JEE Main 20204 Sep, Evening shiftPhysicsMediumElectromagnetic Waves120× since 2002The electric field of a plane electromagnetic wave is given by E→=E0(x^+y^)sin(kz−ωt)\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)E=E0(x+y)sin(kz−ωt) Its magnetic field will be given by :AOption A. E0c(x^+y^)sin(kz−ωt){{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)cE0(x+y)sin(kz−ωt)BOption B. E0c(x^−y^)sin(kz−ωt){{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)cE0(x−y)sin(kz−ωt)COption C. E0c(x^−y^)cos(kz−ωt){{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)cE0(x−y)cos(kz−ωt)DOption D. E0c(−x^+y^)sin(kz−ωt){{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)cE0(−x+y)sin(kz−ωt)CheckSkip · show solution
