# Electric and magnetic fields

1. Nov 28, 2006

### questionmark

hi,

i'm puzzled.

if an observer stands next to a length of wire with a dc current in it, he will detect a circular magnetic field around the wire due to the movement of the electrons, but no electric field due to the cancellation of the electrons' and protons' electric fields.

what happens if the observer now hitches a ride on one of the moving electrons in the wire (cet par), what does he measure now and why?

2. Nov 28, 2006

### Meir Achuz

In practice, it would be a very bumpy ride, because the electrons actually bounce between molecules in the wire, resulting in a "drift velocity" that is much slower than their velocity between collisions.
If she moves outside the wire at this drift velocity, a Lorentz transformation would give the fields she sees in her rest system.
B would change to \gamma B, and there would be an E field given by
E=\gamma(vXB).

3. Nov 28, 2006

### masudr

Hang on: she would see positive charges (the ions) moving in the opposite direction at the drift velocity and so still see a magnetic field, but of the opposite sign.

Meir Achuz's analysis only holds for beams of electrons, where moving with them truly means seeing a stationary E-field.

4. Nov 28, 2006

### pervect

Staff Emeritus
Meir's analysis is completely general - the electromagnetic field will transform according to the standard Faraday tensor transformation laws: see

http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html

I believe the analysis is correct, too, though it's possible we've both made an error.

The transformation laws on the webpage above are given for a boost in the z direction.

If we take a point at some distance x=L, y=0 from the wire, the electromagnetic field in the rest frame of the wire will be

Ex = Ey = Ez = 0 = Bx = Bz = 0, By=B

(depending on the direction of current flow in the wire for the sign of B)

The transformation laws say that the only field components in the boosted frame are

By' = gamma*B, Ex' = -gamma*v*B

There have been a lot of threads on how to explain this, one of the better explanations it that the Lorentz transformation squashes the field of each moving electron so that it is not round, but ellipsoidal. Various other related effects happen as well, the distance between charges gets Lorentz contracted.

If you look, you'll see some past threads on this issue - one of them is

and there have been many others as well.

5. Nov 28, 2006

### masudr

I'm not denying that. All I'm saying is that if you transform to an inertial frame so that the electrons/charge carriers are at rest, there will be a net motion of positive charges in the opposite direction.

If the electrons in a wire did not come from atoms, then indeed a frame moving with the electrons will not perceive any magnetic fields. But most wires are electrically neutral (i.e. every negative charge carrier has a corresponding positive charge), so if we transform to an inertial frame where the electrons are at rest, we will see that there are ions moving in the opposite direction, so we will still get a magnetic field.

I know how that the EM tensor is indeed a tensor, and transforms under a change of inertial co-ordinates with 2 copies of the Lorentz transformation. This is not what is at debate. Instead I am talking about something far more basic -- essentially that a wire is usually neutral, and the consequences of that.

6. Nov 28, 2006

### marcusl

I'm not following your point. Did someone say there wouldn't be a magnetic field if a test charge moved parallel to and at the same speed as the charge carriers?

7. Nov 28, 2006

### cesiumfrog

..but isn't the drift velocity always incredibly small?

8. Nov 28, 2006

### marcusl

Sure. Isn't that why the magnetic force between two current carrying wires is incredibly small compared to that between two electrically charged wires? In fact it's down by order (v_drift / c)^2.

9. Nov 30, 2006

### swimmingtoday

Your method is corect, but you should have tried to explain the discrepancy between your result and Meir's. Two seemingly correct ways of doing the problem gives different results, yet you are not troubled by this. (Se my next post)

10. Nov 30, 2006

### swimmingtoday

11. Nov 30, 2006

### marcusl

No, that's not quite right. The magnetic field must be in the same direction regardless of whether the test charge is moving more slowly or more quickly than the electron drift velocity. It's not the apparent direction of ions that matters; it's the relative charge densities of electrons and ions after Lorentz contractions are applied to each.

I think you got this backwards. There's no E in the lab frame, if there were then free electrons in the circuit would be attracted until the excess charge were neutralized. This agrees with observation, too (your hairs don't stand up when you stand next to house wiring, nor next to a battery cable carrying 200A as you crank you car, nor next to a superconducting magnet carrying thousands of amps). The neutrality in the lab frame is precisely why you see a field once you move.

12. Nov 30, 2006

### pervect

Staff Emeritus
I'm not quite sure I'm following your point, unfortunately.

Note that the total net charge on the wire is a physical quantity that's Lorentz invariant. The total charge on the entire wire must be specified for the problem to be analyzed and for the electric and magnetic fields to be calculated in the rest frame. It is usually assumed that the total charge on the wire is zero in the rest frame of the wire, but it's certainly possible to have a charged loop of wire. However, the initial poster explicitly said that the electric field in his sample problem was zero.

By Gauss's law, no electric field implies no net charge on the wire, by drawing a Gauss surface around the entire loop of wire.

So we know from the problem statement that the wire is on the average, neutral. We can think of this neutral electrical wire as having +N positive charges and -N negative charges, for a total charge of zero.

These charges re-destribute under a Lorentz boost due to the relativity of simultaneity. This allows charges to redistribute like this for a sample square loop of wire (the boost in this diagram is in the left-right direction, one can assume the current flow is a clockwise loop).

Code (Text):

++++++++
x       x
x       x
x       x
----------

There can and will be electric fields after the boost, but this is and must be compatible with the total charge of the wire being constant. There will still be N positive charges and N negative charges in the loop above, but their distribution will be different after a boost.

