Question

16. Very far away from the source of the EM wave, you can often treat it as a plane At the location of your detector, which is in vacuum, suppose that you magnitude and direction of the electric field of an EM wave is given by wave. find that the E Eo cos (ot - kz)X (a) Give an exact expression for the magnitude and direction of the magnetic field in the wave: (b) In wh at direction will your detector be pushed by the radiation pressure? (Circle ans.) -Z NO FORCE +X +Y -X (c) Suppose E, 3x10 V/m. Suppose the wave originates from an isotropic source 1 km ay. Calculate the average power of the source. (Isotropic means emitted equally in all aw directions, so that the source illuminates a sphere of radius R-1 km, area 4TR2.)

Answer 1

E = E_0 cos (wt = kz)x By Using Faraday's - Maxwell nabla times E = -partial differential B/dt Therefore B = -integral (nabla times E) dt Now nabla times E = Vertical-bar x partial differential/partial differential x E y partial differential/partial differential y 0 z partial differential/partial differential z 0 Vertical-bar = partial differential E./ partial differential z y Rightarrow nabla times E = + kE_0 sin(wt - kz)y Now B = -K.E integral sin (wt - kz) dt = kE_0 y/ w cos (wt - kz)y Rightarrow B = k/w E_0 cos(wt - kz)y (b)Radiation pressure = < s > /c = < E times H > /C Therefore Direction of P is direction of (E times H) Rightarrow x times y = K i.e. + Z (c) P_avg = I_avg A = CE_0 E_0^2 times 4 pi (1000)^2 =