Discussion in 'Math' started by BCELECTRIC, Jul 9, 2014.

1. ### BCELECTRIC Thread Starter New Member

Jun 18, 2014
5
0
I have a copper disc that is 10" in diameter and 1" thick if I spin this at 2500 rpm's what kind of voltage and current can I get off this disc.

Thanks BC

2. ### Kermit2 AAC Fanatic!

Feb 5, 2010
4,165
1,120
That depends on the local strength of terrestrial magnetism in your area, the orientation of your disc AND the number and distribution of your current pick off brushes. The Earths magnetic field is an extremely weak source for exciting induction in a generator.

3. ### BCELECTRIC Thread Starter New Member

Jun 18, 2014
5
0
I am open to any suggestions you might have, I am looking for the brushes that would work best this project. Can I add electronics to help with the excitation?

4. ### Kermit2 AAC Fanatic!

Feb 5, 2010
4,165
1,120
Faraday used copper/brass leaf brushes. Something along those lines would be fine.

The formula is also quite simple.

Assuming you are aligned with the north south horizontal component of the earths field and supposing the earths field to be .2 cgs units of magnetic flux.

H(flux) * n(r.p.sec. of disc) * pi * radius(squared)(must be in cm.)
divided by 100,000,000

equals volts generated.

This will be about .005 or less volts.

You REALLY need a stronger magnetic field to generate any kind of measurable voltage.

5. ### LordKelvin001 New Member

Jul 11, 2014
1
0
Hello I am trying to pin down the voltage/current generated With a 10 inch (25.4 cm) Faraday copper disk. Between two magnets with a 2600 Gauss each. It will be moving on the slow end of 2500 RPM; and the fast end would be 5000 RPM. I could really use a complete brake down of the the math processes.Thank you

6. ### Kermit2 AAC Fanatic!

Feb 5, 2010
4,165
1,120
In the old formula I gave here, I used Gaussian (cgs) units.

It is really simple to figure.

Revs per second of disc. RPM's divided by 60.
We all know pi...