Currently I am stuck, trying to derive the self-inductance of a long wire. According to literature it should be


and in literature its derived by looking at the energy of the magnetic field. I tried to derive this formula via the magnetic flux and I am getting $4\pi$ instead of $8\pi$. This are my considerations:

   _ _
 /  |R \
|   '   | a wire with radius R and length l
 \ _ _ /    

The magnetic flux density $B(r)$ is given according to Ampere's law:

$$B(r) 2\pi r=\mu_0\mu_r\frac{r^2}{R^2}I$$

$$B(r)=\mu_0\mu_r\frac{r}{2\pi R^2}I$$

where $r$ is the distance from the center of the wire, $R$ is its radius and $I$ is the total current through the wire. Now I know that the magnetic flux $\phi$ through the upper part of a longitudinal section is

$$\phi = \int_A B dA = \int_0^R B(r)l dr = \frac{\mu_0\mu_rIl}{4\pi}$$

where $l$ is the wire's length. No I use $\phi = LI$ and arrive at $4\pi$.

What am I doing wrong? Where is the mistake in my considerations?

Moreover I have the following problem. If I look at an entire longitudinal section of the wire and not only at its upper half the magnetic flux is zero:

   _ _
 /  |  \
|   |2r | => Magnetic flux is zero (the magnetic field 
 \ _|_ /     penetrating the upper half of the longitudinal cross section is 
             exactly opposite to the magnetic field penetrating the lower

Hopefully I formulated my problem clear enough. If not please ask me for further details.

  • 3
    $\begingroup$ @ftiaronsem: don't worry, Georg just has a bit of an abrasive personality sometimes. I think the ambiguity is whether your ASCII diagrams are showing a cross-section of the wire or a top view of a loop of wire. Other than that, your question seems quite clear, so thanks for constructing a great homework question ;-) $\endgroup$
    – David Z
    Jul 3, 2011 at 4:36
  • 2
    $\begingroup$ A fun note: "its derived by looking at the energy of the magnetic field", this reminds me of my freshman year in college where I took the formulas in the book and found that a wire contains infinite energy per unit length due to the magnetic field around it. I took several instructors and they all agreed it was error free. They said the only reason it wasn't true was likely using $B^2$ for the energy of the magnetic field, which was probably only valid for certain cases of coiled wire interiors. Not the same kind of problem but I couldn't resist sharing my personal story. $\endgroup$ Jul 3, 2011 at 12:43
  • $\begingroup$ @Zassounotsukushi Thanks for this comment. I tried to follow you but for the magnetic energy density term I get $w=\frac{\mu_0i^2}{8\pi^2r^2}$. Integrating this in cylindrical coordinates from $r=a \rightarrow r=\infty$ and over $\theta = 0 \rightarrow \theta=2\pi$ and $l=0 \to l=1$ I get something finite. What am I doing wrong? Thanks in advance $\endgroup$
    – ftiaronsem
    Jul 3, 2011 at 17:43
  • 1
    $\begingroup$ @ftiaronsem Are you including the Jacobian? $\int_0^1 dl \int_0^{2\pi} d\theta \int_a^{+\infty} dr\,r \frac{A}{r^2} = 1 \cdot 2\pi \cdot A \cdot \left[\ln r\right]_a^{+\infty}$, diverging logarithmically. $\endgroup$
    – mmc
    Jul 3, 2011 at 22:16
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    $\begingroup$ The energy as $B^2$ is correct, and the divergence is the consequence of unphysical assumptions: infinite current-carrying wire in the middle of nowhere. Once you consider a return path, the divergence goes away. C.f. Edit 2 of my answer. $\endgroup$ Jul 4, 2011 at 9:05

4 Answers 4


I know this post is old and has been answered I thought I would post the exact derivation to help out anybody in the future.

In order to calculate the the internal inductance of a wire we have to equate the equation for the energy of the magnetic field to the energy from the inductor/inductance.

Energy of the $B$ field: $\frac{1}{2\mu}\int B^2dV$, where $B$ is integrated over all space.

