Undergrad Deriving Curl of B from Biot-Savart Law & Vector Identity

[georg gill]

$$\nabla \times B(r)=\frac{\mu _0}{4\pi} \int \nabla \times J(r') \times \frac{ (r-r')}{|r-r|^3}dV'$$

using the vector identity:

$$\nabla \times (A \times B) = (B \cdot \nabla)A - B(\nabla \cdot A) - (A \cdot \nabla )B + A(\nabla \cdot B)$$

$A=J$ and $B=\frac{r-r'}{|r-r'|^3}$

since current is constant
$$\nabla \cdot J=0$$

And since current is constant we also obtain

$$(\frac{r-r'}{|r-r'|^3} \cdot \nabla)J=0$$

After some calculation one also obtains

$$\nabla \cdot \frac{r-r'}{|r-r'|^3}=0$$

All of the three relations above are undefied when $r-r'=0$.

So I would believe that $(J \cdot \nabla )\frac{r-r'}{|r-r'|^3}$ would be 0 outside $r-r'=0$. But my calculations did obtain that it was not 0 outside $r-r'=0$. In order to obtain that $\nabla \times B=\mu_0 J$ in other derivations one ends up with an integral that is 0 everywhere except when you integrate over the point $r-r'=0$. Can someone show that $(J \cdot \nabla )\frac{r-r'}{|r-r'|^3}=0$? Or explain why this fails? I am not looking for another proof for $\nabla \times B=\mu_0 J$. I am looking for an explantion for this problem described above in the calculation.

 

[Charles Link]

I think for $\nabla \cdot B$ you get $4 \pi \delta(r-r' )$, because $B$ represents the $E$ field from a point charge at $r=r'$.

 

[Charles Link]

and a follow-on: I think $J(r') \cdot \nabla B$ integrates to zero over a large volume because $\nabla (J \cdot B)=J(r') \cdot \nabla B$ once you set all the other terms to zero in the identity. Note $\nabla \times B=0$ appears in one of these terms. Meanwhile the vector operator operates on the unprimed coordinates, and not on the prime coordinates of $J(r')$.
Note: There is a vector identity that $\int \nabla F \, d^3x=\int F \, dS$, which will be zero over a distant boundary that encloses the system.
Note also : $\nabla (a \cdot b)=a \cdot \nabla b+b \cdot \nabla a + a \times \nabla \times b + b \times \nabla \times a$.

 

[hutchphd]

$$\nabla \times B(r)=\frac{\mu _0}{4\pi} \int \nabla \times J(r') \times \frac{ (r-r')}{|r-r|^3}dV'$$

using the vector identity:

$$\nabla \times (A \times B) = (B \cdot \nabla)A - B(\nabla \cdot A) - (A \cdot \nabla )B + A(\nabla \cdot B)$$

$A=J$ and $B=\frac{r-r'}{|r-r'|^3}$

since current is constant
$$\nabla \cdot J=0$$

And since current is constant we also obtain

$$(\frac{r-r'}{|r-r'|^3} \cdot \nabla)J=0$$

After some calculation one also obtains

$$\nabla \cdot \frac{r-r'}{|r-r'|^3}=0$$

All of the three relations above are undefied when $r-r'=0$.

So I would believe that $(J \cdot \nabla )\frac{r-r'}{|r-r'|^3}$ would be 0 outside $r-r'=0$. But my calculations did obtain that it was not 0 outside $r-r'=0$. In order to obtain that $\nabla \times B=\mu_0 J$ in other derivations one ends up with an integral that is 0 everywhere except when you integrate over the point $r-r'=0$. Can someone show that $(J \cdot \nabla )\frac{r-r'}{|r-r'|^3}=0$? Or explain why this fails? I am not looking for another proof for $\nabla \times B=\mu_0 J$. I am looking for an explantion for this problem described above in the calculation.

Sorry but I'm lost. Where does the initial equation come from?
.

 

[Charles Link]

Sorry but I'm lost. Where does the initial equation come from?
.

