Electromagnetic waves waves review

AI Thread Summary
The discussion focuses on solving problems related to a plane electromagnetic wave with a wavelength of 56.0 m and an electric field amplitude of 20.0 V/m. The frequency of the wave is calculated to be 5.35 x 10^6 Hz. Participants clarify that the maximum electric field (Emax) is indeed 20 V/m, which simplifies the calculations for the magnetic field. The equations for the electric and magnetic fields are discussed, with emphasis on their respective amplitudes and wave properties. The conversation concludes with the participant successfully solving the remaining parts of the problem.
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Homework Statement


The figure below shows a plane electromagnetic sinusoidal wave propagating in the x direction. Suppose the wavelength is 56.0 m and the electric field vibrates in the xy plane with an amplitude of 20.0 V/m.

34-p-013.gif
(a) Calculate the frequency of the wave.

(b) Calculate the magnetic field when the electric field has its maximum value in the negative y direction.

(c) Write an expression for with the correct unit vector, with numerical values for Bmax, k, and ω, and with its magnitude in the form
B = Bmax cos (kx − ωt).
(Assume B is measured in nT, x is measured in m and t in s.)

Homework Equations



c = 1/(μ0ε0)1/2

c = 299609581 m/s

f = c / λ

E = Emax cos(kx - ωt)

B = Bmax cos(kx - ωt)

E max / B max = E / B = ck = 2π/λ

ω = 2πf

The Attempt at a Solution



I only found the frequency, which is 5.35 x 106 HzI would like some hints how to find E max then I can just use the third equation listed.I'm thinking that E = Emax cos(kx - wt) and I think

E = (20V/m) cos[(0.1121)(x) - (33615041)(t)]

where 0.1121 is k (k=2pi / λ)

and where 33615041 is ω (ω = 2πf)
What would I do with regards to x and t in the equation?
 
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The amplitude of the wave is given to you, and that is the magnitude of the electric field when it is maximal, isn't it?

Edit: wait, why are you asking about E_\text{max} when you already observed it's 20 \; \text{V/m}
 
I didn't really see that... thanks

I got the rest of the problem as well.
 
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