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Let's say I have a conducting rod passing through a constant magnetic field. I get positive charges in the rod moving with velocity will experience a force (eg. upwards) and move up to the top of the rod and hence the potential difference between the top and bottom of the rod will induce an emf.

The thing is, Faraday's law also states that the induced emf is proportional to the rate of change in magnetic flux linkage. Here's what I don't get. Is flux linkage a property inherent in closed loops or can be used for rods as well? I know that rods technically do not have a 'number of turns' but they do have an area.

If so, then there will be magnetic flux through the rod, but as it moves through the field, the magnetic flux will remain the same, so there will be no induced emf in the rod.

Would it be correct to say that there are 2 sources of induced emf in the rod: one from the separation of positive and negative charges and another one from induced emf via Faraday's law (which is doing nothing).

Does that mean that when a rod first enters a magnetic field, the magnetic flux through the rod will increase and there will be an emf induced? Would it mean there will be 2 contributing emfs in the rod, one from the separation of charges and one from the increase in flux linkage by Faraday's Law?

I can't find answers for this online. For a bit more about my knowledge level on this subject, I do not know anything about Maxwell's equations and what they might imply.

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  • $\begingroup$ "Would it be correct to say that there are 2 sources of induced emf in the rod: one from the separation of positive and negative charges and another one from induced emf via Faraday's law" It wouldn't be correct to say this. Please see my answer. $\endgroup$ – Philip Wood May 30 '19 at 11:19
  • $\begingroup$ > "hence the potential difference between the top and bottom of the rod will induce an emf" No, when we talk about emf in this context, we talk about motional emf, which is due to conductor moving in magnetic field. This emf has nothing to do with separation of charges that is a later equilibrium result of this emf. The electric field due to separated charges then opposes the motional emf. $\endgroup$ – Ján Lalinský May 30 '19 at 14:06
  • $\begingroup$ @JánLalinský Hold on, I don't get this. Isn't conductor moving in magnetic field --> separation of charges --> emf? Are you saying that there is some emf just due to the conductor moving in the field, before the separation of charges even begins? I have not heard of this before. Could you point me to some resources? My understanding of motional emf is due to the separation of charges. $\endgroup$ – Yip Jung Hon May 31 '19 at 0:26
  • $\begingroup$ " Are you saying that there is some emf just due to the conductor moving in the field, before the separation of charges even begins?" I know this question was addressed to JánLalinský, but I'm sure he wouldn't mind me stepping in and saying 'YES'. Do read the first part of my answer again; it explains how the emf arises! $\endgroup$ – Philip Wood May 31 '19 at 10:03
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    $\begingroup$ Philip is right, the answer is yes. Motional emf is due to motion of conductor in magnetic field, not due to separation of charges. This emf pushes charges to the end of the rod, which results in separation of charges. This separation of charges then acts against the emf. If the rod moves with constant velocity in uniform magnetic field, eventually the Coulomb electric field will cancel out effect of the motional emf on the distribution of the charges in the rod, so this distribution will become constant in time, even though the emf alone would act to increase it. $\endgroup$ – Ján Lalinský May 31 '19 at 20:45
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The emf in a moving wire arises from the magnetic Lorentz forces acting on the free electrons, as they are carried along by the wire... $$\vec F = (-e)\ \vec w \times \vec B$$ Here, $\vec w$ is the velocity of the free electron. This is the vector sum of the wire's velocity, $\vec v,$ and the drift velocity, $\vec {v_{dr}},$ of the electron through the wire. So $$\vec F = (-e)\ \vec v \times \vec B\ +\ (-e)\ \vec {v_{dr}} \times \vec B$$ The second term on the right acts at right angles to the wire and implies that we have to do work as we push the wire. The first term on the right gives rise to the emf. If the wire has length $\ell$, and is being moved in a direction at right angles to itself, with $\vec B$ at right angles to plane containing the wire and $\vec B$, then the emf is $$\mathscr E =\frac{\text{work done on charge}}{\text{charge}}=\frac{-evB \ell}{-e}=vB \ell$$ Now let's deal with the flux approach. In my opinion, for a moving conductor this approach is not as fundamental as the Lorentz force approach just given.

For the flux approach we need to include the moving conductor in a circuit. But it's important that the rest of the circuit isn't moving with the conductor, or emf's in the conductor and the rest of the circuit will cancel out. This is why elementary textbooks have the conductor sliding on fixed rails. The circuit might be completed by a fixed voltmeter connected between the rails. As the conductor slides, the area of the circuit changes, so the rate of change of flux through the circuit is $$\frac{d \Phi}{dt}=\frac{\ell v dt\ B}{dt}=\ell v\ B$$

So you see that the flux approach leads to the same emf as the Lorentz force approach. Why use the flux approach? (a) Because it automatically sums the emfs in any parts of the circuit that are moving, (b) It works both for a moving conductor (provided it is thought of as part of a circuit) and for changing flux density in a fixed circuit (the province of the Faraday-Maxwell law): rather neat.

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  • $\begingroup$ What happens if there is no fixed rails though? Will there be any emf induced by the change in magnetic flux? $\endgroup$ – Yip Jung Hon May 30 '19 at 8:48
  • $\begingroup$ If the rest of the circuit is missing, you have to $imagine$ it, in order to use the flux approach! But there's still an emf, as the Lorentz force approach shows, though we have to alter that argument if the conductor isn't part of a circuit. The force will drive free electrons towards one end of the wire leaving the other end depleted. Electrons will stop moving when the electric filed force on them balances the magnetic Lorentz force, so$$(-e)\frac{V}{\ell}=(-e)vB.$$ $\endgroup$ – Philip Wood May 30 '19 at 8:58
  • $\begingroup$ Okay, but if there is just a conducting rod and it just begins to enter the magnetic field, will there be any emf induced that is not due to the movement of charged particles, but due to the increase in flux linkage? $\endgroup$ – Yip Jung Hon May 30 '19 at 9:05
  • $\begingroup$ The emf in a moving conductor is $always$ due to the moving charges in it; they experience a magnetic Lorentz force. The change in flux linkage doesn't give rise to a $separate$ emf; it is another way of calculating the same emf. If you want to use the flux approach then you must imagine the conductor to be part of a circuit.This applies whether the conductor is moving in a uniform field or entering a field. In the latter case the emf will vary with time – whichever way you calculate it! $\endgroup$ – Philip Wood May 30 '19 at 9:27
  • $\begingroup$ Are you clear now? $\endgroup$ – Philip Wood May 30 '19 at 11:38

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