Without the curl operator, the OP has Biot-Savart in integral form. The OP applied the curl operator to both sides. The task is then to show Maxwell's expression for $\nabla \times B$ , without the $\mu_o \epsilon_o \dot{E}$ term.

 

[Charles Link]

@georg gill Please see post 3=I think that solves it.
Without compete mathematical rigor, I think we can take $\nabla \times B =0$, because $B$ (as defined in the OP on the fourth line=slightly clumsy to use $B$ for two different things in the same problem) is basically an electrostatic type inverse square law field.

 

[georg gill]

and a follow-on: I think $J(r') \cdot \nabla B$ integrates to zero over a large volume because $\nabla (J \cdot B)=J(r') \cdot \nabla B$ once you set all the other terms to zero in the identity. Note $\nabla \times B=0$ appears in one of these terms. Meanwhile the vector operator operates on the unprimed coordinates, and not on the prime coordinates of $J(r')$.
Note: There is a vector identity that $\int \nabla F \, d^3x=\int F \, dS$, which will be zero over a distant boundary that encloses the system.
Note also : $\nabla (a \cdot b)=a \cdot \nabla b+b \cdot \nabla a + a \times \nabla \times b + b \times \nabla \times a$.

Sorry but how is

$\nabla (J \cdot \frac{r-r'}{|r-r'|^3})=(J(r') \cdot \nabla) \frac{r-r'}{|r-r'|^3}$?

$$(J(r') \cdot \nabla) \frac{r-r'}{|r-r'|^3}=(J_x\frac{\partial}{\partial x}+J_y\frac{\partial}{\partial y}+J_z \frac{\partial}{\partial z})\frac{(x-x')\textbf{i}+(y-y')\textbf{j}+(z-z')\textbf{k}}{((x-x')^2+(y-y')^2+(z-z')^2)^{\frac{3}{2}}}$$

While

$$\nabla (J \cdot \frac{r-r'}{|r-r'|^3})=(\frac{\partial}{\partial x}\textbf{i}+\frac{\partial}{\partial y}\textbf{j}+ \frac{\partial}{\partial z}\textbf{k})\frac{J_x(x-x')+J_y(y-y')+J_z(z-z')}{((x-x')^2+(y-y')^2+(z-z')^2)^{\frac{3}{2}}}$$

 

[Charles Link]

You integrate the gradient over the large volume=see post 3=the argument of the gradient then becomes a surface integral over that same volume. (Gauss' law for a gradient). The current density $J$ is assumed to occupy a finite region and will vanish on a very large surface. See https://math.stackexchange.com/ques...eorem-for-volume-integral-of-a-gradient-field

See also post 3 at the bottom for the gradient of a dot product. We have the $a \times \nabla \times b=0$ , because $\nabla \times b=0$. Then the two are the same.

 

[georg gill]

and a follow-on: I think $J(r') \cdot \nabla B$ integrates to zero over a large volume because $\nabla (J \cdot B)=J(r') \cdot \nabla B$ once you set all the other terms to zero in the identity. Note $\nabla \times B=0$ appears in one of these terms. Meanwhile the vector operator operates on the unprimed coordinates, and not on the prime coordinates of $J(r')$.
Note: There is a vector identity that $\int \nabla F \, d^3x=\int F \, dS$, which will be zero over a distant boundary that encloses the system.
Note also : $\nabla (a \cdot b)=a \cdot \nabla b+b \cdot \nabla a + a \times \nabla \times b + b \times \nabla \times a$.

How is the term $a \times \nabla \times b$ equal to 0 in

$\nabla (a \cdot b)=a \cdot \nabla b+b \cdot \nabla a + a \times \nabla \times b + b \times \nabla \times a$
?

 

[Charles Link]

$b=B_{second \, one}=\frac{r-r'}{|r-r'|^3}$ is a vector. Because it has the conservative field inverse square law form, we know $\nabla \times b=0$, without a lengthy calculation.

