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Sahay Sir > Question Answers > JEE Mains Physics 2002-2019 Solved Video Solutions > Chapter 15 - Electromagnetic Waves > The magnetic field of a plane electromagnetic wave is given by: B=B0[cos(kz−ωt)]i+B1cos(kz+ωt)j where B0=3×10^−5T and B1=2×10^−6T. The rms value of the force experienced by a stationary charge Q=10^−4C at z=0 is close to:

The magnetic field of a plane electromagnetic wave is given by: B=B0[cos(kz−ωt)]i+B1cos(kz+ωt)j where B0=3×10^−5T and B1=2×10^−6T. The rms value of the force experienced by a stationary charge Q=10^−4C at z=0 is close to:

Velocity of light is equal to