Question

An electromagnetic wave going through vacuum is described by $E=E_o\sin (kx-\omega t)$; $B=B_o\sin (kx-\omega t)$.
Which of the following equation is true?

• A. E₀k = B₀ω
• B. E₀ω = B₀k
• C. E₀B₀ = ωk
• D. None of these

Answer 1

The correct option is A

E₀k = B₀ω

Electromagnetic wave

1. An electromagnetic wave is transverse in nature (example: light).
2. We know. an electromagnetic wave consists of a time-varying electric field $E$and a time-varying magnetic field $B$.
3. The velocity of the electromagnetic wave is $c=3\times {10}^{8}m/s$, which is defined by the form, $c=\nu \lambda$, where, $\lambda$ is the wavelength of the wave and $\nu$ is the frequency of the wave.
4. In the case of electromagnetic wave, velocity, $c$ is defined as the ratio of electric field and magnetic field, i.e, $c=\frac{E}{B}$.

Explanation for the correct option

Option (A) : E₀k = B₀ω

As we know, $c=\frac{E}{B}$

$$\begin{gathered} \frac{E}{B}=c=\nu \lambda \\ or\frac{E}{B}=\frac{\omega }{2\pi }\times \frac{2\pi }{k}\left(\sin ce,\lambda =\frac{2\pi }{k}and\omega =2\pi \nu \right) \\ or\frac{E}{B}=\frac{\omega }{k} \\ orEk=B\omega \end{gathered}$$

So, $E_{o}k=B_o\omega$

Here, $k$ is the propagation vector and $\omega$ is the angular frequency.

So, option (A) is correct.