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If you consider a current carrying conductor, every instant an electron enters the conductor, another electron will be leaving the conductor. Thus, the current carrying conductor will not be charged (i.e, it would not have any net positive or negative charge). Remember dipole has zero net charge, but it does have electric field around it. So, if net charge is zero, it doesn't mean there is no electric field.

It is important to notice that, if assume only electrons to be moving, and kernels (positive nuclei) to be static, magnetic field will be produced only due to electrons.

The speed at which energy or signals travel down a conductor is actually the speed of the electromagnetic wave, not the movement of electrons (this is an modified statement extracted from wiki encyclopedia-speed of electricity).

Does it mean that electric field and magnetic field exists around the current carrying conductor?
Does it mean that only magnetic field exists around the current carrying conductor?

By the discussion until now (2/11/2013), I have found difference in answers with respect to AC and DC. So, from here on wards, I want the question to be discussed on both AC and DC. Every one are suggested to update their answers with respect to both the cases (AC and DC).


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Presumably, since this is not a superconductor, there's a electric field inside the conductor - why are these electrons flowing? That field cannot sharply end at the edge of the conductor. – MSalters Oct 21 '13 at 10:56
a bit Related – Waqar Ahmad Oct 27 '13 at 11:39
I am an experimentalist. There do not exist appreciable electric fields outside current carrying wires. We are all doing the experiment continually just by typing on the computer to communicate on this page. There would be sparks continually. So we can only talk of fields smaller than the ionization energy of air, or our skin. – anna v Nov 2 '13 at 6:05
@annav The computer example is not true for everyone :D . There are continual sparks from my laptop as I type and they sting. But they are of course related to fields generated by lousy earthing :D . – dj_mummy Nov 2 '13 at 6:23
Would there be electric field around the current carrying conductor, even when DC source is used? Here is the answer. The point is that for maintaining $\vec E$ inside wire it requires a variable surface charge density i.e. the quasi-neutrality is destroyed making the wire charged hence a minor $\vec E$ is observed outside the wire too. – user Dec 17 '13 at 13:05

For the magnetic field, the currents are one source of the magnetic, but this problem is more linked to the source of the current in the wire. For a conductor with finite conductivity, an electric field is needed in order to drive a current in the wire.

If we assume your wire is straight, this required field is uniform. One way to realize this field is by taking two oppositely charged particles and send them to infinity while increasing the magnitude of their charge to maintain the correct magnitude of electric field. In this limit, you will obtain a uniform electric field through all space.

Now, put your conductor in place along the axis between the voltage sources--a current will flow. In the DC case, this gives rise to the magnetic field outside of the wire. As for the electric field, a conductor is a material with electrons that can move easily in response to electric fields and their tendency is to shield out the electric field to obtain force balance. Because the electrons can't just escape the conductor, they can only shield the field inside the conductor and not outside the conductor. With this model, we see that the electric field is entirely set up by the source and placing the conductor in the field really just establishes a current. Note here that if you bend the wire or put it at an angle relative to the field, surface charges will form because you now have a field component normal to the surface.

For the limit of an ideal conductor, no electric field is needed to begin with to drive the current and so there isn't one outside the wire.

For the AC case, solving for the fields becomes wildly complicated very fast as now the electric field driving particle currents has both a voltage source and a time-varying magnetic source through the magnetic vector potential. The essential physics is the same, though, as the source will establish the fields (in zeroth order), and the addition of the conductor really just defines the path for particle currents to travel. In the next order, the current feeds back and produces electromagnetic fields in addition to the source(s) and will affect the current at other locations in the circuit.

I guess a short answer to your question is that there are always fields outside of the current-carrying wire and the electric field outside disappears only in the ideal conductor limit. Conductors generally do not require very strong fields to drive currents anyway so that the electric field outside is usually negligible, but don't neglect it for very large potentials in small circuits.

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It seems to be little known, but a static current inside a wire of finite conductivity does produce a static electric field outside. See "Application 9.2" in the new text by Zangwill (which is fantastic and will replace Jackson), or

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There is no E field outside the conductor due to electrons because the conductor is not charged--this means there exist an equal number of protons to every electron in the conductor. If the conductor was charged, then there would be E field outside the conductor due to that charge. In fact, for a conductor to produce an E field outside of itself, it must be statically charged like a capacitor plate.

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A conductor (not considering superconductors) with current flowing through it also has an electric field inside it. Maxwell's curl equation for electric fields guarantee's that the electric field must exist outside as well. Your answer is correct only if you consider conductors without current flowing through them. – David Oct 29 '13 at 21:59
There must be a potential difference along the wire to permit electrons to flow. Doesn't that imply an electric field? A path integral $\int E\cdot ds$ from A to B outside the wire must result in a voltage $V_{AB}$. – Floris Aug 20 '15 at 14:52

This isn't a direct answer to your question, but I believe goes to the heart of the issue.

When talking about energy and electrodynamics in a classical sense, it's often possible to interpret energy in two different ways. The first way is to think about the potential energy of a charge configuration. For example, if you have a fixed positive charge, and bring a second positive charge close, then you're pushing it uphill against the potential energy. The electric field from the first charge would normally repel the second. The work necessary to bring the second charge closer goes into the potential energy of the configuration.

