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Compute flux through rectangular area

  • Thread starter EV33
  • Start date
  • #1
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Homework Statement



There is an aXb rectangular area in the same plane as a wire with current I. The wire is parallel to side b, and a distance d away. Compute the flux through the rectangular area.

Homework Equations



[tex]\Phi[/tex]=[tex]\int[/tex]B*dA

B=([tex]\mu[/tex]I)/(2[tex]\pi[/tex]r)

cylindrical coordinates... rdrd[tex]\theta[/tex]dz

The Attempt at a Solution


([tex]\mu[/tex]I)/(2[tex]\pi)[/tex][tex]\int[/tex][tex]\int[/tex][tex]\int[/tex](1/r)(rdrd[tex]\theta[/tex]dz)

dr is from d to d+a
d[tex]\theta[/tex] is from 0 to 2[tex]\pi[/tex]
dz is from 0 to b

The solution to this ends up having natural logs in it, and to do that I would have to drop r, but I don't feel like that is right. Could someone please help point me in the right direction on setting up this integral.

Thank you.
 

Answers and Replies

  • #2
196
0


I am not sure why some things look like subscripts or exponents so just a warning there are no exponents are subscripts lol
 
  • #3
196
0


I think I see why now. I should be using rectangular coordinates huh?
I don't know what I was thinking... I must have been thinking partially of amperes law
 
  • #4
196
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([tex]\mu[/tex]I)/(2[tex]\pi)[/tex][tex]\int[/tex][tex]\int[/tex](1/r) dr dy

dr from d to d+a
dy from 0 to b

would this be the correct set up?
 
  • #5
ideasrule
Homework Helper
2,266
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Yes, that's the correct setup. Your solution should have a natural log in it.
 
  • #6
459
0


([tex]\mu[/tex]I)/(2[tex]\pi)[/tex][tex]\int[/tex][tex]\int[/tex](1/r) dr dy

dr from d to d+a
dy from 0 to b

would this be the correct set up?
[tex]\frac{ \mu I }{2\pi} \int_0^b \int_d^{d+a} \frac{1}{r} dr dy [/tex]

Code:
\frac{ \mu I }{2\pi} \int_0^b \int_d^{d+a} \frac{1}{r} dr dy
Hope that helps, though it's nothing to do with the actual question.
 

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