# When is the force null between parallel conducting wires?

Consider two long wire with negligible resistance closed at one end of the resistance R (say a light bulb), and the other end connected to a battery (DC voltage). Cross-sectional radius of each wire is in $x = 100$ times smaller than the distance between the axes of the wires.

Question:

At which value of resistance R (in ohms) the resultant force of the interaction of parallel conductors disappears?

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Smells like homework. – mbq Jan 19 '11 at 10:15
I've taken the liberty of marking it as homework. Even if it is not your homework, it is a very homeworky question. But feel free to change it back if you disagree. – Malabarba Jan 19 '11 at 14:00
And I suggest changing your title to something like When is the force null between parallel conducting wires?. – Malabarba Jan 19 '11 at 14:05
I'd recommend "....between a n t i parallel conducting....." This seems to be the "essence" in this homework. – Georg Jan 20 '11 at 10:30
@Martin Gales, I like to know what the solution was? – Georg Feb 11 '11 at 14:50

The magnetic force should be straightforward, you probably have a formula for it, or you can use the field strength from one wire, cross the current in the other to compute the force. The charge is more difficult, there you have two oppositely charged cyllinders separated by a distance. You can need to figure out the voltage difference between them as a function of the charge denisty per unit length of the system. As a first approximation assume the charge is evenly distributed around the circumference of your wire, then you should be able to compute the voltage difference between the two wires by integrating the electric field between them. I'm not going to do this (if it really is homework, you need to do it yourself). You should be able to use symmetry to reduce the amount of detailed math you need. If you are really ambitious, you could calculate the effect of uneven charge distribution on each wire, i.e. there will be more charge on the surface closer to the other wire than on the outside -or at least show the approximate size of the correction factor. Should be a fun exercise.

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+1) I think you described qualitatively correctly. – Martin Gales Jan 20 '11 at 6:27
We know the voltage difference between the wires, and you could use that to find a charge difference if you had more information on the geometry of the resistor. And even if you did, charge difference is not enough, you need absolute charge values to calculate the coulomb force between the wires, and the only way we could know that is if we had more information on how the battery works (like I've been saying). – Malabarba Jan 20 '11 at 13:12
Bruce, we don't need details of about the resistor (I'm assuming it is as illustrated, at the rightmost end, not distributed along the wires). The resistance determines the ratio of current to battery voltage, the capacitance determines the amount of charge on each wire. Residual charge would be evenly ditributed to both wires, and would mess the electrostatic forces up, I think the intent of the problem is to assume a ground at the midpoint of the resister. Then it is matter of equating mag forces to electrostaic forces. Let the student do the math, thats homework... – Omega Centauri Jan 20 '11 at 17:04
@omega: That's exactly what I meant. You need knowledge about the geometry of the resistor to calculate its capacitance, or you just need its capacitance. Either way, the question doesn't assume knowledge of either one of these, that's why I said it's lacking information. If the question had said that the capacitance is $C$ than it would be OK. – Malabarba Jan 27 '11 at 17:29
Bruce, If you take the OP literally, the wires are long enough that we can use an infinite 2wire capacitor approximation (can compute capacitance in 2D coordinate system). And you assume the wires have zero resistance, then the voltage difference on the wires is simply current divided by the resistance. The per unit length capacitance of two cyllindrical conductors (per unit length) is actually a bit interesting, if you want to use more than the principal cyllindrical harmonic. But the capactance is solely determined by the geometry. – Omega Centauri Jan 27 '11 at 19:03

I wrestled with this problem more than a week and got quite an unexpected result, i think. I am using SI system.

A starting point is that there are excessive surface charges at each of the wires, flows a current through them or not. Therefore, besides the magnetic force $F_m$ it is necessary to take into account also the electrical force $F_e$ between wires. Let the unit length of the wires have a charge equal to $\lambda (\frac{C}{m})$. Then the electric force acting on unit length of wire from other wire:
$$F_e=\lambda E=\frac{2\lambda^2}{4\pi\epsilon_0d}$$ where $d$ is distance between wires.

