estimate of loop self inductance using "line current assumption" fails
estimate of loop self inductance using "line current assumption" fails
(OP)
This is a purely theoretical question.
Let's say I want to develop a simple estimate the inductance of a loop consisting of a very long pair of straight unshielded solid-conductor small-diameter wires carrying supply and return current.
Call the conductors S and R for supply and Return. The distance between S and R is d. The length of each conductor is Le.
Since the conductors are very long, we neglect the contributions at the ends of the loop.
One would think that there would not be a huge error for calculating the inductance if we made the simplifying assumption (call it the "line current assumption") that the conductor cross section is 0. i.e. the conductor volume resembles a line, not a cylinder.
We should be able to solve it as a simple problem. Assume a fixed current. Find the flux field from each conductor crossing the area between conductors (neglecting end effects), add them together to get the flux linked by the loop, and divide by current to get inductance.
Let's look at the magnetic field from the single conductor S carrying current Is
ClosedIntegral H dot dl = Is
Choose any path circular path of radius r perpendicular to the wire and centered on the wire. H is constant along that path by symmetry (call it Ht = Htangential).
Ht * 2*pi*r = Is
Ht(r) = Is / (2*pi*r)
Bt(r) = mu0 * Ht =mu0 * Is / (2*pi*r)
Look at the flux crossing the rectangular section of the plane which is bounded by the conductors (the rectangle whose area is d * Le).
The flux crossing that area due to current Is is:
Phi_s = Integral(Bt(r) dA)
Let dA = Le * dr
Phi_s = Integral(Bt(r)* Le * dr, r=r0..d)
(r0 is a dummy variable which should correspond to 0 for the problem as described, but we will not yet call it 0... the reason should become apparent later)
Phi_s = Integral(mu0 * Is / (2*pi*r) Le * dr, r=r0..d)
Phi_s = mu0 * Is Le / (2pi) * Integral((1/r) dr, r=r0..d)
Phi_s = mu0 * Is Le / (2pi) * [ln(r)], r=r0..d
Phi_s = mu0 * Is Le / (2pi) * [ln(d) – ln(r0]
Phi_s = mu0 * Is Le / (2pi) * [ln(d/r0)]
Now we see that limit as r0->0 (Phi_s) = Infinity !
The flux from the 2nd conductor in the area of interest is additive, won't change the conclusion. So we will also have Phi_total->Infinity and inductance ->:Infinity as r0-> 0.
That is the surprising result that I'd like to try to understand. I recognize that 0 wire size is a mathematical abstraction, but we are not talking about only 0, but the limit as r1->0. According to this theory, every time we decrease the wire size, we increase the Phi_s... can continue without bound as long as we continue decreasing the wire size.
That seems very non-intuitive to me. Why should I be able to increase without bound the self-linked flux for a conductor pair carrying fixed current, simply by decreasing the wire size toward 0 ? Is there an intuitive explanation why it should be so?
(by the way, I have expressions for inductance based on the actual conductor radiuses... just wondering why I can't use the approach outlined above).
Let's say I want to develop a simple estimate the inductance of a loop consisting of a very long pair of straight unshielded solid-conductor small-diameter wires carrying supply and return current.
Call the conductors S and R for supply and Return. The distance between S and R is d. The length of each conductor is Le.
Since the conductors are very long, we neglect the contributions at the ends of the loop.
One would think that there would not be a huge error for calculating the inductance if we made the simplifying assumption (call it the "line current assumption") that the conductor cross section is 0. i.e. the conductor volume resembles a line, not a cylinder.
We should be able to solve it as a simple problem. Assume a fixed current. Find the flux field from each conductor crossing the area between conductors (neglecting end effects), add them together to get the flux linked by the loop, and divide by current to get inductance.
Let's look at the magnetic field from the single conductor S carrying current Is
ClosedIntegral H dot dl = Is
Choose any path circular path of radius r perpendicular to the wire and centered on the wire. H is constant along that path by symmetry (call it Ht = Htangential).
Ht * 2*pi*r = Is
Ht(r) = Is / (2*pi*r)
Bt(r) = mu0 * Ht =mu0 * Is / (2*pi*r)
Look at the flux crossing the rectangular section of the plane which is bounded by the conductors (the rectangle whose area is d * Le).
The flux crossing that area due to current Is is:
Phi_s = Integral(Bt(r) dA)
Let dA = Le * dr
Phi_s = Integral(Bt(r)* Le * dr, r=r0..d)
(r0 is a dummy variable which should correspond to 0 for the problem as described, but we will not yet call it 0... the reason should become apparent later)
Phi_s = Integral(mu0 * Is / (2*pi*r) Le * dr, r=r0..d)
Phi_s = mu0 * Is Le / (2pi) * Integral((1/r) dr, r=r0..d)
Phi_s = mu0 * Is Le / (2pi) * [ln(r)], r=r0..d
Phi_s = mu0 * Is Le / (2pi) * [ln(d) – ln(r0]
Phi_s = mu0 * Is Le / (2pi) * [ln(d/r0)]
Now we see that limit as r0->0 (Phi_s) = Infinity !
