I have an in depth question about basic proportionality of resistance. Why doesn't it scale with the square of length? I ask this because what is causing the potential difference on two leads is charged electric fields at each lead, which decrease with the square of distance.
Why doesn’t linear resistance scale proportionally with the square of length
resistanceresistorstheory
Related Solutions
The wire suggested wire is adequate- the resistance is about 130 ohms/km. I got that from kinda a long route. I used the table for copper (AWG) wire resistance per km, adjusted that for the ratio between copper and steel resistivity, and then for the ratio (squared) between the diameters for AWG and the gauge system used for steel wires (whew!) which is "Washburn & Moen; Roebling; or American Steel and Wire".
So only maybe 2.5 ohms for your circuit (based on two wires, 9.1m each, connected at one end.
You might want to use something other than wood for the supports, insulate the wire from the posts, or at least seal the wood with epoxy. If, say, you were in Portland OR, it will (on average) rain 144 days per year and the leakage through the wet wood would tend to drain the battery.
Here are some insulators used on commercial electric fences.
In terms of $ per kWh, AA batteries are a much better deal than 9V batteries, and six AA cells in series would give you 9V. Energy capacity being a lot higher, it also means that you won't need to replace the batteries as often. There is a good reason why 9V batteries are seldom found in modern electronics.
AA cells have a short-circuit current in the amperes, so a current limiting resistor as Andy suggests would be a very good idea, or you could use a blinking LED in series such as this Lumex one, available from distributors such as Digikey for $1 each.
If you see the LED blinking, the wire is shorted.
Hopefully this helps: Because resistance is proportional to material length and inversely proportional to cross sectional area (see below), I think it's safe to say that because your face has significantly lower cross-sectional area compared to your arm-span (probably by a couple orders of magnitude) that the length difference is negligible.
\$ R = \rho * \frac{L}{A_c} \$
Added to that (like what others have said before) a 'sweaty' surface like your hands likely gives excellent probe contact, while the less-moist (and hairier) surface of your face gives less complete contact.
All these factors result in much higher resistance across your face compared to your arm-span
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Best Answer
You are confusing several basic concepts. The E field of two point charges in a 3 dimensional space follows a \$ \frac{1}{r^2}\$ relationship. If instead to two point charges, you use two infinitely long wires running parallel to each other the field is proportional to \$\frac{1}{r}\$. So the dimensionality really matters. In a resistor there is one path, if you double the length of the path it will take twice as long. The electric field is confined to the interior of the resistor.