Harry Veeder wrote: > I don't know. Perhaps you are right. My understanding of the theory > is limited, but please read the introduction here: > > http://www.springerlink.com/content/q6634pp556m08500/fulltext.pdf > > Even Feynman and Maxwell say there is no e-field.
They say nothing about Feynman in the introduction. Later they quote from his description of an idealized wire in the Feynman lectures, which were introductory and targeted toward undergraduates. In the quoted text, Feynman's not considering the exact details of a *resistive* wire carrying a current. Most of the time you can consider \rho, the resistivity of the wire, to be *zero* and in that case there's no E field outside the wire and that is what he's doing here, even though he doesn't explicitly state it. Placing a charge outside the wire will induce a charge on the wire -- but that's not something Feynman's considering, either, in the brief quote they give. This is a total straw man. The E field near a resistive wire is a well known phenomenon and has nothing to do with consideration of the B field as being anything other than a "relativistic effect". Their claim that "some authors *believe* there is no force..." is kind of silly, and they provide no evidence to back it. Some authors, in introductory texts, ignore the E field which drives the charges through the wire, and also ignore the surface charge which would be induced on the wire by the presence of an external charge. Really, so what? This has *nothing* to do with considering the B field as a relativistic effect. Rather, it has to do with casual assumptions regarding ideal wires in simple circuits. A similar assumption is that the potential of two nodes in a circuit connected by a wire is identical. That's false in general, of course, as anyone working with long wires, high currents, or fast edges is well aware, but in many cases it's so close to true that it can be taken as true, and in introductory circuit theory it's usually assumed. They also mangle what Purcell said. They say, > He [Purcell] then considers > two current carrying metallic wires at rest in the frame of the laboratory > and says ( p. 178) , ``In a metal, however, only the positive charges > remain > fixed in the crystal lattice. Two such wires carrying currents in opposite > directions are seen in the lab frame in Fig. 5.23a. The wires being > neutral, Note well: Purcell is working with the question of whether there is an *electrostatic* charge on the wire. To first order, there isn't, and as a result no electrostatic field due to a charge imbalance is present. In particular, as long as resistance is very low, the current in the wire does not result in any RADIAL electric field, and that's what he's concerned with here. > there is no electric force from the opposite wire on the positive ions > which > are stationary in the lab frame.’’ That is, he believes that there > will be no > electric field generated by the stationary current-carrying resistive > wire in > any point outside itself. BS. Their conclusion is a non sequetor. They don't know what Purcell believes. All they know is that in this quote, from his *introductory* E&M book, in which he is NOT TALKING ABOUT THE EFFECTS OF RESISTANCE (despite their inclusion of the word "resistive" in their paraphrase), he asserts that there will be no charge imbalance in the wire due to the presence of the current, and hence no electrostatic E field due to an imbalance in charge carriers in the wire. He is *correct*. A more complete model of a wire doesn't ignore resistance. In saying Purcell *believes* there can't be a field caused by a resistive wire, they are, essentially, asserting that Purcell doesn't know that wires can have resistance or doesn't know what resistance is, which is an outrageous leap from his description in the text. In inserting the word "resistive" in their paraphrase, when Purcell didn't use it in the original quote, they are (apparently intentionally) distorting what he said so they can appear to make a point. As to Maxwell himself, the form of the theory he he initially proposed was incomplete, and at this point his exact statements are of historical interest and not much else. > > Harry > > ----- Original Message ----- From: "Stephen A. Lawrence" > <[email protected]> Date: Tuesday, September 15, 2009 11:46 am Subject: > Re: [Vo]:The Electric Field Outside a Stationary Resistive Wire > Carrying a Constant Current > >> >> Harry Veeder wrote: >>> Maxwell's theory needs the field concept. The theory says and >>> electric force can not be present without an electric field. >>> >>> If we follow Maxwell's theory to the letter, it says there will >> be no >>> electric field outside a current carrying wire. >> I don't know what you're talking about here. If by "Maxwell's >> theory" you mean Maxwell's four equations, as they are normally >> written, and as they are embedded in the model of