Classical Electrodynamics · Problem 11.29

Radiating Systems, Multipole Fields and Radiation — John David Jackson

Physics — E&M · moving-magnetic-dipole-potentials

Problem

Instead of the electric dipole potential of Problem 11.28, consider a point magnetic moment m\mathbf{m} in the moving frame K′K', with its potentials, Φ′=0\Phi' = 0, A′=m×r′/r′3\mathbf{A}' = \mathbf{m} \times \mathbf{r}'/r'^3 (and so only a magnetic field). (a) Show that to first order in β\beta the potentials in KK are Φ=(β×m)⋅RR3,A=(m×R)R3\Phi = \frac{(\boldsymbol{\beta} \times \mathbf{m}) \cdot \mathbf{R}}{R^3}, \quad \mathbf{A} = \frac{(\mathbf{m} \times \mathbf{R})}{R^3} Note that the scalar potential is the same as the static potential of the electric dipole moment of Problem 11.27b. [But this gives only the irrotational part of the electric field.] (b) Calculate the electric and magnetic fields in KK from the potentials and show that the electric field can be expressed alternatively as E=Edipole(peff=β×m)−m×[3n(n⋅β)−β]R3\mathbf{E} = \mathbf{E}_{\rm dipole}(\mathbf{p}_{\rm eff} = \boldsymbol{\beta} \times \mathbf{m}) - \mathbf{m} \times \frac{[3\mathbf{n}(\mathbf{n} \cdot \boldsymbol{\beta}) - \boldsymbol{\beta}]}{R^3} E=Edipole(peff=β2×m)+32n[m(n⋅β)+β(n⋅m)]R3\mathbf{E} = \mathbf{E}_{\rm dipole}\left(\mathbf{p}_{\rm eff} = \frac{\boldsymbol{\beta}}{2} \times \mathbf{m}\right) + \frac{3}{2}\mathbf{n}\frac{[\mathbf{m}(\mathbf{n} \cdot \boldsymbol{\beta}) + \boldsymbol{\beta}(\mathbf{n} \cdot \mathbf{m})]}{R^3} E=B×β\mathbf{E} = \mathbf{B} \times \boldsymbol{\beta} where B\mathbf{B} is the magnetic dipole field. In light of Problem 6.22, comment on the interpretation of the different forms.

Related theory

Browse the problem library