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Force between identical wires in a circuit (1st post)

  1. Dec 9, 2009 #1
    1. Two Identical wires denoted A & B are part of an electric circuit and therefore carry some currents. The wires are characterized by resistance per unit length,r, and both have length, L, each are spaced by a distance, a. What is the magnitude, F, of the force the wires exert on each other? Is this force an attractive force? For the given resistance, R of the resistor, what should be the length, L, to achieve maximum possible, F,?

    Equations i thought might be useful..:
    E=I*((r*L/2) + R)

    My attempt:


    I =emf/2rL+R

    emf=I*((r*L/2) + R)

    Then the force is equal to length, by current, by magnetic field. Because the current is going in the same direction in the two wires it will be an attraction force.

    When you get Force as a function of L,
    F = (µ0 *E^2)/(2*pi*a*r^2*L) where
    pi = 3.141592....

    Im unsure if this is even right, and where I go from here.. all help is appreciated..

    Attached Files:

  2. jcsd
  3. Dec 9, 2009 #2
    sry i may have posted this in wrong section.. please close and I'll repost
  4. Dec 10, 2009 #3
    A pretty comprehensive problem!
    First find the current I in the relevant wires. If we find the current through the external resistor R,

    [tex]I_{\textrm{thru R}}=\frac{\varepsilon }{R+\frac{rL}{2}}=2I[/tex]

    Now calculate the magnetic field a distance a from a wire carrying current I using Ampere's law. Use a circular amperian loop with one of the wires in question going through the center.
    [tex]\oint_{C}\overrightarrow{B}\cdot \overrightarrow{d\l }=\mu _{0}I_{\textrm{thru C}}\Rightarrow \overrightarrow{B(a)}=\frac{\mu _{0}I}{2\pi a}\widehat{\theta }=\frac{\mu _{0}\frac{\varepsilon }{\left 2(R+\frac{rL}{2} \right )}}{2\pi a}\widehat{\theta }=\frac{\mu _{0}\varepsilon }{4\pi a(R+\frac{rL}{2})}\widehat{\theta }[/tex]
    Where the theta direction is counterclockwise looking down the current I (opposite to the I direction.)

    Now let's say that's the formula for the field around the bottom wire (in your diagram, with current flowing to the right). Call the direction along the wire in the direction of current the x direction, and call the direction down the page the z direction. At the top wire, the field due to the bottom wire would be in the y direction. ([tex]\widehat{y}=\widehat{\theta }[/tex])

    Then use the formula for the force on a current I due to the B field we found.
    [tex]\overrightarrow{F}=\int I\overrightarrow{dl}\times \overrightarrow{B}=IB(a)\int_{0}^{L}dl(\widehat{x}\times \widehat{y})=IL\frac{\mu _{0}\varepsilon }{4\pi a(R+\frac{rL}{2})}\widehat{z}[/tex]

    So that's the formula for force on the top wire--the force on the bottom would be equal and opposite, so they are indeed attracting each other.

    Now you can find the L satisfying a maximum F for a given R however you'd like, perhaps by setting dF/dL = 0 and solving.
    Last edited: Dec 10, 2009
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