This page derives the weak formulation for magnetism using the magnetic field intensity H \boldsymbol{H} H
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 The H \boldsymbol{H} H
∇ × E = − ∂ B ∂ t ∇ × H = J \begin{align*}
\nabla \times \boldsymbol{E}
&= -\frac{\partial \boldsymbol{B}}{\partial t} \tag{1} \\[10pt]
\nabla \times \boldsymbol{H}
&= \boldsymbol{J} \tag{2}
\end{align*} ∇ × E ∇ × H = − ∂ t ∂ B = J ( 1 ) ( 2 ) Using Ohm’s law J = σ E \boldsymbol{J} = \sigma \boldsymbol{E} J = σ E
E = 1 σ ( ∇ × H ) . \begin{align*}
\boldsymbol{E}
= \frac{1}{\sigma} (\nabla \times \boldsymbol{H})\ . \tag{3}
\end{align*} E = σ 1 ( ∇ × H ) . ( 3 ) Using the material relation B = μ H \boldsymbol{B} = \mu \boldsymbol{H} B = μ H
∇ × E = − ∂ μ H ∂ t . \begin{align*}
\nabla \times \boldsymbol{E} = -\frac{\partial \mu \boldsymbol{H}}{\partial t}\ . \tag{4}
\end{align*} ∇ × E = − ∂ t ∂ μ H . ( 4 ) Substituting equation (3) \text{(3)} (3) (4) \text{(4)} (4) H \boldsymbol{H} H
∇ × ( 1 σ ∇ × H ) + ∂ ( μ H ) ∂ t = 0 . \begin{align*}
\nabla \times (\frac{1}{\sigma} \nabla \times \boldsymbol{H})
+ \frac{\partial (\mu \boldsymbol{H})}{\partial t}
= 0 \ . \tag{5}
\end{align*} ∇ × ( σ 1 ∇ × H ) + ∂ t ∂ ( μ H ) = 0 . ( 5 ) To obtain the weak form, multiply (5) \text{(5)} (5) H ′ \boldsymbol{H}^{\prime} H ′ Ω \Omega Ω
∫ Ω ( ∇ × E + ∂ ( μ H ) ∂ t ) ⋅ H ′ d Ω = 0 ∫ Ω ( ∇ × E ) ⋅ H ′ d Ω + ∫ Ω ∂ ( μ H ) ∂ t ⋅ H ′ d Ω = 0 . \begin{align*}
\int_{\Omega}\ (\nabla \times \boldsymbol{E}
+ \frac{\partial (\mu \boldsymbol{H})}{\partial t}) \cdot \boldsymbol{H}^{\prime}\ d \Omega
&= 0 \tag{6} \\[10pt]
\int_{\Omega}\ ( \nabla \times \boldsymbol{E}) \cdot \boldsymbol{H}^{\prime}\ d \Omega
+ \int_{\Omega}\ \frac{\partial (\mu \boldsymbol{H})}{\partial t} \cdot \boldsymbol{H}^{\prime}\ d \Omega
&= 0 \ . \tag{7}
\end{align*} ∫ Ω ( ∇ × E + ∂ t ∂ ( μ H ) ) ⋅ H ′ d Ω ∫ Ω ( ∇ × E ) ⋅ H ′ d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω = 0 = 0 . ( 6 ) ( 7 ) Applying the divergence of a cross product on the first term we have
∫ Ω ∇ ⋅ ( E × H ′ ) d Ω + ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ ( μ H ) ∂ t ⋅ H ′ d Ω = 0 . \begin{align*}
\int_{\Omega}\ \nabla \cdot (\boldsymbol{E} \times \boldsymbol{H}^{\prime})\ d\Omega
+ \int_{\Omega}\ \boldsymbol{E} \cdot (\nabla \times \boldsymbol{H}^{\prime})\ d\Omega
+ \int_{\Omega}\ \frac{\partial (\mu \boldsymbol{H})}{\partial t} \cdot \boldsymbol{H}^{\prime}\ d\Omega
= 0 \ . \tag{8}
\end{align*} ∫ Ω ∇ ⋅ ( E × H ′ ) d Ω + ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω = 0 . ( 8 ) Using the divergence theorem on the divergence term we obtain
∫ Γ ( E × H ′ ) ⋅ n d Γ + ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ ( μ H ) ∂ t ⋅ H ′ d Ω = 0 . \begin{align*}
\int_{\Gamma}\ (\boldsymbol{E} \times \boldsymbol{H}^{\prime}) \cdot \boldsymbol{n}\ d\Gamma
+ \int_{\Omega}\ \boldsymbol{E} \cdot (\nabla \times \boldsymbol{H}^{\prime})\ d\Omega
+ \int_{\Omega} \frac{\partial (\mu \boldsymbol{H})}{\partial t} \cdot \boldsymbol{H}^{\prime}\ d\Omega
= 0 \ . \tag{9}
\end{align*} ∫ Γ ( E × H ′ ) ⋅ n d Γ + ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω = 0 . ( 9 ) Making use of the scalar triple product rule on the boundary and rearranging we get
∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ ( μ H ) ∂ t ⋅ H ′ d Ω + ∫ Γ ( n × E ) ⋅ H ′ d Γ = 0 , \begin{align*}
\int_{\Omega}\ \boldsymbol{E} \cdot (\nabla \times \boldsymbol{H}^{\prime})\ d\Omega
+ \int_{\Omega} \frac{\partial (\mu \boldsymbol{H})}{\partial t} \cdot \boldsymbol{H}^{\prime}\ d\Omega
+ \int_{\Gamma} (\boldsymbol{n} \times \boldsymbol{E}) \cdot \boldsymbol{H}^{\prime}\ d\Gamma
= 0 \ , \tag{10}
\end{align*} ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω + ∫ Γ ( n × E ) ⋅ H ′ d Γ = 0 , ( 10 ) and by substituting equation (3) \text{(3)} (3)
∫ Ω 1 σ ( ∇ × H ) ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ ( μ H ) ∂ t ⋅ H ′ d Ω + ∫ Γ ( n × E ) ⋅ H ′ d Γ = 0 . \begin{align*}
\int_{\Omega} \frac{1}{\sigma} (\nabla \times \boldsymbol{H}) \cdot (\nabla \times \boldsymbol{H}^{\prime})\ d\Omega
+ \int_{\Omega} \frac{\partial (\mu \boldsymbol{H})}{\partial t} \cdot \boldsymbol{H}^{\prime}\ d\Omega
+ \int_{\Gamma} (\boldsymbol{n} \times \boldsymbol{E}) \cdot \boldsymbol{H}^{\prime}\ d\Gamma
= 0 \ . \tag{11}
\end{align*} ∫ Ω σ 1 ( ∇ × H ) ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω + ∫ Γ ( n × E ) ⋅ H ′ d Γ = 0 . ( 11 ) Applies a fixed value to the H \boldsymbol{H} H H \boldsymbol{H} H
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 magnetic field strength vector values for each dimension. In the expression editor, write the direct expression. Example [1; 5; 0] applies a 3D magnetic field strength constraint of H x = 1 A m H_x = 1\:\frac{A}{m} H x = 1 m A H y = 5 A m H_y = 5\:\frac{A}{m} H y = 5 m A H z = 0 A m H_z = 0\:\frac{A}{m} H z = 0 m A Unit Magnetic field strength in Amperes per meter (A/m)
Imposes periodic boundary conditions on the magnetic field strength H \boldsymbol{H} H
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:
H-𝜑 coupling
Electrical insulator
