|Jan7-11, 10:49 AM||#1|
(Self)Inductance of a straight current carrying wire
I'd like to calculate the inductance of a straight current carrying wire per unit length of wire.
The wire has radius R. Assume the current is uniformly distributed over the wire's cross section.
The magnetic energy density at every point is given by u = B^2 / (2 * mu-nought).
Knowing the total energy per unit length we can calculate the inductance per unit length of the wire:
U = .5 * L * i^2
But when I calculate it (integral breaks in two pieces: B inside and outside the wire), I find this energy per unit length is +infinity ! This off course, isn't possible.
Now, I have two questions:
1) what am I doing wrong using this approach?
Some sites talk about "internal induction". They only calculate the energy associated with the magnetic field INSIDE the wire. Has it something to do with that?
2) what EMF is causing this inductance?
Greetings to you,
|Jan10-11, 10:15 AM||#2|
Seems an interesting question to me, why don't you try "advanced physics forum"?
|Jan10-11, 10:44 AM||#3|
On second thought
a) I did a calculation and it does not diverge. The internal and external integrals both converge.
b) As the curren increases there is a force that opposes to it(Lenz's Law)
|Jan10-11, 11:09 AM||#4|
(Self)Inductance of a straight current carrying wire
Why not use Maxwell's Eqns and Stokes' Theorem? Should be a far simpler approach with symmetry in mind.
|Jan10-11, 11:18 AM||#5|
That is precisely what one does when appling Ampere's Law to calculate B. I don't know any other approach
|Jan11-11, 09:13 AM||#6|
Strange, my integral diverges... I'll post my entire calculation within in a few days because i'm rather convinced the integral diverges...
within the conductor B is proportional to the distance to the conductor axis.
B_in ~ r => B_in^2 ~ r^2 => u ~ r^2
outside the conductor B is inversely proportional to the distance to the conductor axis.
B_out ~ 1/r => B_out^2 ~1/r^2 => u ~ 1/r^2
Now, to calculate the total energy per unit length conductor, one must integrate u over a plane perpendicular to the conductor axis.
This area integral breaks into two seperate integrals, one for u inside the conductor (1), and one for u outside the conductor (2).
Warning: area integral in polar coordinates: dA = r dr dtheta; so the integrandum for (1) and (2) respectively is r*B_in^2 and r*B_out^2.
Now you see:
u*r ~ 1/r
int(1/r,r=R..+infinity) = ln(+inf)
I don't think i'm doing the math wrong. Maybe i have a wrong interpretation of the energy density?
Thanks for the response so far!
|Jan13-11, 03:28 AM||#7|
You are right I checked again and the energy outside de condcutor seems to diverge.
I go back to my original advice and suggest to move your post to advanced physics forum.
May be your calculation is correct and a possible explanation is that it is an idealized situation. Such a field can only exist with an infinite conductor and we shouldn't be so surprised to find and infinite energy but I'm not sure.
|Jan13-11, 04:37 AM||#8|
Do you have physically reasonable boundary conditions, or idealized boundary conditions such as a infinitely long wire, or not?
Does your straight wire correctly radiate energy?
Where is your return current path? Without it, your answer should be doubtful whether you place it at infinite radius or not.
|Jan13-11, 06:25 AM||#9|
You have a point there about not returning the current.
I don't understand what you mean by radiation of energy... The current is steady (no acceleration of charges, so no EM waves?)
My teacher gave us the following exercise:
A non-magnetic (radius R) wire carries a current I, uniformly distributed across its cross section.
Calculate the magnetic energy per unit length of wire.
I think you're right: an infinitely long wire is assumed here.
The solution would be mu_0 * I^2 / (16 * Pi)
You can find this answer by integrating u (see posts above) over the cross section of the wire, not outside!
|Jan13-11, 12:51 PM||#10|
OK, that clarifies things.
Place a return current path on a cylinder of radius R'. This models an infinitely long coaxial cable. The fields outside R' should cancel. This provides a sanity check for your result.
Take R' to infinity and see if your integral still diverges.
|Jan14-11, 03:59 AM||#11|
FoxBox, I looked at your problem in more detail and obtained the same answer--infinite energy per unit length of wire. Return current at infinity doesn't change the result.
But infinite energy per unit length of conductor is not too surprising, really. It's still an infinite volume per unit length.
If we change the problem around a little by making the current sinusoidal then the we have a resonant cavity between the two cylinders of radius R1 and R2. R1 is the radius of the wire and R2 is the radius of the coaxial shield providing the return current path. I believe the B field intensity is B(r,t) = B1(1/r)sin(wt), with the same form for the associated electric field. Taking R2 to infinity, we have an infinite volume per unit length of wire to drive, even if the field intensity does drop off as 1/r.
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