Discussion
For a localized time varying current distribution, if the Poynting vector, \(\vec{S}\) varies as \(|\vec{S}| \propto \frac{1}{r^{n}}\), where r is the distance from the center of the current distribution, then ..........
Answer: A
The answer is \(n \le 2\), because the surface integral of the Poynting vector must be non-zero for the radiating fields.
A localized time-varying current distribution radiates. The power crossing a large sphere of radius \(r\) is
\[ \begin{aligned} P &= \oint \vec{S}\cdot d\vec{a} \\ &\sim |\vec{S}|\, 4\pi r^{2} \propto r^{2-n} \end{aligned} \]
Radiation means energy that escapes to infinity, so \(P\) must stay finite and non-zero as \(r \to \infty\). That requires the radiation part of \(\vec{S}\) to fall exactly as \(1/r^{2}\), which happens because the radiation fields fall as \(1/r\). A faster fall-off (\(n \gt 2\)) would carry no energy to infinity.
The option claiming \(n \gt 2\) for radiating fields contradicts this, the option with \(n \gt 2\) and non-radiating fields denies that a time-varying localized source radiates, and the general argument above holds without further detail of the source.
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