Energy of an inductor: $\frac{1}{2}LI^2$

To solve for the Magnetic field B we use an Amperian loop - $\oint \vec{B}\cdot d\vec{l}=\mu I_{enc}$
First we make an assumption that the current is uniformly distributed throughout the wire (which is a reason this inductance is usually neglected, at higher frequencies the current is not uniform but is carried on the surface of the wire which creates a more real resistance in the wire and less self inductance). With a uniform current $I_{enc} = I\frac{\pi r^2}{\pi R^2}$ for r$ < R$ where $r$ is a variable distance inside the wire, $R$ is the radius of the wire, and I is the total current running through the wire. This equation is a simple ratio of the variable area to the total area of the wire multiplied by the total current inside the wire to find the current for any variable amount of the wire.

Now plug this back into Ampere's equation $$\oint \vec{B}\cdot d\vec{l}= μI\frac{\pi r^2}{\pi R^2} = μI\frac{ r^2}{ R^2}$$ Now because B is independent of the closed loop itself for any r distance the B can be pulled out of the integral. $$B\oint dl=μI\frac{ r^2}{ R^2}$$ The closed loop integral along dl can be calculated out, but for simplification we will realize that the loop is simply the circumference at any variable r distance. $$B\cdot2\pi r=μI\frac{ r^2}{ R^2}$$ Solving for $$B = B(r) = \frac{\mu Ir}{2\pi R^2}$$ $$B^2 = \frac{\mu^2 I^2 r^2}{4\pi^2R^4} \quad\text{for}\quad r < R$$ Meaning this is only the B field inside of the wire

Now we can equate the two energy equations: $$\frac{1}{2\mu}\int B^2 dV = \frac{1}{2} LI^2$$ Plugging in for $B^2$ that we solved previously we get: $$\frac{1}{\mu}\int\frac{\mu^2 I^2r^2}{4\pi^2R^4}d = LI^2$$Plugging in the volume integral in polar form: $$\iiint\frac{\mu r^2}{4\pi^2R^4}r\mathrm{d}r\mathrm{d}ϕ\mathrm{d}z = L$$

At this point is where some people "screw up" and get an infinite inductance. Because the magnetic field outside of the wire is not coupled to anything and is not bounded it does not contain any real energy (thus why we solved for $B$ inside of the wire only), now our integrations go from $0$ to $R$ in the $r$ direction, $0$ to $2\pi$ in the $\phi$ direction, and $0$ to $l$ in the $z$ direction where $l$ is the length of the wire. This integrates over the entire wire.

$$L = \frac{\mu}{4\pi^2R^4} \int\limits_0^l\mathrm{d}z \int\limits_0^{2\pi}\mathrm{d}\phi \int\limits_0^R r^3\mathrm{d}r$$

$$\therefore L = \frac{\mu}{4π^2R^4}\left(l2\pi\frac{1}{4}R^4\right)$$ $$\therefore L = \frac{\mu l}{8\pi}$$
To match the original questioner's format realize that $\mu=\mu_r\mu_0$

Note that in this equation the $R$'s have cancelled out, which means the inductance is independent of the radius of the wire itself. Hope this helps anyone out there!

  • $\begingroup$ Your calculation is flawed when you take the B field out of the integral because your wire is not infinite. Yes,$ \vec{B}\cdot \vec{dl}$ reduces to |B||dl| and YES you CAN pull out the B field from this due to symmetry. However amperes law is invalid for this situation since the divergence of J is non zero. Even more so, using biotsavart would give you a different B field $\endgroup$ Dec 18, 2021 at 18:49
  • $\begingroup$ > "At this point is where some people "screw up" and get an infinite inductance..." No, that's all false. Magnetic field outside does contain energy, and for infinite wire, it is infinite. Thus the infinite self-inductance per unit length is the correct result for infinite wire. The concept of "internal self-inductance of a wire" due to flux/energy in the inner region of the wire is of limited relevance, as only total self-inductance can be observed, and calculation of that has to take into account the flux outside the wire. $\endgroup$ Feb 25, 2023 at 5:00
  • $\begingroup$ If the current flowed in a thin layer on the surface, magnetic flux inside the wire would be zero, and this method would predict zero "internal self-inductance", while net self-inductance is very much not zero. In flux-based methods of calculating inductance, it is simply not correct to ignore magnetic flux that is outside the wires. Cf. Rosa's article on self-inductance of finite wire; he uses weighing of magnetic flux, but also takes into account the flux outside the wire: g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf $\endgroup$ Feb 25, 2023 at 5:05
  • $\begingroup$ @jensenpaull Divergence of $J$ is assumed zero, a circuit is implicitly assumed. Ampere's law seems valid here, as long as displacement current is negligible. That is certainly the case if electric field is constant in time, i.e. when current is linear function of time and thus $\ddot{I}=0$. If there is current oscillation of high enough frequency so that $\ddot{I}$ is great enough, you may be right that displacement current may be so strong that Ampere's law becomes invalid. $\endgroup$ Feb 25, 2023 at 20:27

