This page derives the weak formulation for magnetostatics using the magnetic vector potential A \boldsymbol{A} A
Gauss’s law \text{Gauss's law} Gauss’s law ∇ ⋅ D = ρ \nabla \cdot \boldsymbol{D} = \rho ∇ ⋅ D = ρ Gauss’s law for magnetism \text{Gauss's law for magnetism} Gauss’s law for magnetism ∇ ⋅ B = 0 \nabla \cdot \boldsymbol{B} = 0 ∇ ⋅ B = 0 Faraday’s law \text{Faraday's law} Faraday’s law ∇ × E = − ∂ B ∂ t \nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t} ∇ × E = − ∂ t ∂ B Ampere-Maxwell law \text{Ampere-Maxwell law} Ampere-Maxwell law ∇ × H = J + ∂ D ∂ t \nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t} ∇ × H = J + ∂ t ∂ D Our starting points along with the material relation are
∇ ⋅ B = 0 ∇ × H = J B = μ H . \begin{align*}
\nabla \cdot \boldsymbol{B} &= 0 \tag{1} \\[5pt]
\nabla \times \boldsymbol{H} &= \boldsymbol{J} \tag{2} \\[5pt]
\boldsymbol{B} &= \mu \boldsymbol{H}. \tag{3}
\end{align*} ∇ ⋅ B ∇ × H B = 0 = J = μ H . ( 1 ) ( 2 ) ( 3 ) Since the magnetic field is divergence-free, we can use the identity for divergence of a curl and write the magnetic field as a curl of some vector A \boldsymbol{A} A
B = ∇ × A . \begin{align*}
\boldsymbol{B} = \nabla \times \boldsymbol{A}. \tag{4}
\end{align*} B = ∇ × A . ( 4 ) Vector A \boldsymbol{A} A (4) \text{(4)} (4) (3) \text{(3)} (3)
H = 1 μ B = 1 μ ( ∇ × A ) . \begin{align*}
\boldsymbol{H}
= \frac{1}{\mu} \boldsymbol{B}
= \frac{1}{\mu} \left( \nabla \times \boldsymbol{A} \right). \tag{5}
\end{align*} H = μ 1 B = μ 1 ( ∇ × A ) . ( 5 ) Substituting (5) \text{(5)} (5) (2) \text{(2)} (2)
∇ × H = 1 μ ( ∇ × A ) = J . \begin{align*}
\nabla \times \boldsymbol{H}
= \frac{1}{\mu} \left( \nabla \times \boldsymbol{A} \right)
= \boldsymbol{J}. \tag{6}
\end{align*} ∇ × H = μ 1 ( ∇ × A ) = J . ( 6 ) To obtain the weak form, multiply (2) \text{(2)} (2) A ′ \boldsymbol{A}^\prime A ′ Ω \Omega Ω
∫ Ω ( ∇ × H − J ) ⋅ A ′ d Ω = 0 ∫ Ω ( ∇ × H ) ⋅ A ′ d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. \begin{align*}
\int_{\Omega}\ (\nabla \times \boldsymbol{H}
- \boldsymbol{J}) \cdot \boldsymbol{A^\prime}\ d \Omega
= 0 \tag{7} \\[5pt]
\int_{\Omega}\ (\nabla \times \boldsymbol{H}) \cdot \boldsymbol{A^\prime}\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0. \tag{8}
\end{align*} ∫ Ω ( ∇ × H − J ) ⋅ A ′ d Ω = 0 ∫ Ω ( ∇ × H ) ⋅ A ′ d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. ( 7 ) ( 8 ) Using the identity for divergence of a cross product on the first term we get
∫ Ω ∇ ⋅ ( H × A ′ ) d Ω + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. \begin{align*}
\int_{\Omega}\ \nabla \cdot (\boldsymbol{H} \times \boldsymbol{A^\prime})\ d \Omega
+ \int_{\Omega}\ \boldsymbol{H} \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0. \tag{9}
\end{align*} ∫ Ω ∇ ⋅ ( H × A ′ ) d Ω + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. ( 9 ) Applying the divergence theorem on the first term we obtain
∫ Γ ( H × A ′ ) ⋅ n d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. \begin{align*}
\int_{\Gamma}\ (\boldsymbol{H} \times \boldsymbol{A^\prime}) \cdot \boldsymbol{n}\ d \Gamma
+ \int_{\Omega}\ \boldsymbol{H} \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0. \tag{10}
\end{align*} ∫ Γ ( H × A ′ ) ⋅ n d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. ( 10 ) Using the scalar triple product rule on the boundary term we get
∫ Γ ( n × H ) ⋅ A ′ d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. \begin{align*}
\int_{\Gamma}\ (\boldsymbol{n} \times \boldsymbol{H}) \cdot \boldsymbol{A^\prime}\ d \Gamma
+ \int_{\Omega}\ \boldsymbol{H} \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0. \tag{11}
\end{align*} ∫ Γ ( n × H ) ⋅ A ′ d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0. ( 11 ) Substituting (5) \text{(5)} (5)
∫ Ω 1 μ ( ∇ × A ) ⋅ ( ∇ × A ′ ) d Ω + ∫ Ω − J ⋅ A ′ d Ω + ∫ Γ ( n × H ) ⋅ A ′ d Γ = 0. \begin{align*}
\int_{\Omega}\ \frac{1}{\mu}\ (\nabla \times \boldsymbol{A}) \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
+ \int_{\Omega}\ -\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
+ \int_{\Gamma}\ (\boldsymbol{n} \times \boldsymbol{H}) \cdot \boldsymbol{A^\prime}\ d \Gamma
= 0. \tag{12}
\end{align*} ∫ Ω μ 1 ( ∇ × A ) ⋅ ( ∇ × A ′ ) d Ω + ∫ Ω − J ⋅ A ′ d Ω + ∫ Γ ( n × H ) ⋅ A ′ d Γ = 0. ( 12 ) Imposes a zero tangential magnetic field n × H = 0 \boldsymbol{n} \times \boldsymbol{H} = 0 n × H = 0
Defines a remanent magnetization B r \boldsymbol{B_r} B r
How to use You can use either the matrix editor or the expression editor . With the matrix editor, provide a 3×1 matrix for a 3D problem (2×1 for 2D, etc.) with remanence field vector values for each dimension. In the expression editor, write the direct expression. Example [1; 5; 0] applies a 3D remanence field of B r , x = 1 T B_{r,x} = 1\:T B r , x = 1 T B r , y = 5 T B_{r,y} = 5\:T B r , y = 5 T B r , z = 0 T B_{r,z} = 0\:T B r , z = 0 T Adding spatially varying remanence field to permanent magnets .Unit Remanence magnetic field in Teslas (T)
Imposes periodic boundary conditions on the magnetic vector potential field A \boldsymbol{A} A
Example Periodicity follows the same principles, regardless of which physics module it belongs to. See how periodicity is used in magnetism as a reference: Periodicity in electric motors
This formulation supports the following couplings:
