Talk:Electromotive force/Archive 1

message: Silly people who abbreviate things in lowercase.  :-)  mph, rms, emf.  When will it end?? 
I know there's some convention for this, but it doesn't make sense to me.  Is there anything in the manual of style about this? - Omegatron 15:38, August 9, 2005 (UTC)
:Actually, it should be e.m.f. as in the old books!--Light current 00:00, 27 February 2006 (UTC)
thread_title:  EMF or emf? 

message: Moved here for Stevenj's talk page for a more general discussion: "Nothing, but there is no reason to put it in a large font. In any case, the most common symbol I've seen is a script \mathcal{E}, not a Roman-font E. —Steven G. Johnson".
The symbol I added is a script E, and the Unicode character for EMF. If it's showing up as a normal E, then it's something to do with the fonts you have installed on your computer. porges 04:40, 12 October 2005 (UTC)
thread_title:  Symbol 

message: From Gauss's law for magnetism, we have:
: \nabla \cdot \mathbf{B} = 0 
From vector calculus, there is an identity which states:
:The divergence of the curl of any vector field  A is always zero:
::\nabla \cdot ( \nabla \times \mathbf{A} ) = 0 
Thus, we can define a vector field that is uniquely defined at all points in space such that the curl of this so-called vector potential A is equal to the magnetic flux density B:
: \mathbf{B} = \nabla \times \mathbf{A}
or equivalently,
:\int_S \mathbf{B} \cdot d \mathbf{a}  =   \oint_C \mathbf{A} \cdot d \mathbf{l} 
From Faraday's law, we have:
:\nabla \times \mathbf{E} = - {\partial \mathbf{B} \over \partial t}
Combining the last equation with the differential form of the prior equation, and simplifying, we have:
:\mathbf{E} = - \nabla \Phi -  {\partial \mathbf{A} \over \partial t}
where Φ is the electric potential, as before, except that we are now allowing time-varying magnetic fields in this region.
If we now take the definition of emf, as cited above, we have
: \mathcal{E} = \oint_C \mathbf{E} \cdot d\mathbf{l} =  - \oint_C { \partial \mathbf{A} \over \partial t } \cdot d\mathbf{l} = - {d \over dt } \int_S \mathbf{B} \cdot d \mathbf{a} = - { d \Phi_B \over dt }
:Comment:  So here again, we have a problem, because this last equation suggests that the emf is always zero in the absence of a time-varying magnetic field.
-- Metacomet 23:31, 26 February 2006 (UTC)
replies: text: Just goes to show you cant always believe the predictions of vector calculus! I find common sense is far more reliable.--Light current 00:04, 27 February 2006 (UTC)
replies: text:  \mathcal{E}(b) - \mathcal{E}(a) =  \int_a^b \mathbf{E} \cdot d\mathbf{l} 
-- Metacomet 00:13, 27 February 2006 (UTC)

text: Yes I think you've just defined potential difference here (but not emf)--Light current 00:18, 27 February 2006 (UTC)

text: I must admit my 'definitions' were confined originally to circuit theory. If we are going into field theory then things may be more complex.--Light current 00:16, 27 February 2006 (UTC)
thread_title:  A more general approach 

message: If Alfred is talking about an isolated battery sitting alone on a shelf with nothing connected to its terminals,
then, in this case:
#there is no external current
#therefore there is no internal current from one terminal to the other (not sure about circulating currents inside)
#there is an external field.
#this must be balanced by an internal field of opposite sense
Is THAT what you are saying? If so I agree.--Light current 01:06, 5 April 2006 (UTC)
replies: text: Ah!  Now I understand where our disconnect was.  Yes, I was referring to a battery that is not connected to an external circuit.  Alfred Centauri 02:16, 5 April 2006 (UTC)
thread_title:  Correction 

message: Ive just read this carefully, and I think I agree with what I think it says (I think!). In fact, Im sure I said something similar myself (but not as well put) further up the page.
i.e 
e.m.f= -pd
We must now move on!--Light current 01:22, 5 April 2006 (UTC)
replies: text: We agree.  The sign of the emf is opposite of the potential difference.  That is what I had hoped to show with my battery example.  So, does professor Steve disagree with us on this?  Alfred Centauri 02:21, 5 April 2006 (UTC)
thread_title:  Your last para 

message: Did the early scientists appreciate this subtlety of distinction between emf and pd when they were bandying around this 'emf' term ? especially as they could only measure pds with potentiometers. Or did they just ignore the sign difference? I think these experiments would have been before the days of Maxwell, and so these scientists may not have had the advantage(?) of  vector calculus to help them.
If they did appreciate the opposite signs, they appear to have been much more perceptive than todays electrical engineers (and certainly wiser than I!). Yet the voltages they measured they always referred to as emfs! So maybe they were using a shortcut or maybe they just referred to the magnitude and didnt give a hoot about the sense?
Does anyone have any comments on this as it may be good to put in the history para of the article.
This subtle difference betweeen emf and pd is in danger of being forgotten unless we get this article right and as Alfred Centauri has pointed out in the past, the concept of emf is essential to understanding the fundaments of electrical engineering--Light current 16:05, 5 April 2006 (UTC)
thread_title:  Early scientists and emf 

message: LC, we still have a disputed tag on this article but the comments by the person who put it here are now archived.  I think we should have a summary of the point(s) that are disputed, address them, remove the disputed tag, and then archive all of the comments regarding the dispute in a separate archive.  What think?  Alfred Centauri 14:01, 6 April 2006 (UTC)
replies: text: I believe the correct procedure would be to summarise Steves comments from the archive page and post the summary here, addess the points, and then, if no further comments, remove the tag. Yes.--Light current 14:08, 6 April 2006 (UTC)
replies: text: Oh boy... yes, this article still needs a lot of work. I just found it after splitting Voltage ("pd" in this context) off from Volt. I think our first task ought to be to collect definitions with cited sources; without them, all this talk is just talk. Melchoir 00:28, 12 April 2006 (UTC)

text: I think you'll find most of the article is correct. But please go ahead and find some refs!--Light current 00:39, 12 April 2006 (UTC)
thread_title:  Status of disputed tag