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Railgun force help

  1. Sep 24, 2004 #1
    Hi all.

    I've been trying to nut out an expression for the force on the projectile

    ( i hope you have an understanding of the basic structure of a railgun, i don't want to draw it)

    Anyway, i is known, w is the rail seperation, and R is the circular rail's radius.

    F(x) = i B(x) dx
    (the scalar magnitudes of F and B, we know what the directions are of course)

    So we need to integrate across the whole length of the projectile, w
    [tex]
    |\vec{B}(x)| = \frac{i \mu_0}{2 \pi}(\frac{1}{x} + \frac{1}{-x+R+w})
    [/tex]

    at any point distance x along the gap

    Thus,

    [tex]

    |\vec{F}| = \frac{i^2 \mu_0}{2 \pi} \int_{w}^{0} \frac{1}{x} + \frac{1}{-x+R+w} dx

    [/tex]

    And ultimately get...

    [tex]

    \frac{i^2 \mu_0}{\pi} log(\frac{r}{r+w})
    [/tex]

    Does this look right?
     
    Last edited: Sep 24, 2004
  2. jcsd
  3. Sep 24, 2004 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    The first part of your setup looks reasonable. You are approximating the current-carrying wire as two infinite "line currents" at the center of the wire. Then you are cutting the result in half, assuming that the line currents are really "half a line".

    This is good for manetostatic applications. For the pulsed environment of the railgun it may not be accurate, but I can't come up with a better approximation.

    I have a bit of a problem with the limits in

    [tex]
    |\vec{F}| = \frac{i^2 \mu_0}{2 \pi} \int_{w}^{0} \frac{1}{x} + \frac{1}{-x+R+w} dx
    [/tex]

    though, it seems to me that the intergal should start at R/2 and go to R/2+w

    This will change ln(r/(r+w)) to ln(r/(r+2w))

    Also, if you want the force to be positive, it should be ln(1+2w/r).
     
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