Compute flux through rectangular area

In summary, the conversation discusses setting up an integral to compute the flux through a rectangular area in the same plane as a wire with current I. The wire is parallel to one side of the rectangle and a distance d away. The correct setup involves using rectangular coordinates and integrating with respect to dr and dy. The solution should have a natural log in it.
  • #1
EV33
196
0

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.
 
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  • #2


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


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


([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


Yes, that's the correct setup. Your solution should have a natural log in it.
 
  • #6


EV33 said:
([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.
 

1. What is flux and how is it related to rectangular areas?

Flux is a measure of the flow of a physical quantity, such as energy or particles, through a given surface or area. In the case of computing flux through a rectangular area, we are interested in determining the amount of a certain quantity that passes through the area in a given time.

2. How do you calculate flux through a rectangular area?

To calculate flux through a rectangular area, we use the formula: Flux = (Magnitude of the quantity) x (Area of the rectangular surface) x (Cosine of the angle between the surface and the direction of the quantity's flow). This formula takes into account the direction and orientation of the surface in relation to the flow of the quantity.

3. What units are used to measure flux?

The units of flux depend on the quantity being measured. For example, if we are calculating the flux of energy through a rectangular area, the units would be watts (W) or joules per second (J/s). If we are calculating the flux of particles, the units would be particles per square meter (particles/m^2).

4. Can flux be negative?

Yes, flux can be negative. This occurs when the quantity being measured is flowing in the opposite direction of the surface being analyzed. In this case, the cosine of the angle between the two will be negative, resulting in a negative flux value.

5. How is computing flux through a rectangular area useful in science?

Computing flux through a rectangular area is useful in many scientific fields, including physics, engineering, and environmental science. It allows us to understand and quantify the flow of quantities, such as energy or particles, through a given surface. This information can help us analyze and predict the behavior of systems and make informed decisions in various applications.

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