Question

The electric field for an electromagnetic wave in free space is $\overrightarrow{E}=\overrightarrow{i}30\cos (kz-5\times {10}^{8}t)$, where magnitude of $\overrightarrow{E}$ is in $V/m$. The magnitude of wave vector, $k$ is:

(velocity of em wave in free space $=3\times {10}^{8}m/s$)

• A. $0.46$ $rad{m}^{-1}$
• B. $3$ $rad{m}^{-1}$
• C. $1.66$ $rad{m}^{-1}$
• D. $0.83$ $rad{m}^{-1}$

Answer 1

The correct option is C $1.66$ $rad{m}^{-1}$

Electric field of an EM wave $\overrightarrow{E}=30(kz-5\times {10}^{8}t)\hat{i}$

$⟹$ $\omega =5\times {10}^8$ rad/s

Given : Velocity of EM wave $v=3\times {10}^8$ $m/s$

Using $v=\frac{w}{k}$ $⟹k=\frac{w}{v}$

$\therefore$ $k=\frac{5\times {10}^8}{3\times {10}^8}=1.66$ rad $m^{-1}$