Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 15C (Electromagnetism)

Differentiate under the integral and use ${\nabla }_x|x-x^{\prime }{|}^{-1}=-{\nabla }_{x^{\prime }}|x-x^{\prime }{|}^{-1}$:The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation ${\nabla }^{\prime }\cdot J=0$, hence

For $r=|x|$ much larger than the source size,Localization and ${\nabla }^{\prime }\cdot J=0$ imply $\int J{d}^3{x}^{\prime }=0$. They also implyby integrating ${\partial }_k^{\prime }(x_i^{\prime }{x}_j^{\prime }{J}_k)$. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole momentConsequently the Coulomb-gauge vector potential of a localized steady current has far fieldThe dimensions are

The total hoop charge is $Q=2\pi R\eta$. One revolution takes $2\pi /\omega$, so the current isMultiplying by the enclosed area gives

An annulus of radius $s$ and width $ds$ has charge $dQ=2\pi s\sigma ds$, currentand dipole moment $dm=\pi s^{2}dI\hat{n}$. Therefore the magnetic dipole moment of a rigidly rotating charge distribution is