Magnetic induction of infinite conducting sheet using Ampere's Law

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The discussion focuses on solving a problem involving the magnetic induction of an infinite conducting sheet using Ampere's Law. Participants express confusion about the appropriate path for the line integral, suggesting a rectangular path in the x-y plane. It is clarified that the integral can be computed line by line, resulting in a closed loop line integral that equates to the enclosed current. The importance of considering symmetry in determining the direction of the magnetic field is emphasized, indicating that some integrals may be straightforward. Additionally, the thickness of the sheet is relevant for calculating the total current enclosed based on the given current density.
fatherdaly
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Homework Statement


http://i.imgur.com/j3uzQ.png" to screenshot of question


Homework Equations


The "Integral Form" the question refers to is the http://upload.wikimedia.org/math/0/3/1/0311484f33c932135c09ab12ca8d1a29.png"


The Attempt at a Solution



The hint hasn't really helped me here, and I don't know what path dl should follow. Obviously it should be a rectangle of sorts but then you would have to do it line by line, meaning its not a closed loop. Also, I'm not sure how the thickness comes into it.
 
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fatherdaly said:
The hint hasn't really helped me here, and I don't know what path dl should follow. Obviously it should be a rectangle of sorts but then you would have to do it line by line, meaning its not a closed loop.

They tell you the path in the problem statement. The path is the perimeter of a rectangle in the x-y plane. Yes, you do have to do the integral line by line, but when you add up the results for each line segment, you still get a closed loop line integral:

\oint\vec{B}\cdot d\vec{l}=\int_1 B_1dx + \int_2 B_2dy + \int_3 B_3dx + \int_4 B_4dy = \mu_o I_{enc}

HINT: If you consider, using symmetry, the direction of the magnetic field, two of the above integrals should be trivial.

As for the thickness, d: You need the total current enclosed. You are given a current density...
 

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