Electronic – Electromagnet design

First, a battery is basically a voltage source. The current rating is how much current the battery can deliver when the load demands it. It is not the current the battery always puts out. So, just because the 5 kV battery can deliver up to 500 mA, it doesn't mean that the coil will draw that much current. That is up to the coil. In steady state, the coil would have to have a resistance of 5kV / 500mA = 10 KOhms to draw the 500mA. More, and it will simply draw less current. Less, and the battery's current capacity is exceeded and its output voltage will probably droop, it could vanish into a puff of greasy black smoke, or whatever.

The magnetic field strength generated by any electromagnet is directly proportional to the current thru it. Let's say your coil had a DC resistance of 10 KOhms and therefore drew 500 mA from the 5 kV battery. The current it would draw from the 500 mV battery would be 500mV / 10kOhms = 50 uA. That means the magnetic field will be 10000 times weaker. The fact that the battery could have delivered 5 kA is irrelevant since the coil isn't trying to draw that much current.

Earlier I said that the magnetic field of a electromagnet is proportional to its current. That is true for any one electromagnet. Different electromagnets can be constructed with different proportionality constants. For example, it's quite possible to make one electromagnet that draws 1A at 10V and another one that draws 2A at 5V that have the same magnetic field strength. The first one would use a longer but thinner wire, but both could be wrapped around the same core.

Here is a formula that is useful

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

With 100 amps through a 100 turn coil of area 1 sq m, the force on a magnetizable metal at 100mm is 6283 newtons (about 1400 pounds).

But what about the permeability of the electromagnet's core and the permeability of the "iron" it's attracting to - air forms the gap and it dictates the flux density.

The input power needed to do this is zero because no work is being done in creating this force. However, if you actually want to know the losses you need to think about how thick the copper wire is to pass this current and maybe trade off number of turns with amps.

Here's an online calculator and here is a page that explains the theory.