JEE Main 20207 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2020
- Date
- 7 January
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2020 · 7 January · Morning Shift · Physics · Electromagnetic Waves
If the magnetic field in a plane electromagnetic wave is given by →...
If the magnetic field in a plane electromagnetic wave is given by $\overset{\to}{B} = 3 \times {10}^{- 8}\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{j}T, then what will be expression for electric field ?
Options
- A\overset{\to}{E} = \left(60\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{k}\dfrac{V}{m}\right)
- B\overset{\to}{E} = \left(9\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{k}\dfrac{V}{m}\right)
- C\overset{\to}{E} = \left(3 \times {10}^{- 8}\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{j}\dfrac{V}{m}\right)
- D\overset{\to}{E} = \left(3 \times {10}^{- 8}\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{i}\dfrac{V}{m}\right)
Correct answer
B. \overset{\to}{E} = \left(9\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{k}\dfrac{V}{m}\right)
Step-by-step solution
(speed of light in vacuum)
So,
\overset{\to}{E} = \left(9\sin \left(1.6 \times {10}^{3}x + 48 \times {10}^{10}t\right)\overset{^}{k}\dfrac{V}{m}\right)