The original question talked about a discrepancy between the result obtained by calculating the flux directly and using the definition $L= \Phi/I$. The confusion arises because of a concept known as "flux linkage". When you calculate the flux enclosed by the region of unit length between r and r+dr, you calculated an expression for flux which you integrated to get the total flux. However, the entire flux calculated by you is not "linked" to this area since the current enclosed by the contour of this radius is a fraction $\pi r^2/R^2$ of the total current. Thus the linked flux is $d\Phi= \dfrac{\mu_0 I r dr}{2\pi R^2} \dfrac{\pi r^2}{\pi R^2}$. If you integrate this expression, you would get the correct result. $$\Phi = \dfrac{\mu_0 I}{2\pi R^4}\int_0^R r^3 dr= \mu_0 I/8\pi$$ "Flux linkage' is not a very easy concept but consider what happens when you have N turns of the wire through which the same flux passes. In order to calculate emf using Faraday's law, you will need N times the flux to get the correct emf.

  • $\begingroup$ Thanks to this response I realized I was not looking at flux integrals as flux-linkage integrals (stupid!). Of course! For a field to be well-defined, 1 to 1, only 1 flux line ought to go through any point. A corollary is that flux lines never cross. In any case, for a correct flux integral, only cross each flux line once! I was counting 'em twice. But what is a flux line?.... Perhaps, a line which follows the infinitesimal displacements dr for an infinitesimal unit of that field (i.e. dm for gravity, dq for elec), across all space. $\endgroup$ Dec 28, 2020 at 17:35
  • $\begingroup$ Hey I know this is a super old post but I had a question about this solution. Didn’t we already take into account the fraction of the total current when we calculated the magnetic field? What is the reason for doing this twice? $\endgroup$ Nov 2, 2022 at 1:53
  • 2
    $\begingroup$ @JosephSanders there is no good reason, the whole idea of taking weighted average of magnetic flux values based on radius is very bizarre and I see no justification in physics. The result is also useless - self-inductance calculation needs the whole magnetic flux, including contribution outside the wire , but this was omitted here which is unjustifiable. See my comment to the OP. $\endgroup$ Feb 25, 2023 at 1:53

Edit 1: I think I just understood you question: you are actually trying to calculate some sort if “internal” inductance, i.e. the contribution to the inductance of only the field inside the conductor.

When calculating the flux, you have to choose a closed path over which you would want the electromotive force, and then integrate the magnetic flux over the surface limited by this path. Normally the path would be the whole electrical circuit, but since you are only interested in the contribution of the internal field, you chose the return path along the edge of the wire, which is fine. Now you have to choose the forward path.

The forward path should be along the lines of current. The problem is that, different lines of current give different fluxes. Then you can calculate the flux as a function of where, in the conductor's cross-section, you take the forward path. But since you are using the low-frequency approximation (no skin effect, then uniform current density), you can just average the forward-path dependence over the whole cross-section. Then you get the missing factor two.

A somewhat different argument is given in this old bulletin of the Bureau of Standards: the author instead weights individual flux lines as per the fraction of the conductor they enclose. This gives the same factor two.

Edit 2: As requested, a few clarifications.

By “integrate the magnetic flux” I really mean “calculate the magnetic flux”. I used “integrate” because the calculation involves an integral: $$ \phi = \int_A \mathbf{B}\cdot\mathbf{n}\; \mathrm{d} A $$ where $\mathbf{n}$ is the unit normal to the surface. It's not exactly the same as “integrate the magnetic field” because of the dot product with $\mathbf{n}$.

