1
$\begingroup$

Suppose a solid metal plate moves trough a homogenous magnetic field $\vec{B}$ as shown in the following image. You will see that the plate will slow down (as long as not the entire plate is in the homogeneous field).

For example in the upper part of the plate in the picture there are eddy-currents induced because the magnetic field is changing in that region (the part of the plate is going to enter the homogenous magnetic field).

To explain that the plate is slowed down often Lenz law is cited which states in this case that the magnetic field produced by the eddy-currents is opposes the changing magnetic field.

But how can one explain that this leads to a force which slows down the plate in detail? In particular the situation is not so easy such that one could speak of magnetic north and south poles...

Is there a general mathematical expression for the force in this case?

(It's clear that one cannot calculate it already because I didn't give any technical details, but I just want a general equation for this situation which illustrates the concepts in this example from a mathematical point of view)

Eddy Currents

Do you have any good references about this point?

Improve this question
$\endgroup$
1
$\begingroup$

A physical explanation........

When currents are setup , the energy lost in heat due to moving current (power lost is $i^2R$ in a where $R$ is resistance of the loop in which current moves.) comes from the Kinetic Energy of the plate. So when heat is dissipated, the KE. decreases.

According to lenz law, it sets up a magnetic field to nullify the effect of changing field , The magnetic field would posess energy that would come from it's source ie. the plate.

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks but that's clear. However the missing link is between "...magnetic field opposed to the changing magnetic field..." and the produced force. Perhaps one can establish the link using the energy argument but I don't see how to... $\endgroup$ – Julia Apr 22 '13 at 17:42
  • $\begingroup$ You write "it sets up a magnetic field to nullify the effect of changing field" up to that point it is clear it's just a statement about the magnetic field. But why does this produce a force? I don't know a general theory which says that opposing magnetic fields produce repulsive forces between the bodies which has created the fields or something like this. So how can one embed it in general theory that the opposing fields in this case generate the repulsive force? $\endgroup$ – Julia Apr 22 '13 at 17:51
  • $\begingroup$ @Julia Can you give me the exact statement of explanation, maybe it's talking of force by magnetic force on the induced current.ie. used in concept of brakes. $\endgroup$ – ABC Apr 22 '13 at 17:59
  • $\begingroup$ Take for example this one: phywe.com/461/apg/301/pid/28904/… $\endgroup$ – Julia Apr 22 '13 at 18:23
  • $\begingroup$ @Julia Yes, that's what I thought.It's the force which the magnetic field exerts on the induced eddy currents.which act like brakes. $\endgroup$ – ABC Apr 23 '13 at 2:28
0
$\begingroup$

Magnetic fields exert a force on moving charges. The Lorentz force law for the force on a charge $q$ is

$$\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B}) $$ Where $\mathbf{E}$ is the electric field, $\mathbf{v}$ is the velocity and $\mathbf{B}$ is the magnetic field.

Some of the electrons in the plate can move freely, so when the neutral plate moves in the magnetic field those neutral charges experience a force. This is what creates the eddy current.

But the induced current is also moving charge, and the motion of that charge is also affected by the magnetic field. This eventually works out such that the electrons are dragged to one side of the plate more than the other, which induces an electric field, which in turn pulls the nuclei in the plate and slows it down.

Improve this answer
$\endgroup$

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.