JEE PYQ: Electromagnetic Waves Question 37
Question 37 - 2019 (10 Jan 2019 Shift 2)
The electric field of a plane polarized electromagnetic wave in free space at time $t = 0$ is given by an expression $\vec{E}(x, y) = 10\hat{j} \cos[(6x + 8z)]$. The magnetic field $\vec{B}(x, z, t)$ is given by: ($c$ is the velocity of light)
(1) $\frac{1}{c}(6\hat{k} + 8\hat{i}) \cos[(6x - 8z + 10ct)]$
(2) $\frac{1}{c}(6\hat{k} - 8\hat{i}) \cos[(6x + 8z - 10ct)]$
(3) $\frac{1}{c}(6\hat{k} + 8\hat{i}) \cos[(6x + 8z - 10ct)]$
(4) $\frac{1}{c}(6\hat{k} - 8\hat{i}) \cos[(6x + 8z + 10ct)]$
Show Answer
Answer: (2)
Solution
$\vec{K} = 6\hat{i} + 8\hat{k}$. $\hat{C} \times \hat{E} = \frac{-4\hat{i} + 3\hat{k}}{5}$. $\vec{B} = \frac{1}{C}(6\hat{k} - 8\hat{i})\cos(6x + 8z - 10ct)$.
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