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JEE Main201910 Jan 2019Evening ShiftPhysicsElectromagnetic WavesActual

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by the expression E → x ,   y = 10 j ^ c o s 6 x + 8 z . The magnetic field B → x ,   z ,   t is given by ( c is the velocity of light.)

Options

  1. A1 c 6 k ^ - 8 i ^ cos 6 x + 8 z + 10   c t
  2. B1 c 6 k ^ + 8 i ^ cos 6 x + 8 z - 10   c t
  3. C1 c 6 k ^ + 8 i ^ cos 6 x - 8 z + 10   c t
  4. D1 c 6 k ^ - 8 i ^ c o s 6 x + 8 z - 10   c t

Correct answer

D. 1 c 6 k ^ - 8 i ^ c o s 6 x + 8 z - 10   c t

Step-by-step solution

E = E 0 cos ω t - K     → ·   r   → ∴ At any time t , K → = 6 i ^ + 8 k ^ B 0 = E 0 c = 10 c λ = 2 π k = 2 π 10 = π 5 Also, E → × B → = c → ∴ ω = 2 π f = 2 π c λ i ^ j ^ k ^ 0 1 0 B x B y B z = 6 i ^ + 8 k ^ 10 ω = 2 π × c π 5 = 10 c ∴ i ^ B z - 0 + j ^ 0 + k ^ - B x = 0.6 i ^ - 0.8 k ^ Comparing the coefficients on both sides, ∴ B z = 0.6 ;   B x = - 0.8 B

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