Question

An electromagnatic wave is traveling in vacuum with frequency 6.1 x 10¹⁴ Hz. The wave has average total energy density of 4.8 x 10^-6 J/m³.

(a) What is the speed of this wave?

(b) What is the wavelength of this wave?

(c) What is the maximum electric field of this wave?

(d) What is the rms magnetic field of this wave?

(e) How much energy does a 1.8 m² flat surface (perpendicular to the wave propagation direction) receive in 7 s?

(f) How long does it take for this wave to travel from the sun to the earth?

(g) How many wavelengths of the wave fit in this distance?

Answer 1

Part A.

Speed of electromagnetic wave is constant and equal to the speed of light

Speed of wave = c = 3*10^8 m/sec

Part B.

wavelength is given by:

lambda = c/f

f = frequency = 6.1*10^14 Hz

lambda = 3*10^8/(6.1*10^14) = 4.918*10^-7 m = 491.8*10^-9 m = 491.8 nm

Part C.

Energy density of EM waves is given by:

U = (1/2)*e0*Emax^2

Emax = sqrt (2*U/e0)

Emax = sqrt (2*4.8*10^-6/(8.85*10^-12))

Emax = 1041.5 V/m

Part D.

Max magnetic field is given by:

Bmax = Emax/c

Bmax = 1041.5/(3*10^8)

Bmax = 3.47*10^-6 T

Now rms magnetic field will be

Brms = Bmax/sqrt 2

Brms = 3.47*10^-6/sqrt 2 = 2.45*10^-6 T

Part E.

Power of energy source will be given by:

P = U*c

Energy received by flat surface will be

E = P*A*t

E = U*c*A*t

E = 4.8*10^-6*3*10^8*1.8*7

E = 18144 J

Part F.

distance between sun and earth = 1.496*10^11 m

Speed of EM wave = c = 3*10^8

time = distance/speed

t = 1.496*10^11/(3*10^8) = 499 sec

t = 499 sec = 8 min 19 sec

Part G

Number of wavelengths in this distance can be:

N = total distance/wavelength

N = 1.496*10^11/(491.8*10^-9)

N = 3.04*10^17 wavelengths

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