Question

An electromagnetic wave going through vacuum is described by
$E=E_0\sin (kx-\omega t)$
$B=B_0\sin (kx-\omega t)$.
Which of the following is true ?

• A. $E_0{B}_0=\omega k$
• B. $E_0\omega =B_{0}k$
• C. $E_{0}k=B_0\omega$
• D. $E_0=B_0$

Answer 1

The correct option is C $E_{0}k=B_0\omega$

The speed of light in vacuum is given by,

$C=\frac{E_0}{B_0}$ and $C=\frac{\omega }{k}$

$\therefore \frac{E_0}{B_0}=\frac{\omega }{k}$

$\therefore E_{0}k=B_0\omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.