Question

The electric field associated with an e. $m$. wave in vacuum is given by $\vec{ E }=\hat{ i } 40 \cos \left( kz -6 \times 10^{8} t \right)$, where $E , z$ and $t$ are in volt $/ m$, meter and seconds respectively. The value of wave vector $k$ is :

• A. $2\, m^{-1}$
• B. $0.5\, m^{-1}$
• C. $6\, m^{-1}$
• D. $3\, m^{-1}$

Answer 1

Answer: (A)

Compare the given equation with
$E=E_{0} \cos (k z-\omega t)$
we get, $\omega=6 \times 10^{x} s ^{-1}$
Wave vector, $k=\frac{\omega}{c}=\frac{6 \times 10^{8} s ^{-1}}{3 \times 10^{8} m s ^{-1}}=2\, m ^{-1}$