Question

The electric field of a wave in free space is given by $E=10\cos ({10}^{7}t+kx)\hat{j}V{m}^{-1}$. It can be inferred that

• A. the wave amplitude is $10V{m}^{-1}$ and is propagating along $+ve$ $x$-direction
• B. the direction and its wave number is $0.66rad{m}^{-1}$
• C. the wave number $k$ is $0.33rad{m}^{-1}$ and its amplitude is $15V{m}^{-1}$
• D. the wavelength $\left(\lambda \right)$ is $188.5m$ and is propagating along $-ve$ $x$-direction

Answer 1

Answer: (D)

the wavelength $\left(\lambda \right)$ is $188.5m$ and is propagating along $-ve$ $x$-direction

Comparing the given equation with standard progressive wave equation, we see that the wave is travelling along -x-axis and
$\omega ={10}^7$.
Now, $v=\frac{\omega }{k}\Rightarrow 3\times {10}^8=\frac{{10}^7}{k}\Rightarrow k=\frac{0.1}{3}=0.033$ rad/m.
Since $k=\frac{2\pi }{\lambda }$, $\lambda =\frac{3\times 2\pi }{0.1}=188.5$ m.
Ans: D