Induced EMF and farady's law

  • Thread starter pardesi
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  • #51
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I just wanted to throw this out there because it is related to the problems we're having with the Flux rule. It is an "exception to the Flux rule" pointed out by Feynman in his Lectures on Physics:
Consider a copper disk rotating with constant angular velocity, as shown in the figure. A bar magnet is directed normal to the surface of the disk, as shown in the following figure:
feynman disk.JPG

If a galvanometer is used to measure the induced current in the outer rim of the disk, a nonzero induced current is detected. Using the Lorentz force law, this is easily explainable. Consider a small charge dq on the outer rim of the disk. At any given time, it has a nonzero velocity because of the rotation of the disk. For this reason, the bar magnet exerts a magnetic force on the charge. This makes the charge move differently than the disk itself. Some of these charge will go through the galvanometer, and thus the galvanometer will indicate the existence of a current.

But wait a minute. Let's try applying the integral form of Faraday's Law: the emf along a closed loop is equal to the rate of change of the magnetic flux through the loop. Take the loop to be the outer rim of the copper disk. But the magnetic flux through surface of the copper disk is constant, and thus the rate of change of magnetic flux is zero. This is because both the magnetic field and the area are both constant. So we have a strange situation in which the induced emf is nonzero even though the rate of change of magnetic flux is zero.
So why doesn't the integral form work in this case? What modification of the integral form would allow us to calculate the induced emf?
 
  • #52
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great example.
yes this proves the point. what is changing in other cases is the area.so it is good that formulaes coincided but fact remains the proof and the assumptions that go behind it should be well appreciated.
 
  • #53
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great example.
yes this proves the point. what is changing in other cases is the area.so it is good that formulaes coincided but fact remains the proof and the assumptions that go behind it should be well appreciated.
As in your original post, this problem is solved with a loop of changing area--one side moves and the other is stationary. If you have access to R. Becker, "Electromagnetic Fields and Interactions", Vol 1, (1964) look on page 378.
 
  • #54
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As in your original post, this problem is solved with a loop of changing area--one side moves and the other is stationary. If you have access to R. Becker, "Electromagnetic Fields and Interactions", Vol 1, (1964) look on page 378.
Yes, but what is wrong with using as the loop the outer rim of the disk? The integral form of
Faraday's law applies to any loop.
 
  • #55
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Does anyone have any answers to the questions I asked in post #50?
 
  • #56
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Yes, but what is wrong with using as the loop the outer rim of the disk? The integral form of
Faraday's law applies to any loop.
I'm beginning to understand your questions. Some loops work, some don't. I don't remember any of the books explaining how to chose which one.
 
  • #57
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I'm beginning to understand your questions. Some loops work, some don't. I don't remember any of the books explaining how to chose which one.
Apparently, there are books explaining how to choose the loop over which the integration is done for Faraday's. I'm not sure what textbooks discuss this, but Giuliani mentions these books. The point is, however, that we shouldn't need to be restricted in our choice of loops. All loops are supposed to work.
 
  • #58
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Yeah but, some loops give different answers than others.
 
  • #59
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but that shouldn't be the case.so hoe do u resolve it probably an example would help
 
  • #60
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My immediate, and incomplete, reaction is that if v=0 all the loops give the same answer, but, maybe not for motion.
 

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