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Force between two short current carrying wires

  1. Apr 23, 2014 #1
    The question is fairly simple but I am afraid the answer will not. I have seen many information about magnetic force between infinite parallel wires but almost nothing about finite ones.

    So, let's have two finite parallel wires, each one 50cm length, in 10cm distance, both have the same current 10A of the same direction.
    What force will be on each wire?
    (we ignore wire begin/end parts connected to power source that can affect force result)

    And how force changes if wires are shifted to half of their length, or
    if we shift them by whole 50cm - will be there any force? If I understand Biot-Savart Law it seams each current carrying element is creating magnetic field that is going not just perpendicular to current direction but in all directions, so it should reach opposite wire even if wires are shifted outside each other...
     
  2. jcsd
  3. Apr 23, 2014 #2

    UltrafastPED

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    You can work it out by means of the Biot-Savart law ... which allows you to take all of these details into account.

    It would make a fine homework problem for an upper level physics student taking a course in electrodynamics.

    The really long wires have a symmetry which simplifies the problem considerably; your first estimate for your short wires can be taken from the long wire approximation. How accurate would this be?
     
  4. Apr 23, 2014 #3

    vanhees71

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    Be careful! Finite wires with a stationary current make no sense physically, I've just written a comment to a corresponding article in Eur. Jour. Phys. which is highly misleading:

    http://fias.uni-frankfurt.de/~hees/publ/ampere-law-discussion.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Apr 23, 2014 #4
    Thank you UltrafastPED, yes I know we can simplify this by using formulas for infinite wires... But I think result may be very inaccurate, it is usable if wires distance to length ratio is small...
     
  6. Apr 23, 2014 #5
    Thank you, maybe you can share your comment?
    What do you mean exactly by "stationary current make no sense physically"?
     
    Last edited by a moderator: May 6, 2017
  7. Apr 23, 2014 #6

    UltrafastPED

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    I read the note ... there was a link to it.

    The point of the note is that you cannot have a finite length of wire with a current in it, though you can define such for an infinite length of wire.

    What has been proposed here is somewhat different: a finite length of wire which is part of a larger loop. When the loop is present the conditions for the Biot-Savart law are fulfilled; you can now have a steady current in your chosen segment of wire.

    However, the presence of the two loops, one for each finite segment, means that there are additional magnetic fields which are not being accounted for. This is actually what Ampere did in his fundamental experiments with "long straight wires" ... he made them long enough, and with the remainder of the loops far enough away that he was able to get good measurements of the forces between the two straight pieces. Of course, Ampere was a clever guy, and did some clever things ...

    For a description and discussion of Ampere's experiments see http://farside.ph.utexas.edu/teaching/302l/lectures/node70.html
     
  8. Apr 23, 2014 #7

    vanhees71

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    Yes, of course, if you consider closed loops of wires, you can calculate the magnetic field due to one loop and then the force it exerts on another such closed currend-carrying loop. There's no problem with that, but you don't get a physically sensible magnetic field if you just assume an open finite-length wire and calculate a "magnetic field" via Biot-Savart's law from it. This triviality is worked out in my comment to a paper in EJP, which in my opinion should not have been published in the first place. Unfortunately I've no free source of this paper, but I think my comment is self-explaning. The DOI to the journal article is

    dx.doi.org/10.1088/0143-0807/35/3/038001
     
  9. Apr 26, 2014 #8
    Force between current-carrying particles of finite wires

    Dear scientists, I am developing application that simulates magnetic fields of current-carrying wires and calculates forces between them.
    Instead of using Biot-Savart formulas I try to implement different approach.

    The wires are divided to small segments, each one produces magnetic field, the magnetic intensity and induction is evaluated in target place of second wire and Lorentz force is computed. All force vectors are summarized to final force in first and second wire.

    Please verify my calculations and computation logic:

    http://www.wish2reality.com/ext_images/problem_statement.jpg [Broken]

    Let me ask two simple questions:

    1.) I assume magnetic field emitted by particle dl is traveling to all directions like a sphere, thus magnetic intensity in target point P can be derived from sphere surface H = l1 /(4⋅π⋅r2). This is basic thing I am not sure - if field emitted by wire has just circular shape (perpendicular to current direction) it will not reach target wire at all...

    2.) Lorentz force in target point P has just single direction, perpendicular to second wire current? Or are here also other forces, i.e. in force along wire axis or magnetic induction vector?

