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Two Infinitely Long Perpendicular Wires

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Wire Ia and wire Ib are infinitely long and perpendicular to each other, separated by distance d. Wire a has current Ia and the direction of current is into the page. Wire b has current Ib and the direction of current is straight up. What is the force per length of Ia on Ib and force per length of Ib on Ia?


    2. Relevant equations
    B=[itex]μ_{0}I/2πr[/itex]
    Magnetic field at a specific location is a function of distance from the source of the field (current).

    F=IdsχB

    3. The attempt at a solution
    The image shows my attempt at finding the magnetic field on wire b due to wire a. This problem has been eating away at me for over a week now - it is how it is worded. I know that due to the right hand rule.

    My initial attempt shown in image 2 which arrives at the form of the Biot-Savart shown above. My idea was that distance ranges from d to infinity. The distance between the two wires will never be less than d (separation distance at the perpendicular) and can increase to infinity. Also, I would imagine that the force acting on a location, say, s above the perpendicular point on Ib would cancel out the force acting at distance s below the point. When I try to take the integral, though, it is not convergent.

    Logic tells me there would be a torque which would act to align the two currents.

    There is a piece of this I am overlooking though

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2013 #2

    rude man

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    Did you draw wire 1 as going into the page? Doesn't look like it.

    Draw an x-y coord. system with wire 1 going into the page at the origin. Then draw wire 2 running in the y direction a distance d to the right of the y axis. Current in wire 2 is along the +y direction.

    Now draw an Amperian circle of the B field set up by wire 1. It will intersect wire 2 in 2 places, above & below the x axis. What is the direction of the B field at those two points? Taking for the moment just the half of wire 2 for which y > 0, you can relate the magnitude and direction of B as a function of y: B = B (y).

    Now take an element of wire 2 = dy j and form the force d F = i1 d l x B and integrate d F from y = 0 to y = ∞. That is the total force on wire 2 above the x axis. Then do the same for the half below the x axis.

    What is the result? A net force on wire 2? A net torque on it, & if so about what axis?
     
  4. Mar 1, 2013 #3

    rude man

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    That is sound logic!
     
  5. Mar 4, 2013 #4
    I have drawn the image as you depicted, and I see the B field, and was able to relate the angle the B field was going off at (90 degrees to the hypotenuse of an imposed triangle in the clockwise direction). I know it seems silly, but I am having an issue relating magnitude and direction to y

    If I draw a triangle and make the hypotenuse r, I know that as y increases, r increases, and B decreases. When y=0, θ=0. When y=∞, θ=90 as per my attached diagram. Force will be 0 at the x-axis , into the page for +y, and out of the page for -y.

    My attempt at the force can be seen on the lower right. I'm having an issue distinguishing between d and r sometimes as well and am not very confident in my answer.

    The effect of wire 2 on wire 1 is essentially the same, except the forces are pointing the opposite direction. The two wires will want to rotate about the x-axis and align themselves.
     

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  6. Mar 4, 2013 #5

    rude man

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    This is a whole lot better!

    You should retain vector math, not go off into scalars the way you did though. And take note that it's dF , not F for the force on an element dy.

    All correct.
    Yes, once you get the force of wire a on wire b, you can just exchange coordinates so as to make wire b be wire a and vice-versa. Or believe in Sir Isaac!

    How about this: B = Bsinθ i - Bcosθ j, i, j unit vectors
    sinθ = y/r, cosθ = d/r
    B = μI/2πr as you stated but r = √(d2 + y2)
    dl = dy j

    I'm using bold type for vectors.

    Now can you form dF = Idl x B and integrate from y = 0 to y = ∞?

    (You have correctly deduced that dF = 0 for y = 0 and for y = ∞. For extra credit, where along y is dF maximum? :smile: )
     
  7. Mar 4, 2013 #6
    You have been a tremendous help so far, I genuinely appreciate it as vector calculus and trig have never been strong points of mine

    I worked through the problem keeping it in vector form but I don't get a convergent integral. I got dF to be in the -k direction which is into the page and what I would expect.

    As per the extra credit, my initial inkling would be where θ=45 degrees, but I'll certainly investigate further once I figure out the force!
     

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  8. Mar 4, 2013 #7

    rude man

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    - k direction is correct. What was your integral?
    That happens to be correct! If you get the right integral you can easily show that that is so (y = d).
     
  9. Mar 4, 2013 #8

    The work is shown in the image in my last post, but essentially I ended up with

    dF=μ0I12ydy/2∏(d2+y2) -k

    dyj crossed with the j component of the magnetic field zeros out, leaving only the -k component
     
  10. Mar 4, 2013 #9

    rude man

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    [/quote]

    That is right and that is the integral I got. And now I see that you're right, the integral does not converge if taken to ∞ as the upper limit. :redface:

    I notice belatedly the problem asks somewhat cryptically for the "force per length of Ia on Ib" and vice-versa. So for that you would not integrate but give the answer as just dF which is finite everywhere and goes to zero at infinity. Somewhat of a strangely posed problem, seemsd to me.

    Anyway, I agree with everything you did at this point.

    EDIT: another thought - maybe by "per unit length" means divide the integrand by y. Then your integral would be ~ dy/(y2 + d2) = (1/d)tan-1y/d which evaluates to π/2d if taken from 0 to ∞.
     
    Last edited: Mar 4, 2013
  11. Mar 5, 2013 #10
    I was able to leave it in the integral form and it was acceptable. Thank you for your help!
     
  12. Mar 5, 2013 #11

    rude man

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    Thanks for letting me know. I was really curious as to what they were looking for.
     
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