JEE Main 201912 AprilEvening Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 12 April
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2019 · 12 April · Evening Shift · Physics · Electromagnetic Waves
A plane electromagnetic wave having a frequency f = 23.9GHz...
A plane electromagnetic wave having a frequency propagates along the positive direction in free space. The peak value of the Electric Field is Which among the following is the acceptable magnetic field component in the electromagnetic wave?
Options
- A\overset{\to}{B} = 2 \times {10}^{- 7}sin \left(0.5 \times {10}^{3}z - 1.5 \times {10}^{11}t\right)\overset{^}{i}
- B\overset{\to}{B} = 2 \times {10}^{- 7}sin \left(1.5 \times {10}^{3}x + 0.5 \times {10}^{11}t\right)\overset{^}{j}
- C\overset{\to}{B} = 2 \times {10}^{7}sin \left(0.5 \times {10}^{3}z + 1.5 \times {10}^{11}t\right)\overset{^}{i}
- D\overset{\to}{B} = 60sin \left(0.5 \times {10}^{3}x + 1.5 \times {10}^{11}t\right)\overset{^}{i}
Correct answer
A. \overset{\to}{B} = 2 \times {10}^{- 7}sin \left(0.5 \times {10}^{3}z - 1.5 \times {10}^{11}t\right)\overset{^}{i}
Step-by-step solution
The magnetic field can be written as,
\overset{\to}{B} = 2 \times {10}^{- 7}\sin (0.5 \times {10}^{3}z - 1.5 \times {10}^{11}t ) \overset{^}{i}