Question

A cell phone in a poor signal strength area is broadcasting a signal of freuquency 1.865 GHz with a power of 2.95 W. The signal is broadcast omnidirectionally. a) Determine the intensity of the signal at a distance of 1.25 km from the cell phone. b) Determine the intensity of the signal at a distance of 5.75 cm from the cell phone (inside the caller's head) Determine the rms magnitude of the electric field at a distance of 5.75 cm from the cel phone c)

Answer 1

A)

I = power /4*pi*r^2

r = 1.25*10^3 m

= 2.95 / (( 4*pi*( 1.25*10^3)^2)

= 1.502*10^-7 watt/m²

B)

r = 0.0575 m

I = power /4*pi*r^2 )

= 2.95 / (( 4*pi*0.0575^2)

= 71 watt/m²

C)

I = c*ε₀ *E₀ ^2

E_rms   = sqrt( I /c*ε₀ )

= sqrt( 71 / (3*10^8*8.85*10^-12))

Input interpretation: 71 V 3 108 8.85 10-12 Result: 163.530...

163.53 V/m

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