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JEE Main20199 Apr 2019Morning ShiftPhysicsElectromagnetic WavesActual

The magnetic field of a plane electromagnetic wave is given by B → = B 0 i ^ c o s k z - ω t + B 1 j ^ c o s k z + ω t , where B 0 = 3 × 10 – 5 T and B 1 = 2 × 10 – 6 T . The RMS value of the force experienced by a stationary charge Q = 10 – 4 C at z = 0 is closest to

Options

  1. A0.1 N
  2. B0.9 N
  3. C3 × 10 – 2 N
  4. D0.6 N

Correct answer

D. 0.6 N

Step-by-step solution

B → = B 0 i ^ c o s k z - ω t + B 1 j ^ c o s k z + ω t B 0 = 3 × 10 - 5   T   &     B 1 = 2 × 10 - 6   T Electric filed associated with it is, E → = - B 0 c j ^ c o s k z - ω t - B 1 c i ^ c o s k z + ω t Here, c is the speed of light in vacuum. At z = 0 , F rms = Q B 0 c 2 + Q B 1 c 2 2 Here c is speed of light in vacuum. = 10 - 4 × 3 × 10 - 5 × 3 × 10 8 2 + 10 - 4 × 2 × 10 - 6 × 3 × 10 8 2 2 = 0.81 + 0.

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