The flux from the 2nd conductor in the area of interest is additive, won't change the conclusion. So we will also have Phi_total->Infinity and inductance ->:Infinity as r0-> 0.
That is the surprising result that I'd like to try to understand. I recognize that 0 wire size is a mathematical abstraction, but we are not talking about only 0, but the limit as r1->0. According to this theory, every time we decrease the wire size, we increase the Phi_s... can continue without bound as long as we continue decreasing the wire size.
That seems very non-intuitive to me. Why should I be able to increase without bound the self-linked flux for a conductor pair carrying fixed current, simply by decreasing the wire size toward 0 ? Is there an intuitive explanation why it should be so?
(by the way, I have expressions for inductance based on the actual conductor radiuses... just wondering why I can't use the approach outlined above).
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RE: estimate of loop self inductance using "line current assumption" fails
Bt(r) =mu0 * Is / (2*pi*r) as r->0 would indicated an infinite magnetic field.
It has been a while since electromagnets, but this simplification is how I tend to think of induction:
1. current flows
2. magnetic field occurs do to current
3. current is influenced by magnetic field
Induction is then a measure of how strong 3 is.
Thus induction goes up when either:
a. the area of the magnetic field gets larger by the conductors getting further apart
b. the magnetic field gets stronger by making the conductor smaller
RE: estimate of loop self inductance using "line current assumption" fails
The flux (integral of flux density) is a quantity of more interest for the circuit since of course V = d/dt (Phi). Flux going to infinity does NOT necessarily imply flux density going to infinity. For example if the flux density was proportaionl to 1/sqrt(r), then we would have a flux density which approaches infinity as r approaches r0, but we would still have a finite total flux NOT appraoching infinity because integral (1/sqrt(r) from r = r0 to 1 = 2sqrt(r) from r=r0 to 1 = 2sqrt(1) - 2(sqrt(r0)... limit as r0 approaches is simply 2.
But with the actual flux density proportional to 1/r, the integral approaches infinite as r0->0 as described above. This would correspond to infinite inductance per unit length of the pair... clearly not realistic by any stretch. It just does not seem intuitive to me that this should be the case. But I guess intuitive is in the eye of the beholder.
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RE: estimate of loop self inductance using "line current assumption" fails
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RE: estimate of loop self inductance using "line current assumption" fails
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RE: estimate of loop self inductance using "line current assumption" fails
http://www.edn.com/article/CA46977.html
Z
RE: estimate of loop self inductance using "line current assumption" fails
RE: estimate of loop self inductance using "line current assumption" fails
Inductance depends on the total flux and current, not the flux and current densities. When integrating a function where both the numerator and denominator approach infinity (flux and current densities per unit area), the math can be confusing. But the limit of these ratios does approach a real value for inductance.
RE: estimate of loop self inductance using "line current assumption" fails
bacon4life - Gabriel's horn came to mind for me as well, which is why I am not surprised by flux density going to infinity, but was surprised about integrated flux and inductance going to infinity
[quote PHovnanian]Inductance depends on the total flux and current, not the flux and current densities. When integrating a function where both the numerator and denominator approach infinity (flux and current densities per unit area), the math can be confusing. But the limit of these ratios does approach a real value for inductance. [quote]
That is the intuitive results that one would expect and I would be intuitively satisfied if that is what the math shows.
My math above shows that we don't get a real (finite) value for inductance. Inductance goes to infinity because total integrated flux goes to 0. Or if I have done the math wrong in my initial post (certainly a possibility), please point out the error.
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RE: estimate of loop self inductance using "line current assumption" fails
So, don't use a zero radius conductor.
The proper answer is that the solution for inductance must be solved in two parts: The inductance due to the flux outside the conductor, which you have correctly solved for, and the flux inside the conductor (from r = 0 out to r = Rc ). Making the assumption of uniform current density (ignoring hf skin effects, etc.), the current within r0 is equal to J * PI * r0^2 . This r0^2 term cancels out the r0 in the denominator of the integrated function and the resulting solution will no longer approach infinity.
RE: estimate of loop self inductance using "line current assumption" fails
I know how to solve the problem using finite radius conductor as stated in last sentence of my original post. In this thread I was just looking to see if there was any further intuitively satisfying comment on why the infinitessimal conductor solution acts this way.
The best I can come up with - our intuition is based on experience with other problems where the flux density at a distance from the conductor is of interest. In those we can obviously shrink the conductor to 0 without affect the result. But we simplyg cannot apply the same simplification for total flux when we are includes the area adjacent to the conductor. It still seems weird to me though.
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