The second way of thinking about the energy is a little more subtle. When you bring the two charges together, you get a very large electric field between them. You can actually consider the energy as stored in the electric field.

The big advantage of understanding the energy as stored in the electric field comes when you talk about light. Light has energy, but is also electromagnetic field. There aren't any charge configurations to store the energy, so the energy must be in the field itself.

Electrical energy and signals really do transmit at the speed of light in a conductor, like the Wikipedia article says. The electrons themselves travel much slower. This is clearer if you consider that the energy is stored in the electric field (and magnetic field), so the signal should travel at the speed of the propagating electric field (aka the speed of light).

A more direct answer to your question: Yes, there are both electric fields and magnetic fields around the conductor while there's current flowing through it. This is guaranteed by Maxwell's equations, and the fact that current is proportional to the electric field in a conductor.

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I am not sure what exactly you are asking. I assume you are asking about the presence of an electrostatic field around a current carrying conductor. By approximating this system to infinite columns of positive charges and negative charges (a crude simplification), I will use relativistic arguments to demonstrate that when the 2 columns are moving with respect to each other in a certain frame, an electric field could very well be generated. I hope that this helps.

In electromagnetism, if the current densities are completely known for all space-time (like in your question), it is a straightforward matter to calculate the electric and magnetic fields. It is given here . The current charge density is given by:

$$j^{a} = {\rho}_0 u^{a}$$

Here ${\rho_0} $is the local charge density at event P which is the charge density around P as seen by an observer moving at the same speed as the charge element at P. $u^{a}$ at event P is the proper velocity of the charge element at P, this is the rate of change of position as measured by you, divided by the proper time of the charges at P. In practice $u^{0}$ is slightly greater than 1 and $u^{i}$ (i stands for x,y and z components of proper velocity) is slightly greater than the $v^{i}$. These are relativistic corrections.

Since Maxwell's equations are linear, we can superpose fields and charge densities. So let us separate the conductor into 2 charge distributions. $$j_{+} = ({\rho}_0,0,0,0)$$ $$j_{-} = (-{\rho}_0,0,0,0)$$ One is a infinite column of electrons and another is an infinite column of positive charges occupying the same space. When no current is flowing and the charge columns are stationary, the electrostatic fields disappear on superstition of the.

But what happens when only the negative column is moving at speed u in your frame? Well here's what happens:

$$j_{+} = ({\rho}_0,0,0,0)$$

Obviously the positive charge density remains the same.

$$j_{-} = (-{\gamma}{\rho}_0,-{\gamma}v_x{\rho}_0,0,0)$$


Here $-{\gamma}$ is $u^{0}$ and is a relativistic correction applied to each point P, in everyday life, it is a little larger than 1. The 3 spatial components contribute solely to the magnetic field (in this case where nothing changes with time). Now the net charge density (solely responsible for electric fields in this case where nothing changes with time) is equal to $(1-{\gamma})\rho_0$. Therefore, we find the existence of both electric and magnetic fields in your frame of reference, although the electric field is very weak. If the electron column were moving close to the speed of light, we would have significantly higher electric fields.

This is due to relativistic length contraction. The negative column undergoes length contraction in the direction of velocity at every point, this raises negative charge density.

Of course, this is a very simplistic rendering of an actual conductor, but it is good enough to illustrate the key ideas.

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Yes there is an electric field outside of a current carrying wire, in a direction along the wire axis (i.e. parallel to the wire). This is true in both the AC and DC case. There is also of course a magnetic field in the azimuthal direction.

For a resistive wire oriented along the z-axis, the electric field inside the wire is given by Ohm's Law $E_y=\eta j_y$ where $\eta$ is the resistivity and $j_y$ is the current density. This current tends to flow near the surface of the wire (this is the skin effect). In order that there is no discontinuity in the electric field parallel to the wire, the electric field $E_y$ in the vacuum at the edge of the wire has to match the field in the wire given by Ohm's Law. Further away from the wire, there must be a radial electric field needed to keep the total electric field curl-free.

The fact that this electric field is parallel to the wire is the reason why electromagnetic energy flows radially into the wire. This is somewhat counterintuitive, but can be seen by noting the direction of the Poynting vector $\bf E \times \bf B$.

This is mentioned briefly in Feynman's Lectures on Physics II chapter 27-5.

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The skin depth for a dc current is infinity. A radial electric field requires there to be a net charge on the wire. Not saying your wrong, but your answer doesn't explain anything. – Rob Jeffries Feb 6 '15 at 16:54
The radial electric field inside the wire can be provided by the local charge imbalance caused by the Hall effect. Outside of the wire, a radial electric field could be provided by either the divergence-free field induced by the EM waves that give rise to the azimuthal magnetic field, or by the surface charges discussed in Jackson's (1996) and Assis's (1997) papers on the subject (mentioned above). – kotozna Jun 13 '15 at 11:13

protected by Qmechanic Feb 17 '14 at 21:39

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