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A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 cm and an electric-field amplitude of5.40×10-2V/m at adistance of 250 m from the phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

Short Answer

Expert verified

a. The frequency of wave is 8.74×108Hz.

b. The amplitude of magnetic-field is 1.8×10-10T.

c. The intensity of wave is 3.87×10-6W/m2.

Step by step solution

01

Define the intensity ( I ) and the formulas.

The power transported per unit area is known as the intensity ( I ).

The formula used to calculate the intensity( I ) is:

I=PA

Where,A is area measured in the direction perpendicular to the energy andP is the power in watts.

The relation between the frequency of wave f, wavelengthλ and speed of lightc is:

c=fλ

And also, I=12ε0cEmax2 . Where, I is the intensity in W/m2and is the speed of light that is equal to 3.0×108m/sand ε0=8.85×10-12C2/N·m2.

The relation between the maximum electric field and maximum magnetic field is:

Bmax=Emaxc

02

Determine frequency of wave.

Given that,λ=35.4×10-2m

The frequency of wave is:

f=cλ=3×10835.4×10-2=8.47×108Hz

Hence, the frequency of wave is 8.47×108Hz.

03

Determine the amplitude of magnetic field.

Given that,Emax=5.4×10-2V/m

The amplitude of magnetic field is:

Bmax=Emaxc=5.4×10-23×108=1.8×10-10T

Hence, the amplitude of magnetic-field is 1.8×10-10T.

04

Determine the intensity of wave.

The intensity of wave:

I=12ε0cEmax2=12×8.854×10-123×1085.4×10-2=3.87×10-6W/m2

Hence, intensity of wave is 3.87×10-6W/m2.

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