A steady current density is confined to a bounded region: outside some finite volume

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

The Boundary-Value Problem, Ch The magnetic multipole expansion and the magnetic scalar potential

Problem

A steady current density J(r′)\mathbf{J}(\mathbf{r}') is confined to a bounded region: J=0\mathbf{J} = 0 outside some finite volume. Its vector potential is given, in the Coulomb gauge, by the solution integral 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'. For field points with r>r′r > r' for every source point, the Legendre generating function gives 1∣r−r′∣=1r∑ℓ=0∞(r′r)ℓPℓ(cos⁡γ),\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{1}{r}\sum_{\ell=0}^{\infty}\left(\frac{r'}{r}\right)^{\ell}P_\ell(\cos\gamma), where γ\gamma is the angle between r\mathbf{r} and r′\mathbf{r}'. ((a)) Insert the generating function into the solution integral and write out the ℓ=0\ell=0 and ℓ=1\ell=1 terms of A(r)\mathbf{A}(\mathbf{r}) explicitly, as integrals over the source. ((b)) Show that the ℓ=0\ell=0 term vanishes for any steady, localised current. Start from the identity Ji=∇′⋅(xi′J)−xi′(∇′⋅J)J_i = \nabla'\cdot(x_i'\mathbf{J}) - x_i'(\nabla'\cdot\mathbf{J}), integrate over a volume enclosing all the current, and apply the divergence theorem. Name the step that requires the current to be steady and the step that requires it to be localised. ((c)) The ℓ=1\ell=1 term can be rewritten using the identity (r′⋅r^)J=12[(r′⋅r^)J+(J⋅r^)r′]+12(r′×J)×r^.(\mathbf{r}'\cdot\hat{\mathbf{r}})\mathbf{J} = \tfrac12\left[(\mathbf{r}'\cdot\hat{\mathbf{r}})\mathbf{J} + (\mathbf{J}\cdot\hat{\mathbf{r}})\mathbf{r}'\right] + \tfrac12(\mathbf{r}'\times\mathbf{J})\times\hat{\mathbf{r}}. The symmetric bracket integrates to zero by the same steady-and-localised argument as in part (b) (you may quote this without repeating the full divergence-theorem steps). Hence the ℓ=1\ell=1 term is A1(r)=μ04πm×r^r2,m=12∫r′×J dτ′.\mathbf{A}_1(\mathbf{r}) = \frac{\mu_0}{4\pi}\frac{\mathbf{m}\times\hat{\mathbf{r}}}{r^2}, \qquad \mathbf{m} = \frac12\int \mathbf{r}'\times\mathbf{J}\,d\tau'. Apply this definition of m\mathbf{m} to a thin spherical shell of radius aa carrying total charge QQ spread uniformly over its surface, spinning with steady angular velocity ω\boldsymbol{\omega} about a diameter. Show that m=13Qa2ω\mathbf{m} = \tfrac13 Qa^2\boldsymbol{\omega}.

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