JEE Main 20178 AprilOnline
- Exam
- JEE Main
- Year
- 2017
- Date
- 8 April
- Mode
- Online
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2017 · 8 April · Online · Physics · Electromagnetic Waves
Magnetic field in a plane electromagnetic wave is given by, → B = B...
Magnetic field in a plane electromagnetic wave is given by,
\overset{\to}{B} = {B}_{0}\sin \left(kx + \omegat\right)\overset{^}{j}T. Expression for corresponding electric field will be:
(Where is speed of light)
\overset{\to}{B} = {B}_{0}\sin \left(kx + \omegat\right)\overset{^}{j}T. Expression for corresponding electric field will be:
(Where is speed of light)
Options
- A\overset{\to}{E} = - {B}_{0}csin\left(kx + \omegat\right)\overset{^}{k}V{m}^{- 1}
- B\overset{\to}{E} = {B}_{0}csin\left(kx - \omegat\right)\overset{^}{k}V{m}^{- 1}
- C\overset{\to}{E} = {B}_{0}csin\left(kx + \omegat\right)\overset{^}{k}V{m}^{- 1}
- D\overset{\to}{E} = \dfrac{{B}_{0}}{c}\sin \left(kx + \omegat\right)\overset{^}{k}V{m}^{- 1}
Correct answer
C. \overset{\to}{E} = {B}_{0}csin\left(kx + \omegat\right)\overset{^}{k}V{m}^{- 1}
Step-by-step solution
For electromagnetic wave speed of light is given by
Electric field is expressed as
The expression of electric field is \overset{\to}{E} = c{B}_{0}\sin ( kx + wt ) \overset{^}{k}V{m}^{- 1}