The electric field associated with an electromagnetic wave in vacuum is given by `vecE=40cos(kz-6xx10^(8)t)hati,` where E, z and t are in volt per met
The electric field associated with an electromagnetic wave in vacuum is given by `vecE=40cos(kz-6xx10^(8)t)hati,`
where E, z and t are in volt per meter, meter and second respectively. The value of wave vector k is
A. `2 m^(-1)`
B. `0.5 m^(-1)`
C. `6 m^(-1)`
D. `3 m^(-1)`
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Correct Answer - A
Electromagnetic wave equation,
`E = E_(0)cos(kz - omegat)`
Speed of electromagnetic wave, `v = (omega)/(k)`
Given, equation,
`E = hati 40 cos(kz - 6 xx 10^(8)t)`
On comparing Eqs. (i) and (ii), we get
`omega = 6 xx 10^(8)` and `E_(0) = 40 hati`
Here, wave factor `k = (omega)/(v) = (6 xx 10^(8))/(3 xx 10^(8)) = 2 m^(-1)`
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