# Does constant magnetic field around a wire induce an electric field?

1. Dec 22, 2011

Hello friends, please help me understand this tricky aspect of Ampere's and Faraday's Laws as they relate to the static, circular magnetic field enclosing a constant current carrying wire.

(image courtesy of Wikipedia)

The existence of this magnetic field is given by the I or J terms of Ampere's equation:

$\oint\textbf{B}\bullet d\textbf{l} = \mu I_{encl} + \mu\epsilon\frac{d\phi_{E}}{dt}$
$\nabla$ x B = μJ + $\mu\epsilon$$\frac{d\textbf{E}}{dt}$

Before this current I is turned on, there is no current and thus no B field.

At the instant current I is turned on, B appears very close to the wire. B does not appear at far away points instantaneously. Rather it propagates out at the speed of light c.

My question concerns this propagation. Let's consider a point very close to the wire. Before current I was turned on, there was no magnetic field, or we could say B = O T. After current I is turned on, there is some magnetic field, say B = B1 with direction given by the right hand rule. This constitutes a change in magnetic field in time: $\frac{d\textbf{B}}{dt} \neq 0$

Then this must generate an electric field by Faraday's law:
$\nabla$ x E = -$\frac{d\textbf{B}}{dt}$

I believe this induced E would be located very near the wire and B = B1 which induced it. This new electric field arising means $\frac{d\textbf{E}}{dt} \neq 0$ which then feeds back into Ampere's law, and now we have an outwardly propagating EM wave.

But we never talk about an electric field when referring to Ampere's law in the constant current situation as shown in the picture. This leads me to believe my understanding of the propagation mechanism for the static magnetic field is mistaken.

Yet this same argument, as far as I can see, is employed by Feynman in his Lectures Vol 2 Chapter 18.4, A Travelling Wave. He uses an infinitely large sheet of constant current moving in one direction, and it results in outwardly propagating E and B.

Please help me understand how this situation is different from our normal Oersted/Ampere law where we only look at the static magnetic field. Thanks very much.

Last edited: Dec 23, 2011
2. Dec 23, 2011

I found the answer to my question in Wikipedia and Hayt, Engineering Electromagnetics p.313

Per Wikipedia:

Ampère's circuital law is now known to be a correct law of physics in a magnetostatic situation: The system is static except possibly for continuous steady currents within closed loops. In all other cases the law is incorrect unless Maxwell's correction is included.

The scenario I described, where the wire starts with no current and then suddenly a current I is turned on, causing fields to propagate as EM waves, is not a magnetostatic situation. Changing currents (ie accelerating charges) are not allowed in magnetostatics. Whereas the scenario in the picture is magnetostatic because the currents are constant, and we analyze it without asking how they originally got that way.

But I'm still not sure how such a static magnetic field without electric field could ever exist in reality. For it to be created in the first place, it would always cause an electric field to also be created. And this new E field would have to be statically maintained, since if it disappeared, that would be a dynamic situation again and create new magnetic fields.

Last edited: Dec 23, 2011
3. Dec 23, 2011

### Per Oni

Can you accept that in a conductor there are free line charges even when no current flows? If so, then you can find the resulting E-fields round a conductor using Gauss’ law. Since both kinds of charges (not all free) are present there’s no net E-field. When a current flows one or sometimes both line charges are moving. The result is that these same fields are moving. The non-relativistic equation shows that B=vxE/c^2.

4. Dec 23, 2011

Yes I see there is zero net charge in the conducting wire due to cancellation of fixed protons and free electrons (whether or not current is moving). As you suggested let's use a cylindrical Gauss surface around the wire:

$\oint \vec{E}\cdot\vec{n}dA = \frac{Q_{encl}}{\epsilon}$ = 0

Certainly we don't expect any $\vec{E}$ sticking radially out from the wire.

(Actually in researching this, I found this topic in which some physicists have experimentally found electric field outside a constant current wire, contrary to what everyone expected. But I think that advanced topic isn't what we're discussing here haha.)

But let's please look at the magnetic field solution you provided, B=vxE/c^2.
I don't know how to obtain that, but our text Hayt tells us to use Ampere's law and we get:

$\vec{B} = \frac{\mu I_{encl}}{2\pi\rho}$

Certainly this field arises, but in creating it from $\vec{B}=0$ now we have $\frac{d\vec{B}}{dt}\neq 0$ and an electric field should be created in the vicinity of the magnetic field (not necessarily in the wire). How does our calculated magnetic field arise without this effect?

5. Dec 23, 2011

### Per Oni

It doesn’t.

The current in the conductor goes from I= 0 to I= I max in time t. During this time also B will be build up to its max value. So as you say time changing B should generate an E-field. This so, and when you hold a parallel conductor next to the original one you should be able to measure a resulting voltage.
That is the principle of the core less air transformer.

But to come back to an E-field sticking radially out from the conductor, that is just the way I visualise it. One field sticking out and another one sticking in, resulting in a zero net field.

6. Dec 23, 2011

That is a good tip to represent 0 net field by two field lines in opposite directions. In studying this topic I've seen several people (including Feynman) use that technique, which I will try to adopt.

