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Question: The electric field associated with an electromagnetic wave in vacuum is given b\(\overrightarrow{E} ...

The electric field associated with an electromagnetic wave in vacuum is given bE→=40cos⁡(kz−6×108t)i^,\overrightarrow{E} = 40\cos(kz - 6 \times 10^{8}t)\widehat{i}, where E, z and t are in volt per meter, meter and second respectively. The value of wave vector k is:

A

2m−12m^{- 1}

B

0.5m−10.5m^{- 1}

C

6m−16m^{- 1}

D

3m−13m^{- 1}

Answer

2m−12m^{- 1}

Explanation

Solution

: Compare the given equation with

E=E0cos⁡(kz−ωt)E = E_{0}\cos(kz - \omega t)

we get ω×6×108s−1\omega \times 6 \times 10^{8}s^{- 1}

wave vector,

K=ωc=6×108s−13×108ms−1=2m−1K = \frac{\omega}{c} = \frac{6 \times 10^{8}s^{- 1}}{3 \times 10^{8}ms^{- 1}} = 2m^{- 1}