The Law of Biot and Savart again

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The discussion centers on the Biot-Savart law and its application to calculating the magnetic field and vector potential of a current loop. The vector potential formula is given as A(r) = (μ₀ I / 4π) ∮ C dr' / |r - r'|, emphasizing that the curve C must be closed to avoid divergence. Participants also explore the implications of having multiple non-linking current loops and the potential for a divergence-free field in such scenarios. The conversation touches on the need for integral formulas for vector potentials in complex magnetic field configurations, such as those confined within a closed tube. Overall, the thread delves into the mathematical intricacies of magnetic fields and their vector potentials.
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The magnetic field of a steady current in a loop is given by the Biot and savart integral which is

1/4pi Integral[((x-y)/|x-y|^3) x dy] = B(x)

What is the corresponding formula for the vector potential?
 
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Although i can't decrypt the formula you have stated, i give it a try


<br /> \vec A(\vec r) = \frac{\mu_0 I}{4\pi} \oint \limits_{\mathcal{C}} \mathrm d\vec r^{\, \prime} \, \frac{1}{|\vec r - \vec r^{\, \prime}|}<br />

The curve \mathcal{C} is parameterized through \vec r^{ \, \, \prime}(t) !

It is necessary to mention, that the curve \mathcal{C} must be closed, otherwise the integral diverges!


Best regards...
 
saunderson said:
Although i can't decrypt the formula you have stated, i give it a try


<br /> \vec A(\vec r) = \frac{\mu_0 I}{4\pi} \oint \limits_{\mathcal{C}} \mathrm d\vec r^{\, \prime} \, \frac{1}{|\vec r - \vec r^{\, \prime}|}<br />

The curve \mathcal{C} is parameterized through \vec r^{ \, \, \prime}(t) !

It is necessary to mention, that the curve \mathcal{C} must be closed, otherwise the integral diverges!


Best regards...

thanks I will try to prove it works.

BTW: how do you do the math notation?
 
saunderson said:
Although i can't decrypt the formula you have stated, i give it a try


<br /> \vec A(\vec r) = \frac{\mu_0 I}{4\pi} \oint \limits_{\mathcal{C}} \mathrm d\vec r^{\, \prime} \, \frac{1}{|\vec r - \vec r^{\, \prime}|}<br />

The curve \mathcal{C} is parameterized through \vec r^{ \, \, \prime}(t) !

It is necessary to mention, that the curve \mathcal{C} must be closed, otherwise the integral diverges!Best regards...

Thanks that works.
What about if you have an arbitrary divergence free field defined in space minus possibly a finite number of loops?

For instance if I have two magnetic fields generated by two non-linking current loops their cross product is divergence free. If there an integral formula for the vector potential of the cross product?

Or - suppose the magnetic field is confined to the interior of a closed tube as in a magnetic filament.
 
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