 

[Charles Link]

The inverse square vector will not vanish with a distant surface integral, (the area increases with the square of the distance), but we can assume the current density $J$ will vanish over a remote surface.

 

[hutchphd]

Well the integral over any closed surface has to vanish (in steady state) and that is sufficient I think.

 

[georg gill]

The inverse square vector will not vanish with a distant surface integral, (the area increases with the square of the distance), but we can assume the current density $J$ will vanish over a remote surface.

Yes. Think I got it. 0 due to dot product with the vector J that does not go out of the surface. Thanks!

 

[georg gill]

Sorry but if r-r'=0 within the current integral that would affect

$\nabla (J \cdot \frac{r-r'}{|r-r'|^3})$?

If one uses l'hospital's on the last element one obtains for the x-element:

$$lim\frac{(x-x')}{((x-x')^2+(y-y')^2+(z-z')^2)^{\frac{3}{2}}}=\frac{1}{3(x-x')((x-x')^2+(y-y')^2+(z-z')^2)^{\frac{1}{2}}}$$
Which goes to infinty when r-r'=0. Same for y and z comp.

How can one the integrate over dV' with $\nabla (J \cdot \frac{r-r'}{|r-r'|^3})$ and obtain that the integral is 0. If one spot is undefined within the integral

Note: There is a vector identity that $\int \nabla F \, d^3x=\int F \, dS$, which will be zero over a distant boundary that encloses the system.

And how can one proof that this vector identity quoted above is valid when there is an undefined spot in the integral?

 

[vanhees71]

I think the problem is already in #1, because you must be careful to distinguish between $\vec{x}$ and the integration variables $\vec{x}'$. So what you want to calculate is $\vec{\nabla} \times \vec{B}(\vec{x})$, where $\vec{\nabla}$ acts on the argument $\vec{x}$. It doesn't act on the integration variable $\vec{x}'$. So what you in fact have is
$$\vec{\nabla} \times \vec{B} = \frac{\mu_0}{4 \pi} \int_{\mathbb{R}} \mathrm{d}^3 x' \vec{\nabla} \left [\vec{J}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|} \right].$$
So $\vec{J}(\vec{x}')$ has to be treated as constant when taking he curl. For that we need with a constant vector $\vec{a}$
$$\left \{ \vec{\nabla} \times [\vec{a} \times \vec{V}(\vec{x})] \right ]_j = \epsilon_{jkl} \partial_k \epsilon_{lmn} a_m V_n=(\delta_{jm} \delta_{kn}-\delta_{jn} \delta_{km}) a_m \partial_k V_n = a_j \vec{\nabla} \cdot \vec{V} - (\vec{a} \cdot \vec{\nabla}) V_j$$
or
$$\vec{\nabla} \times (\vec{a} \times \vec{V})=\vec{a} \vec{\nabla} \cdot \vec{V}-(\vec{a} \cdot \vec{\nabla}) \vec{V}.$$
Now apply this to the curl in the above integral with $\vec{V}=(\vec{x}-\vec{x}')/|\vec{x}-\vec{x}'|^3$
$$\vec{\nabla} \times [\vec{J}(\vec{x}') \times \vec{V}] = \vec{J}(\vec{x}') 4 \pi \delta^{(3)}(\vec{x}) - [\vec{J}(\vec{x}') \cdot \vec{\nabla}] \vec{V}$$.
The first expression comes from
$$\vec{\nabla} \cdot \frac{\vec{x}-\vec{x'}}{|\vec{x}-\vec{x}'|^3}=-\Delta \frac{1}{|\vec{x}-\vec{x}'|} = 4 \pi \delta^{(3)}(\vec{x}-\vec{x}').$$
For the 2nd term note that
$$J_k \partial_k V_j=-J_k \partial_k' V_j = -\partial_k'(J_k V_j)+V_j \partial_k' J_k = - \partial_k' (J_k V_j).$$
The latter follows from $\vec{\nabla}' \cdot \vec{J}(\vec{x}')=0$ (current conservation for a stationary current). The integral over the total divergence vanishes. So you only get the contribution for the first integral, from which Ampere's Law in differential form,
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{J}$$
results.

