0
$\begingroup$

Consider the following situation.

enter image description here

There is a metallic strip situated in an homogeneous electric field. The lines of the electric field go from left to right, so the free electron in the metallic strip are attracted to the left side of the strip. There is a force acting on the free moving electrons of the metal due to the external electric field, F = qE. This force is only capable to move the electrons inside the metallic strip, NOT THE ENTIRE METALLIC STRIP (remember, the electric field is homogeneous).

Now consider this other situation.

enter image description here

There is no electric field anymore, but instead there is a homogeneous magnetic field perpendicular the the sheet of paper. The lines of the magnetic field go from above to below, as it is shown in the picture. There is also an electric current flowing in the metallic strip, the electron move from up to down, as it is shown. Due to the Lorentz force acting on the free electrons of the metallic strip (F = -evB, where v is the electron drift speed), there is an electron buildup on the left side of the strip, just like in the previous situation. This is known as the Hall Effect. But unlike in the previous situation, it is empirically known that THERE IS A FORCE ACTING ON THE ENTIRE STRIP, so the strip would move leftwards.

The question is simple, can you explain why?

Remember, the Lorentz Force should act only on the moving free electrons, not on the ionic lattice of the metallic strip.

Improve this question
$\endgroup$
6
  • $\begingroup$ I believe I can answer my own question. But I'd like to know your opinions about it first. $\endgroup$
    – Eduard Oganesian
    Commented Feb 18, 2023 at 21:59
  • $\begingroup$ If you don't like something, just say it, I really don't understand what it point of voting down my question. $\endgroup$
    – Eduard Oganesian
    Commented Feb 18, 2023 at 22:11
  • 2
    $\begingroup$ I didn't downvote, but I have no idea what your actual question is. You've correctly explained why there is a force in one situation and no force in the other. Where's the question? $\endgroup$
    – Michael Seifert
    Commented Feb 18, 2023 at 22:18
  • $\begingroup$ I am asking why there is a force on the current currying metallic strip when it is immersed in homogeneous magnetic field? That is my question. $\endgroup$
    – Eduard Oganesian
    Commented Feb 18, 2023 at 22:28
  • $\begingroup$ THERE IS NOTHING HOMEWORK-LIKE ABOUT THIS QUESTION. IT IS CLEARLY CONCEPTUAL. $\endgroup$
    – Philip Wood
    Commented Feb 19, 2023 at 0:25

3 Answers 3

Reset to default
3
$\begingroup$

You've correctly explained the Hall effect as due to the magnetic Lorentz force, $\mathbf F_{\text B}$, acting on the moving free electrons. But why aren't these free electrons pushed right out of the left hand side of the wire? It is because they are restrained by another force, $\mathbf F_{\text {e,i}}$ essentially electrostatic, which attracts the electrons to the positive ions of the lattice. This force acts to the right on the free electrons, and is equal and opposite to the magnetic Lorentz force. The Newton's third law partner to this electrostatic force on the free electrons from the ions is an equal force, $\mathbf F_{\text {i,e}}$ to the left exerted by the free electrons on the ions, that is on the wire! Summary: $\mathbf F_{\text {e,i}}=\ –\mathbf F_{\text B}$ (equilibrium condition) and $\mathbf F_{\text {e,i}}=\ –\mathbf F_{\text {i,e}}$ (Newton's third law), so $\mathbf F_{\text {i,e}}=\ \mathbf F_{\text B}$

The force on the wire used to be called the 'ponderomotive force'. You might, loosely, regard it as the magnetic Lorentz force on the moving free electrons that has been 'passed on to' the whole wire by means of electrostatic forces.

