- #1
nonequilibrium
- 1,439
- 2
Hello,
First of all, I have no objections against Faraday's Law in differential form, i.e.
[tex]\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}.[/tex]
But in integral form, I usually encounter it in the form
[tex]\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.[/tex]
But surely this is wrong? (after all when B is constant, but the loop-surface is changing, there is no induced E-field; then it's the B-field delivering the emf)
The correct expression should be
[tex]\oint ( \vec E + \vec v \times \vec B ) \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}. (*)[/tex]
Does everyone agree?
(*) for clarification: with the "v" in the LHS I mean the netto speed of the particles in each point (of the loop) at the time of integration
ADDENDUM: Okay, another point of view is "but I'm looking at the E-field as viewed by the electrons moving along with the possibly changing loop, and then there is purely an induced E-field"; I suppose that could work (honestly, I'm not even sure if that is allowed, because what if the loop is accelerating, then you're using the laws of Newton in an accelerating frame?), but anyhow it sounds like a weird implicit thing to do, and it messes up the idea that the four laws of Maxwell describe the E and B-field in your own reference frame.
NB: I wasn't sure if I should make the partial derivative a total derivative when going to the integral form, but I suppose it's irrelevant
First of all, I have no objections against Faraday's Law in differential form, i.e.
[tex]\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}.[/tex]
But in integral form, I usually encounter it in the form
[tex]\oint \vec E \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}.[/tex]
But surely this is wrong? (after all when B is constant, but the loop-surface is changing, there is no induced E-field; then it's the B-field delivering the emf)
The correct expression should be
[tex]\oint ( \vec E + \vec v \times \vec B ) \cdot \mathrm d \vec l = - \frac{\partial \Phi_B}{\partial t}. (*)[/tex]
Does everyone agree?
(*) for clarification: with the "v" in the LHS I mean the netto speed of the particles in each point (of the loop) at the time of integration
ADDENDUM: Okay, another point of view is "but I'm looking at the E-field as viewed by the electrons moving along with the possibly changing loop, and then there is purely an induced E-field"; I suppose that could work (honestly, I'm not even sure if that is allowed, because what if the loop is accelerating, then you're using the laws of Newton in an accelerating frame?), but anyhow it sounds like a weird implicit thing to do, and it messes up the idea that the four laws of Maxwell describe the E and B-field in your own reference frame.
NB: I wasn't sure if I should make the partial derivative a total derivative when going to the integral form, but I suppose it's irrelevant
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