NTA Abhyas JEE Main2020PhysicsElectromagnetic WavesPractice
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
- A0.1 N
- B0.9 N
- C3 × 10 – 2 N
- 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 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 + 0.0036 2 = 0.6 N