Editing Openai/6897769e-4ee4-800f-aba5-69cca34f701c (section)

==== Integrate ∇×B=μ0J+μ0ε0∂tE\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\partial_t\mathbf{E}∇×B=μ0J+μ0ε0∂tE over the pillbox volume and apply Stokes’ theorem to the tiny loop that runs around the pillbox edge. More simply, take the tangential components and integrate across the small height: ====

In the limit h→0h\to0h→0 the dominant contribution is from the tangential integral across the top and bottom faces. One obtains

n^×(Bout−Bin)=μ0K∥(θ,φ,t)+μ0ε0n^×(Eout−Ein)⋅(order h).\hat n\times(\mathbf{B}_{\rm out}-\mathbf{B}_{\rm in}) = \mu_0 \mathbf{K}_\parallel(\theta,\varphi,t) + \mu_0\varepsilon_0 \hat n\times \big(\mathbf{E}_{\rm out}-\mathbf{E}_{\rm in}\big) \cdot(\text{order }h).n^×(Bout−Bin)=μ0K∥(θ,φ,t)+μ0ε0n^×(Eout−Ein)⋅(order h).
The displacement term is order hhh and drops out as h→0h\to0h→0. Hence,

  n^×(Bout−Bin)  =  μ0 j∥(θ,φ,t)  (Ampeˋre jump)\boxed{\;\hat n\times(\mathbf{B}_{\rm out}-\mathbf{B}_{\rm in}) \;=\; \mu_0\,\mathbf{j}_\parallel(\theta,\varphi,t)\; } \tag{Ampère jump}n^×(Bout−Bin)=μ0j∥(θ,φ,t)(Ampeˋre jump)
where j∥\mathbf{j}_\parallelj∥ is the tangential surface current (the spatial part of jsurfμj^\mu_{\rm surf}jsurfμ).

QAT interpretation: j∥\mathbf{j}_\parallelj∥ is produced by charge motion along the spherical boundary after absorption (decoherence and redistribution); it is the agent that radiates (emission) and rearranges local field lines.