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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.2011-12-07

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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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    @vivaelche05: Derivatives are defined by a limiting process that refers to function values in the neighbourhood of a point. Knowing the function value only at a point tells you nothing about the derivative at that point. I'm wondering whether you understand that the curl is a certain form of derivative?2011-12-09