Divergence (Divergence) and Curl of Magnetic Fields
Theorem
The divergence and curl of a magnetic field are as follows:
$$ \begin{align*} \nabla \cdot \mathbf{B} =&\ 0 \\ \nabla \times \mathbf{B} =&\ \mu_{0} \mathbf{J} \end{align*} $$
Description
Just like the electric field was always a special vector function with a curl of $\mathbf{0}$, so is the magnetic field. Let’s calculate the divergence and curl using the law of Biot-Savart for volume currents.
$$ \mathbf{B} (\mathbf{r}) = \dfrac{\mu_{0}}{4 \pi} \int \dfrac{\mathbf{J} (\mathbf{r}^{\prime}) \times \crH }{\cR^2} d \tau $$
It is important to note here what coordinates each function depends on.
$\mathbf{B}$ is a function of the coordinates $(x, y, z)$ of the observation point, and $\mathbf{J}$ is a function of the coordinates $(x^{\prime}, y^{\prime}, z^{\prime})$ of the source point. The separation vector and volume element are as follows:
$$ \bcR = (x-x^{\prime})\mathbf{x} + (y-y^{\prime})\mathbf{y} +(z-z^{\prime})\mathbf{z} $$
$$ d\tau^{\prime}=dx^{\prime}dy^{\prime}dz^{\prime} $$
The integral is an operation regarding the source point (with prime $^{\prime}$), and divergence and curl are operations regarding the observation point (without prime $^{\prime}$).
Proof
Divergence of the Magnetic Field
According to the Biot-Savart law, the divergence of the magnetic field is as follows:
$$ \nabla \cdot \mathbf{B}=\dfrac{\mu_{0}}{4\pi} \int \nabla \cdot \left( \mathbf{J} \times \dfrac{\crH}{\cR^2}\right)d\tau^{\prime} $$
Multiplication rule involving the del operator
$$ \nabla \cdot ( \mathbf{A} \times \mathbf{B} ) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B} ) $$
Applying the above multiplication rule to the integrand yields
$$ \nabla \cdot \left( \mathbf{J} \times \dfrac{\crH}{\cR^2}\right) = \dfrac{\crH}{\cR^2}\cdot (\nabla \times \mathbf{J} ) -\mathbf{J} \cdot \left( \nabla \times \dfrac{\crH}{\cR^2} \right) $$
In the first term, $\nabla$ is a derivative operation regarding $x, y, z$, and $\mathbf{J}$ is a function regarding $(x^{\prime}, y^{\prime}, z^{\prime})$, hence it equals $\nabla \times \mathbf{J} = \mathbf{0}$. Additionally, since the curl of the separation vector is $\mathbf{0}$, the second term in brackets equals $\mathbf{0}$ as well. Therefore, the divergence of the magnetic field is
$$ \nabla \cdot \mathbf{B} = \dfrac{\mu_{0}}{4\pi} \int 0 d\tau^{\prime} = 0 $$
Curl of the Magnetic Field
According to the Biot-Savart law, the curl of the magnetic field is as follows:
$$ \nabla \times \mathbf{B} = \dfrac{\mu_{0}}{4\pi} \int \nabla \times \left( \mathbf{J} \times \dfrac{\crH}{\cR^2}\right)d\tau^{\prime} $$
Multiplication rule involving the del operator
$$ \nabla \times (\mathbf{A} \times \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A}) $$
Applying the above multiplication rule to the integrand yields:
$$ \nabla \times \left( \mathbf{J} \times \dfrac{\crH}{\cR^2} \right)= \left( \dfrac{\crH}{\cR^2} \cdot \nabla \right) \mathbf{J} - \left( \mathbf{J} \cdot \nabla \right) \dfrac{\crH}{\cR^2} + \mathbf{J} \left( \nabla \cdot \dfrac{\crH}{\cR^2} \right) - \dfrac{\crH}{\cR^2} \left( \nabla \cdot \mathbf{J} \right) $$
Here, the first and fourth terms, when $\nabla$ is applied to $\mathbf{J}$, result in $0$ for the same reason mentioned in the calculation of divergence earlier. Although the second term is not directly $0$, integrating it results in $0$. This process is explained separately below. Finally, the third term is equal to the divergence of the separation vector, thus a Dirac delta function.
