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A plane electromagnetic wave is given by Ex=+Eo sin(ky- wt) and B₂= - Bo sin(ky-ot) with all other components zero. Which equation results upon plugging these electric and magnetic fields into Faraday's law and performing the partial derivatives? Hint: use the third equation of Faraday's law listed above.

A plane electromagnetic wave is given by Ex=+Eo sin(ky- wt) and B₂= - Bo sin(ky-ot) with all other components zero. Which equation results upon plugging these electric and magnetic fields into Faraday's law and performing the partial derivatives? Hint: use the third equation of Faraday's law listed above.

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Faraday's Law дЕz" деу JBx: ду. дz at 0 дЕх дz деу дх дЕz ?x дех ду -- = әв, Ət двz Ət O None of these Ampere-Maxwell law ӘEx JBz дву ду дz at әBх дz aBy ?x JBz Әx JBx ду = Moto = Moto = Moto - k Eo cos(ky - @ t ) = - @ Bo cos(ky-ot) деу Ət A plane electromagnetic wave is given by Ex = + Eo sin( k y - ∞ t ) and Bz = - Bo sin(ky-t) with all other components zero. Which equation results upon plugging these electric and magnetic fields into Faraday's law and performing the partial derivatives? Hint: use the third equation of Faraday's law listed above. JEz at O + 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
Transcribed Image Text:Faraday's Law дЕz" деу JBx: ду. дz at 0 дЕх дz деу дх дЕz ?x дех ду -- = әв, Ət двz Ət O None of these Ampere-Maxwell law ӘEx JBz дву ду дz at әBх дz aBy ?x JBz Әx JBx ду = Moto = Moto = Moto - k Eo cos(ky - @ t ) = - @ Bo cos(ky-ot) деу Ət A plane electromagnetic wave is given by Ex = + Eo sin( k y - ∞ t ) and Bz = - Bo sin(ky-t) with all other components zero. Which equation results upon plugging these electric and magnetic fields into Faraday's law and performing the partial derivatives? Hint: use the third equation of Faraday's law listed above. JEz at O + 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 Law HEz де, ду ду дz - дех де, дEx дz ?x деу дх -- = a Bx at - - дву Ət Ampere-Maxwell law дву ӘEx at дz двz at JBz ду JBx дz 1 aBy ?x JBz ?x дех ду Take 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 magnetic fields into the Ampere-Maxwell law and performing the partial derivatives? Hint: use the first equation of the Ampere-Maxwell law listed above. Moto JBx ду = Moto ӘЕ, у at ?Е, = моеоат o-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 - o O None of these O - o Bo cos( ky - o t ) = - k po €0 Eo cos(ky-at) t ) = - k po 80 Eo cos( ky-at)
Transcribed Image Text:Faraday's Law HEz де, ду ду дz - дех де, дEx дz ?x деу дх -- = a Bx at - - дву Ət Ampere-Maxwell law дву ӘEx at дz двz at JBz ду JBx дz 1 aBy ?x JBz ?x дех ду Take 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 magnetic fields into the Ampere-Maxwell law and performing the partial derivatives? Hint: use the first equation of the Ampere-Maxwell law listed above. Moto JBx ду = Moto ӘЕ, у at ?Е, = моеоат o-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 - o O None of these O - 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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