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Now, I'm not sure if this is enough information to go off, but I'll give it a shot.

I'm trying to find the intensity of an electromagnetic wave using the formula $\frac{1}{2} c \epsilon_0 E_0^2$, however I'm not sure how to find E.

The radio wave has a frequency of 1600 kHz, giving it a period of 625 nanoseconds and wavelength of 187.37 meters. According to my oscilloscope, the radio wave (travelling through air) has an amplitude of 2.5 mV.

Is it possible to find E and the intensity of the wave from this info?

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  • $\begingroup$ What is this formula "$\frac{1}{2} c \epsilon_0 E_0^2$" you are referring to and what is $E$? $\endgroup$
    – hyportnex
    Commented Apr 16, 2017 at 11:54
  • $\begingroup$ Oh sorry. If I understand it correctly, it's related to the Poynting vector. It's explained a bit here. Apparently E is the peak electric field, but I have also heard it described as the complex amplitude of the wave. $\endgroup$
    – Kris Walker
    Commented Apr 17, 2017 at 22:20
  • $\begingroup$ OK, so $c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ then $\frac{1}{2} c \epsilon_0 E_0^2 = \frac{1}{2} \sqrt \frac{\epsilon_0} {\mu_0}E_0^2$ is the average power transmitted per unit area with $E_0$ being the electric field amplitude. To estimate what this has to do with the observed voltage amplitude on the oscilloscope you must know what the radiation resistance and effective area of your receiver antenna are. Are $E$ and $E_0$ the same? $\endgroup$
    – hyportnex
    Commented Apr 17, 2017 at 22:30
  • $\begingroup$ How would I go about calculating those for a handheld transistor radio? Also, I'm not sure if $E$ and $E_0$ are the same. $\endgroup$
    – Kris Walker
    Commented Apr 20, 2017 at 9:27
  • $\begingroup$ An electromagnetic wave has an amplitude measure in V/m, not V. $\endgroup$
    – ProfRob
    Commented Apr 20, 2017 at 17:41

1 Answer 1

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A handheld "transistor" radio receives around 1MHz band, wavelength 300m. All receive antennas in that band are tiny compared to the wavelength for the dominant noise is atmospheric not thermal noise in the amplifier, so it does not matter much how efficient is the receive antenna.

For such signals most receive antennas are ferrite loaded resonant coils whose received open circuit voltage and power in the load are approximately: $$V_{oc}=-\iota \omega \mu_e \mu_0 N (\pi a^2) |H_i| $$ $$P_{rec} = \frac{(Q k_0 Z_0 N (\pi a^2))^2 |H_i|^2 }{8R_L}$$ where $Q=\frac{R}{\omega L}$, the magnetic field $H_i$ points along the axis of the coil of radius $a$ and length $\mathcal l$ whose inductance is $L=\mu_e \mu_0 N^2 \frac{\pi a^2}{\mathcal l}$.

For further details I suggest you read Collin: Antennas and Radiowave Propagation, chapter 5.3

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