Calculating Railgun Force: iBdx

[MaximumTaco]

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?

 

[pervect]

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).