    Please point me to anything I am wrong, if any :blushing:
     
    Last edited by a moderator: May 6, 2017
  10. Apr 26, 2014 #9
  11. Apr 26, 2014 #10

    jtbell

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    You can find this approach e.g. in Griffiths's "Introduction to Electrodynamics", third edition, in Problem 5.49. The student is asked to show that for two current-carrying loops, the force exerted on loop 2 by loop 1 is
    $${\vec F}_2 = -\frac{\mu_0}{4 \pi} I_1 I_2 \oint {\oint {\frac{\hat r}{r^2} d{\vec l}_1 \cdot d{\vec l}_2}}$$

    where r is the distance between segment ##d{\vec l}_1## of loop 1 and segment ##d{\vec l}_2## of loop 2; and ##\hat r## is a unit vector that points along the line connecting the two segments, from loop 1 to loop 2.
     
  12. Apr 26, 2014 #11
    Great thanks jtbel... this F2 force vector on target wire can be divided to two force vectors - one perpendicular to target wire and one axial along current direction, right?
     
  13. Apr 26, 2014 #12
    I would like to add something that the OP might be interested in reflecting on...
    I am quoting from Kip, "Fundamentals of Electricity and Magnetism", 2nd edition, p.245:
    "the equation
    $$d{\vec F} = \frac{\mu_0}{4 \pi} \frac{i_2 d{\vec l}_2 \times ( i_1 d{\vec l}_1 \times \hat r)}{r^2}$$
    is complete only when we integrate the expression over the entire loop of which i1 dl1 and i2 dl2 are only differential elements. It can be shown that the net force on the whole system is then always zero.
    From the experimental point of view there is no difficulty since it is impossible to set up an isolated current element.
    "

    Incidentally one consequence of this 'incompleteness' is that the principle of action and reaction does not work with these forces, i.e., the magnetic force on element dl1 due to the current flowing in dl2 is in general not equal and opposite to the force on element dl2 due to current in dl1.
    After all, if the force is perpendicular to dl1 on one element and to dl2 on the other, all it takes to fool Newton's third law is a couple of non parallel elements. Computing the net forces on the whole circuits will give a result that satisfies Newton, though. Back to Kip:
    "The equations [for dB and dF] are always used in their integral form
    $${\vec B} = \frac{\mu_0}{4 \pi} \oint {\frac{i d{\vec l} \times \hat r}{r^2}}$$
    and
    $${\vec F} = \oint {i d{\vec l} \times {\vec B}}$$
    Another seemingly troublesome case is that of current elements resulting from isolated charges moving with a velocity v [...]"

    This latter case seems to violate the conservation of momentum at least until the momentum associated with the electromagnetic field is considered.
     
    Last edited: Apr 26, 2014
  14. May 1, 2014 #13

    Jano L.

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    Finite-length current element is a valid situation physically. Consider metal rod with oppositely charged bodies attached to its ends. The charge will move from one end of the rod to another where it will stop. If the discharge is slow enough, the electric field will be potential field and the Biot-Savart formula for magnetic field is fine.

    Of course, the electric field is not static in such situation but decreases in time so there is non-zero displacement current throughout the space. The problem with the paper you cited is that the author does not understand limits of validity of the Ampere law.

    In the middle of the page L3, the author admits that the Ampere equation with ##\mathbf J## augmented by the displacement current is valid for his situation with finite current element, yet immediately after that he ignores this fact and proceeds under the assumption that the original static Ampere rule (without the displacement current) is generally thought of as applicable to this situation and thus needs correction. He proceeds to reformulate it with requirement that the current be free of ends. This is of course not sufficient; what is really needed is stationary electric field and vanishing displacement current, which is a stronger requirement.
     
  15. May 2, 2014 #14

    vanhees71

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    Of course, non-stationary sources are more general than stationary ones. Here the "current-conservation constraint" is
    [tex]\partial_{\mu} j^{\mu}=\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.[/tex]
    This admits non-trivial standing-wave solutions for the current along a finite piece of wire, which is a simple model for a "dipole antenna". This you can find in many textbooks on electromagnetics. You simply set
    [tex]\vec{j}=I \exp(-\mathrm{i} \omega t) \sin \left (\frac{kd}{2}-k|z| \right ) \delta(x) \delta(y) \Theta \left(\frac{d}{2}-|z| \right) \delta(x) \delta(y) \vec{e}_z[/tex]
    with [itex]c k=\omega[/itex]. Then, when taking the divergence, the term from the derivation of the [itex]\Theta[/itex] function wrt. [itex]z[/itex], leading to terms [\itex]\propto \delta(z \pm d/2)[/itex] vanishes identically due to the sine factor, and the charge density can be accomodated easily such to fulfill the continuity equation. More general wave forms can then be evaluated by using Fourier integrals.
     
    Last edited: May 2, 2014
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