# (Self) Inductance of a straight current carrying wire, vol. 2

1. Aug 18, 2015

### QuasiParticle

It's a really old thread, but I happened to come across the same problem a while back and since the old thread does not give the correct answer, I decided to create continuation for it. The solution was actually a little bit surprising to me.

If one considers a single infinitely long straight wire with radius $R$ and computes the energy from $1/2 \int B^2/\mu_0$, it indeed gives infinity for the reason already mentioned in the old thread. However, if one calculates the total energy using the vector potential: $E = 1/2 \int \vec{A} \cdot \vec{J} d^3r$, the energy (and inductance) is finite*.

It turns out that the energy density $B^2/2\mu$ is not correct in the case of an infinite wire. The reason I was surprised was that I had always thought $E=1/2\int B^2/\mu$ to be exact in all cases. When deriving this result for the energy, the integration is extended to all space and a surface integral term appearing in the expression is neglected. The term in question is

$$-\frac{1}{2\mu} \oint (\vec{A} \times \vec{B}) \cdot d\vec{S}$$
Usually this can be neglected, when integrating over all space. For a dipole field, for example, $B\sim1/r^3$ and $A\sim1/r^2$, so the surface integrand is $\sim1/r^3$. The infinite wire, however, is a very special case, and the surface integral term becomes infinite.

This also demonstrates how difficult it is to localize electromagnetic energy. The integration of $B^2/2\mu$ must be extended over all space to give the correct total energy. This leads me to state that $B^2/\mu$ does not represent any true energy density of the field, even though it has units $[\mathrm{J/m}^3]$.

*There remains a constant in the vector potential, which I cannot determine and which is a bit strange. The vector potential is:

$$A_z=\left\{ \begin{array}{l l} \frac{\mu_0I}{4\pi}\left(C-\frac{r^2}{R^2} \right) & r\leq R\\ \frac{\mu_0I}{4\pi}\left(C-1-\ln \frac{r^2}{R^2} \right) & r\geq R \end{array}\right.$$
I think the correct answer is found, when we set $C=1$, but I have no justification for this. Help?

Last edited by a moderator: Aug 19, 2015
2. Aug 20, 2015

### Jano L.

Do you mean total energy comes out finite or energy per unit length of wire comes out finite?

3. Aug 21, 2015

### QuasiParticle

Sorry, per unit length, of course.

4. Aug 21, 2015

### Jano L.

Infinite wire is a nice idealization of real wires when calculating magnetic field via the Biot-Savart formula, but its unphysicality manifests when we try to calculate the energy necessary to have infinitely long wire with electric current.

The magnetic energy density formula $\frac{1}{2\mu_0}B^2$ is based on the work-energy interpretation of the Poynting theorem, which is mathematically valid even in case of infinite wire. It gives infinite energy per unit length of wire. That is not really an implausible result considering that to establish current even in the smallest part of the wire, one needs to overcome opposing electromotive forces (self-inductance effect) of an infinitely long wire.

I do not think there is a more valid formula for the energy per unit length giving finite value. Could you elaborate on why do you think that standard formula does not apply (other than the energy comes out infinite)? Why would another formula you mentioned be valid?

Last edited: Aug 21, 2015
5. Aug 21, 2015

### QuasiParticle

Let's begin with the total work (against the EMF) required to create the currents of a system:

$$W=\frac{1}{2} \int \vec{A}\cdot\vec{J}d^3r$$
If you plug in the vector potential of an infinite wire (given in the first post) with a current density

$$\vec{J}=\frac{I}{\pi R^2}\hat{e}_z$$
then the energy per unit length is finite and (if I'm not mistaken with that one constant) equals:

$$W/ \Delta z=\frac{\mu I^2}{16\pi}$$
To see why $\int B^2/2\mu$ is wrong in this case, we write the current density in terms of the field:

$$\vec{J}=(\nabla \times \vec{B})/\mu$$
and get

$$W=\frac{1}{2\mu}\int (\nabla \times \vec{B})\cdot \vec{A}d^3r$$
Now, using $(\nabla \times \vec{B})\cdot \vec{A}=B^2-\nabla \cdot (\vec{A} \times \vec{B})$ and the divergence theorem, the work can be written in the form

$$W=\frac{1}{2\mu}\int B^2 d^3r- \frac{1}{2\mu} \oint (\vec{A} \times \vec{B})\cdot d\vec{S}$$
Then usually the integration volume is extended over all space and the surface integral is neglected, as explained in the first post. However, in the case of an infinite wire, this surface term becomes infinite, and thus cannot be neglected. It cancels the infinity coming from the $B^2$-integral.

Poynting's theorem relates the work done on a charge distribution with the energy change in the electromagnetic field.

6. Aug 21, 2015

### Jano L.

Your math seems OK, but why do you assume that energy is given by

$$W=\frac{1}{2} \int \vec{A}\cdot\vec{J}d^3r$$

and not by

$$W= \int \frac{1}{2\mu_0} B^2\,d^3r$$ ? After all, the latter is a standard expression inferred from the Poynting theorem.

