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Down below is a conductive bar of length $L$ moving on two conductive rails in a magnetic field. I have a couple of questions regarding the flow of current in such a system.

How does the current flow through the closed circuit if the length of the bar $L$ is larger than the width of the rails $l$ ? I'm curious to know what is happening at the ends of the bar in particular. Do the electrons reach them and make a U-turn or something?

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    $\begingroup$ The energy doesn't flow in the electrons. It flows in the electromagnetic field surrounding the conductors (and a little bit of it flows on their inside because of skin effect). Energy flow is one of the limits of the classical electron-flow model in electrical conductors. It gives us the wrong intuition about what is happening. If you want to get the correct understanding, then you have to learn about Maxwell's equations and the Poynting vector. $\endgroup$
    – FlatterMann
    Commented May 6, 2023 at 16:07
  • $\begingroup$ @Flatterman you're just spreading the misconception popularized by youtubers that the energy flux vanishes in a perfect carrying current conductor. If you,want to learn.the truth, have a look at non eq. thermodynamics. There is definiteluy an energy flux due to moving charges. $\endgroup$
    – untreated_paramediensis_karnik
    Commented May 6, 2023 at 17:12
  • $\begingroup$ @untreated_paramediensis_karnik Please note my remark "and a little bit of it flows on their inside because of skin effect". You are welcome to analyze energy flow in a transformer, or, if that is not enough... explain the energy flow of 1kW/m^2 on Earth's surface that is caused by solar radiation. There are certainly no wires involved there. $\endgroup$
    – FlatterMann
    Commented May 6, 2023 at 17:32
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    $\begingroup$ All emf's are "induced." The normal term for the kind of emf you are talking about (due to the Lorentz force of a constant magnetic field on a moving circuit element) is "motional emf." $\endgroup$
    – Buzz ♦
    Commented May 7, 2023 at 6:08
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    $\begingroup$ @untreated_paramediensis_karnik Electrons moving at a few um per second (the drift speed in a typical metal) carry very little kinetic energy. $\endgroup$
    – FlatterMann
    Commented May 8, 2023 at 2:59

2 Answers 2

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The current will take the path of least resistance and so will flow from the bar to the rails. The ends of the bar can essentially be ignored regarding the electron flow.

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All along the moving rod, including the ends beyond the rails, its charges experience the magnetic Lorentz force, $\vec F=q\vec v \times \vec B$. The free electrons are urged towards the end of the rod furthest from us (the viewer).

Some free electrons in the end nearest us will move out of the end into the main circuit, leaving the end positively charged as long as the rod's motion continues. Likewise some free electrons will move out of the main circuit into the end furthest from us making it negatively charged.

These 'chargings of the ends' will happen very quickly. [The unbalancing of charges will give rise to electric fields and electric Lorentz force ($\vec F=q\vec E$) on the free electrons in the ends of the rods that will quickly counterbalance the magnetic Lorentz forces on them so no more charge will accumulate.]

As the rod continues to move, free electrons from the middle section of the rod will escape from the rod itself to process around the circuit mabnm..., their drift speed being limited by the resistance due to their collisions with ions in the metal.

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