 

[Charles Link]

And how can one proof that this vector identity quoted above is valid when there is an undefined spot in the integral?

To prove it rigorously might be somewhat difficult. This problem contains a couple of places where the term being evaluated involves $r-r'$ in the denominator. e.g. $\nabla \cdot b=4 \pi \delta(r-r')$, (notice $\nabla \cdot b=0$ everywhere except $r=r'$), while $\nabla \times b=0$, including supposedly for $r-r'$. (Note: $b=\frac{r-r'}{|r-r'|^3}$ ).

Gauss' law (including the forms for the integral of a gradient and that of the integral of a curl) makes it possible to evaluate these expressions at $r=r'$, even when the function becomes undefined at $r=r'$.

For the case of $\int \nabla (a \cdot b) \, d^3x'$, again Gauss' law gives us an answer. There may be a more rigorous answer. Otherwise, the results of this problem seem to hinge on the different forms of Gauss' law.

Reading @vanhees71 post 15, I think even I have been careless with the unprimed and primes on this last gradient operator. It started off as an unprimed gradient operator, and would pick up a minus sign when changed to $\int \nabla' (J \cdot b ) \, d^3 x'$. The Gauss' law identity only holds if the gradient operates on the same coordinates as the volume integral. For my method to be valid, this last step of switching to $\nabla'$ must be included.

 

[vanhees71]

The proof of
$$\Delta 1/r=-4 \pi \delta^{(3)}(\vec{x})$$
isn't too difficult. Starting from
$$\Delta G=-\delta^{(3)}(\vec{x})$$
by symmetry you get $G=G(r)$, from which
$$G=A/r$$
Then you get
$$-1=\int_K \mathrm{d}^3x \Delta G=\int_{\partial K} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} G =-4 \pi A \;\Rightarrow \; A=1/(4 \pi).$$

 

[Charles Link]

Basically, I think what the above is doing, is to let $E=\frac{r-r'}{|r-r'|^3}$. You find $\nabla \cdot E=0$ except for $r=r'$ where it remains undefined. By Gauss' law, with a sphere centered around $r'$, $\int \nabla \cdot E \, d^3x=\int E \cdot dS=4 \pi$. Because $\nabla \cdot E=0$ everywhere else, we must have $\nabla \cdot E=4 \pi \delta^{(3)}(r-r')$ for the integral to always give $4 \pi$.

 

[georg gill]

Basically, I think what the above is doing, is to let $E=\frac{r-r'}{|r-r'|^3}$. You find $\nabla \cdot E=0$ except for $r=r'$ where it remains undefined. By Gauss' law, with a sphere centered around $r'$, $\int \nabla \cdot E \, d^3x=\int E \cdot dS=4 \pi$. Because $\nabla \cdot E=0$ everywhere else, we must have $\nabla \cdot E=4 \pi \delta^{(3)}(r-r')$ for the integral to always give $4 \pi$.

I guess for gauss law I would unerstand that you can calculate the E-field inside a charged sphere as
$$E=\frac{Q\textbf{r}}{4\pi\varepsilon_0 R^3}$$

by using unit vector and obtaining the divergence inside the charged sphere one obtains:

$$\nabla \cdot E=\nabla \cdot \frac{Qr}{4\pi\varepsilon_0 R^3}[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=3\frac{Q}{4\pi\varepsilon_0 R^3}=\frac{\rho}{\varepsilon_0}$$

Without the result above for the E-field inside a sphere I don't understad how one could use the divergence theorem as the derivative must be continuous in the divergence theorem. The result above quantifies the undefined part at least. I know that you don't like images but I will upload a proof for the divergence theorem for a sphere just for info:

 

[Charles Link]

You might be looking for complete mathematical rigor for these cases, when it may not exist. The best solution I can see is to assign the delta function at the point $r=r'$ to $\nabla \cdot \frac{r-r'}{4 \pi |r-r'|^3}$.

