## EMC Question of the Week: July 27, 2020

What is the wavelength of a 300 MHz electromagnetic plane wave that is represented using phasor notation as,

$\overrightarrow{E}=10\text{\hspace{0.33em}}{e}^{-j3\pi x}\widehat{\text{\hspace{0.17em}}y}\text{V/m}$

- 0.67 meters
- 1.0 meters
- 1.5 meters
- not enough information

## Answer

The correct answer is "a". From the expression we know that the electric field has a peak amplitude of 10 V/m, is polarized in the $\widehat{\text{\hspace{0.17em}}y}$ direction and is propagating in the $\widehat{\text{\hspace{0.17em}}x}$ direction. The phase constant, β, is 3π. The wavelength is 2π/β, which is equal to 0.67 meters.

Note that this plane wave is not propagating in free space. The free space wavelength would be 1.0 meters. The velocity of propagation for this wave is v = ω/β = 2 x 10^{8} m/sec.

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