JEE Main & Advanced Physics EM Waves JEE PYQ-Electro Magnetic Waves

A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by\[\vec{B}=5\times {{10}^{-8}}\hat{j}\text{ }T\]. The corresponding electric field \[\vec{E}\] is (speed of light \[c=3\times {{10}^{8}}\text{ }m{{s}^{1}}\]) [JEE MAIN Held on 08-01-2020 Evening]

A) \[-1.66\times {{10}^{-16}}\hat{i}\text{ }V/m\]

B) \[1.66\times {{10}^{-16}}\hat{i}\text{ }V/m\]

C) \[-15\text{ \hat{i} }V/m\]

D) \[15\,\,\hat{i}\,\,V/m\]

Correct Answer: D

Solution :

[d] \[\vec{E}=\vec{B}=\vec{C}\]

\[\therefore \,\,\,\,\,\,\,\left| E \right|=\left| B \right|\cdot C\]

\[\Rightarrow \,\,\,\,\,\,\,\left| {\vec{E}} \right|=5\times {{10}^{-8}}\times 3\times {{10}^{8}}=15\,\,N/C\]

\[\vec{E}=15\hat{i}\,\,V\,/\,m\]