JEE PYQ: Electromagnetic Waves Question 13
Question 13 - 2020 (03 Sep 2020 Shift 1)
The magnetic field of a plane electromagnetic wave is $\vec{B} = 3 \times 10^{-8} \sin[200\pi(y + ct)]\hat{i}$ T, where $c = 3 \times 10^{8}$ ms$^{-1}$ is the speed of light. The corresponding electric field is:
(1) $\vec{E} = 9 \sin[200\pi(y + ct)]\hat{k}$ V/m
(2) $\vec{E} = -10^{-6} \sin[200\pi(y + ct)]\hat{k}$ V/m
(3) $\vec{E} = 3 \times 10^{-8} \sin[200\pi(y + ct)]\hat{k}$ V/m
(4) $\vec{E} = -9 \sin[200\pi(y + ct)]\hat{k}$ V/m
Show Answer
Answer: (4)
Solution
$E_0 = CB_0 = 9$ V/m. Wave propagates in $-y$ direction. $\hat{B} = \hat{i}$, $\hat{C} = -\hat{j}$, so $\hat{E} = -\hat{k}$. $\vec{E} = -9\sin[200\pi(y + ct)]\hat{k}$ V/m.
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