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Magnetism 𝜑

Strong formulation

Our starting points along with the material relation are

B=0×H=JB=μH.\begin{align} \nabla \cdot \boldsymbol{B} &= 0 \\[5pt] \nabla \times \boldsymbol{H} &= \boldsymbol{J} \\[5pt] \boldsymbol{B} &= \mu\boldsymbol{H}. \end{align}

Under the assumption of no current flow (J=0\boldsymbol{J} = 0), Gauss’s law for magnetism and Ampere’s law state

B=0×H=0.\begin{align} \nabla \cdot \boldsymbol{B} &= 0 \\[5pt] \nabla \times \boldsymbol{H} &= 0. \end{align}

The curl of H\boldsymbol{H} is zero, so H\boldsymbol{H} is a conservative field, meaning there exists a scalar function φ\varphi such that

H=φ.\begin{align} \boldsymbol{H} = -\nabla \varphi. \end{align}

Here, φ\varphi is called the magnetic scalar potential. Substituting (5) and the constitutive relation B=μH\boldsymbol{B} = \mu \boldsymbol{H} into Gauss’s law for magnetism we get

μH=0μφ=0,\begin{align} \nabla \cdot \mu \boldsymbol{H} &= 0 \\[5pt] \nabla \cdot \mu \nabla \varphi &= 0, \end{align}

resulting in Laplace’s equation.

Weak formulation

The partial differential equation (8)(8) is multiplied by the test function φ\varphi^{\prime} and integrated over the entire domain Ω\Omega to get

Ω  (μφ)  φ  dΩ=0.\begin{align} \int_{\Omega}\; \nabla \cdot (\mu \nabla \varphi)\; \varphi^{\prime}\; d \Omega = 0. \end{align}

Using the Leibniz rule for nabla operator on the divergence term we get

Ω  (φ  μφ)  dΩ+Ω  (φμφ)  dΩ=0.\begin{align} \int_{\Omega}\; \nabla \cdot (\varphi^{\prime}\; \mu \nabla \varphi)\; d \Omega + \int_{\Omega}\; -(\nabla \varphi^{\prime} \cdot \mu \nabla \varphi)\; d \Omega = 0. \end{align}

Applying the divergence theorem on the divergence term we get

Γ  μφn φ  dΓ+Ω  (μφφ)  dΩ=0.\begin{align} \int_{\Gamma}\; \mu \nabla \varphi \cdot \boldsymbol{n}\ \varphi^{\prime}\; d \Gamma + \int_{\Omega}\; -(\mu\nabla \varphi \cdot \nabla \varphi^{\prime})\; d \Omega = 0. \end{align}

Rearranging the terms and using relation B=μH=μφ\boldsymbol{B} = \mu \boldsymbol{H} = -\mu \nabla \varphi on the Neumann term, we obtain the final weak formulation

Ω  (μφφ)  dΩ+Γ  (Bn)  φ  dΓ=0.\begin{align} \int_{\Omega}\; (-\mu\nabla \varphi \cdot \nabla \varphi^{\prime})\; d \Omega + \int_{\Gamma}\; (-\boldsymbol{B} \cdot \boldsymbol{n}) \; \varphi^{\prime}\; d \Gamma = 0. \end{align}