Current density as a function of radius inside a wire (DC voltage applied)
Hugo Caballero
I was thinking a couple of days ago about the geometrical distribution
of the charges inside a wire when DC was applied.
In the beginning I thought that all the charges would be distributed
homogeneously throughout the whole section of the wire.
However, I soon discovered something concerning the magnetic field
generated by the accelerated charges that annoyed me.
Inside the wire, the magnetic field generated by the charges is
proportional to the radius. Once we leave the wire, the field decays
with 1/r.
My question was: How does the magnetic field inside the wire affects the
distribution of the charges? Applying Lorentz force, the electrons would
be accelerated inwards, towards the center of the wire. However, this
would also imply an increase of the effective resistance and therefore a
consequent reduction of the current through the wire. This, at the same
time, would imply a reduction of the magnetic field and, consequently, a
reduction of the force acting on the electrons, which would tend to
occupy a wider section. After that the process would repeat.
I suppose the final section occupied by the charges is the equilibrium
between these two effects (the magnetic field yields a reduction of the
effective section of the wire and the increase of the resistance yields
a reduction of the magnetic field). One thing I have not considered is
that concentrating a higher amount of charges in a smaller section
increases the repulsion force exerted by each charge on each other.
How does all these phenomena come together? Which is the best way to
describe the system from a formal point of view?
I suppose these are the most relevant equations for the problem:
R=rho*L/A
rotH=j
I=V/R
j=sigma*E
F_lorentz=q*(E+vxB)
Another question... Since each charge only experiments a small
displacement, how do I determine the dependence of the section as a
function of the Lorentz force exerted on it? Maybe using data such as
mean velocity and mean covered distance? I need to couple all the
previous equations. I will try to dedicate some time to this problem
during the next weeks.
However, even if you don't calculate the whole problem, could you please
tell me if you know the solution? Which is the real distribution of the
charges?
Thank you very much in advance.
Benj
> My question was: How does the magnetic field inside the wire affects the
> distribution of the charges? Applying Lorentz force, the electrons would
> be accelerated inwards, towards the center of the wire. However, this
> would also imply an increase of the effective resistance and therefore a
> consequent reduction of the current through the wire. This, at the same
> time, would imply a reduction of the magnetic field and, consequently, a
> reduction of the force acting on the electrons, which would tend to
> occupy a wider section. After that the process would repeat.
So, does current flow evenly inside a DC high current bus bar?
Everybody is always quick to tell you how much they know about the
"skin effect", but that doesn't apply here. I said DC bus bar!
Fact is that current is NOT evenly distributed in such a conductor as
you have already surmised. However, I'd suggest going back and
recalculating the direction of the force on the electrons. Note that
there is a sign associated with "q" in qV x B!
Once you have time-varying currents things then really get
interesting! Now you have inductance distributed all over the
conductor! And it's not uniformly distributed either! This creates a
conductor that has impedance vary as a function of geometric
position! This is really advanced stuff! See how easy it is to push
the limits of knowledge with just a few simple but pertinent
questions? This is really what science is all about.
Note too that thinking successively about feedback systems is also an
error! Saying that this does that does this other thing that comes
back and does the first thing sounds like it "explains" it but it's
not how things work. Generally unless delays are large, you need to
calculate the transfer function of the feedback system which has a
certain mathematical form. You can study feedback systems to learn
more.
Timo Nieminen
> My question was: How does the magnetic field inside the wire affects the
> distribution of the charges? Applying Lorentz force, the electrons would
> be accelerated inwards, towards the center of the wire.
(A positive current would be accelerated inwards, yes? So, for electrons?)
Basically, yes. And what will stop this? As soon as the conduction charges
are not uniformly distributed, there will be an electric force, and this
will oppose the magnetic force.
So, there will be a small re-arrangement of the conduction charges in the
wire. How large?
At the outer radius of the wire, the field is B = mu I / 2 pi r, the
magnetic force on an electron will be F_m = evB = e v mu I / 2 pi r.
The drift velocity is small. For an overestimate, let us assume 1cm/s.
Will the excess or deficiency of electrons within the wire be large or
small? Let us calculate how much current is needed to produce an
excess/deficiency of 1 electron per cm.
The E field due to this charge will be E = charge/length / 2 pi epsilon r,
and the force that will be exerted on an electron at the outer radius of
the wire will be Ee. For our chosen charge density, in SI units, we have
F_e = 100 e^2 / 2 pi epsilon r.
