Question

The electric field of a plane polarized electromagnetic wave in free space at time $$t=0$$ is given by an expression. $$\overrightarrow{E}(x,y)=10\hat{j}\cos [(6x+8z)]$$. The magnetic field $\overrightarrow{B}(x,z,t)$ is given by (c is the velocity of light):
(a) ${\displaystyle \frac{1}{c}}(4\hat{k}+8\hat{i})\cos [(2x-8z+20ct)]$
(b) ${\displaystyle \frac{1}{c}}(6\hat{k}-8\hat{i})\cos [(6x+8z-10ct)]$
(c) ${\displaystyle \frac{1}{c}}(5\hat{k}+8\hat{i})\cos [(6x+8z-80ct)]$
(d) ${\displaystyle \frac{1}{c}}(4\hat{k}-8\hat{i})\cos [(6x+8z+70ct)]$

Answer 1

Hint: A plane polarized EM wave means an electric field lies in a single plane and amplitude of field would be oscillating in the same plane only unlike unpolarized light. The direction of propagation of waves has the same direction given by the k vector of electromagnetic waves.

Formula used:
1.General expression of Electric field:
   $\overrightarrow{E}(\overrightarrow{r},t)=E_0\cos [(\overrightarrow{k}.\overrightarrow{r}-\omega t)]$ …… (1)
 2. Direction of magnetic field:
  $$\hat{B}=\hat{k}\times \hat{E}$$ …… (2)
3. Magnitude of magnetic field:
  $\left|\overrightarrow{B}\right|={\displaystyle \frac{\left|\overrightarrow{E}\right|}{c}}$ …… (3)
4. Relation between direction of $\overrightarrow{B},\overrightarrow{E}\text{ and }c:$
   $$\begin{gathered} \stackrel{\wedge }{E}\times \stackrel{\wedge }{B}\text{ is in direction of }c\text{. But, }c\text{ has same direction as }\overrightarrow{k.} \\ \text{Therefore, }\stackrel{\wedge }{E}\times \stackrel{\wedge }{B}=\stackrel{\wedge }{c}=\stackrel{\wedge }{k}\text{ (in terms of direction)} \end{gathered}$$
5. $\text{Relation between }\omega \text{ and }k:$
     $\Rightarrow \omega =\left|k\right|c$ …… (4)

Complete step-by-step answer:
Given:
1. Electric field is given as: $$\overrightarrow{E}(x,y)=10\hat{j}\cos [(6x+8z)]$$ at t=0

To find: The magnetic field $\overrightarrow{B}(x,z,t)$.

Step 1 of 3:
Compare the given $\overrightarrow{E}$ with the general equation of $\overrightarrow{E}$ as in eq (1).
$$\overrightarrow{E}(x,y)=10\hat{j}\cos [(6\hat{i}+8\hat{k}).(x\hat{i}+z\hat{k})]$$
This gives:
$$\overrightarrow{k}=6\hat{i}+8\hat{k}$$

Step 2 of 3:
Find the direction of $\overrightarrow{B}$ using eq (2).
$$\hat{k}\times \hat{E}={\displaystyle \frac{-4\hat{i}+3\hat{k}}{5}}$$

Step 3 of 3:
Now, calculating magnitude of Magnetic field using equation (3)
$\left|\overrightarrow{B}\right|={\displaystyle \frac{\left|\overrightarrow{E}\right|}{c}}={\displaystyle \frac{10}{c}}$
Step 4 of 5:
Using equation 4 we get angular frequency of field:
$\Rightarrow \omega =\left|k\right|c=\sqrt{6^2+8^2}.c=10c$

Step 4 of 4:
Putting magnitude, angular frequency and wave-vector together and direction altogether we get $\Rightarrow \overrightarrow{B}={\displaystyle \frac{1}{c}}(6\hat{k}-8\hat{i})\cos [(6x+8z-10ct)]$ as answer.

Correct Answer is Option (b) ${\displaystyle \frac{1}{c}}(6\hat{k}-8\hat{i})\cos [(6x+8z-10ct)]$

Note:
Here, every variable of the Electric field equation is crucial to determine the nature of the EM wave. Hence, we should remember the above relations in formula to deduce equivalent conversion between field equations of EM waves. I have used the Wave vector term above. It is the quantity which is responsible for energy of light and momentum carried by the wave. Hence, knowledge at an early stage is good to read.