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Subject:
Physics question to test service.
Category: Science > Physics Asked by: cleanandoverhead-ga List Price: $10.00 |
Posted:
26 Dec 2004 15:20 PST
Expires: 25 Jan 2005 15:20 PST Question ID: 447557 |
I'd like an equation for the power an electric generator can output. Let's say there's a uniform magnetic field of magnitude B, and the generator is a simple circular coil of N loops and radius r, with a axis of rotation perpendicular to the B field. | |
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There is no answer at this time. |
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Subject:
Re: Physics question to test service.
From: anonoboy-ga on 26 Dec 2004 16:05 PST |
This question is too basic if your motive is to test the service. |
Subject:
Re: Physics question to test service.
From: tardis-ga on 26 Dec 2004 16:10 PST |
Sounds more like a homework question. |
Subject:
Re: Physics question to test service.
From: crythias-ga on 27 Dec 2004 06:47 PST |
In a generator, something moves... So what's moving? And, I guess it's important to know how fast it moves/oscillates. Otherwise you are describing essentially an electromagnet without a power source. I'd be really interested to know (and so would battery manufacturers!) that one could generate electricity with a stationary configuration as described. Sorry, I think the question can't be answered as asked. This is a free comment. |
Subject:
Re: Physics question to test service.
From: crythias-ga on 27 Dec 2004 15:53 PST |
http://www.physics.nmt.edu/~raymond/classes/ph13xbook/node164.html ... As you are the graduate and I am ... well, I have a degree in Electrical Engineering, but I am a computer guy, and have used little of my college physics and diff eq's, etc... Perhaps this link can help. I am not a GA Researcher. This is a free comment. |
Subject:
Re: Physics question to test service.
From: crythias-ga on 30 Dec 2004 17:49 PST |
Why does everything relating to a generator seem to look like an ideal voltage source? As I'm searching, it seems that power is measured but not really designed. I know that's not correct, but there seems to be something that correlates emf to power... |
Subject:
Re: Physics question to test service.
From: richard-ga on 31 Dec 2004 06:09 PST |
Since this was a "physics question to test service" I'd like to note here how much careful and polite attention this questioner has received in exchange for a 50 cent listing fee. |
Subject:
Re: Physics question to test service.
From: dietlein-ga on 03 Jan 2005 09:46 PST |
A generator question, to test the service? Hmmm. Anyway, the one comment about determining the power DOES require the load impedance to be known. Besides that, what you are looking for is this, for flux: Phi(t) = B*N*pi*r^2*cos(w*t) Then, to find the EMF, take the negative time derivative (Thank you Lentz) of Phi(t), which will give you something like: e(t) = w*pi*r^2*B*N*sin(w*t) You can say E_max = w*pi*r^2*B*N, if you like, so the sinusoidal induced EMF is E_max*sin(w*t). EMF is in volts, so, we're about to the end of the line. P = V^2/R, so your question will be answered if you insert numbers for all of these variables, including load impedance R. Do realize we're neglecting all losses, including conductor, dielectric, and mechanical (resistance, efficiency). |
Subject:
Re: Physics question to test service.
From: dietlein-ga on 05 Jan 2005 18:41 PST |
Was this a joke? Really. What are you driving, a 4-ohm speaker? A 50-ohm transmission line? You have asked two questions: (1) Give me an equation... (2) How much power... (1) has been answered. (2) cannot be answered until numbers are provided. The answer to (2) will not be accurate, since you've not included any material properties or mechanical specifications. Nothing is ideal. |
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