The Poynting vector for an electromagnetic wave is given by (500 W/m^2)sin^2[(10

Question

The Poynting vector for an electromagnetic wave is given by (500 W/m^2)sin^2[(1000 m^1)z(5.0×10^11 s^1)t]k^.

Part B

What is the time-averaged energy per unit time radiated through a 1.0 m^2 surface aligned with its normal parallel to the direction of propagation?

Part D

What is the wavelength of this electromagnetic wave?

Part E

What is the frequency of this electromagnetic wave?

Part F

What is the magnitude Emax of the electric field vector?

Part G

What is the magnitude Bmax of the magnetic field vector?

Explanation / Answer

Poynting vector = e0 c E^2 sin^2( kx - wt)

time -averaged energy per unit time per unit area = c e0 E^2 A / 2

and E^2 / u0 c = e0 c E^2 = 500 W /m^2

  
so, time -averaged energy per unit time per unit area = 500A/2 = 250A

through 1 m^2 = 250 W

D) k = 2pi / lambda

lambda = 2pi / ( 1000) = 6.283 x 10^-3 m

E) w = 5 x 10^11

2pif = w = 5 x 10^11

f = 7.96 x 10^10 Hz

F) e0 c E^2 = 500

E^2 = 500 / (8.854 x 10^-12 x 3 x 10^8)

E = 433.786 N/C

G) E = c B

B = (433.786 ) / (3 x 10^8) = 1.45 x 10^-6 T

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