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Loop with radius R and long straight segments

$\int\frac1rdl$ where $r$ is the position vector from each element $dl$ to the center of the loop with radius $R$.

Then I need to take the Curl of that to calculate the magnetic field. I'm having trouble setting up the integral to get anything meaningful.

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Your first sentence seems to be missing its first part. – msh210 Dec 7 '11 at 21:34

The vector potential generated by an infinitesimal current element is proportional to $\mathrm d\vec l/r$, where $\mathrm d\vec l$ is the infinitesimal displacement along the wire and $r$ is the distance to the point at which the vector potential is evaluated. Note that you got this the wrong way around in the question; you described $r$ and not $\mathrm d\vec l$ as a vector.

The magnetic field is the curl of the vector potential. Since only $r$ and not $\mathrm d\vec l$ depends on the point at which the vector potential is evaluated, the magnetic field generated by the infinitesimal current element is proportional to

$$\vec\nabla\times\frac{\mathrm d\vec l}{r}=\left(\vec\nabla\frac{1}{r}\right)\times\mathrm d\vec l=\frac{\vec r}{r^3}\times\mathrm d\vec l\;.$$

If you want to calculate the magnetic field at the centre of a circular current loop, the vector product $\vec r\times\mathrm d\vec l$ is the same everywhere and points out of the plane of the loop, since $\vec r$ and $\mathrm d\vec l$ are always at right angles to each other, so the magnetic field points out of the plane of the loop and its magnitude is proportional to $\frac R{R^3}(2\pi R)=\frac{2\pi}R$.

If you want to read up on this in more detail, this set of notes might be helpful.

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I see what you're saying, but say I wanted to calculate the vector potential on its own how would I set up this integral? – vivaelche05 Dec 9 '11 at 15:59
@vivaelche05: How do you mean, "on its own"? In case you just mean calculating the vector potential at the centre of the loop: It's zero by symmetry. – joriki Dec 9 '11 at 16:53
I don't follow, if the vector potential at the center is 0 its curl (the magnetic field) must also be zero, right? well, see the question is to use Biot-Savart's law to calculate the Magnetic field at the center of the loop then calculate the integral for the vector potential then take its curl, then to compare the two results – vivaelche05 Dec 9 '11 at 17:22
@vivaelche05: That's a fundamental misunderstanding. Just like the derivative of a function isn't zero where the function is zero, the curl of a vector field isn't zero where the vector field is zero. You can't get the magnetic field by calculating the vector potential only at the centre and then taking the curl, since taking the curl requires knowing the vector potential in a neighbourhood of the point, not just the point itself. – joriki Dec 9 '11 at 17:40
I don't quite follow your point about knowing the vector potential in a neighborhood of the point – vivaelche05 Dec 9 '11 at 17:46

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