JEE Main 202430 Jan, Morning shiftPhysicsEasyElectromagnetic Waves120× since 2002The electric field of an electromagnetic wave in free space is represented as E→=E0cos(ωt−kz)i^\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}E=E0cos(ωt−kz)i^. The corresponding magnetic induction vector will be :AOption A. B→=E0Ccos(ωt+kz)j^\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}B=E0Ccos(ωt+kz)j^BOption B. B→=E0Ccos(ωt−kz)j^\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}B=CE0cos(ωt−kz)j^COption C. B→=E0Ccos(ωt−kz)j^\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}B=E0Ccos(ωt−kz)j^DOption D. B→=E0Ccos(ωt+kz)j^\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}B=CE0cos(ωt+kz)j^CheckSkip · show solution
