I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)
Electrical – Does volts or amps increase the strength of an electromagnet
amperageelectromagnetism
Related Solutions
Even you are not specify the holding or attracting “strength” of your electromagnet, increasing the area of your pole tip, you spread the flux and increase the leakage. Usually conical, or sharp pole tips design resulting a flux density gain especially in short distances. You can see this in magnetic tweezers.
The pole tip of your core also highly related with the dimensions of the object you are going to lift, because the ferromagnetic object acts as a mirror of your electromagnet core. If your core tip area is much larger than your object, then it will shift to the edge of the core. Good performance starts with an analogy of 0.8:1.
In your case you can construct a “jacketed” electromagnet in case of a holding electromagnet desired, or try to make the coil longer (longer core) increasing the magnetic lines return path and strengthen the attractive force.
So I think the answer is no.
EDIT The traction force of the electromagnet will be
where P is pull in Kgr, Ag the cross sectional area of gap, lg the length of the air gap in cm, and NI is the ampere-tirns. (Charles R. Underhill "Solenoids Electromagnets and Electromagnetic Windings" 2nd edition NY 1914)
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.
Best Answer
Firstly, voltage plays no part in the strength of an electromagnet, it's only the current through the windings that generate the field. Consider a super-conducting magnet with zero resistance windings. There's no voltage, no power dissipation, and a large magnetic field.
However, most of us don't have the luxury of using a super-conductor to wind our magnets, we have to make do with a good-conductor, like copper. As it has resistance, we need to apply a voltage across it to push a current through it. This results in bad things like heating in the coil, whose temperature needs to be limited if it's to keep working.
For any given wound magnet, we do not have independent control over voltage and current. The magnet coil defines the ratio of those, it's called its resistance. If we apply a voltage, then (voltage/resistance) current flows. If we connect a current source, then the voltage across the coil responds, and becomes (resistance * current).
The limit for both voltage and current is given by the coil's cooling. If we operate the coil for a very short time, we can increase the power supply, and allow the heat to be stored as a temperature rise in the windings, as long as we switch off before the coil overheats, to let it cool down before the next pulse.