A railgun is a fairly straightforward electromagnetic device that converts electrical potential into linear mechanical motion.
The general principle of a railgun is very simple: two rails have a voltage difference applied to them. The device being accelerated rests between the rails and has a conducting cross-link of some sort called the armature that allows a current to flow from the high rail into the low. The current through the rails produces a circulating magnetic field around the two rails, which results in a net magnetic field normal to the plane of the rails and device under acceleration. The current through the armature contains moving charges, and these are accelerated by the magnetic field according to the Lorentz force. The result is that the armature and the affixed device experience a constant force parallel to the rails, and in the absence of other forces, the result is an acceleration in that direction. It is worth noting that no matter whether the potential difference across the rails is positive or negative, the device and armature will be accelerated away from the voltage source end of the rails.
Railguns are pretty cool devices, but they tend to be very high-current and high-power constructions, which means they have to deal with a lot of heat dissipation, not to mention finding awesome power sources that can provide enormous currents. As an example, let's look at a sample system.
But first, a bit of electromagnetism for the sake of calculations:
The magnetic field produced by a semi-infinite wire at the end of that wire is $B(x)=\frac{\mu_0 i}{2\pi r}$, where $\mu_0$ is the magnetic constant, $i$ is the current through the wire, and $r$ is the distance from the wire. Define a coordinate $x$ to start at the center of one rail and be positive in the direction of the other rail. We see that if the rails of radius $r$ are separated by a distance $d$, the magnetic field in the region between the wires is given by
\[B(x)=\frac{\mu_0 i}{2\pi}\left( \frac{1}{x}+\frac{1}{d-x} \right).\]
The force on a current-carrying wire carrying current $i$ in a perpendicular magnetic field $B$ is $F=i\ell B$, where $\ell$ is the length of the wire. We therefore see by integrating over the length of the wire that
\begin{align*}
F &= \int_r^{d-r}i B(x)\,dx\\
&= \frac{\mu_0 i^2 d}{2\pi}\int_r^{d-r}\frac{d}{x(d-x)}\,dx\\
&= \frac{\mu_0 i^2}{\pi}\ln\left( \frac{d-r}{r} \right)\\
&\approx \frac{\mu_0 i^2}{\pi}\ln\left( \frac{d}{r} \right)
\end{align*}
where we have assumed that $r\ll d$ in the last step. Based on this, we can calculate the current necessary to generate a given force in a system:
\[
i=\sqrt\frac{\pi F}{\mu_0 \ln(d/r)}
\]
The resulting power dissipation is given by $P=i^2 R$, where $R$ is the overall resistance of the rail-armature system, which is at most $\rho(2\ell+d)$, where $\ell$ is the length of the rails and $\rho$ is the resistance per length of the wire and armature. We see
\[
P=\frac{\pi F}{\mu_0 \ln(d/r)}\rho(2\ell+d)
\]
Let's choose some arbitrary parameters for our system:
$\ell = 40$ cm (length of the rails)
$d = 10$ cm (distance between the rails)
$r = 0.5$ cm (radius of the rails and armature)
$\rho = 1.608 \Omega/km$ (resistance per length of wire)
If we want to generate a force of $F=200$ N, then the current necessary is on the order of 13 kA (eek, that's a lot of current!), and the power generated is roughly 2 kW, which is around half the power generation of a microwave. The upside is that with this kind of wire, it takes a mere 2 V to generate these currents, so the voltage requirements are reasonable. The limiting factor in this power source will be the current draw necessary, which will be limited by any battery's internal resistance. A chief expense of this device will therefore be the use of a high-quality battery certified for absurdly high currents.