In equlibrium, F_e = - F_m, so
e 0.01 mu I / 2 pi r = 100 e^2 / 2 pi epsilon r
I = 10^4 e / epsilon mu
= 10^4 c^2 e
= 10^4 * 10^17 * 10^-19
= 100A.
So, for 100A, a large current, we should have an excess/deficiency of
about 1 electron per cm. Since the density of conduction charges is about
10^29/m^3, this doesn't appear to be a significant effect on the charge
distribution.
So, you should be able to safely ignore the change in resistance due to
redistribution of the conduction charges.
From the conductivity (or resistivity) and the density of conduction
charges, you can calculate the drift velocity for any given current
density. From the above, just ignore the change in conductivity due to
redistribution. This leaves you with just the equilibrium between the
magnetic and electric forces. Find the steady-state solution. You will
have some charge density that will give you F_e = -F_m. For a
rotationally symmetric wire, you can reduce it to a 1D problem. I don't
know if there's an analytical solution, but it should be easy to do
numerically.
The biggest approximation in that will be assuming that the charge
distibution can be represented as a continuous charge density - it looks
from the above that the excess electron density will be so low that this
isn't a very good approximation. But it will do.
This brings me to another question: how much charge do you need on the
surface to get the conduction charges to go around a corner when the wire
is bent?
There's a nice paper on the electrostatics of circuits by Sherwood and
Chabay, which you should be able to find via google or google scholar.
--
Timo
Don Kelly
cross out to reply
"Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.1008181021320.3040-100000@localhost...
Agreed
Another factor of interest is that the magnetic field is not as Hugo
indicates -(which is equivalent to the field at the conductor radius due to
a filament of current at the center). Inside the conductor the magnetic
field varies from 0 to this value due to partial flux linkages. A circular
conductor is assumed to have a "geometric mean radius" r*e*-1/4. on the
basis of constant J throughout. The result is that B at the outer radius
is mu I/2 pi (r e^-1/4. Shapes other than circular or situations where skin
effects are not negligable or magnetic materials involved are messier and
often need actual tests to determine the effective radius.
This is all pretty standard stuff in powerline modeling.
Don Kelly
Salmon Egg
Hugo Caballero <hugoca...@coit.es> wrote:
You are making things too complicated. The charge density inside the
conductor is essentially ZERO! It is true that it may be that there are
positive charges, atomic nuclear core, that are balanced out by mobile
electrons. Current can flow when an electric field is applied to such a
neutral plasma. Mobile charges respond to the field.
There is a small Hall effect produced by magnetic field acting on the
mobile carriers.
Bill
--
An old man would be better off never having been born.
Szczepan Bialek
"Hugo Caballero" <hugoca...@coit.es>wrote
news:i4f3o2$bt8$1...@speranza.aioe.org...
> Hi!
>
> I was thinking a couple of days ago about the geometrical distribution of
> the charges inside a wire when DC was applied.
>
> In the beginning I thought that all the charges would be distributed
> homogeneously throughout the whole section of the wire.
>
> However, I soon discovered something concerning the magnetic field
> generated by the accelerated charges that annoyed me.
>
> Inside the wire, the magnetic field generated by the charges is
> proportional to the radius. Once we leave the wire, the field decays with
> 1/r.
>
> My question was: How does the magnetic field inside the wire affects the
> distribution of the charges?
There is the pinch effect:
http://en.wikipedia.org/wiki/Pinch_(plasma_physics)
S*
Hugo Caballero
how to solve problems where high frequencies are implied, distributed
systems, antennas, and so on, but nobody has ever introduced the concept
of how charges are distributed inside a wire when a dc voltage is
applied, what I find a much more fundamental concept than the rest of
the theory. At least regarding how we are taught at university.
Thanks for the answer!
Hugo Caballero
of the force! =)
I have found this paper:
http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=PHTEAH000047000002000103000001&idtype=cvips&gifs=yes&ref=no
I will first calculate what you propose, try to find the paper you have
mentioned and the go through it!
Thanks again!
Salmon Egg
Hugo Caballero <hugoca...@coit.es> wrote:
> What I find curious is that I have been taught how to use smith charts,
> how to solve problems where high frequencies are implied, distributed
> systems, antennas, and so on, but nobody has ever introduced the concept
> of how charges are distributed inside a wire when a dc voltage is
> applied, what I find a much more fundamental concept than the rest of
> the theory. At least regarding how we are taught at university.
