Hi all.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Railgun force help

**Physics Forums | Science Articles, Homework Help, Discussion**