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Faradays Law, Calculate E field 
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#1
Jun2711, 12:57 PM

P: 126

1. The problem statement, all variables and given/known data
I have been asked to calculate the magnitude of the induced electric field using Faraday's law. I am unable to provide the picture that accompanies the problem and must describe it with words. A circle has a magnetic field (perpendicular to the circle) directed into the page throughout it, that varies with time. This circle has a radius R. I am asked to calculate the E field strength at a point outside this circle, located at a distance 2R from the center of the original circle. There is no field outside the circle (past distance R). I have seen the solution to the problem and Faraday's law is used to solve it. For the line integral they use the circle's radius Line integral becomes E*2pi*R But for the surface integral (to get flux) they use the distance to the point (2R). negative change in flux becomes (d/dt) (B*pi*(2R)^2) Can someone please help me. I thought that for Faraday's law, the line integral should be bounding the surface. 2. Relevant equations EMF=−N ΔΦBΔt or −N dΦBdt 


#2
Jun2711, 01:23 PM

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According to MaxwellFaraday Law the line integral around a closed loop is equal to the negative of the time derivative of the magnetic flux enclosed in the loop:
∫E_{s}d_{s}=∂Φ/∂dt http://en.wikipedia.org/wiki/Maxwell...c.27_equations Regarding symmetry, E is tangential to any circle around the centre of the one with magnetic field. So the line integral can be taken along a circle of any radius. ehild 


#3
Jun2711, 05:54 PM

P: 126

I believe I am getting hung up on whether the line integral must be the boundary of the surface over which the flux integral is preformed. I have seen it presented this way twice, one of which is on the Wikipedia page below under the heading "The MaxwellFaraday Equation".
http://en.wikipedia.org/wiki/Faraday...w_of_induction 


#4
Jun2711, 06:09 PM

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Faradays Law, Calculate E field
The magnetic field is nonzero only within the smaller circle, radius R.
You want the E field long the boundary of the larger circle, radius 2R. The area of the larger circle which is outside of the smaller circle has no magnetic flux, so the flux enclosed by the larger circle is the same as the flux enclosed by the smaller circle. 


#5
Jun2711, 08:46 PM

P: 126

Ok, thank you. So flux is the same. I am still confused about the notation (with the wikipedia article above as an example) as to why they show the line integral path to be the boundary of the surface.



#6
Jun2711, 09:10 PM

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At what location do you want to find E?



#7
Jun2811, 11:10 AM

P: 126

The point at which I want to find E is at a distance 2*R from the center of the circle containing the magnetic firld, which has a radius of R. In the plane of the circle, with B perpendicular to it.
I think I am getting hung up on whether the line integral's path is "coupled" (i.e. the boundary) of the surface for the surface integral. 


#8
Jun2811, 01:11 PM

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The path form the line integral is indeed the boundary for the surface integral, but in this case, remember that the magnetic flux through the surface area of both circles is the same.



#9
Jun2811, 04:17 PM

P: 126

So that means that if I used the radius of the outer circle enclosing the point in question region (2R), instead of just R, for the surface of the surface integral I should get the same result. It makes intuitive sense that the flux through both but looking at the reduced formula made it seem as if I would get a different answer for a different radius (for the flux). I will recheck my work.



#10
Jun2811, 04:54 PM

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To quote the last sentence of the second paragraph of your Original Post:
"There is no field outside the circle (past distance R)."I assume that refers to the magnetic field. Therefore, the magnetic flux through the surface of the larger circle is the same as the magnetic flux through the surface of the smaller circle. 


#11
Jun2811, 09:25 PM

P: 126

Yes, that is correct. I will recheck my calculations tomorrow, and try using the same R radius in both integrals to see where it gets me.



#12
Jun2911, 09:32 AM

P: 126

Thank you all very much. I now understand my error. I was substituting the same radius for flux and the line integral, but that of course would imply flux through through an area in which there is none. Flux through the circle is = flux through the path of the line integral. But I can't use the same radius.



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