Improve this answer
$\endgroup$
6
  • $\begingroup$ That was a great answer! But now consider the Faraday Disk (the so called Homopolar Motor). There is no metallic strip, but a metallic round disk immersed in a magnetic field applied externally. And an electric current flown between the center and the edge of the disk. There are no longer left or right sides (as in the previous case). Still there is a ponderomitive force acting on the disk, which creates torque, which makes the disk spin. How the torque is created in that case? I believe that the answer is in the non-homogeneous magnetic field created by the current inside the disk. $\endgroup$
    – Eduard Oganesian
    Commented Feb 19, 2023 at 3:40
  • $\begingroup$ (a) This set-up is not, I think, called a Faraday disc unless the disc is being rotated by an external motor (such as a hand-crank) and is being used to generate an emf. In your case it's being used with an external source of emf as a homopolar motor. (b) I think that the analysis here has the same basic form as for the wire, except that $\mathbf F_{\text{e,i}}$ is the mean tangential resistive force on free electrons as they move 'sideways' through the disc under the action of the Lorentz force. [The free electrons are also moving radially owing to the external emf.] $\endgroup$
    – Philip Wood
    Commented Feb 19, 2023 at 9:36
  • $\begingroup$ Although the very same machine can work as a motor and as a generator, I want to discuss its "motor part" only. As we apply a current between the axle and the edge of the disk while it is immersed in a magnetic field, disk would spin. I believe that this situation is different than for the wire. As you said, the electrons get to the side the the wire, but can't jump out because of the electrostatic forces of the atoms. So the electrons drag the atoms with them, which creates a ponderomotive force. The disk has no left side (as the wire), so the electron can move all the way around the disk. $\endgroup$
    – Eduard Oganesian
    Commented Feb 19, 2023 at 10:10
  • $\begingroup$ (a) I did understand that you wished only to discuss the motor function; the point I was making is that the term Faraday disc is used only for the generator mode. (b) As for how, in motor mode, the wheel is turned, I think that what you've said about the electrons dragging the disc with them accords fully with what I wrote in my comment above. The quasi-static force pair, $\mathbf F_{\text {e,i}}$ and $\mathbf F_{\text {i,e}}$ are replaced by a resistive force on the free electrons from the ions and an equal and opposite partner force on the ions – your drag force, i.e. the ponderomotive f. $\endgroup$
    – Philip Wood
    Commented Feb 19, 2023 at 11:45
  • $\begingroup$ (a) To the best of my knowledge the term Faraday Disk is used to depict the motor function, not the generator. To depict the generator, the term Barlow Disk is used instead, as the generator function was discovered by Barlow. Michael Faraday first discovered its motor by placing a conductive magnet in a mercury cup, and observed it spinning by setting a current between the center and the edge of the magnet. $\endgroup$
    – Eduard Oganesian
    Commented Feb 19, 2023 at 16:34
2
$\begingroup$

In your first picture the electric forces act on every charged particle (the free electrons and the atomic cores) because electric force is proportional to charge $q$. Each free electron (because of its negative charge) is pulled to the left. And each atomic core (because of its positive charge) is pulled to the right. So the sum of these forces is exactly zero, because the sum of the charges is zero. And hence there is no net force on the metallic body as a whole.

In your second picture the magnetic forces act on the moving charges only (i.e. only the moving free electrons, but not the resting atomic cores) because magnetic force is proportional to $qv$. Each free electron is pulled to the left. And there is no pull on the atomic cores. So the sum of these forces is to the left, and the metallic body as a whole is pulled to the left.

Improve this answer
$\endgroup$
-2
$\begingroup$

Ok. While I was writing my question, two different answers came to my mind. The first one is as follows. In the first case, where an homogeneous electric field is applied to me metallic strip, the electric force is acting on both: the free moving negatively charged electrons and positively charged nuclei. The electrons tend to move to the left, the nuclei tend to move to the right. These two forces cancel each other out, so there is no net force on the metal. While in the second case, where magnetic field is applied on the free moving electrons of the metal, the resulting Lorentz force is acting only on the moving electrons, not on the static nuclei. This creates an unbalanced force to the left, which tend to move the metallic strip the te left. An alternative explanation could be the following: As the electric current flows in the metallic strip, it creates it's own magnetic field. This adds to the externally applied magnetic field. The resulting magnetic field is no longer homogeneous. The nuclei of the metallic lattice have magnetic moments. A magnetic moment in a inhomogeneous magnetic field experiences a net force. So that's the reason the metallic body moves to the left.

Improve this answer
$\endgroup$
0

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