# A plane electromagnetic wave having a frequency

Question:

A plane electromagnetic wave having a frequency $v=23.9 \mathrm{GHz}$ propagates along the positive $z$-direction in free space. The peak value of the Electric Field is $60 \mathrm{~V} / \mathrm{m}$. Which among the following is the acceptable magnetic field component in the electromagnetic wave ?

1. (1) $\vec{B}=2 \times 10^{7} \sin \left(0.5 \times 10^{3} z+1.5 \times 10^{11} t\right) \hat{i}$

2. (2) $\vec{B}=2 \times 10^{-7} \sin \left(0.5 \times 10^{3} z-1.5 \times 10^{11} t\right) \hat{i}$

3. (3) $\vec{B}=60 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{k}$

4. (4) $\vec{B}=2 \times 10^{-7} \sin \left(1.5 \times 10^{2} x+0.5 \times 10^{11} t\right) \hat{j}$

Correct Option: , 2

Solution:

(2) $B_{0}=\frac{E_{0}}{C}=\frac{60}{3 \times 10^{8}}$

$=20 \times 10^{-8} \mathrm{~T}=2 \times 10^{-7} \mathrm{~T}$

$K=\frac{\omega}{v}=\frac{2 \pi f}{v}=\frac{2 \pi \times 23.9 \times 10^{9}}{3 \times 10^{8}}=500$

Therefore, $\vec{B}=B_{0} \sin (k z-\omega t)$

$=2 \times 10^{-7} \sin \left(0.5 \times 10^{3} z-1.5 \times 10^{11} t\right) i$