The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression
`vec(E) (x,y) = 10 hat(j) cos [(6 x + 8z)]`
The magnetic field `vec(B) (x, z, t)` is given by: (c is the velocity of light)

Options

A. `(1)/(c) (6hatk+8hati)cos[(6x-8z+10ct)]`
B. `(1)/(c)(6hatk+8hati)cos[(6x+8z-10ct)]`
C. `(1)/(c)(6hatk-8hati)cos[(6x+8z-10ct)]`
D. `(1)/(c)(6hatk-8hati)cos[(6x+8z+10ct)]`

Answer

C

Explanation

`vecE=E_(0)(veck*vecr-omegat)`
`vecE(x,y)=10hatjcos(6x+8z)=10j cos[(6hati+8hatk)*(x hati+z hatk)]`
`veck=6 hati+8hatk, omega=|veck|c=10c`
`vecE(x,y,z,t)=10hatj cos[(6hati+8hatk)*(xhati+z hatk)-10ct]`
Now, `vecE xx vecB= (E^(2))/(c)hatk=(E^(2))/(10c)(6hati+8hatk)`
`10hatj xx (B_(x)hati+B_(y)hatj+B_(z)hatk)=(10)/(c) (6hati+8hatk)`
Looking at the options we can say
`vecB=(1)/(c)(6 hatk-8hati)`
`vecB=(x,y,z,t)=(1)/(c)(6hatk-8hati)cos[(6x+8z)-10ct]`