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In the following circuit there is a power supply applying a voltage(+$V$) to a circuit with resistance ($R$), current($I$) is now flowing in the circuit, and there is a movable part like so: enter image description here

The white part is able to move freely. Now I've introduced a magnetic field($B$) to that movable part:

enter image description here

Due to the Lorentz force($F_L$) that movable part will accelerate: enter image description here

(Ignoring self-inductance) Due to that Lorentz force, there is a change in magnetic flux($\phi$) and by Faraday & Lenz law, there must be a phenomenon that resist's that change. My confusion is with the following, intuitively, there is an induced EMF that oppose the applied voltage(that is causing the motion) like so: enter image description here

However, would that mean... that there is also Eddy currents that resist the change? By creating a drag force like so: enter image description here

Are both phenomenons coexistent in this case to resist the cause of change? Eddy current's and induced EMF? I think that Eddy current's is a form of opposition to the change the induces it, like dropping a magnet in a copper bar, or moving a piece of copper slab into a magnetic field, but when the change is caused like the circuit above... the form of "opposition" would purely be a negative induced EMF.

Also to note, this is quite similar to the basics of an electric motor, which I found having no eddy current losses(besides iron-core related). Only back/counter EMF that resist the change that is caused from motion that is caused by the applied power source.

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  • $\begingroup$ I recommend that you first think about the situation when the white region is a narrow wire - so the eddy currents are negligible. Do you understand the main mechanisms at play in that scenario? It's always a good idea to go from "simplified" to "more complex" to "real life" in small steps. $\endgroup$
    – Floris
    Jul 30, 2015 at 17:01
  • $\begingroup$ @Floris Yes, that's way I neglected the narrow wire because it's easy to analyze and simply conclude that $-\epsilon$ the "opposer" of change. Without considering Eddy currents,however, here I'm focusing about larger plates of copper 5cm wide 50cm long as such(1-3mm thick). In powerful magnetic fields($0.8T$+) Jumping to complex at the moment, before real life(considering friction and many other factors). $\endgroup$
    – Pupil
    Jul 30, 2015 at 22:15
  • $\begingroup$ It might be time to invest in some multi physics simulation tools. $\endgroup$
    – Floris
    Jul 30, 2015 at 22:16
  • $\begingroup$ Well, sure. Any recommendations? Also, could you possibly answer the question if you have the time please? It's been bugging me for that last day or two, my assumption is that there are Eddy currents but much much smaller in effect Vs. induced -$\epsilon$. $\endgroup$
    – Pupil
    Jul 30, 2015 at 22:18

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Attempting an answer:

Eddy currents are induced where there is a change in the field. The way you have drawn the situation, there is no place where the field changes while the white rectangle moves: $\frac{dB}{dt}=0$ everywhere in the conductor. So while there is a current flowing around the loop, there is no eddy current induced (that I can see).

The same is not true if the B field doesn't have such a nice uniform shape / boundary...

I must admit that without doing a proper simulation, I am not sure that my analysis is correct - but right now I can't see anything wrong with it.

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  • $\begingroup$ Maybe this is my confusion: There can't be induced Eddy's if there is already current flowing in the conductor like the diagram I've presented? Or is it due to the change in magnetic field is $0$? So I am correct to assume that the only form of opposition is due to -$\epsilon$? $\endgroup$
    – Pupil
    Jul 31, 2015 at 4:41
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    $\begingroup$ There is no eddy current if the field across the conductor is not changing. So the force in that case is due only to the emf induced current in the loop (since flux changes across the loop). $\endgroup$
    – Floris
    Jul 31, 2015 at 11:45
  • $\begingroup$ Thanks for clearing that out, it's quite interesting... never knew Eddy current's rely on the change in $B$ only. $\endgroup$
    – Pupil
    Jul 31, 2015 at 14:20

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