# Does a current carrying wire produce electric field outside?

In the modern electromagnetism textbooks, electric fields in the presence of stationary currents are assumed to be conservative,$$\nabla \times E~=~0 ~.$$ Using this we get$$E_{||}^{\text{out}}~=~E_{||}^{\text{in}} ~,$$which means we have the same amount of electric field just outside of the wire!

Is this correct? Is there any experimental proof?

Outside a current carrying conductor, there is, in fact, an electric field. This is discussed for example, in "Surface charges on circuit wires and resistors play three roles" by J. D. Jackson, in American Journal of Physics – July 1996 – Volume 64, Issue 7, pp. 855.

To quote Norris W. Preyer quoting Jackson:

Jackson describes the three roles of surface charges in circuits:

1. to maintain the potential around the circuit,
2. to provide the electric field in the space around the circuit,
3. and to assure the confined flow of current.

Experimental verification was provided by Jefimenko several decades ago. A modern experimental demonstration is provided by Rebecca Jacobs, Alex de Salazar, and Antonio Nassar, in their article "New experimental method of visualizing the electric field due to surface charges on circuit elements", in American Journal of Physics – December 2010 – Volume 78, Issue 12, pp. 1432.

• As an electronics engineer, this seems trivially true. Consider HV power lines... What's the big deal? – user56903 Feb 6 '15 at 15:41
• @DirkBruere, it is very common to think of the DC live wire as uncharged, perhaps because of the theorem which says that in stationary situation, charge density in a conductor vanishes. Of course, the electric field inside the wire must be due to some charges and the only place they can be is the surface of the wire. Once we realize surface charges must be on the wire, the presence of electric field outside seems natural. – Ján Lalinský Jun 1 '15 at 16:32

That equation is true for electrostatics. Inside an electrically neutral current-carrying wire, the electric parallel to the wire is zero. So outside the wire it's also zero.

More importantly, Gauss's law will tell you that the components perpendicular to the wire must also be zero.

So the electric field is zero everywhere for an electrically neutral current-carrying wire.

• then what is the $E$ in the equation $J=\sigma E$ always we have inside the current which by the way has zero flux through every closed surface inside the wire!! – richard Apr 22 '13 at 15:53
• For a short wire for length $L$, $EL=V$. But for a long wire, $L \rightarrow \infty$ so $E \rightarrow 0$. For a short wire the electric field isn't zero everywhere. – Pricklebush Tickletush Apr 22 '13 at 17:24
• Alec, for long wires, one would use higher voltage to maintain the electric field, and hence the current, the same. Your answer is incorrect; there is electric field both inside and outside the wire. – Ján Lalinský Oct 31 '13 at 14:17
• In order to maintain the flow of electrons in a current-carrying wire, there must be a potential difference. Going from A to B on a long wire, an electron must experience the same potential drop regardless of path. So if it leaves the wire it must experience $\Delta V = \int E\cdot ds$ - and if the line integral of the field is finite, the field must be finite. – Floris Jun 23 '14 at 5:42

I think is that on the outer diameter for a distance tending to zero, the electric field will be same as inside but when you move further outside of the cable towards larger distance, the field will be reducing.

If electrons are ejected out of an atom to create a constant current in a wire, then the nuclei of the atoms that lost the electron become positively ionised, which creates a positive radial electric field. Conceptually in electrostatics theory, this field must permeate throughout and beyond the confines of the wire. However, the released electron itself is negatively ionised, which in turn creates a negative radial electric field permeating within and outside of the current carrying wire.

The combined electrostatic field created both inside and outside of the wire has potentially neutralising positive and a negative components. In the concept of electrostatics, the electron and a proton have exactly the same magnitude of charge and so the fields both inside and outside the wire should always be neutral.

However, the assumption behind this logic is that when particles with equal positive and negative charges meet, all their radial line of force inter-connect in a neutralising manner. However, if the pair can still connect to other charges in the vicinity equally strongly, the answer is the opposite and that an electrostatic field does always exist both inside and outside of the current carrying wire.

I expect that someone, somewhere must have done experiments with a single proton and multiple electrons to verify this dilemma. If so they must have created a hydrogen atom with multiple electrons filling one or more of its s,p,d,f shells, but I have not heard of such an atom being created yet!

Let $r_0$ be the wire radius. For $r>r_0$ we have magnetic field$$B\left(t\right)= \frac{\mu_0}{2 \pi r}I\left(t\right) ~.$$ Construct the thin wire rectangle $r{\times}Z$ : sides $r$ are perpendicular to the wire axis, one of $Z$'s is inside the wire close to surface (it is no electric field there).

The induced voltage (in this rectangle!) then is $$U_i=\frac{d}{dt} \int {dr \frac{dB}{dt}} ~.$$Integration is made from $r_0$ to $r$. It is only outer $Z$ - side where the voltage is induced. Then $U_i= E(r)\,Z$, one has $E(r)$ ($Z$ disappears!), Lenz theorem gives the direction of $E$.

The experimental proof of the presence of the electric field outside a current carrying wire can be achieved by measuring the voltage between any two points along the wire.

Non zero voltage will be a proof of the existence of the electric field.

Your problem is clearly and comprehensively treated by Hans De Vries in:

http://chip-architect.com/physics/Magnetism_from_ElectroStatics_and_SR.pdf

The quintessence is that a current carrying wire appears electrostatically charged to an observer in relative motion to that wire, even when the same current carrying wire appears uncharged to an observer at rest relative to that wire. The observed electric field is identical to that given by the Lorentz-formula E = v X B. For simplicity, disregard the resistance of the wire and assume a superconducting wire.

• This doesn't seem to answer the question, which is about the experimental proof of $E_\parallel^{out}=E_\parallel^{in}$ in a wire. – Kyle Kanos Aug 27 '14 at 16:47