An em wave going through vacuum is described by `E=E_0sin(kx-omegat)`
`B=B_0sin(kx-omegat)`

Options

A. `E_(0)k = B_(0)omega`
B. `E_(0) omega = B_(0)k`
C. `E_(0) B_(0) = omegak`
D. None of these

Answer

A

Explanation

From `k = (2pi)/(lambda)` and `omega = 2piv`
we get, `k/(omega) = ((2pi)/(lambda))/(2piv) = v/(lambda) = 1/c`
As, `(E_(0))/(B_(0)) = c rArr (k)/(omega) = (B_(0))/(E_(0))`
`E_(0)k = B_(0)omega`