The electric field of a sinusoidal electromagnetic wave obeys the equationE=(375V/m)cos[1.99×107rad/m×x+5.97×1015rad/st] . (a) What is the speed of the wave? (b) What are the amplitudes of the electric and magnetic fields of this wave? (c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?

Short Answer

Expert verified
  1. The speed of a sinusoidal electromagnetic wave is3×108m/s .
  2. The amplitudes of the electric and magnetic fields of the wave are 375 V/m and1.25×10-6T respectively.
  3. The frequency, wavelength and period of the wave are9.50×1014Hz,3.15×10-7m and1.05×10-15s respectively.

Step by step solution

01

Step 1: Define frequency and write formulas.

The number of waves that moves a fixed point in unit time is known as frequency.

f=1Torω2π

Where,T =Time period in seconds andω is angular frequency in .

The frequency, wavelength, and speed of light of any wave are related by the wavelength-frequency relationship.

The wavelength-frequency relationship is:

c=λf

Where, is the speed of light which is equal to3×108m/s.λ , is the wavelength and is the frequency.

The wavelength is inversely proportional to the frequency.

λ=cforvf

Where, v is velocity of wave in .

The relation between wave number and wavelength is:

k=2πλ

Where, k is wavenumber in .

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

Emax=2Iε0CBmax=Emaxc

The equations of sinusoidal electromagnetic wave for electric and magnetic field are:

E=Emaxcos(kxωt)B=Bmaxcos(kxωt)

02

Determine the amplitudes of electric and magnetic fields.

Given that,E=(375V/m)cos1.99×107rad/mx+5.97×1015rad/st

So,Emax=375V/m

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

Bmax=Emaxc

Substitute the values of variables

Bmax=3753×108=1.25×106T

Hence, the amplitudes of the electric and magnetic fields of the wave are 375 V/m and1.25×10-6T respectively.

03

Determine the frequency and wavelength of wave.

From equationE=(375V/m)cos1.99×107rad/mx+5.97×1015rad/st

The values of angular frequency is and the wavenumber is .

The frequency of wave is:

f=ω2π=5.97×10152π=9.50×1014Hz

The wavelength of wave is:

λ=2πk=2π1.99×107=3.15×107m

Hence the frequency and wavelength of the wave are9.50×1014Hz and3.15×10-7m respectively.

04

Determine the period and speed of wave.

The period of wave is:

T=1f=19.50×1014=1.052×1015s

The speed of the wave is:

C=fλ=9.50×10143.15×107=3×108m/s

Hence, the period and speed of wave are1.05×10-15s and3×108m/s respectively.

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