JEE Main 201510 AprilOnline
- Exam
- JEE Main
- Year
- 2015
- Date
- 10 April
- Mode
- Online
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2015 · 10 April · Online · Physics · Electromagnetic Waves
An electromagnetic wave travelling in the x - direction has...
An electromagnetic wave travelling in the direction has frequency of and electric field amplitude of oscillates in direction. From the options given below, which one describes the magnetic field for this wave?
Options
- A\overset{\to}{B}\left(x, t\right) = ( 9 \times {10}^{- 8}T ) \overset{^}{j} \sin \left[1.5 \times {10}^{- 6}x - 2 \times {10}^{14}t\right]
- B\overset{\to}{B}\left(x, t\right) = ( 9 \times {10}^{- 8}T ) \overset{^}{i}\sin \left[2\pi\left(1.5 \times {10}^{- 8}x - 2 \times {10}^{14}t\right)\right]
- C\left(x, t\right) = \left(3 \times {10}^{- 8}T\right)\overset{^}{j}\sin 2 \pi\left[\left(\dfrac{x}{1.5 \times {10}^{- 8}}\right) - 2 \times {10}^{14}t\right]
- D\left(x, t\right) = \left(9 \times {10}^{- 8}T\right)\overset{\text{^}}{k}\sin 2 \pi\left[\left(\dfrac{x}{1.5 \times {10}^{- 6}}\right) - 2 \times {10}^{14}t\right]
Correct answer
D. \left(x, t\right) = \left(9 \times {10}^{- 8}T\right)\overset{\text{^}}{k}\sin 2 \pi\left[\left(\dfrac{x}{1.5 \times {10}^{- 6}}\right) - 2 \times {10}^{14}t\right]
Step-by-step solution
When undefined
Then
Of light in travelling along \overset{^}{i} then\overset{\to}{B}\overset{^}{j}\overset{^}{k}∴C = \dfrac{{E}_{0}}{{B}_{0}} ⇒ {B}_{0} = \dfrac{{E}_{0}}{C}⇒ {B}_{0} = \dfrac{27}{3 \times {10}^{8}} = 9 \times {10}^{- 8}T\omega = 2\pif = 2\pi \times 2 \times {10}^{14} = 4\pi \times {10}^{14}k = \dfrac{\omega}{c} = \dfrac{4\pi \times {10}^{14}}{3 \times {10}^{8}}= ( \dfrac{2\pi}{1.5 \times {10}^{- 6}} )\overset{\to}{B} = 9 \times {10}^{- 8}\sin 2\pi\left(\dfrac{x}{1.5 \times {10}^{- 6}} - 2 \times {10}^{14}t\right)\overset{^}{k}
Then
Of light in travelling along \overset{^}{i} then\overset{\to}{B}\overset{^}{j}\overset{^}{k}∴C = \dfrac{{E}_{0}}{{B}_{0}} ⇒ {B}_{0} = \dfrac{{E}_{0}}{C}⇒ {B}_{0} = \dfrac{27}{3 \times {10}^{8}} = 9 \times {10}^{- 8}T\omega = 2\pif = 2\pi \times 2 \times {10}^{14} = 4\pi \times {10}^{14}k = \dfrac{\omega}{c} = \dfrac{4\pi \times {10}^{14}}{3 \times {10}^{8}}= ( \dfrac{2\pi}{1.5 \times {10}^{- 6}} )\overset{\to}{B} = 9 \times {10}^{- 8}\sin 2\pi\left(\dfrac{x}{1.5 \times {10}^{- 6}} - 2 \times {10}^{14}t\right)\overset{^}{k}