0
$\begingroup$

Faraday Disk Dynamo

How can there be an induced emf? According to Faraday's Law change in the magnetic flux produces an induced emf but B (magnetic field), A (area) and the angle between them (area vector and the magnetic field) is constant then how come there is an induced emf?

Please help me with this. I am already so confused in electromagnetism. Consider like I don't know nothing about motional emf and faraday's law and then answer the question because I know that I've developed so many misconceptions.

$\endgroup$
4
  • 1
    $\begingroup$ See physics.stackexchange.com/questions/443933/… on this site. $\endgroup$
    – John Darby
    Jan 23 at 21:52
  • $\begingroup$ Your diagram is seems to be set up to derive the motional emf using the magnetic Lorentz force (aka motor effect force) on charge carriers in the disc. In my opinion, this is the best way to proceed. Trying to understand what's going on in terms of flux linkage is problematic. Does this make any sense to you? $\endgroup$ Jan 23 at 23:01
  • $\begingroup$ Yeah. Thanks. But if both (the Faraday's law and motional emf) are defined for the same quantity that is magnetic flux then why is there no explanation in terms of Faraday's Law? I just want to confirm if I've any misconception or is it just how the things are. $\endgroup$
    – Physics
    Jan 24 at 4:01
  • $\begingroup$ Both are defined for the same quantity that is (Induced emf)* at the place of Magnetic Flux. I made an error while typing. $\endgroup$
    – Physics
    Jan 24 at 4:26
1
$\begingroup$

This should have been explained in the course or a textbook that gave you the exercise. You have found a well-known example where the standard formulation of Faraday's law using flux through a loop isn't directly applicable - a motional EMF for a path with open endpoints. Motional EMF for any open path in a conductive body is not due to induced electric field, but due to motion in magnetic field(magnetic force pushes on charge carriers, they push on the conductor and the conductor pushes back on the carriers, giving them energy for current and thus slowing down its motion).

This is not described by standard Faraday's law for loops, as you correctly observe, because for any stationary loop, magnetic flux is constant and induced emf for such stationary circuit is zero (as it should be, since there is no induced electric field).

The question is asking about motional emf for a path $\gamma$ joining point A to point B, which is a little different thing than the induced emf in a loop that Faraday's law talks about.

We can find it however from definition of this motional EMF: it was found that this is simply integral of magnetic force per unit charge along the given path from A to B:

$$ emf_{\gamma} = \int_{\gamma} \mathbf v \times \mathbf B \cdot d\mathbf s~~(*) $$ where $\mathbf v$ is velocity vector of the conductor (not that of charge carriers!). You have to take this as independent law for motional EMF at this point, there is a way to derive it from other knowledge but it gets tricky.

There is a way to formulate Faraday's law for moving closed paths and then apply this to the situation in the exercise (when the path is completed outside the wheel into a closed loop). But this is a little confusing if you do not know about (*) and when you know about it, it is pointless complication.

If your book did not explain this, get a better book (Feynman's lectures and some older american, british and german textbooks are nice).

So for your example of rotating disk, the vector product $\mathbf v \times \mathbf B $ on all points of the path $\gamma$ points in radial direction away from the center, so dot product with $d\mathbf s$ is trivial. We get the integral

$$ emf_\gamma = \int_0^{R} \omega r B dr = \omega B \frac{R^2}{2}. $$

$\endgroup$
1
  • $\begingroup$ Thanks a lot. I got it now. By the way I am using "University Physics by Hugh D. Young and Roger Freedman". Also I thought earlier that I've had a misconception about Faraday's Law. Thanks again! $\endgroup$
    – Physics
    Jan 24 at 4:06

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.