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Divergence (Divergence) and Curl of Magnetic Fields 📂Electrodynamics

Divergence (Divergence) and Curl of Magnetic Fields

Theorem

The divergence and curl of a magnetic field are as follows:

∇⋅B= 0∇×B= μ0J \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 0\mathbf{0}, so is the magnetic field. Let’s calculate the divergence and curl using the law of Biot-Savart for volume currents.

B(r)=μ04π∫J(r′)×2dτ \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.

B\mathbf{B} is a function of the coordinates (x,y,z)(x, y, z) of the observation point, and J\mathbf{J} is a function of the coordinates (x′,y′,z′)(x^{\prime}, y^{\prime}, z^{\prime}) of the source point. The separation vector and volume element are as follows:

=(x−x′)x+(y−y′)y+(z−z′)z \bcR = (x-x^{\prime})\mathbf{x} + (y-y^{\prime})\mathbf{y} +(z-z^{\prime})\mathbf{z}

dτ′=dx′dy′dz′ 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:

∇⋅B=μ04π∫∇⋅(J×2)dτ′ \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

∇⋅(A×B)=B⋅(∇×A)−A⋅(∇×B) \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

∇⋅(J×2)=2⋅(∇×J)−J⋅(∇×2) \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,zx, y, z, and J\mathbf{J} is a function regarding (x′,y′,z′)(x^{\prime}, y^{\prime}, z^{\prime}), hence it equals ∇×J=0\nabla \times \mathbf{J} = \mathbf{0}. Additionally, since the curl of the separation vector is 0\mathbf{0}, the second term in brackets equals 0\mathbf{0} as well. Therefore, the divergence of the magnetic field is

∇⋅B=μ04π∫0dτ′=0 \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:

∇×B=μ04π∫∇×(J×2)dτ′ \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

∇×(A×B)=(B⋅∇)A−(A⋅∇)B+A(∇⋅B)−B(∇⋅A) \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:

∇×(J×2)=(2⋅∇)J−(J⋅∇)2+J(∇⋅2)−2(∇⋅J) \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 J\mathbf{J}, result in 00 for the same reason mentioned in the calculation of divergence earlier. Although the second term is not directly 00, integrating it results in 00. This process is explained separately below. Finally, the third term is equal to the divergence of the separation vector, thus a Dirac delta function.

∇⋅(^2)=4πδ2() \nabla \cdot \left( \dfrac{ \hat { \bcR} }{\cR^2} \right) = 4\pi \delta^2(\bcR )

Therefore, the curl of the magnetic field is as follows:

∇×B=μ04π∫J(r′)4πδ3(r−r′)dτ′ \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:

∇×B=μ0J(r) \nabla \times \mathbf{B} = \mu_{0} \mathbf{J}(\mathbf{r})

What remains is to verify why the integration of the second term equals 00. 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 =(x−x′)x^+(y−y′)y^+(z−z′)z^\bcR = (x-x^{\prime})\hat{\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}, the following equation holds:

−∂∂x=−x^=∂∂x′and−∂∂y=−y^=∂∂y′and−∂∂z=−z^=∂∂z′ -\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:

−(J⋅∇)2= −(Jx′∂∂x+Jy′∂∂y+Jz′∂∂z)2= (Jx′∂∂x′+Jy′∂∂y′+Jz′∂∂z′)2=(J⋅∇′)2 \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 x\mathbf{x} component of this vector is as follows:

[(J⋅∇′)2]x=(J⋅∇′)(x−x′3) \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 ∇⋅(fA)=f(∇⋅A)+A⋅(∇f)\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f) yields:

(J⋅∇′)(x−x′3)=∇′⋅[x−x′3J]−(x−x′3)(∇′⋅J) ( \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 J\mathbf{J} is 00 (reference), the second term equals 00. Integrating the remaining terms, by the divergence theorem:

∫V∇′⋅[x−x′3J]dτ=∮Sx−x′3J⋅da′ \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, J=0\mathbf{J}=0 at the region’s boundary. Therefore, as J=0\mathbf{J}=0 over the entire area for the right-hand side’s area integral, the integral result is 00. By the same logic, the y\mathbf{y} and z\mathbf{z} components also equal 00, so the integration of the second term equals 00.

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