Q10 / 15 · NTA Abhyas (JEE Main) 2020 · Physics · Electromagnetic Waves

The magnetic field of a plane electromagnetic wave is given by B→=B0i^coskz-ωt+B1j^coskz+ωt, where B0=3×10–5 T and B1=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

  • A
    0.1 N
  • B
    0.9 N
  • C
    3×10–2 N
  • Correct answer
    D
    0.6 N

Solution

B→=B0i^coskz-ωt+B1j^coskz+ωt
B0=3×10-5 T     B1=2×10-6 T
Electric filed associated with it is,
E→=-B0cj^coskz-ωt-B1ci^coskz+ωt
Here, c is speed of light in vacuum.
At z=0,
Frms=QB0c2+QB1c22
Here c is speed of light in vacuum.
=10-4×3×10-5×3×1082+10-4×2×10-6×3×10822
=0.81+0.00362=0.6 N

Hence, the correct answer is option: 'D'.

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