# Homework Help: Flux through a square loop

1. Nov 10, 2005

### WhoWeAre

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?

2. Nov 10, 2005

### lightgrav

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?

Last edited: Nov 10, 2005
3. Nov 10, 2005

### WhoWeAre

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. Nov 10, 2005

### asrodan

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. Nov 10, 2005

### lightgrav

your final result is proportional to I*b*ln[(d+a)/a]
those "a"'s do NOT cancel! (integral of dxdr/r ...)

6. Nov 10, 2005

### asrodan

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. Nov 10, 2005

### WhoWeAre

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. Nov 10, 2005

### WhoWeAre

Oh, I see, I think. Are you saying that dA is equal to b*dx, where dx is the height of the loop?

9. Nov 10, 2005

### lightgrav

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.

Last edited: Nov 10, 2005
10. Nov 10, 2005

### WhoWeAre

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. Nov 10, 2005

### lightgrav

do you know how to integrate dr/r?

hint: do you remember the derivitive of ln[r]?

12. Nov 10, 2005

### WhoWeAre

Ya, that's pretty easy integral. But I guess you don't understand my question. Oh well, thanks for try'in guy.

Later

13. Nov 10, 2005

### lightgrav

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. May 17, 2010

### v_pino

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)?