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A satellite 575 km above the earth’s surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0 kW. (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring 15.0 cm by 40.0 cm oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

Short Answer

Expert verified

A)l=6.02×10-9W/m2B)Emax=2.12×10-3V/m,Bmax=7.07×10-12TC)F=12×10-10N

Step by step solution

01

Concept of the intensity of a sinusoidal electromagnetic wave in vacuum

The intensity I is proportional to Emax2and it represents the incident power P per area A is given as l=PAThe intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude E the amplitude of magnetic field B and it is given by equation in the form l=12ε0cEmax2Where ε0is the electric constant, c is the speed of light

02

Calculate the intensity

The radius represents the distance from the satellite to the Earth r=575×103m. So, the area of the is calculated byA=4πr2=4π575×103m2=4.15×1012m2

Substitute the values in equation I=PAwe have,

I=PA=25×103W4.15×1012m2=6.02×10-9W/m2

03

STEP 3Calculate the amplitudes of the electric and magnetic fields at the receiver

The intensity of a sinusoidal electromagnetic wave in vacuum is given asI=12ε0cEmax2which can be written as Emax=2lε0cSubstitute the values in equation we have

Emax=26.02×10-9W/m28.85×10-12C2/Nm2×3×108m/s=2.12×10-3V/m

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation Bmax=EmaxcSubstitute the values in equation we have

Bmax=2.12×10-3V/m3×108m/s=7.07×10-12T

04

Calculate the force

The wave exerts an average force F per unit area and this is the radiation pressure Prad and it is the average value of dp/dt divided by the area. So, the radiation pressure of the wave that totally absorbed is given by equation prad=IcSubstitute the values in equation we have

prad=6.02×10-9W/m23×108m/s=2×10-17PaF=pradA=2×10-17Pa0.15m×0.40m=12×10-10N

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