Biot-Savart law My proof includes a negative sign

AI Thread Summary
The discussion revolves around the application of the Biot-Savart law to calculate the magnetic field at a point due to a square coil of wire. A participant encounters a negative sign in their solution, leading to confusion about the correctness of the given answer. They debate the choice of the differential length element, dl, and its direction, concluding that the integration limits should be from negative to positive a when using dl = dy. Ultimately, it is emphasized that the integration must respect the defined limits and that the direction of current is inherently managed by the choice of dl. The conversation highlights the importance of careful integration and adherence to the mathematical framework of the Biot-Savart law.
etothey1
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Homework Statement


A square coil of wire of side length 2a lies in the yz plane. A current I flows through the loop. The x-axis is defined such that it passes through the centre of the loop, with the loop boundaries being at z±a and y=±a respectively. Given that I flows in a direction anticlockwise with respect to the x-axis use the Biot Savart law to show that the B field at a point P along the positive x-axis due to the side of the loot at z=a is given by

B=\frac{I*u_{0}}{4* pi}∫(x\hat{k} + a\hat{i})dy/(y^{2} + a^{2} + x^{2})^{3/2} . This integral goes from negative to positive a.









Homework Equations


Biot Savart law, written in the solution. Note that dl has direction, same as current.



The Attempt at a Solution



Shown in the image

Issues:
My issue is that I get a negative sign in my solution, thus I believe that the negative sign was forgotten in the given answer of the problem.

http://img263.imageshack.us/img263/8818/solutiondj.jpg

Homework Statement

 
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I should remove the sign in dl=-dly, that will fix things, and also makes the magnetic field at point p be appropriate.
 
The given answer is correct:
r = x i - y j - a k
dl = -dy j
dl x r = a*dl i + x*dl k
dB = (μ0/4π)(a*dy i + x*dy k)
B = (μ0/4π)(a i + x k){∫dy/r^3}

where r^2 = x^2 + a^2 + z^2
and the integration is from y= -a to y = +a.

Note that in the answer the term a i + x k can be taken outside the integration sign.
 
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Your error was integrating from a to -a. Why did you do that?
 
rude man said:
Your error was integrating from a to -a. Why did you do that?
Hi! thanks for you're response. I integrated from a to negative a because it is a line integral and the curve is from a to negative a (because the current flows in that direction)
therefore I integrated along the direction of the current ( also because my problem is set up to be integrated like that)

However, i believe the problem lies in dl and the integration. If i choose that dl=-dyj then i have to integrate form negative
a to a to get it correct. But if I choose dl=dyj and i integrate it from negative a to a then it becomes correct.

I read about the current density which the biot-savart law comes from. The current density has a direction along the current,
by changing a volume integral into line integral, dl and changing current density to current, dl is in direction of current density, current, which gives me -dyj.
but i believe i should have dyj and the direction of the integral takes care of the sign for dyj.
 
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etothey1 said:
Hi! thanks for you're response. I integrated from a to negative a because it is a line integral and the curve is from a to negative a (because the current flows in that direction)
therefore I integrated along the direction of the current ( also because my problem is set up to be integrated like that)

However, i believe the problem lies in dl and the integration. If i choose that dl=-dyj then i have to integrate form negative
a to a to get it correct. But if I choose dl=dyj and i integrate it from negative a to a then it becomes correct.

I read about the current density which the biot-savart law comes from. The current density has a direction along the current,
by changing a volume integral into line integral, dl and changing current density to current, dl is in direction of current density, current, which gives me -dyj.
but i believe i should have dyj and the direction of the integral takes care of the sign for dyj.

Well, you have to be careful with your integration.

You're integrating wrt y, not l. So you have to integrate from - to +. There is no choice.

The direction of current is taken care of when you defined dl = - dy. From then on it's strictly math; the physics is left in the dust!

Anyway - all's well ...
 
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