{\displaystyle {\begin{aligned}H_{\phi }&=i{\frac {\ I\ \delta \ell \ k\ }{4\pi r}}\ e^{-ikr}\ \sin \theta \ ,\\[2pt]E_{\theta }&=i{\frac {\ \zeta _{0}\ I\ \delta \ell \ k\ }{4\pi r}}\ e^{-ikr}\ \sin \theta ~.\end{aligned}}}