I talked about “forward path” and “return path” because, if it's not an antenna (as the low-frequency approximation suggests), a wire is usually part of a transmission line which consists of at least two conductors. Assume for example that you use a pair of wires to connect a source to a load, like in the figure below (I hope everyone can see Box Drawing characters):

╔════════╗                 ╔════════╗
║        ╟→→→→→→→→→→→→→→→→→╢        ║
║ source ║   (flux here)   ║  load  ║
║        ╟←←←←←←←←←←←←←←←←←╢        ║
╚════════╝                 ╚════════╝

where the arrows represent the electric current. I assume the wire you are interested in is the top one, which I called “forward path”. The bottom wire, which I called “return path”, brings the current back to the source. Taken together, these two wires form a loop and the current will make some magnetic flux through the loop. Then, if you try to change the current, some electromotive force will appear because of this flux, and you will be able to model this as the effect of an inductor along the transmission line, as below:

╔════════╗                 ╔════════╗
║        ╟────(inductor)───╢        ║
║ source ║                 ║  load  ║
║        ╟─────────────────╢        ║
╚════════╝                 ╚════════╝

This is the self inductance of the transmission line, and is what I first thought you where trying to calculate.

The self inductance of a bare wire is somewhat ill-defined. Well, it is defined, but with some assumptions about the surface over which to integrate the flux, and it scales as $l\log\frac{l}{r}$, which makes it's value per unit length diverge logarithmically when considering an arbitrarily long wire, as pointed out by Zassounotsukushi and mmc. Once you add the second wire, the surface over which you have to integrate the flux is clearly defined, and the inductance of the line scales like $l\log\frac{d}{r}$, where $d$ is the distance between the wires. No more logarithmic divergence with respect to $l$. On the other hand, it depends logarithmically on the distance between the wires, therefore you cannot just assume that the return path is just far enough to be ignored. BTW, the return path is not necessarily a wire, it could be, e.g., a ground plane.

For the particular calculation you are doing (only the contribution of the field inside the conductor), you use a very narrow loop where the return path is replaced by a line along the edge of the conductor, in order to enclose only the internal field.

Original answer below, which is somewhat bogus, as I thought you where after the total self-inductance (including external field) per unit length of an infinite wire. The comments of Georg refer to this original version.

You cannot assign an inductance to a long wire alone: you have to consider the whole circuit. The current carried by the wire has to come back in some way, and you need to know how far from your wire is the way back.

Assume for a moment that the wire is actually the inner conductor of a coaxial cable. You can easily calculate the linear inductance of the cable as a function of the inner an outer conductor radii. Now make the outer radius go to infinity and you have a diverging self-inductance! This means that in practice you can never assume that the way back is “far enough” to ignore it.

  • $\begingroup$ ""You cannot assign an inductance to a long wire alone: "" Yes You can! Look at a antenna like the on at Grimeton in Sweden. Even a simple lambda/4 vertical has a inductance without any "return". $\endgroup$
    – Georg
    Jul 2, 2011 at 18:13
  • $\begingroup$ Oh, right, I did not think about antennas! Maybe I misunderstood the question, but I thought the OP was after the inductance per unit length of an infinite wire. $\endgroup$ Jul 2, 2011 at 18:59
  • $\begingroup$ You wil find formulas in electronics textbooks even for such a case. $\endgroup$
    – Georg
    Jul 2, 2011 at 22:24
  • 1
    $\begingroup$ Yes, but with a logarithmic divergence of the linear inductance. $\endgroup$ Jul 3, 2011 at 11:48
  • 1
    $\begingroup$ Thanks very much for this answer Edgar, however I have some difficulties following. The first part I do not understand is: "When calculating the flux, you have to choose a closed path over which you would want the electromotive force, and then integrate the magnetic flux over the surface limited by this path." What do I get by integrating the magnetic flux over a surface? Did you mean to integrate the magnetic field? The other thing I do not understand is the forward/backward path thing. What is a forward path? What is a backward path? It would be great if you could clarify that. Thanks $\endgroup$
    – ftiaronsem
    Jul 3, 2011 at 17:30

In case of uniform current distribution inside of a long wire, magnatic field increases linearly inside of the conductor and decreases inversely with radius outside of the conductor. For outside of the conductor, we can assume a coaxial cable with infinite outer radius. In this case the square of the magnetic flux decreases with inverse square of r and the volume integration from Ri to Ro(infinite) gives finite value per unit length of wire.


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