(i) Comparing the given equation with the standard equation of electromagnetic wave,
`B = B_(0) sin (kx + omegat)`
We have, `k = (2pi)/(lambda) = 0.5 xx 10^(2)`
`:. Lambda = (2pi)/(0.5 xx 10^(3)) m = 1.26 cm`
`omega = 2piv = 1.5 xx 10^(11)`
`:. v = (1.5 xx 10^(11))/(2pi) Hz = 2.38 xx 10^(10) Hz`
Given, `B_(0) = 2 xx 10^(-7)T`
`:. E_(0) = B_(0)c = 2 xx 10^(-7) xx 3 xx 10^(8) = 60 NC^(-1)`
The electric field is perpendicular to both, the magnitude field and the direaction of propagation of wave, Thus
`E_(z) = (60 NC^(-1))sin (0.5 xx 10^(3) x + 1.5 xx 10^(11)t)`