This problem works through three of the structural results behind the vector potential

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Magnetostatics · Maxwell's equations and conservation laws

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

Problem

This problem works through three of the structural results behind the vector potential, using only ∇⋅B=0\nabla\cdot\mathbf{B}=0, the definition of the curl and divergence, and the free-space Green function of Poisson's equation. ((a)) Existence. A steady field B(r)\mathbf{B}(\mathbf{r}) satisfies ∇⋅B=0\nabla\cdot\mathbf{B}=0 everywhere. Consider the trial potential, with x0=y0=z0=0x_0=y_0=z_0=0 taken as the lower limits of every integral, Ax(x,y,z)=∫0zBy(x,y,z′) dz′−∫0yBz(x,y′,0) dy′,Ay(x,y,z)=−∫0zBx(x,y,z′) dz′,Az=0.\begin{aligned} A_x(x,y,z) &= \int_0^z B_y(x,y,z')\,dz' - \int_0^y B_z(x,y',0)\,dy', \\ A_y(x,y,z) &= -\int_0^z B_x(x,y,z')\,dz', \\ A_z &= 0. \end{aligned} Show, using ∇⋅B=0\nabla\cdot\mathbf{B}=0 only where it is needed, that ∇×A=B\nabla\times\mathbf{A}=\mathbf{B} at every point, so that a vector potential of B\mathbf{B} exists. ((b)) Reaching the Coulomb gauge. In a region where the true field is the uniform B=B0z^\mathbf{B}=B_0\hat{\mathbf{z}} (B0B_0 a constant, unit T\mathrm{T}), someone proposes A=Cx x^+B0x y^,\mathbf{A} = Cx\,\hat{\mathbf{x}} + B_0x\,\hat{\mathbf{y}}, with CC a constant (unit T\mathrm{T}). (i) Verify that ∇×A=B0z^\nabla\times\mathbf{A}=B_0\hat{\mathbf{z}}, so this is a legitimate vector potential of the field. (ii) Show that ∇⋅A=C≠0\nabla\cdot\mathbf{A}=C\ne0 in general, so this A\mathbf{A} is not in the Coulomb gauge. (iii) Find a gauge function λ(x,y,z)\lambda(x,y,z) solving ∇2λ=−∇⋅A\nabla^2\lambda=-\nabla\cdot\mathbf{A} that removes the divergence, and give the resulting A′=A+∇λ\mathbf{A}'=\mathbf{A}+\nabla\lambda. Confirm ∇×A′=B0z^\nabla\times\mathbf{A}'=B_0\hat{\mathbf{z}} still. ((c)) Recovering Biot-Savart. A steady current density J(r′)\mathbf{J}(\mathbf{r}'), confined to a bounded region and satisfying ∇⋅J=0\nabla\cdot\mathbf{J}=0, produces 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 of this integral with respect to the field point r\mathbf{r} (the integrand's r′\mathbf{r}'-dependence is untouched by ∇\nabla) and show that the result is the Biot-Savart law, B(r)=μ04π∫J(r′)×r−r′∣r−r′∣3 dτ′.\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}')\times\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\,d\tau'. State where the identity ∇×[fJ(r′)]=(∇f)×J(r′)\nabla\times[f\mathbf{J}(\mathbf{r}')] = (\nabla f)\times\mathbf{J}(\mathbf{r}') is used and why the term f(∇×J)f(\nabla\times\mathbf{J}) that the product rule would otherwise add is absent.

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