PhysicsMedium120×since 2002Q6763The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is E→=E0j^cos(ωt−kx)\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)E=E0jcos(ωt−kx). The magnetic field B→\overrightarrow BB , at the moment t = 0 is :AB→=E0μ0∈0cos(kx)j^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat jB=μ0∈0E0cos(kx)jBB→=E0μ0∈0cos(kx)k^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat kB=μ0∈0E0cos(kx)kCB→=E0μ0∈0cos(kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat kB=E0μ0∈0cos(kx)kDB→=E0μ0∈0cos(kx)j^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat jB=E0μ0∈0cos(kx)jCheck answerSkip