o 1/4 points | Previous Answers Tipler6 30.P.046. My An electromagnetic wave has

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o 1/4 points | Previous Answers Tipler6 30.P.046. My An electromagnetic wave has a frequency of 110 MHz and is traveling in a vacuum. The magnetic field is given by B(z, t)- (4.00 x 10-8 T) cos (kz ot)i. (Assume the electromagnetic wave is travelling in the k direction.) | (a) Find the wavelength of this wave. 2.73 (b) Find the electric field vector Elz, t). (Use the following as necessary: z, and t) (c) Determine the Poynting vector. (Use the following as necessary: z, and t) mW/m2 (d) Using this Poynting vector, find the intensity of the wave. mW/m2 eBook

Explanation / Answer

Given,

f = 110 M Hz = 110 x 10^6 Hz

a)we know that,

c = f lambda

lambda = c/f = 3 x 10^8/110 x 10^6 = 2.73 m

Hence, lambda = 2.73 m

b)We know that electric and magnetic field of an electromagentic wave is related as:

c = E/B => E = B c

E = 4 x 10^-8 x 3 x 10^8 = 12 V/m

w = 2 pi f

w = 2 x 3.14 x 110 x 10^6 = 6.91 x 10^8 rad/s

k = 2 pi/lambda

k = 2 x 3.14/2.73 = 2.30

E(z,t) = -(12 V/m) cos[(2.3 m^-1)z - (6.91 x 10^8)t]j

c)Poynting vector is given by:

S = 1/u0 (E x B)

S = (12)(4 x 10^-8) cos^2 [[(2.3 m^-1)z - (6.91 x 10^8)t]]/4 pi x 10^-7

S = (382 mW/m^2)cos^2 [[(2.3 m^-1)z - (6.91 x 10^8)t]]

d)the average value of cos^2(f) = 0.5

S = 0.5 x 382 mW/m^2 = 191 mW/m^2

Hence, S = 191 mW/m^2

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