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Why ampere's law cannot be applied to certain situations

  1. Apr 3, 2007 #1
    I know this question is asked a lot, but I am confused as to why ampere's law cannot be applied to certain situations. In particular: why can't it be used to calculate the field from a wire of finite length?
  2. jcsd
  3. Apr 3, 2007 #2
    Is it perhaps because as the current moves through the wire, there is changing electric flux due to the fact that the charge on one side is going away (even at the instant when the charge is still uniform)?
  4. Apr 3, 2007 #3
    No, its because Ampere's law (in the integral form i.e. amperian loop that you are familiar with) involves solving for an unknown function B(x,y,z) which is inside a path integral. I don't know of any numerical method for doing this, and it only works symbollically in the toroid, solenoid, slab of current, and infinite wire cases.

    The equation is pretty to look at, but in almost all cases e.g. (finite wire) no one has steps for trying to find a solution or an approximation to one...

    ...because we just transform it into a different equation which we can solve beatuifully: we define the MAGNETIC VECTOR POTENTIAL [tex] A [/tex] so that its CURL is the magnetic field [tex] B[/tex], and then [tex] A[/tex] satisfies LAPLACES EQUATION, which is a standard (easy) PARTIAL DIFFERENTIAL EQUATION.

    If numerical integration of laplaces equation cannot proceed for some reason then you can always integrate the field of current elements according to the biot-savaart law.
  5. Apr 3, 2007 #4
    I was thinking about this too since we just learned Ampere's law...I had the idea that it didn't work perhaps because some "information" about the B-field configuration is not included in the calculation for the path integral. In other words, it's hard to "undo" the integral to get the proper B.

    There's a good explanation by nrqed in this thread:
  6. Apr 4, 2007 #5
    What is the shape of your finite length wire? It cannot be just a straight segment because current must come from somewhere and go elsewhere (This is not a problem if the wire is infinite, infinity is so far!). You must add other bits of wire to carry the coming and the going current. But, unless the new wires are aligned, you break the symmetry and you now know that B will no more be the same at constant distance and you do not know any more the direction of B.

    In fact, out of some elementary and high symmetric problems, you must use Biot-Savart.
  7. Apr 4, 2007 #6
    Thanks guys, but I figured it out.
  8. Apr 5, 2007 #7
    Care to post your solution? I'm interested.
  9. Apr 11, 2007 #8
    rather than posting the full solution, I'll just explain the concept of each step.
    First, we must consider that the meaning of a current in a finite wire is the same as that of a staticaly charged rod moving with uniform velocity. With this in mind, one can determine the rate of change of electric flux through some surface by first finding the electric flux from a point charge a distance x from the surface (a circle). Then one can integrate over the length of the rod to find an expression from the flux of a full rod of length "L" and with distance "a" from the cross sectional area of the circular path. After doing this, one can find the derivative of this with respect to time (noting that this is an implicit differentiation). Now you have the rate of change of electric flux with respect to time. Plugg it into the Ampere-Maxwell law, and you will find that you can use it to determine the magnetic field.
  10. Apr 12, 2007 #9
    There is a small problem: the electric field created by a static charge is not the same when the charge is moving.

    This is the electric field created by a charge q in movement:

    [tex]\vec E={-q\over 4\pi \epsz}\left[{\vec e_{r'}\over r'^2}+
    {r'\over c}{d\ \over dt}\left({\vec e_{r'}\over r'^2}\right) +
    {1\over c^2}{d^2\ \over dt^2}\left(\vec e_{r'}\right)\right][/tex]

    This formula and its user manual can be found in tome II chap. 21-1 of the Feynman Lectures on Physics.

    You are ignoring the second term (the wake field). The third is zero if speed is constant.
  11. Apr 12, 2007 #10
    uh oh...............
    I'm sort of in high school...get back to me in a year or two...lol
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