Magnetic field of long parallel straight wires

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sweetdion
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Homework Statement


Two infinitely long parallel straight wires carry currents in the +- z direction as shwn in the figure below. Each wire is located on the x-axis a distance of a from the origin.

a) Determine B as a function of y along the line x=0
b) Sketch a graph of B vs. y along the line x=0, including all values of y.


Homework Equations


B=u0I/2PiR


The Attempt at a Solution


for part a i get u0I/PiR
for part b as y increases, B decreases because they're inversely related?
 
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sweetdion said:

Homework Equations


B=u0I/2PiR


The Attempt at a Solution


for part a i get u0I/PiR
Not quite. It looks like you're calculating B at the origin (x=y=0) only, but you need to find B all along the y-axis, as a function of y. So you should get some expression that has y in it.

for part b as y increases, B decreases because they're inversely related?
For this, you'll need to get part (a) done correctly first, then you can sketch the function you get for (a).
 
well if one current is flowing in the +z direction and one is flowing in the -z direction the magnetic field is the following:

B=μI/2∏ {[(-y/(x-a)²+y²)+(y/(x+a)²+y²)]ihat + [(x-a/(x-a)²+y²)+(x+a/(x+a)²+y²)]jhat

If we let x equal 0 and reduce, we get B = μIa/∏(a²+y²)

I just know this from the answer in the back of my book, and I have no idea how they get it. :(

From this equation you can deduce the graph. As y goes to infinity it seems as if B goes to zero. Correct?
 
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sweetdion said:
well if one current is flowing in the +z direction and one is flowing in the -z direction the magnetic field is the following:

B=μI/2∏ {[(-y/(x-a)²+y²)+(y/(x+a)²+y²)]ihat + [(x-a/(x-a)²+y²)+(x+a/(x+a)²+y²)]jhat

If we let x equal 0 and reduce, we get B = μIa/∏(a²+y²)

I just know this from the answer in the back of my book, and I have no idea how they get it. :(
How about B due to a single wire, do you know that?
From this equation you can deduce the graph. As y goes to infinity it seems as if B goes to zero. Correct?
Yes.