This problem works through the three steps that make useful: its existence

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Magnetostatics · Magnetostatics

The Boundary-Value Problem, Ch The vector potential and the magnetostatic boundary-value problem

Problem

This problem works through the three steps that make A\mathbf{A} useful: its existence, the freedom to fix its divergence, and the recovery of the Biot-Savart law from its solution integral. ((a)) Starting only from ∇⋅B=0\nabla\cdot\mathbf{B}=0 (true everywhere, for any current distribution), construct an explicit vector potential by defining Ax(x,y,z)=∫z0zBy(x,y,z′) dz′−∫y0yBz(x,y′,z0) dy′,Ay(x,y,z)=−∫z0zBx(x,y,z′) dz′,Az=0.\begin{aligned} A_x(x,y,z) &= \int_{z_0}^{z}B_y(x,y,z')\,dz'-\int_{y_0}^{y}B_z(x,y',z_0)\,dy', \\ A_y(x,y,z) &= -\int_{z_0}^{z}B_x(x,y,z')\,dz', \\ A_z &= 0. \end{aligned} Show that ∇×A=B\nabla\times\mathbf{A}=\mathbf{B}, using ∇⋅B=0\nabla\cdot\mathbf{B}=0 at the one step where it is needed. State why the electrostatic route E=−∇V\mathbf{E}=-\nabla V has no analogue here. ((b)) A uniform field B0z^B_0\hat{\mathbf{z}} has vector potential A0=B0x y^\mathbf{A}_0=B_0x\,\hat{\mathbf{y}}, which satisfies ∇⋅A0=0\nabla\cdot\mathbf{A}_0=0. A second, equally valid potential for the same field is A1=A0+∇χ,χ=κx2y,\mathbf{A}_1=\mathbf{A}_0+\nabla\chi, \qquad \chi=\kappa x^2y, with κ\kappa a constant. Write out A1\mathbf{A}_1 in components, verify that ∇×A1=B0z^\nabla\times\mathbf{A}_1=B_0\hat{\mathbf{z}} still, and show that ∇⋅A1≠0\nabla\cdot\mathbf{A}_1\neq0. Find the gauge function λ\lambda that brings A1\mathbf{A}_1 back into the Coulomb gauge, so that A1+∇λ\mathbf{A}_1+\nabla\lambda is divergence-free, i.e. solve ∇2λ=−∇⋅A1\nabla^2\lambda=-\nabla\cdot\mathbf{A}_1, and identify the resulting potential. ((c)) For a localised steady current with ∇⋅J=0\nabla\cdot\mathbf{J}=0, the Coulomb-gauge solution is A(r)=μ04π∫J(r′)∣r−r′∣ dτ′.\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,d\tau'. Take the curl with respect to r\mathbf{r} under the integral sign and show that the result is the Biot-Savart law.

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