The electric field of an electromagnetic wave in free space is given by $\vec{E}=10 \cos \left(10^7 t+k…

The electric field of an electromagnetic wave in free space is given by $\vec{E}=10 \cos \left(10^7 t+k x\right) \hat{j} V / m$, where $t$ and $x$ are in seconds and metres respectively. It can be inferred that (1) The wavelength $\lambda$ is $188.4 \mathrm{~m}$. (2) The wave number $k$ is $0.33 \mathrm{rad} / \mathrm{m}$. (3) The wave amplitude is $10 \mathrm{~V} / \mathrm{m}$. (4) The wave is propagating along.t $x$ direction Which one of the following pairs of statements is correct?
  1. (3) and (4)
  2. (1) and (2)
  3. (2) and (3)
  4. (1) and (3)

Solution

The electric field of electromagnetic wave $\begin{aligned} & \overrightarrow{\mathrm{E}}=10 \cos \left(10^7 \mathrm{t} \pm \mathrm{kx}\right) \hat{\mathrm{j}} \\ & \text { Amplitude }=10 \mathrm{~V} / \mathrm{m} \\ & \because \quad \mathrm{c}=\frac{\omega}{\mathrm{k}} \\ & \therefore \quad 3 \times 10^8=\frac{10^7}{\mathrm{k}} \\ & \text { or } \quad k=\frac{1}{30} \\ & \text { or } \quad \frac{2 \pi}{\lambda}=\frac{1}{30} \\ & \text { or } \quad \lambda=188.4 \mathrm{~m} \\ & \end{aligned}$ So, (1) and (3) option are correct.

Asked in: NEET 2010 (Mains)

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