JEE MainPhysicsElectromagnetic Waves
The magnetic field of an electromagnetic wave in free space is given by B = 2 10⁻⁸ [ 10^7 t + 30 (x + 3 z) ] j Tesla. The corresponding electric field vector E in V/m is
Options
- A(3 3 i - 3 k ) [ 10^7 t + 30 (x + 3 z) ]
- B(3 i - 3 3 k ) [ 10^7 t + 30 (x + 3 z) ]
- C(-3 3 i + 3 k ) [ 10^7 t + 30 (x + 3 z) ]
- D(-3 i + 3 3 k ) [ 10^7 t + 30 (x + 3 z) ]
Correct answer
C. (-3 3 i + 3 k ) [ 10^7 t + 30 (x + 3 z) ]
Step-by-step solution
The phase of the wave is = t + k r . Comparing this with the given argument 10^7 t + 30 (x + 3 z) , the wave propagates in the - k direction. The unit vector along the direction of propagation is n = - i + 3 k 1^2 + ( 3 )^2 = - 1 2 i - 3 2 k . The amplitude of the electric field is E₀ = c B₀ = (3 10^8) (2 10⁻⁸) = 6 V/m. The direction of the electric field is given by E = B n . E = j (- 1 2 i - 3 2 k ) = - 1 2 ( j i ) - 3 2 ( j k ) . Since j i = - k and j k = i , we get E = - 3 2 i + 1 2 k . Therefore, the electric