The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is `vecE = E_0 hatj cos (omegat - kx)` . The magnetic field `vecB`, at the moment t=0 is :

Options

A. `(2E_0 ) /(c ) hatj sin kz cos omega t `
B. `-(2 E_0 ) /( c ) hatj sin kz sin omega t `
C. `(2E_0 )/( c ) hatj sin kz sin omega t`
D. `(2E_0 )/(c ) hatj cos kz cos omega t `

Answer

A

Explanation

`(dE)/(dz ) =- (dE)/( dt ) `
` (dE )/( dz ) = - 2E _0 k sin kz cos omega t `
` (dE )/( dz ) =- ( dE )/( dt ) =-(dB )/(dt) `
thus ` dB = 2E_0 k sin kz cos omega t dt `
`or , b= 2E_0 k sin kz int cos omega t dt = ( 2E_0 k ) /( omega ) sin kz sin omega t `
the relation beween the amplitude of the electric and the magnetic field is given b y:
` (E_0 )/( B_0 ) = C `
the speed o flight can also be expressed in terms of the angular frequency ` ( omega )` and the wave number (k) as :
` c= (omega ) /( k ) `
therefore
` (E_0 ) /( B_0 ) = ( omega)/(k) =C `
Hence , the magnetic field is given by :
` B= ( 2E_0 )/(C ) hatj sin kz sin omega t `