Confusion with electromagnetism and Faraday's law

  • #1
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Hi- Sorry if this is a silly question, but by definition the magnetic flux is given by integral B dot dA. But From Gauss' law for magnetostatics is this not zero around a closed loop? So would that not then imply that the EMF around any closed loop is zero? Obviously I'm missing something, so I would be really grateful for any clarification.

Thanks in advance :) you guys rock
 

Answers and Replies

  • #2
jtbell
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In Gauss's law, the surface integral is over a closed surface. In Faraday's law, the surface integral for the flux is over an open surface.

Also, note that Faraday's law uses the rate of change of the flux through the surface (i.e. the time derivative of the surface integral), not the flux itself. At a particular point in time, it's possible for the flux to be instantaneously zero, but increasing or decreasing. And it's possible for the flux to be nonzero but constant (time derivative is zero).
 
  • #3
247
11
In Gauss's law, the surface integral is over a closed surface. In Faraday's law, the surface integral for the flux is over an open surface.

Also, note that Faraday's law uses the rate of change of the flux through the surface (i.e. the time derivative of the surface integral), not the flux itself. At a particular point in time, it's possible for the flux to be instantaneously zero, but increasing or decreasing. And it's possible for the flux to be nonzero but constant (time derivative is zero).
Thanks for the reply. That makes sense, but my real problem is that the time derivative can be taken out of the integral meaning that 0 is being integrated over a closed circuit ( I get that Faradays Law is for an open surface, but for arguments sake consider a closed circuit) then this implies that the rate of change of flux, and hence the EMF is constant regardless of how the magnetic field is changing. This seems counter intuitive since if it were changing rapidly and randomly I'd expect a different EMF to when the field is changing slowly and periodically.

Thanks again :)
 
  • #4
jtbell
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I get that Faradays Law is for an open surface, but for arguments sake consider a closed circuit
In Faraday's law, the EMF is the integral of the induced ##\vec E## around the boundary (edge) of an open surface. If the surface is closed, where's the boundary?
 

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