The magnetic force acting on the unit length of wire:
$$F_m=\frac{\mu_0}{4\pi}\frac{2I^2}{d}$$ where I is a direct current through wires. It follows from $F_m=F_e$ that $$\left(\frac{\lambda}{I}\right)^2=\epsilon_0\mu_0$$ Next we notice that $$\lambda=CU=CIR_0$$ or $$\frac{\lambda}{I}=CR_0$$ and thus $$CR_0=\sqrt{\epsilon_0\mu_0}$$ where C is a capacitance per unit length for a pair of parallel wires, U is a voltage between wires (due to the battery), $R_0$ is the resistance, $\epsilon_0=8.85*10^{-12}\frac{F}{m}$ and $\mu_0=4\pi*10^{-7}\frac{H}{m}$

Capacitance per unit length for a pair of parallel wires is

$$C=\frac{\pi \epsilon_0}{ln\left(\frac{d}{r}\right)}= \frac{\pi \epsilon_0}{ln(x)}$$ Here $r<<d$ is assumed. $r$ is a radius of wires and $d$ is a distance between wires.

Finally, required resistance:
$$R_0=\sqrt{\frac{\mu_0}{\epsilon_0}}\frac{ln(x)}{\pi}=552.4\Omega$$ An unexpected result is that this resistance does not depend on the applied voltage.

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You wrote (Q/m) as dimension of that lambda, I think that should be C/m ? And: why do You name that charge "excessive" ? In excess to what? That is nothing but the charge due to the voltage of the battery and the capacity of that Lecher system. – Georg Feb 15 '11 at 11:08
@Georg You are right. I fixed this. Yes "excessive" is unnecessary. – Martin Gales Feb 16 '11 at 8:46
@Martin: it obviously doesn't depend on the voltage--- the magnetic force is linear in the voltage and the electric force too. – Ron Maimon Nov 1 '11 at 19:13

Hmmmm...,

the same formula is for the impedance of a Lecher line!

Z = 1/pi sqrt(µ0/eps0) ln(d/r)

Sometimes the formula is given with arcosh(d/2r) instead of ln(d/r), I assume that is a variant more precise for small d/r = x.

Funny isn't it?

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+1 for connecting to the Lecher line. – Martin Gales Feb 16 '11 at 8:47
@Martin, that is "obvious" for an old radio tinkerer :=) – Georg Feb 16 '11 at 9:59
@Martin : First I thought about some connection to the "impedance of vacuum" 377 ohms. After some more thinking I stumbled on the Lecher line. – Georg Feb 16 '11 at 10:20

It should be equal when the characteristic impedance of the line is equal to the load impedance, so there is no reflected wave. You can verify this most readily in the simple case of parallel sheets. A freely propagatin e-m wave, with no reflection, generates currents and charges on these sheets. The net forces of these currents and charges cancel each other out. I am guessing that this principle generalized to all types of transmission lines.

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I think I was pretty clear about this. I said you can verify it pretty easily for a parallel plate transmission line, and I'm guessing that the result generalizes to all transmission lines. Is there a problem with that? – Marty Green Nov 1 '11 at 22:40
Do you doubt that the resistance calculated by Gales after one week of effort is not simply the characteristic impedance of the line, as I was able to conjecture at once without a single line of mathematics? – Marty Green Nov 2 '11 at 0:50
@MartyGreen: By definition, the characteristic impedance of a DC line is $Z_0=\sqrt{\frac{R}{G}}$ where $R$ is the resistance per unit length and $G$ is the conductance of the dielectric per unit length. In given case R is supposed to be zero. G is supposed to be zero also. So that the characteristic impedance is not determined in given case, actually. The load impedance is 552.4 Ohms. So that your answer is not correct. – Martin Gales Nov 2 '11 at 6:49
No, this is wrong. The characteristic impedance depends only on the geometry of the line, and it doesn't depend on frequency. Every transmission line has a characteristic impedance. I don't remember the formula but it has L and C in it, not R and G. For any transmission line, when the load impedance is equal to the line impedance, the reflected wave is zero. It has nothing to do with the ohms in the copper lines. – Marty Green Nov 2 '11 at 12:01
@MartyGreen: The exact formula for the characteristic impedance is $Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$. Here $\omega=0$ because we have a direct current transmission line, not an alternate current line. I repeat again: this is a dc line not ac line. There is no waves in dc current lines at stationary state. Marty, you do not have even the most basic things about transmission lines clear. – Martin Gales Nov 3 '11 at 6:42