If the explanation for the B-field due to a current in a wire that I've heard (length contraction by moving electrons in conductor) is true, would you really need to be moving to feel the "B-field" of this wire? What I mean is, if the electrons within the conductor are moving quickly enough, then they will already be (barely) length-contracted along their path. Since there is no "moving positive-charge", as electrons are the only charges that are moving, wouldn't this length-contraction of electrons immediately come off as an increase in negative-charge density?
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asked Oct 30, 2015 at 17:04
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Assume there are a fixed number of electrons in the circuit. If their number increased in the wire, where would it decrease?
You might be thinking of the electrons in the wire as cars in a train. The cars are coupled to each other. So if they become length-contracted, the total length of the train is shorter and the cars are closer together.
The electrons are not coupled. Even if you apply length contraction to a single electron, that doesn't mean the charges have to move closer together.
answered Oct 30, 2015 at 17:13
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$\begingroup$ Typically this argument is done under the consideration of an infinitely long wire (so an infinite number of charges, positive and negative, are at the model's disposal). Also, if two objects are separated by a finite distance and co-moving, the distance measured between them from another inertial frame will certainly be length-contracted. In that sense, they are coupled (i.e. co-moving with the same drift speed). $\endgroup$ Commented Nov 2, 2015 at 0:01