0
\$\begingroup\$

Let's say I want to calculate the electric field strength from an AC conducting wire, with a certain frequency and current. The wire is in the form av a loop. I want to calculate the E-field along the symmetry axis (z-axis) of that loop of wire.

How would I go about it?


OLD POST BELOW:

Let's say I want to calculate the electric field strength from an AC conducting loop of wire, with a certain frequency.

The total charge Q will be

$$Q=\int i(t) dt$$

Since the current entering in one end of the wire, and at the same time exits at the other end, the total charge Q will be zero? Is this correct? Therefore the electric field strength will be zero?

\$\endgroup\$
  • \$\begingroup\$ How can current enter and exit a wire at the exact same time? Are you asking something like \$\displaystyle Q=\int^t_t{i(t)}dt?\$ \$\endgroup\$ – KingDuken Dec 11 '18 at 19:17
  • \$\begingroup\$ Have I perhaps phrased it wrong? But consider a flow of water in a tube; the same amount of water that is entering will be pushed out through the other end? So the net amount of water in the tube will remain the same? I now realize that the change in current will still be present since it's AC? I think. But how? \$\endgroup\$ – Defindun Dec 11 '18 at 19:20
  • \$\begingroup\$ In a perfect world, yes, the amount of water that would enter a tube would contain the same amount going out of the tube... But the charge \$Q\$ would be the total amount of electron charge that pass from the beginning of the wire to the end of the wire in \$t\$ amount of time. So if no electron charge go through that wire, \$Q=0\$, indicating there is no current. \$\endgroup\$ – KingDuken Dec 11 '18 at 19:23
  • \$\begingroup\$ I'm relating to E, since you need the total charge Q to calculate it according to this site: hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c2 And what will determine Q, is the current. Therefore I would like to find the relation between Q and i(t). \$\endgroup\$ – Defindun Dec 11 '18 at 19:33
  • 1
    \$\begingroup\$ @KingDuken I think your comments are misleading. The current (the \$i(t)\$ term) flowing through a wire is the net charge flowing past any given point on the wire, not the difference between charge in and charge out. As long as the wire is short compared to the wavelength then the current out is the same as the current in for practical purposes, meaning that as a whole the metal remains charge neutral. At very high frequencies the situation is different, of course, and the question (field strength of a loop of wire?) is also confusing. \$\endgroup\$ – Elliot Alderson Dec 11 '18 at 19:40
0
\$\begingroup\$

If same amount of Q is flowing in the loop wire as out of it (I_in = I_out), then net charge of the loop is Q=0 and there is no E-field (But there is a H/B-field).

There can only be Q in the wire if I_in != I_out for some time. Then the loop wire is behaving like a capacitor and there is some Q != 0 left in the loop of wire.

The site you linked shows a setup where there is "a ring of charge". So there is no current I involved. (There maybe was a current I_in to bring that charge Q onto the ring).

\$\endgroup\$
  • \$\begingroup\$ Side note: != means, "not equal to". Not everyone is a programmer :) \$\endgroup\$ – KingDuken Dec 11 '18 at 20:15
  • \$\begingroup\$ Okay, so for a wire in form av loop (coil of you will), with AC-current between, let's say 0 and 300 mA oscillating in a square or sinusoidal wave, there will be no E-field? Then my hunch was correct. Doesn't feel very intuitive, since I knew there was an H/B-field, but perhaps it doesn't have to mean that there must be an E-field. Thank you! EDIT: What would happen if you put capacitors in series with the wire/loop? \$\endgroup\$ – Defindun Dec 11 '18 at 20:21
  • \$\begingroup\$ Not quite correct. There is an E-field involved, but only in the direction of current flowing. Because there would not flow any current if there was no E-field (or voltage) along the wire. (BTW, I do not understand your capacitor question). \$\endgroup\$ – Stefan Wyss Dec 11 '18 at 20:29
  • \$\begingroup\$ Yes, that's true. Is it possible to determine an expression for that E-field as a function of the distance from the loop/coil, along with the z-axis? Regarding the capacitors; in some cases, it's desirable to have resonance with the use of capacitors. Since you made a reference to capacitors above: "then the loop wire is behaving like a capacitor" I wondered if there would be some charge left in the wire if you were to put capacitors in series with the loop/coil? Thank you! \$\endgroup\$ – Defindun Dec 11 '18 at 20:35
  • \$\begingroup\$ @StefanWyss You have repeated a popular misconception that there must be a voltage difference between two points for there to be a current flowing between those same two points. There does not need to be a voltage drop across a wire for current to flow through it. Consider superconductors for example. An electric field is necessary to accelerate charge carriers. \$\endgroup\$ – Elliot Alderson Dec 11 '18 at 23:07
0
\$\begingroup\$

I view a loop of wire, at low frequencies, as an excellent provider of displacement currents. I don't view the shape of the wire as a major influence (at low frequencies) on the interference generated.

Two parallel wires, radius 1mm and 1meter long and 1meter apart, have capacitance (ignoring the finite length) of

C = Eo * Er * 2 * PI * Length / natural_log[X * sqrt(X^2 -1)]

where

X = Num/Denom, Num is (Center-to-Center-separation)^2 for large separation, Denom is 2 * Radius2 * Radius2 [notice this is dimensionally neutral, thus is a fine argument to feed into a log-function]

Poking in the numbers, you'll see about 4pF between the 2 wires.

You start out with question about general shapes, then move to a specific (loop) shape.

Does this answer give you some insight?

I use this method to provide WORST CASE displacement currents from AC_power_cords that couple into sensitive nodes of signal chains. Or from switching power supply nodes (the drain of the MOSFET, switching 200 volts in 200 nanoSeconds).

\$\endgroup\$
  • \$\begingroup\$ While I agree with all of this, I do not see a relation to the OPs question? \$\endgroup\$ – Stefan Wyss Dec 12 '18 at 5:26
  • \$\begingroup\$ I'm suggesting the Efield may not need to be precisely computed, for some very useful interference predictions, if the system is small compared to the wavelength. \$\endgroup\$ – analogsystemsrf Dec 12 '18 at 15:51
-1
\$\begingroup\$

without a core you won't have a magnetic force to define a charge constants. the magnetization will propagate at the speed of light thus produce the desired effects of the loop which could be a near zero drop of E through the field. thus you can't possibly measure E within a free space loop without several orders of magnitude of currents. lets say dozens to hundred of amps. in that case the conductor itself produces such a resistance that you can measure E from the magnetic field it creates in the conductor itself.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.