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Q: Physics question to test service. ( No Answer,   8 Comments )
Question  
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.

Clarification of Question by cleanandoverhead-ga on 26 Dec 2004 17:03 PST
Let me clarify.  
1. This is not a homework question.  I graduated from univesity 14 years ago.
2. This is not a test of "how hard a question this service can
answer." Having never used the service before, I'm simply posting a
basic question in order to test the speed and quality of response.

Clarification of Question by cleanandoverhead-ga on 27 Dec 2004 10:20 PST
In response to the last comment, the question states that the circular
coil has an axis of rotation perpendicular to the the B field.  Hence
assume it's rotating about this axis.  Let's say it has an angular
rate of rotation = w.

Request for Question Clarification by hedgie-ga on 27 Dec 2004 20:28 PST
Hi
      The equation for generated voltage is here

http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/induction/rotatingcoil.html

  Yo just insert the area A = pi * r * r 

The power produced depends on the load attached to the coil.
That will also affect the torque needed to turn the coil.

Would that answer pass your expectations?

Hedgie

Clarification of Question by cleanandoverhead-ga on 29 Dec 2004 15:53 PST
Another clarification spured by Hegie's comment: Getting the voltage
is pretty direct from Maxwell's equations.  The objective is to
determine how much power can be generated given the setup.  So assume
you've got all the torque and load you can handle.  The question: how
much power can be produced.

Given the folks that complained about how simple the question is, I'm
curious why no one has supplied an anwser?

Request for Question Clarification by hedgie-ga on 31 Dec 2004 03:30 PST
Hello again cleanandoverhead-ga 

 and thanks for answering my RFC.  We make special allowances for new
customers - and so here is a bit of guessing on your latest querry:
 "why your questions is still outstanding":

Most questions in this category do not deal with dificult science but
with popular aspects of science and even FAQs. Difficult and important
part often is to match the facts to the interest and level of the
asker. To be able to do that, it helps us to know a bit about the
asker, her motives, what she already knows and what what research she
has already done.

Well asked question is a key to good answer.

Here is a quote:
" Good questioning skills may be the world's most unsung talent. Ask
the right questions in the right way, and you'll engage people; do it
differently, and you'll put them off..".
http://www.youthlearn.org/learning/teaching/questions.asp
  
 
rule 1: http://blogs.msdn.com/toub/archive/2004/03/18/91982.aspx

rule 2: http://www.infotoday.com/newsbreaks/nb020422-3.htm

Finally: 
          Generated power is  V*V/R 
 
where V is EMF given by my previous equation
and R is resistance of the load.
  
           In general, researchers are not interested in arguing about
the facts - that is one difference between the discussion group and a research
service.
           So - in conclusion : Your question is not clear.

 If you are really interested in  'can R go to zero' ?  What are the 
 material and other limitations on produced EMF and internal resistance
of the loop? - then you would have to indicate that interest in the question.

  Without a clue of what, if anything, you really want to know, many
researchers may steer away from your questions. I will.

Hedgie
Answer  
There is no answer at this time.

Comments  
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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