JEE PYQ: Electromagnetic Waves Question 31
Question 31 - 2019 (10 Apr 2019 Shift 1)
The electric field of a plane electromagnetic wave is given by $\vec{E} = E_0 \hat{i} \cos(kz) \cos(\omega t)$. The corresponding magnetic field $\vec{B}$ is then given by:
(1) $\vec{B} = \frac{E_0}{c} \hat{j} \sin(kz) \sin(\omega t)$
(2) $\vec{B} = \frac{E_0}{c} \hat{j} \sin(kz) \cos(\omega t)$
(3) $\vec{B} = \frac{E_0}{c} \hat{j} \cos(kz) \sin(\omega t)$
(4) $\vec{B} = \frac{E_0}{c} \hat{k} \sin(kz) \cos(\omega t)$
Show Answer
Answer: (1)
Solution
$\vec{E} = \frac{E_0}{2}[\cos(kz-\omega t) - \cos(kz+\omega t)]\hat{i}$. Correspondingly $\vec{B} = \frac{E_0}{C} \sin(kz)\sin(\omega t) \hat{j}$.
BETA
