# Magnetostatics of Current-Carrying wire

A question has been nagging at me about Faraday's Law as related to a wire with a constant current:

If you have a circular loop of wire with some small resistivity, connected to a battery so that it has a constant DC current, the electric field inside the wire is going to be longitudinal and proportional to the current density. In particular, is going to be something like $\vec{E}=E_0\hat{\phi}$, choosing cylindrical coordinates with the origin at the center of the loop. The curl of this field is obviously nonzero, so applying Faraday's Law there should be a linearly growing magnetic field in the Z direction, at every point inside the wire. But this seems like nonsense; we're supposed to be in a steady-state, and it doesn't seem physically plausible to me that there is a field that can be growing arbitrarily large as we keep our battery plugged in, or that its growth should be affected by the shape of the wire loop. So I expect I am missing some contribution that would cancel this effect. But what it is? I feel like I must be missing something obvious, but can't for the life of me see what it is.

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If you apply Faraday's law here, the net emf around the loop is zero (as expected). You forget that there is a battery in the loop, which has a large, opposite electric field across it. So $\varepsilon=\varepsilon_{loop}+\varepsilon_{battery}=0$. Which means that $-\frac{d\Phi_B}{dt}=0$, so $B=constant$, which means that the loop on its own will not induce any magnetic field, but it can survive in a system with a constant magnetic field without any change.

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You are certainly correct that the total magnetic flux is zero, but this doesn't quite resolve my question. Applying the differential form of Faraday's law to the battery in the same way as I did to the wire above, it seems to me that one would get a time-changing magnetic field in each, such that their contributions to the path integral cancel each other out. A continuously increasing magnetic field in the battery, whose total flux is opposite that of the continuously increasing field in the wire, still seems physically suspect to me. – Rococo Mar 28 '13 at 23:51
@Rococo: How are you using the differential form? You don't know what the electric field inside the loop looks like. You have assumes the electric field on the wire to have some value, but you need to know what it looks like in the near neighborhood to be able to calculate a curl. The differential and integral forms are consistent, if the integral form gives an answer, the differential form will too. – Manishearth Mar 29 '13 at 6:00
Hi Manishearth, I was applying Ohm's Law, in the form $\vec{J}=\sigma\vec{E}$. In this case, assuming that the current density in the wire is uniform, the electric field also will be. – Rococo Mar 29 '13 at 23:05
@Rococo: That's only in the wire. (a) you can't assume that $\vec J$ is uniform, it may vary as you move radially outwards on the wire. (b) If you're taking a thin wire, you still need to know $\vec E$ outside. – Manishearth Mar 30 '13 at 6:29
@Rococo: I gues the best way to explain this is to consider the fringing fields on the battery as well. These will most likely set the curl back to zero. – Manishearth Mar 30 '13 at 6:59

First off, the self-inductance of any closed loop like this inhibits current flow; it doesn't swell it infinitely. That's back-emf. Second, while the total magnetic field of the loop will be in the $Z$ direction at the origin, it won't be so in the wires. Remember that the magnetic field of any small section of current-carrying wire is a cylinder. The sum of all these little contributions is what gives us the total field that you described, which looks like this.

So, summing up, when the circuit is first connected, self-inductance in the wires themselves and in the loop as a whole prevent the current from reaching its final state instantaneously, and the resistance of the wire and the internal resistance of the battery cause the current to reach equilibrium at its steady state. At that point, the loop has a steady magnetic field, but doesn't have to worry about self-inductance anymore.

I may have misunderstood you question, but I think that will help.

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My question has nothing to do with self-inductance. Basically I am asking how an electric field that is clearly physically curving, due to the influence of a wire, avoids creating a magnetic field. – Rococo Mar 29 '13 at 0:01
oops, sorry about that then. I thought I was missing something. – krs013 Mar 29 '13 at 3:16