A magnetic induction question

[schattenjaeger]

Homework Statement

A square of edge a lies in the xy plane with the origin at its center. Find the value of the magnetic induction at any point on the z axis when a current I' circulates around the square.

Homework Equations

B=u/4pi * lineint[(I'*ds' x R)/R^3)]

The Attempt at a Solution

I guess you've got the 4 different sides of the square, I assumed by "circulating" it meant it goes around, so opposite sides the current is going opposite directions. The way I did this, let's say it's going around counter-clockwise, for the side on the right running parallel to the x-axis with the current traveling in the positive x direction...
(capital letters = unit vectors)
ds'=dx'X
R=-x'X-a/2Y+zZ
ds' X R = -(zY+a/2*Z)dx'

R^3 = (x'^2+a^2/4+z^2)^(3/2)

so when I integrate that, I'm just going to say k=uI'/4pi and...
-k{4a(zY+a/2Z)sqrt(2)/[sqrt(a^2+2z^2)*(a^2+4z^2)]}

is my approach even right? I carefully repeated that for all 4 sides and added them, but didn't get the right answer(I don't have the "answer" per se, rather just the value in the middle of the square, which I didn't get right)

 

[OlderDan]

Homework Statement

A square of edge a lies in the xy plane with the origin at its center. Find the value of the magnetic induction at any point on the z axis when a current I' circulates around the square.

Homework Equations

B=u/4pi * lineint[(I'*ds' x R)/R^3)]

The Attempt at a Solution

I guess you've got the 4 different sides of the square, I assumed by "circulating" it meant it goes around, so opposite sides the current is going opposite directions. The way I did this, let's say it's going around counter-clockwise, for the side on the right running parallel to the x-axis with the current traveling in the positive x direction...
(capital letters = unit vectors)
ds'=dx'X
R=-x'X-a/2Y+zZ
ds' X R = -(zY+a/2*Z)dx'

R^3 = (x'^2+a^2/4+z^2)^(3/2)

so when I integrate that, I'm just going to say k=uI'/4pi and...
-k{4a(zY+a/2Z)sqrt(2)/[sqrt(a^2+2z^2)*(a^2+4z^2)]}

is my approach even right? I carefully repeated that for all 4 sides and added them, but didn't get the right answer(I don't have the "answer" per se, rather just the value in the middle of the square, which I didn't get right)

You are starting in the right place. I'm having a bit of trouble sorting out your substitutions, so I'm going to leave the work up to you. I did a similar problem recently, so I know what the integral looks like. You should have the equivalent of

B = \frac{{\mu _o IR}}{{4\pi }}\int_{ - a/2}^{a/2} {\frac{{ds}}{{\left( {R^2 + s^2 } \right)^{3/2} }}}

for each wire, where my R is the distance from the center of the wire to the point on the z axis where B is being calculated. R is independent of s. The integral gives you

B = \frac{{\mu _o I}}{{4\pi }} {\frac{a}{{R \sqrt {R^2 + \left( {a/2} \right)^2 } }}}