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

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SUMMARY

The discussion focuses on applying the Biot-Savart Law to calculate the magnetic field generated by a square loop of wire carrying a steady current I. The participants confirm that the magnetic field B(r) at any point r in the plane P, but not within the loop, is perpendicular to the plane. The calculation of the magnetic field at the center of the loop involves integrating the contributions from each side of the square, utilizing symmetry and the vector product inherent in the Biot-Savart Law. The formula used includes the constant C and integrates over the dimensions of the loop.

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  • Understanding of the Biot-Savart Law
  • Familiarity with vector calculus, particularly line integrals
  • Knowledge of magnetic fields generated by current-carrying conductors
  • Basic principles of electromagnetism
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  • Study the derivation and applications of the Biot-Savart Law in different geometries
  • Learn how to perform line integrals involving vector products
  • Explore symmetry arguments in electromagnetic field calculations
  • Investigate the relationship between current, magnetic fields, and forces in electromagnetism
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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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