Electronic – Why don’t electrons take the shorter path in coils

Generating 200 Gauss using a 9 Volt battery is a pretty tall order.

The formula for magnetic flux density for a long solenoid coil is:

B = n x I / (2.02 x L),  where 
    B = flux density in Gauss
    L = coil length in inches
    I = current in Amperes
    n = number of turns in the coil

*Applicable only if length of coil >> diameter, by a factor of 5 or more

Since the coil, battery and current specifications are not provided, taking some assumptions here:

• Length of coil = 1 inch
• Using Lithium 9 Volt battery

The Lithium battery can supply a maximum rated current of 1 Ampere (see linked datasheet). Other 9 Volt batteries will deliver less than this.

Number of turns needed per inch:

n = (B x 2.02 x L) / I
  = 404

Thus, your coil would need to have at least 404 turns per inch to approximately generate 200 Gauss at the center of the coil, and the coil itself must be less than 0.2 inches in diameter. That is evidently impractical for a self-made coil.

So let us approach the problem from the opposite direction: How much current is needed for a coil of 100 turns per inch to achieve 200 Gauss?

Flipping the above formula around, we get a required current of 4.0424 Amperes per inch of coil length. If the coil is 2 inches long, that figure doubles.

Again, it is impractical to draw anywhere close to 4.0425 Amperes from a 9 Volt battery.

In addition, what wire gauge would you be using, to allow 4 Amperes of current to flow through without heating up the coil or melting the insulation / enamel?

The reason the battery voltage is dropping is the internal resistance of the battery:

It simply is not designed to support the kind of current you are trying to draw from it, and this is causing the voltage drop due to current across that internal resistance, dropped voltage V = I x R(internal).

Solution:

Revisit your requirements, including how much current the wire can sustain, wind a fresh coil of suitable turn density, and then use a high current power supply, not a 9 Volt battery.

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.