Faraday's Disk Dynamo: why is there an emf?

  • Context: Undergrad 
  • Thread starter Thread starter jpas
  • Start date Start date
  • Tags Tags
    Disk Dynamo Emf
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 6K views
jpas
Messages
45
Reaction score
0
Consider a rotating disk on a uniform magnetic field. Faraday's law states that

[tex]\epsilon = - \frac{d\phi}{dt}[/tex]

In this situation, [tex]\vec B[/tex] is constant and the area of the disk is constant. Hence, the magnetic flux is constant and there should be no emf. What am I missing?
 
Physics news on Phys.org
jpas said:
Consider a rotating disk on a uniform magnetic field. Faraday's law states that

[tex]\epsilon = - \frac{d\phi}{dt}[/tex]

In this situation, [tex]\vec B[/tex] is constant and the area of the disk is constant. Hence, the magnetic flux is constant and there should be no emf. What am I missing?

This question is common and has been discussed in several threads previously.

This is one of those tricky situations that appears paradoxical, but really isn't. This paper provides a good explanation.

Basically, even though the field could be considered constant, the total flux includes area. The area also appears constant at first, but as this paper shows, the conduction path usually chosen is not valid. In reality, the area is changing if a valid conduction path is chosen. And, changing area with constant field gives rise to a flux change.
 

Attachments