JEE Main 201910 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 10 April
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2019 · 10 April · Morning 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}\overset{^}{i}\cos \left(kz\right)\cos ( \omegat )
The corresponding magnetic field is then given by:
\overset{\to}{E} = {E}_{0}\overset{^}{i}\cos \left(kz\right)\cos ( \omegat )
The corresponding magnetic field is then given by:
Options
- A\overset{\to}{B} = \dfrac{{E}_{0}}{C}\overset{^}{j}\cos \left(kz\right)\sin ( \omegat )
- B\overset{\to}{B} = \dfrac{{E}_{0}}{C}\overset{^}{k}\sin \left(kz\right)\cos ( \omegat )
- C\overset{\to}{B} = \dfrac{{E}_{0}}{C}\overset{^}{j}\sin \left(kz\right)\sin ( \omegat )
- D\overset{\to}{B} = \dfrac{{E}_{0}}{C}\overset{^}{j}\sin \left(kz\right)\cos ( \omegat )
Correct answer
C. \overset{\to}{B} = \dfrac{{E}_{0}}{C}\overset{^}{j}\sin \left(kz\right)\sin ( \omegat )
Step-by-step solution
\overset{\to}{E} = {E}_{0}\overset{^}{i}\cos \left(kz\right)\cos \left(\omegat\right)
\overset{\to}{E} = \dfrac{{E}_{0}}{2}\overset{^}{i}\cos \left(kz - \omegat\right) + \dfrac{{E}_{0}}{2}\overset{^}{i}\cos \left(kz + \omegat\right)
\overset{\to}{B} = \dfrac{{E}_{0}}{2C}\overset{^}{j}\cos \left(kz - \omegat\right) + \dfrac{{E}_{0}}{2C}\overset{^}{j}cos\left(kz + \omegat\right)= \dfrac{{E}_{0}}{2C}\overset{^}{j}\left[2\sin \left(kz\right)\sin \left(\omegat\right)\right]= \dfrac{{E}_{0}}{C}\overset{^}{j}\left[\sin \left(kz\right)\sin \left(\omegat\right)\right]
\overset{\to}{E} = \dfrac{{E}_{0}}{2}\overset{^}{i}\cos \left(kz - \omegat\right) + \dfrac{{E}_{0}}{2}\overset{^}{i}\cos \left(kz + \omegat\right)
\overset{\to}{B} = \dfrac{{E}_{0}}{2C}\overset{^}{j}\cos \left(kz - \omegat\right) + \dfrac{{E}_{0}}{2C}\overset{^}{j}cos\left(kz + \omegat\right)= \dfrac{{E}_{0}}{2C}\overset{^}{j}\left[2\sin \left(kz\right)\sin \left(\omegat\right)\right]= \dfrac{{E}_{0}}{C}\overset{^}{j}\left[\sin \left(kz\right)\sin \left(\omegat\right)\right]