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An Intiative by Sri Chaitanya
a
E→(x,t) =36 sin1× 103x+ 0.5 × 1011t j^Vm
b
E→(x,t) =36 sin 0.5 × 103x+1.5× 1011t i^Vm
c
E→(x,t) =−36 sin 0.5 × 103x+1.5× 1011t j^Vm
d
E→(x,t) =36 sin 0.5 × 103x+1.5× 1011t k^Vm
answer is C.
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