On Sat, 23 Jul 2005 17:27:26 +0100, Polymath wrote:
> A well-reasoned response, but yours is based on the
> "fluid flow" analysis which assumes that charge is smoothly
> flowing through a pipe. Admittedly, this is the model used
> when developing Maxwell's Equations, as it relies upon there
> being a "Current Density Vector", "J" with flows represented
> as you do below.
Sure, average electron velocity.
> Now, the difficulty with learning electricity is that there are many
> models to go by, each of which is wrong in some respect when
> you learn a little more. Now, in the Drude model (which I erroneously
> remembered as the Druse model), the electrons are treated as little balls
> which leap forward under the influence of the applied electric field
> and then collide with atoms and come to a halt, before leaping onwards
> again. In order to match the average speeds that you analyse below,
> the Drude model results in a much higher instantaneous speed between
> the collisions.
It would seem that the energy needed to accelerate the electrons (non-zero
mass), which would then be dissipated when the collision occured. What
is the source/sink of this energy? What is the carrier? Nah, Occam seems
to apply here.
> Whatever the speed you wish to believe, if you take the distance
> traversed in relation to the size of your electron, you find that they
> are, indeed, whizzing along.
You're changing your definitions here. Your definition of "wizzing
along" in the article I quoted was 10M/s and compared it to a bicycle.
"certainly with respect to any speed that you are I might obtain on a
bicycle!"
Of course, the parenthetical before that "(assuming you accept the
figure)" is wrong. I don't accept the figure and showed (poly)math.
> However, electrons are not little balls, but wave packets, so the Drude
> model favoured by the 13-year-old pedants such as Mrs.Nugatory and Brian
> Reay has its faults. Do I sneer at them in an infantile manner and
> suggest that theirs' are the attitudes of "total beginners" because they
> have chosen a model which is wrong? No. Mrs.Nugatory is as Mrs.Nugatory
> does.
Please!
> Now, the actual model you use at any one time must be one that suits you
> and gets you the right answer in any engineering problem, so, setting
> aside the pedantic-and-sometimes-wrong-straw-men that are flying around,
> returning to the original question - do you ever imagine yourself inside
> your circuitry as an assistance to analysis and diagnosis?
No. I try not to make electronics personal. ;-)
> "keith" <(E-Mail Removed)> wrote in message
> news
(E-Mail Removed)...
>> On Sat, 23 Jul 2005 10:17:31 +0100, Polymath wrote:
>>
>>> So, the electron whizzes along at a few metres per second.
>>
>> Don't think so, but I havent' done this in a long time. ;-)
>>
>> Current = (electrons/m^3) * (area in m^2) * (velocity in m/s) *
>> (electron charge)
>>
>> That gives us I(Amps) = n*m^-3 * A*m^2 v*ms-1 Q*C (the units work out
>> A=C/s) so I=nAvQ solving for v:
>>
>> v=I/nAQ
>>
>> Let's pick some numbers:
>>
>> Let's assume a wire with a cross section of 1mm^2 with a current of 5A.
>> I= 5
>> A= 1E-6
>> n= ~8.5E28 (copper)
>> Q= 1.6E-19
>>
>> v= 5/(8.5E28 * 1E-6 * 1.6E-19)
>> v= .37E-3 or less than a half a millimeter per second.
>>
>> That's "wizzin", alright. ;-)
>>
>>
>>> How many would you like? 10, say?
>>
>> A "little" high. ;-)
>>
>>> OK 31 feet /sec. Now, I remember from my "A" Level maths that 60MPH is
>>> 88 feet per second, so 31 ft/second (assuming you accept the figure)
>>> is fair "whizzing along", certainly with respect to any speed that you
>>> are I might obtain on a bicycle!
>>
>> I think I could beat a half millimeter per second. ;-)
>>
--
Keith
P.S. Your newsreader is broken, but what's new (Outhouse Excess).