The X radio station broadcasts at a frequency of 104.9 times 10^6Hz. At a reason

Question

The X radio station broadcasts at a frequency of 104.9 times 10^6Hz. At a reasonable distance from the transmitter the magnitude of the electric component of the light wave is 1.2N/C Compute the displacement current as a function of time through a surface of area 0.3m2 whose normal is parallel to the direction of the electric component of the wave. Let the location of the loop be the zero of position in the light wave equation. Assume the loop is small compared to the wavelength of the radio wave. Select One of the Following: (a) -(3.3 times 10^-4 A) cos(-1.0 times 10^8 Hz) t) (b) -(5.3 times 10^-4 A) cos (-1.0 times 10^8 Hz)t) (c) -(2.1 times 10^-3 A) cos (-6.6 times 10^8 Hz)t) (d) -(7.1 times 10^-2 A) cos (-(6.6 times 10^8 Hz)t) (e) Toby

Explanation / Answer

Given

frequency f = 104.9*10^6 Hz

electric field component of the light is E = 1.2 N/C

   the area A = 0.3 m2

we knwo that the electric flux phi_e = E*A

and the displacement current as a function of time is

   Id(t) = epsilon not*dphi_e/dt

   = epsilon not*(d(A*Esin wt)/dt)

   = -epsilon not *A*W*E cos wt

   = -8.854*10^-12*0.3*2pi*104.9*10^6*1.2 cos(2pi*104.9*10^6 t)

   = -0.0021008612708 A cos(-cos 6.6*10^8*t)
   = -(2.1*10^-3 A) cos (-(6.6*10^8 Hz)t)

the answer is option c

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