From what I have read, as a conducting rod moves perpendicular to a uniform magnetic field, this will lead to a change in magnetic flux and therefore an emf will be induced (area being swept as the conducting rod moves).

However, in terms of eddy currents, I have been told that when the metal plate is moving perpendicular to and within the uniform magnetic field, there is not change in magnetic flux and therefore no induced emf.

As the conducting rod moves and there is an induced emf (and therefore change in magnetic flux), how is it that there is no induced emf in the form of eddy currents as the metal plate is within the uniform magnetic field.

To me, I don't see where the difference lies between the two scenarios, they are both conductors moving within a uniform magnetic field.

Could someone please explain to me where my understanding is flawed.

Conducting rod moving through uniform magnetic field

  • $\begingroup$ If an electric charge moves in a magnetic field the Lorenz forces (see your formula above) acts upon it. However, this is only true, if the motion is not parallel to the magnetic field. Therefore, could you please clarify the relative directions of the B-field and the motion of the metal plate. $\endgroup$
    – Semoi
    Commented Jun 15, 2019 at 10:45
  • $\begingroup$ @Semoi please see it now $\endgroup$
    – Max604
    Commented Jun 16, 2019 at 12:33
  • $\begingroup$ Thank you for clarifying the directions. My second guess is, that the problem is due to wording. We usually use the term eddy currents if we like to address the induced currents which are produced by a change in the B-field, $dB/dt \propto I$. However, in the situation described above the B-field is constant. Therefore, although we obtain a voltage across the moving conductor, we usually do not speak of eddy currents. Nevertheless, the voltage across the plate exists. $\endgroup$
    – Semoi
    Commented Jun 16, 2019 at 18:20

1 Answer 1


The flux of the magnetic field through any surface whose border lies inside the rod is defined as $$\int_S\vec{B}\cdot\hat{n}dS$$ where the integral is calculated over the whole surface and $\hat{n}$ is the versor perpendicular to the surface (point by point). For example if you consider a rectanguar border and the rectangle as the surface, the flux is simply given by $$\int_S\vec{B}\cdot\hat{n}dS=\vec{B}\cdot\hat{n}\cdot A$$ where $A$ is the area of the rectangle and $\vec{B}\cdot\hat{n}$ is the same in any point of the rectangle because the field is uniform. For this particular surface the flux is constant. Without actually calculating it but just by looking at the flux definition we can conclude that this is true for any surface with any border that moves with the rod because the field is uniform and if the rod moves perpendicularly to the field, the angle between $\vec{B}$ and $\hat{n}$ does not change over time.


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.