 

[georg gill]

For $\nabla \times B=J\mu_0$ From Griffiths one can obtain that the curl of B inside a rotating sphere is $\nabla \times B=J\mu_0$ but that is for an arbitrary large sphere. So it seems that the current outside r-r' is not taken into account for that that value of $\nabla \times B$ since a larger sphere would have more current outside r-r'. So since the outside of the point r-r' can be arbitrary one could say that it is in fact in r-r' that the value $\nabla \times B=J\mu_0$ is obtained. So in fact it seems that one always have the relation $\nabla \times B=J\mu_0$ inside current and from the proof in this thread $\nabla \times B=0$ outside current?

 

[Charles Link]

You seem to be getting sidetracked, or overthinking the original problem you presented in the OP. The proofs are relatively straightforward, perhaps lacking complete mathematical rigor, using the vector identities that were presented. I think it has been shown with a reasonable amount of mathematical precision that starting with the integral form of Biot-Savart, the differential form of ampere's law $\nabla \times B=\mu_o J$ is the result.

 

[vanhees71]

For full mathematical rigor but still very easy to understand, see

M. Lighthill, Introduction to Fourier analysis and generalised
functions, Cambridge University Press (1958).

 

[georg gill]

I have pondered a bit further:
By using Leibniz integral rule
$$\nabla \times B(r)=\frac{\mu _0}{4\pi} \int \nabla \times \frac{ J(r') \times (r-r')}{|r-r|^3}dV'$$

$$ \frac{J(r') \times (r-r')}{|r-r'|^3}=\frac{(J_y[z-z']-J_z[y-y'])\textbf{i}+(J_z[x-x']-J_x[z-z'])\textbf{j}+(J_x[y-y']-J_y[x-x'])\textbf{k}}{((x-x')^2+(y-y')^2+(z-z')^2)^{3/2}}$$
the $\textbf{i}$'th component of

$$\nabla \times \frac{ J(r') \times (r-r')}{|r-r|^3}$$

is
$$\frac{\partial}{\partial y}\frac{(J_x[y-y']-J_y[x-x'])}{(((x-x')^2+(y-y')^2+(z-z')^2)^{3/2}}-\frac{\partial}{\partial z}\frac{(J_z[x-x']-J_x[z-z'])}{(((x-x')^2+(y-y')^2+(z-z')^2)^{3/2}}$$

Now this should be 0 outside where r-r'=0. But my calculations says it is not? How can then $$\nabla \times B(r)= \mu _0 J(r')$$

I mean if the component above is not 0 outside r-r' I could add more and more current outside r-r' to alter the value of $$\nabla \times B(r)$$

 

[Charles Link]

Upon further study of the problem, I think I have an error in how I assumed the $J \cdot \nabla b$ term integrates to zero. I need to do some additional study of the problem.

The problem I'm having is that we have $\nabla (J(x') \cdot b )$. That is not $-\nabla' (J(x') \cdot b )$, unless we have some additional identity. Perhaps $\nabla' \cdot J(x')=0$ will get us there, but I'm not yet convinced.

Edit: I think I see the solution to this: It can be written as $\nabla \int J(x') \cdot b \, d^3x'$, with the gradient operator outside of the integral.
Now $\int J(x') \cdot b \, d^3 x'= \int J(x') \cdot \nabla' \frac{1}{|x-x'|} \, d^3 x' =\int \nabla' \cdot (\frac{J(x')}{|x-x'|}) \, d^3 x'$, if $\nabla \cdot J=0$, and use Gauss' law to make this last integral a surface integral, which will vanish at a large distance. (Edit: Note the primes on these last operators which I had to edit in).

 

[Charles Link]

@georg gill Note one additional Edit I did to the above, where I made the last operators with primes. That needs to be a primed operator when the integration is over $d^3 x'$ if Gauss' law is going to be employed.

 

[georg gill]

@georg gill Note one additional Edit I did to the above, where I made the last operators with primes. That needs to be a primed operator when the integration is over $d^3 x'$ if Gauss' law is going to be employed.

