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I have encountered the problem:
Find the magnetic energy stored inside a 1m length of wire carrying 10 amp. The wire is 1 mm in radius, and the current density is uniform.
I have found a solution where the energy density is used, or:

$$ \frac{U}{\textrm{ vol}}=\frac{B^2}{2μ_0} $$

Now I have two questions:
1) The energy density is derived for the case of magnetic field of an inductor (solenoid), but the single wire is not a solenoid and how could the same formula for energy density be used?;
2) I think it could be that the problem is about the wire of the solenoid, I mean is it possible to be implied that the wire in question is wound in the form of solenoid but it is not written explicitly?

After all, I solved the problem without knowing what exactly is meant. I applied the energy density expression derived for a solenoid (as given above) and got wrong answer. In matter of fact the final formula I obtained, and also find it in another solution in the web, is this: $U=\frac{μ_0i^2}{16π}=2.5\times 10^{-6}$ joules. The correct answer in the book is claimed to be $U=3.3(3)\times 10^{-3}$ joules!

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closed as off-topic by Rob Jeffries, ACuriousMind, Daniel Griscom, Gert, user36790 Jan 31 '16 at 14:51

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  • $\begingroup$ Voting to close as homework. I would be working in cylindrical coordinates, assuming that the wire was "long", using Ampere's law to calculate the field assuming the wire has a uniform current density, using this to calculate the (position dependent) energy density in the magnetic field and then integrating this over your required volume. I get your answer. $\endgroup$ – Rob Jeffries Jan 30 '16 at 15:28
  • $\begingroup$ That is not how your question reads. The formula for energy density of a magnetic field is a general one and applies to all geometries/situations. $\endgroup$ – Rob Jeffries Jan 30 '16 at 18:04
  • $\begingroup$ "Although we have derived the formula for the magnetic energy density for the special case of a very long solenoid, its derivation is valid for any arbitrary magnetic field". This is written in the book I read, but it is not proven. Also similar statements are given in another sources, but it is not obvious to me. Can you give me a literature source where this statement is proven at least for a single long wire? $\endgroup$ – Pekov Jan 30 '16 at 19:25
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    $\begingroup$ You can find derivation of the formula based on work of non-electromagnetic forces during quasi-static process in textbooks on EM theory, like Griffiths or Jackson. A general process, however, involves radiation and energy of EM field does not have unique value. There is infinity of possibilities for how energy can be distributed. It is customary to use Poynting theorem then and fix this arbitrariness by defining EM energy as to be given by the Poynting expression. Magnetic energy is then given by the mentioned formula. $\endgroup$ – Ján Lalinský Jan 30 '16 at 20:10
  • $\begingroup$ Thanks a lot Ján the book of Griffiths explained well even it has similar example! $\endgroup$ – Pekov Jan 31 '16 at 8:41
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The formula is local, so it is correct for any geometry (I did not check your factors).

To apply the formula to the case of a wire, you have to calculate the distribution of magnetic field in the wire depending on the distance from the axis (as far as I remember, the dependence is linear for uniform current distribution). Then you need to integrate the expression in your formula over the volume of the wire.

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  • $\begingroup$ I did exactly this but the answer given in the book is $3.3(3)\times10^{-3}$ and it is not the same as mine! $\endgroup$ – Pekov Jan 30 '16 at 17:41
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    $\begingroup$ I don't have time to do the calculations, but the answer in the book may be wrong, according to Rob Jeffries. $\endgroup$ – akhmeteli Jan 30 '16 at 17:54
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To find the distribution of B-field inside the wire $B(r)$ just use Ampere's Law.
Although there is an integration to do it is a very simple one.

Once you knoe how the B-filed depends on the distance from the centre of the wire you need to do an integration (again a fairly straight forward one) to eventuate the energy stored.

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