voltage
Newbie: I have a signal generator set (I hope!) to output a 60htz, 12v sine wave. When I put a DMM set for AC across the output I measure < 5v?? I expected either 12v peak to peak, or 8.5 RMS. What am I missing?
DMM AC 0.421 V RMS
Since you are making the coil yourself, you can wind it to various voltage versus current tradeoffs. It seems you used too many turns of too thin wire, which therefore has a high DC resistance and needs high voltage to generate the large magnetic field you want.
Use less turns of a thicker wire. That will require more current to generate the same magnetic field, but a lower voltage to cause the current.
You can maybe achieve this with the wire you have by breaking down the "several thousand turns" into several winding and connecting them in parallel. For example, let's say you have one winding of 3000 turns now, which a DC resistance of 300 Ω. 10 A would require 3 kV, and provide 30 kAmp-turns of magnetic field. If you make that into three windings of 1000 turns each, then each winding has 100 Ω DC resistance. Put in parallel, the combined coil has 33 Ω resistance. It would require 1 kV and draw 30 A to produce the same 30 kAmp-turns magnetic field. This is actually still unrealistic. You want to go lower resistance and higher current overall. I just used these numbers as example to show how to calculate equivalent magnetic field.
I would find the largest capacitors you can get your hands on, and design the coil for that. This means putting enough current thru the wire for a short time to heat it just below damage level. The time will be so short that there won't be opportunity to conduct the heat away. Therefore, figure all the energy goes into heating the copper. Figure the maximum copper temperature you are willing to tolerate, use the specific heat of copper to find the energy per unit length of wire, then figure how much of that wire you need to use to absorb all the energy of the capacitor.
The answer should be easy to see if you look at the complete circuit you created:
The battery is modeled by the 9 V voltage source in series with R1. R1 represents the internal resistance of the batter, and V1 is its open circuit voltage. V1 and R1 together form a Thevenin source, which is a better approximation of a battery than just being a perfect voltage source.
R2 is the resistor you put in series with the battery, and R3 is the resistance of your voltmeter. The meter will show whatever voltage is accross R3.
Now put some plausible values on R1 and R3 (you already said what R2 is). R1 for a 9 V battery is probably in the 10-20 Ω range. Let's say 20 Ω, because as you will see shortly, R1 doesn't really matter in this case. A typical modern voltmeter will have about 10 MΩ resistance, so set R3 to that.
Now you have a voltage divider with R1+R2 forming the top resistor and R3 the bottom. What is the attenutation ratio of a divider with a top resistor of 240 Ω and a bottom resistor of 10 MΩ? From the fact that the bottom resistor is much much larger than the top, you should be able to see this is very nearly 1. In reality it is so close to 1 that you don't notice the difference on the voltmeter between the top resistor being 20 Ω (just the battery) and 240 Ω (the battery with your extra resistor in series).
Best Answer
There can be two problems:
Let's see what RMS results from a peak or p-p voltage specification. A sine with 12 V peak is 8.5 V RMS. 12 V peak to peak would be half that, or 4.2 V RMS. Considering your vague 12 V spec, a 4.2 V reading from a good meter is a possibility.