JEE Main2015PhysicsElectromagnetic WavesActual
An electromagnetic wave travelling in the x - direction has frequency of 2 × 10 14 H z and electric field amplitude of 27 V m – 1 oscillates in Y - direction. From the options given below, which one describes the magnetic field for this wave?
Options
- AB → x ,   t = ( 9 × 10 - 8   T ) j ^ sin 1.5 × 10 - 6   x - 2 × 10 14 t &#
- BB → x ,   t = ( 9 × 10 - 8 T ) i ^ sin 2 π 1.5 × 10 - 8   x - 2 × 10 14 t
- CB → x , t = 3 × 10 - 8 T j ^ sin 2π x 1.5 × 10 - 8 - 2 × 10 14 t
- DB → x , t = 9 × 10 - 8 T k ^ sin2π x 1.5 × 10 - 6 - 2 × 10 14 t
Correct answer
D. B → x , t = 9 × 10 - 8 T k ^ sin2π x 1.5 × 10 - 6 - 2 × 10 14 t
Step-by-step solution
When undefined Then   B   =   B 0 sin ⁡   ( k x − ω t ) Of light in travelling along i ^ then B → in either along j ^ or k ^ . ∴ speed of light   C =   E 0 B 0   ⇒ B 0 =   E 0 C ⇒           B 0 =   27 3 × 10 8 = 9   × 10 - 8   T also, ω = 2 π   f = 2 π   × 2   × 10 14 = 4   π   × 10 14 k = ω c = 4 π × 10 14 3