# Farday's Disc - Diameter of Disc

Hi

Does anyone know how changing the diameter of the conducting disc used in Faraday's disc (a homopolar generator) would affect the current and voltage induced in the disc.

e.g. would a disc with a greater diameter produce a greater current and voltage than that of a disc with a smaller diameter?

tpoulter

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ranger
Gold Member
e.g. would a disc with a greater diameter produce a greater current and voltage than that of a disc with a smaller diameter?
Yup.

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How's this for a late reply?

If your generator is what I think it is, then the potential of the circuit, E, is found as a function of the angular velocity of the disc, w, when the magnetic flux density is known, B, as well as the working radius of the disc, r by E = 1/2 B r^2 w. Current, I, given the potential, V, and net resistance, R, is found simply by Ohm's Law I = V/R, where V = E. As far as I know, no resistance is had by electrical components in a magnetic field. So resistance of the electrical components "hooked up" to your Faraday disc must be found.

Making your life easier, resistances of electrical components are usually defined by the manufacture. If not, you'll have to find it on your own. Electrical resistivity is the product of the electrical resistance and the cross-sectional area of the material divided by the length of the material. For example, copper has an electrical resistivity of 1.678e-8 Ohmmeters at twenty degrees Celsius. As one kcmil is equivalent to 5.0671e-7 square meters, 2000 kcmil wire has a cross sectional area of 1.0134e-3 square meters. Supposing the current traveled through a 2-meter wire, the resistance of a single 2000 kcmil wire from end to end would be 3.312e-5 Ohms. The current could then travel to a motor (or other resister like a light bulb) that resisted 1.000e-2 Ohms, and back through a 2-meter 2000 kcmil copper wire. As such, the total resistance of the circuit would be 1.007e-2 Ohms. If your generator put out a current with a potential of 3 volts, the current of the system would be 2.979e2 Ampere.

Secondly, it would be mean not to tell you about brushes. As your dynamo spins and produces current, which would probably be picked up by brushes. The manufacture usually defines a "permissible current density" of the brush. It is the amount of current that the brush can pick up per unit of surface area, such as Amperes per square inch. Supposing your circuit draws 297.9 Ampere, but your pickup brush only has 1 sq. mm. of contact with your dynamo and a permissible current density of 250000 A / sq. m., your brush would pick up a maximum of 250 Ampere. Likewise, if your drop off brush only has one half of a square millimeter of contact with your dynamo, only 125 can return. So, the current in the circuit would, in this case, be limited to the smallest of these values, or 125 Ampere.

Going one step further, take note that a current in transmission experiences a voltage drop. Most times, this is small and is not considered. Other times, it is huge and very critical. The drop is equivalent to the product of the current, the resistance, and the distance traveled. For example, a 125 Ampere current that runs over a 2-meter 2000 kcmil copper wire (resists 3.312e-5 Ohms) will experience a voltage drop of about 8.28e-3 Volts. If your dynamo produced a current at 3 volts, the motor in this described circuit would receive the current at 2.99 Volts. Noting this voltage drop to be one third of one percent, it really doesn't have to be considered. Then again, 2000 kcmil transmission wire is pretty big (but good for this example).

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