1
$\begingroup$

Consider a system of two oppositely charged plates (a capacitor) connected by a wire like

enter image description here

allowing current to flow (say electrons to right). For convenience, let's assume that the wire is rigid. Now say there exists a magnetic field into your screen. I have two questions:

  1. What is the motion of the system? Remember that i consider the wire to be rigid. I feel that it should move downward. Initial the velocity is to the right and hence force is downward. Later on velocity is downward and right cause force to be leftward and downward. I feel that the leftward force just slows down the electrons (like more resistance) and hence the system moves downward. I'm not sure. Is a quantitative analysis possible?

  2. What is the work done by magnetic field on the system? I know that the answer to this question is zero. If you consider all the electrons and if you carefully see the forces and all, you'll get zero work done by magnetic field since everything is Lorrentz force. But I'm not very satisfied with this "explanation". Can anyone think of better way to explain why the work done by magnetic field is zero? Also is it correct to say that macroscopically work done by magnetic field is non-zero here?

$\endgroup$

3 Answers 3

1
$\begingroup$

I'm not giving a full answer, just suggestions hoping they will help in making the problem more clear

If you consider just one free charge moving between the two plates the equation of motion will be \begin{gather*} m\frac{\text{d}\boldsymbol{v}}{\text{d}t} = q \left( \boldsymbol{E} +\boldsymbol{v}\times\boldsymbol{B} \right) \end{gather*} where $\boldsymbol{E}$ is directed to the right and $\boldsymbol{B}$ pointing out the page towards the observer. The single velocity-component equation will be \begin{gather*} m\frac{\text{d}v^i}{\text{d}t} = q\left( E^i +\epsilon_{jki}v^jB^k \right) \end{gather*} where you can consider $\boldsymbol{E},\boldsymbol{B}$ constants and where $\epsilon_{jki}$ are the components of Levi-Civita tensor. As you can see the problem is complicated by the fact that every component depends on all the others and so is not immediate to solve this ordinary differential equation (ODE).

Anyway for a rigid wire the motion is unidimensional and so \begin{gather*} m\frac{\text{d}v^i}{\text{d}t} = q E^i \end{gather*} Consider anyway that in a microscopic view the presence of magnetic field may disturb the motion of electron inside the wire and increase the viscous force to which they are subjected

About the work done by the magnetic field, this is $0$ in every physical situation. Remember that you can define the work exerted by a force $\boldsymbol{F}$ in a certain time instant as $\boldsymbol{F}\cdot\boldsymbol{v}$, but for magnetic force you will have $q(\boldsymbol{v}\times\boldsymbol{B})\cdot\boldsymbol{v}=\epsilon_{ijk}v^j B^k v^i=0$

$\endgroup$
1
$\begingroup$

Here is an analysis which I encountered somewhere (assuming your computer screen is vertical): Electrons moving to the right are pushed down by the magnetic field. This puts an excess of negative charge on the bottom of the conductor and leaves the top positive ( the Hall effect). The resulting electric field in the conductor tries to move the electrons up, but the magnetic force prevents them from going up and they push back on the wire. Bottom line; it is an electric force, from the electrons, pushing in the direction of motion of the wire that does the work.

$\endgroup$
0
$\begingroup$

The conductor moves forwards. This is because Lorentz Force postulates that a current (I) carrying conductor with a length l and a magnetic field B experiences a force

$F$ = $I(l×B)$

Clearly the conductor move downward which you can easily find out doing the above cross product. You can also do this, for ease using Fleming's Left Hand Rule.

What qualitatively causes this force on a conductor? Well no one knows. An interesting reason could be that a moving charge itself causes a magnetic field, and the net resulting magnetic field is the cause? Note that this is far from being correct, but it could help you understand, or rather remember intuitively. I'm sorry to say this, but there exists, at least not to my knowledge any foolproof qualitative analysis of this force, it's prediction is purely theoretical in nature.

As for the work done, magnetic force on a charged particle is always perpendicular to the velocity of the charge, implying the work done by it is zero. All the kinetic energy that the charge acquires is due to the induced Electric field as seen from EM theory. But it still doesn't answer the question how the conductor itself gains kinetic energy? Well, any macroscopic body can gain kinetic energy due to it's internal energy (you do that everyday while running, or walking, or even moving). The charges feel a force, increasing its energy which in turn pushes the wire forward. Therefore,the magnetic field itself never really does any work whatsoever.

$\endgroup$
2
  • $\begingroup$ If your computer screen is vertical, and B is into the screen, then the “left hand rule”, (which is used with negative current) gives the force on the wire as down on your screen. $\endgroup$
    – R.W. Bird
    Aug 11, 2020 at 14:43
  • $\begingroup$ Thank you for pointing out the error. $\endgroup$ Aug 11, 2020 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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