JEE Main 20203 SeptemberMorning Shift
- Exam
- JEE Main
- Year
- 2020
- Date
- 3 September
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2020 · 3 September · Morning Shift · Physics · Electromagnetic Waves
The magnetic field of a plane electromagnetic wave is → B = 3 × 10...
The magnetic field of a plane electromagnetic wave is \left[200\pi ( y + ct )\right]\overset{^}{i}Tc = 3 \times {10}^{8}m{s}^{- 1} is the speed of light the corresponding electric filed is :
Options
- A\overset{\to}{E} = 9sin [ 200\pi ( y + ct ) \overset{^}{k}V / m
- B\overset{\to}{E} = - {10}^{- 6}sin [ 200\pi ( y + ct ) \overset{^}{k}V / m
- C\overset{\to}{E} = 3 \times {10}^{- 8}sin [ 200\pi ( y + ct ) \overset{^}{k}V / m
- D\overset{\to}{E} = - 9sin [ 200\pi ( y + ct ) \overset{^}{k}V / m
Correct answer
A. \overset{\to}{E} = 9sin [ 200\pi ( y + ct ) \overset{^}{k}V / m
Step-by-step solution
∴ \overset{\to}{E} \times \overset{\to}{B} ‖ \overset{\to}{C} ⇒ \overset{^}{k} \times \overset{^}{i} = \overset{^}{j}
\overset{\to}{E} = 3 \times {10}^{8} \times 3 \times {10}^{- 8}sin [ 200\pi\left(y + ct\right) ( - \overset{^}{k} ) = 9sin [ 200\pi\left(y + ct\right) ] ( - \overset{^}{k} )