What is the Flux through a Square Loop with a Current-Carrying Wire?

In summary: The reason you are doing a line integral instead of a surface integral is because you are only interested in the flux through the loop, and not the area of the loop.
  • #1
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I'm confused. Here is the problem I'm working on. Imagine a long, straight wire crossing the screen or page horizontally. It is carrying a current i. Now imagine a rectangular loop underneath the wire a distance of d away. The loop has a height of a and a width of b. I am supposed to find the the flux that passes through the loop. i is a constant, as are b, a, and d. My confusion arrises from that fact that the flux = ((mu(naught)*i*b)/2*pi)(ln((d+a/a)). Since ampere's law is (flux)= integral(B dot dA) (where B is the B of the wire), shouldn't there be an a in the formula for flux? I know that there shouldn't be, but not why. Could some one please help me?
 
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  • #2
You HAVE an "a" in your correct result for flux (2 of them!)

Do you mean, why Doesn't Ampere's general law label the Area
as Length x width ? well, LW doesn't have an "a" in it either!
More to the point, somebody will want to use A = pi R^2 ...

Can you try to be a little more specific?
 
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  • #3
I should have asked:

Why is only the width of the loop included in the final formula. Shouldn't there be a (width*height) in the final formula?
 
  • #4
You need to write B in terms of d and a because it's value depends on the distance from the wire, when you evaluate the resulting integral you don't get any b*a
 
  • #5
your final result is proportional to I*b*ln[(d+a)/a]
those "a"'s do NOT cancel! (integral of dxdr/r ...)
 
  • #6
Actually it's been awhile since I've done any problems like this, but it seems to me that it would, instead of re-writing B, be better to make a change of variables, from dA to dx where x is the distance from the wire then integrate "toward" the wire.
 
  • #7
I don't see how to do that. It doesn't seem like I need to do that because I am integrating over a. That should take care of B in terms of d and a. But what happens to dA in Ampere's Law. Does it become just A because it is constant or does something else happen to it?
 
  • #8
Oh, I see, I think. Are you saying that dA is equal to b*dx, where dx is the height of the loop?
 
  • #9
you are NOT integrating over "a"!
you're integrating dr, from d to (d+a).
"a" is a constant parameter, labeled in the diagram,
while "r" is the variable of integration.

The functional form of Ampere's Law, in cylindrically symmetric situations,
would usually be written in terms of radial distance "r":
proportional to dxdr/r , where x is the traditional horizontal direction.
 
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  • #10
Right. I'm saying that it seems to me that the area of the loop (b*a) should be in the final formula, but it is not. Only the width of the loop (b) is. Why?
 
  • #11
do you know how to integrate dr/r?

hint: do you remember the derivitive of ln[r]?
 
  • #12
Ya, that's pretty easy integral. But I guess you don't understand my question. Oh well, thanks for try'in guy.

Later
 
  • #13
the "a" is there, inside of the logarithm.

integral (I dA / r) = I * integral( dx dr / r) = I X integral ( dr / r)
where "X" has to be evaluated at "x" and at "x+b" .

Since one of the dimensions of dA is in the denominator,
why aren't you surprized that that dimension DOES show up in the end,
rather than cancelling out.
 
  • #14
I'm doing a similar question to this. From lightgrav's method, I got the right answer. But I don't understand why we seem to be doing a line integral instead of a surface integral. How did you change from integrating dA (surface integral) to dr (line integral)?
 

1. What is flux through a square loop?

Flux through a square loop is a measure of the flow of an electric or magnetic field through a square-shaped loop of wire. It is a fundamental concept in electromagnetism and is used to describe the strength of the field passing through the loop.

2. How is flux through a square loop calculated?

The flux through a square loop is calculated by multiplying the strength of the electric or magnetic field by the area of the loop and the cosine of the angle between the field and the normal vector of the loop. This can be expressed mathematically as Φ = B*A*cos(θ).

3. What factors affect the flux through a square loop?

The flux through a square loop can be affected by several factors, such as the strength of the electric or magnetic field, the size and orientation of the loop, and the angle between the field and the normal vector of the loop. The material of the loop can also affect the flux, as some materials may have higher permeability, making them more susceptible to magnetic fields.

4. What is the unit of flux through a square loop?

The unit of flux through a square loop is weber (Wb) in the International System of Units (SI). It can also be expressed in units of tesla (T) multiplied by square meters (m²).

5. How is flux through a square loop used in practical applications?

The concept of flux through a square loop is used in many practical applications, such as in electric generators and motors, where it is necessary to control the strength of the magnetic field passing through a loop of wire to generate or convert electrical energy. It is also used in magnetic sensors and detectors, as well as in various engineering and scientific experiments involving electromagnetic fields.

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