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Ampere's Law, must the current extend to infinity?

  1. Apr 21, 2006 #1
    Ampere's law states that,
    the closed integral of B over the loop enclosed it equals uI, where u = permeability of the material and I = current "passing" through the loop.

    I feel confused, because, should the current be extended to infinity?
    I mean, when we have an infinite wire of current I, everything works very good; on the other hand, when we have a finite wire, this law FAILs(hm...I think it fails, am I right? please correct my viewpoint)

    The above example seems to imply that the current should be extended to infinity.

    However, I have another example which shows that the current no need to be extended to infinity!

    Let' look at a solenoid, we can calculate the magnetic field B along the central of the solenoid by using Ampere's law, yet the wire (current) is not infinitely long!
    The method we use is: we construct a rectangular loop that one of the sides lies along the central of the solenoid, 1 side outside the solenoid and parallel to the previos one, and the remaining 2 sides connect the previos 2 sides and perpendicular to them.
    This example shows that,
    we can use ampere's law here even though the current is not extended to infinity!

    So, should the current extended to infinity so that we can use Ampere's law?
    Notice that for each side (for and against) there is always a contradiction which is highlightened in the above 2 examples!

  2. jcsd
  3. Apr 21, 2006 #2


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    In a steady-state situation, there is no such thing as a "finite current": a current that starts at one point, flows to another point, and stops there. Steady-state currents always form closed loops. In some situations, we can place part of the loop very far away ("at infinity") so that we can ignore its effects. That's what we're actually doing when we apply Ampere's Law to a long straight wire: we imagine the wire looping back around, perhaps in a semicircle of very large radius.

    In a non-steady-state situation, you can have a "finite current" in your sense, for a short period of time. For example, when you discharge a capacitor by connecting a wire between the two plates. But in those situations, Ampere's Law is not complete! This led Maxwell to introduce the "displacement current" which is related to the electric field that is produced by the capactor, and which gave him his complete set of electromagnetic field equations.
  4. Apr 21, 2006 #3

    Andrew Mason

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    No. In fact you can bend the wire back through the Ampere loop n times and you will increase the current value of the path integral, I, n times. This is why coils are used to create a large magnetic field.

  5. Apr 26, 2006 #4
    well actually about that solenoid example.. the two sides of the rectangular loop parallel to the axis of the solenoid.. now the side outside the solenoid.. why is B zero on this side.. plz dunt tell me as a result of the right hand rule or and other assumptions... if we r using amperes law to find B. how can we use the result from another method?!?!
  6. Apr 26, 2006 #5


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    Ampere's law is valid (well, if you include the term added by Maxwell), no matter what. *BUT* the integral over [itex] \vec B \cdot \dl [/itex] is in general difficult to do. It is only in a simple case like an (idealized) infinite wire that you can say that (with a circular closed loop centered around the wire) that [itex] vec B \cdot \dl = B dl [/itex] and that, moreover, the magnitude of the B field is a constant which may be taken out of the integral.

    It's the same as for Gauss' law for the electric field. It is *always* valid but the integral is easy to do only in specific cases with a lot of symmetry (spherical, cylindrical or planar symmetry). When a system is more complex and there is no obviosu symmetry, it is not that the law fails, it is rather than it is not very useful because it involves an integral very difficult to do. The reason books look at those special cases (infinte planes, infinite wires, infinite cylinders, etc) is that these are the only cases where the integrals involved in Gauss' and Ampere's laws are easy to carry out. It does not mean that the laws are not valid al the time, simply that they are not terribly useful.

    Hope this makes sense

  7. Apr 26, 2006 #6


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    Imagine another rectangular loop that is completely outside the solenoid, with two sides parallel to the axis and two sides perpendicular. There's no current through this loop, so the integral of [itex]\vec B \cdot d \vec l[/itex] around it must be zero. This means that [itex]\vec B[/itex] must be equal along the two "parallel" sides. But we can put those sides at any distance from the solenoid, and make the same argument. Therefore [itex]\vec B[/itex] must have the same magnitude everywhere outside the solenoid. Obviously it must be zero when we're infinitely far away, so it must also be zero everywhere else (outside the solenoid).

    Of course this argument works only if the solenoid is infinitely long. Finite-length solenoids do have a nonzero [itex]\vec B[/itex] outside.
  8. Apr 28, 2006 #7
    Lemme just throw in the Biov-Savart law: Helps you calculte the differential amount of magnetic field produced by a short length of wire...a sort of short hand case of Ampere's law. Unforunately, don;t know too many detail about it, but I know that it's useful in those circumstances. Think of Ampere's law as a limiting case.
  9. Sep 2, 2010 #8
    the fact is the magnetic field outside the solenoid is very littlein magnitude and hence negligible it has nothing to do with any law or assumptions....

    correct me if i am wrong
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