The electric field part of an electromagnetic wave in vacum is given by `E=6 cos [ 1.2("rad")/(m)y+3.6xx10^(8)("rad")/(s)]hati (N)/(C)`, then find
(a) Frequuency of propagation (f)
Wavelength `(lambda)`
(c) Direction of propagation
(d) Amplitude of magnetic field in electromagnetic wave

Explanation

`(a)2nf=3.6xx10^(8)`
`f=(3.6xx10^(8))/(2xx3.14)=5.7xx10^(7)Hz`
`(b) lambda =(2pi)/(1.2)=5.23 m`
(c) Negative y-axis
(d) `B_(0)=(E_(0))/(c)=(6)/(3xx10^(8))=2xx10^(-8)T`
`(e) vecB=2xx10^(-8)T cos (1.2y+3.6xx10^(8)t)hatj`