From what I know, I would assume the latter is correct and use your steps backwards to show the former formula does not apply for infinite wire, since the surface integral of $A\times B$ does not vanish.

7. Aug 23, 2015

### QuasiParticle

I think the vector potential version of the energy is more fundamental. It can be derived with some rather general arguments. And then from that one finds the $B^2$-energy by neglecting that one term.

I'm not convinced that the $B^2$-energy term can be derived easily from Poynting's theorem; without making the same trick as in the vector potential case.

8. Aug 24, 2015

### Jano L.

I do not think so. Could you post a reference to source that does that?
The magnetic energy is defined as the $B^2$-energy expression; it is not derived. The definition is based on the work-energy interpretation of the Poynting theorem. Check out Panofsky, Melba Phillips, Classical electricity and magnetism,1962 section 10-5 for its derivation.

9. Aug 24, 2015

### QuasiParticle

See, for example:

Griffiths, Introduction to Electrodynamics (1999), section 7.2.4
Vanderlinde, Classical Electrodynamic Theory (2005), section 4.1.4
Jackson, Classical Electrodynamics (1999), section 5.16

The $B^2$-term might be defined as the magnetic energy term, but it gives the correct total energy only if the already-mentioned surface term can be neglected.

If you look at section 10-1 in Panofsky & Phillips, they derive the $B^2$-energy term and they have to make the same assumption about the surface term. Consequently they proceed to derive the vector potential energy expression from this. In the process they again neglect a surface integral term; I think they are basically restoring the term neglected earlier.

With Poynting's theorem I have doubts as to what is the total energy of a system with some given currents. Poynting's theorem states

$$\frac{dW}{dt}=-\frac{d}{dt} \int \frac{1}{2}\left(\epsilon_0 E^2+B^2/\mu_0 \right) d^3r-\oint (\vec{E} \times \vec{B})/\mu_0 \cdot d\vec{S}$$
I'm not sure how to derive the total energy from this. I suspect the surface term might give a contribution equivalent to the one often neglected.

10. Aug 25, 2015

### Jano L.

Now I think both formulae are OK and give the same infinite energy per unit length.

Check out Griffiths' section 7.2.4. He derives the formula 7.31 for a wire in form of a closed circuit under the assumption that magnetic flux $\Phi$ is finite and can be expressed as
$$\Phi = \oint \mathbf A\cdot d\mathbf l.$$

Now we can take the circuit and expand its size to infinity, and the formula is valid throughout the process. The inductance of the circuit $L$ goes as $R(\ln \frac{8R}{a} - 2)$ where $R$ is size of the circuit and $a$ is radius of the wire circular cross-section.This was first calculated by Maxwell I think, and can be found for example in the document

Self inductance of a wire loop as a curve integral
R. Dengler
http://arxiv.org/pdf/1204.1486.pdf

Vector potential in the wire thus depends on $R$ as $\ln \frac{8R}{a}$.

This means its magnitude diverges into infinity as the wire changes into the infinitely long straight wire via the limiting process. This makes sense, as the vector potential in this quasi-static process is given by the Coulomb-like integral

$$\mathbf A(\mathbf x)=\frac{\mu_0}{4\pi}\int \frac{\mathbf j}{|\mathbf x - \mathbf x'|} d^3\mathbf x'.$$

This vector potential, evaluated inside the wire, is finite for finite sized circuit but infinite for infinitely large circuit, due to infinite contribution of distant parts of the wire. It is the same behaviour as with electric potential of an infinitely long charged wire; inside the wire, it is infinitely high due to distant parts of the wire.

Thus the net energy per unit length of wire is infinite, per both formulae.

Generally, the $B^2$ variant of magnetic energy expression is better in the sense it follows naturally from Poynting's theorem which is valid even in highly violent processes, when the fields change quickly like near radiating antenna or radiating particle beams. In such processes, the assumptions behind derivation of the variant 7.31 do not hold and this latter formula cannot be relied upon.

11. Sep 12, 2015

### QuasiParticle

I agree that a loop (or two parallel wires) in the limit of infinite radius (distance) has an infinite inductance per unit length. However, I think the case of a single infinitely long wire is different.

It is true that in deriving the vector potential form for the energy, one considers the flux through a loop. However, I still feel that the energy expression is valid even for an infinite wire, even though the loop in the derivation may not be perfectly defined. If the vector potential expression for the energy is not correct in the case of an infinite wire, then I don't know how to compute the energy, since the $B^2$ expression is not correct.

I still disagree that the $B^2$ version follows naturally from Poynting's theorem; additional assumptions has to be made (e.g. neglecting surface integral terms).

12. Sep 13, 2015

### Jano L.

The surface integral in the Poynting equation vanishes if the surface containing the circuit is expanded to infinity and the process of current increase is quasi-static. Both energy formulas are valid, and both give +infinity.

13. Sep 13, 2015

### QuasiParticle

Right, but my suspicion is that the surface integral cannot be neglected for the infinite wire.

I don't understand. By using the vector potential of an infinite wire given in the first post, the energy (per unit length) is clearly finite.