special >> relativity (which is how thisis normally applied), then what you >> said is simply false. >> >> If there is a current flowing through a resistive wire, then there >> is an E field within the wire directed parallel to the wire. That >> E field is what drives the current, and its value is proportional >> to \rho*I where \rho is the resistivity per unit length of the >> wire. The curl of the E field within the wire and near the surface >> of the wire is zero, since >> >> Del x E = -dB/dt >> >> in rationalized CGS units. >> >> Since the curl is zero, if the field points along the wire just >> withinthe wire, it must also point along the wire just outside the >> wire. Otherwise you'd get a nonzero integral of the E vector >> around a small loop which is partly inside the wire and partly >> outside the wire, whichwould imply the curl was nonzero. >> >> For points near the wire, that field runs parallel to the wire. >> This field is independent of the presence or absence of a charge >> outside thewire. >> >> Arguments straight out of Purcell, based directly on Maxwell's >> equationsand the Lorentz force law, lead to the conclusion that a >> point charge located close to the wire will also induce a local >> charge on the wire, which will result in a local field which is >> perpendicular to the surfaceof the wire. This field vanishes if we >> remove the external charge. But did you perhaps mean something else >> by "Maxwell's theory? >> >> (Incidentally I said "As embedded in the model of SR" because >> without that extra bit of icing you have no way of transforming the >> equations from one frame of reference to another, and no answer to >> the question of what happens when moving a uniform velocity.) >> >> >>> Consequently, the theory leads one to expect an electric force is >>> absent as well. >>> >>> Weber's theory is not built on the field concept, so this curious >>> expectation does not arise. >>> >>> My analysis is based on reading of this preface to the book >> suggested> by Taylor J. Smith. >>> http://www.ifi.unicamp.br/~assis/Preface-Webers-Electrodynamics.pdf >>> >>> >>> >>> Harry >>> >>> ----- Original Message ----- From: "Stephen A. Lawrence" >>> <[email protected]> Date: Monday, September 14, 2009 6:18 pm >>> Subject: Re: [Vo]:The Electric Field Outside a Stationary >>> Resistive Wire Carrying a Constant Current >>> >>>> Harry Veeder wrote: >>>>> fyi Harry >>>>> >>>>> Foundations of Physics © Plenum Publishing Corporation 1999 >>>>> 10.1023/A:1018874523513 >>>>> >>>>> The Electric Field Outside a Stationary Resistive Wire >>>>> Carrying a Constant Current >>>>> >>>>> A. K. T. Assis, W. A. Rodrigues Jr. and A. J. Mania >>>>> >>>>> Abstract We present the opinion of some authors who believe >>>> there is >>>>> no force between a stationary charge and a stationary >>>>> resistive wire carrying a constant current. >>>> That's stated a lot, but it's just sloppiness. Anyone who >>>> knows electronics realizes it's not really true. >>>> >>>> A good conductor carrying small current has *nearly* zero >>>> voltage drop along any small length, and calling the drop >>>> "zero" is usually "good enough". But really the voltage drop >>>> along any segment is equal to I*Rwhere R is the resistance of >>>> that segment. >>>> >>>> When the voltage drop along a (resistive) wire is nonzero, then >>>> *any* path which leads from a higher voltage point on the wire >>>> >> to a >>>> lower voltage point on the same wire must traverse the same >>>> exact potential change, which means that there must be an >>>> electric field *outside* the wire, running parallel to the >>>> wire. >>>> >>>> This is well known but, as I said, usually neglected, because >>>> it's usually too small to matter in real-world problems. >>>> >>>> The fact that there's an "image charge" induced in the wire as >>>> well, which consequently must be having its effect on the >>>> charge sitting outside the wire, is certainly the case and >>>> could even be called "obvious", but it's not something I ever >>>> thought of until I saw it mentioned in the abstract. :-) >>>> >>>> >>>>> We show that this force is different from zero and present >>>>> its main components: the force due to the charges induced in >>>>> the wire by the test charge and a force proportional to the >>>>> current in the resistive wire. We also discuss briefly a >>>>> component of the force proportional to the square of the >>>>> current which should exist according to some models and >>>>> another component due to the acceleration of the conduction >>>>> electrons in a curved wire carrying a dc current (centripetal >>>>> acceleration). Finally, we analyze experiments showing the >>>>> existence of the >>>> electric> field proportional to the current in resistive wires. >>>> >>>>> complete paper available here: >>>>> http://www.springerlink.com/content/q6634pp556m08500/fulltext.html >>>>> >>>>> >>>>> >>>>> >>> >> > >
