Faraday's LawдЕz" деу JBx:ду.дzat0дЕхдzдеудхдЕz?xдехду--=әв,ƏtдвzƏtO None of theseAmpere-Maxwell lawӘExJBz двудудzatәBхдzaBy?xJBzӘxJBxду= Moto= Moto= Moto-k Eo cos(ky - @ t ) = - @ Bo cos(ky-ot)деуƏtA plane electromagnetic wave is given byEx = + Eo sin( k y - ∞ t ) and Bz = - Bo sin(ky-t)with all other components zero. Which equation results upon plugging these electric and magneticfields into Faraday's law and performing the partial derivatives? Hint: use the third equation ofFaraday's law listed above.JEzatO + o Eo cos( ky - w t ) = - k Bo cos(ky-ot)O + к E cos(ky - @t) = - @ Bo cos(ky-wt)O - w E. cos(ky - @ t ) = - k Bo cos(ky-wt)=4 Faraday's LawHEzде, дудудz-дех де,дExдz?xдеудх--=a Bxat- -двуƏtAmpere-Maxwell lawдвуӘExatдzдвzatJBzдуJBxдz1aBy?xJBz?xдехдуTake the same plane electromagnetic wave given in the previous question,Ex = + Eo sin( ky-t) and Bz = - Bo sin(ky-ot)with all other components zero. Which equation results upon plugging these electric and magneticfields into the Ampere-Maxwell law and performing the partial derivatives? Hint: use the first equationof the Ampere-Maxwell law listed above.MotoJBxду= MotoӘЕ,уat?Е,= моеоатo-k Bocos(ky - w t ) = - @ μlo €o Eo cos(ky-at)O+ k Bo cos(ky - w t ) = - @ po 80 Eo cos(ky-ot)O + @ Bo cos(ky - oO None of theseO - o Bo cos( ky - o t ) = - k po €0 Eo cos(ky-at)t ) = - k po 80 Eo cos( ky-at)

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Answered: A plane electromagnetic wave is given…

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