Given a current density distribution $\mathbf J(\mathbf x)$ inside a closed bounded region $\Omega$, the magnetic field at any point $\mathbf y$ outside of $\Omega$ can be expressed as $$ \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\mathbf J(\mathbf x)\times\nabla_{\mathbf x}\frac{1}{|\mathbf x-\mathbf y|}d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\left[\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)-\nabla_{\mathbf x}\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)\right]d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x-\frac{\mu_0}{4\pi}\int_{\partial\Omega}\mathbf n(\mathbf x)\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)d^2 S(\mathbf x) \end{aligned}$$ where $\partial\Omega$ is the boundary of $\Omega$, $n(\mathbf x)$ is the unit normal of $\partial \Omega$ and $S(\mathbf x)$ is the area of the surface element. Now, if the current density $\mathbf J(\mathbf x)$ is zero at the boundary $\partial\Omega$ (this can be achieved by slightly enlarging $\Omega$ if $\mathbf J(\mathbf x)$ is not zero at $\partial\Omega$) we can then drop the second term on the last line. Now we simply have $$ \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x \end{aligned}.$$

If the current density $\mathbf J(\mathbf x)$ is continuous and differentiable, the above conclusion should be correct. However, $\mathbf J(\mathbf x)$ might not be continuous in $\Omega$, e.g., infinite thin coils inside $\Omega$ carrying electrical current. Is the above derivation correct for $\mathbf J(\mathbf x)$ containing delta functions? What kind of singularities in $\mathbf J(\mathbf x)$ is permitted?

  • $\begingroup$ I think that the above is always true, simply because the the definition of the derivative of a distribution (such as a delta-function or a step function, which is how we describe the current configurations you're talking about) is done via a similar formula. The Mathworld article on distributions (aka "generalized functions") might be worth a look on your part. $\endgroup$ – Michael Seifert Oct 6 '15 at 23:47
  • $\begingroup$ Thanks for bringing the reference. As you said, it is correct even if $\mathbf J$ contains delta functions, since it can be verified that $\int_\Omega f(\mathbf x)\nabla\delta(\mathbf x-\mathbf x_0)d^3\mathbf x=-\nabla f(\mathbf x)|_{\mathbf x=\mathbf x_0}$ for $\mathbf x_0$ in the interior of $\Omega$ for any differentiable function $f(\mathbf x)$. $\endgroup$ – Jasper Oct 7 '15 at 18:32

Interesting observation. As you have stated, the second equation only valid when the boundary contains all the current distribution inside. But is this what you are asking? You should open this question for objections as well.

  • $\begingroup$ Thanks for your comments. I have updated my question to make a little clearer. It is easy to make the current at the boundary vanished, but my concern is that there might be requirements on the current inside the boundary. $\endgroup$ – Jasper Oct 6 '15 at 21:33
  • $\begingroup$ Can you give an example of singularity of current? The current is considered as a result of charge movement. As long as you can still talking about the continuous movements of charges, $\mathbf{J}$ may always be treated as continuous. Unless you are thinking of quantum processes when sudden jumps happen at the microscopic level and quantum electrodynamics should serve your purpose, otherwise I think you are safe. $\endgroup$ – Xiaodong Qi Oct 6 '15 at 21:44
  • $\begingroup$ Sure. Let's assume $\mathbf J(\mathbf x)$ is a segment of line current, e.g., $\mathbf J(\mathbf x)=I_c\int_{\tau_1}^{\tau_2}\delta(\mathbf x-\mathbf x^\prime(\tau))\frac{d\mathbf x^\prime(\tau)}{d\tau}d\tau$, where $\delta(\mathbf x-\mathbf x^\prime)$ is the Dirac delta function, $I_c$ is the amplitude of the current, $\mathbf x^\prime(\tau)\in c\in \mathbb R^3$ is the parametric representation of the line segment $c$ with parameter $\tau\in[\tau_1,\tau_2]$. $\endgroup$ – Jasper Oct 6 '15 at 21:50
  • $\begingroup$ In your case, the integral $\int_{t_1}^{t_2}\delta(x-x'(\tau))\frac{dx'}{d\tau}d\tau=\int_{x'(t_1)}^{x'(t_2)}\delta(x-x'(\tau))dx'=\mathrm{constant}$. Is that correct? $\endgroup$ – Xiaodong Qi Oct 6 '15 at 22:02
  • $\begingroup$ No, I don't think it is a constant. How do you perform a integral that has starting and ending points of vectors? What is the integral domain? $\endgroup$ – Jasper Oct 6 '15 at 22:40

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