So now we have B generating E and vice versa, propagating out from the wire. In the constant current magnetostatic case shown here:

Is the generated electric field simply omitted to emphasize the Ampere relationship?
$\oint \vec{B}\cdot d\vec{l} = \mu I_{encl}$

The electric field can't have disappeared because if it did, that would be $\frac{d\vec{E}}{dt} \neq 0$ which would cause changes to $\vec{B}$, which we know to be static.

This is, as far as I can see, the same argument used by Feynman in section 18.4 here. A constant current causes both E and B to propagate into space at speed c.

7. Dec 24, 2011

### Per Oni

The way I see it is that once I max has been reached, there’s no changing B field anymore. There’s therefore also no secondary induced E-field anymore. But the radial E-field sticking out (and/or in) is still travelling with the same drift velocity as the charge carriers responsible for the current I.

So I disagree that a constant current causes both E and B to propagate into space at speed c. Which formula or part in 18.4 are you referring to?

8. Dec 24, 2011

Definitely I expect you are right, not only because your argument makes sense, but also because this is a cornerstone of physics for a long time haha.

But please help me figure out what I'm misinterpreting in Feynman's article. He has two infinitely large sheets of charges superposed on top of each other in the YZ plane, one sheet positive, one negative. He says this is to prevent complicated electrostatic effects. It's like our copper wire in the cancellation of net charge. Then he instantaneously accelerates one sheet to some constant velocity in the y direction. Now we have a constant sheet current. This results in a wavefront of E and B propagating outward. (E and B are constant in magnitude because the current sheet J is infinite.)

The current sheet J is viewed on its edge as it moves upward along the y axis. The wave front of constant E and B has partially filled the little rectangle, which he uses to take a line integral.

http://img824.imageshack.us/img824/8859/sidekb.png [Broken]

Similar situation with J viewed coming out of the page:

http://img545.imageshack.us/img545/5976/topsz.png [Broken]

Feynman explains:

http://img443.imageshack.us/img443/3284/setupvb.png [Broken]

Last edited by a moderator: May 5, 2017
9. Dec 24, 2011

### Per Oni

At this moment in time I completely agree with you. I cannot see how Feynmann can draw a constant E-field in the –ve y-direction next to the conductor even though the wave front has passed.

However my head is feeling a bit funny at the moment and that could be the reason. Hopefully some other PF’ers will come to our help. Just like you I do want to get this sorted out.

Happy X-mas all.

10. Dec 24, 2011

Merry Christmas Per Oni and thank you for the help. Yes please other forum members, give us your perspectives.

11. Dec 27, 2011

### Per Oni

Now with a clear head, I can see where the text goes wrong.

Halfway page 18-6:
This is wrong since E is only present during the time that B changes. One of the problems with 18-4 is that the current is almost instantly up to the full value. There’s no dt to speak off, so there’s no way to show where and when E is present.

There’s a Feynman lectures website where mistakes can be reported. That’s definitely worth doing.

12. Dec 27, 2011

Thank you for continuing to look into this question! I checked out the Feynman Lectures Errata website you suggested. Our situation is not listed there. (They mention that only 2 errors in physics have ever been found, while there have been dozens of typos and such.) Frankly I don't feel comfortable criticizing Feynman haha.

For sure your explanation is right for a wire. Not only does it make sense, but also it took experimentalists until recently to detect even a tiny electric field outside a wire (and the theorized mechanism for that strange finding is totally different from what we're talking about.)

But do you think maybe Feynman's infinite sheet scenario has some special characteristic that makes it different from a wire (other than just shape obviously)? He goes out of his way to emphasize the sheet is infinite.

In the setup of 18.4 he gives his reasoning why the infinite sheet current causes B to be uniform out to infinity:

http://img577.imageshack.us/img577/1540/vol2ch16inducedcurrents.png [Broken]

Then he notes that B coming into existence would cause a brief dB/dt, and hence E comes into existence. Then perhaps by the same reasoning, E must also be uniform to infinity because the sheet is infinite. In contrast, this uniformity definitely doesn't apply for B or E in the case of a wire.

Perhaps this bridges the gap?

Last edited by a moderator: May 5, 2017
13. Dec 28, 2011

### Per Oni

Yeah, but you def have a case. This will enable you to remove that “lowly” prefix in your name and replace it with *+.
If you think that, then let’s make a dc transformer.

Duplicate the infinite sheet approximately with a big rectangular coil. Send a dc current through the coil and wait till it reaches its max value. Have a big secondary test loop with one leg v close to this coil and its return leg v far out. According to RF, the leg close in “sees” a strong permanent E-field, but the return leg sees a smaller E-field because we only have an approximation of an infinite sheet. Hence a dc current will flow. Or not. Don’t waste your time. It will not.

Last edited by a moderator: May 5, 2017
14. Dec 28, 2011

Oh wow, it seems we are not the first to consider this problem on physicsforum.com:

Poster djy came up with a derivation to support Feynman. I haven't studied magnetic vector potential A, but I will have to now! I edited it here only to make the latex show up properly:

15. Dec 28, 2011

### Per Oni

From Born2bwire
I couldn’t put it better.

Infinite energy, more energy than in all multi verses combined. Hoe silly of us not to think of that.
What is RF trying to teach?