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} \\[10pt]
\nabla \times \boldsymbol{H}
&= \boldsymbol{J}
\end{align} ∇ × E ∇ × H = − ∂ t ∂ B = J 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})\ .
\end{align} E = σ 1 ( ∇ × H ) . 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}\ .
\end{align} ∇ × E = − ∂ t ∂ μ H . Substituting equation (3) into equation (4) and rearranging we derive the 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 \ .
\end{align} ∇ × ( σ 1 ∇ × H ) + ∂ t ∂ ( μ H ) = 0 . The partial differential equation (4) is multiplied by the test vector 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 \\[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 \ .
\end{align} ∫ Ω ( ∇ × E + ∂ t ∂ ( μ H ) ) ⋅ H ′ d Ω ∫ Ω ( ∇ × E ) ⋅ H ′ d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω = 0 = 0 . 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 \ .
\end{align} ∫ Ω ∇ ⋅ ( E × H ′ ) d Ω + ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω = 0 . Using the divergence theorem on the divergence term we obtain
∫ Γ ( E × H ′ ) ⋅ n d Γ + ∫ Ω E ⋅ ( ∇ × H ′ ) d 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 \frac{\partial (\mu \boldsymbol{H})}{\partial t} \cdot \boldsymbol{H}^{\prime}\ d\Omega
= 0 \ .
\end{align} ∫ Γ ( E × H ′ ) ⋅ n d Γ + ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ ∂ t ∂ ( μ H ) ⋅ H ′ d Ω = 0 . 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 \ ,
\end{align} ∫ Ω E ⋅ ( ∇ × H ′ ) d Ω + ∫ Ω ∂ t ∂ ( μ H ) ⋅ H ′ d Ω + ∫ Γ ( n × E ) ⋅ H ′ d Γ = 0 , and by substituting equation (3) into the first term we derive the weak formulation
∫ Ω 1 σ ( ∇ × H ) ⋅ ( ∇ × H ′ ) d 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 \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 \ .
\end{align} ∫ Ω σ 1 ( ∇ × H ) ⋅ ( ∇ × H ′ ) d Ω + ∫ ∂ t ∂ ( μ H ) ⋅ H ′ d Ω + ∫ Γ ( n × E ) ⋅ H ′ d Γ = 0 . Applies a fixed value to the H \boldsymbol{H} H H \boldsymbol{H} H
Imposes periodic boundary conditions for the H \boldsymbol{H} H
This formulation supports the following couplings:
H-𝜑 coupling (Magnetism 𝜑)
Electrical insulator (Magnetism 𝜑)