$$ \nabla \cdot \left( \dfrac{ \hat { \bcR} }{\cR^2} \right) = 4\pi \delta^2(\bcR ) $$
Therefore, the curl of the magnetic field is as follows:
$$ \nabla \times \mathbf{B} = \dfrac{\mu_{0}}{4\pi} \int \mathbf{J} (\mathbf{r}^{\prime} ) 4\pi \delta^3(\mathbf{r} - \mathbf{r}^{\prime} ) d\tau^{\prime} $$
By the definition of the Dirac delta function:
$$ \nabla \times \mathbf{B} = \mu_{0} \mathbf{J}(\mathbf{r}) $$
What remains is to verify why the integration of the second term equals $0$. If this part is difficult, it can be understood as being so due to physical meaning rather than a mathematical one. Since the separation vector equals $\bcR = (x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}$, the following equation holds:
$$ -\dfrac{\partial \bcR}{\partial x} = - \hat{\mathbf{x}} = \dfrac{\partial \bcR}{\partial x^{\prime}} \quad \text{and} \quad -\dfrac{\partial \bcR}{\partial y} = - \hat{\mathbf{y}} = \dfrac{\partial \bcR}{\partial y^{\prime}} \quad \text{and} \quad -\dfrac{\partial \bcR}{\partial z} = - \hat{\mathbf{z}} = \dfrac{\partial \bcR}{\partial z^{\prime}} $$
Thus, the following equality holds:
$$ \begin{align*} -( \mathbf{J} \cdot \nabla) \dfrac{\crH}{\cR^2} =&\ -\left( J_{x^{\prime}} \dfrac{\partial }{\partial x} + J_{y^{\prime}} \dfrac{\partial }{\partial y} + J_{z^{\prime}} \dfrac{\partial }{\partial z}\right) \dfrac{\crH}{\cR^2} \\ =&\ \left( J_{x^{\prime}} \dfrac{\partial }{\partial x^{\prime}} + J_{y^{\prime}} \dfrac{\partial }{\partial y^{\prime}} + J_{z^{\prime}} \dfrac{\partial }{\partial z^{\prime}}\right) \dfrac{\crH}{\cR^2} \\ =&( \mathbf{J} \cdot \nabla^{\prime} ) \dfrac{\crH}{\cR^2} \end{align*} $$
The $\mathbf{x}$ component of this vector is as follows:
$$ \left[ ( \mathbf{J} \cdot \nabla^{\prime} ) \dfrac{\crH}{\cR^2} \right]_{x}=( \mathbf{J} \cdot \nabla^{\prime} ) \left( \dfrac{ x-x^{\prime}}{\cR^3} \right) $$
Expanding this using the multiplication rule $\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$ yields:
$$ ( \mathbf{J} \cdot \nabla^{\prime} ) \left( \dfrac{ x-x^{\prime}}{\cR^3} \right) = \nabla^{\prime} \cdot \left[ \dfrac{x-x^{\prime}}{\cR^3} \mathbf{J} \right] - \left( \dfrac{x-x^{\prime}} {\cR^3} \right) (\nabla^{\prime} \cdot \mathbf{J} ) $$
Additionally, if there’s steady current, since the divergence of $\mathbf{J}$ is $0$ (reference), the second term equals $0$. Integrating the remaining terms, by the divergence theorem:
$$ \int_\mathcal{V} \nabla^{\prime} \cdot \left[ \dfrac{x-x^{\prime}}{\cR^3}\mathbf{J} \right]d\tau=\oint_\mathcal{S} \dfrac{x-x^{\prime}}{\cR^3}\mathbf{J} \cdot d\mathbf{a}^{\prime} $$
The reason for changing $\nabla$ to $\nabla^{\prime}$ above is due to this integration. The left-hand side’s integral range must be large enough to contain all currents, and in this case, $\mathbf{J}=0$ at the region’s boundary. Therefore, as $\mathbf{J}=0$ over the entire area for the right-hand side’s area integral, the integral result is $0$. By the same logic, the $\mathbf{y}$ and $\mathbf{z}$ components also equal $0$, so the integration of the second term equals $0$.
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