An electromagnetic plane wave has an electric field that is parallel to the y ax

Question

An electromagnetic plane wave has an electric field that is parallel to the y axis, and has a Poynting vector given by (x,t) = (190 W/m2) cos2(kx - ?t) where x is in meters, k = 12.0 rad/m, ? = 3.60 109 rad/s, and t is in seconds.
(a) What is the direction of propagation of the wave?
1 Direction +x
(b) Find the wavelength and the frequency of the wave.
? = 0.52 m
f = 573 MHz

(c) Find the electric and magnetic fields of the wave as functions of x and t. (Use the following as necessary: x and t. To avoid rounding issue, enter the value for k and ? to two significant figures.)
E(x,t) = ____ V/m
B(x,t) = ______ nT

I only need part C

Explanation / Answer

speed of the light c = 2.998*108 m/s permiability 0 = 4*10-7 N/A2 poynting vector S(x,t) = (190 W/m2) cos2(kx - t) i^ .................. (1) partc : poynting vector S = E2/0c therefore , electric field is             E = [S0c]1/2                = [(190 W/m2)(4*10-7 N/A2)(2.998*108 m/s)]1/2            E = 267.47 V/m according to given formula ,         S = E*B/0    , where * = cross product we know that, poynting vector (x-direction) is perpendicular to the electric field (y -direction) and magnetic field (z-direction) direction. so , electric field E(x,t) = E cos2(kx - t) j^                   E(x,t) = (267.47 V/m) cos2(kx - t) j^ ................... (2) here , k = 12 rad/m and = 3.6*109 rad/s substitute the given data in eq (2) , we get     E(x,t) = (267.47 V/m) cos2((12 rad/m)x - (3.6*109 rad/s)t) j^ magnetic field B = E/c                          =  (267.47 V/m) / (2.998*108 m/s)                      B = 892.18 nT so, magnetic field B(x,t) = (892.18 nT) cos2(kx - t) k^ .............. (3) here , k = 12 rad/m and = 3.6*109 rad/s substitute the given data in eq (3) , we get      B(x,t) = (892.18 nT) cos2((12 rad/m)x - (3.6*109 rad/s)t) K^ here , k = 12 rad/m and = 3.6*109 rad/s substitute the given data in eq (3) , we get      B(x,t) = (892.18 nT) cos2((12 rad/m)x - (3.6*109 rad/s)t) K^ substitute the given data in eq (3) , we get      B(x,t) = (892.18 nT) cos2((12 rad/m)x - (3.6*109 rad/s)t) K^                           

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