13. Nov 30, 2006

### swimmingtoday

<<I think you got this backwards. There's no E in the lab frame, if there were then free electrons in the circuit would be attracted until the excess charge were neutralized.>>

The E field is perpendicular to the wire, so it would not attract charges from the battery.

:<<This agrees with observation, too (your hairs don't stand up when you stand next to house wiring,>>

That is probably due to the quantitative weakness of the effect.

If I walk past a magnet there will be an effective eelectric field (for example viewed in the inertial frame where I am at rest), but my hairs have never stood up when walking past a magnet--the effect is too quantitatively small.

<<The neutrality in the lab frame >>

Consider a wire with no current. The density of the positive charges is equal to the density of the negative ones. So the densities of the negative charges must be different from the density of the positive charges when the negative charges start moving. For the densities to be the same in the wire with current, you must either claim that they are not the same in the wire with no current, or you must reject the Lorentz contraction--neither of which I think you would want to do.

14. Nov 30, 2006

### Staff: Mentor

Consider a loop of wire that initially has no current. Current carrying wires must ultimately be part of a complete circuit, at least in steady-state current situations. Induce a current in the loop by using a changing magnetic field. If the number density of the moving electrons increases because of Lorentz contraction, we need to add more electrons in order to maintain that increased density all around the wire. Where do those electrons come from?

This seems to be related to the sometimes-posed question that starts with a series of stationary objects laid out along a line, not connected to each other. Accelerate the objects up to a relativistic speed. Does the distance between them shrink because of Lorentz contraction, or does it stay constant? The answer turns out to depend on the details of how the objects are accelerated, because of the relativity of simultaneity.

15. Dec 1, 2006

### swimmingtoday

<<Consider a loop of wire that initially has no current. Current carrying wires must ultimately be part of a complete circuit, at least in steady-state current situations. Induce a current in the loop by using a changing magnetic field. If the number density of the moving electrons increases because of Lorentz contraction, we need to add more electrons in order to maintain that increased density all around the wire. Where do those electrons come from?>>

Consider a loop spinning around an axis perpendicular to the plane of the loop going thru the center. The ratio of the circumference to the radius is no longer 2 pi, because of Lorentz contractions of the pieces of the moving circumference. Now back to the scenario you propoesed. Treat the wire as two loops, a loop of positive charge and a loop of negative charge. When there is no current the loops have equal circumferences. When there is current, the negative loop is spinning and has a decreased circumference. So the two loops occupy the same space, having the same radius, yet one has a smaller circumference. Freaky, eh?

So the negative loop has a greater magnitude of charge density because it has the same magnitude of charge but a shorter length

Now we argue that the charge density is the thing determining the electric field. This is a bit uncomfortable for me, and for regions far from the loop, it actually might not even be true but it is true "close" to the loop, which is the region the matters in the original problem of making the Lorentz transformation to a frame where the electrons are at rest. So we see that, at least in the relevant region (close to the wire), the E field from the positive charges is not large enough to cancel out the field from the negative charges.

If you are uncimfortable with this, consider that if what I said does not work out we have a REAL unresolvable problem, the original paradox I pointed out several posts ago--that making a Lorentz transformation of the charges in a wire so as to have the electrons at rest gives a different magnetic field than one would get by performing that Lorentz transafornation directly on the electric and magnetic fields.

16. Dec 1, 2006

### ZapperZ

Staff Emeritus
swimmingtoday: Could you please use the QUOTE function that is available in each post (see lower right-hand corner of each post)? This will make your replies a lot easier and clearer to read, especially when you make multiple replies to a number of people.

Zz.

17. Dec 1, 2006

### swimmingtoday

I'm quite ignorant about computer stuff, so I'll tell you what I do, and you can tell me specifically how to do it better.

I hit the quote button whenever I respond to a post, so the person I respond to will get an email notifying him. Then when I want to post an actual quote, I usually do not want to quote the *whole* post I am referring to, but rather specific excerpts. So I just delete out everything but the specific excerpts.

If I do not delete out certain things, like the word "QUOTE" will that make it work the way you want? Do I need to write out "QUOTE" both before and after something I want to quote? As I said, I know almost nothing about computer stuff.

18. Dec 1, 2006

### ZapperZ

Staff Emeritus
You seem to have done it corectly here. However, I want to correct something. The person you are replying to will NOT get an e-mail notification that you have replied, unless he/she chosed to have e-mail notification (which I believe is not the default setting).

If you simply want to make specific quotes of a part of the post, then make sure you bookend the phrase with a [...quote] and [.../quote] (no periods). If you want to make it clear it was a quote from someone, then start with [..quote=ZapperZ] and end it with [../quote] as usual. This will put that member's nickname in the quotation.

Is that all clear now?

Zz.

19. Dec 1, 2006

### marcusl

Untrue for the case you consider of a localized region where the charges don't cancel.

20. Dec 1, 2006

### pervect

Staff Emeritus
The electric field inside a perfect conductor is zero pretty much by defintion. We'd have to jump through some hoops to handle the case of a real wire with an ohmic drop - I'm not really in the mood to do that, though, hopefully you aren't either.

Which result is that? COuld you quote the post in question?

There is no net electric field in the rest frame of an uncharged wire when the wire is a perfect conductor. The fact that there is no radial electric field is obvious from Gauss's law. See for instance:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c2

where the Gauss surface is a cylinder enclosing the entire wire loop. When the net enclosed charge is zero, the radial electric field around the wire is zero. This relies on symmetry - for a symmetrical circualr loop, the radial electric field must be independent of the angle theta.

So if we have an uncharged wire with no current flowing through it, then we complete the circuit, the wire will still be uncharged, and there will be no net radial electric field after we complete the circuit.