JEE Main 20204 SeptemberEvening Shift
- Exam
- JEE Main
- Year
- 2020
- Date
- 4 September
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2020 · 4 September · Evening Shift · Physics · Electromagnetic Waves
The electric field of a plane electromagnetic wave is given by → E...
The electric field of a plane electromagnetic wave is given by \overset{\to}{E} = {E}_{0}\left(\overset{^}{x} + \overset{^}{y}\right)\sin \left(kz - \omegat\right). Its magnetic field will be given by
Options
- A\dfrac{{E}_{0}}{c}\left(- \overset{^}{x} + \overset{^}{y}\right)\sin \left(kz - \omegat\right)
- B\dfrac{{E}_{0}}{c}\left(\overset{^}{x} + \overset{^}{y}\right)\sin \left(kz - \omegat\right)
- C\dfrac{{E}_{0}}{c}\left(\overset{^}{x} - \overset{^}{y}\right)\sin \left(kz - \omegat\right)
- D\dfrac{{E}_{0}}{c}\left(\overset{^}{x} - \overset{^}{y}\right)\cos \left(kz - \omegat\right)
Correct answer
A. \dfrac{{E}_{0}}{c}\left(- \overset{^}{x} + \overset{^}{y}\right)\sin \left(kz - \omegat\right)
Step-by-step solution
As in electromagnetic wave electric field is always perpendicular to the magnetic field and both the fields oscillates perpendicularly to the direction pf propagation so