What is the electric field in a circular region using Faraday's law?

AI Thread Summary
The discussion focuses on calculating the electric field in a circular region using Faraday's law. The process involves integrating the magnetic field over the area of the circle while considering the time-varying nature of the magnetic field. A key point raised is whether the magnetic field, described by a specific function, can be treated as constant across the area, which complicates the calculations. The integration steps lead to an expression for the induced electromotive force (emf) and ultimately for the electric field. The conclusion emphasizes the importance of clarifying whether the magnetic field is constant or variable when applying Faraday's law.
Ashley1nOnly
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Homework Statement


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The region is a circle with radius=a

Homework Equations


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Using Faraday's law to find E.

The Attempt at a Solution


Dealing with the Left side of the second equation first
1.) Pull out the Partial/partial t because it is a constant in this equation
2.) Now I have B dot (n-hat) da
3.) Using the dot product I know that I have |B||da|cos(theta)
4.) the angle between them is zero so I have B*da
5.) I can pull B out and integrate over the area of the circle which gives me A=2(pi)(r)^2 =2(pi)(a)^2
6.) Now I have
-[2(pi)(a)^2 * Partial(B)/partial t] this is also equal to equation 1 our emf induced
7.) Taking the partial of B with respect to t gives me
-[2(pi)(a)^2 *
upload_2017-2-24_11-21-14.png
]

Now I deal with the right side of equation 2 which is equal to 7.)

[/B]
8.) I know that E is parallel to dl and that I can pull E out
9.) Now I integrate over dl which just gives me

E*2(pi)(r)= -[2(pi)(a)^2 *
upload_2017-2-24_11-21-14.png
]

of course excluding the B(x,y,t).

Then I divide by 2(pi)r)
Which will give me E if I did everything correctly

 

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Ashley1nOnly said:
5.) I can pull B out and integrate over the area of the circle ...
Is B constant over the surface area of the circle?
 
Yes
 
Ashley1nOnly said:
Yes
I think not. The question says ##\vec{B}(x,y,t) = B_0 \cos (\pi x/L) \cos(\pi y /L) \sin (\omega t) \hat{k}##. At different values of x and y inside the circle the field has different values.
 
Well we don't usually do anything that's not constant because it would make it complicated. We always deal with constant things. So my assumption since it was not stated was that B was constant
 
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