JEE Main 201910 Apr, Morning shiftPhysicsMediumElectromagnetic Waves120× since 2002The electric field of a plane electromagnetic wave is given by E→=E0i^cos(kz)cos(ωt)\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)E=E0icos(kz)cos(ωt) The corresponding magnetic field B→\overrightarrow BB is then given byAOption A. B→=E0Cj^sin(kz)sin(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)B=CE0jsin(kz)sin(ωt)BOption B. B→=E0Cj^sin(kz)cos(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)B=CE0jsin(kz)cos(ωt)COption C. B→=E0Cj^cos(kz)sin(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)B=CE0jcos(kz)sin(ωt)DOption D. B→=E0Ck^sin(kz)cos(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)B=CE0ksin(kz)cos(ωt)CheckSkip · show solution
