I wound two coils (A & B) using 27 AWG winding wire. Each coil was wound on a ¾ inch thick x 3 inch steel rod. Coil A had 7000 windings. Coil B had 4600 windings. Coil A measured 128 ohms. Coil B measured 76 ohms. I connected a 6 volt lantern battery to each coil thinking coil A would produce a much stronger magnetic core but it turned out to be the opposite. I don’t have a gauss or flux meter. I simply measured the difference in strength with a piece of flat iron. I’m guessing it has to do with volts vs. resistance. My question is…is there a “sweet spot” with volts & ohms when winding a coil for maximum magnetic core strength?


2 Answers 2


It's all about Amp-turns and how well those Amp-turns couple to the rod. If it's a fixed voltage source and fixed wire-gauge and all the turns are tightly wound on the rod, then the resistance is proportional to the number of turns, the current is inversely proportional to the number of turns, and the Amp-turns doesn't change at all. In your case, the extra turns you added were further from the rod (less tightly coupled) and had a larger circumference (adding disproportionate resistance) so you ended up with fewer Amp-turns more poorly coupled.
To get the maximum magnetization, you need to know the internal resistance of the battery and have freedom to choose the wire gauge. You also need to make some estimate of the size of the "winding window". For a rod that's not easily defined, but you can probably usefully stack the wire up to a thickness of about 1/3 the length of the rod at most.
The coil resistance should match the battery internal resistance to get the maximum power into the coil. Given the coil resistance and the winding window area, you can then figure out what wire gauge to use (or you can put wires in parallel to create an effectively thicker gauge).

Be careful! With this recipe it is very likely you will cause both the battery and the coil to overheat. This high-power approach may also be unnecessary anyway because the steel core will be driven well into saturation.

  • $\begingroup$ I spent the evening researching what a batteries internal resistance is. I found that a 6 volt battery has an internal resistance of 0.3 ohms. To find the voltage going through the coil I used this formula from the internet. [6V - (128.3 ohm x .046A) = 6V – 5.9V] = 0.1V Is this correct? Also you mentioned a rough rule of thumb for determining number of windings is that they should roughly be 1/3 the length of the rod. Could that be 1/3 the diameter? I've learned a lot tonight. Hope I got it right...Thanks $\endgroup$
    – user278631
    Nov 12, 2020 at 10:02
  • $\begingroup$ I think the 0.1 volt is the voltage drop inside the battery, but you're definitely on the right track! The winding window is well defined for a ring magnet but not for a rod. It's more a function of length than diameter. In your example, I think you're ok piling turns up to a thickness of about 1 inch in the middle of the rod, but drop that down to 1/2-inch as you approach the ends of the rod. PS. if you've only got one wire gauge available, just put several separate windings and connect them in parallel. $\endgroup$
    – Roger Wood
    Nov 12, 2020 at 17:31

Here is your problem in a nutshell:

When you add a turn to the coil, you add to its inductance but you also add to its DC resistance. When you drive it with the same power supply voltage (6 volts), the incremental increase in inductance is cancelled by the incremental increase in resistance, because it reduces the current through the coil.

The fundamental strength of the electromagnet depends on the product of amps through the coil times number of turns. The optimum electromagnet design for a fixed value of the driving voltage is then the one which maximizes this product, which occurs when

(number of amps) = (number of turns).


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