 Solution

ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2022/section-b/5/a/solution

Only the

1/ r

fields transport a nonzero outgoing power to infinity. Retain those terms from the Hertzian dipole fields:

E_{θ}^{rad} = \frac{s i n θ}{4 π ϵ _{0} c ^{2} r} [\overset{p}{¨}], B_{ϕ}^{rad} = \frac{μ _{0} s i n θ}{4 π cr} [\overset{p}{¨}] = \frac{E _{θ}^{rad}}{c} .

[\overset{p}{¨}] \equiv \overset{p}{¨} (t - r / c),

for a point source at the origin in vacuum. They are not a time average. The radial Poynting vector in the radiation zone is

S_{r} = \frac{1}{μ _{0}} E_{θ}^{rad} B_{ϕ}^{rad} = \frac{μ _{0}}{16 π ^{2} c r ^{2}} [\overset{p}{¨}]^{2} sin^{2} θ .

Integrating over a sphere uses

\int sin^{2} θ d Ω = 2 π \int_{0}^{π} sin^{3} θ d θ = \frac{8 π}{3} .

Therefore

P (t, r) = \int r^{2} S_{r} d Ω = \frac{μ _{0}}{6 π c} [\overset{p}{¨}]^{2} .

Near-field terms either give inverse powers of

r

after the surface integral or describe reversible exchange with the near field. The emitted power labelled by source time

t_{src}

μ_{0} \overset{p}{¨} (t_{src})^{2} / (6 π c)

; the enclosing surface receives it one light-travel time later. For a changing dipole orientation, the general form is

P = μ_{0} ∣ \ddot{p} ∣^{2} / (6 π c)

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