(a) Comparing the given equation with
`B_(y)=B_(0)sin[2pi((2)/(gamma)+(i)/(T))]`
We get, `gamma=(2pi)/(0.5xx10^(3))m=1.26cm`.
and `(1)/(T)=v=(1.5xx10^(11))//2pi=23.9GHz`
`E_(0)=B_(0)C=2xx10^(-7)Txx3xx10^(8)m//s=6xx10^(1)V//m` lt brgt The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore the electric field component along the z-axis is obtained as
`E_(z)=60sin (0.5xx10^(3)x+1.5xx10^(11)t)V//m`