The electrical field for an electromagnetic wave in free space `E=i30 (kz -5 xx10^(8)t)` ,where magnitude of E is in V/m. The magnitude of wave vector .k is (velocity of em wave in free space `=3xx 10^(8) m//s`

Options

A. `0.46 rad m^(-1)`
B. `3 rd m^(-1)`
C. `1.66 rad m^(-1)`
D. `0.83 rad m(-1)`

Answer

C

Explanation

`E=i30cos(kx-(5xx10^(8))t)`
Comparing with,
`E=iVcos(kz-wt)`
`V=30V,omega=5xx10^(8)rad//s`
`K=(2pi)/(lamda),omega=2pin`
`(omega)/(K)=(2pin)/((2pi)/lambda)=nlambda=b`
`b=3xx10^(8)m//s(Given)`(omega)/(K)=3xx10^(8)`
`impliesK=(5xx10^(8))/(3xx10^(8))=1.66rad//m.`