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\[ U=\frac{1}{2} L I^{2}=\frac{1}{2} \Phi I=\frac{\Phi^{2}}{2 L} . \] Now let's express this magnetic energy in terms of the magnetic field \( \mathbf{B}(\mathbf{x}) \) inside the coil. The magnetic flux though the coil can be expressed in terms of the vector potential as \[ \Phi=\oint_{\text {coil }} d \mathbf{x} \cdot \mathbf{A}(\mathbf{x}) \] hence the magnetic energy \[ U=\frac{1}{2} I \Phi=\frac{1}{2} \oint_{\text {coil }} I d \mathbf{x} \cdot \mathbf{A}(\mathbf{x}) \text {. } \] This formula assumes a coil made of thin wires; if we replace them with thicker conductors carrying some volume currents \( \mathbf{J}(\mathbf{x}) \), then in the integral (8) we replace \( I d \mathbf{x} \rightarrow d^{3} \mathbf{x} \mathbf{J}(\mathbf{x}) \), hence \[ U=\frac{1}{2} \iiint d \mathbf{x} \mathbf{J}(\mathbf{x}) \cdot \mathbf{A}(\mathbf{x}) \] Now let's integrate by parts using \( \nabla \times \mathbf{A}=\mathbf{B} \) and the Ampere's Law \( \nabla \times \mathbf{H}=\mathbf{J} \). For any vector fields \( \mathbf{f} \) and \( \mathbf{g} \), we have \[ \nabla \cdot(\mathbf{f} \times \mathbf{g})=\mathrm{g} \cdot(\nabla \times \mathbf{f})-\mathbf{f} \cdot(\nabla \times \mathbf{g}), \] 2 hence \[ \nabla \cdot(\mathbf{H} \times \mathbf{A})=\mathbf{A} \cdot(\nabla \times \mathbf{H})-\mathbf{H} \cdot(\nabla \times \mathbf{A})=\mathbf{A} \cdot \mathbf{J}-\mathbf{H} \cdot \mathbf{B} \] and therefore \[ \iiint_{\mathcal{V}} d^{3} \mathbf{x} \mathbf{J} \cdot \mathbf{A}=\oiint(\mathbf{H} \times \mathbf{A}) \cdot d^{2} \mathbf{a}+\iiint_{\mathcal{V}} d^{3} \mathbf{x} \mathbf{H} \cdot \mathbf{B} . \] for any integration volume \( \mathcal{V} \) and its surface \( \mathcal{S} \). In eq. (9), we integrate over the conductor's volume, but we may just as well extend the integration volume \( \mathcal{V} \) to the whole space. Consequently, in eq. (12) the surface integral on the RHS goes away, and the magnetic energy (9) becomes \[ U=\frac{1}{2} \iiint_{\substack{\text { whole } \\ \text { space }}} d^{3} \mathrm{x} \mathbf{H} \cdot \mathbf{B} . \]

          \[
U=\frac{1}{2} L I^{2}=\frac{1}{2} \Phi I=\frac{\Phi^{2}}{2 L} .
\]

Now let's express this magnetic energy in terms of the magnetic field \( \mathbf{B}(\mathbf{x}) \) inside the coil. The magnetic flux though the coil can be expressed in terms of the vector potential as
\[
\Phi=\oint_{\text {coil }} d \mathbf{x} \cdot \mathbf{A}(\mathbf{x})
\]
hence the magnetic energy
\[
U=\frac{1}{2} I \Phi=\frac{1}{2} \oint_{\text {coil }} I d \mathbf{x} \cdot \mathbf{A}(\mathbf{x}) \text {. }
\]

This formula assumes a coil made of thin wires; if we replace them with thicker conductors carrying some volume currents \( \mathbf{J}(\mathbf{x}) \), then in the integral (8) we replace \( I d \mathbf{x} \rightarrow d^{3} \mathbf{x} \mathbf{J}(\mathbf{x}) \), hence
\[
U=\frac{1}{2} \iiint d \mathbf{x} \mathbf{J}(\mathbf{x}) \cdot \mathbf{A}(\mathbf{x})
\]

Now let's integrate by parts using \( \nabla \times \mathbf{A}=\mathbf{B} \) and the Ampere's Law \( \nabla \times \mathbf{H}=\mathbf{J} \). For any vector fields \( \mathbf{f} \) and \( \mathbf{g} \), we have
\[
\nabla \cdot(\mathbf{f} \times \mathbf{g})=\mathrm{g} \cdot(\nabla \times \mathbf{f})-\mathbf{f} \cdot(\nabla \times \mathbf{g}),
\]
2
hence
\[
\nabla \cdot(\mathbf{H} \times \mathbf{A})=\mathbf{A} \cdot(\nabla \times \mathbf{H})-\mathbf{H} \cdot(\nabla \times \mathbf{A})=\mathbf{A} \cdot \mathbf{J}-\mathbf{H} \cdot \mathbf{B}
\]
and therefore
\[
\iiint_{\mathcal{V}} d^{3} \mathbf{x} \mathbf{J} \cdot \mathbf{A}=\oiint(\mathbf{H} \times \mathbf{A}) \cdot d^{2} \mathbf{a}+\iiint_{\mathcal{V}} d^{3} \mathbf{x} \mathbf{H} \cdot \mathbf{B} .
\]
for any integration volume \( \mathcal{V} \) and its surface \( \mathcal{S} \). In eq. (9), we integrate over the conductor's volume, but we may just as well extend the integration volume \( \mathcal{V} \) to the whole space. Consequently, in eq. (12) the surface integral on the RHS goes away, and the magnetic energy (9) becomes
\[
U=\frac{1}{2} \iiint_{\substack{\text { whole } \\ \text { space }}} d^{3} \mathrm{x} \mathbf{H} \cdot \mathbf{B} .
\]

U=(1)/(2) L I^2=(1)/(2)Φ I=(Φ^2)/(2 L) .

Now let's express this magnetic energy in terms of the magnetic field 𝐁(𝐱) inside the coil. The magnetic flux though the coil can be expressed in terms of the vector potential as

Φ=∮coil d 𝐱·𝐀(𝐱)

hence the magnetic energy

U=(1)/(2) I Φ=(1)/(2)∮coil I d 𝐱·𝐀(𝐱) .

This formula assumes a coil made of thin wires; if we replace them with thicker conductors carrying some volume currents 𝐉(𝐱), then in the integral (8) we replace I d 𝐱→ d^3𝐱𝐉(𝐱), hence

U=(1)/(2)∭ d 𝐱𝐉(𝐱) ·𝐀(𝐱)

Now let's integrate by parts using ∇×𝐀=𝐁 and the Ampere's Law ∇×𝐇=𝐉. For any vector fields 𝐟 and 𝐠, we have

∇·(𝐟×𝐠)=g·(∇×𝐟)-𝐟·(∇×𝐠),

2
hence

∇·(𝐇×𝐀)=𝐀·(∇×𝐇)-𝐇·(∇×𝐀)=𝐀·𝐉-𝐇·𝐁

and therefore

∭𝒱 d^3𝐱𝐉·𝐀=(𝐇×𝐀) · d^2𝐚+∭𝒱 d^3𝐱𝐇·𝐁 .

for any integration volume 𝒱 and its surface 𝒮. In eq. (9), we integrate over the conductor's volume, but we may just as well extend the integration volume 𝒱 to the whole space. Consequently, in eq. (12) the surface integral on the RHS goes away, and the magnetic energy (9) becomes

U=(1)/(2)∭ whole
space d^3x𝐇·𝐁 .

Added by Tiffany P.

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