Magnetic Fields from Two Infinite Sheets of Current Problem

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Gee Wiz
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Homework Statement



Two infinite sheets of current flow parallel to the y-z plane as shown. The sheets are equally spaced from the origin by xo = 4.2 cm. Each sheet consists of an infinite array of wires with a density n = 16 wires/cm. Each wire in the left sheet carries a current I1 = 2.3 A in the negative z-direction. Each wire in the right sheet carries a current I2 = 4.2 A in the positive z-direction.

What is
∫B *dl
where the integral is taken along the dotted line shown, from a to b. H is 11.7cm.

Homework Equations



∫B *dl (it's a dot product)

The Attempt at a Solution



I'm not really sure how to start this problem. I thought about finding the current enclosed, because that is what i did for a similar problem, but there isn't any current enclosed. Would both sheets of current be affecting this segment? I think so, and i know they would be acting in opposite directions (using the right hand rule at those locations). I also thought that i could ignore the horizontal distance between a and b since i thought that would be perpendicular to dl.
https://www.smartphysics.com/Content/Media/Images/EM/15/h15_sheets.png
 
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Of course there is current enclosed, just look at those x inside the trapezoid! But you are correct that this method won't work. Because the other sheet is there, you won't be able to find B by a symmetry argument.

However, if there was only one sheet then you may be able to do it.

I suggest you use something called superposition. In other words, B = B from sheet 1 + B from sheet 2. So first figure the B field from just one sheet.
 
But, this question isn't talking about the trapezoid. It only wants the segment from a to b. I tried doing 4.2(current from right sheet) minus 2.3(the current from the left sheet) times 16*11.7, to get a current enclosed. But the enclosed terminology didn't seem to apply here.
 
Also, I think that all the contributions to the segment would be vertical components since the sheets are infinite.
 
Like I wrote before, first consider only one sheet and find the B field everywhere outside that single sheet.
 
Okay, so then would i use Iu/(2pi*r)? to find the b field at that distance away. But then if i do that what do i select as my r, since a and b are not at an equal distance. I'm thinking a, because if i draw a triangle the vertical component is at h, which is the same distance as a is from the sheet.
 
Try to find the B field of a single sheet. This is due to a bunch of currents at different distances. You could try and use the Biot–Savart law, but there is an easier way. You can just assume that the B field is parallel to the sheet everywhere (and perpendicular to the current).
 
Okay so i guess I'm blanking on how to find the b-field for a single sheet. I thought my only two equations to find the b field were biot-savart and the integral of b*dl
 
So, try the second one, ∫bdl! I suggest you try a rectangular path.
 
MisterX said:
Of course there is current enclosed, just look at those x inside the trapezoid! But you are correct that this method won't work. Because the other sheet is there, you won't be able to find B by a symmetry argument.

The line integral is independent of whatever current lies outside the loop. The answer is μI where I is the current enclosed.
please refer to Fundamentals of Physics by Halliday and Resnick
 
So, i tried to do make the rectangle with h as my height for both sheets. I did 11.7*16 to get the number of wires enclosed times 2.3 and 4.2 for the respected sheets. I subtracted the two values and then multiplied by u, but this wasn't correct. So I am thinking that i didn't make the correct rectangle..
 
Gee Wiz said:
I did 11.7*16 to get the number of wires enclosed times 2.3 and 4.2 for the respected sheets.

you don't need to include the current [itex]I_2[/itex] as it is not within the loop. Is the answer [itex]5.4\times10^{-4}[/itex]?
 
That was the answer to an earlier question, when it asked for the B-field within the trapezoid, but this one is only asking for the field along the line segment. As far as i can tell, it doesn't really include and current.