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Force on 2 power lines

  1. Apr 28, 2009 #1
    Why is it that two power lines carrying anti parallel currents exert force away from each other as oppose to to each other?
  2. jcsd
  3. Apr 28, 2009 #2


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    Just take a look at the Lorentz force. If you have a power line that is feeding current into the page on the left and a second one feeding out of the page on the right, then the magnetic field from the right-hand wire is pointing to the bottom of the page at the left-hand wire. The current, or velocity of the charges, is going into the page and the magnetic field is pointing down. Velocity cross the magnetic field gives you the direction of the force and it is to the left, away from the adjacent wire.

    A similar exercise will show you that the right-hand wire experiences a force to the right.
  4. Apr 29, 2009 #3
    I find all that right-hand-rule stuff unconvincing. But you can draw a picture which makes things pretty obvious. Take a constant magnetic field (straight lines) and then superimpose the field of a wire (concentric circles). If you draw it right, you will see than the lines of the external field are bent around the wire in such a way that they seem to push it one way or the other.
  5. Apr 29, 2009 #4


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    Wait... what? What are the physics here? The fields don't interact with each other.
  6. Apr 29, 2009 #5
    I couldn't find the exact picture I wanted but this website has something similar:

    https://rdl.train.army.mil/soldierPortal/atia/adlsc/view/public/8865-1/accp/mm0703/lsn3.htm [Broken]

    Ironically in the example shown here you need to use the right hand rule to get the field lines, but if you do the same thing with just two currents instead of a current and a magnet, you don't need to use any special rules. I find the picture to be quite convincing in showing that like currents attract.
    Last edited by a moderator: May 4, 2017
  7. Apr 30, 2009 #6


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    Again, what are the physics? Just because a field line is bent has nothing to do with the force unless you can state the relationship.

    With just currents, how are you determining the fields? What is interacting with the fields to produce a force? How do you determine the direction of the force?
  8. Apr 30, 2009 #7
    The original question wasn't concerned with how to calculate the force between two wires. We all know about the vxB's etc. The question was about the sign of the force. You can't explain the sign of the force by pointing to the equation and saying "the term in vxB is positive". That doesn't explain why it's positive.

    In electrostatics its a little easier. You can account for the field energy by calculating E-squared and show that the total field energy is reduced when opposites attract. It's a pretty convincing argument but it doesn't work so well for magnetism. Or gravity for that matter. It's not always that easy to say why things go the way they do. But for the case of the two wires, I still find the picture quite convincing. Have you looked at it by the way?
  9. Apr 30, 2009 #8
    The inductance of a bifilar lead, two conductors of radius a separated by a distance b, carrying a current uniformly distributed throughout the conductor in opposite directions has an inductance (Smythe "Static and Dynamic Electricity" 2nd Ed, page 342)

    L = (u0/4 pi) [1 + 4 Ln(b/a)] Henrys per meter
    So the stored magnetic energy per unit length for both conductors with current I is

    W = (1/2) L I2 = (u0/8 pi )[1 + 4 Ln(b/a)] I2 Newton meters per meter

    The force between the two conductors at constant current is then (dW/db is a partial derivative)

    F = dW/db = (u0/8 pi) (a/b) 4 (1/a) I2= (u0/(2 pi b)) I2 Newtons per meter
    So the force is positive (repulsive), and decreases as 1/b.

    Note: I2 has units of amps2 = Newton-meters per Henry.

    For two cables carrying 1000 amps and separated by 10 cm, the force is 2 Newtons per meter. I have observed this in cables for big magnets.
  10. Apr 30, 2009 #9
    Well that's the whole problem...is the force repulsive?? Are you sure you've got the sign right? Because it seems that as the wires get farther apart, the magnetic energy INCREASES. Doesn't that make the force should be attractive? That's how it works in electrostatics: the forces move so as to minimize the field energy. Yes, I agree with you that the force is in fact repulsive. But then the field energy would INCREASE as the wires move apart. So I don't see how the partial derivative gives you the right answer. Is this not a problem?
  11. Apr 30, 2009 #10
    I think you are right; the force is attractive, not repulsive. Here is the same problem with a parallel plate capacitor with charge Q, area A, and separation x. I think the force between the plates in this case is attractive:

    stored energy = W= (1/2) Q2/C = (1/2)Q2 x/e0A

    dW/dx = (1/2) Q2/e0A >0

    So dW/dx > 0 for attractive force
  12. May 1, 2009 #11
    I think you were actually right the first time: the force between opposite currents is repulsive. But differentiating the energy gives you the wrong sign for the force. The reason for this discrepancy is suttle. You cannot let the currents be constant, because once the wires start moving voltages are induced in the loop.
  13. May 1, 2009 #12
    This youtube experiment seems to be rather definitive:

    Smythe "Static and Dynamic Electricity" 3rd Edition, in Section 7.18 proves that the force between two parallel wires with opposing currents is repulsive, using the Biot Savart Law, the Lorentz force, and vector potentials. The inductance of a bifilar wire is given in section 8.12. In Section 8.13, Smythe shows that the force along any coordinate is the positive derivative of the total magnetic energy (e.g., 1/2 LI2) along that coordinate.
    Last edited by a moderator: Sep 25, 2014
  14. May 1, 2009 #13
    Okay. I think the reason its confusing is that the same logic which tells you that capacitor plates attract (decreasing the field energy) gives you the opposite answer when you find that opposite currents repel (INCREASING the total field energy!).

    What happens physically to make it different is that once the wires start moving in them, there are voltages induced which create new currents. So it's really not the same as the moving capacitor plates where the charges stay constant.
  15. May 1, 2009 #14
    Smythe holds all currents constant when he does the differentiation. He specifically states in Section 7.18 that "this [force] is exactly the reverse of the electric case where the force on equal and opposite charges tends to bring them together and destroy the field".
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