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Magnetism H

Strong formulation

The H\boldsymbol{H}-formulation is derived based on the magnetostatic approximation. Our starting equations are Faraday’s law and Ampere’s law.

∇×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}

Using Ohm’s law J=σE\boldsymbol{J} = \sigma \boldsymbol{E} and plugging in the Ampere’s law we get

E=1σ(∇×H) .\begin{align} \boldsymbol{E} = \frac{1}{\sigma} (\nabla \times \boldsymbol{H})\ . \end{align}

Using the material relation B=μH\boldsymbol{B} = \mu \boldsymbol{H} and substituting it into Faraday’s law we end up with

∇×E=−∂μH∂t .\begin{align} \nabla \times \boldsymbol{E} = -\frac{\partial \mu \boldsymbol{H}}{\partial t}\ . \end{align}

Substituting equation (3) into equation (4) and rearranging we derive the H\boldsymbol{H}-formulation

∇×(1σ∇×H)+∂(μH)∂t=0 .\begin{align} \nabla \times (\frac{1}{\sigma} \nabla \times \boldsymbol{H}) + \frac{\partial (\mu \boldsymbol{H})}{\partial t} = 0 \ . \end{align}

Weak formulation

The partial differential equation (4) is multiplied by the test vector H′\boldsymbol{H}^{\prime} and integrated over the entire domain Ω\Omega.

∫Ω (∇×E+∂(μH)∂t)⋅H′ dΩ=0∫Ω (∇×E)⋅H′ dΩ+∫Ω ∂(μH)∂t⋅H′ dΩ=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}

Applying the divergence of a cross product on the first term we have

∫Ω ∇⋅(E×H′) dΩ+∫Ω E⋅(∇×H′) dΩ+∫Ω ∂(μH)∂t⋅H′ dΩ=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}

Using the divergence theorem on the divergence term we obtain

∫Γ (E×H′)⋅n dΓ+∫Ω E⋅(∇×H′) dΩ+∫∂(μH)∂t⋅H′ dΩ=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}

Making use of the scalar triple product rule on the boundary and rearranging we get

∫Ω E⋅(∇×H′) dΩ+∫Ω∂(μH)∂t⋅H′ dΩ+∫Γ(n×E)⋅H′ dΓ=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}

and by substituting equation (3) into the first term we derive the weak formulation

∫Ω1σ(∇×H)⋅(∇×H′) dΩ+∫∂(μH)∂t⋅H′ dΩ+∫Γ(n×E)⋅H′ dΓ=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}