14. Sep 14, 2015

### Jano L.

Infinite wire does not exist, it is only a result of a limiting process where circuit size expands to infinity. If the enclosing surface is much greater than the circuit in the process, then the surface term vanishes.

The vector potential is infinite in the wire due to its distant parts. I think the formulae you gave in the first post do not obey the condition that $\mathbf A$ is given by the standard Coulomb-like integral, which I think is necessary if the derivation of the A.j formula for energy is to hold (the vector potential needs to decay with distance).

15. Sep 22, 2015

### QuasiParticle

So are you saying that we should not even try to compute the inductance (or energy) of an infinite wire?

I disagree that an infinite wire is the limiting case of a large circuit. In the infinite circuit case the vector potential at the other wire due to the other becomes infinite (like you mention). This doesn't happen for the infinite wire.

Wouldn't an infinite wire be approximated well by a very long wire with an initial charge imbalance between the ends? This would not form a loop at all.

Not sure I'm following your argument. The vector potential for an infinite wire is obtained directly using Maxwell's equations (see e.g. Vanderlinde p. 78).

By the way, in Vanderlinde the case of two infinite parallel wires is calculated. He says about the resulting inductance that "The constant term, $\mu_0 l/4\pi$, represents the self inductance of the current in each wire on itself." This is precisely the result one obtains for the single infinite wire using the vector potential energy expression and the vector potential of the first post.

16. Sep 22, 2015

### Jano L.

No, I am not; we may try to compute energy per unit length of an infinite wire, but we will find that the integral diverges and the only reasonable result we can assign it is +infinity.

Infinite straight wire is not necessarily limit of a circuit, but limiting finite circuit to infinity it's the only physically sensible way to evaluate the infinite case. Laws of electromagnetism are experimentally confirmed for finite bodies, everything about hypothetical infinite bodies is extrapolation.

Wire with charge imbalance would be hard to force into obeying the condition of quasistaticity for which the A.j formula was derived.

His vector potential is fine as far as the role of potential for $\mathbf B$ goes. However, derivation of the A.j formula is based on the idea of finite-sized circuits where loop integrals $\oint \mathbf{A}\cdot d\mathbf{l}$ have finite values. For infinite wire, there is no loop and even if we tried to put it there, the integral would be infinite. There are no grounds to use this diverging A in the A.j formula.

Moreover, his potential diverges as $r\rightarrow\infty$. I think this may make it invalid to use in the general derivation of the A.j formula.

Last edited: Sep 22, 2015
17. Sep 24, 2015

### QuasiParticle

I think more or less static situation could be created with an appropriate RC circuit. But in any case, we can try to compute the result based on the electromagnetic theory, regardless of whether it is possible to construct an infinite wire in reality.

Here is an argument for the vector potential energy expression without reference to a circuit. Let's start with the Lorentz force
$$\vec{F} = q\left( \vec{E}+\vec{v}\times\vec{B} \right)$$
Only the electric field can do work on the charges. Let the work done against the EMF of a group of charges $\sum q_i$, which are displaced by $\vec{r}_i=\vec{v}_i dt$ be
$$dW = -\sum\vec{F}_i\cdot d\vec{r}_i = -\sum q_i\vec{E}\cdot \vec{v}_i dt$$
Let's replace the charges with the current density $\vec{J}(\vec{r})=\sum q_i \vec{v}_i \delta(\vec{r}-\vec{r}_i)$:
$$dW = -\int d^3r ( \vec{J}\cdot\vec{E}) dt$$
The electric field in terms of the vector potential is $\vec{E}=-\partial \vec{A}/\partial t$. Let's integrate to get the total work:
$$W = \int d^3r \int_{0}^{\vec{A}} \vec{J}\cdot d\vec{A}$$
By using the linearity between the current density and the vector potential, this integral yields
$$W = \frac{1}{2} \int \vec{J}\cdot\vec{A}~d^3r$$
I believe it is completely valid to use this energy expression for the infinite wire. And since the vector potential of an infinite wire is finite, the energy per unit length is also finite.

18. Sep 25, 2015

### Jano L.

This formula gives work done by electromotive forces (e.g. in a battery) against electric forces.

This is not a valid transformation of variables, because $\mathbf J(\mathbf x)$ is not a function of $\mathbf A(\mathbf x)$ (in words, current density at a point is not a function of vector potential at this point).

19. Sep 25, 2015

### QuasiParticle

But the work done to produce the system is exactly the energy of the system.

Which transformation of variables are you talking about? Replacing the electric field with the derivative of the vector potential? What does this have to do with the current density?

20. Sep 26, 2015

### Jano L.

I agree with that. My point was that the work of electromotive force (chemical, mechanical or due to temp. gradient) is stored into the EM energy of the system, not the work of electric force.

The transformation where the integral over time is replaced by the integral over components of $\mathbf A$. This is not a valid transformation, because, while $\mathbf J(\mathbf x)$ is a function of $t$, it is not a function of $\mathbf A(\mathbf x)$.