Question

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

• A. $\overrightarrow{B}=2\times {10}^7\sin (0.5\times {10}^{3}z+1.5\times {10}^{11}t)\hat{i}$
• B. $\overrightarrow{B}=2\times {10}^{-7}\sin (0.5\times {10}^{3}z-1.5\times {10}^{11}t)\hat{i}$
• C. $\overrightarrow{B}=2\times {10}^{-7}\sin (0.5\times {10}^{3}x+1.5\times {10}^{11}t)\hat{j}$
• D. $\overrightarrow{B}=60\sin (0.5\times {10}^{3}x+1.5\times {10}^{11}t)\hat{k}$

Answer 1

The correct option is B $\overrightarrow{B}=2\times {10}^{-7}\sin (0.5\times {10}^{3}z-1.5\times {10}^{11}t)\hat{i}$

$B_0=\frac{E_0}{c}=\frac{60}{3\times {10}^8}=2\times {10}^{-7}\text{T}$

$K=\frac{2\pi f}{v}=\frac{2\pi \times 23.9\times {10}^9}{3\times {10}^8}\approx 500{\text{m}}^{-1}$

$\omega =2\pi f=2\times 3.14\times 23.9\times {10}^9=1.5\times {10}^{11}{\text{sec.}}^{-1}$

Therefore, $\overrightarrow{B}=B_0\sin (kz-\omega t)$

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

Hence, $\left(B\right)$ is the correct answer.