The magnetic field of an electromagnetic wave is given by:
`oversetB=1.6xx10^(-6)cos(2xx10^(7)z+6xx10^(15)t)(2hati+hatj)(Wb)/m^(2)`
The associated electric field will be:

Options

A. `vecE = 4.8 xx 10^(2) cos ( 2 xx 10^7 z + 6 xx 10^(15) t ) (-1 hati + 2 hatj ) (V) /(m)`
B. `vecE =4.8 xx 10^2 cos (2 xx 10^7 z + 6 xx 10^(15) t) (-2hatj + 2 hati ) (V )/(m)`
C. ` vecE = 4.8 xx 10^2 cos ( 2 xx 10^7 z + 6 xx 10^(15) t) ( hati + 2 hatj ) (V )/(M)`
D. ` vecE = 4.8 xx 10^2 cos ( 2 xx 10^7 z + 6 xx 10^(15) t) ( 2hati + hatj ) (V )/(m)`

Answer

A

Explanation

From the given equation of ` vecB ` we can see that EM wave we can see that wave is propagating along negative z- direction Direction of propagation of electromagnetic wave is given by
`vecE xx vecB `
Further we know that `E_0 = cE_0` hence
` E_0 = 3 xx 10^8 xx 1.6 xx 10^(-6) sqrt(5) = 4.8 sqrt(5) xx 10^(2) V//m `
Further we know that `vecE and vecB ` are perpendicular to each other hence
` vecE . vecB =0`
from equation (i ) and (ii )
magnetic of a ` vecE_0` is along `(- hati + 2hati ) ` so ,` vec E_0 = 4.8 xx 10^2 ( - hati + 2 hatj`
all above concepts indicate that option (a) is correct .