# Electric Field of a Current

This is a question, I think, of "unknown unknowns." I know I don't understand something but I am not quite sure what I don't understand and I am trying to work through it.

So...

1)
I have been following some of the threads about current in wires and electrical fields and, if I understand it correctly, the electrical field in a current carrying wire is confined to the wire. Is that correct?

2)
If the above is correct and I consider a current carrying wire, then at any point outside the wire, the electric field is zero. Is that also correct?

3)
However, the magnetic field outside the wire is non-zero. Therefore I can consider a Poynting vector for this system. But the Poynting vector will, one again, be zero.

4)
Now, for a solenoid with constant current. Once again, the electric field will be confined to the wire wrapping. Is that correct?

5)
However, if the solenoid has a time-varying current, does that give rise to an electric field within the solenoid? I think Maxwell's equations to say it does. Am I correct?

6)
And what about outside the solenoid? Is there an electric field there for time varying currents?

Thanks for any help.

## Answers and Replies

I might be wrong, but I believe assertions 1 and 2 are incorrect.

The electric field inside a current carrying wire is created by a generally complex distribution of charges on the surface of the wire. Therefore, these charges also create an electric field outside the wire.

Born2bwire
Gold Member
For DC circumstances, we assume that the electric field that induces the current is confined to the wire's volume. I believe that you can use the Poynting vector for the DC case, noting that the wire has a non-zero cross-section and that the magnetic and electric fields will be non-zero on the interior and surface of the wire because of this. I have seen a few derivations of this and you get the P=IV relationship back out again. However, for any AC current you will have electric and magnetic fields outside the wire or whatever medium you are using to transmit the signal. For any AC signal, you actually use a waveguide to transfer the signal because now the signal is contained in electromagnetic waves. It is these electromagnetic waves that excite the currents but they are guided by the shape and structure of the waveguide. How the fields will look fore a given system depends on the geometry and electrical properties of the system.

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If the current in the wire is coupled to a return current in another wire, and the wires are transporting electric power (real or reactive) to a load, then there is a voltage between the two wires as well as current in the wires. The combined azimuthal magnetic field around each wire, and the electric field (voltage drop) between the two wires then provide the necessary E x H for the Poynting vector. This is even true for dc power.

Also, if the wire has resistance R, there is an IR electric field due to the voltage drop parallel to and outside the wire. (E-parallel is continuous at the wire/air boundary).

Smythe "Static and Dynamic Electricity" 3rd edition page 297 solves the B field for a solenoid in the form of a helix with a pitch. There is an azimuthal B field outside the solenoid.
Bob S

Hi.

Wires are originally neutral in electricity, say minus of conducting electrons and plus of ion lattice cancel. But when the end of wire is connected to a excess charged body like a terminal of battery, the wire is also charged like gold-leaf electroscope. Excess charges of wire make external electric field appear outside the wire. This electric field and the magnetic field by current in wire make Poynting vector EXH appear around the wire corresponding to the flow of energy.

Regards.

PS

As Bob S said, coaxial cable contains electric field within it because plus excess charge of say go-wire and minus excess charge of come-wire make sealed condenser. There is no leak of electric field outside. Therefore, Poynting vector or the flowing energy is also confined within the coaxial cable.