JEE Main On 8 April 2017 Question 15
Question: Magnetic field in a plane electromagnetic wave is given by $ \vec{B}=B _0\sin (kx+\omega t)\hat{j}T $ Expression for corresponding electric field will be [JEE Online 08-04-2017]
Options:
A) $ \vec{E}=-B _0c\sin (kx+\omega t)\hat{k} , \text{V/m}$
B) $ \vec{E}=B _0c\sin (kx-\omega t)\hat{k}T/m $
C) $ \vec{E}=B _0c\sin (kx+\omega t)\hat{k}T/m $
D) $ \vec{E}=\frac{B _0}{c}\sin ,(kx+\omega t)\hat{k},V/m $
Show Answer
Answer:
Correct Answer: C
Solution:
$ C=\frac{E _0}{B _0} $ $ E=cB _0 $ $ =cB _0\sin (kx+\omega t)\hat{k} $ $ =cB _0\sin (kx+\omega t)\hat{k} $
BETA