Do you mean this:

Upon further study of the problem, I think I have an error in how I assumed the $J \cdot \nabla b$ term integrates to zero. I need to do some additional study of the problem.

The problem I'm having is that we have $\nabla (J(x') \cdot b )$. That is not $-\nabla' (J(x') \cdot b )$, unless we have some additional identity. Perhaps $\nabla' \cdot J(x')=0$ will get us there, but I'm not yet convinced.

Edit: I think I see the solution to this: It can be written as $\nabla \int J(x') \cdot b \, d^3x'$, with the gradient operator outside of the integral.
Now $\int J(x') \cdot b \, d^3 x'= \int J(x') \cdot \nabla' \frac{1}{|x-x'|} \, d^3 x' =\int \nabla' \cdot (\frac{J(x')}{|x-x'|}) \, d^3 x'$, if $\nabla \cdot J=0$, and use Gauss' law to make this last integral a surface integral, which will vanish at a large distance. (Edit: Note the primes on these last operators which I had to edit in).

If it is post 25 I am having a bit of a problem getting the relevance to post 24. Thanks

 

[Charles Link]

@georg gill I haven't worked yet at resolving the difficulties in post 24. When I revisited this problem by reading your post 24, I did spot an error in my previous calculations, which I have now worked through, and have an amended solution per post 25.

 

[Charles Link]

One other item: The condition $\nabla \cdot J=0$ is logical, because by the continuity equation, $\nabla \cdot J +\frac{\partial{\rho}}{\partial{t}}=0$ and we must have no build-up of charge.

If you include cases where $\nabla \cdot J \neq 0$, you get an additional term in this last expression (post 25) of $- b \nabla' \cdot J(x')=\frac{\partial{\rho}(x')}{\partial{t}}/|x-x'|$.

I believe if you carry out the whole calculation, this would yield the additional term in ampere's law of $\mu_o \epsilon_o \dot{E}$.

 

[vanhees71]

The Biot-Savart Law,
$$\vec{B}(\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}$$
only holds for magnetostatics, i.e., $\vec{j}$ must be time-independent. Since further
$$\vec{\nabla} \times \vec{B}=\mu_0 \vec{j}$$
you must necessarily have
$$\vec{\nabla} \cdot \vec{j}=0.$$
You also must have
$$\vec{\nabla} \cdot \vec{B}=0.$$
That in this (and ONLY in this) case the Biot-Savart Law provides the correct answer we transform the Biot-Savart Law a bit. First of all
$$\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}|^3}=-\vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}$$
and thus
$$\vec{B}(\vec{x})=-\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \times \vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}=+\frac{\mu_0}{4 \pi} \vec{\nabla} \times \int_{\mathbb{R}^3} \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Of course the integral is nothing else than a vector potential,
$$\vec{B}=\vec{\nabla} \times \vec{A}$$
with
$$\vec{A}(\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
From this it's immediately clear that
$$\vec{\nabla} \cdot \vec{B}=0,$$
i.e., the first magnetostatic Maxwell equation is fulfilled.

Further we show that also
$$\vec{\nabla} \cdot \vec{A}=0,$$
i.e., $\vec{A}$ obeys the Coulomb-gauge condition. Indeed
$$\vec{\nabla} \cdot \vec{A} =\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \cdot \vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}=- \frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \cdot \vec{\nabla}' \frac{1}{|\vec{x}-\vec{x}'|} = + \frac{\mu_0}{4 \pi}\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{1}{|\vec{x}-\vec{x}'|} \cdot \vec{\nabla}' \cdot \vec{j}(\vec{x}')=0.$$
From this we find
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\Delta \vec{A} = -\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \Delta \frac{1}{|\vec{x}-\vec{x}'|} = +\mu_0 \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \delta^{(3)}(\vec{x}-\vec{x}') = \mu_0 \vec{j}(\vec{x}).$$
So it fulfills also the 2nd magnetostatic Maxwell equation.