JEE Main20198 Apr 2019Evening ShiftPhysicsElectromagnetic WavesActual
The magnetic field of an electromagnetic wave is given by: B → = 1.6 × 10 - 6 cos ⁡ 2 × 10 7 z + 6 × 10 15 t 2 i ^ + j ^ W b m 2 The associated electric field will be:
Options
- AE → = 4.8 × 10 2 c o s 2 × 10 7 z - 6 × 10 15 t - 2 j ^ + i ^ V m
- BE → = 4.8 × 10 2 c o s 2 × 10 7 z + 6 × 10 15 t i ^ - 2 j ^ V m
- CE → = 4.8 × 10 2 c o s 2 × 10 7 z + 6 × 10 15 t - i ^ + 2 j ^ V m
- DE → = 4.8 × 10 2 c o s 2 × 10 7 z - 6 × 10 15 t 2 i ^ + j ^ V m
Correct answer
C. E → = 4.8 × 10 2 c o s 2 × 10 7 z + 6 × 10 15 t - i ^ + 2 j ^ V m
Step-by-step solution
Given magnetic field, B → = 1.6 × 10 - 6 c o s 2 × 10 7 z + 6 × 10 15 t 2 i ^ + j ^ W b m 2 The relation between magnetic field and electric field is E = c B = ( 3 × 10 8 ) × ( 1.6 × 10 - 6 × 5 ) = 4.8 × 10 2 × 5 The direction of light propagation will be along E → × B → . E → = 4.8 × 10 2 c o s 2 × 10 7 z + 6 × 10 15 t - i ^ + 2 j ^ V m