1
$\begingroup$

When a conductor (e.g. a wire) is moving through a uniform magnetic field, it will create a electric field in the wire as the magnetic field exerts a force on the moving conduction electrons in the wire and create a potential difference.

the wire is so thick!

And here is my questions:

  1. (If we can get the magnetic field to exert a super strong force on the electrons) Is it ever possible for the electrons in the wire to get separated from the wire like what happens when high energy photon hits an atom and knock off its electrons?

  2. Since $F_e$ and $F_b$ always act in the opposite direction, why don't they cancel off one another (so the net force is 0) when the charges are in the electromagnetic equilibrium ($qE = qvB$) ?

  3. What's so significant about the this equilibrium ($qE = qvB$)? The formula for motional emf $emf = vBl$ is derived from the fact that $E = vb$ and that was because the charges are in the electromagnetic equilibrium. But wouldn't this formula $emf = vBl$ be inaccurate if we have a motional emf that are caused by charges not in the electromagnetic equilibrium ($qE = qvB$)? Why when there is a motional emf we assume that the chargers are in this equilibrium?

I'm still learning secondary school physics and I'm very new to this topic and so please explain it in laymen terms if possible. Thank you so much.

$\endgroup$
2
$\begingroup$
  1. yes if the generated electric field is high enough, field emission can occur--electrons are forced out the negatively charged end purely from the high electric field..

  2. note that in this case the electrons dont move down the conductor because of an electric force, but the magnetic force (aka Lorentz force). it is because they have settled at the lower end that an electric field occurred. (electrons in this position cause electric field, not vice versa). note that the electrons are all forced to the bottom, creating a large negative charge there. naturally, electrons will want to spread back out, but they are kept in that uncomfortable position because of the magnetic force. you are right to say they are in equilibrium--as long as the rod is not being accelerated, the electric force wanting to push the electrons up the conductor is matched by the magnetic force pushing them down. therefore the electrons stay stuck at the lower end.

  3. note that a net force of zero doesnt mean that the electrons are all spread out normally, it just means they stopped moving (no net force, no acceleration). where they stopped is a matter of how strong the Lorentz force (how fast the rod is being moved) is. the higher the force, the more strongly it can pack all the electrons at one end. the moment the rod starts to slow down, the Lorentz force decreases lower than the electric force (electric force is not dependent on how fast the rod is moving), and the forces are no longer in equilibrium--the charges spread out accordingly. until equilibrium is reached again and the charges stop moving.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ in school you will likely encounter a free electron moving to the right instead of it being in a conductor and the conductor moved to the right. in that case the electron still encounters the Lorentz force, but because it is not confined in a aconductor, it curves south. if there is a huge negative APPLIED electric field at the bottom, it wont curve as much. there is a sweet spot where the electric field is just right that the electron moves through the magnetic field in a straight line. same concept! $\endgroup$ – gregsan Oct 2 '13 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.