What I read in several places, tells me that, the fact Coulomb's Law follows inverse-square law and gives a force which is radial, implies that a static electric field must be conservative.(In short, Coulomb's law is conservative!) But the electric field in presence of a time-changing magnetic flux (either time changing fields or a change of area of the loop, Motional EMF) is non-conservative, since closed-loop line integral of electric field is no longer zero but is $-\frac{d\phi}{dt}$ where $\phi$ is the magnetic flux through the surface bound by the loop along which the line integral is carried out. I have the following question:-

Consider a situation for the sake of presenting my argument, where there are only steady currents and no accelerated charges, and the changing flux is due to a wire sliding over another U-shaped wire and thus changing the area bound by the loop formed by the U-shaped end and the moving wire.enter image description here

In this case, all the components and the charges (metallic free electrons and the positive kernels) are unaccelerated (i.e. on an average) and hence, there should be no radiation of electromagnetic waves. Therefore, whatever the electric field produced, which is responsible for the existence of current in the loop, must be produced by a particular non-accelerated charge distribution in the apparatus. But, whatever the field produced by any complicated arrangement of charges, through superposition must follow the inverse-square and the radial dependence of Coulomb's law and must be conservative, since all the superposed fields are conservative in themselves as the individual charges follow the conservative coulomb's law.

So, how can a non-conservative configuration of the field, result from the superposition of fields produced by an arrangement of individual charges which follow a law (Coulomb's) which is conservative , assuming the absence of electromagnetic waves?


3 Answers 3


First, consider the positive and negative charges in your moving wire. Since they are moving in a (obviously non-conservative) magnetic field, they experience Lorentz's force $q \ \mathbf{v} \times \mathbf{B}$ which is, in your picture, upwards for positive (and downwards for negative) charges. So they will be accelerated in exactly the same way (for whatever movement your wire gets) as if they were experiencing the electric field that you could calculate using the flux integral variation.

On the other hand, by changing to a moving reference frame, you transform any magnetic field into an electric field - and vice versa (Lorentz transformation). So, in a frame moving with the wire you see the magnetic field as a non-conservative electric field, and this E field accelerates your charges. That's what creates the current in your circuit.

Of course, after a short transient phase where your charges accelerate, you get (because of collisions) a constant current - that's basic Ohm's law here.

And the important point is, whatever your point of view, you will always find the exact same motion for the charges.

Now, neither the magnetic field nor the electric field that appears in the moving frame are conservative (the latter does not appear from Coulomb's law, which in this case states $\nabla \cdot E=0$, but from induction)

  • $\begingroup$ A constant current in this case is not an example of Ohm's law. $\endgroup$
    – BMS
    Sep 30, 2013 at 15:56
  • $\begingroup$ So once the constant current is established, accelerations cease to exist in the stationary frame? how can we then justify non-conservative field by charges obeying a law which predicts a conservative field under non-accelerated conditions? $\endgroup$ Sep 30, 2013 at 16:24
  • $\begingroup$ @BMS : In this case, it actually is an example of the Ohm's law, stated as the current is proportionnal to the voltage between two points. You just have to extend the "voltage difference" definition to be the integral of electric field along the wire. Of course there is obviously no correct definition of voltage here as the electric field is inductive, therefore is not the gradient of a scalar field ; however this is the case for most generators. Indeed the basic mechanism (electric field acceleration compensating collisions) is at work here. $\endgroup$
    – Nicolas
    Oct 1, 2013 at 11:40
  • $\begingroup$ @SatwikPasani More precisely, all charges get permanently accelerated, but collisions within the conductive material slow them down (and heat said material) so in average their velocity is constant. That's quite similar to the fall of a body in a fluid, where collisions with air molecules (called viscosity at a large scale) compensate a constant gravity force, leading to a constant fall velocity. I edited the answer with respect to the conservative/non conservative nature of the electric field. $\endgroup$
    – Nicolas
    Oct 1, 2013 at 11:47
  • $\begingroup$ @Nicolas but then this acceleration which averages to a constant drift velocity is also present in normal cases where the field is conservative, as in a current in a simple DC circuit. Why then here the field is non-conservative? $\endgroup$ Oct 17, 2013 at 13:30

The electrons and nuclei in the moving bar are definitely accelerating - you move them through the magnetic field together and they experience opposite deflection forces.

If the bar were not connected to other conductors, the deflection just charge up the opposite ends of the bar as a capacitor and the process would terminate. But it is connected to other conductors, so the motion induces a current as long as the motion continued.

This charge acceleration produces electromagnetic waves. This situation is not as static as you said it is. Once you realize that, the question disappears.

  • $\begingroup$ How then do we calculate the various parameters($E_{max},\omega, \vec k$) of the electromagnetic waves produced in this case and calculate the energy dissipated? $\endgroup$ Oct 17, 2013 at 13:31
  • $\begingroup$ Well, if the motion of the bar is constant velocity, I don't think any wave would be produced. Of course, during the initial (and final) acceleration that's a different matter. $\endgroup$
    – Nicolas
    Oct 21, 2013 at 12:53

in the problem that you describe the electric field is indeed conservative as you suggest.

We have here a steady-state current, which means that $\text{div} \;\mathbf J = 0$ and therefore, according to the continuity equation $\frac{\partial \rho}{\partial t}+ \mathbf {\nabla} \cdot \mathbf{j}= \sigma$, the charge density is constant in time; this is an electrostatic situation and the integral of the electric field over a closed loop is zero. the Motional EMF comes from the magnetic part of the Lorentz force to the closed-loop integral. see D.J. Grifiths "Introduction to Electrodynamics" Ch. 7 and in particular 7.1.3


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