# need help with faraday engine calcs

Discussion in 'Physics' started by algae, Jun 18, 2008.

1. ### algae Thread Starter New Member

Jan 14, 2008
6
0
Here is a pic of my setup (top view):

As you can see the magnet is 15mm wide, and the inner diameter of the tube is about 18mm, leaving a gap of 1.5mm on either side of the magnet. (The magnet is about 19mm long cylinder).

I know that V = N(dB/dt) where N is the number of turns of wire, and B is the magnetic flux (in webers). The magnetic flux is the product of the magnetic flux density times the area.

My best guess of the magnetic flux density would be 12,700 Gauss (Br max) and/or 40 MGOe (Bh max).

How can I figure out the most ideal N given the above if each turn of wire is .021 Ohms? Specifically, when we talk about magnetic flux being the magnetic flux density time area, which area is being referred to...the area of the magnet, or the area of the coil around the tube?

thank you very much for your help

2. ### mik3 Senior Member

Feb 4, 2008
4,846
70
The voltage does not equal V = N(dB/dt) but it equals V=NA(dB/dt)

The area A is the area the coil makes around the tube.