That is because it is basically a non problem. See Smythe "Static and
Dynamic Electricity." Look at the chapter on eddy currents. DC current
distribution is almost never a practical problem except at transitions
such as between wires of different sizes. The lifetime (relaxation time)
of a net charge inside a conductor is extremely short. Hall effect is
usually a subject for solid plasma physicists and not ordinary
electrical engineers.
glen herrmannsfeldt
(snip)
> Hall effect is usually a subject for solid plasma physicists
> and not ordinary electrical engineers.
Hall effect is commonly taught as part of semiconductor physics,
which is pretty much ordinary electrical engineering by now.
You can also learn some interesting things about metals.
Consider which of the following statements is true?
A) Most electrical power is conducted by electrons.
B) Most electrical power is conducted by holes.
C) Electrical power is conducted by both electrons
and holes, in approximately equal amounts.
-- glen
Benj
> That is because it is basically a non problem. See Smythe "Static and
> Dynamic Electricity." Look at the chapter on eddy currents. DC current
> distribution is almost never a practical problem except at transitions
> such as between wires of different sizes. The lifetime (relaxation time)
> of a net charge inside a conductor is extremely short. Hall effect is
> usually a subject for solid plasma physicists and not ordinary
> electrical engineers.
What you say is true, but it's also a good thing you used to words
"Almost never a practical problem". The way I came to know about this
was when I was working for a company that made current sensors for
submarines. In case you don't know, submarine bus bars handle some
SERIOUS DC currents. At the time I was amazed to find that DC currents
were NOT uniformly distributed in the conductors as I had been taught.
In fact, it affected the accuracy of the sensors. It was all a
surprise to me, but clearly makes sense if you thought about it for a
bit.
glen herrmannsfeldt
(snip)
> What you say is true, but it's also a good thing you used to words
> "Almost never a practical problem". The way I came to know about this
> was when I was working for a company that made current sensors for
> submarines. In case you don't know, submarine bus bars handle some
> SERIOUS DC currents. At the time I was amazed to find that DC currents
> were NOT uniformly distributed in the conductors as I had been taught.
> In fact, it affected the accuracy of the sensors. It was all a
> surprise to me, but clearly makes sense if you thought about it for a
> bit.
The current density before melting decreases as the wire (bar)
gets larger. Integrated circuit interconnects easily run at
more than 1e5 A/cm**2, but you can't do that with bus bars.
There are microprocessors that run with Icc more than 100A,
which has to get into, and distribute across, the chip.
I am not sure how the distribution scales with size at constant
current density, though.
-- glen
Szczepan Białek
"glen herrmannsfeldt" <g...@ugcs.caltech.edu> wrote
news:i4ili7$7md$1...@speranza.aioe.org...
> Benj <bja...@iwaynet.net> wrote:
> (snip)
>
>> What you say is true, but it's also a good thing you used to words
>> "Almost never a practical problem". The way I came to know about this
>> was when I was working for a company that made current sensors for
>> submarines. In case you don't know, submarine bus bars handle some
>> SERIOUS DC currents. At the time I was amazed to find that DC currents
>> were NOT uniformly distributed in the conductors as I had been taught.
>> In fact, it affected the accuracy of the sensors. It was all a
>> surprise to me, but clearly makes sense if you thought about it for a
>> bit.
>
> The current density before melting decreases as the wire (bar)
> gets larger.
So the current flow only in the centre (pinch effect). The rest of the bar
is to dissipate the heat.
> Integrated circuit interconnects easily run at
> more than 1e5 A/cm**2, but you can't do that with bus bars.
The all switches have the contact points almost like point. The currents
like that.
> There are microprocessors that run with Icc more than 100A,
> which has to get into, and distribute across, the chip.
>
> I am not sure how the distribution scales with size at constant
> current density, though.
The answer is here: " The current density before melting decreases as the
wire (bar) gets larger."
S*
Salmon Egg
<03986e61-a4b3-41eb...@y11g2000yqm.googlegroups.com>,
Benj <bja...@iwaynet.net> wrote:
Ordinarily, I do not read posts by Benj abytmore. This time, because of
the subject, I took a peek.
It is not clear whether you are talking about the interior of a
(copper?) bus on a long run or not. I suppose there could be a small
effect on distribution resulting from Hall effect, temperature
inhomogeneity, or different degrees of work hardening in the bus that
cause conductivity inhomogeneity,
Measurement using Ayrton shunt ammeters could give a problem from weird
current distribution around the connection. Ordinarily however, such
shunts are trimmed with a file or similar way of removing material so
tat current distribution in the shunt is not a problem.
Smythe also has a section on calculating current distribution.