Question

The electric field associated with an e.m. wave in vacuum is given as $\overrightarrow{E}=40\cos (kz-6\times {10}^{8}t)\hat{i}$, where all quantities are in SI units. The value of the propagation constant $k$ is,

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

Answer 1

The correct option is C $2{\text{m}}^{-1}$

Given:
$\overrightarrow{E}=40\cos (kz-6\times {10}^{8}t)\hat{i}$

From the given equation, we can say that the electric field is pointed along the $x-$axis.
Comparing the given equation with the standard equation,
$\overrightarrow{E}=E_0\cos (kz-\omega t)\hat{i}$

We get, $E_0=40\text{N/C}$
and, $\omega =6\times {10}^8\text{rad/s}$

The speed of electromagnetic wave is given by,
$v=\frac{\omega }{k}$

It is given that the wave is travelling in the vacuum, so, $v=3\times {10}^8\text{m/s}$

$\Rightarrow 3\times {10}^8=\frac{6\times {10}^8}{k}$
$\therefore k=2{\text{m}}^{-1}$

Hence, $\left(C\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{This question was asked in the NEET 2012 exam.}\end{array}$