A plane e.m. wave travelling along the x-direction has a wavelength of 3mm. The variation in the electric field occurs in the y-direction with an amplitude `66Vm^-1`. The equation for the electric and magnetic fields as a function of x and t are respectively

Options

A. `E_y=33cos pixx10^(11)(t-x/c)`
`B_z=1.1 xx10^-7cos pixx10^(11) (t-x/c)`
B. `E_y=11cos 2pixx10^(11)(t-x/c)`
`B_y=11 xx10^-7cos 2pixx10^(11) (t-x/c)`
C. `E_x=33cos pixx10^(11)(t-x/c)`
`B_y=11 xx10^-7cos pixx10^(11) (t-x/c)`
D. `E_y=66cos pixx10^(11)(t-x/c)`
`B_z=2.2 xx10^-7cos 2pixx10^(11) (t-x/c)`

Answer

d

Explanation

`lambda=3mm=3xx10^-3m , E_0=66Vm^-1`
`B_0=(E_0)/c =66/(3xx10^8)=2.2 xx10^-7T`
As em wave is propagating along
x-axis and electric field variation is along
y-direction, the magnetic field variation is along
z-direction.
Using the relation for harmonic wave
`E_y=E_0cos .(2pi)/lambda(ct-x)`
`E_y=E_0cos .(2pic)/lambda(t-x/c)`
`:. E_y=66cos .(2pixx3xx10^8)/(3xx10^-3)(t-x/c)`
`=66 cos 2pixx10^(11)(t-x/c)`
and `B_z= B_0cos .(2pic)/lambda(t-x/c)`
`=2.2xx10^-7cos 2pixx10^(11)(t-x/c)`
Thus, option (d) is correct.