Question

Choose the correct answer from the alternatives given.
An electromagnetic wave travelling along $Z$-axis is given as $E=E_{0}cos(kz-\omega t)$.Choose the correct options from the following :

• A. The associated magnetic field is given as$\overrightarrow{B}=\frac{1}{c}(\hat{k}\times \overrightarrow{E}).$
• B. The electromagnetic field can be written in terms of the associated magnetic field as$\overrightarrow{E}=c(\overrightarrow{B}\times \hat{j}).$
• C. $\hat{k}\cdot \overrightarrow{E}=0,\hat{k}\cdot \overrightarrow{B}\ne 0.$
• D. $\hat{k}\cdot \overrightarrow{E}=0,\hat{k}\times \overrightarrow{B}=0.$

Answer 1

Answer: (A), (B), (C)

The correct options are
A The associated magnetic field is given as$\overrightarrow{B}=\frac{1}{c}(\hat{k}\times \overrightarrow{E}).$
B The electromagnetic field can be written in terms of the associated magnetic field as$\overrightarrow{E}=c(\overrightarrow{B}\times \hat{j}).$
C $\hat{k}\cdot \overrightarrow{E}=0,\hat{k}\cdot \overrightarrow{B}\ne 0.$

Given: An electromagnetic wave traveling along Z-axis is given as $E=E_0\cos (kz-\omega t)$

Since the direction of propagation of the wave is along the Z-axis, the Electric field will be perpendicular to the Z-axis will be acting in negative Y-axis

Then the associated magnetic field will be along the direction $(\hat{k}\times \hat{E})$.

$\overrightarrow{B}=\frac{1}{c}(\hat{k}\times \hat{E})=\frac{1}{\omega }(\hat{k}\times \hat{E})$

Now, the electric field will be along the direction $(\overrightarrow{B}\times \hat{k})$ i.e. perpendicular to the magnetic field and direction of motion.

The electromagnetic field can be written in terms of the associated magnetic field as

$\overrightarrow{E}=c(\overrightarrow{B}\times \hat{k}).$

The angle between the $\hat{k}$ and $\overrightarrow{E}$ will be ${90}^{\circ }$ whereas between the $\hat{k}$ and $\overrightarrow{B}$ will be ${180}^{\circ }$.

So, here $\hat{k}.\overrightarrow{E}=0,\hat{k}.\overrightarrow{B}\ne 0.$

Option $\left(A\right),\left(B\right)$ and $\left(C\right)$ are correct.