Question

An electromagnetic wave passing through vacuum is described by the equations $E=E_0\sin (kx-\omega t)$ and $B=B_0\sin (kx-\omega t)$. Then,
A. $E_{0}k=B_0\omega$
B. $E_0\omega =B_{0}k$
C. $E_0{B}_0=\omega k$
D. $E_0=B_0$

Answer 1

Hint: Find the relation between electric field and magnetic field. They are related to each other by c i.e. speed of light, Write the equation for c in terms of wavelength. Then, get the expression for c in terms of angular frequency and wave number. Finally, substitute that value in the relation of electric field and magnetic field.

Complete answer:
Given: Equations of electromagnetic waves: $E=E_0\sin (kx-\omega t)$ and $B=B_0\sin (kx-\omega t)$
The relation between Electric field $E_0$ and Magnetic field $B_0$ is given by,
$E_0=c{B}_0$ …(1)
We know, $\nu ={\displaystyle \frac{c}{\lambda }}$
Rearranging the above equation we get,
$c=\nu \lambda$ …(2)
We also know, $\nu ={\displaystyle \frac{\omega }{2\pi }}$
Thus, substituting this value in the equation. (2) we get,
$c={\displaystyle \frac{\omega }{2\pi }}\lambda$
Now, substituting $\frac{2\pi }{\lambda }=k$ we get,
$c={\displaystyle \frac{\omega }{k}}$ …(2)
Then, substituting the equation. (3) in equation. (1) we get,
$E_0={\displaystyle \frac{\omega }{k}}{B}_0$
Rearranging the above equation we get,
$E_{0}k=B_0\omega$

So, the correct answer is “Option A”.

Note:
The direction of electromagnetic waves is found by vector cross product of the electric field and magnetic field. Electromagnetic radiations also show dual nature i.e. have the properties of a wave as well as the properties of a particle.