Firstly, I do not think you can use a neodymium magnet as a projectile.
Coilgun
A coilgun (or Gauss gun, in reference to Carl Friedrich Gauss, who formulated mathematical descriptions of the magnetic effect used by magnetic accelerators) is a type of projectile accelerator consisting of one or more coils used as electromagnets in the configuration of a linear motor that accelerate a ferromagnetic or conducting projectile to high velocity.
And you should be more focused on the kinetic energy of the projectile as it leaves the tube. A 400 farad 2.7 v capacitor stores 1450 joules of energy, which given a efficiency of 1 percent translates to 14.5 joules of projectile energy. Plugging this in to the kinetic energy equation \$E = \frac{1}{2}mv^2\$, gives us a muzzle velocity of 170 m/s for a 1 gram projectile.
Now let's come to the question of force. The force depends on how much power you can pump through the coil. And this minimum force will be huge, because the projectile must reach muzzle velocity by the middle of the coil. (The coil must also be discharged by this time, or the projectile will slow down) For a one gram projectile, this will be 729 newtons or 73 kgs of force, assuming a coil length of 4 cm. (You can calculate acceleration from this equation - \$ a = \frac{v^2}{s}\$ where s is the barrel length and v is velocity)
And what does this mean? You need a strong projectile, definitely. And your capacitor must provide a vast amount of power. Taking the above parameters, the time before the projectile hits the middle of the barrel is 0.23 milliseconds, which you can calculate using the kinematic equation \$x = \frac{1}{2}at^2\$. Dividing the total energy by the time gives us a power requirement of 6.5 MW to be discharged. That's right. 6.5 MW.
With a 2.7 volt capacitor, the resistance must be below 1 micro-ohm. Definitely not possible. With a 400 volt capacitor, the minimum will be 24 milliohms. This is possible.
Now, your question specifies you are not interested in maximizing velocity. In that case, you can go through these calculations for your specific use case. The wire gauge depends on the current going through the wire and the voltage. Once you have that you can calculate the number of turns needed, and that gives you the resistance of the coil. You can add this to the resistance of the capacitor and the diode to give you the total resistance. This must be lower than the minimum resistance you calculated.
And of course, exercise caution. 1500 joules is a lot of energy.
Iron is a poor conductor compared to copper so making an electromagnet from iron wire is off to a bad start. Next, an electromagnet has basically a massive air gap and this is the bit that attracts i.e. the working end of an electromagnet is the air gap. This is generally why the attractive force produced by a solenoid electromagnet has nothing to do with the permeability of the core: -
Force = \$(N\cdot I)^2\cdot 4\pi 10^{-7}\cdot \dfrac{A}{2g^2}\$
- F = Force
- I = Current
- N = Number of turns
- g = Length of the gap between the solenoid and the magnetizable metal
- A = Area
Best Answer
If you look at the force generated by a solenoid you will not normally find a term in there for the permeability of the solenoid core: -
\$F = \dfrac{(N*I)^2 μ_0 A }{ (2 g^2)}\$
This is because the flux produced to attract another piece of metal is mainly flowing through a large air gap. This air gap totally dominates and should not be confused with dimension g in the formula above - g above is the gap between one end of the solenoid and the metal to be attracted BUT flux then has to flow out of that metal and back to the other end of the solenoid (a much bigger gap).
However, should you have a solenoid that is attracting the metal with both pole pieces i.e. one constructed from a simple transformer core then this is improved. However, the path that the mag field takes is still thru a significant air gap and the magnetic reluctance of this air gap will still likely dominate the low reluctance of the core except for very small gaps.
Reluctance adds in series like resistors in a conventional circuit and clearly 1k in series with 1 ohm is still largely 1k.
If you are using a single ended solenoid (I have no idea what a MOT is) then abandon the MOT and make your own from much fatter gauge wire so you can push the current through it. Make the cross sectional area (A) as big as possible and stack the windings as close to each other as possible.