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An EM wave radiates outwards from a dipole antenna, with $\mathrm{E}_0$ as the amplitude of its electric field vector. The electric field $\mathrm{E}_0$ which transports significant energy from the source falls off as
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The correct answer is:
$\frac{1}{r}$
$\frac{1}{r}$
As we know that, the electric field is inversly proportional to $\mathrm{r}$, so $\left(\mathrm{E}_0 \propto \frac{1}{\mathrm{r}}\right)$
From a diode antenna, an electromagnetic waves are radiated outwards from dipole antenna with the amplitude of electric field vector $\left(\mathrm{E}_0\right)$ which transports significant energy from the source falls off intensity inversely as the distance (r) from the antenna, i.e.,
$$
\text { radiated energy }\left(E_0 \propto \frac{1}{r}\right)
$$
From a diode antenna, an electromagnetic waves are radiated outwards from dipole antenna with the amplitude of electric field vector $\left(\mathrm{E}_0\right)$ which transports significant energy from the source falls off intensity inversely as the distance (r) from the antenna, i.e.,
$$
\text { radiated energy }\left(E_0 \propto \frac{1}{r}\right)
$$
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