Magnetic Dipole Radiation

Let's look at the antisymmetric bit first, as it is somewhat simpler and we can leverage our existing results. The second term is the magnetization (density) due to the current $\mbox{\boldmath$J$}$:

$$\vec{\cal M} = \frac{1}{2} (\mbox{\boldmath$x$}\times \mbox{\boldmath$J$})$$ (11.143)

(see J5.53, 5.54) so that

$$\mbox{\boldmath$m$}= \int \vec{\cal M}({\bf x'}) d^3x'$$ (11.144)

where $\mbox{\boldmath$m$}$ is the magnetic dipole moment of the (fourier component of) the current.

Considering only this antisymmetric term, we see that:

$$\mbox{\boldmath$A$}_{\rm M1} (\mbox{\boldmath$x$}) = \frac{... ...f m}) \frac{e^{ikr}}{r} \left ( 1 - \frac{1}{ikr} \right ).$$ (11.145)

HMMMMMMM, (you had better say)! This looks ``just like'' the expression for the magnetic field that resulted from the electric dipole vector potential. Sure enough, when you (for homework) crank out the algebra, you will show that

$$\mbox{\boldmath$B$}= \frac{\mu_0}{4\pi} \left\{k^2 ({\bf n}... ...ft ( \frac{1}{r^3} - \frac{ik}{r^2} \right ) e^{ikr}\right\}$$ (11.146)

and

$$\mbox{\boldmath$E$}= -\frac{1}{4\pi} \sqrt{\frac{\mu_0}{\ep... ...bf m}) \frac{e^{ikr}}{r} \left ( 1 - \frac{1}{ikr} \right ).$$ (11.147)

Clearly, we don't need to discuss the behavior of the fields in the zones since they are completely analogous. The electric field is always transverse, and the total field arises from a harmonic magnetic dipole. For this reason, this kind of radiation is called either magnetic dipole (M1) radiation or transverse electric radiation. For what it's worth, electric dipole radiation is also called (E1) radiation.

However, this is only ONE part of the contribution from $\ell = 1$ terms in the Green's function expansion. What about the other (symmetric) piece? Oooo, ouch.