heaviside-Lorentz units and in the sense of microscopic electromagnetism)
    [tex]\vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}.[/tex]
    Then you can apply Stokes's integral theorem, i.e., you integrate this over an oriented open surface [itex]S[/itex] with boundary [itex]\partial S[/itex]. The orientation is such that the direction of the boundary curve and the surface-normal vectors are related according to the right-hand rule. Then Stokes's theorem tells you that
    [tex]\int_{\partial S} \mathrm{d} \vec{x} \cdot \vec{B}=\frac{1}{c} \int_{S} \mathrm{d} \vec{S} \cdot \vec{j}.[/tex]
    The left-hand side is the circulation of the magnetic field, and according to the just derived Ampere Law in integral form, it's equal to the total electric current (in the here applied units divided by [itex]c[/itex]) flowing through the surface. For a finite surface there's nothing infinitesimal here!
     
  6. Oct 3, 2014 #5

    Redbelly98

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    The current is equivalent to the amount of charge that has moved through that surface in a unit (finite) amount of time, and so has a finite value.

    Are you perhaps thinking of the actual amount of charge contained within the surface? That is infinitesimal, but is not the same as the current.
     
  7. Oct 3, 2014 #6

    jtbell

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    Don't take the word "enclosed" literally here. You're correct, the surface is not closed, so it can't enclose anything, strictly speaking. Nevertheless, introductory textbooks often use the word "enclosed" here anyway. It's sloppy language, but common.

    That is, ##I \approx \Delta Q / \Delta t##. More precisely, we take the limit as ##\Delta t \rightarrow 0## to get what is essentially a derivative: ##I = dQ/dt##.
     
  8. Oct 3, 2014 #7
    Ok thanks guys I think I get it now... So is it possible to derive amperes law from biot savarts law?
     
  9. Oct 3, 2014 #8

    jtbell

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    Yes. See for example section 5.3.2 of Griffiths, Introduction to Electrodynamics (3rd ed.). Other intermediate and advanced E&M books probably have it, too.
     
  10. Oct 3, 2014 #9

    OldEngr63

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    Just to confuse the issue a bit, if you are up for that sort of thing, take a look at Inductance and Force Calculations in Electrical Circuits by Bueno & Assis (Nova, 2001). This will tell you that what you thought was Ampere's Law is not Ampere's Law at all. Rather, they say that Ampere's Law, in its original expression gives the force between current elements, a really complex expression (that I can't write out because I don't know LaTeX well enough). One of the really interesting aspects is that, according to B&A, this force does obey Newton's 3rd Law, something that is usually thought not to be true. I'll be very interested to hear your thoughts if any of you dig into this.
     
  11. Oct 4, 2014 #10

    vanhees71

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    Electromagnetism is a relativistic theory, and the em. field carries momentum. That implies that in general Newton's 3rd Law does not hold, if you consider particles, interacting electromagnetically but there's momentum in the field, and thus has to be taken into account for the momentum bilance. Of course, for the whole system, field + matter, momentum conservation holds, and Newton's 3rd Law is nothing than momentum conservation for action-at-a-distance forces.

    As I've stressed above, it is utmost important to keep in mind that Ampere's and Biot-Savart's Law are valid for stationary currents and fields. For time-dependent situations, it's an approximation that is only valid if you consider only regions in space that are small compared to the typical wavelength of the electromagnetic field involved. Otherwise you have to take into account the "displacement current", Maxwell's greatest discovery in the physics of electromagnetism, and thus the retardation effects which are due to the finite signal velocity, which is the speed of light.
     
  12. Oct 4, 2014 #11

    OldEngr63

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    "Electromagnetism is a relativistic theory ..."? Really? I had no idea!

    Of course, Ampere, who came in the early 19th century probably did not know this either, so he found a force law that does obey the 3rd law. Maxwell himself acknowledged this as a great achievement. It was not until the dawn of the 20th century that the Lorentz law suddenly became the dominant thinking.
     
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