JEE Main 201610 Apr, Morning shiftPhysicsEasyElectromagnetic Waves120× since 2002Consider an electromagnetic wave propagating in vacuum. Choose the correct statement :AOption A. For an electromagnetic wave propagating in +x direction the electric field is E⃗=12Eyz (x,t)(y^−z^)\vec E = {1 \over {\sqrt 2 }}{E_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y - \hat z} \right)E=21Eyz(x,t)(y^−z^) and the magnetic field is B⃗=12Byz (x,t)(y^+z^)\vec B = {1 \over {\sqrt 2 }}{B_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y + \hat z} \right)B=21Byz(x,t)(y^+z^)BOption B. For an electromagnetic wave propagating in +x direction the electric field is E⃗=12Eyz (y,z,t)(y^+z^)\vec E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)E=21Eyz(y,z,t)(y^+z^) and the magnetic field is B⃗=12Byz (y,z,t)(y^+z^)\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)B=21Byz(y,z,t)(y^+z^)COption C. For an electromagnetic wave propagating in + y direction the electric field is E→=12Eyz (x,t)y^\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat yE=21Eyz(x,t)y and the magnetic field is B⃗=12Byz (x,t)z^\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat zB=21Byz(x,t)zDOption D. For an electromagnetic wave propagating in + y direction the electric field is E→=12Eyz (x,t)z^\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat zE=21Eyz(x,t)z and the magnetic field is B→=12Bz (x,t)y^\overrightarrow B = {1 \over {\sqrt 2 }}{B_{z{\mkern 1mu} }}\left( {x,t} \right)\widehat yB=21Bz(x,t)yCheckSkip · show solution
