How Does the Biot-Savart Law Apply to a Square Loop in a Plane?

AI Thread Summary
The discussion focuses on applying the Biot-Savart law to a square loop of wire carrying a steady current, specifically to show that the magnetic field at any point in the plane but outside the loop is perpendicular to that plane. Participants express uncertainty about calculating line integrals involving vector products and suggest using symmetry to simplify the problem. The calculation of the magnetic field at the center of the loop is addressed, with one participant proposing a method involving integration over the loop's sides. Confirmation is given that the proposed method is correct, despite the initial doubts about the integral's complexity. The conversation highlights the challenges faced in understanding and applying the Biot-Savart law in this context.
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Homework Statement



A square loop of wire C, with side length 2a lies in a plane P and carries a steady electic current I. By using the Biot-Savart law show that the magnetic field B(r) at any point r in P but not in C is perpendicular to the plane P.

Calculate the magnetic field at the centre of the wire.


The Attempt at a Solution



Dotting with a would seem the normal way to do the first bit but I don't think this works so I think it must be to do with the vector product always being 0.

I don't recall ever being taught how to calculate a line integral with a vector product inside it so really not sure where to begin. I expect there is a way to use symmetry to simpliify it.

If my guess is correct then the component of one side of the square (x=a) is going to be (assuming centre is origin):

(u*I/4pi)*(0,0, Integral -a and a of (a/((a^2+(y')^2)^3/2)dy')

I think all 4 sides can be summed to make it 4 times that. I don't really have any idea what I'm doing though so if someone could explain it would be appreciated.
 
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There is nothing to calculate. Since the law says that: \vec{dB} = C\vec{dl} \times \hat{r}, where C is a constant that contains all the relevant factors, we can see that dB will always be perpnedicular to both dl and r. In this case they both lie on the plane P, so dB is always normal to P.
 
Thanks a lot, what about calculating the magnetic field at the centre?
 
sebb1e said:
If my guess is correct then the component of one side of the square (x=a) is going to be (assuming centre is origin):

(u*I/4pi)*(0,0, Integral -a and a of (a/((a^2+(y')^2)^3/2)dy')

I think all 4 sides can be summed to make it 4 times that. I don't really have any idea what I'm doing though so if someone could explain it would be appreciated.

Yes, that's exactly how you calculate the field at the center. Why do you suspect it isn't correct?
 
No reason, just never been directly taught how to do an integral of this form, this just seemed the logical thing to do. Frustrating as this was an exam question last week and I didn't bother to try it as assumed I wouldn't be able to :(
 

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