- #1
QuasiParticle
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This thread is continuation to:
https://www.physicsforums.com/threads/self-inductance-of-a-straight-current-carrying-wire.461622/
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?
https://www.physicsforums.com/threads/self-inductance-of-a-straight-current-carrying-wire.461622/
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?
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