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An Intiative by Sri Chaitanya
a
B→=CE0cos(ωt−kz)j^
b
B→=(E0C)sin(ωt+kz)j^
c
B→=CE0sin(ωt+kz)(−j^)
d
B→=(E0C)sin(ωt−kz)j^
answer is D.
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