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I am trying to compare the "relation" conventions used in Kirchoff's Loop Rule with Faraday's Loop Rule.
Kirchoff
Please go to this MIT OCW link on Kirchoff's Rule and go to page 8/29. Of the four boxes, I would like to point this one
Note that the yellow electric field was added by me. This picture also follows the integral
[tex]-\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s} = \int_{a}^{b} -Eds = \Delta V = \varepsilon = -IR [/tex]
Since ds and E are parallel.
Nonconservative Fields with Faraday
Now if I go to this video (I take you to EXACTLY where I want you to watch, so don't worry about searching which part of the video does this happen)
http://www.youtube.com/watch?v=UpO6t00bPb8#t=10m22s
Now quoting him
Note that his circuit arrangement is exactly like mine, he went from a high potential to a low potential. The current, electric field in the wire, and the traveling direction are all the same, yet he gets +IR instead of -IR
is he still using this "E dot dl" [tex]-\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s}= \Delta V[/tex]? How does Faraday's Law apply for non-closed paths? Because it seems like Lewin is using [tex]\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s}= \Delta V[/tex] (no minus sign)
The same confusion goes when he talks about the electric field in the battery.
Could someone please clarify for me? Thank you
Kirchoff
Please go to this MIT OCW link on Kirchoff's Rule and go to page 8/29. Of the four boxes, I would like to point this one
Note that the yellow electric field was added by me. This picture also follows the integral
[tex]-\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s} = \int_{a}^{b} -Eds = \Delta V = \varepsilon = -IR [/tex]
Since ds and E are parallel.
Nonconservative Fields with Faraday
Now if I go to this video (I take you to EXACTLY where I want you to watch, so don't worry about searching which part of the video does this happen)
http://www.youtube.com/watch?v=UpO6t00bPb8#t=10m22s
Now quoting him
Now I go through the resistor, so I get +IR, since E and dl are in the same direction, Ohm's Law tells me I get +IR
Note that his circuit arrangement is exactly like mine, he went from a high potential to a low potential. The current, electric field in the wire, and the traveling direction are all the same, yet he gets +IR instead of -IR
is he still using this "E dot dl" [tex]-\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s}= \Delta V[/tex]? How does Faraday's Law apply for non-closed paths? Because it seems like Lewin is using [tex]\int_{a}^{b} \mathbf{E}\cdot d\mathbf{s}= \Delta V[/tex] (no minus sign)
The same confusion goes when he talks about the electric field in the battery.
Could someone please clarify for me? Thank you